Generalized kernels and bargaining sets for families of coalitions Текст научной статьи по специальности «Математика»

Generalized Kernels and Bargaining Sets for Families of Coalitions 1

Natalia Naumova

St. Petersburg State University,

Faculty of Mathematics and Mechanics,

28, Universitetsky pr., St. Petersburg, 198504, Russia E-mail: Nataliai.Naumova@mail. ru

Abstract. For a fixed collection of subsets of the player set, two generalizations of Aumann-Maschler theory of the bargaining set for cooperative TU-games, where objections and counter-objections are permitted only between elements of this collection, and corresponding generalizations of the kernel are considered. We describe conditions on the fixed collection of coalitions that ensure existence of corresponding sets of imputations for all n-person games.

All sufficient conditions are based on a generalization of [Peleg]. Here relations are defined not on the player set, but on the set of coalitions, and acyclicity is not assumed. Obtained sufficient conditions are also necessary for both generalized bargaining sets if the number of players is no more than five and for one of generalized kernels.

Keywords: Cooperative games, kernel, bargaining set. Introduction

The theory of the bargaining set, the kernel, and the nucleolus for cooperative TU-games develops more than forty years (see [Maschler, 1992]). The existence theorems for the kernel and the bargaining set are well known. See [Davis, 1963], [Davis, 1967], [Peleg, 1967], [Maschler, 1966].

In [Naumova, 1976] the bargaining set is generalized. For each family of coalitions A, she permits objections and counter-objections only between the members of this family. The resulting A-bargaining set is denoted . In this spirit a

1 This research was supported by NWO (The Netherlands Organization for Scientific Research), grant NL-RF 047.017.017 and RFBR (Russian Foundation for Basic Research), grant 05-01-89005-HBO-a.

A-kernel, Ka, which is contained in M\, are also defined. If A is the set of all singletons, then M\ = Ml and Ka is the kernel. Here another generalization of Ml, denoted by M\ and the corresponding generalization of the kernel Ka are defined.

Ka C Ma C MiA but neither Ka C Ka nor Ka C Ka.

Papers [Naumova, 1976] and [Naumova, 1978] contain sufficient conditions on A which guarantee that each game would have a non-empty A-kernel for every coalition structure.

Sufficient conditions on A for existence of M\ and new sufficient conditions on A for existence of the generalized bargaining set M\ are obtained.

All these sufficient conditions are formulated in terms of special directed A-admissible (or weakly A-admissible) graphs, where A is the set of vertices. For fixed A, they can be verified in a finite number of steps.

If the number of players is no more than 5, for both generalizations of bargaining sets their sufficient conditions on A for existence result are also necessary.

Sufficient condition on A for existence of Ka which was obtained in [Naumova, 1978], is also necessary.

The new existence theorems are based on a generalization of Peleg’s theorem in [Peleg]. Peleg proved the existence of equilibrium imputation for open acyclic relations on the set of players. This generalization guarantees the existence of equilibrium imputations for open relations on the considered family of coalitions under some additional conditions without acyclicity assumption.

1. Definitions

A cooperative TU-game is a pair (N, v), where N = {1,...,n} is a set of players, v is a map that takes each S C N to a number v(S), v(0) = 0.

For cooperative TU-game (N,v), an imputation is a vector x = {xi}ieN such that J2ieN Xi = v(N) and Xi > v({i}) for all i G N.

Let r0 be the set of games (N, v) such that v({i}) = 0 for all i G N and v(S) > 0 for all S C N. (Such games are 0-normalizations of games (N, v) with ^ieS v({i}) < v(S) for all S C N.)

In what follows, we consider only (N, v) G r0.

Let (N,v) be a cooperative TU-game, K,L C N, x be an imputation for (N,v). An objection of K against L at x is a pair (C, yc), such that K C C C N, Ln C = 0,

Vo = {yi}iec,2ieC yi = v(C), yi > xi for all i G K, and yi > xi for all i G C.

A counter-objection of L against K to this objection is a pair (D,zd) such that L C D C N, K C D, YIzi = v(D), zi > xi for all i G D, zi > yi for all i G CnD.

A strong counter-objection of L against K to this objection is a pair (D, zd) such that L C D C N, K n D = 0, ^ieD zi = v(D), zi > xi for all i G D, zi > yi for all

i G C n D.

An objection is justified if there is no counter-objection to it. An objection is weakly justified if there is no strong counter-objection to it.

Let A be a set of subsets of N. An imputation x of (N, v) belongs to the bargaining set M\(N, v) if for all K,L gA there are no justified objections of K against L at x.

