Circulant games

Abstract

We study a class of two-player normal-form games with cyclical payoff structures. A game is called circulant if both players’ payoff matrices fulfill a rotational symmetry condition. The class of circulant games contains well-known examples such as Matching Pennies, Rock-Paper-Scissors, as well as subclasses of coordination and common interest games. The best response correspondences in circulant games induce a partition on each player’s set of pure strategies into equivalence classes. In any Nash Equilibrium, all strategies within one class are either played with strictly positive or with zero probability. We further show that, strikingly, a single parameter fully determines the exact number and the structure of all Nash equilibria (pure and mixed) in these games. The parameter itself only depends on the position of the largest payoff in the first row of one of the player’s payoff matrix.

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Notes

  1. 1.

    The simplest example of a matching game is Heads and Tails. If both players match the strategy of the other player each player gets a payoff of 1, otherwise each player receives a payoff of zero.

  2. 2.

    We choose to label players’ strategies from 0 to \(n_i-1\) as this will later simplify notation significantly.

  3. 3.

    Table 1 in the appendix contains the remaining Nash equilibrium profiles.

  4. 4.

    Table 2 in the appendix shows the Nash equilibria and the equivalence classes for the two counter-circulant games we introduced in Example 4 and 5.

  5. 5.

    Many results on the number of equilibria in n-person, normal-form games hold only for an open set of payoff vectors, see e.g., McKelvey and McLennan (1997) and McLennan (1997).

  6. 6.

    As players’ matrices cycle in opposite directions, counter-circulant games with \(n>2\) are never symmetric.

  7. 7.

    Note that in the limit case when \(a=(b+c)/2\), \(\sigma ^*\) is a neutrally stable strategy (see e.g., Weibull 1995).

  8. 8.

    The original form of the game can be obtained by relabeling the strategies of player 2 for both A and the transpose of A, \(A^T\). Strategy 3 in the symmetric representations corresponds to strategy 0 in the original game, strategy 4 to 1, 0 to 2, 1 to 3, and 2 to 4, respectively. Since \(a_{n-1}^2=a_4^2=6\) is the largest payoff in player 2’s payoff matrix, the original game has a shift of \(k=1\).

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Acknowledgments

We would like to thank Carlos Alós-Ferrer, Tanja Artiga Gonzalez, Wolfgang Leininger, participants at the SAET13 conference in Paris and seminar participants in Cologne and Innsbruck for helpful comments and discussions. We also thank two anonymous referees and the coordinating editor for helpful comments and suggestions. Johannes Kern gratefully acknowledges financial support from the German Research Foundation through research projects AL-1169/1-1 and AL-1169/1-2. Ɖura-Georg Granić also gratefully acknowledges financial support from the German Research Foundation (DFG) through research project AL-1169/2-1.

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Appendices

Appendix 1: Transformation of games

Lemma 5

  1. (i)

    Let \(\varGamma _n\) be an iso-circulant game in which both players’ payoff matrices are circulant. There is a permutation of row vectors that fixes the first row in both matrices and transforms both players’ payoff matrices into anti-circulant matrices.

  2. (ii)

    Let \(\varGamma _n\) be a counter-circulant game in which player 1’s payoff matrix is circulant. There is a permutation of row vectors that fixes that first row in both matrices and transforms player 1’s payoff matrix into an anti-circulant matrix and player 2’s matrix into a circulant matrix.

Proof

(Lemma 5)

  1. (i)

    A matrix A is anti-circulant if and only if \(A=P C\), where C is a circulant matrix and

    $$\begin{aligned} P= \left( \begin{matrix} 1&{}\quad 0&{}\cdots &{}0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\cdots &{}0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\cdots &{}0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\cdots &{}1&{}\quad 0&{}\quad 0\\ \vdots &{}\quad \vdots &{}&{}\vdots &{}\quad \vdots &{}\quad \vdots \\ 0&{}\quad 1&{}\cdots &{}0&{}\quad 0&{}\quad 0 \end{matrix} \right) \end{aligned}$$

    (Davis 1979, p. 162, Corollary). The matrix P switches rows i and \(n+1-i\) and fixes the first row. Using this result, we obtain that \(PA_1\) and \(P A_2\) are anti-circulant matrices since both \(A_1\) and \(A_2\) are circulant matrices.

  2. (ii)

    Using the matrix P defined as in (i), we obtain that \(PA_1\) is anti-circulant (Davis 1979, p. 162, Corollary). As \(A_2\) is anti-circulant, \(A_2=P C\) for some circulant matrix C (p. 162, Corollary Davis 1979). Hence \(P A_2=P(PC)\) and since \(P=P^{-1}\) (Davis 1979, p.28, equ. 2.4.22), we obtain that \(P A_2\) is a circulant matrix.

\(\square \)

Appendix 2: Proofs of main results

We remind the reader that the sum and the difference of strategies in a circulant game as well as multiplications of integers with strategies are read modulo n. Central to the proofs of our main results is Proposition 6 below.

Proposition 6 identifies sufficient conditions under which the number of Nash equilibria of a finite two-player normal-form game can be calculated by merely identifying one parameter of the game. Under the hypotheses of Proposition 6, each Nash equilibrium strategy of a player corresponds to one specific combination of elements of a partition of that player’s strategy set. Moreover, for each possible combination of elements of the partition there exists exactly one corresponding Nash equilibrium strategy. The parameter necessary to determine the number of Nash equilibria is the cardinality of the partition.

