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. Boros, V. Gurvich Mathematical Social Sciences 39 2000 175 –194
23, is stable whenever is stable. In other words, if G9 has a CARS then G also
G 9 G
has one, i.e. G9 is kernel-solvable whenever G is kernel-solvable. Analogously, we can define ‘induced subEFFs’ of an arbitrary EFF. Given an EFF
9 9
5 kC , S ; I, A; Jl and a subset J9 J, the subEFF 5 kC , S ; I9, A9; J9l induced
i a
J 9 i
a
by J9 is defined by setting I9 5 hi [ IuC J9 ± 5j, and A9 5 ha [ AuS J9 ± 5j, and
i a
9 9
defining C 5 C J9 for all i [ I9, and S 5 S J9 for all a [ A9. Obviously,
i i
a a
J 9
for any J9 J, thus the following is implied by Lemma 23.
Lemma 24. An induced subEFF is stable whenever the EFF is stable.
J 9
As an application, let us consider an EFF 5 kC , S ; I, A; Jl for which I 5 hi , i j,
i a
1 2
A 5 ha , a j, and J 5 h1, 2j, and where C 5 S 5 h1j, and C 5 S 5 h2j. It is easy to
1 2
1 1
2 2
verify that T 5 hhi j, ha j, ha j; hi j, ha j, ha jj is a rejecting table for , with no
1 1
2 2
2 1
common player cycles, and hence is not stable, by Theorem 16. Together with Lemma 24 this implies Corollary 9 of Theorem 8.
5. Proofs of main theorems and their applications Proof of Theorem 8. Given an EFF 5
kC , S ; I, A; Jl, let us suppose, in contrast to
i a
i that there exists a subset J9 J such that uC J9u 1 for every i [ I but there exists
i
no a [ A for which J9 S . We shall prove that under these conditions is not stable.
a
We can assume that J9 is an inclusion-minimal subset with these properties, i.e. that for every J0 J9 there exists an a [ A such that J0 S . This implies immediately that
a
uJ9u ± 1, since otherwise J9 S would hold for some a [ A. Let us consider the subEFF
a
9 9
5 kC , S ; I9, A9; J9l induced by the set J9. By its definition and by the minimality
J 9 i
a
of the set J9, the equation h jj, J9 2 j 5 1 holds for all j [ J9. Since such an EFF
J 9
cannot be stable whenever uJ9u . 1, we obtain that is not stable, by Lemma 24.
otherwise subEFF would be stable, too.
J 9
The second part ii of the theorem is dual to the first part. h
Let us note that in fact we have shown a bit more. The conditions of Theorem 8 are
d
necessary not only for the stability of but also for the stability of its dual .
Proof of Theorem 11. Let us note first that conditions ii and ii9 are equivalent with the relation . Since G is assumed to be perfect by i, is stable by Proposition
G G
18 and Theorem 20 see Boros and Gurvich, 1994, and thus is stable, too, by Lemma 24.
h Let us remark that again we have shown a somewhat stronger result. The conditions
of Theorem 11 are sufficient not only for the stability of but also for the stability of its
d c
dual . Indeed, let us consider the complementary graph G instead of G. Cliques of
c d
G are stable sets of G and vice versa, and by the definition, we have 5 . It is
c
G c
known that a graph G is perfect iff its complement G is perfect see Lovasz, 1972b,
E . Boros, V. Gurvich Mathematical Social Sciences 39 2000 175 –194
185
thus if conditions i, ii and ii9 of Theorem 11 hold for and G then they hold for
d c
and G , as well. Let us note that the conditions of Theorem 8 are necessary but not sufficient for
stability and the conditions of Theorem 11 are sufficient but not necessary. This is so, for example, because otherwise the stability of would be equivalent to the stability of
d
, which is not the case, as we shall see in the next section. The following example will clarify these limits of Theorems 8 and 11.