An imputation x of (N,v) belongs to the strong bargaining set MMA(N,v) if for all K,L gA there are no weakly justified objections of K against L at x.

If A is the set of all singletons, then MiA(N,v) = MA(N,v) = Ml(N,v).

Note that MMA(N,v) C MA(N,v).

For a set A of subsets of N consider the following two generalizations of the kernel.

Let K,L C N and x be an imputation of (N, v). K overweights L at x if KnL = 0,

J2ieL xi > v(L), and sk,l(x) > sl,k(x), where

sp,q(x) = max{v(S) — xi : S C N,P C S,Q C S}.

ies

The set Ka(N,v) is the set of all imputations x of (N,v) such that no K G A overweights any L gA.

K prevails L at x if K n L = 0, ^ieL xi > v(L), and tK,L(x) > tL,K(x), where

tP,Q(x) = max{v(S) — xi : S C N,P C S,Q n S = 0}.

ies

The set Ka (N, v) is the set of all imputations x of (N, v) such that no K gA prevails any L gA.

If A is the set of all singletons, then Ka and Ka coincide with the kernel. It was proved in [Naumova, 1976] that Ka(N, v) C MA(N, v). It will be proved below that K a(N,v) C MM A (N,v). It will be shown below (Examples 4 and 14) that neither Ka(N, v) C Ka(N,v) nor Ka(N,v) D Ka(N,v). In what follows for each x G Rn, we denote x(S) = Y xi.

ieS

2. Existence results

2.1. Equilibrium points for general relations on A

Let N = {1,... ,n}, X C Rn, A be a collection of subsets of N, {yx}xeX be a collection of binary relations yx defined on A. Then x0 G X is an equilibrium vector on A if K /xo L for all K,L gA.

For b > 0, K gA denote

X(b) = {x G Rn : xi > 0, x(N) = b}, FK(b) = {x G X(b) : L fx K for all L G A}.

In this subsection sufficient conditions for existence of equilibrium vectors in X(b) are described. The following result of [Peleg, 1963] will be used in the proof. Let P be a partition of the set N, w be a non-negative function defined on P,

X(P, w) = {x G Rn : x(S) = w(S) VS gP, xi > 0 Vi G N}.

Peleg’s Lemma. Let c\,...,cn be nonnegative continuous functions defined on X (P ,w). If for each x G X (P ,w), S G P, there exists io = io(x) G S such that cio (x) > xio, then there exists x0 G X(P, w) such that ci(x0) > x0 for all i G N.

Theorem 1. Let a family of binary relations {^x}xex(b) on A satisfy the conditions:

1) for all K gA, the set FK (b) is closed;

2) if xi = 0 for all i G K, then x G FK (b);

3) for each x G X (b), the set of coalitions {L gA : K yx L for some K G A} does not cover N.

Then there exists an equilibrium vector x0 G X(b) on A.

Proof.

Denote Ai = {K gA: i G K}, N0 = {i G N : nK eA, FK (b) = 0}.

If x G X(b), then by condition 3, there exists i0 = i0(x) G N such that the set

{L G A : K yx L for some K G A}

does not cover i0, i.e. x G FK (b) for all K G Ai0. This implies x G nKeAio FK (b) and i0 G N0, hence N0 = 0.

Let P be the partition of N that consists of N0 and {j} for all j G N \ N0, w(N0) = b, w({j}) = 0 for all j G N \ N0. Then

X(P, w) = {x G X(b) : xj =0 for all j G N \ N0}.

Define functions ci (i G N) on the set X(P, w) by

ci(x) = xi — xip(x, nKeAiFK(b))/b for i G N0,

ci(x) = xi = 0 otherwise, where p(x, y) = maxieN \xi — yi \, p(x, Y) = infyey p(x, y) for all Y C Rn.

Then ci are nonnegative continuous functions. Let x G X(P,w), then x G X(b)

and by the proved above, there exists i0 = i0(x) G N0 such that x G nKeAio FK(b). Then cio (x) = xio. By Peleg’s Lemma, there exists x0 G X(P,w) such that ci(x0) > xi0 for all i G N.

We prove that x0 G FM (b) for all M gA. If x0 = 0 for all i G M, then x0 G FM (b) by condition 2. If xj0 > 0 for some j G M, then j G N0 and p(x°, nKe_jLjFK(b)) = 0. Since the sets FK (b) are closed by condition 1, x0 G FK (b) for all K G Aj and, in particular, x0 G FM (b).