The proof of Theorem 1 (Theorem 2) first establishes that iso-circulant (counter-circulant) games satisfy the hypotheses of Proposition 6. Once this is done, determining the cardinality of the partitions is merely a counting exercise.

Proposition 6

For the two-player normal-form game \(\varGamma _n\), let \(\overline{S}_1\) and \(\overline{S}_2\) be partitions of \(S^n\) such that \(|\overline{S}_1|=|\overline{S}_2|\). Let \([s]_i \in \overline{S}_i\) be the equivalence class represented by the pure strategy s. If \(\varGamma _n\), \(\overline{S}_1\), and \(\overline{S}_2\) satisfy

  1. (a)

    for all Nash equilibria \((\sigma _1,\sigma _2)\), and all \(s,s' \in S^n\), if \(s' \in [s]_i\) then \(\sigma _i(s)=0\) if and only if \(\sigma _i(s')=0\),

  2. (b)

    for all \(\sigma _i \in \varSigma _i\), \(i=1,2\), \({\text {supp}}(\sigma _{i})\cap [s]_{i}=\emptyset \) for \([s]_{i} \in \overline{S}_{i}\) implies \(BR_{-i}(\sigma _{i}) \cap [-s]_{-i}=\emptyset \),

  3. (c)

    for all \(s \in S^n\), \(\varGamma _n\) has a Nash equilibrium \((\sigma _1,\sigma _2)\) with \({\text {supp}}(\sigma _1)=[s]_1\) and \({\text {supp}}(\sigma _2)=[-s]_2\),

then

  1. (i)

    for any \(M \subseteq \overline{S}_1\) \(\varGamma _n\) has a unique Nash equilibrium \((\sigma _1,\sigma _2)\) with \({\text {supp}}(\sigma _1)=\bigcup _{[s]_1 \in M} [s]_1\);

  2. (ii)

    \(\varGamma _n\) has exactly \(2^{|\overline{S}_1|}-1\) Nash equilibria.

Proof

(Proposition 6) (i) Given \(\emptyset \ne M \subseteq \overline{S}_1\) let \(-M:=\{[-s]_2|[s]_1 \in M\} \subseteq \overline{S}_2\) and let \(\varGamma _n^M\) be the reduced game where player 1’s set of strategies is \(\bigcup _{[s]_1 \in M} [s]_1\) and player 2’s set of strategies is \(\bigcup _{[s]_1 \in M} [-s]_{2}\) (and the payoff functions are restricted accordingly).

Claim A: Let \(M'\subseteq M \subseteq \overline{S}_1\) be a non-empty subset of \(\overline{S}_1\) and let \((\sigma ^{M'}_1,\sigma ^{M'}_2)\) be a completely mixed Nash equilibrium of \(\varGamma _n^{M'}\). Then \((\sigma ^M_1,\sigma ^M_2)\) defined by \(\sigma ^M_1(s)=\sigma ^{M'}_1(s)\) if \([s]_1 \in M'\) and \(\sigma ^M_1(s)=0\) otherwise, and \(\sigma ^M_2(s)=\sigma ^{M'}_2(s)\) if \([s]_2 \in -M'\) and \(\sigma ^M_2(s)=0\) otherwise is a Nash equilibrium in \(\varGamma ^M_n\).

Since \((\sigma ^{M'}_1,\sigma ^{M'}_2)\) is a completely mixed Nash equilibrium of \(\varGamma _n^{M'}\), all strategies in \(\bigcup _{[s]_1 \in M'} [-s]_2\) yield the same payoff for player 2 against \(\sigma _1^M\). By hypothesis (b), since \({\text {supp}}(\sigma _1^M)=\bigcup _{[s] \in M'} [s]\), no strategy outside \(\bigcup _{[s]_1 \in M'} [-s]_2\) can be a best response for player 2 against \(\sigma _1^M\). Analogously all strategies in \(\bigcup _{[s]_1 \in M'} [s]_1\) yield the same payoff for player 1 against \(\sigma _2^M\), and since \({\text {supp}}(\sigma _2^M)=-\bigcup _{[s]_1 \in M'} [-s]_2\), no strategy outside \(\bigcup _{[s]_1 \in M'} [s]_1\) is a best response for player 1 against \(\sigma _2^M\). Hence, by Proposition 1, \((\sigma ^M_1,\sigma ^M_2)\) is a Nash equilibrium in \(\varGamma ^M_n\). This proves the claim.

Claim B: For any \(\emptyset \ne M \subseteq \overline{S}_1\), the reduced game \(\varGamma _n^M\) has exactly one completely mixed Nash equilibrium.

Let \(\emptyset \ne M \subseteq \overline{S}_1\) be such that \(|M|=m\). We will prove the claim by induction over m. Note first, that by hypothesis (b), in any Nash equilibrium \((\sigma _1,\sigma _2)\) of \(\varGamma ^M_n\), \({\text {supp}}(\sigma _1)\) is a union of elements of M.

For \(m=1\), this follows by hypothesis (c). For \(m>1\), by induction hypothesis we obtain that for all \(\emptyset \ne M' \subsetneq M\) the reduced game \(\varGamma _n^{M'}\) has a unique completely mixed Nash equilibrium. By Claim A, for every \(\emptyset \ne M' \subsetneq M\) there is a Nash equilibrium \((\sigma ^{M}_1,\sigma ^{M}_2)\) in \(\varGamma ^M_n\) with \({\text {supp}}(\sigma ^{M}_1)=\bigcup _{[s] \in M'}[s]\). As by Proposition 2(ii), for any \(\emptyset \ne M' \subsetneq M\) there can be at most one Nash equilibrium \((\sigma _1,\sigma _2)\) in \(\varGamma _n^M\) with \({\text {supp}}(\sigma _1)=M'\) we obtain that there is exactly one such Nash equilibrium. This implies that \(\varGamma _n^M\) has at least \(2^m-2\) Nash equilibria.