Example 25. For every positive integer k let us define an EFF by setting J 5
k
h1, . . . , kj, and I 5 A 5 h p, qu1 p , q kj, and by defining C 5 S
5 h p, qj
p, q p, q
for all p, q [ I 5 A. Direct computations show that is stable if k 6, but it is
k
not stable for k 5 7. A corresponding CARS for is given by the following three
7
directed cycles 1 →
2 →
3 →
4 →
5 →
6 →
7 →
1, 1 →
3 →
5 →
7 →
2 →
4 →
6 →
1, and 1
→ 5
→ 2
→ 6
→ 3
→ 7
→ 4
→ 1. Since is an induced subEFF of for any k . 7, it
7 k
follows that is not stable for all k 7.
k
Let us note that the conditions of Theorem 8 hold automatically for every k, but is
k
not stable if k 7. On their turn, the conditions of Theorem 11 hold only for k 4, while and are still stable.
5 6
For example , if k 5 5 the only graph satisfying ii and ii9 of Theorem 11 is the hole
c
C but it is not perfect, while if k 5 6 then either G or G must contain a triangle,
5
consequently either ii or ii9 must be violated. Thus for and no
‘perfect split’ exist, but still these EFFs are stable.
5 6
Let us consider now an important class of EFFs for which Theorems 8 and 11 are applicable.
Given an EFF 5 kC , S ; I, A; Jl 5 kK , B ; J; I, Al, let us assume that
i a
j j
uC S u 1 for all i [ I and a [ A P
i a
or in dual terms,
d
K K 5 5 or
B B 5 5 or both for every j ± j9 [ J
P
j j 9
j j 9
Let us suppose first that there exists a pair j ± j9 [ J such that both sets K K and
j j 9
B B are empty, or in dual terms, that there exists a pair j, j9 [ J, j ± j9 which does
j j 9
not belong to the same sets C , i [ I or S , a [ A. Then, according to Corollary 9 of
i a
d
Theorem 8, neither nor is stable.
So let us now assume that for every pair j ± j9 [ J exactly one of the following two options holds
i K K ± 5 or
j j 9
ii B B ± 5,
j j 9
or in dual terms,
d
i h j, j9j C for some i [ I, or
i
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. Boros, V. Gurvich Mathematical Social Sciences 39 2000 175 –194
d
ii h j, j9j S for some a [ A.
a c
c
Let us denote by G 5 kJ, El and G 5 kJ, E l the two complementary graphs having
the same set of vertices J, and the edges of which are defined by i and ii,
d
respectively. Theorem 8 claims that neither nor its dual can be stable unless
d
5 and 5 . So let us now assume that both these equalities hold.
c
G G
d
Furthermore, Theorem 11 claims that both and are stable if the graph G is perfect. Finally, let us suppose that the graph G is not perfect. This is the only open case.
According to Conjecture 5, in this case the EFF 5 is not stable. According to the
G c
results of Lovasz 1972a, the complementary graph G is not perfect either, in this case,
d
thus according to Conjecture 5, the dual EFF 5 is not stable either.
c
G
Proposition 26. Conjectures 5 and 13 are equivalent.
Proof. The above arguments prove that if the assumption in Eq. P and Conjecture 5
d
hold then for the dual pair of EFFs and , either both or neither of them are stable, i.e. Conjecture 5 implies Conjecture 13.
Let us suppose that Conjecture 5 does not hold then the equivalent Conjecture 6 fails
c
too, i.e. there exists a complementary pair of graphs G and G such that G is
c
kernel-solvable while G is not. Then, according to Propositions 1–3, the EFF is
G d
stable while its dual 5 is not. However, the assumption in Eq. P holds for
c
G
EFFs generated by graphs, hence Conjecture 13 fails. h
Proposition 27. SPGC and Conjecture 15 are equivalent.
Proof. SPGC
⇒ Conjecture 15 Let us suppose that the graph G has neither odd holes
nor odd antiholes. SPGC claims that G is perfect. According to Theorem 4, G is then kernel-solvable, and according to Propositions 1–3, the EFF
is stable.
G
Conjectures 5 and 15
⇒
SPGC. Let us suppose that the graph G contains neither odd
holes nor odd antiholes. According to Conjecture 15, the EFF is stable. According to
G
Propositions 1–3, the graph G is then kernel-solvable, implying finally that, according to Conjecture 5, the graph G is perfect.