Corollary. Let G be an undirected graph with the set of vertices A and a family of binary relations {^x}xex(b) on A satisfy the conditions:

1) for all K gA the set FK (b) is closed;

2) if xi = 0 for all i G K, then x G FK(b);

3) for each x G X (b) the relation yx is acyclic;

4) if a single vertex is taken in each connected component of G, then the remaining elements of A do not cover N.

Then there exists an equilibrium vector x0 G X(b) on A.

This corollary coincides with Theorem 2 in [Naumova, 1978], which provides sufficient condition for existence of the generalized kernel Ka and the bargaining set

ma.

2.2. Necessary and sufficient condition for existence of generalized kernels

A set of coalitions A generates the undirected graph G = G(A), where A is the set of vertices and K, L gA are adjacent iff K n L = 0.

Theorem 2. Let A be a set of subsets of N. Then Ka(N,v) = 0 for all (N,v) if and only if A satisfies the following condition .

C0) If a single vertex is taken in each connected component of G(A), then the remaining elements of A do not cover N.

Proof.

It was proved in [Naumova, 1978] that if A satisfies C0, then Ka(N,v) = 0 for all (N, v).

Now suppose that A does not satisfy C0. We can assume without loss of generality that each connected component of G(A) contains at least two vertices. Indeed, let A be obtained from A by deleting all isolated vertices of G(A). Then A does not satisfy C0 and Ka(N,v) = Ka(N, v).

Let G(A) have k connected components. Let S\,...,Sk be taken in all different connected components of G(A) and the union of the remaining elements of these components cover N. We construct a game (N,w) such that Ka(N,w) = 0. We consider two cases.

Case 1. The family A does not contain singletons. We can assume that each Si does not include the other elements of A. Indeed, let Q gA, Q C Si. There exists P gA such that P n S\ = 0. Then P n Q = 0, hence, S\ and Q belong to the same connected component and it is possible to take Q instead of Si. Define (N,w) as follows:

w(N) = w(Si) = 1 for i = 1,...,k,

w(Q) = 0 otherwise.

Suppose that there exists x G Ka(N,w). It will be proved that for each i = 1,...,k, if Q and S\ belong to the same connected component, Q = Si, then x(Q) = 0. This implies x(N) = 0 and contradicts x(N) = w(N) = 1.

Fix i < k. Since x(Si) < 1 = w(Si), v(Si) — x(Si) > 0. Let L gA, L n Si = 0.

Then Sj D L for all j = 1, ...,k, so v(P) — x(P) < 0 for all P D L, P = N. Thus, sSi,L(x) > 0, sL,Si(x) < 0 and x G Ka(N,w) implies x(L) = 0. Let T G A,

T = Si, T n L = 0. Since x(L) = 0, sl,t(x) > 0. Since Sj D T for all j = 1,..., k, st,l(x) < 0, therefore, x G Ka(N,w) implies x(T) = 0. Thus, by induction on the distance from S'i in G(A) we prove that x(P) = 0 for all P gA \ {Si, ..., Sk}.

Case 2. The family A contains a singleton. Denote it by {1}. Then all connected components of G(A) that do not contain {1}, contain only one element. Therefore, we can assume that G(A) is a connected graph.

Let S be a coalition in A such that the remaining coalitions cover N and S is minimal in A. If \S\ > 1, then take the same game as in case 1.

Let \S\ = 1. Assume that S = {1}. We can not take w(S) = 1 because w({i}) = 0 for all i G N. There exist Si, Qi gA such that 1 G Si, \Si\ > 1, and Si n Qi = 0. Define (N,w) as follows:

w(N) = w(Qi U {1}) = 1, w(P) = 0 otherwise.

Let x G Ka(N, w). For each imputation x of (N, w),

SQi,Si (x) > w(Qi U {1}) — xi — x(Qi) > 0, ssltQ! (x) < 0.

Therefore, x G Ka(N,w) implies x(Si) = 0, hence, xi = 0. Let L G A, 1 G L, then S{i}jL(x) > v({1}) — xi = 0 and s^j^x) < 0, hence, x G Ka(N,w) implies x(L) = 0. In particular, x(Qi) = 0.

Let T G A, 1 G T. If T C (Qi U {1}), then x(T) = 0 by the proved above. Let T C (Qi U {1}). There exists M G A such that M n T = 0. Since x(M) = 0, sM,T(x) > v(M) — x(M) = 0. Since T C (Qi U {1}), st,m(x) < 0. Therefore, x G Ka(N,w) implies x(T) = 0. Thus, x(P) = 0 for all P in the main connected component, hence, x(N) = 0 and this contradicts x(N) = w(N) = 1.

