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On Matrix Method of Symmetric Games

Lei Wang [email protected]    Xinyun Liu [email protected]    Ting Li [email protected]    Jiandong Zhu [email protected] Institute of Mathematics, School of Mathematical Sciences, Nanjing Normal University, Nanjing, 210023, China School of Mathematics and Information Sciences, Weifang University, Weifang, 261061, China
Abstract

This paper provides a new version of matrix semi-tensor product method based on adjacent transpositions to test symmetric games. The advantage of using adjacent transpositions lies in the great simplification of analysis of symmetric games. By using the new method, new necessary and sufficient conditions for symmetric games are proposed, and a group of bases of a symmetric game space can be easily calculated. Moreover, the testing equations with the minimum number can be concretely determined. Finally, two examples are displayed to show the effectiveness of the proposed method.

keywords:
Finite game, Symmetric game, Semi-tensor product of matrices, Adjacent transpositions.
thanks: This paper was not presented at any IFAC meeting. Corresponding author J. Zhu. Tel. +86-13851781823.

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1 Introduction

Game theory studies mathematical models of strategic interactions among rational agents. Modern game theory was formally established after John von Neumann proved the basic principles of game theory [19, 25], which has been paid wide attention and applied to many fields such as economics, biology, finance, computer science and so on. The concept of symmetric game was first proposed by von Neumann in [25], and the details can be seen in [27], which is a game where the payoffs for playing a particular strategy depend only on the other strategies employed, not on who is playing them. In other words, a symmetric game means that all players have the same set of strategies and fair payoffs. In real world, symmetric games are universal. For example, the well-known games rock-paper-scissors, prisoner’s dilemma are symmetric games. Moreover, the importance of symmetric game lies in the discussion on existence of Nash equilibria [16, 23]. Therefore, how to check whether a game is a symmetric game is an important problem. In [4], symmetric games are classified into three types, including ordinary symmetric games, renaming symmetric games and name-irrelevant games. The symmetric games discussed in this paper are actually ordinary symmetric games.

In the past decade, Cheng and his collaborators applied the semi-tensor product (STP) of matrices to game theory [11, 13], which enable a game to be expressed as a linear representation. Based on the linear representations of games, some problems about potential games, networked games and evolutionary games have been solved [6, 13, 21]. It should be noted that the matrix method based on STP is also powerful for symmetric games. In [8], algebraic conditions for symmetric games have been proposed by using the linear representation of symmetric group in structure vectors. Compared with the results in [8], some simpler algebraic conditions for symmetric games are given in [9] via some STP operations. Furthermore, for Boolean games, a group of bases of the symmetric game subspace is constructed via a recursive procedure in [9]. For general finite games, construction algorithms of bases of symmetric game subspaces are given in [15] and [20], respectively. It should be noted that the concept of strategy multiplicity vector and an equivalent condition for symmetric games in [4] play an important role in [9] [15] and [20].

In this paper, we still use the matrix method based on STP to investigate symmetric games. However, due to the introduction of adjacent transpositions, some new equivalent conditions for symmetric games are derived, and the proof is greatly simplified via a purely algebraic approach without using the concept of strategy multiplicity vector [4]. Based on our results, a group of bases of the symmetric game subspace can be easily constructed, and the testing equations with the minimum number can be concretely determined.

The rest of this paper is organized as follows: Section 2 gives some preliminaries and problem statement. In Section 3, the main results are presented. Section 4 is a brief conclusion.

2 Preliminaries and problem statement

In this section, we give some necessary preliminaries and problem statement. Firstly, we list the following notations.

  • \bullet

    𝒟={0,1}:\mathcal{D}=\{0,1\}: the set of values of logical variables;

  • \bullet

    δki:\delta_{k}^{i}: the ii-th column of IkI_{k};

  • \bullet

    Δk:={δki:i=1,2,,k}\Delta_{k}:=\{\delta_{k}^{i}:i=1,2,\cdots,k\};

  • \bullet

    δk[i1i2in]:=[δki1δki2δkin]\delta_{k}[i_{1}~{}i_{2}~{}\cdots i_{n}]:=[\delta_{k}^{i_{1}}~{}\delta_{k}^{i_{2}}~{}\cdots~{}\delta_{k}^{i_{n}}];

  • \bullet

    m×n:{\mathcal{M}}_{m\times n}: the set of m×nm\times n matrices;

  • \bullet

    m×n:={Lm×n|Col(L)Δm}{\mathcal{L}}_{m\times n}:=\{L\in{\mathcal{M}}_{m\times n}|\textrm{Col}(L)\subseteq{\Delta}_{m}\};

  • \bullet

    :\ltimes: the left semi-tensor product of matrices;

  • \bullet

    1n\textbf{1}_{n}: the nn-dimensional column vector of ones;

  • \bullet

    0m×n0_{m\times n}: the m×nm\times n matrix with zero entries;

  • \bullet

    Sn\textbf{S}_{n}: the nn-th order symmetric group, i.e., a permutation group that is composed of all the permutations of nn things;

  • \bullet

    \mathbb{R}: the set composed of all the real numbers.

Definitoin 1.

[22] A finite game is a triple G=(N,S,C)G=(N,S,C), where

(1)

N={1,2,,n}N=\{1,2,\cdots,n\} is the set of nn players;

(2)

S=S1×S2××SnS=S_{1}\times S_{2}\times\cdots\times S_{n} is the set of strategy profiles, where Si={s1i,s2i,,skii}S_{i}=\{s_{1}^{i},s_{2}^{i},\cdots,s_{k_{i}}^{i}\} is the set of strategies of player ii;

(3)

C={c1,c2,,cn}C=\{c_{1},c_{2},\cdots,c_{n}\} is the set of payoff functions, where ci:Sc_{i}:S\rightarrow\mathbb{R} is the payoff function of player ii.

Denote the set composed of all the game above by 𝒢[n;k1,,kn]\mathcal{G}_{[n;k_{1},\cdots,k_{n}]}. When |Si|=k|S_{i}|=k for each i=1,2,,ni=1,2,\cdots,n, we denote it by 𝒢[n;k]\mathcal{G}_{[n;k]}.

