Result: On the Set of Balanced Games

Title:

On the Set of Balanced Games

Authors:

Garcia-Segador, Pedro, Grabisch, Michel, Miranda, Pedro

Contributors:

Grabisch, Michel, Centre d'économie de la Sorbonne (CES), Université Paris 1 Panthéon-Sorbonne (UP1)-Centre National de la Recherche Scientifique (CNRS), Universidad Complutense de Madrid = Complutense University of Madrid Madrid (UCM)

Source:

Mathematics of Operations Research. 50:2047-2072

Publication Status:

Preprint

Publisher Information:

Institute for Operations Research and the Management Sciences (INFORMS), 2025.

Publication Year:

2025

Subject Terms:

Secondary 52B05 OR/MS subject classification: Games/group decisions:Cooperative, balanced games, cooperative TU-games, 0211 other engineering and technologies, core, combinatorial polytope. MSC2000 subject classification: Primary 91A12, 0102 computer and information sciences, 02 engineering and technology, 01 natural sciences, [MATH.MATH-CO] Mathematics [math]/Combinatorics [math.CO], [MATH.MATH-CO]Mathematics [math]/Combinatorics [math.CO], Mathematics:Sets:Polyhedra, FOS: Mathematics, Mathematics - Combinatorics, convex polyhedra, Combinatorics (math.CO)

Document Type:

Academic journal Article

File Description:

application/pdf

Language:

English

ISSN:

1526-5471
0364-765X

DOI:

10.1287/moor.2023.0379

DOI:

10.48550/arxiv.2501.14341

Access URL:

http://arxiv.org/abs/2501.14341
https://hal.science/hal-04908637v1/document
https://doi.org/10.1287/moor.2023.0379
https://hal.science/hal-04908637v1

Rights:

arXiv Non-Exclusive Distribution

Accession Number:

edsair.doi.dedup.....d9cf03e220a61a98ce056d2d896d34f6

Database:

OpenAIRE

Further Information

We study the geometric structure of the set of cooperative transferable utility games having a nonempty core, characterized by Bondareva and Shapley as balanced games. We show that this set is a nonpointed polyhedral cone, and we find the set of its extremal rays and facets. This study is also done for the set of balanced games whose value for the grand coalition is fixed, which yields an affine nonpointed polyhedral cone. Finally, the case of nonnegative balanced games with fixed value for the grand coalition is tackled. This set is a convex polytope, with remarkable properties. We characterize its vertices and facets, study the adjacency structure of vertices, develop an algorithm for generating vertices in a random uniform way, and show that this polytope is combinatorial and its adjacency graph is Hamiltonian. Last, we give a characterization of the set of games having a core reduced to a singleton. Funding: This work was supported by the Spanish Government [Grant PID2021-124933NB-I00].

AN0187697216;mor01aug.25;2025Sep04.08:19;v2.2.500

On the Set of Balanced Games

Keywords: Primary 91A12; secondary 52B05; cooperative TU games; balanced games; core; convex polyhedra; combinatorial polytope

1. Introduction

Given a set of players N, a game with transferable utility, abbreviated hereafter as TU game or simply game, is a mapping v assigning to each subset <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊆</mo><mi>N</mi></mrow></math> (called a coalition) a quantity v(S), representing, for example, the benefit of cooperation of the players in S. The core of a game appears to be a fundamental concept, introduced by Gillies [[6]]: It is the set of payment vectors <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><msup><mrow><mi mathvariant="double-struck">R</mi></mrow><mi>N</mi></msup></mrow></math> to players, such that every coalition S receives at least the quantity v(S), under the constraint that the total payment which is distributed is equal to v(N). This set, however, may be empty. Bondareva [[1]], and independently Shapley [[19]], found a sufficient and necessary condition for a game to have a nonempty core. Games satisfying this condition are called balanced by Shapley, because it is based on particular arrangements of players, called (minimal) balanced collections, which generalize the notion of partition of a set. As the number of minimal balanced collections is finite, this condition amounts to check a finite number of linear inequalities, one for each possible minimal balanced collection. This shows that the set of balanced games is a (closed convex) polyhedron. To the best of our knowledge, there is only one article studying related polyhedra (Kroupa and Studený [[10]]), where the facets of the polyhedra of totally balanced, balanced, and exact games are studied, but in that paper, it is assumed that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mo>Ø</mo><mo stretchy="false">)</mo></mrow></math> might attain a nonnull value, and hence the structures of the polyhedra are different.

The main aim of our paper is precisely to fill this gap. The game-theoretic motivation behind this mathematical study, apart its own interest, is to solve (in a future study) the problem of finding the closest balanced game to a nonbalanced game, as that would provide a new solution concept for games with an empty core. This problem amounts to an orthogonal projection on a polyhedron, a problem that has no analytical solution in general (Rutkowski [[17]]). The only hope to solve it is a deep understanding of the structure of the polyhedron under consideration, which is exactly what the present paper does.

Before elaborating on our findings, we emphasize that this topic is not limited to (cooperative) game theory. Games as defined above are merely set functions vanishing on the empty set and are encountered in many domains linked to operations research, in particular decision theory, voting theory, combinatorial optimization and reliability (see Grabisch [[7]] for details). In decision theory, capacities (Choquet [[2]]) are a particular class of games that is monotonic and represents uncertainty on the set of states of nature (Schmeidler [[18]]). In this context, the core is the set of probability measures that are "compatible with" (i.e., eventwise dominating) a given capacity. Voting theory considers only 0-1–valued monotone games, which are called simple games (Peters [[16]]). They represent the power of coalitions of players to win an election. They are in fact special Boolean functions, whereas games are nothing other than pseudo-Boolean functions, as introduced by Hammer (Foldes and Hammer [[4]]). In combinatorial optimization, submodular games are considered, as they are encountered for example as rank function of a polymatroid (Edmonds [[3]]), and the core corresponds to the base polyhedron of a matroid (see the monograph of Fujishige devoted to this topic [[5]]). Simple games are also used in reliability theory where they indicate the state of a system (functioning or not), depending on the states of its components. We mention finally that our study is also linked to combinatorics and the study of polytopes and polyhedra. Indeed, the concept of minimal balanced collection generalizes the concept of partitions, and their enumeration remains an open problem. Also, they are closely related to many geometrical properties of the games and the core (see Laplace Mermoud et al. [[11]] for details). As a conclusion, our results are not limited to (cooperative) game theory but concern many fields of operations research, as well as combinatorics and polyhedra.

We split our study in three parts, considering three different sets of balanced games. The first one is simply the set of all possible balanced games on N, which we denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi>N</mi><mo stretchy="false">|</mo><mo>=</mo><mi>n</mi></mrow></math> . For the second one we impose the restriction that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mi>α</mi></mrow></math> for some fixed <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>α</mi><mo>∈</mo><mi mathvariant="double-struck">R</mi></mrow></math> . We denote this set by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Last, we impose in addition that v should be a nonnegative function, with α = 1, without loss of generality (w.l.o.g.). We denote the set of such balanced game as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . It follows that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a polytope, and we give the complete characterization of its vertices, as well as an algorithm to randomly generate them in a uniform way. Interestingly, the number of vertices appears to be a known integer sequence related to Boolean functions. We also characterize the adjacency of vertices, and finally we show that this polytope is combinatorial (in the sense of Naddef and Pulleyblank [[14]]), that is, the adjacency graph of its vertices is Hamilton connected.

We show that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a nonpointed cone and identify its lineality space, all its extremal rays, and all its facets, obtaining thus a complete description of the polyhedron. The study of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is similar and shows that it is an affine nonpointed cone. Again, its structure is completely elucidated.

Last, we address the following question: For which balanced games is the core reduced to a singleton? The question is of interest because a core reduced to a singleton provides a very simple and handy solution to a game. We solve completely this question for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Some open questions remain for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

The paper is organized as follows. Section 2 introduces the necessary material on games and balanced collections. In Section 3, we study the cone <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , its lineality space, and its extremal rays and facets, whereas Section 4 is devoted to the study of the affine cone <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Section 5 is devoted to the study of the polytope <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> : dimension, vertices and their enumeration, adjacency, and facets. We also provide a random procedure to generate vertices of this polytope. In Section 6, we address the problem of finding balanced games whose core is reduced to a singleton. Section 7 is devoted to point some possible applications of these results. Section 8 gives some concluding remarks.

2. Basic Concepts

We refer the reader to Peleg and Sudhölter [[15]] and Grabisch [[7]] for full details, proofs, and references. We limit here ourselves to the essential. Throughout the paper, we consider a (fixed) set N of n players, simply denoted by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>N</mi><mo>=</mo><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mo>...</mo><mo>,</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></math> . Coalitions are nonempty subsets of N, denoted by capital letters S, T, and so on. A TU game (N, v) (or simply a game v) is a function <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>:</mo><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup><mo>→</mo><mi mathvariant="double-struck">R</mi></mrow></math> satisfying <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mo>Ø</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn><mo>.</mo></mrow></math> The value v(S) represents the maximal value (benefit) that the coalition S can guarantee, no matter what players outside S might do. We will denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo></mrow></math> (or simply <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , as no subgame will be considered) the set of games v on N.

For further use, we introduce <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , the set of games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>⩾</mo><mn>0</mn></mrow></math> (i.e., <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩾</mo><mn>0</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup></mrow></math> ) and v(N) = 1, and for every <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>α</mi><mo>∈</mo><mi mathvariant="double-struck">R</mi></mrow></math> we introduce <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , the set of games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mi>α</mi></mrow></math> . In addition, we will often use the following families of games, which are bases of the <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">(</mo><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></math> -dimensional vector space <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> :

The Dirac games δ<subs>S</subs>, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>Ø</mo><mo>≠</mo><mi>S</mi><mo>⊆</mo><mi>N</mi></mrow></math> , defined by

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>δ</mi></mrow><mi>S</mi></msub><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mrow><mo>{</mo><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>1</mn><mo>,</mo></mrow></mtd><mtd columnalign="left"><mrow><mtext>if</mtext><mo /><mi>T</mi><mo>=</mo><mi>S</mi></mrow></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>0</mn><mo>,</mo></mrow></mtd><mtd columnalign="left"><mrow><mtext>otherwise</mtext><mo>.</mo></mrow></mtd></mtr></mtable></mrow></mrow></mrow></math>

The unanimity games u<subs>S</subs>, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>Ø</mo><mo>≠</mo><mi>S</mi><mo>⊆</mo><mi>N</mi></mrow></math> , defined by

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>u</mi></mrow><mi>S</mi></msub><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mrow><mo>{</mo><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>1</mn><mo>,</mo></mrow></mtd><mtd columnalign="left"><mrow><mtext>if</mtext><mo /><mi>T</mi><mo>⊇</mo><mi>S</mi></mrow></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>0</mn><mo>,</mo></mrow></mtd><mtd columnalign="left"><mrow><mtext>otherwise</mtext><mo>.</mo></mrow></mtd></mtr></mtable></mrow></mrow></mrow></math>

Assuming that all players agree to form the grand coalition N, we look for a way to share the benefit v(N) among all players, that is, for an allocation or payment vector <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><msup><mrow><mi mathvariant="double-struck">R</mi></mrow><mi>N</mi></msup></mrow></math> , where coordinate x<subs>i</subs> indicates the payoff given to player i. For any coalition S, we denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>≔</mo><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>i</mi><mo>∈</mo><mi>S</mi></mrow></msub><mrow><msub><mrow><mi>x</mi></mrow><mi>i</mi></msub></mrow></mstyle></mrow></math> the total payoff given to the players in S. An allocation is efficient if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo></mrow></math> . A systematic way of assigning a set of allocations to a game is called a solution concept.

In this paper we focus on one of the best known solution concepts, which is the core (Gillies [[6]]). The core is the set of efficient allocations satisfying coalitional rationality, which means that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩾</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo></mrow></math> for all coalitions S. Under this condition, no coalition S has an incentive to leave the grand coalition N to form a subgame on S. The core of a game (N, v) is denoted by C(N, v) (or C(v) for short if N is fixed):

The core is a closed convex polytope that may be empty, as illustrated in the following example.

Example 1.

The unanimity game u<subs>S</subs> has a nonempty core for any <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>Ø</mo><mo>≠</mo><mi>S</mi><mo>⊆</mo><mi>N</mi></mrow></math> . Indeed, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>=</mo><mn>1</mn><mrow><mo>/</mo><mo>|</mo><mi>S</mi><mo>|</mo></mrow><msup><mrow><mn mathvariant="bold">1</mn></mrow><mi>S</mi></msup></mrow></math> is a core allocation, where <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn mathvariant="bold">1</mn></mrow><mi>S</mi></msup></mrow></math> is the characteristic vector of S, that is, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mn mathvariant="bold">1</mn></mrow><mi>i</mi><mi>S</mi></msubsup><mo>=</mo><mn>1</mn></mrow></math> if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>S</mi></mrow></math> and zero otherwise. Moreover, it can be easily checked that

However, any Dirac game δ<subs>S</subs> has an empty core, except when S = N, as it is easy to check.

A condition for nonemptiness of the core has been given by Bondareva [[1]] and Shapley [[19]], which we detail below.

A balanced collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> on N is a family of nonempty subsets of N such that there exist positive (balancing) weights <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup><mo>,</mo><mo /><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></math> , satisfying

This notion is an extension of the notion of partition, as any partition is a balanced collection with balancing weights all equal to one. A balanced collection is minimal if it contains no balanced proper subcollection. We denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> the set of all minimal balanced collections on a set N of cardinality n, excluding the collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>N</mi><mo stretchy="false">}</mo></mrow></math> . It can be shown that minimal balanced collections (abbreviated hereafter by m.b.c.) have a unique set of balancing weights and that their cardinality is at most n. Moreover, if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> is a m.b.c. with balancing weights <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mrow><mo stretchy="false">(</mo><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup><mo stretchy="false">)</mo></mrow></mrow><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></msub></mrow></math> , then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mover accent="true"><mi mathvariant="script">B</mi><mo stretchy="true">¯</mo></mover><mo>≔</mo><mo stretchy="false">{</mo><mi>N</mi><mo>\</mo><mi>S</mi><mo /><mo>:</mo><mo /><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi><mo stretchy="false">}</mo></mrow></math> is also a m.b.c. with balancing weights

A game v is balanced if

The following result holds (Bondareva [[1]], Shapley [[19]]).

Theorem 1.

Consider a game (N, v). Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo><mo>≠</mo><mo>Ø</mo></mrow></math> if and only if v is balanced.

We will denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo></mrow></math> the set of games in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo></mrow></math> being balanced, that is, the set of games having a nonempty core. For the same reasons as before, we simplify the notation to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Hence, applying Theorem 1 and (1), we get

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>=</mo><mo stretchy="true">{</mo><mi>v</mi><mo>∈</mo><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo /><mo>:</mo><mo /><mstyle displaystyle="true"><munder><mo>∑</mo><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></munder><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>−</mo><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>⩽</mo><mn>0</mn><mo>,</mo><mo /><mo>∀</mo><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="true">}</mo><mo>.</mo></mrow></math> (2)

For application purposes and further studies, the following subsets of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are of interest:

The set of balanced games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> for some <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>α</mi><mo>∈</mo><mi mathvariant="double-struck">R</mi></mrow></math> . We denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> the set of such games, that is,

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="true">=</mo><mo stretchy="true">{</mo><mi>v</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo /><mo>:</mo><mo /><mstyle displaystyle="true"><munder><mo>∑</mo><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></munder><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩽</mo><mi>α</mi><mo>,</mo><mo /><mo>∀</mo><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="true">}</mo><mo>.</mo></mrow></math> (3)

The study of this set is motivated as follows: When searching for the closest balanced game <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo></mrow></math> to a given nonbalanced game v, it is natural to impose that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo></mrow></math> , as this is the total benefit which has to be distributed among players.

The set of balanced games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . We denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> the set of such games, that is,

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>=</mo><mo stretchy="true">{</mo><mi>v</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo /><mo>:</mo><mo /><mstyle displaystyle="true"><munder><mo>∑</mo><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></munder><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩽</mo><mn>1</mn><mo>,</mo><mo /><mo>∀</mo><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="true">}</mo><mo>.</mo></mrow></math> (4)

Studying this set is motivated by the fact that many applications concern games with nonnegative values. Therefore, it would make no sense to find a closest balanced game taking negative values.

Observe that if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>⩾</mo><mn>0</mn></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mi>α</mi><mo>≠</mo><mn>0</mn></mrow></math> , then v is balanced if and only if (iff) <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>(</mo><mn>1</mn><mo>/</mo><mi>α</mi><mo>)</mo><mi>v</mi><mo>∈</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . This shows that studying the balancedness of nonnegative games is equivalent to study the balancedness of nonnegative games with v(N) = 1, hence the set <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , except if v(N) = 0, but the latter case is trivial as only v = 0 is balanced.

Note that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>,</mo><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are convex polyhedra. The next sections are devoted to the study of the structure of these polyhedra.

3. Polyhedron BG(n)

Because for any game <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>,</mo><mo /><mi>v</mi><mo stretchy="false">(</mo><mo>Ø</mo><mo stretchy="false">)</mo></mrow></math> is fixed, we consider v as being a vector in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="double-struck">R</mi></mrow><mrow><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup><mo>\</mo><mo stretchy="false">{</mo><mo>Ø</mo><mo stretchy="false">}</mo></mrow></msup></mrow></math> . We start by showing a technical result, which will be useful in the sequel.

Lemma 1.