Suppose there is no completely mixed Nash equilibrium in \(\varGamma _n^M\). Then \(\varGamma _n^M\) has exactly \(2^m-2\) Nash equilibria. From hypotheses (a) and (b), it follows that \(\varGamma _n\) is non-degenerate and hence that \(\varGamma _n^M\) is non-degenerate. By Proposition 2(i), \(\varGamma _n^M\) must have an odd number of Nash equilibria, which contradicts the fact that \(2^m-2\) is even. Hence there is at least one completely mixed Nash equilibrium and again because \(\varGamma _n^M\) is non-degenerate by Proposition 2(ii) there is exactly one. This proves the claim.

By Claim B, for \(\emptyset \ne M \subseteq \overline{S}_1\), \(\varGamma ^M_n\) has exactly one completely mixed Nash equilibrium \((\sigma _1^M,\sigma _2^M)\). By Claim A, this induces a Nash equilibrium \((\sigma _1,\sigma _2)\) in \(\varGamma _n\) with \({\text {supp}}(\sigma _1)=\bigcup _{[s]_1 \in M} [s]_1\). Any Nash equilibrium \((\sigma _1',\sigma _2') \ne (\sigma _1,\sigma _2)\) with \({\text {supp}}(\sigma _1')=\bigcup _{[s]_1 \in M} [s]_1\) would induce a completely mixed Nash equilibrium in \(\varGamma _n^M\) different from \((\sigma _1^M,\sigma _2^M)\), a contradiction. Hence \(\varGamma _n\) has exactly one Nash equilibrium \((\sigma _1,\sigma _2)\) with \({\text {supp}}(\sigma _1)=\bigcup _{[s]_1 \in M} [s]_1\).

(ii) From (i) it follows that for any \(\emptyset \ne M \subseteq \overline{S}_1\) there is a unique Nash equilibrium \((\sigma _1,\sigma _2)\) in \(\varGamma _n\) such that \({\text {supp}}(\sigma _1)=\bigcup _{[s]_1 \in M}[s]_1\). Further, by hypothesis (a), for any Nash equilibrium \((\sigma _1,\sigma _2)\) of \(\varGamma _n\) there is \(\emptyset \ne M \subseteq \overline{S}_1\) such that \({\text {supp}}(\sigma _1)=\bigcup _{[s]_1 \in M}[s]_1\). As \(\overline{S}_1\) has \(2^{|\overline{S}_1|}-1\) non-empty subsets, \(\varGamma _n\) has exactly \(2^{|\overline{S}_1|}-1\) Nash equilibria. \(\square \)

The following lemma is required in the proofs of Lemmata 2 and 3.

Lemma 6

Let \(\varGamma _n\) be a circulant game with shift k in which player 1’s payoff matrix is anti-circulant.

  1. (i)

    For all \(\sigma _2 \in \varSigma _2\) and all \(s \in S^n\) if \(\sigma _2(s)=0\) then \(-s \notin BR_1(\sigma _2)\) .

  2. (ii)

    If \(\varGamma _n\) is iso-circulant, then for all \(\sigma _1 \in \varSigma _1\) and all \(s \in S^n\) if \(\sigma _1(s)=0\) then \((-s-k) \notin BR_2(\sigma _1)\) .

  3. (iii)

    If \(\varGamma _n\) is counter-circulant, then for all \(\sigma _1 \in \varSigma _1\) and all \(s \in S^n\) if \(\sigma _1(s)=0\) then \((s-k) \notin BR_2(\sigma _1)\) .

Proof

(Lemma 6)

  1. (i)

    Let \(\sigma _2 \in \varSigma _2\) be such that \(\sigma _2(s)=0\) for some \(s \in S^n\). Since player 1’s payoff matrix is anti-circulant \(\pi _1(s|s')=a^1_{s+s'}\). We will show that there exists a strategy for player 1 that yields a strictly higher payoff against \(\sigma _2\) than strategy \(-s\). Let \(l:=\min \{s<l'\le s+n-1|\sigma _2(l')>0\}\). Since \(n>1\) the set \(\{s<l'\le s+n-1|\sigma _2(l')>0\}\) is non-empty and l exists. By construction of l, \(\sigma _2(s)=\dots =\sigma _2(l-1)=0\). We claim that \(\pi _1(-s|\sigma _2)<\pi _1(-l|\sigma _2)\). To see this, note that

    $$\begin{aligned} \pi _1(-s|\sigma _2)=\sum _{t=l}^{s+n-1} \sigma _2(t)a^1_{t-s} \end{aligned}$$

    and

    $$\begin{aligned} \pi _1(-l|\sigma _2)=\sum _{t=l}^{s+n-1} \sigma _2(t)a^1_{t-l}. \end{aligned}$$

    Comparing these payoffs for \(t=l\), we obtain that \(a^1_{t-l}=a^1_0>a^1_{t-s}=a^1_{l-s}\), where the strict inequality holds by part (ii) of Definition 2. Further, for \(l < t \le s+n-1\) we have \(0\le t-l<t-s \le n-1\) and hence that \(a^1_{t-l} \ge a^1_{t-s}\) again by part (ii) of Definition 2. Since by construction of l, \(\sigma _2(l)>0\) we obtain \(\pi _1(-s|\sigma _2)<\pi _1(-l|\sigma _2)\) which proves the claim. Hence \(-s \notin BR_1(\sigma _2)\).