Let us consider the collections of subsets A that satisfy the condition C0 of Theorem 2. If N is not covered by the elements of A, then this condition is obviously fulfilled. Let A0 be the set of isolated vertices of graph G(A), A = A\A0. Then the collections of coalitions A and A satisfy the condition C0 simultaneously.

A player i is called a fanatic for A if it belongs to precisely one element of A. Let us describe for some n all collections of coalitions A that cover N and satisfy

C0.

If G(A) has only one connected component, then each element of A has a fanatic. Example 1. For n = 3, if A satisfies C0, then either A = {{1}, {2}, {3}} or A= {{i,j}, {k}}.

Example 2. For n = 4, if A satisfies C0, then either each element of A contains a fanatic of A or A = {{i,j}, {k, l}, {i, k}, {j, l}}. Indeed, if A contains 1-person or 3-person coalition, then G(A) has only one connected component. In the remaining case, if G(A) has 3 connected components, then C0 is not fulfilled, and only the case when G(A) has 2 connected components, is non-trivial.

Example 3. For n = 5, there are two non-trivial cases of A that satisfy C0.

1) A = j}, {k, ^ m}, {i, k}, {j, l}};

2) A = j}, {k, ^ m}, {i, k}, {j, ^ m}}.

Example 4. Ka C Ka- Indeed, let N = {1, 2, 3}, A = {{1}, {2}, {1, 3}}. By Theorem 2, there exists a game (N, v) such that v({i}) = 0 for i = 1, 2, 3, 0 < v(S) < v(N) = 1 for S C N, and Ka(N, v) = 0. We prove that Ka(N, v) = 0.

Let xi = (1 — v({2, 3}))/1, x2 = (1 — v({1, 3}))/2, x3 = 1 — xi — x2. Then x is an imputation for (N, v) and

v({1}) — xi = (v({2, 3}) — 1)/2, v({2}) — x2 = (v({1, 3}) — 1)/2,

v({1, 3}) — xi — x3 = (v({1, 3}) — 1)/2, v({2, 3}) — x2 — x3 = (v({2, 3}) — 1)/2.

Therefore,

t{i},{2}(x) = max{(v({2, 3}) — 1)/2, (v({1, 3}) — 1)/2} = t{2},{l}(x), t{2},{i,3}(x) = (v({1, 3}) — 1)/2 = t{1,3},{2}(x),

so x G Ka(N, v).

2.3. Existence conditions for generalized bargaining sets

The condition C0 of Theorem 2 is sufficient for non-emptyness of M\ but is not necessary.

Example 5. Let A consists of all 1-person and (n — 1)-person coalitions. Then MiA = Mi.

Indeed, the objections of 1-person coalition against (n — 1)-person coalition are impossible and if (n — 1)-person coalition has a justified objection against 1-person coalition, then each singleton, contained in (n — 1)-person coalition, has the same justified objection against 1-person coalition.

Example 6. Let A = {K, L, M}, where K C L, K = L, M n L = 0, M U L = N. Then A does not satisfy the condition C0, but MA(N, v) = 0 for each (N, v) G r0. Indeed, for (N, v), let x0 be an imputation such that

x0(M) = min{v(N), v(M)}, x0 = 0 for all i G K.

Then x0 G M\(N, v). Indeed, if there exists a justified objection (Q, vq) at x0, then, since x0(K) < v(K) and x0(M) < v(M), (Q,vq) is an objection of M against L and Q = M. If v(M) < v(N), then x0(M) > v(M) and such objection is impossible. If v(M) > v(N), then x0(L) =0 < v(L) and there exists a counter-objection (L, zL).

Now we describe more weak conditions on A that ensure the non-emptyness of M\ for all games in r0. The following lemmas generalize well known results on the bargaining set Mi.

Lemma 1. Let (Q,vq) be an objection of K against L at x. Let S C N, L C S, K C S. Then there exists a counter-objection (S, zs) to this objection if (y — x)(Q n

S) < v(S) — x(S).

Proof.

Let (y — x)(Q n S) > v(S) — x(S). Suppose that there exists a counter-objection (S,zs) to (Q,vq). Then

z(S) = z(Q n S)+ z(S \ Q) > y(Q n S)+ x(S \ Q) = (y — x)(Q n S)+ x(S),

this contradicts z(S) = v(S).

Let (y — x)(Q n S) < v(S) — x(S). Then we can take zs such that

zQnS = VQnS, zS\Q > xS\Q,

(z — x)(S \ Q) = v(S) — x(S) — (y — x)(Q n S).