Let v be such that

for all partitions Π of N of the following form: either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="normal">Π</mi><mo>=</mo><msup><mrow><mi>N</mi></mrow><mo>⊥</mo></msup><mo>≔</mo><mo stretchy="false">{</mo><mo stretchy="false">{</mo><mn>1</mn><mo stretchy="false">}</mo><mo>,</mo><mo>...</mo><mo>,</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo stretchy="false">}</mo></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="normal">Π</mi><mo>=</mo><mo stretchy="false">{</mo><mi>S</mi><mo>,</mo><msup><mrow><mo stretchy="false">(</mo><mi>N</mi><mo>\</mo><mi>S</mi><mo stretchy="false">)</mo></mrow><mo>⊥</mo></msup><mo stretchy="false">}</mo></mrow></math> for any <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> , with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi>T</mi></mrow><mo>⊥</mo></msup><mo>=</mo><mo stretchy="false">{</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo>,</mo><mi>i</mi><mo>∈</mo><mi>T</mi><mo stretchy="false">}</mo></mrow></math> . Then

Proof.

Assume v satisfies the assumption. Take any m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> with balancing weights <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mrow><mo stretchy="false">(</mo><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup><mo stretchy="false">)</mo></mrow></mrow><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></msub></mrow></math> and denote <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mover accent="true"><mi mathvariant="script">B</mi><mo stretchy="true">¯</mo></mover><mo>=</mo><mo stretchy="false">{</mo><mi>N</mi><mo>\</mo><mi>S</mi><mo>,</mo><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi><mo stretchy="false">}</mo></mrow></math> . Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mover accent="true"><mi mathvariant="script">B</mi><mo stretchy="true">¯</mo></mover></mrow></math> is a m.b.c. with weights <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mrow><mover accent="true"><mi mathvariant="script">B</mi><mo stretchy="true">¯</mo></mover></mrow></msubsup><mo>=</mo><mfrac><mrow><msubsup><mrow><mi>λ</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow><mi mathvariant="script">B</mi></msubsup></mrow><mrow><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>T</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></msub><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>T</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mo>−</mo><mn>1</mn></mrow></mfrac></mrow></math> (see Section 2). It follows that

Remark that the reciprocal holds as all partitions are minimal balanced collections with balanced weights equal to one.

Let us first establish the structure of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

3.1. Structure of BG(n)

Theorem 2.

Let <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>⩾</mo><mn>2</mn></mrow></math> . Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">(</mo><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></math> -dimensional polyhedral cone, which is not pointed. Its lineality space has dimension n, with basis <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mrow><mo stretchy="false">(</mo><msub><mrow><mi>w</mi></mrow><mi>i</mi></msub><mo stretchy="false">)</mo></mrow></mrow><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></msub></mrow></math> , with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>w</mi></mrow><mi>i</mi></msub><mo>=</mo><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>S</mi><mo>∋</mo><mi>i</mi></mrow></msub><mrow><msub><mrow><mi>δ</mi></mrow><mi>S</mi></msub></mrow></mstyle></mrow></math> , where δ<subs>S</subs> is the Dirac game centered on S (seeSection 2).

Proof.

We make the proof in two steps.

First, observe that the zero game is balanced, and if v is balanced, then αv with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>α</mi><mo>⩾</mo><mn>0</mn></mrow></math> is also balanced. Then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a cone. Moreover, it is a polyhedral cone because it is defined by a finite number of linear inequalities. Next, observe that no equality can be implied by System (2), as for any <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>≠</mo><mo>Ø</mo><mo>,</mo></mrow></math> all coefficients of v(S) are of the same sign. Because, in addition there is no equality, the cone is full dimensional.

Denoting by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><mi>v</mi><mo>⩽</mo><mn>0</mn></mrow></math> the system of inequalities in (2), the lineality space is the set of solutions of Av = 0. From Lemma 1, we infer that this is equivalent to solve the system

for all partitions Π of N of the following form: either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="normal">Π</mi><mo>=</mo><msup><mrow><mi>N</mi></mrow><mo>⊥</mo></msup><mo>≔</mo><mo stretchy="false">{</mo><mo stretchy="false">{</mo><mn>1</mn><mo stretchy="false">}</mo><mo>,</mo><mo>...</mo><mo>,</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo stretchy="false">}</mo></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="normal">Π</mi><mo>=</mo><mo stretchy="false">{</mo><mi>S</mi><mo>,</mo><msup><mrow><mo stretchy="false">(</mo><mi>N</mi><mo>\</mo><mi>S</mi><mo stretchy="false">)</mo></mrow><mo>⊥</mo></msup><mo stretchy="false">}</mo></mrow></math> for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> .

Therefore, we obtain that for any <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn><mo>,</mo><mo /><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>−</mo><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>i</mi><mo>∈</mo><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub><mi>v</mi></mstyle><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></msub><mi>v</mi></mstyle><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo></mrow></math> . It follows that its set of solutions is, expressing all coordinates in terms of those of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>,</mo><mo /><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mo>...</mo><mo>,</mo><mi>n</mi></mrow></math> ,

A basis for this subspace is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msub><mrow><mi>w</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi>w</mi></mrow><mi>n</mi></msub><mo stretchy="false">}</mo></mrow></math> with

The reader may compare this result with lemma 3.1 in Kroupa and Studený [[10]], where it is proved that modular games form the lineality space of the cone of balanced games, totally balanced games, and exact games.

As <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is not pointed, it can be decomposed as the direct sum of its lineality space of dimension n (that we will denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mtext mathvariant="sans-serif">Lin</mtext><mo stretchy="false">(</mo><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow></math> ) and a supplementary subspace (not necessarily orthogonal) of dimension <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mi>n</mi><mo>−</mo><mn>1</mn></mrow></math> , which is a pointed cone and whose extremal rays can be found in the usual way. However, because there is no unique way to choose a supplementary space, there is no unique representation of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> by extremal rays. It seems that in the case of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> the orthogonal supplement does not yield simple results, and we will use instead the supplement where the coordinates corresponding to singletons are zero. We denote this set as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msup><mrow><mi mathvariant="script">G</mi></mrow><mn>0</mn></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , and hence we have

We study in the next section the extremal rays of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

3.2. Extremal Rays of BG(n)

Theorem 3.

Let <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>⩾</mo><mn>2</mn></mrow></math> . The extremal rays of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are

The 2n extremal rays corresponding to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mtext mathvariant="sans-serif">Lin</mtext><mo stretchy="false">(</mo><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow></math> : <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>w</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi>w</mi></mrow><mi>n</mi></msub><mo>,</mo><mo>−</mo><msub><mrow><mi>w</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><mo>−</mo><msub><mrow><mi>w</mi></mrow><mi>n</mi></msub></mrow></math> ;

The <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mi>n</mi><mo>−</mo><mn>2</mn></mrow></math> extremal rays of the form <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo>=</mo><mo>−</mo><msub><mrow><mi>δ</mi></mrow><mi>S</mi></msub><mo>,</mo><mo /><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> ;

The n extremal rays of the form

This yields in total <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>+</mo><mn>2</mn><mi>n</mi><mo>−</mo><mn>2</mn></mrow></math> extremal rays.

Proof.

We study each part separately.

Let us consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> and show that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo>=</mo><mo>−</mo><msub><mrow><mi>δ</mi></mrow><mi>S</mi></msub></mrow></math> is an extremal ray. Let us denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo></mrow></math> the coordinate of r<subs>S</subs> corresponding to subset T and similarly for all rays. First, r<subs>S</subs> is a ray of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msup><mrow><mi mathvariant="script">G</mi></mrow><mn>0</mn></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> because it satisfies System (2) of inequalities and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> .

Suppose that it is not extremal. Then, there exist two rays <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo>,</mo><mi>r</mi><mo>′</mo><mo>∈</mo><mi mathvariant="script">B</mi><msup><mrow><mi mathvariant="script">G</mi></mrow><mn>0</mn></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> not proportional to r<subs>S</subs> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo>=</mo><mi>r</mi><mo>+</mo><mi>r</mi><mo>′</mo></mrow></math> . Suppose that r(T) > 0 (w.l.o.g. r(T) = 1) for some <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>≠</mo><mi>S</mi><mo>,</mo><mo /><mn>1</mn><mo><</mo><mo stretchy="false">|</mo><mi>T</mi><mo stretchy="false">|</mo><mo><</mo><mi>n</mi></mrow></math> . Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo>′</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mo>−</mo><mn>1</mn></mrow></math> . Using the partition <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>T</mi><mo>,</mo><msup><mrow><mo stretchy="false">(</mo><mi>N</mi><mo>\</mo><mi>T</mi><mo stretchy="false">)</mo></mrow><mo>⊥</mo></msup><mo stretchy="false">}</mo></mrow></math> , because <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> , the corresponding inequality in (2) yields <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>⩾</mo><mn>1</mn></mrow></math> because the weights for a m.b.c. being a partition are one, and because <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo>′</mo><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mo>−</mo><mi>r</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo></mrow></math> , we get <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo>′</mo><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>⩽</mo><mo>−</mo><mn>1</mn></mrow></math> . Taking now the partition <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi>N</mi></mrow><mo>⊥</mo></msup></mrow></math> , we obtain

which is impossible. Letting r(T) < 0 yields the same contradiction by inverting the roles of r and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo>′</mo></mrow></math> . We conclude that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn><mo>,</mo><mo>∀</mo><mi>T</mi><mo>≠</mo><mi>S</mi><mo>.</mo></mrow></math> Finally, observe that r(S) > 0 is not possible as the inequality for partition <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>S</mi><mo>,</mo><msup><mrow><mo stretchy="false">(</mo><mi>N</mi><mo>\</mo><mi>S</mi><mo stretchy="false">)</mo></mrow><mo>⊥</mo></msup><mo stretchy="false">}</mo></mrow></math> would not be satisfied. This finishes the proof.

Let us take <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> and show that r<subs>i</subs> is an extremal ray. First, as

it follows that r<subs>i</subs> satisfies all inequalities in (2); moreover, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>i</mi></msub><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>j</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>j</mi><mo>∈</mo><mi>N</mi></mrow></math> and hence, r<subs>i</subs> is a ray of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msup><mrow><mi mathvariant="script">G</mi></mrow><mn>0</mn></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math> To show that it is extremal, we need to show that the set of solutions of the subsystem of (2) formed by tight inequalities has dimension 1 in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msup><mrow><mi mathvariant="script">G</mi></mrow><mn>0</mn></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Consider a m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> and its corresponding inequality:

We obtain

Hence, the inequality is tight iff <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo>∉</mo><mi mathvariant="script">B</mi></mrow></math> . Let us call (2)′ the subsystem of tight inequalities in v, with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><mi mathvariant="script">B</mi><msup><mrow><mi mathvariant="script">G</mi></mrow><mn>0</mn></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Consider the partition <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>S</mi><mo>,</mo><msup><mrow><mo stretchy="false">(</mo><mi>N</mi><mo>\</mo><mi>S</mi><mo stretchy="false">)</mo></mrow><mo>⊥</mo></msup><mo stretchy="false">}</mo></mrow></math> with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo><</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo><</mo><mi>n</mi></mrow></math> . In terms of (2)′, the corresponding equality is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo></mrow></math> . Now, consider any m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo>∉</mo><mi mathvariant="script">B</mi></mrow></math> . The corresponding equality reads

Substituting v(S) by v(N) yields <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>−</mo><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> . Therefore, the subspace of solutions is given by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mo stretchy="false">(</mo><mi>α</mi><mo>,</mo><mo>...</mo><mo>,</mo><mi>α</mi><mo stretchy="false">)</mo><mo>,</mo><mi>α</mi><mo>∈</mo><mi mathvariant="double-struck">R</mi><mo stretchy="false">}</mo></mrow></math> , of dimension 1.

It remains to prove that there is no other extremal ray. Consider a ray w in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msup><mrow><mi mathvariant="script">G</mi></mrow><mn>0</mn></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , hence satisfying <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>w</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> , and (2). Suppose w is not a conic combination of the extremal rays r<subs>S</subs> and r<subs>i</subs>, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn><mo>,</mo><mo /><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> ; that is, the following system has no solution in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>α</mi></mrow><mi>S</mi></msub><mo>,</mo><msub><mrow><mi>α</mi></mrow><mi>i</mi></msub></mrow></math> :

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mtable><mtr><mtd><mrow><mstyle displaystyle="true"><munder><mo>∑</mo><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo>|</mo><mi>S</mi><mo>|</mo><mo>></mo><mn>1</mn></mrow></munder><mrow><msub><mrow><mi>α</mi></mrow><mi>S</mi></msub></mrow></mstyle><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo>+</mo><mstyle displaystyle="true"><munder><mo>∑</mo><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></munder><mrow><msub><mrow><mi>α</mi></mrow><mi>i</mi></msub></mrow></mstyle><msub><mrow><mi>r</mi></mrow><mi>i</mi></msub></mrow></mtd><mtd><mrow><mo>=</mo><mi>w</mi></mrow></mtd></mtr><mtr><mtd><mrow><msub><mrow><mi>α</mi></mrow><mi>S</mi></msub></mrow></mtd><mtd><mrow><mo>⩾</mo><mn>0</mn><mo>,</mo><mo /><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo>|</mo><mi>S</mi><mo>|</mo><mo>></mo><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><msub><mrow><mi>α</mi></mrow><mi>i</mi></msub></mrow></mtd><mtd><mrow><mo>⩾</mo><mn>0</mn><mo>,</mo><mo /><mi>i</mi><mo>∈</mo><mi>N</mi><mo>.</mo></mrow></mtd></mtr></mtable></mrow></math> (5)

Using definitions of r<subs>S</subs> and r<subs>i</subs> and omitting coordinates for singletons in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo>,</mo><msub><mrow><mi>r</mi></mrow><mi>i</mi></msub><mo>,</mo><mi>w</mi></mrow></math> as they are all zero, we obtain that the previous system can be written as

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mtable><mtr><mtd><mrow><mo>−</mo><msub><mrow><mi>α</mi></mrow><mi>S</mi></msub><mo>+</mo><mstyle displaystyle="true"><munder><mo>∑</mo><mrow><mi>i</mi><mo>∈</mo><mi>S</mi></mrow></munder><mrow><msub><mrow><mi>α</mi></mrow><mi>i</mi></msub></mrow></mstyle></mrow></mtd><mtd><mrow><mo>=</mo><mi>w</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>,</mo><mo /><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo>|</mo><mi>S</mi><mo>|</mo><mo>></mo><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><mstyle displaystyle="true"><munder><mo>∑</mo><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></munder><mrow><msub><mrow><mi>α</mi></mrow><mi>i</mi></msub></mrow></mstyle></mrow></mtd><mtd><mrow><mo>=</mo><mi>w</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo></mrow></mtd></mtr><mtr><mtd><mrow><msub><mrow><mi>α</mi></mrow><mi>S</mi></msub></mrow></mtd><mtd><mrow><mo>⩾</mo><mn>0</mn><mo>,</mo><mo /><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo>|</mo><mi>S</mi><mo>|</mo><mo>></mo><mn>1</mn></mrow></mtd></mtr><mtr><mtd><mrow><msub><mrow><mi>α</mi></mrow><mi>i</mi></msub></mrow></mtd><mtd><mrow><mo>⩾</mo><mn>0</mn><mo>,</mo><mo /><mi>i</mi><mo>∈</mo><mi>N</mi><mo>.</mo></mrow></mtd></mtr></mtable></mrow></math> (6)

We may denote (with some abuse) the whole system (6) by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><mi>α</mi><mo>⩾</mo><mi>b</mi></mrow></math> in matrix notation. If this system has no solution, then, by Farkas' lemma, this is equivalent to say that there exists a vector <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">[</mo><mi>y</mi><mo /><mo /><mi>z</mi><mo /><mo /><mi>t</mi><mo stretchy="false">]</mo></mrow></math> with coordinates <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>y</mi></mrow><mi>S</mi></msub><mo>∈</mo><mi mathvariant="double-struck">R</mi></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊆</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn><mo>,</mo><mo /><msub><mrow><mi>z</mi></mrow><mi>T</mi></msub><mo>⩾</mo><mn>0</mn><mo>,</mo><mo /><mi>T</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>T</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>t</mi></mrow><mi>i</mi></msub><mo>⩾</mo><mn>0</mn><mo>,</mo><mo /><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> , such that

Observe that the only vectors <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">[</mo><mi>y</mi><mo /><mo /><mi>z</mi><mo /><mo /><mi>t</mi><mo stretchy="false">]</mo></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mo stretchy="false">[</mo><mi>y</mi><mo /><mo /><mi>z</mi><mo /><mo /><mi>t</mi><mo stretchy="false">]</mo></mrow><mo>⊤</mo></msup><mi>A</mi><mo>=</mo><mn>0</mn></mrow></math> have the form:

up to a multiplicative factor K > 0.

Then, we obtain

Observe that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn><mo stretchy="false">}</mo></mrow></math> is a balanced collection (not minimal if n > 3) with balancing weights <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mfrac><mn>1</mn><mrow><msup><mrow><mn>2</mn></mrow><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msup><mo>−</mo><mn>2</mn></mrow></mfrac></mrow></math> . It follows that

Therefore, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mo stretchy="false">[</mo><mi>y</mi><mo /><mi>z</mi><mo /><mi>t</mi><mo stretchy="false">]</mo></mrow><mo>⊤</mo></msup><mi>b</mi><mo>⩽</mo><mo stretchy="false">(</mo><msup><mrow><mn>2</mn></mrow><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msup><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo><mi>w</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>−</mo><mo stretchy="false">(</mo><msup><mrow><mn>2</mn></mrow><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msup><mo>−</mo><mn>2</mn><mo>+</mo><mi>t</mi><mo stretchy="false">)</mo><mi>w</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mo>−</mo><mi>t</mi><mi>w</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>⩽</mo><mn>0</mn></mrow></math> . The last inequality follows from <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>t</mi><mo>⩾</mo><mn>0</mn></mrow></math> and (6). Hence, this system has no solution, and consequently, System (6) always has a solution. Therefore, w is not extremal and the result follows. □

Compare also this result with lemma 5.4 in Kroupa and Studený [[10]], giving condition for a ray to be extreme in the dual cone, and corollary 5.1. mentioning the extremal rays <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>−</mo><msub><mrow><mi>δ</mi></mrow><mi>S</mi></msub></mrow></math> .

