Periodic Attractor in the Discrete Time Best-Response Dynamics of the Rock-Paper-Scissors Game

Abstract

The Rock-Paper-Scissors (RPS) game is a classic non-cooperative game widely studied in terms of its theoretical analysis as well as in its applications, ranging from sociology and biology to economics. In this work, we show that the attractor of the discrete time best-response dynamics of the RPS game is a finite union of periodic orbits. Moreover, we also describe the bifurcations of the attractor and determine the exact number, period and location of the periodic orbits.

Appendix A. Monotonicity Lemmas

Recall that

$$\begin{aligned} m=m_\alpha (\lambda ):=\min \{k\in {\mathbb {N}}:\lambda ^k<\alpha \}. \end{aligned}$$

and that ${\hat{B}}$ is the set of regular strategies in B, i.e. ${\hat{B}}=B\cap {\hat{\Delta }}$.

Lemma A.1

Let $x\in {\hat{B}}$. If $n(x)\ge m$, then $n(P(x))\ge m$.

Proof

We suppose that $m\ge 2$. Otherwise, there is nothing to prove. Let $x\in B_k$ for some $k\ge m$. Then,

$$\begin{aligned} S^2(u_\alpha )\cdot P(x)=\lambda ^{k} S(u_\alpha )\cdot x+(1-\lambda ^{k-1})(2+\alpha ). \end{aligned}$$

Since $x\in B_k$, we have

$$\begin{aligned} S(u_\alpha )\cdot x&= 3-u_\alpha \cdot x-S^2(u_\alpha )\cdot x\\&> 3-(\lambda ^{-1}\alpha -\alpha +1)-b_k. \end{aligned}$$

Hence,

$$\begin{aligned} S^2(u_\alpha )\cdot P(x)> 2+\alpha -\lambda ^{k-1}-\alpha \lambda ^{-1}. \end{aligned}$$

Therefore, to prove that $n(P(x))\ge m$, it is sufficient to show that

$$\begin{aligned} g(\lambda ):=2+\alpha -\lambda ^{k-1}-\alpha \lambda ^{-1}-b_{m-1}>0. \end{aligned}$$

Notice that

$$\begin{aligned} g(\lambda )=\lambda ^{-1}(\alpha +1-\lambda ^{k}-\lambda ^{-m+1}\alpha ). \end{aligned}$$

Since $\lambda ^m< \alpha$ and $\lambda ^k\le \lambda ^m$, we get

$$\begin{aligned} g(\lambda )>\lambda ^{-1}(1-\lambda ^{-m+1}\alpha ). \end{aligned}$$

Again, by the definition of m, we have $\lambda ^{m-1}\ge \alpha$. This shows that $g(\lambda )>0$ as we wanted to prove. $\square$

Lemma A.2

Let $x\in {\hat{B}}$ such that $n(P(x))\ge m$. If $n(P(x))\le n(x)$, then $n(P^2(x))\le n(P(x))$.

Proof

We want to prove that given $k\in {\mathbb {N}}$, if $x\in B_k$ and $P(x)\in B_j$ for some $j\le k$, then $P^2(x)\in B_i$ for some $i\le j$, for any $\alpha >0$ and $\lambda \in (0,1)$ satisfying $\lambda ^j< \alpha$.

So it is enough to see that

$$\begin{aligned} S^2(u_\alpha )\cdot P^2(x)<b_j=\lambda ^{-j-1}\alpha -\lambda ^{-1}(\alpha -1+(1-\lambda )(\alpha +2)). \end{aligned}$$

Since $x\in B_k$,

$$\begin{aligned} P(x)=\lambda ^k S(x)+\lambda ^{k-1}(1-\lambda )e_1+(1-\lambda ^{k-1})e_2 \end{aligned}$$

and $P(x)\in B_j$ implies that

$$\begin{aligned} P^2(x)&=\lambda ^j S(P(x))+\lambda ^{j-1}(1-\lambda )e_1+(1-\lambda ^{j-1})e_2 \\&=\lambda ^{j+k} S^2(x)+\lambda ^{j+k-1}(1-\lambda )e_3+\lambda ^j(1-\lambda ^{k-1})e_1+\lambda ^{j-1}(1-\lambda )e_1\\&\quad +(1-\lambda ^{j-1})e_2. \end{aligned}$$

Because $S^2(u_\alpha )=(0,2+\alpha ,1-\alpha )$, we have that

$$\begin{aligned} S^2(u_\alpha )\cdot P^2(x) = \lambda ^{j+k} u_\alpha \cdot x-\lambda ^{j+k-1}(1-\lambda )(\alpha -1)+(1-\lambda ^{j-1})(2+\alpha ). \end{aligned}$$