  2. (ii)

    Let \(\sigma _1 \in \varSigma _1\) and \(s \in S^n\) be such that \(\sigma _1(s)=0\). Since player 2’s payoff matrix is anti-circulant, \(\pi _2(s|s')=a^2_{s'+s}\) for \(s,s' \in S\). Since \(\varGamma _n\) is a circulant game, by part (iii) of Definition 2 either \(a_{n-k}^2>a_{n-k+1}^2\ge \dots \ge a_{n-1}^2\ge a_0^2\ge a^2_1\ge \dots \ge a_{n-k-1}^2\) or \(a_{n-k}^2>a_{n-k-1}^2\ge \dots \ge a_{1}^2\ge a_0^2\ge a^2_{n-1}\ge \dots \ge a_{n-k+1}^2\). We will only prove the result for the former case as the proof for the latter works analogously. Let \(l:=\min \{s<l'\le s+n-1|\sigma _1(l')>0\}\) which exists since \(\{s<l'\le s+n-1|\sigma _1(l')>0\} \ne \emptyset \). Then \(\sigma _1(s)=\dots =\sigma _1(l-1)=0\). We claim that \(\pi _2(-s-k|\sigma _1)<\pi _2(-l-k|\sigma _1)\). To see this, note that

    $$\begin{aligned} \pi _2(-s-k|\sigma _2)=\sum _{t=l}^{s+n-1} \sigma _1(t)a^2_{t-s-k} \end{aligned}$$

    and

    $$\begin{aligned} \pi _2(-l-k|\sigma _2)=\sum _{t=l}^{s+n-1} \sigma _1(t)a^2_{t-l-k}. \end{aligned}$$

    For \(t=l\), we have \(a^2_{t-l-k}=a^2_{n-k}>a^2_{t-s-k}=a^2_{l-s-k}\), where the strict inequality holds by part (iii) of Definition 2. Further, for \(l < t \le s+n-1\) we have \(a^2_{t-l-k} \ge a^2_{t-s-k}\) by part (iii) of Definition 2 since \(t-l-k<t-s-k\), \(-k\le t-l-k<n-k-1\), and \(-k<t-s-k \le n-k-1\). Since by construction of l, \(\sigma _1(l)>0\) we obtain that \(\pi _2(-s-k|\sigma _1)<\pi _2(-l-k|\sigma _1)\) which proves the claim. Hence \((-s-k) \notin BR_2(\sigma _1)\).

  3. (iii)

    Let \(\sigma _1 \in \varSigma _1\) and \(s \in S^n\) be such that \(\sigma _1(s)=0\). Since player 2’s payoff matrix is circulant, \(\pi _2(s|s')=a^2_{s-s'}\) for \(s,s' \in S\). Since \(\varGamma _n\) is a circulant game, by definition either \(a_{n-k}^2>a_{n-k+1}^2\ge \dots \ge a_{n-1}^2\ge a_0^2\ge a^2_1\ge \dots \ge a_{n-k-1}^2\) or \(a_{n-k}^2>a_{n-k-1}^2\ge \dots \ge a_{1}^2\ge a_0^2\ge a^2_{n-1}\ge \dots \ge a_{n-k+1}^2\). We will only prove the result for the former case as the proof for the latter works analogously. Let \(l:=\min \{s<l'\le s+n-1|\sigma _1(l')>0\}\) which exists since \(\{s<l'\le s+n-1|\sigma _1(l')>0\} \ne \emptyset \). Then \(\sigma _1(s)=\dots =\sigma _1(l-1)=0\). We claim that \(\pi _2(s-k|\sigma _1)<\pi _2(-l-k|\sigma _1)\). To see this, note that,

    $$\begin{aligned} \pi _2(s-k|\sigma _2)=\sum _{t=l}^{s+n-1} \sigma _1(t)a^2_{s-k-t} \end{aligned}$$

    and

    $$\begin{aligned} \pi _2(l-k|\sigma _2)=\sum _{t=l}^{s+n-1} \sigma _1(t)a^2_{l-k-t}. \end{aligned}$$

    For \(t=l\), we have \(a^2_{l-k-t}=a^2_{n-k}>a^2_{s-k-t}=a^2_{s-k-l}\). Further, for \(l < t \le s+n-1\) we have \(a^2_{l-k-t} \ge a^2_{s-k-t}\) by part (iii) of the definition of circulant game since \(l-k-t>s-k-t\), \(-k\ge l-k-t>-n-k+1\), and \(-k>s-k-t \ge -n-k+1\). Since by construction of l, \(\sigma _1(l)>0\) we obtain that \(\pi _2(s-k|\sigma _1)<\pi _2(l-k|\sigma _1)\) which proves the claim. Hence \((s-k) \notin BR_2(\sigma _1)\).

\(\square \)

Lemma 6 allows us to rule out certain strategies as best responses for player i if player \(-i\) plays some strategy with zero probability in the case that player 1’s payoff matrix is anti-circulant. By (i) if player 2 plays a strategy s with probability 0 then for player 1 strategy \(-s\) cannot be a best response. Similarly, (ii) and (iii) state that if in an iso-circulant (counter-circulant) game player 1 places probability 0 on strategy s then \(-s-k\) (\(s-k\)) cannot be a best response for player 2.