The following result is immediate.

Lemma 2.

The cores of w<subs>i</subs>, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>−</mo><msub><mrow><mi>w</mi></mrow><mi>i</mi></msub></mrow></math> , r<subs>i</subs>, and r<subs>S</subs> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi><mo>,</mo><mo /><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> are singletons (respectively, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mrow><mo>{</mo><mi>i</mi><mo>}</mo></mrow></msup><mo stretchy="false">}</mo><mo>,</mo><mo /><mo stretchy="false">{</mo><mo>−</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mrow><mo>{</mo><mi>i</mi><mo>}</mo></mrow></msup><mo stretchy="false">}</mo></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mrow><mo>{</mo><mi>i</mi><mo>}</mo></mrow></msup><mo stretchy="false">}</mo></mrow></math> , {0}).

3.3. Facets

Theorem 4.

Each inequality in (2) defines a facet, that is, minimal balanced collections in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are in bijection with the facets of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

Proof.

Let us consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and the inequality

It suffices to show that this face contains <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> independent extremal rays. We already know by definition of the lineality space that any face contains the n rays <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>w</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi>w</mi></mrow><mi>n</mi></msub></mrow></math> of the lineality space.

Among the extremal rays r<subs>S</subs>, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> , it is easy to check that only those such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∉</mo><mi mathvariant="script">B</mi></mrow></math> satisfy (7). There are <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mi>n</mi><mo>−</mo><mn>2</mn><mo>−</mo><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo><mo>+</mo><msub><mrow><mi>k</mi></mrow><mi mathvariant="script">B</mi></msub></mrow></math> such rays, where <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>k</mi></mrow><mi mathvariant="script">B</mi></msub></mrow></math> is the number of singletons in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> .

Next, define <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>B</mi><mo>=</mo><mi>N</mi><mo>\</mo><mstyle displaystyle="false"><msub><mo>∪</mo><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo>∈</mo><mi mathvariant="script">B</mi></mrow></msub><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></mstyle></mrow></math> . Then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi>B</mi><mo stretchy="false">|</mo><mo>=</mo><mi>n</mi><mo>−</mo><msub><mrow><mi>k</mi></mrow><mi mathvariant="script">B</mi></msub></mrow></math> . Now, observe that any r<subs>i</subs> with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></math> is satisfying (7). Indeed,

This makes another set of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>−</mo><msub><mrow><mi>k</mi></mrow><mi mathvariant="script">B</mi></msub></mrow></math> rays. Therefore, we have in total <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn><mo>+</mo><mi>n</mi><mo>−</mo><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo><mo>⩾</mo><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> , as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> is minimal (see Section 3). It remains to prove independence. Observe that every <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>δ</mi></mrow><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></msub><mo>,</mo><mo /><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> , is used by w<subs>i</subs>, every δ<subs>S</subs>, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn><mo>,</mo><mo /><mi>S</mi><mo>∉</mo><mi mathvariant="script">B</mi></mrow></math> , is used by r<subs>S</subs>, and every δ<subs>S</subs> for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></math> is used in some of the r<subs>i</subs>'s with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>B</mi></mrow></math> , unless <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊆</mo><mi>N</mi><mo>\</mo><mi>B</mi></mrow></math> . However, then consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>′</mo><mo>=</mo><mi mathvariant="script">B</mi><mo>\</mo><mo stretchy="false">{</mo><mi>S</mi><mo stretchy="false">}</mo></mrow></math> and weights

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>T</mi><mrow><mi mathvariant="script">B</mi><mo>′</mo></mrow></msubsup><mo>=</mo><mrow><mo>{</mo><mrow><mtable><mtr><mtd><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>i</mi><mi mathvariant="script">B</mi></msubsup><mo>+</mo><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup><mo>,</mo></mrow></mtd><mtd><mrow><mi>T</mi><mo>=</mo><mo>{</mo><mi>i</mi><mo>}</mo><mo>,</mo><mi>i</mi><mo>∈</mo><mi>S</mi></mrow></mtd></mtr><mtr><mtd><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>T</mi><mrow><mi mathvariant="script">B</mi><mo>′</mo></mrow></msubsup><mo>,</mo></mrow></mtd><mtd><mrow><mtext>otherwise</mtext><mo>,</mo></mrow></mtd></mtr></mtable></mrow></mrow></mrow></math>

and hence, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> is not minimal, a contradiction. If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo><mo>=</mo><mi>n</mi></mrow></math> , then we have exactly <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> extremal rays, which are independent by the above argument. If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo><mo><</mo><mi>n</mi></mrow></math> , then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>−</mo><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo></mrow></math> rays r<subs>i</subs> have to be removed to make the family independent. □

4. Polyhedron BGα(n)

Let us now study the set <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , defined by (see (3)):

We follow the same notation as in Section 3, except that now v is a vector in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="double-struck">R</mi></mrow><mrow><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup><mo>\</mo><mo stretchy="false">{</mo><mo>Ø</mo><mo>,</mo><mi>N</mi><mo stretchy="false">}</mo></mrow></msup></mrow></math> . As the proof techniques are similar to the case of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , the proofs are relegated to the appendix or omitted. We begin by expliciting the structure of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

Theorem 5.

Let <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>⩾</mo><mn>2</mn><mo>,</mo><mo /><mi>α</mi><mo>∈</mo><mi mathvariant="double-struck">R</mi></mrow></math> . Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">(</mo><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></math> -dimensional affine cone,[1]which is not pointed. Its lineality space has dimension <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>−</mo><mn>1</mn><mo>,</mo></mrow></math> with base <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mrow><mo stretchy="false">(</mo><msub><mrow><mi>w</mi></mrow><mi>i</mi></msub><mo stretchy="false">)</mo></mrow></mrow><mrow><mi>i</mi><mo>∈</mo><mi>N</mi><mo>\</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> , where

The affine space is given by

In Theorem 5, element n plays a particular role, but this choice is arbitrary. Indeed, the affine space in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> contains in particular the games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>α</mi><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> , for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> .

To find the extreme rays of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , we proceed as for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Using the notation from the proof of Theorem 5, we write

where <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a cone defined by System (12). Now,

where <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mtext mathvariant="sans-serif">Lin</mtext><mo stretchy="false">(</mo><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow></math> is the lineality space of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi><mn>0</mn></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is its supplement where the coordinates of the singletons <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>1</mn><mo stretchy="false">}</mo><mo>,</mo><mo>...</mo><mo>,</mo><mo stretchy="false">{</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">}</mo></mrow></math> are zero.

Theorem 6.

Let <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>⩾</mo><mn>2</mn></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>α</mi><mo>∈</mo><mi mathvariant="double-struck">R</mi></mrow></math> . The extremal rays of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are

The <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mi>n</mi><mo>−</mo><mn>2</mn></mrow></math> extremal rays corresponding to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mtext mathvariant="sans-serif">Lin</mtext><mo stretchy="false">(</mo><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow></math> : <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>w</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi>w</mi></mrow><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msub><mo>,</mo><mo>−</mo><msub><mrow><mi>w</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><mo>−</mo><msub><mrow><mi>w</mi></mrow><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msub></mrow></math> ;

The n extremal rays of the form

and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>n</mi></msub><mo>=</mo><mo>−</mo><msub><mrow><mi>δ</mi></mrow><mrow><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> .

Observe that by definition of the lineality space, the extremal rays <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>w</mi></mrow><mi>i</mi></msub><mo>,</mo><mo>−</mo><msub><mrow><mi>w</mi></mrow><mi>i</mi></msub></mrow></math> belong to every facet of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . We also have the following result (proof is omitted).

Lemma 3.

The core of w<subs>i</subs>, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>−</mo><msub><mrow><mi>w</mi></mrow><mi>i</mi></msub></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi><mo>\</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></math> , r<subs>i</subs> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> , r<subs>S</subs> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>=</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo>,</mo></mrow></math> are reduced to singletons, which are <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mrow><mo>{</mo><mi>i</mi><mo>}</mo></mrow></msup><mo>−</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mrow><mo>{</mo><mi>n</mi><mo>}</mo></mrow></msup><mo stretchy="false">}</mo><mo>,</mo><mo>−</mo><mo stretchy="false">{</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mrow><mo>{</mo><mi>i</mi><mo>}</mo></mrow></msup><mo>+</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mrow><mo>{</mo><mi>n</mi><mo>}</mo></mrow></msup><mo stretchy="false">}</mo></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mrow><mo>{</mo><mi>i</mi><mo>}</mo></mrow></msup><mo>−</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mrow><mo>{</mo><mi>n</mi><mo>}</mo></mrow></msup><mo stretchy="false">}</mo></mrow></math> , and {0}, respectively.

Theorem 7.

Each inequality in (3) defines a facet; that is, minimal balanced collections in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are in bijection with the facets of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

The proof is similar to the one of Theorem 4 and is omitted.

5. Polytope BG+(n)

Finally, let us study <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Recall that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is defined by (see (4)):

5.1. Dimension, Boundedness

Proposition 1.

The set <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">(</mo><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></math> -dimensional polytope.

Proof.

First, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is defined by a set of linear equations. Hence, it is a polyhedron of dimension at most <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> as v(N) = 1 is fixed. Moreover, the <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>1</mn></mrow></math> unanimity games u<subs>S</subs>, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>≠</mo><mo>Ø</mo></mrow></math> belong to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> (see Example 1), and they are affinely independent because the <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>u</mi></mrow><mi>S</mi></msub><mo>−</mo><msub><mrow><mi>u</mi></mrow><mi>N</mi></msub><mo>,</mo><mo /><mi>S</mi><mo>≠</mo><mo>Ø</mo><mo>,</mo><mi>N</mi></mrow></math> , are linearly independent, so that the polyhedron is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> -dimensional.

Second, the polyhedron is bounded. To see this, if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>,</mo></mrow></math> it follows that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩽</mo><mn>1</mn><mo>,</mo><mo>∀</mo><mi>S</mi><mo>≠</mo><mo>Ø</mo><mo>,</mo><mi>N</mi><mo>.</mo></mrow></math> Indeed, if v(S) > 1, as v is balanced, there exists <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><msup><mrow><mi mathvariant="double-struck">R</mi></mrow><mi>N</mi></msup></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>⩾</mo><mi>v</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>⊆</mo><mi>N</mi></mrow></math> , and x(N) = 1. However, as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>⩾</mo><mn>0</mn></mrow></math> , we have <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mi>i</mi></msub><mo>⩾</mo><mn>0</mn></mrow></math> , which implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩽</mo><mn>1</mn></mrow></math> , a contradiction. □

5.2. Vertices

Theorem 8.

Consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math> Then, v is a vertex of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> if and only if v is balanced and 0-1 valued.

Proof.

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>⇐</mo><mo stretchy="false">)</mo></mrow></math> Suppose v is balanced and 0-1 valued. Assume v is not extremal. Then, there exist <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo>,</mo><mi>v</mi><mo>″</mo><mo>∈</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mo stretchy="false">(</mo><mi>v</mi><mo>′</mo><mo>+</mo><mi>v</mi><mo>″</mo><mo stretchy="false">)</mo><mo>/</mo><mn>2</mn></mrow></math> . Take (if it exists) <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>≠</mo><mi>N</mi></mrow></math> such that v(S) = 1. We have already shown in the proof of Proposition 1 that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>></mo><mn>1</mn></mrow></math> is impossible. Similarly, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo><</mo><mn>1</mn></mrow></math> is impossible as it imposes <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>″</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>></mo><mn>1</mn></mrow></math> . It follows that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo>″</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊆</mo><mi>N</mi></mrow></math> such that v(S) = 1.

Take now <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>Ø</mo><mo>≠</mo><mi>S</mi><mo>⊂</mo><mi>N</mi></mrow></math> such that v(S) = 0. Taking <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>></mo><mn>0</mn></mrow></math> forces <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>″</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo><</mo><mn>0</mn></mrow></math> , which is impossible. Therefore, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo>″</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi></mrow></math> s.t. v(S) = 0. We conclude that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo>=</mo><mi>v</mi><mo>″</mo><mo>=</mo><mi>v</mi></mrow></math> ; that is, v is an extreme point.

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>⇒</mo><mo stretchy="false">)</mo></mrow></math> Take v balanced and extremal and suppose by contradiction that there exists <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>0</mn><mo><</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo><</mo><mn>1</mn></mrow></math> . We distinguish two cases. Suppose first that there exists some core element <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>></mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo></mrow></math> . Then, consider the two games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo>,</mo><mi>v</mi><mo>″</mo></mrow></math> that differ from v only inasmuch as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>−</mo><mi>ϵ</mi></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>″</mo><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>+</mo><mi>ϵ</mi></mrow></math> , with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>0</mn><mo><</mo><mi>ϵ</mi><mo><</mo><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>−</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo></mrow></math> . Then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo>,</mo><mi>v</mi><mo>″</mo></mrow></math> are balanced because <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo>′</mo><mo stretchy="false">)</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo>″</mo><mo stretchy="false">)</mo></mrow></math> . Therefore, v is not extremal as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mo stretchy="false">(</mo><mi>v</mi><mo>′</mo><mo>+</mo><mi>v</mi><mo>″</mo><mo stretchy="false">)</mo><mo>/</mo><mn>2</mn></mrow></math> .

Suppose now that no such core element exists, that is, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> . As <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo><</mo><mn>1</mn><mo>,</mo></mrow></math> there exists <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>j</mi><mo>∈</mo><mi>N</mi><mo>\</mo><mi>S</mi></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mi>j</mi></msub><mo>></mo><mn>0</mn></mrow></math> . Similarly, as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>0</mn><mo><</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>,</mo></mrow></math> we conclude that there exists <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>S</mi></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mi>i</mi></msub><mo>></mo><mn>0</mn><mo>.</mo></mrow></math> We define the two games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo>,</mo><mi>v</mi><mo>″</mo></mrow></math> by

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mtable columnalign="left"><mtr><mtd><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>+</mo><mi>ϵ</mi><mo>,</mo><mo /><mi>v</mi><mo>″</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>−</mo><mi>ϵ</mi><mo>,</mo><mo /><mo>∀</mo><mi>T</mi><mo /><mtext>such</mtext><mo /><mtext>that</mtext><mo /><mi>i</mi><mo>∈</mo><mi>T</mi><mo>,</mo><mi>j</mi><mo>∉</mo><mi>T</mi></mtd></mtr><mtr><mtd><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>−</mo><mi>ϵ</mi><mo>,</mo><mo /><mi>v</mi><mo>″</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>+</mo><mi>ϵ</mi><mo>,</mo><mo /><mo>∀</mo><mi>T</mi><mo /><mtext>such</mtext><mo /><mtext>that</mtext><mo /><mi>j</mi><mo>∈</mo><mi>T</mi><mo>,</mo><mi>i</mi><mo>∉</mo><mi>T</mi></mtd></mtr><mtr><mtd><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>,</mo><mo /><mi>v</mi><mo>″</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>,</mo><mo /><mo /><mtext>otherwise</mtext><mo>,</mo></mtd></mtr></mtable></math>

with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>ϵ</mi><mo>></mo><mn>0</mn></mrow></math> small enough so that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>,</mo><mi>v</mi><mo>″</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>∈</mo><mo stretchy="false">[</mo><mn>0</mn><mo>,</mo><mn>1</mn><mo stretchy="false">]</mo></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>0</mn><mo><</mo><msub><mrow><mi>x</mi></mrow><mi>i</mi></msub><mo>−</mo><mi>ϵ</mi><mo><</mo><msub><mrow><mi>x</mi></mrow><mi>i</mi></msub><mo>+</mo><mi>ϵ</mi><mo><</mo><mn>1</mn><mo>,</mo><mo /><mn>0</mn><mo><</mo><msub><mrow><mi>x</mi></mrow><mi>j</mi></msub><mo>−</mo><mi>ϵ</mi><mo><</mo><msub><mrow><mi>x</mi></mrow><mi>j</mi></msub><mo>+</mo><mi>ϵ</mi><mo><</mo><mn>1</mn></mrow></math> . Clearly, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mo stretchy="false">(</mo><mi>v</mi><mo>′</mo><mo>+</mo><mi>v</mi><mo>″</mo><mo stretchy="false">)</mo><mo>/</mo><mn>2</mn></mrow></math> . Observe that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo>,</mo><mi>v</mi><mo>″</mo></mrow></math> are balanced, as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>′</mo><mo>,</mo><mi>x</mi><mo>″</mo><mo>∈</mo><msup><mrow><mi mathvariant="double-struck">R</mi></mrow><mi>N</mi></msup></mrow></math> defined by

are core elements of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>″</mo></mrow></math> , respectively. Hence, v is not extremal. □

Hence, a vertex v of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is characterized in terms of the subsets <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup><mo>\</mo><mo stretchy="false">{</mo><mo>Ø</mo><mo>,</mo><mi>N</mi><mo stretchy="false">}</mo></mrow></math> such that v(S) = 1. Let us denote the family of such subsets by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">D</mi><mo>.</mo></mrow></math> In next result, we treat the reciprocal problem; that is, we give the necessary and sufficient conditions for a family of subsets <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup><mo>\</mo><mo stretchy="false">{</mo><mo>Ø</mo><mo>,</mo><mi>N</mi><mo stretchy="false">}</mo></mrow></math> to determine a vertex of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

Theorem 9.