Definitoin 2.

[11] Let Am×nA\in\mathcal{M}_{m\times n}, Bp×qB\in\mathcal{M}_{p\times q}. The left semi-tensor product of AA and BB is defined as

AB=(AIαn)(BIαp),A\ltimes B=(A\otimes I_{\frac{\alpha}{n}})(B\otimes I_{\frac{\alpha}{p}}), (1)

where \otimes is the Kronecker product and α=lcm(n,p)\alpha=\operatorname{lcm}(n,p) is the least common multiple of nn and pp. When no confusion may arise it is usually called the semi-tensor product (STP).

If nn and pp in Definition 2 satisfy n=pn=p, the STP is reduced to the traditional matrix product. So, the STP is a generalized operation of the traditional matrix product. Therefore, one can directly write ABA\ltimes B as ABAB.

Given a game G𝒢[n;k]G\in\mathcal{G}_{[n;k]}, by using the semi-tensor product method [6], one can write each strategy xix_{i} into a vector form xiΔkx_{i}\in\Delta_{k}, and express every payoff function cic_{i} as

ci(x1,x2,,xn)=Vicj=1nxj,i=1,2,,n,c_{i}(x_{1},x_{2},\cdots,x_{n})=V_{i}^{c}\ltimes_{j=1}^{n}x_{j},~{}i=1,2,\cdots,n, (2)

where j=1nxjΔkn\ltimes_{j=1}^{n}x_{j}\in\Delta_{k^{n}} is called the STP form of the strategy profile, and VicV_{i}^{c} is called the structure vector of cic_{i}.

Definitoin 3.

(Definition 2.8 of [11]) A swap matrix W[m,n]=(wijIJ)W_{[m,n]}=(w^{IJ}_{ij}) is an mn×mnmn\times mn matrix, defined as follows:

Its rows and columns are labeled by double indices. The columns are arranged by the ordered multi-index Id(i1,i2;m,n)(i_{1},i_{2};m,n), and the rows are arranged by the ordered multi-index Id(i2,i1;n,m)(i_{2},i_{1};n,m). The element at the position with row index (I,J)(I,J) and column index (i,j)(i,j) is

wijIJ={1,I=i and J=j,0,otherwise.w^{IJ}_{ij}=\left\{\begin{array}[]{l}1,\ \ I=i\mbox{ and }J=j,\\ 0,\ \ otherwise.\end{array}\right.

For example, letting m=2m=2, n=3n=3, we get the swap matrix

(11)(12)(13)(21)(22)(23)W[2,3]=[100000000100010000000010001000000001](11)(21)(12)(22)(13)(23).\begin{array}[]{rc}(11)(12)(13)(21)(22)(23){}&\\ W_{[2,3]}=\left[\begin{array}[]{cccccc}1&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}0\\ 0&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}1&~{}~{}~{}0&~{}~{}~{}0\\ 0&~{}~{}~{}1&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}0\\ 0&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}1&~{}~{}~{}0\\ 0&~{}~{}~{}0&~{}~{}~{}1&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}0\\ 0&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}0&~{}~{}~{}1\end{array}\right]&\!\!\begin{array}[]{c}(11)\\ (21)\\ (12)\\ (22)\\ (13)\\ (23)\end{array}\end{array}.

When m=nm=n, W[m,n]W_{[m,n]} is denoted by W[m]W_{[m]}.

Lemma 1.

[11] Swap matrices have the following properties:

1) W[m,n]T=W[m,n]1=W[n,m]W_{[m,n]}^{\mathrm{T}}=W_{[m,n]}^{-1}=W_{[n,m]};

2) W[pq,r]=(W[p,r]Iq)(IpW[q,r])W_{[pq,\ r]}=(W_{[p,\ r]}\otimes I_{q})(I_{p}\otimes W_{[q,\ r]});

3) W[m,p](AB)W[q,n]=BAW_{[m,\ p]}(A\otimes B)W_{[q,\ n]}=B\otimes A for m×nm\times n matrix AA and p×qp\times q matrix BB;

4) Let xx and x~\tilde{x} be the mnmn-dimensional column vectors with each entry xijx_{ij} arranged by Id(i1,i2;m,ni_{1},i_{2};m,n) and Id(i2,i1;n,mi_{2},i_{1};n,m), respectively. Then x~=W[m,n]x\tilde{x}=W_{[m,n]}x. In particular, for any mm-dimensional vector xx and nn-dimensional vector yy, one has yx=W[m,n]xyyx=W_{[m,n]}xy.

By Lemma 2, one can easily get

(IkW[k])(W[k]Ik)(IkW[k])\displaystyle(I_{\!k}\otimes W_{\![k]}\!)(W_{\![k]}\otimes I_{\!k})(I_{\!k}\otimes W_{\![k]})\!
=\displaystyle= (W[k]Ik)(IkW[k])(W[k]Ik).\displaystyle(W_{\![k]}\otimes I_{\!k})(I_{\!k}\otimes W_{\![k]})(W_{\![k]}\otimes I_{\!k}). (3)

3 Symmetric Games and Testing Conditions

This section will present a simple matrix approach to test symmetric games.

Definitoin 4.

[26] A game G𝒢[n;k]G\in\mathcal{G}_{[n;k]} is called a symmetric game if for any permutation σ𝐒n\sigma\in\mathbf{S}_{n}

ci(x1,,xn)=cσ(i)(xσ1(1),,xσ1(n))c_{i}(x_{1},\cdots,x_{n})=c_{\sigma(i)}(x_{\sigma^{-1}(1)},\cdots,x_{\sigma^{-1}(n)}) (4)

for any i=1,2,,n.i=1,2,\cdots,n.