But $u_\alpha \cdot x<\alpha \left( \lambda ^{-1}-1\right) +1$, thus

$$\begin{aligned} S^2(u_\alpha )\cdot P^2(x)<Q_{j,k,\alpha }(\lambda ), \end{aligned}$$

where

$$\begin{aligned} Q_{j,k,\alpha }(\lambda ):=\lambda ^{j+k} \left( \lambda ^{-1}\alpha -\alpha +1 \right) -\lambda ^{j+k-1}(1-\lambda )(\alpha -1)+(1-\lambda ^{j-1})(2+\alpha ). \end{aligned}$$

Now, it is easy to see that $Q_{j,k,\alpha }(\lambda )<b_j$ for every $\alpha >0$ and $\lambda \in (0,1)$ satisfying $\lambda ^j<\alpha$. Indeed,

$$\begin{aligned} \lambda ^{j+1}( Q_{j,k,\alpha }(\lambda )-b_j)&=\lambda ^j \left( \lambda ^{j+k}-2 \lambda ^j+1\right) -\alpha \left( \lambda ^j-1\right) ^2\\&\le \lambda ^j \left( \lambda ^{2j}-2 \lambda ^j+1\right) -\alpha \left( \lambda ^j-1\right) ^2\\&= (\lambda ^j-1)^2(\lambda ^j-\alpha )<0. \end{aligned}$$

$\square$

Lemma A.3

Let $x\in {\hat{B}}$ such that $n(P(x))<m$. If $n(P(x))\ge n(x)$, then $n(P^2(x))\ge n(P(x))$.

Proof

The strategy of the proof is the same as that of the previous lemma. We want to prove that given $k\in {\mathbb {N}}$, if $x\in B_k$ and $P(x)\in B_j$ for some $k\le j<m$, then $P^2(x)\in B_i$ for some $i\ge j$, for any $\alpha >0$ and $\lambda \in (0,1)$ satisfying $\lambda ^j\ge \alpha$.

It is enough to see that

$$\begin{aligned} S^2(u_\alpha )\cdot P^2(x)> b_{j-1}=\lambda ^{-j}\alpha -\lambda ^{-1}(\alpha -1+(1-\lambda )(\alpha +2)). \end{aligned}$$

As in the proof of the previous lemma, we have that

$$\begin{aligned} S^2(u_\alpha )\cdot P^2(x)&= \lambda ^{j+k} u_\alpha \cdot x-\lambda ^{j+k-1}(1-\lambda )(\alpha -1)+(1-\lambda ^{j-1})(2+\alpha ). \end{aligned}$$

But $u_\alpha \cdot x>1$, so

$$\begin{aligned} S^2(u_\alpha )\cdot P^2(x)>Q_{j,k,\alpha }(\lambda ), \end{aligned}$$

where

$$\begin{aligned} Q_{j,k,\alpha }(\lambda ):=\lambda ^{j+k} -\lambda ^{j+k-1}(1-\lambda )(\alpha -1)+(1-\lambda ^{j-1})(2+\alpha ). \end{aligned}$$

Now, it is easy to see that $Q_{j,k,\alpha }(\lambda )>b_{j-1}$ for every $\alpha >0$ and $\lambda \in (0,1)$ satisfying $\lambda ^j\ge \alpha$. Indeed,

$$\begin{aligned} g_{j,k,\alpha }(\lambda )&:=\lambda ^{j+1}(Q_{j,k,\alpha }(\lambda )-b_{j-1}) \\&= \alpha \lambda ^{2j+k+1}-(\alpha -1)\lambda ^{2j+k}-(2+\alpha )\lambda ^{2j}+(1+2\alpha )\lambda ^j-\alpha \lambda \\&=\alpha (1-\lambda )(1-\lambda ^{2j+k})+ \lambda ^{2j+k}-(2+\alpha )\lambda ^{2j}+(1+2\alpha )\lambda ^j-\alpha \\&>\lambda ^{2j+k}-(2+\alpha )\lambda ^{2j}+(1+2\alpha )\lambda ^j-\alpha \\&\ge \lambda ^{3j}-(2+\alpha )\lambda ^{2j}+(1+2\alpha )\lambda ^j-\alpha \\&= \lambda ^{3j}-3\lambda ^{2j}+3\lambda ^j-1-(\alpha -1)(\lambda ^{2j}-2\lambda ^j+1)\\&= (\lambda ^j-1)^3-(\alpha -1)(\lambda ^j-1)^2\\&= (\lambda ^j-1)^2\left( \lambda ^j-\alpha \right) \ge 0. \end{aligned}$$

$\square$