Let <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> be a family of subsets <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup><mo>\</mo><mo stretchy="false">{</mo><mo>Ø</mo><mo>,</mo><mi>N</mi><mo stretchy="false">}</mo><mo>.</mo></mrow></math> Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> defines a vertex of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> iff either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">D</mi><mo>=</mo><mo>Ø</mo></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>≠</mo><mo>Ø</mo><mo>.</mo></mrow></math>

Proof.

The case <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">D</mi><mo>=</mo><mo>Ø</mo></mrow></math> defines <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn><mo>,</mo><mo>∀</mo><mi>S</mi><mo>⊂</mo><mi>N</mi></mrow></math> , which is clearly balanced. For the rest of the proof, we consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>≠</mo><mo>Ø</mo></mrow></math> .

Choose some family <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>∋</mo><mi>i</mi></mrow></math> for some <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> , and construct the corresponding v. By definition, v is 0-1-valued. Hence, by Theorem 8, we just have to check balancedness. Take <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><msup><mrow><mi mathvariant="double-struck">R</mi></mrow><mi>N</mi></msup></mrow></math> s.t. x<subs>i</subs> = 1, and x<subs>j</subs> = 0 for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>j</mi><mo>≠</mo><mi>i</mi></mrow></math> . Then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> .

Conversely, suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">D</mi><mo>≠</mo><mo>Ø</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>=</mo><mo>Ø</mo></mrow></math> . We show that the game v corresponding to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> is not balanced. Consider a maximal family <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>B</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi>B</mi></mrow><mi>r</mi></msub><mo>∈</mo><mi mathvariant="script">D</mi></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>B</mi></mrow><mn>1</mn></msub><mo>∩</mo><mo>⋯</mo><mo>∩</mo><msub><mrow><mi>B</mi></mrow><mi>r</mi></msub><mo>=</mo><mi>T</mi><mo>≠</mo><mo>Ø</mo></mrow></math> . Suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo><mo>≠</mo><mo>Ø</mo></mrow></math> and take a core element <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> . As <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><msub><mrow><mi>B</mi></mrow><mi>i</mi></msub><mo stretchy="false">)</mo><mo>⩾</mo><mi>v</mi><mo stretchy="false">(</mo><msub><mrow><mi>B</mi></mrow><mi>i</mi></msub><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn><mo>,</mo><mo /><mi>x</mi><mo>⩾</mo><mn>0</mn></mrow></math> and x(N) = 1, it follows that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><msub><mrow><mi>B</mi></mrow><mi>i</mi></msub><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn></mrow></math> for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mo>...</mo><mo>,</mo><mi>r</mi></mrow></math> . Therefore,

Summing up we find

where the first inequality comes from the fact that each <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi><mo>\</mo><mi>T</mi></mrow></math> belongs at most to r − 1 sets in the family <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>B</mi></mrow><mn>1</mn></msub><mo>\</mo><mi>T</mi><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi>B</mi></mrow><mi>r</mi></msub><mo>\</mo><mi>T</mi></mrow></math> . If r > 1, we obtain

which implies x(N) > 1, a contradiction. Therefore, it must be that r = 1; that is, there exists <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>,</mo><mi>S</mi><mo>′</mo><mo>∈</mo><mi mathvariant="script">D</mi></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∩</mo><mi>S</mi><mo>′</mo><mo>=</mo><mo>Ø</mo></mrow></math> . As <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo>′</mo><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn></mrow></math> , a core element x should satisfy <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo>′</mo><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn></mrow></math> , which implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>⩾</mo><mn>2</mn></mrow></math> , a contradiction. As a conclusion, no core element x exists. □

The vertex corresponding to the empty collection is u<subs>N</subs>, the unanimity game centered on N (equivalently, the Dirac game δ<subs>N</subs>).

The next result gives explicitely the core of each vertex.

Proposition 2.

Let v be a vertex of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , with associate collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> . If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> is not the empty collection, then

which implies that the dimension of the core of v is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mo>∩</mo><mi mathvariant="script">D</mi><mo stretchy="false">|</mo><mo>−</mo><mn>1</mn></mrow></math> . If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> is the empty collection, then v = u<subs>N</subs>, whose core is the simplex <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="normal">Δ</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>≔</mo><mo stretchy="false">{</mo><mi>x</mi><mo>∈</mo><msubsup><mrow><mi mathvariant="double-struck">R</mi></mrow><mo>+</mo><mi>N</mi></msubsup><mo /><mo>:</mo><mo /><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></msub><mrow><msub><mrow><mi>x</mi></mrow><mi>i</mi></msub></mrow></mstyle><mo>=</mo><mn>1</mn><mo stretchy="false">}</mo></mrow></math> .

Proof.

x is a core element iff <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mo>∩</mo><mi mathvariant="script">D</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn></mrow></math> , from which the result follows. □

Proposition 2 generalizes some known results for simple games. Recall that a simple game v is a 0-1–valued game which is monotonic; that is, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊆</mo><mi>T</mi></mrow></math> implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩽</mo><mi>v</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo></mrow></math> . The collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> is called the collection of winning coalitions, and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi></mrow></math> is the set of veto players. It is well known that the core of a simple game is empty if and only if there is no veto player, and when nonempty, it is expressed by (8) (Peters [[16]]). Our result is more general as not all vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are simple games. We can also deduce the following (a simple game is proper if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">D</mi></mrow></math> implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>N</mi><mo>\</mo><mi>S</mi><mo>∉</mo><mi mathvariant="script">D</mi></mrow></math> ).

Corollary 1.

If a voting game is balanced, then it is proper.

Proof.

Suppose there exists a balanced voting game v that is not proper. Then, there exists S s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>,</mo><mi>N</mi><mo>\</mo><mi>S</mi><mo>∈</mo><mi mathvariant="script">D</mi><mo>.</mo></mrow></math> However, then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>=</mo><mo>Ø</mo><mo>,</mo></mrow></math> contradicting Theorem 9. □

5.3. Vertex Enumeration and Generation

We deduce from Theorem 9 that the enumeration of vertices amounts to the enumeration of the collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> whose intersection is nonempty. In the next result, we obtain a recursive formula to compute the number of vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , which we denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>b</mi></mrow><mi>n</mi></msub><mo>.</mo></mrow></math> Also, for further use, we introduce a number of notations. We denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">A</mi></mrow><mi>N</mi></msub></mrow></math> the set of all collections of sets in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup><mo>\</mo><mo stretchy="false">{</mo><mo>Ø</mo><mo>,</mo><mi>N</mi><mo stretchy="false">}</mo></mrow></math> , including the empty collection. The cardinality of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">A</mi></mrow><mi>N</mi></msub></mrow></math> is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>t</mi></mrow><mi>n</mi></msub><mo>=</mo><msup><mrow><mn>2</mn></mrow><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></msup><mo>.</mo></mrow></math> We introduce also <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">F</mi></mrow><mi>N</mi></msub></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">S</mi></mrow><mi>N</mi></msub></mrow></math> the set of all nonempty collections in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">A</mi></mrow><mi>N</mi></msub></mrow></math> with a nonempty intersection, and empty intersection, respectively. Their cardinalities are denoted by f<subs>n</subs> and s<subs>n</subs>, respectively. We have by definition <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>t</mi></mrow><mi>n</mi></msub><mo>=</mo><msub><mrow><mi>f</mi></mrow><mi>n</mi></msub><mo>+</mo><msub><mrow><mi>s</mi></mrow><mi>n</mi></msub><mo>+</mo><mn>1</mn></mrow></math> .

Theorem 10.

The number of vertices b<subs>n</subs> of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is given by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>b</mi></mrow><mi>n</mi></msub><mo>=</mo><msub><mrow><mi>f</mi></mrow><mi>n</mi></msub><mo>+</mo><mn>1</mn></mrow></math> , where f<subs>n</subs> can be obtained recursively as follows:

Proof.

Vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are in bijection with collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">D</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">A</mi></mrow><mi>N</mi></msub></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>≠</mo><mo>Ø</mo></mrow></math> , plus the empty collection (associated to u<subs>N</subs>).

Now, observe that f<subs>n</subs> can be split by considering the different choices for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>.</mo></mrow></math> Suppose that the intersection is the set S, that is, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>≔</mo><mi>S</mi><mo>,</mo></mrow></math> with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>=</mo><mi>k</mi><mo>.</mo></mrow></math> There are two possibilities: either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">D</mi></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∉</mo><mi mathvariant="script">D</mi><mo>.</mo></mrow></math> If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∉</mo><mi mathvariant="script">D</mi><mo>,</mo></mrow></math> then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> is associated to a nonempty collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> given by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>=</mo><mo stretchy="false">{</mo><mi>B</mi><mo>\</mo><mi>S</mi><mo>:</mo><mi>B</mi><mo>∈</mo><mi mathvariant="script">D</mi><mo stretchy="false">}</mo></mrow></math> . Observe that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">S</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub></mrow></math> . If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">D</mi><mo>,</mo></mrow></math> then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> is associated to a collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">A</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub></mrow></math> given by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>=</mo><mo stretchy="false">{</mo><mi>B</mi><mo>\</mo><mi>S</mi><mo>:</mo><mi>B</mi><mo>∈</mo><mi mathvariant="script">D</mi><mo stretchy="false">}</mo></mrow></math> if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">D</mi><mo>≠</mo><mo stretchy="false">{</mo><mi>S</mi><mo stretchy="false">}</mo></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>=</mo><mo>Ø</mo></mrow></math> otherwise. This way, summing over all S, we obtain

Making the change <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>′</mo><mo>=</mo><mi>n</mi><mo>−</mo><mi>k</mi></mrow></math> and using <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mrow><mo stretchy="true">(</mo><mtable><mtr><mtd><mi>n</mi></mtd></mtr><mtr><mtd><mi>k</mi></mtd></mtr></mtable><mo stretchy="true">)</mo></mrow><mo>=</mo><mrow><mo stretchy="true">(</mo><mtable><mtr><mtd><mi>n</mi></mtd></mtr><mtr><mtd><mi>n</mi><mo>−</mo><mi>k</mi></mtd></mtr></mtable><mo stretchy="true">)</mo></mrow><mo>,</mo></mrow></math> the result holds. □

The first values of b<subs>n</subs> can be seen in Table 1. As we can see, the number of vertices grows exponentially.

Table 1. Number of vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

Remark 1.

The integer sequence of the number of vertices given in Table 1 happens to be already known: It appears as sequence A051381 in the Online Encyclopedia of Integer Sequences (OEIS) (Sloane [[20]]) and is referred to as the "number of Boolean functions of n variables from Post class F(5, _I_inf_i_)." The founding paper is by Jojović and Kilibarda [[9]]. It gives an explicit (nonrecursive) formula for this sequence, shown in a very general context and with a long proof. We have kept our recursive formula with its short proof, as we need it hereafter for random vertex generation.

Table 2 enumerates the 19 vertices for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>=</mo><mn>3</mn><mo>.</mo></mrow></math> Vertices for n = 4 are available as supplementary material. For simplicity, braces and commas are omitted for writing sets, for example, 12 instead of {1, 2}, and so on. Also, we used the following notation: For any collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">C</mi><mo>⊆</mo><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup><mo>\</mo><mo stretchy="false">{</mo><mo>Ø</mo><mo>,</mo><mi>N</mi><mo stretchy="false">}</mo></mrow></math> , we define the game

Table 2. List of vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> for n = 3.

<table><thead valign="bottom"><tr><th align="left" rowspan="2" colspan="1" /><th align="center" colspan="6" rowspan="1">Vertices</th></tr><tr><th align="center" rowspan="1" colspan="1">1</th><th align="center" rowspan="1" colspan="1">2</th><th align="center" rowspan="1" colspan="1">3</th><th align="center" rowspan="1" colspan="1">12</th><th align="center" rowspan="1" colspan="1">13</th><th align="center" rowspan="1" colspan="1">23</th></tr></thead><tbody valign="top"><tr><td rowspan="1" colspan="1"><italic>u</italic><sub>123</sub></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><italic>u</italic><sub>1</sub></td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><italic>u</italic><sub>2</sub></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td></tr><tr><td rowspan="1" colspan="1"><italic>u</italic><sub>3</sub></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">1</td></tr><tr><td rowspan="1" colspan="1"><italic>u</italic><sub>12</sub></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><italic>u</italic><sub>13</sub></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><italic>u</italic><sub>23</sub></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td></tr><tr><td rowspan="1" colspan="1"><p><math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow xmlns=""><msub><mrow><mi>u</mi></mrow><mrow><mn>12</mn></mrow></msub><mo>∨</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>13</mn></mrow></msub></mrow></math></p></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><p><math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow xmlns=""><msub><mrow><mi>u</mi></mrow><mrow><mn>12</mn></mrow></msub><mo>∨</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>23</mn></mrow></msub></mrow></math></p></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td></tr><tr><td rowspan="1" colspan="1"><p><math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow xmlns=""><msub><mrow><mi>u</mi></mrow><mrow><mn>13</mn></mrow></msub><mo>∨</mo><msub><mrow><mi>u</mi></mrow><mrow><mn>23</mn></mrow></msub></mrow></math></p></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">1</td></tr><tr><td rowspan="1" colspan="1"><italic>d</italic><sub>1</sub></td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><italic>d</italic><sub>2</sub></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><italic>d</italic><sub>3</sub></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><p><math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow xmlns=""><msub><mrow><mi>d</mi></mrow><mrow><mn>1</mn><mo>,</mo><mn>12</mn></mrow></msub></mrow></math></p></td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><p><math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow xmlns=""><msub><mrow><mi>d</mi></mrow><mrow><mn>1</mn><mo>,</mo><mn>13</mn></mrow></msub></mrow></math></p></td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><p><math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow xmlns=""><msub><mrow><mi>d</mi></mrow><mrow><mn>2</mn><mo>,</mo><mn>12</mn></mrow></msub></mrow></math></p></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><p><math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow xmlns=""><msub><mrow><mi>d</mi></mrow><mrow><mn>2</mn><mo>,</mo><mn>23</mn></mrow></msub></mrow></math></p></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td></tr><tr><td rowspan="1" colspan="1"><p><math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow xmlns=""><msub><mrow><mi>d</mi></mrow><mrow><mn>3</mn><mo>,</mo><mn>13</mn></mrow></msub></mrow></math></p></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td></tr><tr><td rowspan="1" colspan="1"><p><math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow xmlns=""><msub><mrow><mi>d</mi></mrow><mrow><mn>3</mn><mo>,</mo><mn>23</mn></mrow></msub></mrow></math></p></td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">0</td><td align="center" rowspan="1" colspan="1">1</td></tr></tbody></table>

1 Note. " <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∨</mo><mo>″</mo></mrow></math> indicates the maximum.

for example, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>d</mi></mrow><mrow><mn>1</mn><mo>,</mo><mn>12</mn></mrow></msub><mo>=</mo><msub><mrow><mi>δ</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>δ</mi></mrow><mrow><mn>12</mn></mrow></msub><mo>+</mo><msub><mrow><mi>δ</mi></mrow><mrow><mn>123</mn></mrow></msub></mrow></math> .

Based on the result of Theorem 10 and its proof, we develop an algorithm that generates vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> uniformly at random. For this, consider the recursive expression (see proof of Theorem 10):

This formula comes from the following fact. Suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>=</mo><mi>S</mi></mrow></math> , with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>=</mo><mi>k</mi></mrow></math> . If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">D</mi></mrow></math> , removing S from <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> and from each set in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> yields a collection in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">A</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub></mrow></math> , while if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∉</mo><mi mathvariant="script">D</mi></mrow></math> , removing S from each set in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> yields a collection in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">S</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub></mrow></math> . Generating a collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> randomly in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">A</mi></mrow><mi>N</mi></msub></mrow></math> is very simple. It suffices to sort all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> sets (for example, lexicographically), and generate a vector of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> zeros and ones, the ones marking the sets that are in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">D</mi><mo>.</mo></mrow></math>

The algorithm goes as follows. First, we consider the possibility of drawing the vertex u<subs>N</subs>. As <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>b</mi></mrow><mi>n</mi></msub><mo>=</mo><msub><mrow><mi>f</mi></mrow><mi>n</mi></msub><mo>+</mo><mn>1</mn><mo>,</mo></mrow></math> we choose u<subs>N</subs> with probability

Table 3 shows the first values for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>p</mi></mrow><mn>0</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math> As it can be seen, this value is almost zero for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>></mo><mn>4</mn><mo>.</mo></mrow></math>

Table 3. First values of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>p</mi></mrow><mn>0</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

If u<subs>N</subs> is not chosen, then we must generate a random element <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">F</mi></mrow><mi>N</mi></msub><mo>.</mo></mrow></math> Using (9), we start by choosing the number of elements in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi><mo>.</mo></mrow></math> The probability of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mo>∩</mo><mi mathvariant="script">D</mi><mo stretchy="false">|</mo><mo>=</mo><mi>k</mi></mrow></math> is given by

Table 4 shows the first values of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi>p</mi></mrow><mn>1</mn><mi>k</mi></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

Table 4. First values of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi>p</mi></mrow><mn>1</mn><mi>k</mi></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

As we can see, by far the most likely is that the intersection of the sets of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> has cardinal 1. For <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>≥</mo><mn>5</mn><mo>,</mo></mrow></math> we may consider the probability distribution <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi>p</mi></mrow><mn>1</mn><mi>k</mi></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> to be approximately the Dirac distribution at k = 1.