By Definition 4, to test whether G𝒢[n;k]G\in\mathcal{G}_{[n;k]} is symmetric, it seems that (4) should be checked for every permutation σ\sigma. But actually it is only needed to check (4) for the permutations of a generator of G𝒢[n;k]G\in\mathcal{G}_{[n;k]}. This is due to the fact that the compound permutation σ2σ1\sigma_{2}\circ\sigma_{1} satisfies (4) if both permutations σ1\sigma_{1} and σ2\sigma_{2} satisfy (4). For the readability, we write the demonstration as follows:

For any given xiΔkx_{i}\in\Delta_{k}, i=1,2,,ni=1,2,\dots,n, let yi=xσ11(i)y_{i}=x_{\sigma_{1}^{-1}(i)}. Then

ci(x1,,xn)\displaystyle c_{i}(x_{1},\cdots,x_{n}) =\displaystyle= cσ1(i)(xσ11(1),,xσ11(n))\displaystyle c_{\sigma_{1}(i)}(x_{\sigma_{1}^{-1}(1)},\cdots,x_{\sigma_{1}^{-1}(n)})
=\displaystyle= cσ1(i)(y1,,yn)\displaystyle c_{\sigma_{1}(i)}(y_{1},\cdots,y_{n})
=\displaystyle= cσ2σ1(i)(yσ21(1),,yσ21(n))\displaystyle c_{\sigma_{2}\circ\sigma_{1}(i)}(y_{\sigma_{2}^{-1}(1)},\cdots,y_{\sigma_{2}^{-1}(n)})
=\displaystyle= cσ2σ1(i)(xσ11(σ21(1)),,xσ11(σ21(n))),\displaystyle c_{\sigma_{2}\circ\sigma_{1}(i)}(x_{\sigma_{1}^{-1}(\sigma_{2}^{-1}(1))},\cdots,x_{\sigma_{1}^{-1}(\sigma_{2}^{-1}(n))}),

which implies that σ2σ1\sigma_{2}\circ\sigma_{1} satisfies (4).

Lemma 2.

[17, 18] The set of all the adjacent transpositions (r,r+1)(r,r+1), 1rn11\leq r\leq n-1 is a generator of the symmetric group 𝐒n\mathbf{S}_{n}.

In the following, we will use μr\mu_{r} to represent the adjacent transposition (r,r+1)(r,r+1). From Definition 4 and Lemma 2, the following lemma follows:

Lemma 3.

A game G𝒢[n;k]G\in\mathcal{G}_{[n;k]} is a symmetric game if and only if

ci(x1,,xn)=cμr(i)(xμr(1),,xμr(n))c_{i}(x_{1},\cdots,x_{n})=c_{\mu_{r}(i)}(x_{\mu_{r}(1)},\cdots,x_{\mu_{r}(n)}) (5)

for any adjacent transposition μr\mu_{r}, 1rn11\leq r\leq n-1, i=1,2,,n.i=1,2,\cdots,n.

Now we can use the semi-tensor product method to investigate symmetric games.

Proposition 1.

Consider a finite game G𝒢[n;k]G\in\mathcal{G}_{[n;k]}. GG is a symmetric game if and only if

Vic=Vμr(i)cTμr,iN,1rn1,V_{i}^{c}=V_{\mu_{r}(i)}^{c}T_{\mu_{r}},~{}\forall i\in N,~{}1\leq r\leq n-1, (6)

where Tμr=Ikr1W[k]Iknr1T_{\mu_{r}}\!=\!I_{k^{r-1}}\otimes W_{[k]}\otimes I_{k^{n-r-1}}.

{pf}

By (2), we rewrite the right hand side of (6) into the matrix form as follows:

Vμr(i)cxμr(1)xμr(2)xμr(n)\displaystyle V_{\mu_{r}(i)}^{c}x_{\mu_{r}(1)}x_{\mu_{r}(2)}\cdots x_{\mu_{r}(n)} (7)
=\displaystyle= Vμr(i)cx1xr1(xr+1xr)xr+2xn\displaystyle V_{\mu_{r}(i)}^{c}x_{1}\cdots x_{r-1}(x_{r+1}x_{r})x_{r+2}\cdots x_{n}
=\displaystyle= Vμr(i)cTμrx1xn.\displaystyle V_{\mu_{r}(i)}^{c}T_{\mu_{r}}x_{1}\cdots x_{n}.

From (2) and (7), it follows that (6) is equivalent to(5). Therefore, by Lemma 2, the proposition is proved. \blacksquare Now, we present our main result on testing symmetric games. For the readability of this paper and without loss of generality, we first consider the case of n=4n=4.

Theorem 1.

Consider G𝒢[4;k]G\in\mathcal{G}_{[4;k]}. GG is a symmetric game if and only if

[Ik4Tμ1000Ik4Tμ2000Ik4Tμ3000Ik4Tμ1000Ik4Tμ2](VG)T=0,\begin{bmatrix}I_{{k^{4}}}&-T_{\mu_{1}}&0&0\\ 0&I_{{k^{4}}}&-T_{\mu_{2}}&0\\ 0&0&I_{{k^{4}}}&-T_{\mu_{3}}\\ 0&0&0&I_{{k^{4}}}\!-\!T_{\mu_{1}}\\ 0&0&0&I_{{k^{4}}}\!-\!T_{\mu_{2}}\end{bmatrix}(V_{G})^{\mathrm{T}}=0, (8)

where

Tμ1=W[k]Ik2,Tμ2=IkW[k]Ik,Tμ3=Ik2W[k],T_{\mu_{1}}\!\!=\!W_{[k]}\!\otimes\!I_{k^{2}},\ T_{\mu_{2}}\!\!=\!I_{k}\!\otimes\!W_{[k]}\!\otimes\!I_{k},\ T_{\mu_{3}}\!\!=\!I_{k^{2}}\!\otimes\!W_{[k]}, (9)

and

VG=[V1cV2cV3cV4c].V_{G}=[V_{1}^{c}~{}V_{2}^{c}~{}V_{3}^{c}~{}V_{4}^{c}]. (10)
{pf}

First of all, by the properties of Kronecker product and the property (2) of swap matrices, we have the properties of TμiT_{\mu_{i}} as follows:

Tμi2=Ik4,Tμ1Tμ3=Tμ3Tμ1,Tμ1Tμ2Tμ1=Tμ2Tμ1Tμ2,Tμ2Tμ3Tμ2=Tμ3Tμ2Tμ3.\begin{split}&T_{\!\mu_{i}}^{2}\!\!=\!I_{{k^{4}}},\ T_{\!\mu_{1}}T_{\!\mu_{3}}\!\!=\!T_{\!\mu_{3}}T_{\!\mu_{1}},\\ &T_{\!\mu_{1}}T_{\!\mu_{2}}T_{\!\mu_{1}}\!\!=\!T_{\!\mu_{2}}T_{\!\mu_{1}}T_{\!\mu_{2}},\ T_{\!\mu_{2}}T_{\!\mu_{3}}T_{\!\mu_{2}}\!\!=\!T_{\!\mu_{3}}T_{\!\mu_{2}}T_{\!\mu_{3}}.\end{split} (11)

We write (6) of Proposition 1 for every r=1,2,3r=1,2,3 as follows:

V1c=V2cTμ1,V3c=V3cTμ1,V4c=V4cTμ1;\displaystyle V_{1}^{c}=V_{2}^{c}T_{\mu_{1}},\ V_{3}^{c}=V_{3}^{c}T_{\mu_{1}},\ V_{4}^{c}=V_{4}^{c}T_{\mu_{1}}; (12)
V1c=V1cTμ2,V2c=V3cTμ2,V4c=V4cTμ2;\displaystyle V_{1}^{c}=V_{1}^{c}T_{\mu_{2}},\ V_{2}^{c}=V_{3}^{c}T_{\mu_{2}},\ V_{4}^{c}=V_{4}^{c}T_{\mu_{2}}; (13)
V1c=V1cTμ3,V2c=V2cTμ3,V3c=V4cTμ3.\displaystyle V_{1}^{c}=V_{1}^{c}T_{\mu_{3}},\ V_{2}^{c}=V_{2}^{c}T_{\mu_{3}},\ V_{3}^{c}=V_{4}^{c}T_{\mu_{3}}. (14)

Let

A1=[Ik4Tμ2Ik4Tμ3],A2=Ik4Tμ3,A_{1}\!=\!\begin{bmatrix}I_{{k^{4}}}-T_{\mu_{2}}\\ I_{{k^{4}}}-T_{\mu_{3}}\end{bmatrix}\!,\ \ A_{2}=I_{{k^{4}}}-T_{\mu_{3}}, (15)
A3=Ik4Tμ1,A4=[Ik4Tμ1Ik4Tμ2].A_{3}=I_{{k^{4}}}-T_{\mu_{1}},\ A_{4}\!=\!\begin{bmatrix}I_{{k^{4}}}-T_{\mu_{1}}\\ I_{{k^{4}}}-T_{\mu_{2}}\end{bmatrix}. (16)

Then Eqs. (12)-(14) can be rewritten as

[Ik4Tμ1000Ik4Tμ2000Ik4Tμ3A10000A20000A30000A4](VG)T=0.\begin{bmatrix}I_{{k^{4}}}&-T_{\mu_{1}}&0&0\\ 0&I_{{k^{4}}}&-T_{\mu_{2}}&0\\ 0&0&I_{{k^{4}}}&-T_{\mu_{3}}\\ A_{1}&0&0&0\\ 0&A_{2}&0&0\\ 0&0&A_{3}&0\\ 0&0&0&A_{4}\end{bmatrix}(V_{G})^{\mathrm{T}}=0. (17)

By using an elementary row transformation, one can easily get the equivalent form of (17) as follows:

[Ik4Tμ1000Ik4Tμ2000Ik4Tμ3000A~3000A4](VG)T=0,\begin{bmatrix}I_{{k^{4}}}&-T_{\mu_{1}}&0&0\\ 0&I_{{k^{4}}}&-T_{\mu_{2}}&0\\ 0&0&I_{{k^{4}}}&-T_{\mu_{3}}\\ 0&0&0&\tilde{A}_{3}\\ 0&0&0&A_{4}\end{bmatrix}(V_{G})^{\mathrm{T}}=0, (18)

where

A~3\displaystyle\tilde{A}_{3}\! =\displaystyle= [A1000A2000A3][Ik4Tμ100Ik4Tμ200Ik4]1[00Tμ3]\displaystyle-\!\begin{bmatrix}A_{1}&0&0\\ 0&A_{2}&0\\ 0&0&A_{3}\end{bmatrix}\!\!\begin{bmatrix}I_{{k^{4}}}&-T_{\mu_{1}}&0\\ 0&I_{{k^{4}}}&-T_{\mu_{2}}\\ 0&0&I_{{k^{4}}}\end{bmatrix}^{\!-1}\!\!\begin{bmatrix}0\\ 0\\ -T_{\mu_{3}}\end{bmatrix} (19)
=\displaystyle= [A1000A2000A3][Ik4Tμ1Tμ1Tμ20Ik4Tμ200Ik4][00Tμ3]\displaystyle\ \begin{bmatrix}A_{1}&0&0\\ 0&A_{2}&0\\ 0&0&A_{3}\end{bmatrix}\!\!\begin{bmatrix}I_{{k^{4}}}&T_{\mu_{1}}&\ T_{\mu_{1}}\!T_{\mu_{2}}\\ 0&I_{{k^{4}}}&T_{\mu_{2}}\\ 0&0&I_{{k^{4}}}\end{bmatrix}\!\!\begin{bmatrix}0\\ 0\\ T_{\mu_{3}}\end{bmatrix}
=\displaystyle= [A1Tμ1Tμ2Tμ3A2Tμ2Tμ3A3Tμ3].\displaystyle\ \begin{bmatrix}A_{1}T_{\mu_{1}}\!T_{\mu_{2}}T_{\mu_{3}}\\ A_{2}T_{\mu_{2}}T_{\mu_{3}}\\ A_{3}T_{\mu_{3}}\end{bmatrix}.