Let us verify that z(S) = v(S). z(S) = z(S\Q) + z(S nQ) = z(S\Q)+y(S nQ) = (z —x)(S\Q) + (y — x)(S nQ)+x(S) = v(S) — x(S) — (y — x)(Q n S) + (y — x)(S n Q)+ x(S) = v(S).

Thus (S, zs) is a counter-objection to (Q,vq).

Lemma 2. Let x be an imputation for (N, v), K,L C N, K C Q C N, L n Q = 0.

Let S C N, L C S, K C S, x(Q) < v(Q).

If either QnS = 0 and x(S) < v(S), or QnS = 0 and v(Q) — x(Q) < v(S) — x(S), then for each objection (Q,Vq) of K against L at x there exists a counter-objection (S, zs) of L against K.

Proof.

Let yQ be an objection of K against L at x.

If Q n S = 0 and x(S) < v(S), then take zs such that zs(S) = v(S) and zs > xs. Suppose that Q n S = 0 and v(Q) — x(Q) < v(S) — x(S). Note that

v(Q) — x(Q) = y(Q) — x(Q) = (y — x)(Q n S) + (y — x)(Q \ S),

so (y — x)(Q \ S) > 0 implies

(y — x)(Q n S) < v(Q) — x(Q) < v(S) — x(S).

By Lemma 1, there exists a counter-objection (S,zs) to (Q,vq).

Let A be a family of subsets of N. A directed graph Gr is called A-admissible if A is the set of its vertices and there exists a map f defined on the set of the edges of Gr, that takes each oriented edge (K, L) to a pair f (K, L) = (Q, r) (Q C N, r G R1) and satisfies the following 3 conditions.

C1. If f (K, L) = (Q, r), then K C Q, Q n L = 0, \Q\ > 1.

C2. If f (K, L) = (Q, r), f (R, P) = (S,t), L C S, K C S, then Q n S = 0.

C3. If f (K, L) = (q, r), f (r, P) = (s, t), L C S, K C S, then r > t.

Example 7. Let Ai = {K, L,M}, where K C L, K = L, M n L = 0, M U L = N. Let Gri be a digraph, where Ai is the set of vertices and {(K, M), (M, L)} is the set of edges.

Then Gri is not Ai-admissible. Indeed, if Gri is Ai-admissible and f is the corresponding map, then, by C1, f (M, L) = (M, r), f (K, M) = (Q, t), Q n M = 0, but this contradicts C2.

Example 8. Let N = {1, 2, 3, 4}, A2 = {{1}, {2}, {3}, {3, 4}}. Let Gr2 be a digraph with the set of vertices A2 and the set of edges {({3}, {2}), ({2}, {3, 4}), ({3,4}, {1})}.

Then Gr2 is A2 -admissible. Indeed, take f ({3}, {2}) = ({1, 3}, 2), f ({2}, {3, 4})=({1, 2}, 1), f ({3,4}, {1})=({2, 3,4}, 3).

Theorem 3. Let A be a set of subsets of N. If for each A-admissible graph Gr the set of the ends of its edges does not cover N, then M\(N, v) = 0 for all (N, v) G r0.

Proof.

Let (N,v) G r0, x be an imputation for (N, v). For K,L G A, denote K yx L if there exists a justified objection of K against L at x. Let b = v(N), then X(b) is the set of all imputations for (N, v). We prove that the family of binary relations {yx}xex(b) satisfies all conditions of Theorem 1.

It follows from Lemma 1 that the set FK (b) is closed for all K G A. If xi = 0 for all i G K, then x(K) < v(K) and, by Lemma 2, there exists a counter-objection (K, zK) to each objection against K at x.

Let us verify that for each imputation x the set of coalitions {L G A : K yx L for some K G A} does not cover N. Let Gr(x) be the digraph, where A is the set of vertices and (K, L) is the edge of Gr(x) iff K yx L. Let us prove that Gr(x) is A-admissible.

Define the map f as follows. f (K, L) = (C,r) if there exists a justified objection (C,yc) of K against L at x. (If there exist several justified objections, then one of them is fixed arbitrary.) The numbers r must satisfy the following: if f (K, L) = (C, r), f (M, P) = (Q, t), then r < t iff e(C, x) < e(Q, x), where e(S, x) = v(S) — x(S).

Let us check that f satisfies the conditions C1-C3. C1 follows from the definition of objection; \ fi(K,L)\ > 1 because xi > v(i) for all i G N.

C2. Suppose that there exist the edges (K, L) and (M, P) of Gr(x) such that f (K,L) = (Q, r), f (M,P) = (S,t), L C S, K C S, and QnS = 0. Then x(S) < v(S) by the definition of objection, hence, by Lemma 2, there exists zs such that (S, zs) is the counter-objection of L against K to any objection (Q, vq) of K against L at x, this contradicts the definition of f.