The next step in the algorithm is to choose a set S of cardinal k at random among the <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">(</mo><mtable><mtr><mtd><mi>n</mi></mtd></mtr><mtr><mtd><mi>k</mi></mtd></mtr></mtable><mo stretchy="false">)</mo></mrow></math> possibilities and then to decide if either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">D</mi></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∉</mo><mi mathvariant="script">D</mi></mrow></math> . For a given selected k, the probability of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">D</mi></mrow></math> is given by

Table 5 gives the first values of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>p</mi></mrow><mn>2</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

Table 5. First values of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>p</mi></mrow><mn>2</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

As we can see, for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>≥</mo><mn>5</mn><mo>,</mo></mrow></math> the probabilities of whether the set S is in the selected collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> or not are approximately the same. If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">D</mi><mo>,</mo></mrow></math> it suffices to generate uniformly an element <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">A</mi></math> in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">A</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub></mrow></math> with the procedure described above, and the final collection will be

If the set <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∉</mo><mi mathvariant="script">D</mi></mrow></math> , it suffices to generate a random element <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">S</mi></math> of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">S</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub><mo>.</mo></mrow></math> To achieve this task, we use the fact that the quotient <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>s</mi></mrow><mi>n</mi></msub><mo>/</mo><msub><mrow><mi>t</mi></mrow><mi>n</mi></msub></mrow></math> is almost one for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>≥</mo><mn>5</mn><mo>,</mo></mrow></math> as it can be seen in Table 6.

Table 6. First values of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>s</mi></mrow><mi>n</mi></msub><mo>/</mo><msub><mrow><mi>t</mi></mrow><mi>n</mi></msub></mrow></math> .

Hence, for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>≥</mo><mn>3</mn><mo>,</mo></mrow></math> an element <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">S</mi></math> in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">S</mi></mrow><mi>N</mi></msub></mrow></math> can be generated uniformly at random via a rejection sampler on <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">A</mi></mrow><mi>N</mi></msub><mo>.</mo></mrow></math> That is, we generate an element <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">A</mi></math> of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">A</mi></mrow><mi>N</mi></msub><mo>.</mo></mrow></math> If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">A</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">S</mi></mrow><mi>N</mi></msub><mo>,</mo></mrow></math> then we return <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">A</mi></math> . Otherwise <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">A</mi><mo>∉</mo><msub><mrow><mi mathvariant="script">S</mi></mrow><mi>N</mi></msub></mrow></math> and we repeat the procedure until we obtain an element in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">S</mi></mrow><mi>N</mi></msub><mo>.</mo></mrow></math> For <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>≤</mo><mn>2</mn><mo>,</mo></mrow></math> we have <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>s</mi></mrow><mn>1</mn></msub><mo>=</mo><mn>0</mn></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>s</mi></mrow><mn>2</mn></msub><mo>=</mo><mn>1</mn><mo>,</mo></mrow></math> so these cases are trivial. Once <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">A</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">S</mi></mrow><mi>N</mi></msub></mrow></math> is generated, the chosen collection is

Here, we provide a summary of this algorithm.

Algorithm 1

(Vertex Sampler for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> )

<bold> Step 1: </bold> With probability <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>p</mi></mrow><mn>0</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> <bold>return</bold> <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">D</mi><mo>=</mo><mo>Ø</mo></mrow></math> (vertex u<subs>N</subs>). With probability <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>−</mo><msub><mrow><mi>p</mi></mrow><mn>0</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> go to Step 2.

<bold> Step 2: </bold> Choose some <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>∈</mo><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mo>...</mo><mo>,</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">}</mo></mrow></math> with probability <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi>p</mi></mrow><mn>1</mn><mi>k</mi></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Also, generate a set <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi></mrow></math> with k elements. Go to Step 3.

<bold> Step 3: </bold> With probability <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>p</mi></mrow><mn>2</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mi>k</mi><mo stretchy="false">)</mo></mrow></math> go to Step 4 and with probability <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>−</mo><msub><mrow><mi>p</mi></mrow><mn>2</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mi>k</mi><mo stretchy="false">)</mo></mrow></math> to Step 5.

<bold> Step 4: </bold> Generate at random some <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">A</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">A</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub><mo>.</mo></mrow></math> Then <bold>return</bold> <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> such that

<bold> Step 5: </bold> Generate at random (by rejection) some <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">A</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">S</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub><mo>.</mo></mrow></math> Then <bold>return</bold> <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> such that

Lemma 4.

The previous algorithm generates a vertex of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> uniformly at random.

Proof.

It suffices to show that any family <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> defining a vertex of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> has probability <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>/</mo><msub><mrow><mi>b</mi></mrow><mi>n</mi></msub><mo>.</mo></mrow></math> Suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>=</mo><mo>∩</mo><mi mathvariant="script">D</mi></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">D</mi><mo>.</mo></mrow></math> If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>=</mo><mi>k</mi><mo>,</mo></mrow></math> the probability of selecting <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> in the algorithm is given by

Similarly if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∉</mo><mi mathvariant="script">D</mi></mrow></math> , we get

As the probability of selecting the empty family is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>/</mo><msub><mrow><mi>b</mi></mrow><mi>n</mi></msub><mo>,</mo></mrow></math> the result follows. □

Let us finally deal with the computational complexity of generating a vertex according to the previous algorithm. First, let us establish a result about the growth of the number of vertices in terms of n.

Proposition 3.

The asymptotic growth rate of the number of vertices, b<subs>n</subs>, in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>b</mi></mrow><mi>n</mi></msub><mo>=</mo><mi>O</mi><mo stretchy="false">(</mo><mi>n</mi><msup><mrow><mn>2</mn></mrow><mrow><msup><mrow><mn>2</mn></mrow><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msup></mrow></msup><mo stretchy="false">)</mo><mo>.</mo></mrow></math> (10)

Proof.

As we know <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>b</mi></mrow><mi>n</mi></msub><mo>=</mo><msub><mrow><mi>f</mi></mrow><mi>n</mi></msub><mo>+</mo><mn>1</mn></mrow></math> where f<subs>n</subs> can be written as

Because <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>s</mi></mrow><mi>k</mi></msub><mo>≤</mo><msub><mrow><mi>t</mi></mrow><mi>k</mi></msub></mrow></math> , we get

Therefore,

Considering that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>t</mi></mrow><mi>k</mi></msub><mo>=</mo><msup><mrow><mn>2</mn></mrow><mrow><msup><mrow><mn>2</mn></mrow><mi>k</mi></msup><mo>−</mo><mn>2</mn></mrow></msup></mrow></math> , the dominant term in the expression <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mstyle displaystyle="false"><msubsup><mo>∑</mo><mrow><mi>k</mi><mo>=</mo><mn>1</mn></mrow><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msubsup><mrow><mo stretchy="false">(</mo><mtable><mtr><mtd><mi>n</mi></mtd></mtr><mtr><mtd><mi>k</mi></mtd></mtr></mtable><mo stretchy="false">)</mo></mrow></mstyle><msub><mrow><mi>t</mi></mrow><mrow><mi>n</mi><mo>−</mo><mi>k</mi></mrow></msub></mrow></math> is the one associated with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>=</mo><mi>n</mi><mo>−</mo><mn>1</mn></mrow></math> because the binomial coefficients are polynomial. Thus,

The following result refers to the computational complexity of the worst-case computation time required to generate a vertex by the previous algorithm.

Proposition 4.

The computational complexity of Algorithm 1 is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

Proof.

As usual, we will assume that generating a random number has a complexity of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>O</mi><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo><mo>.</mo></mrow></math> We also assume that the probabilities <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>p</mi></mrow><mn>0</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>,</mo><msubsup><mrow><mi>p</mi></mrow><mn>1</mn><mi>k</mi></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>,</mo><msub><mrow><mi>p</mi></mrow><mn>2</mn></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are known a priori. With these considerations, we know that steps 1 and 3 have a complexity of O(1) because only a random number needs to be generated. For step 2, we can choose k with complexity O(n), and we can generate the set S with complexity O(n) as well. For step 4, we can generate an element of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">A</mi></mrow><mrow><mi>N</mi><mo>\</mo><mi>S</mi></mrow></msub></mrow></math> by listing all possible subsets and generating a zero or one for each case, which means a complexity of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo stretchy="false">)</mo><mo>.</mo></mrow></math> Finally, for step 5, as the rejection rate tends to zero very quickly, the rejection rate will only affect small n, in which case it would multiply the computational complexity by a constant, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>O</mi><mo stretchy="false">(</mo><mi>C</mi><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo stretchy="false">)</mo><mo>.</mo></mrow></math> Taking all this into account, the total computational complexity would be <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>O</mi><mo stretchy="false">(</mo><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo stretchy="false">)</mo><mo>.</mo></mrow></math> □

We can see that in relation to the asymptotic number of vertices (10), the complexity is very reduced.

5.4. Adjacency of Vertices and Related Properties

Recall that two vertices are adjacent if they both belong to the same edge (one-dimensional face of the polytope). The aim of this section is twofold: first, we will see that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a combinatorial polytope and, as a consequence, that there exists a Hamiltonian path connecting each pair of vertices. Second, we will characterize adjacency of vertices corresponding to collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> such that their intersection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mi mathvariant="script">D</mi></mrow></math> is a singleton, which is by far the most common case, according to Table 4. In addition, these vertices have a core reduced to a singleton. This characterization is given in Theorems 13–15.

To prove that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is Hamilton connected, we first show that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a combinatorial polytope.

Definition 1

(Nadder and Pulleyblank [[14]]). A polytope <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">P</mi></math> is said to be combinatorial if the two following conditions hold:

All vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">P</mi></math> are 0-1 valued.

Given two vertices v<subs>1</subs>, v<subs>2</subs> of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">P</mi></math> , if they are not adjacent, then there exist two other different vertices v<subs>3</subs>, v<subs>4</subs> such that

For a combinatorial polytope, the following can be shown.

Theorem 11

(Naddef and Pulleyblank [[14]]). Let G be the adjacency graph of a combinatorial polytope. Then G is either a hypercube or is Hamilton connected.

Now, the following holds.

Lemma 5.

Let v<subs>1</subs>, v<subs>2</subs> be two distinct vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , with associate collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> . Then v<subs>1</subs>, v<subs>2</subs> are not adjacent if and only if there exist vertices v<subs>3</subs>, v<subs>4</subs> distinct from v<subs>1</subs>, v<subs>2</subs>, such that

In addition, the associated collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> satisfy

Proof.

Let us consider two vertices v<subs>1</subs>, v<subs>2</subs> with collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> . They are not adjacent if and only if there exist <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>λ</mi><mo>,</mo><mi>λ</mi><mo>′</mo><mo>∈</mo><mo stretchy="false">[</mo><mn>0</mn><mo>,</mo><mn>1</mn><mo stretchy="false">]</mo></mrow></math> and vertices v<subs>3</subs>, v<subs>4</subs> distinct from v<subs>1</subs>, v<subs>2</subs> such that

Let us denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> the collections associated to v<subs>3</subs>, v<subs>4</subs>. We have the following, using the decomposition into v<subs>1</subs>, v<subs>2</subs>:

Doing similarly with v<subs>3</subs>, v<subs>4</subs>, we deduce that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> . As <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> are distinct from <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> , this imposes that

Similar when <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>⊂</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> .

In any case, this implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>λ</mi><mo>=</mo><mi>λ</mi><mo>′</mo><mo>=</mo><mn>1</mn><mo>−</mo><mi>λ</mi><mo>′</mo><mo>=</mo><mn>1</mn><mo>−</mo><mi>λ</mi></mrow></math> , that is, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>λ</mi><mo>=</mo><mi>λ</mi><mo>′</mo><mo>=</mo><mn>1</mn><mo>/</mo><mn>2</mn></mrow></math> . □

Hence, as a consequence of Theorem 8 and Lemma 5, the following holds.

Theorem 12.

The polytope <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is combinatorial. Moreover, the adjacency graph of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is Hamilton connected.

Proof.

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is combinatorial as a straight consequence of Theorem 8 and Lemma 5. Now, from Theorem 11, the adjacency graph of this polytope is either Hamilton connected or a hypercube. For n = 1, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></math> is reduced to a singleton; therefore, the result holds trivially. For n = 2, the number of vertices is three, and therefore <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mn>2</mn><mo stretchy="false">)</mo></mrow></math> is not a hypercube. For <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>⩾</mo><mn>3</mn></mrow></math> , observe that for any distinct <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>,</mo><mi>j</mi><mo>∈</mo><mi>N</mi></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mover accent="true"><mi>δ</mi><mo stretchy="false">^</mo></mover></mrow><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></msub><mo>≔</mo><msub><mrow><mi>δ</mi></mrow><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></msub><mo>+</mo><msub><mrow><mi>δ</mi></mrow><mi>N</mi></msub></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mover accent="true"><mi>δ</mi><mo stretchy="false">^</mo></mover></mrow><mrow><mo stretchy="false">{</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> are vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> ; however, the game <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>δ</mi></mrow><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></msub><mo>+</mo><msub><mrow><mi>δ</mi></mrow><mrow><mo stretchy="false">{</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></msub><mo>+</mo><msub><mrow><mi>δ</mi></mrow><mi>N</mi></msub></mrow></math> is not a vertex of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . □

Let us now look more closely to adjacency. For further use, we illustrate on Figure 1 the condition of nonadjacency when <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> .

Graph: Figure 1. Nonadjacency of v1, v2, with associated collections D1,D2. Case (a): D3=B1∪(D1∩D2)∪B3, D4=B2∪(D1∩D2)∪B4; Case (b): D3=D1∪B3,D4=(D1∩D2)∪B4 (similar when D1,D2 exchanged); Case (c): D3=D1∪D2,D4=D1∩D2.

We now focus on the vertices whose associated collections have intersection reduced to a singleton.

Theorem 13.

Consider two vertices v<subs>1</subs>, v<subs>2</subs> of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , associated to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> respectively, and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo>=</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> . Then v<subs>1</subs> and v<subs>2</subs> are adjacent iff either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> or the converse, and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mi mathvariant="normal">Δ</mi><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo stretchy="false">|</mo><mo>=</mo><mn>1</mn></mrow></math> .

Proof.

Consider two vertices v<subs>1</subs>, v<subs>2</subs> as above and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub><mo stretchy="false">)</mo></mrow></math> . By definition, v(S) = 1 iff <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn><mo>/</mo><mn>2</mn></mrow></math> iff <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mi mathvariant="normal">Δ</mi><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> , and v(S) = 0 otherwise.

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>⇒</mo><mo stretchy="false">)</mo></mrow></math> Assume v<subs>1</subs>, v<subs>2</subs> are adjacent and suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>∋</mo><msub><mrow><mi>S</mi></mrow><mn>1</mn></msub></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∋</mo><msub><mrow><mi>S</mi></mrow><mn>2</mn></msub></mrow></math> . Consider v<subs>3</subs> generated by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><mo stretchy="false">{</mo><msub><mrow><mi>S</mi></mrow><mn>2</mn></msub><mo stretchy="false">}</mo></mrow></math> and v<subs>4</subs> generated by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><mo stretchy="false">{</mo><msub><mrow><mi>S</mi></mrow><mn>2</mn></msub><mo stretchy="false">}</mo></mrow></math> . Then v<subs>3</subs>, v<subs>4</subs> differ from v<subs>1</subs>, v<subs>2</subs> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>4</mn></msub><mo stretchy="false">)</mo><mo>/</mo><mn>2</mn></mrow></math> , contradicting that v<subs>1</subs>, v<subs>2</subs> are adjacent. Consequently, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> (or the converse).

Assuming the former, let us prove that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo stretchy="false">|</mo><mo>=</mo><mn>1</mn></mrow></math> . Suppose on the contrary that there exist distinct <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>S</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi>S</mi></mrow><mn>2</mn></msub><mo>⊆</mo><mi>N</mi></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>S</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi>S</mi></mrow><mn>2</mn></msub><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> . Hence, we can consider the games v<subs>3</subs>, v<subs>4</subs> generated by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><mo stretchy="false">{</mo><msub><mrow><mi>S</mi></mrow><mn>1</mn></msub><mo stretchy="false">}</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><mo stretchy="false">{</mo><msub><mrow><mi>S</mi></mrow><mn>1</mn></msub><mo stretchy="false">}</mo></mrow></math> , respectively. Thus, defined, v<subs>3</subs>, v<subs>4</subs> differ from v<subs>1</subs>, v<subs>2</subs> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>4</mn></msub><mo stretchy="false">)</mo><mo>/</mo><mn>2</mn></mrow></math> , a contradiction.

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>⇐</mo><mo stretchy="false">)</mo></mrow></math> Suppose by contradiction that v<subs>1</subs>, v<subs>2</subs> are not adjacent. Then by Lemma 5 there exist vertices <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi>v</mi></mrow><mn>4</mn></msub><mo>∈</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> different from v<subs>1</subs>, v<subs>2</subs> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub><mo>=</mo><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>4</mn></msub></mrow></math> , with associated collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> satisfying <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> and similarly for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> .

If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>⊈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>⊈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> we are done considering <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>=</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub><mo>=</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> . Otherwise, assume <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> . The above constraints resume to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> and the same for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> . As v<subs>3</subs>, v<subs>4</subs> differ from v<subs>1</subs>, v<subs>2</subs>, strict inclusion must hold throughout, which implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mi mathvariant="normal">Δ</mi><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> . The case <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> is similar. Hence, the result holds. □

Lemma 6.

Let v<subs>1</subs>, v<subs>2</subs> be two vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> with associated collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo>,</mo><mo /><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></math> . If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>≠</mo><mi>j</mi></mrow></math> , then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>≠</mo><mo>Ø</mo><mo>,</mo><mo /><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> .