We are now ready to prove Lemmata 2 and 3. It follows from Lemma 2(i) and Lemma 3(i) that iso-circulant games satisfy hypotheses (a) and (b) in Proposition 6. Analogously, Lemma 2(ii) and Lemma 3(ii) establish that counter-circulant games fulfill (a) and (b) in Proposition 6.

Proof

(Lemma 2)

  1. (i)

    The “if” part is trivial. To see the “only if” part let \((\sigma _1,\sigma _2)\) be a Nash equilibrium of \(\varGamma _n\) and let \(s \in S^n\) be such that \(\sigma _1(s)=0\). By Lemma 6(ii), \(\sigma _2(-s-k)=0\) and consequently by Lemma 6(i) \(\sigma _1(s+k)=0\). Iterating this argument yields \(\sigma _1(s+mk)=0\) for all \(m=0,\dots ,\frac{n}{d}-1\). If \(\sigma _2(s)=0\) the argument works analogously.

  2. (ii)

    By Lemma 6(i) and (iii) for any Nash equilibrium \((\sigma _1,\sigma _2)\) and any \(s \in S^n\) we obtain

    $$\begin{aligned} \sigma _1(s)=0 \Rightarrow \sigma _2(s-k)=0 \Rightarrow \sigma _1(-s+k)=0 \end{aligned}$$

    and

    $$\begin{aligned} \sigma _1(-s+k)=0 \Rightarrow \sigma _2(-s)=0 \Rightarrow \sigma _1(s)=0. \end{aligned}$$

Analogously, for player 2, we obtain

$$\begin{aligned} \sigma _2(s)=0 \Rightarrow \sigma _1(-s)=0 \Rightarrow \sigma _2(-s-k)=0 \end{aligned}$$

and

$$\begin{aligned} \sigma _2(-s-k)=0 \Rightarrow \sigma _1(s+k)=0 \Rightarrow \sigma _2(s)=0. \end{aligned}$$

\(\square \)

Proof

(Lemma 3)

  1. (i)

    First, let \(s \in S^n\) be such that \({\text {supp}}(\sigma _1)\cap I(s)=\emptyset \). By Lemma 6(ii), \(-s-(m+1)k \notin BR_2(\sigma _1)\) for all \(0 \le m \le n/d-1\). As \(\{-s-(m+1)k|0 \le m \le n/d-1\}=I(-s)\) we obtain \(BR_2(\sigma _1) \cap I(-s)=\emptyset \).

    Next, let \(s \in S^n\) be such that \({\text {supp}}(\sigma _2)\cap I(s)=\emptyset \). By Lemma 6(i), \(-s-mk \notin BR_1(\sigma _2)\) for all \(0 \le m \le n/d-1\). As \(\{-s-mk|0 \le m \le n/d-1\}=I(-s)\) we obtain \(BR_1(\sigma _2) \cap I(-s)=\emptyset \).

  2. (ii)

    If \({\text {supp}}(\sigma _{-i})\cap C_{-i}(s)=\emptyset \) for \(C_{-i}(s) \in C_{-i}(S^n)\), then, since \(C_{-i}(s)=\{s,-s+(-1)^{i-1}k\}\), by Lemma 6(i) and (iii), \(-s,s+(-1)^{i-1}k \notin BR_{i}(\sigma _{-i})\). Hence \(BR_i(\sigma _{-i}) \cap C_i(-s)=\emptyset \). \(\square \)

The following Lemma 7 establishes that iso-circulant games fulfill hypothesis (c) in Proposition 6 and is used in the proofs of Theorem 1 and Proposition 3.

Lemma 7

Let \(\varGamma _n\) be an iso-circulant game in which both players’ payoff matrices are anti-circulant. For every \(s \in S^n\), there is a Nash equilibrium \((\sigma _1,\sigma _2)\) such that \({\text {supp}}(\sigma _1)=I(s)\) and \({\text {supp}}(\sigma _2)=I(-s)\).

Proof

(Lemma 7) Given \(\overline{s} \in S^n\), define \(\sigma _1(s)=d/n\) for all \(s \in I(\overline{s})\) and \(\sigma _2(s)=d/n\) for all \(s \in I(-\overline{s})\). By construction, \({\text {supp}}(\sigma _1)=I(\overline{s})\) and \({\text {supp}}(\sigma _2)=I(-\overline{s})\). By Lemma 3(i), no strategy outside \(I(\overline{s})\) can be a best response for player 1 against \(\sigma _2\) and no strategy outside \(I(-\overline{s})\) can be a best response for player 2 against \(\sigma _1\). Further, \(\pi _1(s|\sigma _{2})=\sum _{m=0}^{n/d-1}\frac{d}{n}a_{s+\overline{s}+mk}=\pi _1(s'|\sigma _{2})\) for all \(s,s' \in I(\overline{s})\) and analogously \(\pi _2(s|\sigma _{1})=\pi _2(s'|\sigma _{1})\) for all \(s, s' \in I(-\overline{s})\). Proposition 1 yields that \((\sigma _1,\sigma _2)\) is a Nash equilibrium of \(\varGamma _n\). \(\square \)

We are now ready to prove Theorem 1 and Proposition 3.