Applying a group of elementary row transformations to A~3\tilde{A}_{3} yields

A^3\displaystyle\hat{A}_{3} =\displaystyle= [(I2Tμ3Tμ2Tμ1)A1Tμ1Tμ2Tμ3Tμ3Tμ2A2Tμ2Tμ3Tμ3A3Tμ3]\displaystyle\begin{bmatrix}(I_{2}\otimes T_{\mu_{3}}\!T_{\mu_{2}}T_{\mu_{1}})A_{1}T_{\mu_{1}}T_{\mu_{2}}T_{\mu_{3}}\\ T_{\mu_{3}}T_{\mu_{2}}A_{2}T_{\mu_{2}}T_{\mu_{3}}\\ T_{\mu_{3}}A_{3}T_{\mu_{3}}\end{bmatrix} (20)
=\displaystyle= [Ik4Tμ3Tμ2Tμ1Tμ2Tμ1Tμ2Tμ3Ik4Tμ3Tμ2Tμ1Tμ3Tμ1Tμ2Tμ3Ik4Tμ3Tμ2Tμ3Tμ2Tμ3Ik4Tμ3Tμ1Tμ3].\displaystyle\begin{bmatrix}I_{k^{4}}-T_{\mu_{3}}\!T_{\mu_{2}}T_{\mu_{1}}T_{\mu_{2}}T_{\mu_{1}}T_{\mu_{2}}T_{\mu_{3}}\\ I_{{k^{4}}}-T_{\mu_{3}}\!T_{\mu_{2}}T_{\mu_{1}}T_{\mu_{3}}T_{\mu_{1}}T_{\mu_{2}}T_{\mu_{3}}\\ I_{{k^{4}}}-T_{\mu_{3}}T_{\mu_{2}}T_{\mu_{3}}T_{\mu_{2}}T_{\mu_{3}}\\ I_{{k^{4}}}-T_{\mu_{3}}T_{\mu_{1}}T_{\mu_{3}}\end{bmatrix}.

Using the properties of TμiT_{\mu_{i}} shown by (LABEL:eqn12), we have

Tμ3Tμ1Tμ3=Tμ3Tμ3Tμ1=Tμ1;\displaystyle T_{\mu_{3}}T_{\mu_{1}}T_{\mu_{3}}=T_{\mu_{3}}T_{\mu_{3}}T_{\mu_{1}}=T_{\mu_{1}}; (21)
Tμ3(Tμ2Tμ3Tμ2)Tμ3=Tμ3(Tμ3Tμ2Tμ3)Tμ3=Tμ2;\displaystyle T_{\mu_{3}}(T_{\mu_{2}}T_{\mu_{3}}T_{\mu_{2}})T_{\mu_{3}}=T_{\mu_{3}}(T_{\mu_{3}}T_{\mu_{2}}T_{\mu_{3}})T_{\mu_{3}}=T_{\mu_{2}}; (22)
Tμ3Tμ2(Tμ1Tμ3Tμ1)Tμ2Tμ3=Tμ3Tμ2Tμ3Tμ2Tμ3\displaystyle T_{\mu_{3}}\!T_{\mu_{2}}(T_{\mu_{1}}T_{\mu_{3}}T_{\mu_{1}})T_{\mu_{2}}T_{\mu_{3}}=T_{\mu_{3}}\!T_{\mu_{2}}T_{\mu_{3}}T_{\mu_{2}}T_{\mu_{3}}
=Tμ3(Tμ3Tμ2Tμ3)Tμ3=Tμ2;\displaystyle=T_{\mu_{3}}\!(T_{\mu_{3}}T_{\mu_{2}}T_{\mu_{3}})T_{\mu_{3}}=T_{\mu_{2}}; (23)
Tμ3Tμ2(Tμ1Tμ2Tμ1)Tμ2Tμ3=Tμ3Tμ2(Tμ2Tμ1Tμ2)Tμ2Tμ3\displaystyle T_{\mu_{3}}\!T_{\mu_{2}}(T_{\mu_{1}}T_{\mu_{2}}T_{\mu_{1}})T_{\mu_{2}}T_{\mu_{3}}=T_{\mu_{3}}\!T_{\mu_{2}}(T_{\mu_{2}}T_{\mu_{1}}T_{\mu_{2}})T_{\mu_{2}}T_{\mu_{3}}
=Tμ3Tμ1Tμ3=Tμ1.\displaystyle=T_{\mu_{3}}T_{\mu_{1}}T_{\mu_{3}}=T_{\mu_{1}}. (24)

So, (18) is equivalent to (8). The proof is complete. \blacksquare

By Theorem 1, symmetric games have been clearly described by a simple form. Then the problem is how to simply determine the dimension and bases of the symmetric game subspace. From (8), we see that the key to solve the problem is the following linear equations:

[Ik3W[k]IkIk3IkW[k]]x=0,\left[\begin{array}[]{c}I_{k^{3}}-W_{[k]}\!\otimes\!I_{k}\\ I_{k^{3}}-I_{k}\!\otimes\!W_{[k]}\end{array}\right]x=0, (25)

where xx is the k3k^{3}-dimensional unknown vector. For the linear equation (25), we give a lemma as follows:

Lemma 4.

The dimension of the solution space of linear equations (25) is Ck+23=16k(k+1)(k+2)C_{k+2}^{3}=\frac{1}{6}k(k+1)(k+2).

{pf}

Let x=(xpqr)x=(x_{pqr}) be a k3k^{3}-dimensional column vector arranged by the ordered multi-index Id(i1,i2,i3;k,k,k)(i_{1},i_{2},i_{3};k,k,k), that is,

x=(x111,,x11k,x121,,x12k,,xkk1,,xkkk)T.x\!=\!(x_{111},\ldots,x_{11k},x_{121},\ldots,x_{12k},\ldots,x_{kk1},\ldots,x_{kkk})^{\mathrm{T}}. (26)

Then, by the property of W[k]W_{[k]}, xx is a solution of (25) if and only if

xpqr=xqpr,xpqr=xprq, 1p,q,rk,x_{pqr}=x_{qpr},\ \ x_{pqr}=x_{prq},\ \ \forall\ 1\leq p,q,r\leq k, (27)

which is equivalent to

xpqr=xτ(pqr),τ𝐒3.x_{pqr}=x_{\tau(pqr)},\ \forall\tau\in\ \mathbf{S}_{3}. (28)

So all the free variables of the linear equations are

xpqr, 1pqrk,x_{pqr},\ \forall\ 1\leq p\leq q\leq r\leq k, (29)

whose number is Ck+23C_{k+2}^{3} by Theorem 2.5.1 of [2] on counting number of combinations of a multiset. \Box

The proof of Lemma 4 actually provides an effective approach to construct a group of bases of the solution space 𝒳3\mathcal{X}_{3} of (25). Based on the bases, we also can easily get the minimum number of equivalent equations to (25). See the details as follows:

For every given repeatable combination pqrpqr (1pqrk)(1\leq p\leq q\leq r\leq k), denote by PpqrP_{pqr} the set composed of all the repeatable permutation of pqrpqr. For example, P111={111}P_{111}=\{111\}, P112={112,121,211}P_{112}=\{112,121,211\}, P123={123,231,312,132,321,213}P_{123}=\{123,231,312,132,321,213\}.