C3. Suppose that there exist the edges (K,L) and (M,P) of Gr(x) such that f (K, L) = (Q, r), f (M, P) = (S, t), L C S, K C S, and r < t. Then v(Q) — x(Q) <

v(S) — x(S). By Lemma 2, there is a counter-objection to each objection (Q,vq) of K against L at x, this contradicts the definition of f.

Thus, Gr(x) is A-admissible, hence the ends of its edges do not cover the set N.

By Theorem 1, there exists an equilibrium imputation x0 on A, i.e. x0 GMA(N,v).

The following propositions help to use Theorem 3.

Proposition 1. Each A-admissible digraph Gr generates an acyclic binary relation on A.

Proof.

Suppose that for some N, A, and A-admissible digraph Gr there exist K1, ...,Kl G A such that (if®, Kt+1) for i = 1,..., t — 1 and (K4, K1) are edges of Gr.

For each integer j, let j G {1,..., t}, j = j (mod. t).

Let f(K\Kt+1) = (Qi,n), rio = maxj=iv..it r*. If Qio ~b Klthen as Qio D Kl°, the condition C3 implies rj0 < rio_ 1, thus, Qi0 D K*0_1. Similarly, Qi0 D Kl°-2,..., and Qio D Kl0+1, but this contradicts Cl.

Proposition 2. Let A, B be families of subsets of N, B C A and A\B contains only (n — 1)-person coalitions. Then Mg(N,v) = 0 for all (N,v) G r0 implies MiA(N, v) = 0 for all (N, v) G T°.

Proof.

Suppose that there exists i0 G N such that i0 G S for all S gB. Then G(A) may contain only one non-trivial connected component {{i0}, N\{i0}}. By Proposition 1, for each A-admissible digraph Gr the ends of its edges do not cover N and, by Theorem 3, MA(N,v) = 0 for all (N,v) G r0. Otherwise, MiB(N,v) = MA(N,v). Indeed, let x G Mg(N, v). Suppose that there exists a justified objection (Q,vq) of K against L at x. Then \Q\ > 1, hence, \L\ < n — 1 and \K\ = n — 1, L = {i0}, and Q = K. In considered case, there exists T G B, T C K. Then (Q,vq) is an objection of T against L at x, hence, there exists a counter-objection (D, zD) to this objection. By the definition of counter-objection, (D, zD) is also a counter-objection of L against K to (Q, vq), and the objection is not justified.

Example 9. Let Ai = {K, L,M}, where K C L, K = L, M n L = 0, M U L = N. If the ends of the edges of digraph Gr cover N, then, by Proposition 1, Gr = Gri (see Example 7), where {(K, M), (M, L)} is the set of edges of Gri. Since Gri is not Ai-admissible, MiAl (N,v) = 0 for all (N,v) G r0.

The next theorem states that for \N \ < 5 the sufficient condition on A for existence of MA(N,v) is also necessary.

Theorem 4. Let A be a set of subsets of N and \N\ < 5. Then MA(N,v) = 0 for all (N,v) G r0 if and only if for each A-admissible graph Gr the set of the ends of its edges does not cover N.

Proof.

Let \N\ <5. Suppose that there exists a set of subsets of N A and an A-admissible digraph Gr such that the ends of its edges cover N. We construct a cooperative game

(N, v) such that M\(N, v) = 0. Let End(Gr) be the set of the ends of the edges of Gr.

We reduce the graph Gr. For each L G End(Gr), we delete all edges (K, L) except one. The reduced graph is also A-admissible, the set End(Gr) is preserved, and each edge of new Gr is defined by its end.

Let f be a map defined on the set of the edges of Gr that takes each edge (K, L) to a pair f (K,L) = (Q(L),r(L)), satisfies the conditions C1-C3 and the following condition. If f is a map defined on the set of the edges of Gr that takes each edge (K,L) to a pair f(K, L) = (Q(L),r(L)), satisfies the conditions C1-C3 and Q(L) C Q(L) for each L G End(Gr), then f = f. (This means that it is impossible to extend Q(L) with preserving r(L) and C1-C3.)

Define (N,v) as follows:

v(N) = 1, v(Q(L)) = 1 for each L G End(Gr),

v(T) = 0 otherwise.

Note that (N,v) is well defined since \Q(L)\ > 1 by C1. We shall prove that

MA(N, v) = 0.