Proof.

Suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>≠</mo><mi>j</mi></mrow></math> . This implies that in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> there must exist S such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∌</mo><mi>j</mi></mrow></math> (otherwise we would have <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>⊇</mo><mo stretchy="false">{</mo><mi>i</mi><mo>,</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></math> ). Hence, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> . Similarly, there must exist <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>∌</mo><mi>i</mi></mrow></math> , which implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> . □

Theorem 14.

Consider two vertices v<subs>1</subs>, v<subs>2</subs> of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , associated to families <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> respectively, such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo>≠</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></math> and suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><mo>Ø</mo></mrow></math> . Then v<subs>1</subs> and v<subs>2</subs> are adjacent iff there do not exist collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> distinct from <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> satisfying

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub><mo>=</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> ,

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> .

Proof.

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>⇒</mo><mo stretchy="false">)</mo></mrow></math> Suppose there exist <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> satisfying the above conditions. Then, because <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> are nonempty by condition 3, they determine vertices v<subs>3</subs> and v<subs>4</subs>, respectively. By conditions 1 and 2, it follows that

showing that v<subs>1</subs> and v<subs>2</subs> are not adjacent.

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>⇐</mo><mo stretchy="false">)</mo></mrow></math> Suppose v<subs>1</subs> and v<subs>2</subs> are not adjacent. By Lemma 5, it follows that there exist vertices v<subs>3</subs> and v<subs>4</subs> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub><mo>=</mo><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>4</mn></msub></mrow></math> , determined by collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> , respectively, satisfying conditions 1 and 2. Now, condition 3 is implied by the fact that v<subs>3</subs> and v<subs>4</subs> are vertices. □

Lemma 7.

Proof.

It suffices to note that if there exists <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>j</mi><mo>∈</mo><mi>S</mi></mrow></math> , then we can consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>≔</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><mo stretchy="false">{</mo><mi>S</mi><mo stretchy="false">}</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub><mo>≔</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>∪</mo><mo stretchy="false">{</mo><mi>S</mi><mo stretchy="false">}</mo></mrow></math> , which fulfill the three conditions of Theorem 14. □

Theorem 15.

Consider two vertices v<subs>1</subs>, v<subs>2</subs> of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , associated to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> respectively, and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo>≠</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></math> . Suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> and denote <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mo stretchy="false">(</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo stretchy="false">)</mo><mo>=</mo><mi>T</mi><mo>∪</mo><mo stretchy="false">{</mo><mi>i</mi><mo>,</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></math> , with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>⊆</mo><mi>N</mi><mo>\</mo><mo stretchy="false">{</mo><mi>i</mi><mo>,</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></math> . Then, v<subs>1</subs> and v<subs>2</subs> are adjacent iff the following two conditions are satisfied:

For all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>,</mo><mo /><mi>j</mi><mo>∉</mo><mi>S</mi></mrow></math> , and for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∉</mo><mi>S</mi></mrow></math>

For every disjoint <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>⊆</mo><mi>T</mi></mrow></math> and disjoint <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>⊆</mo><mi>T</mi></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> , there exists either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>⊈</mo><mi>S</mi></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>⊈</mo><mi>S</mi></mrow></math> , or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>′</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>⊈</mo><mi>S</mi><mo>′</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>⊈</mo><mi>S</mi><mo>′</mo></mrow></math> .

Proof.

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>⇒</mo><mo stretchy="false">)</mo></mrow></math> Let us consider two adjacent vertices v<subs>1</subs>, v<subs>2</subs> as above, and let <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub><mo stretchy="false">)</mo><mo>/</mo><mn>2</mn></mrow></math> . By Lemma 6, we know that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> . Suppose there exists <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>S</mi></mrow><mn>1</mn></msub><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>j</mi><mo>∈</mo><msub><mrow><mi>S</mi></mrow><mn>1</mn></msub></mrow></math> . Consider v<subs>3</subs>, v<subs>4</subs> generated by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><mo stretchy="false">{</mo><msub><mrow><mi>S</mi></mrow><mn>1</mn></msub><mo stretchy="false">}</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>∪</mo><mo stretchy="false">{</mo><msub><mrow><mi>S</mi></mrow><mn>1</mn></msub><mo stretchy="false">}</mo></mrow></math> , respectively. Then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>4</mn></msub><mo stretchy="false">)</mo><mo>/</mo><mn>2</mn></mrow></math> , a contradiction. The argument is the same with the existence of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>S</mi></mrow><mn>2</mn></msub><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><msub><mrow><mi>S</mi></mrow><mn>2</mn></msub></mrow></math> . This proves the first condition.

We prove the second condition. Our strategy is to show that the nonexistence of a partition of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> like in Figure 1 (which is equivalent to adjacency) implies the second condition. We first observe that the first condition implies that a partition of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> like in cases (b) and (c) of Figure 1 can never occur. Indeed, in these cases, by the first condition, it follows that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub></mrow></math> contains neither i nor j, and because <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>≠</mo><mo>Ø</mo><mo>,</mo></mrow></math> it must contain some other element, say k. But then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>∈</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo></mrow></math> a contradiction. Consequently, we only have to consider case (a). Let us call (a)-partition a possible partition like in Case (a) and show that it is not possible to build such a partition. We distinguish different cases in terms of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> .

Suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><mo stretchy="false">(</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo stretchy="false">)</mo><mo>=</mo><mi>T</mi><mo>∪</mo><mo stretchy="false">{</mo><mi>i</mi><mo>,</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></math> with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi>T</mi><mo stretchy="false">|</mo><mo>⩾</mo><mn>2</mn></mrow></math> . There exists an (a)-partition iff one can have <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">(</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>1</mn></msub><mo>∩</mo><mo stretchy="false">(</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>3</mn></msub><mo stretchy="false">)</mo><mo stretchy="false">)</mo><mo>∩</mo><mi>T</mi><mo>≠</mo><mo>Ø</mo><mo>,</mo><mo /><mo stretchy="false">(</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>2</mn></msub><mo>∩</mo><mo stretchy="false">(</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>4</mn></msub><mo stretchy="false">)</mo><mo stretchy="false">)</mo><mo>∩</mo><mi>T</mi><mo>≠</mo><mo>Ø</mo></mrow></math> (to ensure nonemptiness of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> ), and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>1</mn></msub><mo>∩</mo><mo stretchy="false">(</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>2</mn></msub><mo stretchy="false">)</mo><mo>∩</mo><mi>T</mi><mo>=</mo><mo>Ø</mo></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>3</mn></msub><mo>∩</mo><mo stretchy="false">(</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>4</mn></msub><mo stretchy="false">)</mo><mo>∩</mo><mi>T</mi><mo>=</mo><mo>Ø</mo></mrow></math> (to ensure <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></math> ). By letting <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mi>i</mi></msub><mo>≔</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mi>i</mi></msub><mo>∩</mo><mi>T</mi></mrow></math> for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mo>...</mo><mo>,</mo><mn>4</mn></mrow></math> , this is equivalent to: <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∃</mo><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>⊆</mo><mi>T</mi><mo>,</mo><mo /><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>=</mo><mo>Ø</mo><mo>,</mo><mo /><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>=</mo><mo>Ø</mo><mo>,</mo><mo /><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>≠</mo><mo>Ø</mo><mo>,</mo><mo /><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> such that for every <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> , either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>⊆</mo><mi>S</mi></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>⊆</mo><mi>S</mi></mrow></math> , and for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>′</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> , either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>⊆</mo><mi>S</mi><mo>′</mo></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>⊆</mo><mi>S</mi><mo>′</mo></mrow></math> . Therefore, there is no (a)-partition iff: <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∀</mo><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>⊆</mo><mi>T</mi></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>=</mo><mo>Ø</mo><mo>,</mo><mo /><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>=</mo><mo>Ø</mo><mo>,</mo><mo /><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>≠</mo><mo>Ø</mo><mo>,</mo><mo /><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>∩</mo><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> , either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∃</mo><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>1</mn></msub><mo>⊈</mo><mi>S</mi><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>2</mn></msub><mo>⊈</mo><mi>S</mi></mrow></math> , or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∃</mo><mi>S</mi><mo>′</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>K</mi></mrow><mn>3</mn></msub><mo>⊈</mo><mi>S</mi><mo>′</mo><mo>,</mo><msub><mrow><mi>K</mi></mrow><mn>4</mn></msub><mo>⊈</mo><mi>S</mi></mrow></math> .

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>⇐</mo><mo stretchy="false">)</mo></mrow></math> Suppose v<subs>1</subs>, v<subs>2</subs> are not adjacent. Then by Lemma 5, there exist <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>4</mn></msub></mrow></math> such that either an (a)-partition, a (b)-partition, or a (c)-partition of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> exists (Figure 1).

Suppose there exists an (a)-partition. If the first condition is not satisfied, we are done. Hence, assume that the first condition holds. Then, proceeding as in cases 1, 2, 3 in the <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>⇒</mo><mo stretchy="false">)</mo></mrow></math> part, we deduce that if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi>T</mi><mo stretchy="false">|</mo><mo><</mo><mn>2</mn></mrow></math> , no (a)-partition can exist, and if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi>T</mi><mo stretchy="false">|</mo><mo>⩾</mo><mn>2</mn></mrow></math> , the existence of a (a)-partition implies the negation of the second condition, as desired.

Suppose there exists a (b)-partition. Then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>≔</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>3</mn></msub></mrow></math> satisfies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> , say it contains k. Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>∈</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> , which implies k = i. This in turn implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>3</mn></msub></mrow></math> , which violates the first condition.

Suppose there exists a (c)-partition. Observe this cannot occur. Indeed, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>=</mo><mo>∩</mo><mo stretchy="false">(</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo stretchy="false">)</mo><mo>≠</mo><mo>Ø</mo></mrow></math> , say <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>∈</mo><mo>∩</mo><mo stretchy="false">(</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo stretchy="false">)</mo></mrow></math> . In particular, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>∈</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> , which implies k = i. But then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> , a contradiction. □

Observe that if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>=</mo><mo>Ø</mo></mrow></math> in the above theorem, then only the first condition remains.

In Figure 2, the adjacency graph of vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> can be seen. This graph has the vertices as nodes, and there is an edge between two nodes if they are adjacent. Finally, to illustrate Theorem 12, in Figure 3, a Hamilton path joining u<subs>123</subs> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>d</mi></mrow><mrow><mn>2</mn><mo>,</mo><mn>23</mn></mrow></msub></mrow></math> is given.

Graph: Figure 2. Adjacency graph of BG+(n) for n=3. Vertices are defined in Table 2.

Graph: Figure 3. (Color online) Adjacency graph of BG+(n) for n=3. The Hamiltonian path joining u123 and d2,23 is indicated in thick line.

5.5. Facets

The following result gives the facets of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

Theorem 16.

The following holds:

Each equality v(S) = 0, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>≠</mo><mo>Ø</mo><mo>,</mo><mi>N</mi></mrow></math> , defines a facet.

Each equality <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></msub><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn><mo>,</mo><mo /><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , defines a facet.

The number of facets is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>+</mo><mi>b</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>−</mo><mn>3</mn><mo>,</mo></mrow></math> with b(n) the number of minimal balanced collections on N.

Proof.

Recall from (4) that

All the facets of a polytope can be obtained by converting into equality one of the inequalities that define it. However, some inequalities might define a lower-dimensional face.

We claim that every equality <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn><mo>,</mo><mi>S</mi><mo>≠</mo><mi>N</mi><mo>,</mo><mo>Ø</mo></mrow></math> , leads by intersection with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> to a face of dimension <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>3</mn></mrow></math> and hence a facet. To see this, consider the games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mover accent="true"><mi>δ</mi><mo stretchy="false">^</mo></mover></mrow><mi>T</mi></msub><mo>=</mo><msub><mrow><mi>δ</mi></mrow><mi>T</mi></msub><mo>+</mo><msub><mrow><mi>δ</mi></mrow><mi>N</mi></msub><mo>,</mo><mi>T</mi><mo>≠</mo><mi>S</mi><mo>,</mo><mi>N</mi><mo>,</mo><mo>Ø</mo></mrow></math> , that is,

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mover accent="true"><mi>δ</mi><mo stretchy="false">^</mo></mover></mrow><mi>T</mi></msub><mo stretchy="false">(</mo><mi>S</mi><mo>′</mo><mo stretchy="false">)</mo><mo>=</mo><mrow><mo stretchy="true">{</mo><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>1</mn><mo>,</mo></mrow></mtd><mtd columnalign="left"><mrow><mtext>if</mtext><mo /><mi>S</mi><mo>′</mo><mo>=</mo><mi>T</mi><mo /><mtext>or</mtext><mo /><mi>S</mi><mo>′</mo><mo>=</mo><mi>N</mi></mrow></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>0</mn><mo>,</mo></mrow></mtd><mtd columnalign="left"><mrow><mtext>otherwise</mtext><mo>.</mo></mrow></mtd></mtr></mtable></mrow></mrow></mrow></math>

Clearly, the games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mover accent="true"><mi>δ</mi><mo stretchy="false">^</mo></mover></mrow><mi>T</mi></msub></mrow></math> for every <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>≠</mo><mo>Ø</mo><mo>,</mo><mi>S</mi><mo>,</mo><mi>N</mi></mrow></math> are vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and satisfy v(S) = 0. In addition, the game u<subs>N</subs> also belongs to the face v(S) = 0. These <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> games are affinely independent because the <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>3</mn></mrow></math> games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mover accent="true"><mi>δ</mi><mo stretchy="false">^</mo></mover></mrow><mi>T</mi></msub><mo>−</mo><msub><mrow><mi>u</mi></mrow><mi>N</mi></msub><mo>=</mo><msub><mrow><mi>δ</mi></mrow><mi>T</mi></msub></mrow></math> are linearly independent, which proves the claim.

Consider any m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and the corresponding inequality. For any <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> define the game v<sups>i</sups> by

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi>v</mi></mrow><mi>i</mi></msup><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mrow><mo stretchy="true">{</mo><mrow><mtable columnalign="left"><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>1</mn><mo>,</mo></mrow></mtd><mtd columnalign="left"><mrow><mtext>if</mtext><mo /><mi>S</mi><mo>∋</mo><mi>i</mi><mo /><mtext>and</mtext><mo /><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi><mo>,</mo><mo /><mtext>or</mtext><mo /><mi>S</mi><mo>=</mo><mi>N</mi></mrow></mtd></mtr><mtr columnalign="left"><mtd columnalign="left"><mrow><mn>0</mn><mo>,</mo></mrow></mtd><mtd columnalign="left"><mrow><mtext>otherwise</mtext><mo>.</mo></mrow></mtd></mtr></mtable></mrow></mrow></mrow></math>

Observe that each v<sups>i</sups> is a vertex of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , and v<sups>i</sups> satisfies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></msub><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><msup><mrow><mi>v</mi></mrow><mi>i</mi></msup><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn></mrow></math> . Moreover, the number of distinct v<sups>i</sups> is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo><mo>≕</mo><mi>p</mi></mrow></math> . Let us call <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msup><mo>,</mo><mo>...</mo><mo>,</mo><msup><mrow><mi mathvariant="script">D</mi></mrow><mi>n</mi></msup></mrow></math> the corresponding collections defining <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi>v</mi></mrow><mn>1</mn></msup><mo>,</mo><mo>...</mo><mo>,</mo><msup><mrow><mi>v</mi></mrow><mi>n</mi></msup></mrow></math> . Define successively the collections

And so on...

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="script">D</mi></mrow><mi>n</mi></msup><mo>∪</mo><mo stretchy="false">{</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo stretchy="false">}</mo></mrow></math> (if {n} not already present in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="script">D</mi></mrow><mi>n</mi></msup></mrow></math> )

Observe that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msup><mo>,</mo><mo>...</mo><mo>,</mo><msup><mrow><mi mathvariant="script">D</mi></mrow><mi>n</mi></msup></mrow></math> plus all the above ones yields <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> distinct collections, where each <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msup><mrow><mn>2</mn></mrow><mi>N</mi></msup><mo>\</mo><mo stretchy="false">{</mo><mo>Ø</mo><mo>,</mo><mi>N</mi><mo stretchy="false">}</mo></mrow></math> is present at least once. Moreover, each collection defines a game that is a vertex belonging to the face defined by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></msub><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mn>1</mn></mrow></math> . It follows that these <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>2</mn></mrow></math> games are affinely independent (as, for example, the family of the <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mn>2</mn></mrow><mi>n</mi></msup><mo>−</mo><mn>3</mn></mrow></math> games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>−</mo><msup><mrow><mi>v</mi></mrow><mn>1</mn></msup></mrow></math> with v any game as defined above forms a linearly independent family), and therefore the inequality <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></msub><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩽</mo><mn>1</mn></mrow></math> defines a facet.

3. This is an immediate consequence of 1 and 2. □

6. When Is the Core Reduced to a Singleton?

We address in this section the question of which balanced game in any of the three sets of balanced games under consideration in this paper has a core reduced to a singleton. Let us call this for simplicity a point core.

A first observation is the following: Because <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> are subsets of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , it suffices to find all games in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> that have a point core and then to check if they belong to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> or satisfy <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mi>α</mi></mrow></math> . Therefore, we put our effort into finding all balanced games in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> having a point core.

6.1. Case of BG(n)

For a given balanced game v in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , let us denote by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">E</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> the collection

The set <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">E</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> is called the set of effective coalitions for v. Obviously, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>N</mi><mo>∈</mo><mi mathvariant="script">E</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">E</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo><mo>=</mo><mo stretchy="false">{</mo><mi>N</mi><mo stretchy="false">}</mo></mrow></math> if and only if the core is full-dimensional, that is, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">(</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">)</mo></mrow></math> -dimensional. The following result of Laplace Mermoud et al. [[11]] is central in our investigation.

