On the structure of stable tournament solutions

Abstract

A fundamental property of choice functions is stability, which, loosely speaking, prescribes that choice sets are invariant under adding and removing unchosen alternatives. We provide several structural insights that improve our understanding of stable choice functions. In particular, (1) we show that every stable choice function is generated by a unique simple choice function, which never excludes more than one alternative, (2) we completely characterize which simple choice functions give rise to stable choice functions, and (3) we prove a strong relationship between stability and a new property of tournament solutions called local reversal symmetry. Based on these findings, we provide the first concrete tournament—consisting of 24 alternatives—in which the tournament equilibrium set fails to be stable. Furthermore, we prove that there is no more discriminating stable tournament solution than the bipartisan set and that the bipartisan set is the unique most discriminating tournament solution which satisfies standard properties proposed in the literature.

Appendices

Appendix 1: Examples for Remark 2

Brandt and Harrenstein (2011, p. 1729) mention that ${\widehat{\alpha }}$ and ${\widehat{\gamma }}$ are independent from each other in the context of general choice functions. Here, we prove that the same holds even in the context of tournament solutions.

Proposition 1

There exists a tournament solution that satisfies ${{\widehat{\alpha }}}$, but not ${{\widehat{\gamma }}}$.

Proof

Let S be a stable tournament solution. As mentioned in Sect. 3, $\lceil {S}\rceil$ satisfies ${\widehat{\alpha }}$. However, it is easily seen that, unless S is trivial, $\lceil {S}\rceil$ violates ${\widehat{\gamma }}$. Hence, the statement follows from the existence of non-trivial stable tournament solutions (such as $BP$). $\square$

Proposition 2

There exists a tournament solution that satisfies ${{\widehat{\gamma }}}$, but not ${{\widehat{\alpha }}}$.

Proof

Let S be a stable tournament solution. Define the tournament solution $S'$ such that for each tournament $T=(A,\succ )$,

$$\begin{aligned} S'(T) = {\left\{ \begin{array}{ll} S(T) &{}\text {if } |A \setminus S(T)|>1 \\ A &{}\text {otherwise.} \end{array}\right. } \end{aligned}$$

It can be shown that $S'$ satisfies ${{\widehat{\gamma }}}$, but may violate ${{\widehat{\alpha }}}$. For the latter, let $S= TC$ and consider a transitive tournament $(\{a,b,c\},\succ )$ such that $a \succ b$, $b \succ c$, and $a \succ c$. By definition, $S'(\{a,b,c\})=\{a\}$, but $S'(\{a,b\})=\{a,b\}$. $\square$

Appendix 2: Examples for Remark 3

$\textit{BA}$ satisfies local ${\widehat{\alpha }}$, but ${\widehat{\textit{BA}}}={ ME}$ violates ${\widehat{\alpha }}$ (Brandt et al. 2016b).

Similarly, there exists a tournament solution S for which ${\widehat{S}}$ is well-defined, but ${\widehat{S}}$ is not stable. For a stable tournament solution S, we have by definition that $S={\widehat{S}}$ and hence that ${\widehat{S}}$ is also stable. The following proposition shows that ${\widehat{\alpha }}$ does not carry over from S to ${\widehat{S}}$ even if S is simple and ${\widehat{S}}$ is well-defined.

Proposition 3

There exists a simple tournament solution S satisfying ${\widehat{\alpha }}$ such that ${\widehat{S}}$ is well-defined but ${\widehat{S}}$ does not satisfy ${\widehat{\alpha }}$.

Proof

Let S be the tournament solution that always chooses all alternatives, with two exceptions:

• If the tournament is of order 2, then S chooses only the Condorcet winner.

• If the tournament is the tournament $T_4$ given in Fig. 6, then S chooses alternatives a, b,  and c.

Fig. 6

Tournament $T_4$

Clearly, S is simple and satisfies ${\widehat{\alpha }}$. Since ${\widehat{S}}$ chooses alternatives a, b,  and c from $T_4$, but chooses only the Condorcet winner from the transitive tournament of order 3, it does not satisfy ${\widehat{\alpha }}$.

It remains to show that ${\widehat{S}}$ is well-defined. One can check that every tournament contains an S-stable set. Suppose for contradiction that some tournament T contains two distinct minimal S-stable sets, which we denote by B and C. Then B and C are also S-stable in $B\cup C$. If B is a singleton, then B is the Condorcet winner in $B\cup C$, which means C cannot be S-stable, a contradiction. Hence both B and C are transitive tournaments of order 3, and $4\le |B\cup C|\le 6$. One can check all the possibilities of $B\cup C$ to conclude that this case is also impossible. $\square$

For the tournament solution S defined in the proof of Proposition 3, we have that $\widehat{{\widehat{S}}}$ is not well-defined. Even though no tournament contains two distinct minimal ${\widehat{S}}$-stable sets, $T_4$ does not contain any ${\widehat{S}}$-stable set. This example also shows that for a tournament solution $S'$, $\widehat{S'}$ may fail to be well-defined not because it allows two distinct minimal $S'$-stable sets in a tournament but because some tournament contains no $S'$-stable set.

Appendix 3: Example for Remark 6

We show that there exists a tournament solution different from $BP$ that satisfies $LRS$, monotonicity, regularity, and Condorcet consistency.

To this end, we define a new tournament solution called $POS$ which chooses all alternatives with positive relative degree. More precisely, an alternative is chosen by $POS$ if it dominates strictly more than half of the remaining alternatives, is not chosen if it dominates strictly less than half of the remaining alternatives, and goes to a “tie-break” to determine whether it is chosen if it dominates exactly half of the remaining alternatives.