Lemma 5.

For every combination pqrpqr (1pqrk)(1\leq p\leq q\leq r\leq k), define a vector ηpqr=x\eta^{pqr}=x with the form (26) by

xabc={1,abcPpqr,0,otherwise.x_{abc}=\left\{\begin{array}[]{l}1,\ \ abc\in P_{pqr},\\ 0,\ \ \mbox{otherwise}.\end{array}\right.

Then the set of ηpqr\eta^{pqr} (1pqrk)(1\leq p\leq q\leq r\leq k) is a group of bases of the solution space 𝒳3\mathcal{X}_{3} of (25). For every pqr{pqr} (1pqrk)(1\leq p\leq q\leq r\leq k) that does not satisfy p=q=rp=q=r, we define |Ppqr|1|P_{pqr}|-1 number of vectors ζpqruvw=x\zeta_{pqr}^{uvw}=x (uvwPpqr,uvwpqr)(uvw\in P_{pqr},uvw\neq{pqr}) by

xabc={1,abc=pqr,1,abc=uvw,0,otherwise.x_{abc}=\left\{\begin{array}[]{l}1,\ \ abc=pqr,\\ -1,\ \ abc=uvw,\\ 0,\ \ \mbox{otherwise}.\end{array}\right.

Then the set of ζpqruvw\zeta_{pqr}^{uvw} (uvwPpqr,uvwpqr)(uvw\in P_{pqr},uvw\neq{pqr}) is a group of bases of the orthogonal complementary space 𝒳3\mathcal{X}_{3}^{\perp}. Denote by M𝒲M_{\mathcal{W}} the matrix composed by a group of bases of the subspace 𝒲\mathcal{W}. Then the linear system (25) is equivalent to M𝒳3Tx=0M_{\mathcal{X}_{3}^{\perp}}^{\mathrm{T}}x=0.

{pf}

From the equivalent equations (28) and the free variables shown by (29), it follows that the set of ηpqr\eta^{pqr} (1pqrk)(1\leq p\leq q\leq r\leq k) is a group of bases of the solution space 𝒳3\mathcal{X}_{3}. By the construction of ζpqruvw\zeta_{pqr}^{uvw}, it is straightforward to see that each ζpqruvw\zeta_{pqr}^{uvw} is orthogonal to 𝒳3\mathcal{X}_{3}. Since the total number of ζpqruvw\zeta_{pqr}^{uvw} is 1pqrk(|Ppqr|1)=1pqrk|Ppqr|Ck+23=k3Ck+23=k3dim(𝒳3)\sum_{1\leq p\leq q\leq r\leq k}(|P_{pqr}|-1)=\sum_{1\leq p\leq q\leq r\leq k}|P_{pqr}|-C_{k+2}^{3}=k^{3}-C_{k+2}^{3}=k^{3}-\mbox{dim}(\mathcal{X}_{3}), we conclude that the set of ζpqruvw\zeta_{pqr}^{uvw} (uvwPpqr,uvwpqr)(uvw\in P_{pqr},uvw\neq{pqr}) is a group of bases of 𝒳3\mathcal{X}_{3}^{\perp}. For example, when k=2k=2, we have

M𝒳3=[10000100010000100100001000100001],M𝒳3=[00001100100000110100001000010000](111)(112)(121)(122)(211)(212)(221)(222)M_{\mathcal{X}_{3}}\!\!=\!\!\left[\begin{array}[]{cccc}1&~{}0&~{}0&~{}0\\ 0&~{}1&~{}0&~{}0\\ 0&~{}1&~{}0&~{}0\\ 0&~{}0&~{}1&~{}0\\ 0&~{}1&~{}0&~{}0\\ 0&~{}0&~{}1&~{}0\\ 0&~{}0&~{}1&~{}0\\ 0&~{}0&~{}0&~{}1\\ \end{array}\right]\!\!,\ M_{\mathcal{X}_{3}^{\perp}}\!\!=\!\!\left[\begin{array}[]{cccc}0&0&0&0\\ 1&1&0&0\\ -1&0&0&0\\ 0&0&1&1\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1\\ 0&0&0&0\\ \end{array}\right]\ \ \begin{array}[]{c}(111)\\ (112)\\ (121)\\ (122)\\ (211)\\ (212)\\ (221)\\ (222)\end{array}

Based on Theorem 1, Lemma 4 and Lemma lem6, one can investigate the dimension and bases of the symmetric game subspace.

Theorem 2.