Lemma 8.

<math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">E</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> is the union of all minimal balanced collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> such that

An immediate consequence of this lemma is the following general result.

Proposition 5.

If v belongs to the interior of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>,</mo></mrow></math> then v has not a point core.

Indeed, for such a game v, no inequality in (2) is tight, which implies by Lemma 8 that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">E</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> is reduced to N, which means that the core is full-dimensional.

We begin with a simple result.

Lemma 9.

Any game in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mtext mathvariant="sans-serif">Lin</mtext><mo stretchy="false">(</mo><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow></math> has a point core.

Proof.

Take <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><mtext mathvariant="sans-serif">Lin</mtext><mo stretchy="false">(</mo><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo stretchy="false">)</mo></mrow></math> . Then, this game can be written as

As each <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>w</mi></mrow><mi>i</mi></msub><mo>=</mo><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> is an additive game, so is v, and therefore v has a point core. □

We turn to the examination of facets, which by Theorem 4 corresponds to minimal balanced collections in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

Theorem 17.

Consider a minimal balanced collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and its corresponding facet <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">F</mi></math> in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . The following holds:

If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo><mo>=</mo><mi>n</mi></mrow></math> , every game in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">F</mi></math> has a point core.

Otherwise, no game in the relative interior of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">F</mi></math> has a point core.

Proof.

Consider a minimal balanced collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> and its corresponding facet <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">F</mi></math> in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

Take v in the relative interior of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">F</mi></math> . Any core element <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> satisfies the system

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>,</mo><mo /><mi>S</mi><mo>∈</mo><mi mathvariant="script">E</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math> (11)

By Lemma 8, we have that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">E</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo><mo>=</mo><mi mathvariant="script">B</mi><mo>∪</mo><mo stretchy="false">{</mo><mi>N</mi><mo stretchy="false">}</mo><mo>.</mo></mrow></math> Observe that the inequality <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo></mrow></math> is redundant with the others as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> is balanced. Therefore, in (11), we can remove S = N. Consider now the system

As <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> is minimal, the above system has a unique solution. Hence, the rank of the matrix M of this linear system is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo></mrow></math> . Conversely, the matrix of System (11) (without N) is the transpose of M, and therefore has rank <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo></mrow></math> . It follows that the solution of (11) is unique iff <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo><mo>=</mo><mi>n</mi></mrow></math> .

It remains to prove that games on the frontier of a facet <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">F</mi></math> defined by a m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo><mo>=</mo><mi>n</mi></mrow></math> have also a point core. Take v in the frontier of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">F</mi></math> . Then, v also belongs to other facets, say <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mi>r</mi></msub></mrow></math> , associated with m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mi>r</mi></msub><mo>,</mo></mrow></math> respectively. It follows from Lemma 8 that any core element <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> satisfies the system

As the system <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo></mrow></math> for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></math> has a unique solution, it follows that either the above system has no solution or a unique one. However, the system has solution by the fact that v is balanced. □

Remark 2.

As the lineality space is the intersection of all facets, Lemma 9 is obtained as a corollary of the above theorem.

Part 2 of the previous proof allows for a characterization of games with a point core in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math>

Theorem 18.

Proof.

Using the same argument as in the proof of Theorem 17, any core element x of a game v in the relative interior of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">F</mi></math> satisfies the system

and the conclusion follows immediately. Now, if v is in the frontier, proceed as in the proof of Theorem 17. □

The above results plus the result on the interior of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> completely characterizes the set of games with a point core.

Example 2.

Consider n = 4, and the facets <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>2</mn></msub></mrow></math> associated with the m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>1</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mn>12</mn><mo>,</mo><mn>34</mn><mo stretchy="false">}</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>2</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>234</mn><mo stretchy="false">}</mo></mrow></math> . Then, games in the relative interior of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>2</mn></msub></mrow></math> have no point core. Although <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>1</mn></msub><mo stretchy="false">|</mo><mo>+</mo><mo stretchy="false">|</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>2</mn></msub><mo stretchy="false">|</mo><mo>=</mo><mn>4</mn><mo>=</mo><mi>n</mi></mrow></math> , the rank of the matrix <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msup><mrow><mn>1</mn></mrow><mi>S</mi></msup><mo>,</mo><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>2</mn></msub><mo stretchy="false">}</mo></mrow></math> is three; therefore, games in the face <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>2</mn></msub></mrow></math> have no point core but a core of dimension 1.

Consider now the face <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>3</mn></msub></mrow></math> associated with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>3</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mn>13</mn><mo>,</mo><mn>124</mn><mo>,</mo><mn>23</mn><mo stretchy="false">}</mo></mrow></math> . Games in the relative interior of this facet have no point core. However, games in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">F</mi><mo>′</mo><mo>=</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>3</mn></msub></mrow></math> have a point core because the rank of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mi>S</mi></msup><mo>,</mo><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>1</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>3</mn></msub><mo stretchy="false">}</mo></mrow></math> is four.

Example 3.

Let us give a complete analysis with n = 3. The lineality space has basis <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mn>1</mn><mo stretchy="false">}</mo></mrow></msub><mo>,</mo><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mn>2</mn><mo stretchy="false">}</mo></mrow></msub><mo>,</mo><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mn>3</mn><mo stretchy="false">}</mo></mrow></msub><mo stretchy="false">}</mo></mrow></math> . The extremal rays are <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>−</mo><msub><mrow><mi>δ</mi></mrow><mrow><mn>12</mn></mrow></msub><mo>,</mo><mo>−</mo><msub><mrow><mi>δ</mi></mrow><mrow><mn>13</mn></mrow></msub><mo>,</mo><mo>−</mo><msub><mrow><mi>δ</mi></mrow><mrow><mn>23</mn></mrow></msub></mrow></math> , and

Table 7 gives the facets and which extremal rays (except those of the lineality space that all belong to every facet) belong to them.

Table 7. Structure of the facets and extremal rays.

This shows the neighborhood relations between facets (two facets are neighbors if they have a common extremal ray that is not in the lineality space), and consequently all faces. Interestingly, the figure below of the five facets gives a faithful representation of the neighborhood relation, although it is not a correct geometrical representation. The part in blue indicates where are the games with a point core.

Graph

6.2. Case of BGα(n)

Because the facet of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> corresponding to some m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is the intersection of the facet of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> corresponding to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> and the hyperplane <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>N</mi><mo stretchy="false">)</mo><mo>=</mo><mi>α</mi></mrow></math> , every result established for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> applies without any change to <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

6.3. Case of BG+(n)

We know by Theorem 16 that every m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> determines a facet of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , which is therefore a subset of the corresponding facet in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Therefore, once again, Theorem 17 applies without any change.

On the other hand, we know from Theorem 16 that each equality v(S) = 0, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>≠</mo><mo>Ø</mo><mo>,</mo><mi>N</mi></mrow></math> , defines a facet. Taking a game v in the relative interior of this facet, as no inequality <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mstyle displaystyle="false"><msub><mo>∑</mo><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></msub><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩽</mo><mn>1</mn></mrow></math> is tight, v is in the interior of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , and by Proposition 5, v has not a point core. Summarizing, we have found the following.

Theorem 19.

The following holds for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> :

Every game in a facet defined by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mi mathvariant="script">B</mi><mo stretchy="false">|</mo><mo>=</mo><mi>n</mi></mrow></math> has a point core.

No game in the relative interior of any other facet has a point core.

Consider a face <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">F</mi><mo>=</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>1</mn></msub><mo>∩</mo><mo>⋯</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mi>p</mi></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mrow><mi>p</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>∩</mo><mo>⋯</mo><mo>∩</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mi>r</mi></msub></mrow></math> , with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">F</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mi>p</mi></msub></mrow></math> being facets associated to m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>1</mn></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mi>p</mi></msub></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">F</mi></mrow><mrow><mi>p</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi mathvariant="script">F</mi></mrow><mi>r</mi></msub></mrow></math> being facets associated to sets <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>S</mi></mrow><mrow><mi>p</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi>S</mi></mrow><mi>r</mi></msub></mrow></math> . Then any game in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">F</mi></math> has a point core iff the rank of the matrix <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msup><mrow><mn mathvariant="bold">1</mn></mrow><mi>S</mi></msup><mo>,</mo><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mn>1</mn></msub><mo>∪</mo><mo>⋯</mo><mo>∪</mo><msub><mrow><mi mathvariant="script">B</mi></mrow><mi>p</mi></msub><mo stretchy="false">}</mo></mrow></math> is n.

In 3 of the above theorem, the facets associated to sets <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>S</mi></mrow><mrow><mi>p</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>,</mo><mo>...</mo><mo>,</mo><msub><mrow><mi>S</mi></mrow><mi>r</mi></msub></mrow></math> play no role.

Another approach is to address the question under the point of view of vertices, instead of faces. Proposition 2 gives immediately the answer to our question as far as vertices are concerned.

Corollary 2.

A vertex <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> associated to collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">D</mi></math> has a point core iff <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">|</mo><mo>∩</mo><mi mathvariant="script">D</mi><mo stretchy="false">|</mo><mo>=</mo><mn>1</mn></mrow></math> .

Another point is to investigate whether games in a face whose all vertices have a point core have also a point core. Surprisingly, this is false in general already for edges, that is, faces of dimension 1. The next theorem clarifies the situation for edges.

Theorem 20.

Consider two adjacent vertices v<subs>1</subs>, v<subs>2</subs> of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , with associated collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> , respectively, and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mi>j</mi><mo stretchy="false">}</mo></mrow></math> . Consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mi>λ</mi><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>+</mo><mo stretchy="false">(</mo><mn>1</mn><mo>−</mo><mi>λ</mi><mo stretchy="false">)</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub></mrow></math> with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>λ</mi><mo>∈</mo><mrow><mo stretchy="false">]</mo><mrow><mn>0</mn><mo>,</mo><mn>1</mn></mrow><mo stretchy="false">[</mo></mrow></mrow></math> , that is, a game in the edge between v<subs>1</subs>, v<subs>2</subs>. Then:

If i = j, then C(v) is a singleton, that is, v has a point core.

If <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>≠</mo><mi>j</mi></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>⩽</mo><mn>4</mn></mrow></math> , then v has a point core.

Proof.

We prove the result in two steps.

Suppose i = j. Then by Theorem 13, we may suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> . Take <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> . Then for any <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> , v(S) = 1, therefore <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩾</mo><mn>1</mn></mrow></math> , which implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo stretchy="false">)</mo></mrow></math> . As v is balanced and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>C</mi><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo stretchy="false">)</mo></mrow></math> is a singleton, so is C(v).

Suppose w.l.o.g. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mi>j</mi><mo>=</mo><mn>2</mn></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>⩽</mo><mn>4</mn></mrow></math> .

2.1 Suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><mo>Ø</mo></mrow></math> and consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> . Suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>1</mn><mo stretchy="false">}</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> . Then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>⩾</mo><mi>λ</mi></mrow></math> . Similarly, if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>2</mn><mo stretchy="false">}</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> , <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>⩾</mo><mn>1</mn><mo>−</mo><mi>λ</mi></mrow></math> . As x(N) = 1 and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>⩾</mo><mn>0</mn></mrow></math> , both facts force <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>=</mo><mi>λ</mi></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>=</mo><mn>1</mn><mo>−</mo><mi>λ</mi></mrow></math> , and therefore x<subs>k</subs> = 0 for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>≠</mo><mn>1</mn><mo>,</mo><mn>2</mn></mrow></math> , proving that v has a point core. Otherwise, if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>2</mn><mo stretchy="false">}</mo><mo>∉</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> , then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> must contain at least two sets, whose intersection yields {2}. This is possible with n = 4 and then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> must contain <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>2</mn><mo>,</mo><mn>3</mn><mo stretchy="false">}</mo><mo>,</mo><mo stretchy="false">{</mo><mn>2</mn><mo>,</mo><mn>4</mn><mo stretchy="false">}</mo></mrow></math> by Lemma 7. Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>+</mo><msub><mrow><mi>x</mi></mrow><mn>3</mn></msub><mo>⩾</mo><mn>1</mn><mo>−</mo><mi>λ</mi></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>+</mo><msub><mrow><mi>x</mi></mrow><mn>4</mn></msub><mo>⩾</mo><mn>1</mn><mo>−</mo><mi>λ</mi></mrow></math> , which together with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>⩾</mo><mi>λ</mi></mrow></math> forces equality everywhere, from which we deduce x<subs>3</subs> = x<subs>4</subs>, then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>+</mo><msub><mrow><mi>x</mi></mrow><mn>3</mn></msub><mo>=</mo><mn>1</mn></mrow></math> , and so <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>4</mn></msub><mo>=</mo><msub><mrow><mi>x</mi></mrow><mn>3</mn></msub><mo>=</mo><mn>0</mn></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>=</mo><mn>1</mn><mo>−</mo><mi>λ</mi></mrow></math> . Finally, if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>1</mn><mo stretchy="false">}</mo><mo>∉</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>2</mn><mo stretchy="false">}</mo><mo>∉</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> , we get <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> contains <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>3</mn><mo stretchy="false">}</mo><mo>,</mo><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>4</mn><mo stretchy="false">}</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> contains <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>2</mn><mo>,</mo><mn>3</mn><mo stretchy="false">}</mo><mo>,</mo><mo stretchy="false">{</mo><mn>2</mn><mo>,</mo><mn>4</mn><mo stretchy="false">}</mo></mrow></math> , which contradicts adjacency (see case (a) of Figure 1).

2.2 Suppose <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>≠</mo><mo>Ø</mo></mrow></math> and consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>∈</mo><mi>C</mi><mo stretchy="false">(</mo><mi>v</mi><mo stretchy="false">)</mo></mrow></math> . Any set in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> must contain {1, 2}. Suppose first <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo stretchy="false">}</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> . Then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>⩾</mo><mn>1</mn></mrow></math> , and because x(N) = 1 and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo>⩾</mo><mn>0</mn></mrow></math> , we obtain <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>=</mo><mn>1</mn></mrow></math> and x<subs>k</subs> = 0 for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>≠</mo><mn>1</mn><mo>,</mo><mn>2</mn></mrow></math> . Because there must exist <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mo>∉</mo><mi>S</mi></mrow></math> , this implies that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>⩾</mo><mi>λ</mi></mrow></math> . Similarly, there exists <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>∉</mo><mi>S</mi></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>x</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>⩾</mo><mn>1</mn><mo>−</mo><mi>λ</mi></mrow></math> . Combining this with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>=</mo><mn>1</mn></mrow></math> , we obtain equality throughout, which proves that v has a point core.

Suppose now with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>2</mn><mo><</mo><mi>n</mi><mo>⩽</mo><mn>4</mn></mrow></math> that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mn>3</mn><mo stretchy="false">}</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> . Proceeding as above we get <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>+</mo><msub><mrow><mi>x</mi></mrow><mn>3</mn></msub><mo>=</mo><mn>1</mn></mrow></math> and x<subs>k</subs> = 0 for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>k</mi><mo>≠</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mn>3</mn></mrow></math> . Now, there must exist <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>3</mn><mo>∉</mo><mi>S</mi></mrow></math> : either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>=</mo><mo stretchy="false">{</mo><mn>1</mn><mo stretchy="false">}</mo></mrow></math> or, if n = 4, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>=</mo><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>4</mn><mo stretchy="false">}</mo></mrow></math> . This yields <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>1</mn></msub><mo>⩾</mo><mi>λ</mi></mrow></math> . Similarly with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> , we get <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>x</mi></mrow><mn>2</mn></msub><mo>⩾</mo><mn>1</mn><mo>−</mo><mi>λ</mi></mrow></math> , and we can conclude as above. The case <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mn>4</mn><mo stretchy="false">}</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> is similar.

Finally, with n = 4, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mn>3</mn><mo>,</mo><mn>4</mn><mo stretchy="false">}</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> is possible. Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> must contain {1} or both <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>3</mn><mo stretchy="false">}</mo><mo>,</mo><mo stretchy="false">{</mo><mn>1</mn><mo>,</mo><mn>4</mn><mo stretchy="false">}</mo></mrow></math> , and similarly for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>\</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub></mrow></math> . It suffices to proceed as for the case <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><mo>Ø</mo></mrow></math> . □

The following counterexample shows that the result is no more true for n > 4.

Example 4.

Let us take n = 5 and two collections <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> defined by

We have <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mn>1</mn><mo stretchy="false">}</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo>∩</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>=</mo><mo stretchy="false">{</mo><mn>2</mn><mo stretchy="false">}</mo></mrow></math> . Therefore, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> determine vertices, which we denote by v<subs>1</subs>, v<subs>2</subs>, respectively. It can be checked (via Theorem 15 or simply Figure 1) that these two vertices are adjacent.

Let us consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>=</mo><mo stretchy="false">(</mo><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub><mo stretchy="false">)</mo><mo>/</mo><mn>2</mn></mrow></math> . One can check that x, y are two distinct core elements of v, and hence it is not reduced to a singleton:

7. Applications

The fact that the polytope <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is combinatorial has profound implications on optimization issues associated with the polytope. In Matsui and Tamura [[12]], the importance of a polytope being combinatorial when performing optimization on it is explained. In this work, the property of being combinatorial is referred to as Property C. Some results of this work require a stronger property called Property B (Property B implies Property C). Next, we can see that the polytope <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> also satisfies Property B, and therefore all the results of Matsui and Tamura [[12]] are applicable to our case.

Proposition 6.