Conjecture 15
⇒
Conjecture 5. Since Conjecture 5 is equivalent to Conjecture 6 see
Boros and Gurvich, 1994, we shall instead show that Conjecture 15 implies Conjecture 6.
Odd holes and odd antiholes are complementary. Thus the graph G contains one of
c
these iff the complementary graph G contains the complementary one. Let us suppose
c
that both graphs G and G contain an odd hole or an odd antihole, implying that they both are not kernel-solvable see, for example, Berge and Duchet, 1983 or Boros and
Gurvich, 1994. On the other hand, if both graphs contain neither odd holes nor odd antiholes then both graphs are kernel-solvable, according to Conjecture 14. Thus, in all
E . Boros, V. Gurvich Mathematical Social Sciences 39 2000 175 –194
187
c
cases the graph G is kernel solvable iff its complement G is also kernel solvable, which is exactly Conjecture 6.
h Now we will derive Theorem 12 from Theorems 8 and 11 and from Lemma 28 below.
Given an integer b [ N, let us define a graph G 5 kV , E l consisting of 2n vertices
n n
n
V 5 h1, 2, . . . , 2nj, and n pairwise non-adjacent edges E 5 hi, n 1 iu1 i nj.
n n
Lemma 28. Given a graph G 5
kV , E l, and given k subsets } 5 hM , . . . , M j of the
n n
n 1
k
vertex-set V , it is an NP-complete problem to check the validity of the following
n
statement: Every maximal stable set of G is contained in some subset M [ }
Q
n i
Proof. We shall prove the lemma by reducing the well-known NP-complete problem of tautology for DNFs see, for example, Garey and Johnson, 1979 to Eq. Q.
Obviously, the problem of testing the validity of Eq. Q belongs to NP. Let us consider an arbitrary DNF on n variables consisting of k terms
k
] DX , . . . , X 5k
n X n X
1 n
j j
S D
i 51 j [P
j [N
i i
] where X , j 5 1, . . . , n denote the Boolean variables, and X denotes the negation of X.
j
The tautology problem for D consists in deciding if D evaluates to true for every true-false assignments to the variables X , . . . , X .
1 n
Let us now associate to such a DNF a family } 5 } of k subsets of V , as in the
D n
lemma. Let us define M 5
h ju1 j n, j [
⁄ P
j h j 1 nu1 j n, j [
⁄ N
j
i i
i
for i 5 1, . . . , k. It is then straightforward to verify that D is a tautology i.e. it evaluates always to true iff Eq. Q holds true for the family } .
D
Thus, the general problem of tautology for DNFs is polynomially reduced to testing the validity of Eq. Q.
h
Proof of Theorem 12. We shall prove the theorem by reducing the problem of testing the validity of Eq. Q to the stability testing of effectivity functions.
Obviously, testing the stability of an effectivity function belongs to NP by Theorem 16 Keiding, 1985.
Let us consider the graph G 5 kV , E l and a family } 5 hM , . . . , M j of k subsets
n n
n 1
k
of V , and let us associate an EFF 5 G , } 5 kC , S ; I, A; Jl to this pair, as
n n
i a
follows. The EFF G , } consists of n players I 5 h1, . . . , nj and k outcomes
n
A 5 h1, . . . , kj, and we have J 5V , C 5 hi, i 1 nj for i 5 1, . . . , n, and S 5 M , for
n i
a a
a 5 1, . . . , k i.e. 5 E and 6 5 } .
n
For this EFF G , } we have that if Eq. Q holds for }, then it is stable by
n
Theorem 11, since G is a perfect graph, and if Eq. Q does not hold for }, then it
n
cannot be stable by Theorem 8.
188 E
. Boros, V. Gurvich Mathematical Social Sciences 39 2000 175 –194
Thus, we have reduced Eq. Q to the problem of testing the stability of EFFs, and thus the theorem follows by Lemma 28.
h
Remark 29. Let us observe that the condition in Eq. Q is equivalent to condition i of Theorem 8 for the EFF G , } , and it is equivalent with the fact that G , } is
n n
playing-minor, see Remark 10. Thus in fact, it is an NP-complete problem to check condition i i
.e. playing-minority even in a special case when it is sufficient and not only necessary for stability.
6. Stability is not selfdual