Proof

(Theorem 1) If \(\varGamma _n\) is an iso-circulant game in which both players’ payoff matrices are anti-circulant then by Lemma 2(i), Lemma 3(i), and Lemma 7, \(\varGamma _n\) and \(\overline{S}_1=\overline{S}_2=I(S^n)\) as defined in section 3.1 then satisfy the hypotheses of Proposition 6. As \(|I(S^n)|=d\), it follows that \(\varGamma _n\) has \(2^d-1\) Nash equilibria. If \(\varGamma _n\) is an iso-circulant game in which both players’ payoff matrices are circulant, there is a permutation of row vectors that transforms both players’ payoff matrices into anti-circulant matrices while fixing the first row in both matrices (Lemma 5(i)). This permutation, which is essentially a relabeling of the players’ strategies, does not affect the number of equilibria. Hence, the proof is complete. \(\square \)

Proof

(Proposition 3) Note first that if both players’ payoff matrices are circulant then by Lemma 5(i) the game can be transformed into a different version of the same game in which both players’ payoff matrices are anti-circulant by a permutation of row vectors. Since such a permutation does not affect the number of pure strategy Nash equilibria, we assume wlog that both players’ payoff matrices are anti-circulant.

To see the “if” part suppose \(k=n\). Then by construction, each class I(s) is a singleton set and there are n disjoint classes. Hence by Lemma 7, \(\varGamma _n\) has at least n pure strategy Nash equilibria. By Lemma 2(i), in any pure strategy Nash equilibrium \((\sigma _1,\sigma _2)\), \({\text {supp}}(\sigma _1)=I(s)\) for some \(s \in S\) and hence \(\varGamma _n\) has exactly n pure strategy Nash equilibria.

To prove the “only if” part let \(\varGamma _n\) have n pure strategy Nash equilibria and let \((s_1,s_2)\) be one of them. By Lemma 2(i), \(I(s_1)\) must be a singleton set. By construction, \(I(s_1)\) is a singleton set if and only if \(k=n\).

This proves the first part of the theorem.

To see the second part, note that by construction of the classes, I(s) is a singleton set if and only if \(k=n\) for any \(s \in S\). Further by Lemma 2(i) and Lemma 7, \(\varGamma _n\) has a pure strategy Nash equilibrium if and only if there is a singleton equivalence class I(s). Hence, \(\varGamma _n\) has no pure strategy Nash equilibrium if and only if \(k \ne n\). \(\square \)

Before we can turn to the proofs of Theorem 2 and Proposition 4, we require a couple more preliminary lemmata. One hypothesis in Proposition 6 requires the sets \(\overline{S}_1\) and \(\overline{S}_2\) to be partitions of the strategy set. While this is true by construction for \(I(S^n)\) in the case of iso-circulant games, the following Lemma 8 shows that the \(C_1(S^n)\) and \(C_2(S^n)\) form a partition of \(S^n\).

Lemma 8

Let \(\varGamma _n\) be a counter-circulant game. For \(i=1,2\) the set \(C_i(S^n)\) is a partition of \(S^n\).

Proof

(Lemma 8) We will prove the result for \(i=1\) as the proof for \(i=2\) works analogously. Since \(s \in C_1(s)\) for all \(s \in S^n\), it follows that \(\bigcup _{s \in S^n} C_1(s)=S^n\). If there is \(\overline{s} \in C_1(s) \cap C_1(s')\) for some \(s,s' \in S^n\), then since \(\overline{s} \in C_1(s)\) either \(\overline{s}=s\) or \(\overline{s}=-s+k\). If \(\overline{s}=s\) then \(C_i(s)=C_i(\overline{s})\). If \(\overline{s}=-s+k\) then \(-\overline{s}+k=s-k+k=s\). In any case it follows that \(C_1(\overline{s})=C_1(s)\). Using the same argument one obtains \(C_1(\overline{s})=C_1(s')\) and hence that \(C_1(s)=C_1(s')\). \(\square \)

The following Lemma 9 establishes that counter-circulant games fulfill property (c) in Proposition 6.

Lemma 9

Let \(\varGamma _n\) be a counter-circulant game in which player 1’s payoff matrix is anti-circulant and let \(\sigma =(\sigma _1,\sigma _2) \in \varSigma _1 \times \varSigma _2\).

  1. (i)

    \(C_{i}(s)\) is a singleton set if and only if \(C_{-i}(-s)\) is a singleton set.

  2. (ii)

    For every \(s \in S^n\), there is a Nash equilibrium \((\sigma _1,\sigma _2)\) such that \({\text {supp}}(\sigma _1)=C_{1}(s)\) and \({\text {supp}}(\sigma _2)=C_{2}(-s)\).

Proof

(Lemma 9)

  1. (i)

    Suppose that \(C_i(s)\) is a singleton. By construction, \(s\equiv -s+(-1)^{i-1}k \mod n\) which is equivalent to \(-s\equiv s+(-1)^ik \mod n\). This holds if and only if \(C_{-i}(-s)\) is a singleton.

  2. (ii)

    Note that this follows from (i) and Lemma 3(ii) if \(C_1(s)\) is a singleton set. Hence, suppose that \(C_1(s)=\{s,-s+k\}\) contains two elements. Then, by (i), \(C_2(-s)=\{-s,s-k\}\) contains two elements and neither \(2s=k\) nor \(2s=n+k\). Choose \(\sigma _1(s)\) as the solution to \(xa^2_{-2s}+(1-x)a^2_{-k}=xa^2_{-k}+(1-x)a^2_{2s-2k}\), i.e.,

    $$\begin{aligned} \sigma ^s_1(s)=\frac{a_{2s-2k}^2-a_{n-k}^2}{a_{2s-2k}^2-a_{n-k}^2+a^2_{n-2s}-a_{n-k}^2}. \end{aligned}$$

    By definition \(a^2_{n-k}\) is player 2’s largest payoff implying that \(a_{2s-2k}^2-a_{n-k}^2<0\) since \(2s \ne n+k\) and that \(a^2_{n-2s}-a_{n-k}^2<0\) since \(2s \ne k\). Hence \(\sigma _1(s) \in [0,1]\).