Let v<subs>1</subs>, v<subs>2</subs> and v<subs>3</subs> three vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>⩽</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub><mo>⩽</mo><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub></mrow></math> (coordinate-wise), then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub><mo>−</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub></mrow></math> is a vertex of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> (Property B).

Proof.

Let <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>,</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub></mrow></math> be the collections associated with each vertex. Because <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>⊆</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo></mrow></math> it follows that the sets attaining value 1 in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub><mo>−</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub></mrow></math> are exactly those in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>\</mo><mo stretchy="false">(</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>2</mn></msub><mo>∪</mo><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>1</mn></msub><mo stretchy="false">)</mo><mo>.</mo></mrow></math> As this set is contained in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">D</mi></mrow><mn>3</mn></msub><mo>,</mo></mrow></math> it follows that

and therefore <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>v</mi></mrow><mn>1</mn></msub><mo>+</mo><msub><mrow><mi>v</mi></mrow><mn>3</mn></msub><mo>−</mo><msub><mrow><mi>v</mi></mrow><mn>2</mn></msub><mo>∈</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>.</mo></mrow></math> □

For combinatorial polytopes, more efficient vertex enumeration algorithms can be defined. In Merino and Mütze [[13]], one can appreciate how the presence of Hamiltonian paths between vertices can be leveraged to make enumeration algorithms more efficient.

Many interesting problems in cooperative game theory can be expressed as an optimization problem on <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , especially approximation problems. As explained above, the fact that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is combinatorial and even satisfies the stronger Property B permits to benefit from more efficient vertex enumeration algorithms and therefore to obtain better performance in linear optimization.

Many operators on games are linear, for example, the Harsanyi dividends (a.k.a. Möbius transform), the Shapley value, and its generalization the interaction transform <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>I</mi><mo>:</mo><mi>v</mi><mo>↦</mo><msup><mrow><mi>I</mi></mrow><mi>v</mi></msup></mrow></math> defined by

(see Grabisch et al. [[8]] and Grabisch [[7]] for many other linear operators on games). Note that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi>I</mi></mrow><mi>v</mi></msup><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo></mrow></math> is the Shapley value for player i. It follows that the maximization/minimization of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mi>I</mi></mrow><mi>v</mi></msup><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo></mrow></math> over <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> for some S, or any linear combination of such terms, is a LP problem that can be solved efficiently, taking advantage of the fact that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is combinatorial, and even satisfies property B.

Another class of optimization problems is the approximation problem. As explained in the Introduction, the main motivation behind this work is to be able to solve the projection problem on <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , or on <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>,</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , which is a quadratic optimization problem. One can also consider to minimize the L<subs>1</subs> norm instead of the L<subs>2</subs> norm, which leads to a LP problem (after standard linearization):

with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>w</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . However, the linearization implies to introduce new variables and constraints; therefore, it is not obvious if one can still benefit from the properties of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Some further investigation is needed here.

8. Concluding Remarks

Our study has permitted to have a complete description of the polyhedral structure of the set of balanced games <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , as well as of its subsets <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , in terms of extremal rays, vertices, facets, and adjacency relations between vertices. In addition, we provided an algorithm for the uniform random generation of the vertices of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Unexpectedly, the polytope <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> of nonnegative normalized balanced games seems to be related to some well-known combinatorial structures, as its number of vertices is a known sequence in OEIS, related to the number of Boolean functions in some Post classes. Moreover, this polytope is combinatorial, which means that the adjacency graph of its vertices is Hamiltonian. In the last part of the paper, we have given a characterization of faces of these polyhedra that contain games with a core reduced to a singleton.

However, some issues would need a deeper analysis, especially in adjacency relations. Although we provided a characterization of adjacent vertices for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , we did not perform this analysis for the extremal rays of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>,</mo><mo /><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> or for the facets of these polyhedra. This would help to solve the projection problem we intend to address in a future work: finding the closest balanced game for a given nonbalanced game.

Another topic of future research would be to study the set of monotone balanced games. Monotone games are games satisfying the following property: if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊆</mo><mi>T</mi></mrow></math> , then <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>⩽</mo><mi>v</mi><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo></mrow></math> . As <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mo>Ø</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> , these games are nonnnegative and therefore form a subset of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mo>+</mo></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . This class of games has a great importance: They are known under the name of capacities (Choquet [[2]]) in decision theory and nonadditive integral theory (Schmeidler [[18]], and Grabisch [[7]]). Unfortunately, the analysis of the set of balanced capacities reveals to be extremely difficult: with n = 4, the number of vertices is already equal to 9,002, and most of them are not 0-1 valued. Finding an analytical characterization of them seems to be challenging.

Appendix A. Proofs of Section 4

A.1. Proof of Theorem 5

Let us prove that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is an affine cone, that is, the translation of a cone. Observe that a particular game in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>α</mi><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> , that is, the unanimity game centered on {n} and multiplied by α.

Define <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>≔</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo>−</mo><mi>α</mi><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> and let us show that

Pick <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>′</mo><mo>≔</mo><mi>v</mi><mo>−</mo><mi>α</mi><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> . Then, for any <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> ,

Hence, the claim is proved. Conversely, suppose that a game <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>″</mo></mrow></math> satisfies

Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>″</mo><mo>+</mo><mi>α</mi><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> satisfies

for any <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , that is, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>″</mo><mo>+</mo><mi>α</mi><msub><mrow><mi>u</mi></mrow><mrow><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></msub><mo>∈</mo><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and thus <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>″</mo><mo>∈</mo><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> .

Finally, take <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>β</mi><mo>⩾</mo><mn>0</mn></mrow></math> . We have <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>β</mi><mi>v</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> because it satisfies (A.1). Therefore, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is a cone.

Proceeding as for the proof of Theorem 2, the affine space contained in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> is obtained by replacing v(N) by α in the equations giving the lineality space of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mi mathvariant="script">G</mi><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . This yields the system

To obtain the corresponding vector space, we just replace α by 0, thus obtaining

The set of solutions for this system is, expressed in terms of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mn>1</mn><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>,</mo><mo>...</mo><mo>,</mo><mi>v</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>n</mi><mo>−</mo><mn>1</mn><mo stretchy="false">}</mo><mo stretchy="false">)</mo></mrow></math> ,

This yields the basis <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mrow><mo stretchy="false">(</mo><msub><mrow><mi>w</mi></mrow><mi>i</mi></msub><mo stretchy="false">)</mo></mrow></mrow><mrow><mi>i</mi><mo>∈</mo><mi>N</mi><mo>\</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> given above.

A.2. Proof of Theorem 6

We follow the same steps as for Theorem 3.

Observe that because <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>n</mi></msub><mo>=</mo><mo>−</mo><msub><mrow><mi>δ</mi></mrow><mrow><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></msub></mrow></math> , it could be considered as a ray of type r<subs>S</subs> with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>=</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></math> . We follow this way in steps 2 and 3 of the proof.

Consider <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>=</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></math> , and let us show that r<subs>S</subs> is extremal. Clearly (A.1) is satisfied and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi><mo>\</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></math> , hence r<subs>S</subs> is a ray of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi><mn>0</mn></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . Suppose it is not extremal. Then there exist two rays <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo>,</mo><mi>r</mi><mo>′</mo><mo>∈</mo><msubsup><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi><mn>0</mn></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo>=</mo><mi>r</mi><mo>+</mo><mi>r</mi><mo>′</mo></mrow></math> . Suppose that r(T) > 0, say, r(T) = 1 for some <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>≠</mo><mi>S</mi><mo>,</mo><mo /><mn>1</mn><mo><</mo><mo stretchy="false">|</mo><mi>T</mi><mo stretchy="false">|</mo><mo><</mo><mi>n</mi></mrow></math> or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>T</mi><mo>=</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></math> . Then, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo>′</mo><mo stretchy="false">(</mo><mi>T</mi><mo stretchy="false">)</mo><mo>=</mo><mo>−</mo><mn>1</mn></mrow></math> .

Using the partition <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>T</mi><mo>,</mo><msup><mrow><mo stretchy="false">(</mo><mi>N</mi><mo>\</mo><mi>T</mi><mo stretchy="false">)</mo></mrow><mo>⊥</mo></msup><mo stretchy="false">}</mo></mrow></math> , the corresponding inequality in (A.1) for r becomes either <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>⩽</mo><mn>0</mn></mrow></math> if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>∈</mo><mi>T</mi></mrow></math> , or <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>1</mn><mo>+</mo><mi>r</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>⩽</mo><mn>0</mn></mrow></math> if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>n</mi><mo>∉</mo><mi>T</mi></mrow></math> . The first case being a contradiction, let us study the second case. For this case, it follows that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>⩽</mo><mo>−</mo><mn>1</mn></mrow></math> , which in turn implies <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo>′</mo><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>⩾</mo><mn>1</mn></mrow></math> . Considering then the partition <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><msup><mrow><mi>N</mi></mrow><mo>⊥</mo></msup><mo stretchy="false">}</mo></mrow></math> , the corresponding inequality in (A.1) for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>r</mi><mo>′</mo></mrow></math> becomes <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mn>0</mn><mo>+</mo><mi>r</mi><mo>′</mo><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>⩽</mo><mn>0</mn></mrow></math> , a contradiction.

The case r(T) < 0 can be treated similarly. We conclude that r must have zero coordinates, possibly excepted for T = S. Observe that as before, r(S) > 0 is not possible because the inequality for <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>S</mi><mo>,</mo><msup><mrow><mo stretchy="false">(</mo><mi>N</mi><mo>\</mo><mi>S</mi><mo stretchy="false">)</mo></mrow><mo>⊥</mo></msup><mo stretchy="false">}</mo></mrow></math> would not be satisfied (proceed as above with T). Hence, r<subs>S</subs> is extremal.

Let us take <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi><mo>\</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></math> and show that r<subs>i</subs> is an extremal ray. First, we prove it is a ray in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi><mn>0</mn></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> . By definition, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>i</mi></msub><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>j</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>j</mi><mo>≠</mo><mi>n</mi></mrow></math> . It remains to prove that r<subs>i</subs> is a solution of (A.1). Taking any m.b.c. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="script">B</mi></math> , the inequality becomes

This is obviously satisfied as <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mstyle displaystyle="false"><msub><mo>∑</mo><mtable columnalign="left"><mtr><mtd><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></mtd></mtr><mtr><mtd><mrow><mi>S</mi><mo>∋</mo><mi>i</mi></mrow></mtd></mtr></mtable></msub><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mo>=</mo><mn>1</mn></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup><mo>⩾</mo><mn>0</mn><mo>.</mo></mrow></math> Observe that the inequality is tight iff <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>∌</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo></mrow></math> .

To show that r<subs>i</subs> is extremal, we need to show that the set of solutions of the system of tight inequalities has dimension 1. We have already observed that the inequality is tight iff <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo>∉</mo><mi mathvariant="script">B</mi></mrow></math> . Let us consider this system in v, that is,

<math display="block" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mstyle displaystyle="true"><munder><mo>∑</mo><mrow><mi>S</mi><mo>∈</mo><mi mathvariant="script">B</mi></mrow></munder><mrow><msubsup><mrow><mi>λ</mi></mrow><mi>S</mi><mi mathvariant="script">B</mi></msubsup></mrow></mstyle><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn><mo>,</mo><mo>∀</mo><mi mathvariant="script">B</mi><mo>∈</mo><msup><mrow><mi mathvariant="fraktur">B</mi></mrow><mo>*</mo></msup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo><mo /><mtext>s</mtext><mo>.</mo><mtext>t</mtext><mo>.</mo><mo /><mo>{</mo><mi>i</mi><mo>}</mo><mo>∉</mo><mi mathvariant="script">B</mi><mo>,</mo></mrow></math> (A.2)

with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo>∈</mo><msub><mrow><mi mathvariant="script">G</mi></mrow><mi>α</mi></msub><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> s.t. <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>i</mi><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn><mo>,</mo><mi>i</mi><mo>≠</mo><mi>n</mi><mo>.</mo></mrow></math> Take the partition <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi mathvariant="script">B</mi><mo>=</mo><mo stretchy="false">{</mo><mi>S</mi><mo>,</mo><msup><mrow><mi>S</mi></mrow><mo>⊥</mo></msup><mo stretchy="false">)</mo></mrow></math> with <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>S</mi></mrow></math> . The corresponding equality in (A.2) reads v(S) = 0 if <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>∋</mo><mi>n</mi></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>+</mo><mi>v</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> otherwise, that is, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>v</mi><mo stretchy="false">(</mo><mi>S</mi><mo stretchy="false">)</mo><mo>=</mo><mo>−</mo><mi>v</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo></mrow></math> . It follows that the set of solutions has the form <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mo stretchy="false">(</mo><munder><munder><mrow><mi>β</mi></mrow><mo stretchy="true">︸</mo></munder><mrow><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></munder><mo>,</mo><munder><munder><mrow><mn>0</mn><mo>⋯</mo><mn>0</mn></mrow><mo stretchy="true">︸</mo></munder><mrow><mi>S</mi><mo>∋</mo><mi>n</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></munder><mo>,</mo><munder><munder><mrow><mo>−</mo><mi>β</mi><mo>⋯</mo><mo>−</mo><mi>β</mi></mrow><mo stretchy="true">︸</mo></munder><mrow><mi>S</mi><mo>∌</mo><mi>n</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></munder><mo stretchy="false">)</mo><mo>,</mo><mi>β</mi><mo>∈</mo><mi mathvariant="double-struck">R</mi><mo stretchy="false">}</mo></mrow></math> , and hence has dimension 1.

It remains to prove that there is no other extremal ray. Consider a ray <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>w</mi><mo>∈</mo><msubsup><mrow><mi mathvariant="script">C</mi></mrow><mi>α</mi><mn>0</mn></msubsup><mo stretchy="false">(</mo><mi>n</mi><mo stretchy="false">)</mo></mrow></math> , hence satisfying <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>w</mi><mo stretchy="false">(</mo><mo stretchy="false">{</mo><mi>i</mi><mo stretchy="false">}</mo><mo stretchy="false">)</mo><mo>=</mo><mn>0</mn></mrow></math> for all <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>i</mi><mo>∈</mo><mi>N</mi><mo>\</mo><mo stretchy="false">{</mo><mi>n</mi><mo stretchy="false">}</mo></mrow></math> and (A.1). If w is not a conic combination of r<subs>S</subs> and r<subs>i</subs>, <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn><mo>,</mo><mo /><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> , the system (5) has no solution in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>α</mi></mrow><mi>S</mi></msub><mo>,</mo><msub><mrow><mi>α</mi></mrow><mi>i</mi></msub></mrow></math> . Using definitions of r<subs>S</subs> and r<subs>i</subs> and omitting coordinates for singletons (except n) and N in <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>r</mi></mrow><mi>S</mi></msub><mo>,</mo><msub><mrow><mi>r</mi></mrow><mi>i</mi></msub><mo>,</mo><mi>w</mi></mrow></math> , we obtain

We may denote the whole system by <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mi>A</mi><mi>α</mi><mo>⩾</mo><mi>b</mi></mrow></math> in matrix notation (with some abuse). Then, by Farkas' lemma, this system has no solution iff there exists a vector <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">[</mo><mi>y</mi><mo /><mi>z</mi><mo /><mi>t</mi><mo stretchy="false">]</mo></mrow></math> with coordinates <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>y</mi></mrow><mi>S</mi></msub><mo>∈</mo><mi mathvariant="double-struck">R</mi><mo>,</mo><mo /><mi>S</mi><mo>⊆</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn><mo>,</mo><mo /><msub><mrow><mi>z</mi></mrow><mi>T</mi></msub><mo>⩾</mo><mn>0</mn><mo>,</mo><mo /><mi>T</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>T</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn></mrow></math> , and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msub><mrow><mi>t</mi></mrow><mi>i</mi></msub><mo>⩾</mo><mn>0</mn><mo>,</mo><mo /><mi>i</mi><mo>∈</mo><mi>N</mi></mrow></math> , such that <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mo stretchy="false">[</mo><mi>y</mi><mo /><mi>z</mi><mo /><mi>t</mi><mo stretchy="false">]</mo></mrow><mo>⊤</mo></msup><mi>A</mi><mo>=</mo><mn>0</mn></mrow></math> and <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mo stretchy="false">[</mo><mi>y</mi><mo /><mi>z</mi><mo /><mi>t</mi><mo stretchy="false">]</mo></mrow><mo>⊤</mo></msup><mi>b</mi><mo>></mo><mn>0</mn></mrow></math> . Observe that the only vector <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">[</mo><mi>y</mi><mo /><mi>z</mi><mo /><mi>t</mi><mo stretchy="false">]</mo></mrow></math> solution of <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><msup><mrow><mo stretchy="false">[</mo><mi>y</mi><mo /><mi>z</mi><mo /><mi>t</mi><mo stretchy="false">]</mo></mrow><mo>⊤</mo></msup><mi>A</mi><mo>=</mo><mn>0</mn></mrow></math> is

For this solution, we obtain

Considering again the balanced collection <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mo stretchy="false">{</mo><mi>S</mi><mo>⊂</mo><mi>N</mi><mo>,</mo><mo stretchy="false">|</mo><mi>S</mi><mo stretchy="false">|</mo><mo>></mo><mn>1</mn><mo stretchy="false">}</mo></mrow></math> with balancing weights <math display="inline" overflow="scroll" xmlns="http://www.w3.org/1998/Math/MathML"><mrow><mfrac><mn>1</mn><mrow><msup><mrow><mn>2</mn></mrow><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow></msup><mo>−</mo><mn>2</mn></mrow></mfrac><mo>,</mo></mrow></math> it follows that

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Footnotes

That is, a cone plus a point.

By Pedro Garcia-Segador; Michel Grabisch and Pedro Miranda

Reported by Author; Author; Author