Step 1. Denote

B(i) = USEEnd(Gr):r(S)<iS.

We prove that for all L G End(Gr),

N \ Q(L) C B(r(L)). (1)

Suppose that there exists L G End(Gr) such that N \ Q(L) C B(r(L)). Let q G N \ (Q(L) U B(r(L)). Let Q = Q(L) U {q}.

Let us replace Q(L) by Q. Since it is impossible to extend Q(L) with preserving C1-C3, one of these conditions violates.

Since L G B(r(L)), q G L, hence, the condition C1 preserves. If C2 or C3 violate, then there exists an edge (R,P) such that P C Q, P C Q(L), R C Q and either C2 violates, i.e. Q n Q(P) = 0, or C3 violates, i.e. r(L) > r(P).

Let QnQ(P) = 0, then KnQ(P) = 0. If L C Q(P), then, by C2, Q(L)nQ(P) = 0, hence, Q n Q(P) = 0. Therefore, L C Q(P)_,\L \ Q(P)\ > 1. By C1, \Q(P)\ > 2, \Q\ > 3. Since the sets L \ Q(P), Q(P), Q are mutually disjoint, \N\ > 6, this contradicts \N\ <5.

Let r(L) > r(P), then P C B(r(L)). Since P C Q(L), P C Q, we have q G P, hence, q G B(r(L)), and this contradicts the definition of q. Thus, (1) is proved.

Step 2. Suppose that there exists x0 G MA(N,v). We shall prove by induction on r(L) that x0(L) = 0 for all L G End(Gr). Since the set End(Gr) covers N, this would imply x°(N) = 0, but x0(N) = v(N) = 1.

Let r(L) = r0 be minimal, (K,L) be an edge of Gr. Suppose that x0(L) > 0. Then x0(Q(L)) < 1 and there exists an objection of K against L at x0. There exists

a counter-objection (D,zD) to this objection. Since v(D) = zD(D) > x0(L) > 0, v(D) = 1 and D = Q(P) for some P G End(Gr). By the definition of counterobjection, L C D, K C D and C3 implies r(P) < r(L) = r0, but r0 is minimal. Thus, x0(L) = 0 for all L with r(L) = r0 and x0(B(r0)) = 0. In view of (1), this implies

x0(Q(L)) = 1 for all L G End(Gr) with r(L) = r0.

Suppose that x0(L) = 0 and x0(Q(L)) = 1 for r(L) < i. Let r(L) = i, (K,L) be an edge of Gr. Assume that x0(L) > 0, then x0(Q(L)) < 1, v(Q(L)) — x0(Q(L)) > 0 and there is an objection (Q(L),vq(L) of K against L at x0. There exists a counterobjection (S,zs) to this objection. Since v(S) = zs(S) > x0(L) > 0, v(S) = 1 and S = Q(T) for some T G End(Gr). By the definition of counter-objection, L C S, K C S and C3 implies r(T) < r(L). By induction assumption, x0(S) = 1, hence, v(S) — x0(S) = 0. By Lemma 1,

v(S) — x0(S) > vq(L)(S n Q(L)) — x0(S n Q(L)).

By C2, S n Q(L) = 0, and by the definition of objection,

Vq(l)(S n Q(L)) — x0(S n Q(L)) > 0.

This contradicts v(S) — x0(S) = 0, hence, x0(L) = 0. Thus,

x0(L) = 0 for all L G End(Gr) with r(L) = i.

In view of (1), this implies

x°(Q(L)) = 1 for all L G End(Gr) with r(L) = i.

Example 10. Let N = {1, 2, 3,4}, A2 = {{1}, {2}, {3}, {3,4}}.

Let Gr2 be a digraph with the set of vertices A2 and the set of edges

{({3}, {2}), ({2}, {3, 4}), ({3,4}, {1})}.

The set of the edges of Gr2 covers N, and Gr2 is A2-admissible (see Example 8), hence, MA2 (N,v) = 0 for some (N,v) G r0.

The conditions on A that ensure the non-emptyness of M\(N, v) for all (N, v) G r0 are similar.

Let A be a family of subsets of N. A directed graph Gr is called weakly A-admis-sible if A is the set of its vertices and there exists a map g defined on the set of the edges of Gr, that takes each oriented edge (K, L) to a pair g(K, L) = (Q, r) (Q C N, r G R1) and satisfies the following 3 conditions.

C1. If g(K,L) = (Q, r), then K C Q, Q n L = 0, \Q\ > 1.

C2’. If g(K, L) = (Q, r), g(R, P) = (S, t), L C S, K n S = 0, then Q n S = 0.