Choose \(\sigma _2^s(-s)\) as the solution to \(xa^1_0+(1-x)a^1_{2s-k}=xa^1_{-2s+k}+(1-x)a^1_0\), i.e.,

$$\begin{aligned} \sigma ^s_2(-s)=\frac{a_{0}^1-a_{2s-k}^1}{a_{0}^1-a_{2s-k}^1+a_{0}^1-a_{-2s+k}^1}. \end{aligned}$$

By definition \(a_0^1\) is player 1’s largest payoff. Hence as \(2s \ne k\) \(a_{0}^1-a_{2s-k}^1>0\) and \(a_{0}^1-a_{-2s+k}^1>0\) implying that \(\sigma _2(-s) \in [0,1]\). By Lemma 3(ii) and Proposition 1, \((\sigma _1,\sigma _2)\) is a Nash equilibrium. \(\square \)

The set \(C_1(S^n)\) is a partition of the strategy set for player 1, while \(C_2(S^n)\) is a partition of the strategy set for player 2. By Lemma 3(ii) a class \(C_1(s)\) of player 1 “corresponds” to a class \(C_2(-s)\) of player 2 in the sense that if player 1 puts probability 0 on all strategies in \(C_1(s)\) then none of the strategies in \(C_2(-s)\) are a best response for player 2 and vice versa. Part (i) of Lemma 9 states that two corresponding classes contain the same number of elements. By (ii) for every class \(C_1(s)\) there is always a Nash Equilibrium such that player 1’s strategy has this class as its support while player 2’s strategy has support \(C_2(-s)\). The equilibrium constructed to prove (ii) is such that player 1 chooses his strategy (with support \(C_1(s)\)) such that player 2 is indifferent between all strategies in \(C_2(-s)\) (and vice versa). As \(\varGamma _n\) is a non-degenerate game, by Proposition 2(ii) this is the unique equilibrium \((\sigma _1,\sigma _2)\) such that \({\text {supp}}(\sigma _1)=C_1(s)\) and \({\text {supp}}(\sigma _2)=C_2(-s)\).

We are now ready to prove Theorem 2 and Proposition 4.

Proof

(Theorem 2) If \(\varGamma _n\) is a counter-circulant game in which player 1’s payoff matrix is anti-circulant and player 2’s payoff matrix is circulant then by Lemma 8, \(C_1(S^n)\) and \(C_2(S^n)\) as defined in section 3.1 are partitions of \(S^n\). Further, by Lemma 9(i), \(|C_1(S^n)|=|C_2(S^n)|\) and by Lemmata 2(ii), 3(ii), and 9(ii), \(\varGamma _n\), \(\overline{S}_1=C_1(S^n)\), and \(\overline{S}_2=C_2(S^n)\) satisfy properties (a)-(c) in Proposition 6 and hence \(\varGamma _n\) has \(2^{|C_1(S^n)|}-1\) Nash equilibria.

To prove (i)–(iii) it hence suffices to determine \(|C_1(S^n)|\). Note that any class \(C_1(s)\) contains either one or two elements. It contains one element if and only if \(-s+k \equiv s\) which occurs if and only if either \(2s=k\) or \(2s=n+k\). Further, there are at most two singleton classes.

  1. (i)

    If n is odd, then either \(n-k\) is odd (if k is even) or \(2n-k\) is odd (if k is odd). Hence there is one singleton class in \(C_1(S^n)\) and since all other elements of \(C_1(S^n)\) contain two elements, \(|C_1(S^n)|=(n-1)/2+1=(n+1)/2\).

  2. (ii)

    If both n and k are even, then both k and \(n+k\) are even and \(k/2,(n+k)/2 \in S^n\). Hence there are two singleton classes in \(C_1(S^n)\) and since all other elements of \(C_1(S^n)\) contain two elements, \(|C_1(S^n)|=(n-2)/2+2=(n+2)/2\).

  3. (iii)

    If n is even and k is odd, then \(n+k\) is odd and hence neither \(k/2 \in S^n\) nor \((n+k)/2 \in S^n\). Hence there is no singleton class and hence all elements of \(C_1(S^n)\) contain 2 elements, implying that \(|C_1(S^n)|=n/2=n/2\).

If \(\varGamma _n\) is a counter-circulant game in which player 1’s payoff matrix is circulant and player 2’s payoff matrix is anti-circulant, then there is a permutation of row vectors that transforms player 1’s payoff matrix into an anti-circulant matrix. Applying the same permutation of row vectors to player 2’s payoff matrix yields a different version of the same game in which strategies are differently labeled, and player 1’s payoff matrix is anti-circulant and player 2’s payoff matrix is circulant (Lemma 5(ii)). This permutation does not affect the number of Nash equilibria and hence the proof of Theorem 2 is complete. \(\square \)

Proof

(Proposition 4) Note first that if player 1’s payoff matrix is circulant then by Lemma 5(i) the game can be transformed into a different version of the same game in which player 1’s payoff matrix is anti-circulant by a permutation of row vectors. Since such a permutation does not affect the number of pure strategy Nash equilibria, we assume wlog that player 1’s payoff matrix is anti-circulant.