For tournaments of even size, $POS$ chooses exactly the alternatives that dominate at least (or equivalently, more than) half of the remaining alternatives. Hence we do not need a tie-break for tournaments of even size. The tie-breaking rule for tournaments of odd size $2n+1$ is as follows: For any (unlabeled) tournament T of order 2n and any partition of it into two sets B and C of size n, consider two tournaments $T_1$ and $T_2$ of order $2n+1$. The tournament $T_1$ contains T and another alternative a that dominates B but is dominated by C, while the tournament $T_2$ contains T and another alternative a that dominates C but is dominated by B. If $T_1$ or $T_2$ is regular, $POS$ chooses a in that tournament and not in the other one. Otherwise, $POS$ arbitrarily chooses a in exactly one of $T_1$ and $T_2$.

Proposition 4

$POS$ satisfies $LRS$, monotonicity, and regularity.

Proof

We need to show that the tie-breaking rule in the definition of $POS$ is well-defined. First, we show that if we perform a local reversal on alternative a, we do not get an isomorphic tournament with alternative a mapped to itself. Indeed, if a were mapped to itself, it would mean that no tournament solution satisfies $LRS$, which we know is not true since $BP$ satisfies $LRS$. Secondly, the tournament obtained by performing a local reversal on an alternative in a regular tournament is not regular. Hence we do not obtain a conflict within the tie-breaking rule.

It follows directly from the definition that $POS$ satisfies $LRS$, monotonicity, and regularity. $\square$

Fig. 7

Tournament $T_7$. $g\succ \{a,b,c\}$, $\{d,e,f\}\succ g$, and all omitted edges point downwards

The tournament $T_4$ given in Fig. 6 shows that $POS$ violates composition-consistency and ${\widehat{\alpha }}$ (and hence stability).

Interestingly, $BP$ (and all of its coarsenings) always intersect with $POS$ while there exists a tournament for which $BA$ (and all of its refinements such as $TEQ$ and $ME$) do not overlap with $POS$. This follows from results on the Copeland value by Laffond et al. (1993b, 1994).

Appendix 4: Example for Remark 7

We construct a tournament solution that satisfies monotonicity and stability, but violates regularity and composition-consistency.

Every tournament solution has to be regular on tournaments of order 5 or less because of non-trivial automorphisms. Consider the tournament $T_7$ shown in Fig. 7. $T_7$ admits a unique non-trivial automorphism that maps each of the six alternatives in the two 3-cycles to the next alternative in its 3-cycle and maps alternative g to itself.¹⁴

Now, define the simple tournament solution $S_7$, which always returns all alternatives unless the tournament is $T_7$ or it can be modified from $T_7$ by weakening alternative g. In the latter case, $S_7$ returns all alternatives except g.

We check that this definition is sound. First, we know that in $T_7$, there is no automorphism that maps alternative g to another alternative. When we weaken g, it is the unique alternative with the smallest out-degree and hence cannot be mapped by an automorphism to another alternative. Now, the alternatives a, b, c form an orbit, and $S_7$ excludes g whenever it is dominated by d, e, f (and has any dominance relationship to a, b, c). This yields four non-isomorphic tournaments for which $S_7$ excludes g.

Proposition 5

$\widehat{S_7}$ satisfies stability and monotonicity.

Proof

First, observe that $S_7$ trivially satisfies local ${\widehat{\alpha }}$ because $S_7$ only excludes an alternative in tournaments of order 7. By virtue of Theorem 3, it therefore suffices to show that $\widehat{S_7}$ is well-defined.

One can check that every tournament contains an $S_7$-stable set. Let $T_6$ denote the tournament obtained by removing alternative g from $T_7$. Suppose for contradiction that there exists a tournament T that contains two distinct minimal $S_7$-stable sets, which we denote by B and C. Then B and C are also $S_7$-stable in $B\cup C$. Moreover, $T|_{B}$ must correspond to the tournament $T_6$, and each alternative in $C\backslash B$ either has the same dominance relation to B as the alternative g does to $T_6$ or has a dominance relation that is a weakening of g. The same statement holds for C. We consider the following cases.

Case 1: $10\le |B\cup C|\le 11$. The tournament $T|_{B}$ has one of its alternatives corresponding to alternatives d, e, and f in Fig. 7 outside of $B\cap C$, and this alternative must dominate all of the alternatives in C. Similarly, there exists an alternative in $C\backslash B$ that dominates all of the alternatives in B. But this implies that some two alternatives dominate each other, a contradiction.

Case 2: $7\le |B\cup C|\le 9$. At least one of the two tournaments $T|_{B}$ and $T|_{C}$ must have all of its alternatives corresponding to alternatives d, e, and f in Fig. 7 in the intersection $B\cap C$, for otherwise we obtain a contradiction in the same way as in Case 1. Assume without loss of generality that $T|_{B}$ has its alternatives corresponding to d, e, and f in the intersection. Hence three alternatives in $B\cap C$ that form a cycle dominate the same alternative in C. But this does not occur in $T_6$, a contradiction.

It follows from Theorem 4 that $\widehat{S_7}$ satisfies monotonicity. $\square$

Clearly, $\widehat{S_7}$ is not regular since it excludes an alternative from the regular tournament $T_7$. Moreover, it is not a coarsening of $BP$ since $BP$ selects all of the alternatives in $T_7$. Hence we have that stable and monotonic tournament solutions are not necessarily coarsenings of $BP$.