The dimension of the symmetric game subspace 𝒮[4;k]\mathcal{S}_{[4;k]} is kCk+23kC_{k+2}^{3}. A group of bases of 𝒮[4;k]\mathcal{S}_{[4;k]} is composed of the columns of matrix

[W[k3,k]IkW[k2,k]Ik2W[k]Ik4](M𝒳3Ik),\begin{bmatrix}W_{[k^{3},k]}\\ I_{k}\otimes W_{[k^{2},k]}\\ I_{k^{2}}\otimes W_{[k]}\\ I_{k^{4}}\end{bmatrix}(M_{\mathcal{X}_{3}}\otimes I_{k}), (30)

where M𝒳3M_{\mathcal{X}_{3}} is composed of a bases of linear equations (25). Moreover, the linear equations with the minimum numbers to test symmetric games in 𝒢[4;k]\mathcal{G}_{[4;k]} are

[Ik4Tμ1000Ik4Tμ2000Ik4Tμ3000M𝒳3TIk](VG)T=0.\begin{bmatrix}I_{{k^{4}}}&-T_{\mu_{1}}&0&0\\ 0&I_{{k^{4}}}&-T_{\mu_{2}}&0\\ 0&0&I_{{k^{4}}}&-T_{\mu_{3}}\\ 0&0&0&M_{\mathcal{X}_{3}^{\bot}}^{\mathrm{T}}\otimes I_{k}\end{bmatrix}(V_{G})^{\mathrm{T}}=0. (31)
{pf}

It is easy to see that

[Ik4Tμ1Ik4Tμ2]=[Ik3W[k]IkIk3IkW[k]]Ik.\begin{bmatrix}I_{{k^{4}}}\!-\!T_{\mu_{1}}\\ I_{{k^{4}}}\!-\!T_{\mu_{2}}\end{bmatrix}=\begin{bmatrix}I_{k^{3}}-W_{[k]}\!\otimes\!I_{k}\\ I_{k^{3}}-I_{k}\!\otimes\!W_{[k]}\end{bmatrix}\otimes I_{k}. (32)

By Theorem 1 and Lemma 4, we can easily get the dimension of the symmetric game subspace 𝒮[4;k]\mathcal{S}_{[4;k]} is kCk+23kC_{k+2}^{3}. Using M𝒳3M_{\mathcal{X}_{3}} composed of the bases of (25), we get a group of bases of the solution space of (8):

[Tμ1Tμ2Tμ3(M𝒳3Ik)Tμ2Tμ3(M𝒳3Ik)Tμ3(M𝒳3Ik)M𝒳3Ik].\begin{bmatrix}T_{\mu_{1}}T_{\mu_{2}}T_{\mu_{3}}(M_{\mathcal{X}_{3}}\otimes I_{k})\\ T_{\mu_{2}}T_{\mu_{3}}(M_{\mathcal{X}_{3}}\otimes I_{k})\\ T_{\mu_{3}}(M_{\mathcal{X}_{3}}\otimes I_{k})\\ M_{\mathcal{X}_{3}}\otimes I_{k}\end{bmatrix}. (33)

By the properties of swap matrices, we have that

Tμ1Tμ2Tμ3=W[k3,k],Tμ2Tμ3=IkW[k2,k].T_{\mu_{1}}T_{\mu_{2}}T_{\mu_{3}}=W_{[k^{3},k]},\ \ T_{\mu_{2}}T_{\mu_{3}}=I_{k}\otimes W_{[k^{2},k]}.

So, (33) is just (30). Moreover, from (32) and Lemma 5, it follows that (8) is equivalent to (31). Since the coefficient matrix of (31) has a full row rank, the testing equations have the minimum number. \blacksquare

In the rest of this paper, we consider the general finite game G𝒢[n;k]G\in\mathcal{G}_{[n;k]}.

Theorem 3.

Consider G𝒢[n;k]G\in\mathcal{G}_{[n;k]}. GG is a symmetric game if and only if

[IknTμ1IknTμ2IknTμn1IknTμ1IknTμ2IknTμn2]VGT=0,\begin{bmatrix}I_{k^{n}}&-T_{\mu_{1}}&&&&\\ &I_{k^{n}}&-T_{\mu_{2}}&&&\\ &&\ddots&\ddots&&\\ &&&I_{k^{n}}&-T_{\mu_{n-1}}\\ &&&&I_{k^{n}}\!-\!T_{\mu_{1}}\\ &&&&I_{k^{n}}\!-\!T_{\mu_{2}}\\ &&&&\vdots\\ &&&&I_{k^{n}}\!-\!T_{\mu_{n-2}}\end{bmatrix}V_{G}^{\rm T}=0, (34)

where

Tμr=Ikr1W[k]Iknr1, 1rn1T_{\mu_{r}}=I_{k^{r-1}}\otimes W_{[k]}\otimes I_{k^{n-r-1}},\ 1\leq r\leq n-1\\ (35)

and

VG=[V1cV2cVnc].V_{G}=[V_{1}^{c}~{}V_{2}^{c}~{}\cdots~{}V_{n}^{c}]. (36)
{pf}

According to Proposition 1, G𝒢[n;k]G\in\mathcal{G}_{[n;k]} is a symmetric game if and only if

[IknTμ1IknTμ2IknTμn1A1A2An1An]VGT=0,\begin{bmatrix}I_{k^{n}}&-T_{\mu_{1}}&&&&\\ &I_{k^{n}}&-T_{\mu_{2}}&&&\\ &&\ddots&\ddots&&\\ &&&I_{k^{n}}&-T_{\mu_{n-1}}\\ A_{1}&&&&\\ &A_{2}&&\\ &&\ddots&&\\ &&&A_{n-1}&\\ &&&&A_{n}\end{bmatrix}V_{G}^{\rm T}=0, (37)

where

A1=[IknTμ2IknTμ3IknTμn1],A2=[IknTμ3IknTμ4IknTμn1],A_{1}=\begin{bmatrix}I_{k^{n}}-T_{\mu_{2}}\\ I_{k^{n}}-T_{\mu_{3}}\\ \vdots\\ I_{k^{n}}-T_{\mu_{n-1}}\end{bmatrix},\ A_{2}=\begin{bmatrix}I_{k^{n}}-T_{\mu_{3}}\\ I_{k^{n}}-T_{\mu_{4}}\\ \vdots\\ I_{k^{n}}-T_{\mu_{n-1}}\end{bmatrix}, (38)
Ar=[IknTμ1IknTμr2IknTμr+1IknTμn1]( 3rn2),A_{r}\!=\!\!\begin{bmatrix}I_{k^{n}}\!-\!T_{\mu_{1}}\\ \vdots\\ I_{k^{n}}\!-\!T_{\mu_{r-2}}\\ I_{k^{n}}\!-\!T_{\mu_{r+1}}\\ \vdots\\ I_{k^{n}}\!-\!T_{\mu_{n-1}}\end{bmatrix}\!(\ 3\!\leq\!r\!\leq\!n-2),\ (39)
An1=[IknTμ1IknTμ2IknTμn3],An=[IknTμ1IknTμ2IknTμn2].A_{n-1}\!=\!\!\begin{bmatrix}I_{k^{n}}\!-\!T_{\mu_{1}}\\ I_{k^{n}}\!-\!T_{\mu_{2}}\\ \vdots\\ I_{k^{n}}\!-\!T_{\mu_{n-3}}\\ \end{bmatrix},A_{n}\!=\!\!\begin{bmatrix}I_{k^{n}}\!-\!T_{\mu_{1}}\\ I_{k^{n}}\!-\!T_{\mu_{2}}\\ \vdots\\ I_{k^{n}}\!-\!T_{\mu_{n-2}}\end{bmatrix}. (40)