  1. (i)

    By Lemmata 2(ii) and 9(ii), \(\varGamma _n\) has one pure strategy Nash equilibrium if and only if one of the classes \(C_1(s)\) is a singleton set, which by construction happens if and only if n is odd.

  2. (ii)

    By Lemmata 2(ii) and 9(ii), \(\varGamma _n\) has two pure strategy Nash equilibria if and only if two of the classes \(C_1(s)\) are singleton sets, which by construction happens if and only if both n and k are even.

  3. (iii)

    By Lemmata 2(ii) and 9(ii), \(\varGamma _n\) has no pure strategy Nash equilibrium if and only if none of the classes \(C_1(s)\) is a singleton set, which by construction happens if and only n is even and k is odd.

\(\square \)

Finally, we prove Lemma 4 and Proposition 5.

Proof

(Lemma 4)

  1. (i)

    To see the first part, let \(M=\bigcup _{j=1}^m I(s^j)\) be a union of elements of \(I(S^n)\). By Lemma 2(i) and Lemma 7, \(\varGamma _n\) and \(\overline{S}_1=\overline{S}_2=I(S^n)\) as defined in section 3.1 then satisfy the hypotheses of Proposition 6. Hence, there is a unique Nash equilibrium \((\sigma _1,\sigma _2)\) with \({\text {supp}}(\sigma _1)=M\).

    To prove the second part, let \((\sigma _1, \sigma _2)\) be a Nash equilibrium. By Lemma 2(i), \({\text {supp}}(\sigma _1)\) is a union of elements in \(I(S^n)\).

  2. (ii)

    To see the first part, let \(M=\bigcup _{j=1}^m C_1(s^j)\) be a union of elements of \(C_1(S^n)\). By Lemma 8, \(C_1(S^n)\) and \(C_2(S^n)\) as defined in section 3.1 are partitions of \(S^n\). Further, by Lemma 9(i), \(|C_1(S^n)|=|C_2(S^n)|\) and by Lemma 2(ii), Lemma 3(ii), and 9(ii), \(\varGamma _n\), \(\overline{S}_1=C_1(S^n)\), and \(\overline{S}_2=C_2(S^n)\) satisfy properties (a)-(c) in Proposition 6. It follows that there is a unique Nash equilibrium \((\sigma _1,\sigma _2)\) with \({\text {supp}}(\sigma _1)=M\).

To prove the second part, let \((\sigma _1, \sigma _2)\) be a Nash equilibrium. By Lemma 2(ii), \({\text {supp}}(\sigma _1)\) is a union of elements in \(C_1(S^n)\). \(\square \)

Proof

(Proposition 5) Given \(\sigma ^*\), the payoff of any strategy \(\sigma \in \varSigma \) against \(\sigma ^*\) is constant and given by \(\pi (\sigma |\sigma ^*)= (a+b+c)/3\). In particular, \(\pi (\sigma |\sigma ^*)=\pi (\sigma ^*|\sigma ^*)\) for all \(\sigma \in \varSigma \). Since \((\sigma ^*,\sigma ^*)\) is a completely mixed Nash equilibrium, every strategy \(\sigma \) is an alternative best response and condition (i), and the hypothesis of condition (ii) in Definition 5 holds true for any \(\sigma \in \varSigma \). Further, the payoff of \(\sigma ^*\) against any strategy \(\sigma \) is also \(\pi (\sigma ^*|\sigma )=(a+b+c)/3\). Thus, \(\sigma ^*\) is an ESS if and only if \(\pi (\sigma |\sigma )<(a+b+c)/3\) for all \(\sigma \ne \sigma ^*\). The latter inequality follows for any \(\sigma \ne \sigma ^*\) if \(\sigma ^*\) is the unique maximizer of \(\pi (\sigma |\sigma )\), i.e., \(\sigma ^* = \text {argmax}_{\sigma \in \varSigma } \pi (\sigma |\sigma )\). The first-order conditions for a maximum of \(\pi (\sigma |\sigma )\) are \( (b+c-2a)(1-2\sigma (i)-\sigma (j))=0\) for \(i\ne j\) and \(i,j =1,2\) and \(\sigma ^*\) is the unique solution if \((b+c-2a) \ne 0\). The Hessian is negative semidefinite if \(a < (b+c)/2\). Thus, if \(a < (b+c)/2\), then \(\sigma ^*\) is an ESS. Analogously, if \(a > (b+c)/2\), \(\sigma ^*\) is the unique minimizer of \(\pi (\sigma |\sigma )\) and \(\sigma ^*\) is not an ESS. Finally, if \(a=(b+c)/2\), every pure strategy s can successfully ‘invade’ \(\sigma ^*\) as \(\pi (\sigma ^*|s)=\pi (s|s)=a\) and \(\sigma ^*\) is not an ESS. \(\square \)

Appendix 3: Tables

Table 1 Examples of iso-circulant games
Table 2 Examples of counter-circulant games

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Granić, ƉG., Kern, J. Circulant games. Theory Decis 80, 43–69 (2016). https://doi.org/10.1007/s11238-014-9478-4

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Keywords

  • Bimatrix games
  • Circulant games
  • Circulant matrix
  • Number of Nash equilibria
  • Rock-Paper-Scissors