C3’. If g(K, L) = (Q, r), g(R, P) = (s, t), L C S, K n S = 0, then r > t.

Example 11. Let N = {1, 2, 3,4}, A3 = {{1}, {2}, {1, 3}, {1,4}}.

Then digraph Gr3 with the set of edges {a,$,Y}, where

a = ({1}, {2}), $ =({2}, {1, 3}), y =({2}, {1,4})

is not weakly A3-admissible.

Indeed, suppose that Gr is weakly A3-admissible. Then, by C1, g($) = ({2, 4}, i), g(l) = ({2, 3},j) and, by C1 and C2’, g(a) = ({1, 3,4},k). Since {2} C {2,4} and {1} n {2,4} = 0, C3’ implies i < k. Since {1, 3} C {1, 3, 4} and {2} n {1, 3,4} = 0, C3’ implies k < i, this contradicts i < k.

Note that weakly A-admissible digraph Gr does not generate an acyclic binary relation on A.

Example 12. Let N = {1, 2, 3,4}, A4 = {{1, 2}, {3}, {4}}. Then digraph with the set of edges {a,$,Y}, where a = ({1, 2}, {3}),$ = ({3}, {4}), y = ({4}, {1, 2}), is weakly A4-admissible because we can take

g(a) = ({1,2,4}, 2),

g($) = ({1,3}, 3),

g(Y) = ({3,4}, 1).

Lemma 3. Let (Q,y)Q) be an objection of K against L at x. Let S C N, L C S, K n S = 0. Then there exists a strong counter-objection (S, zs) to this objection iff

(y — x)(Q n S) < v(S) — x(S).

Proof.

The proof is similar to the proof of Lemma 1.

Lemma 4. Let x be an imputation for (N, v), K,L C N, K C Q C N, L n Q = 0. Let S C N, L C S, K n S = 0, x(Q) < v(Q).

If either QnS = 0 and x(S) < v(S) or QnS = 0 and v(Q) — x(Q) < v(S) — x(S), then for each objection (Q,yQ) of K against L at x there exists a strong counterobjection (S,zs) of L against K.

Proof.

The proof is similar to the proof of Lemma 2.

Corollary. Ka(N, v) C MiA(N,v).

Proof.

Let x G Ka(N, v). Suppose that there exists an objection (C, yC) of K against L at x for some K, L gA. We have either x(L) < v(L) (and then there exists a strong counter-objection (L, zL)) or tK,L(x) < tL,K(x) and then there exists D C N such

that L C D, K n D = 0, v(D) — x(D) > v(C) — x(C). Since v(C) — x(C) > 0, by Lemma 4, there exists a strong counter-objection to (C,yC).

Theorem 5. Let A be a set of subsets of N. If for each weakly A-admissible graph Gr the set of the ends of its edges does not cover N, then MA(N,v) = 0 for all (N,v) G r0.

Proof.

The proof is similar to the proof of Theorem 3, Lemmas 1 and 2 are changed by Lemmas 3, 4.

Example 13. Let N = {1, 2, 3, 4}, A = {{1}, {2}, {1, 3}, {1, 4}, {2, 3, 4}}. Then MiA(N, v) = 0 for all (N, v) G r0.

Let us check the condition of Theorem 5. Suppose that there exists a weakly admissible digraph Gr such that the ends of its edges cover N. Then {2, 3,4} is not the end of an edge by C1 and Gr contains the edges {a, $, y}, where

a =({1}, {2}), $ =({2}, {1, 3}), y = ({2}, {1,4}).

It was proved (see Example 11) that such Gr is not weakly A-admissible.

Theorem 6. Let A be a set of subsets of N and |N| < 5. Then MA(N, v) = 0 for all (N, v) G r0 if and only if for each weakly A-admissible graph Gr the set of the ends of its edges does not cover N.

Proof.

The proof is the same as the proof of Theorem 4.

Example 14. Ka C Ka-

Let N = {1,... ,4}, A = A = {{1, 2}, {3}, {4}}. There exists A4-admissible digraph such that the ends of its edges cover N (see Example 12), hence, by Theorem 6, there exists a game (N,v) with M\i (N,v) = 0 and then by Corollary to Lemma 4, Ka4 (N,v) = 0. But by Theorem 2, Ka4 (N,v) = 0.

3. Conclusion

The paper considers conditions on the families of coalitions that ensure the existence of two types of generalized bargaining sets with respect to these families. Sufficient conditions are valid for each finite number of players and, for a fixed family, can be verified in finite number of steps. It is proved that these conditions are also necessary if the number of players is no more than 5.

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