Let F1=In2(Tμn1Tμn2Tμ1)F_{1}=I_{n-2}\otimes(T_{\mu_{n-1}}T_{\mu_{n-2}}\cdots T_{\mu_{1}}) and Fr=In3(Tμn1Tμn2Tμr)F_{r}=I_{n-3}\otimes(T_{\mu_{n-1}}T_{\mu_{n-2}}\cdots T_{\mu_{r}}) for every 2rn12\leq r\leq n-1. Then a series of elementary row transformations to (37)(\ref{eqn37}) results in

[IknTμ1IknTμ2IknTμn1F1A1Tμ1Tμ2Tμn1F2A2Tμ2Tμ3Tμn1Fn1An1Tμn1An]VGT=0.\begin{bmatrix}I_{k^{n}}&-T_{\mu_{1}}&&&&\\ &I_{k^{n}}&-T_{\mu_{2}}&&&\\ &&\ddots&\ddots&&\\ &&&I_{k^{n}}&-T_{\mu_{n-1}}\\ &&&&\!\!F_{1}A_{1}T_{\mu_{1}}T_{\mu_{2}}\!\!\cdots\!\!T_{\mu_{n-1}}\\ &&&&\!\!F_{2}A_{2}T_{\mu_{2}}T_{\mu_{3}}\!\!\cdots\!\!T_{\mu_{n-1}}\\ &&&&\vdots\\ &&&&\!\!F_{n-1}A_{n-1}T_{\mu_{n-1}}&\\ &&&&A_{n}\!\!\end{bmatrix}\!\!V_{G}^{\rm T}\!=\!0. (41)

From the properties of swap matrix W[k]W_{[k]}, it follows that

Tμn1TμrTμiTμrTμn1={Tμi, 1ir2,Tμi1,r+1i<n.T_{\mu_{n-1}}\cdots T_{\mu_{r}}T_{\mu_{i}}T_{\mu_{r}}\!\cdots\!T_{\mu_{n-1}}\!=\!\left\{\!\!\begin{array}[]{l}T_{\mu_{i}},\!~{}~{}~{}~{}~{}\forall\ 1\!\leq\!i\!\leq\!r\!-\!2,\\ T_{\mu_{i-1}},~{}~{}\forall\ r\!+\!1\!\leq\!i\!<\!n.\end{array}\right. (42)

Substituting (42) into (41) yields the equivalent linear equations (34). Thus the proof is complete. \blacksquare To get the dimension and bases of a general symmetric game subspace, we consider the following linear equations:

[Ikn1W[k]Ikn3Ikn1IkW[k]Ikn4Ikn1Ikn3W[k]]x=0.\begin{bmatrix}I_{k^{n-1}}-W_{[k]}\otimes I_{k^{n-3}}\\ I_{k^{n-1}}-I_{k}\otimes W_{[k]}\otimes I_{k^{n-4}}\\ \vdots\\ I_{k^{n-1}}-I_{k^{n-3}}\otimes W_{[k]}\end{bmatrix}x=0. (43)

Let x=(xp1p2pn1)x=(x_{p_{1}p_{2}\cdots p_{n-1}}) be a kn1k^{n-1}-dimensional column vector arranged by the ordered multi-index Id(i1,i2,,in1;k,k,,k)(i_{1},i_{2},\ldots,i_{n-1};k,k,\ldots,k). Then, by the property of W[k]W_{[k]}, xx is a solution of (43) if and only if

xp1p2pn1=xμr(p1p2pn1)x_{p_{1}p_{2}\cdots p_{n-1}}=x_{\mu_{r}(p_{1}p_{2}\cdots p_{n-1})} (44)

for any adjacent transposition μr\mu_{r}, 1rn21\leq r\leq n-2 and 1p1,p2,,pn1k1\leq p_{1},p_{2},\ldots,p_{n-1}\leq k, which is equivalent to

xp1p2pn1=xσ(p1p2pn1)x_{p_{1}p_{2}\cdots p_{n-1}}=x_{\sigma(p_{1}p_{2}\cdots p_{n-1})} (45)

for any σ𝐒n1\sigma\in\mathbf{S}_{n-1} by Lemma 2. Therefore, the free variables of the linear equations can be chosen as

xp1p2pn1(1p1p2pn1k).x_{p_{1}p_{2}\cdots p_{n-1}}\ \ (1\!\leq\!p_{1}\!\leq\!p_{2}\!\leq\!\cdots\!\leq\!p_{n-1}\!\leq k). (46)
Lemma 6.

The dimension of the solution space of linear equations (43) is Ck+n2n1C_{k+n-2}^{n-1}. For every repeatable combination p1p2pn1p_{1}p_{2}\cdots p_{n-1} (1p1p2pn1k)(1\!\leq\!p_{1}\!\leq\!p_{2}\!\leq\!\cdots\!\leq\!p_{n-1}\!\leq k), let Pp1p2pn1P_{p_{1}p_{2}\cdots p_{n-1}} be the set composed of all the repeatable permutations of p1p2pn1p_{1}p_{2}\cdots p_{n-1}. Define a vector ηp1p2pn1=x\eta^{p_{1}p_{2}\cdots p_{n-1}}=x by

xa1a2an1={1,a1a2an1Pp1p2pn1,0,otherwise.x_{a_{1}a_{2}\cdots a_{n-1}}=\left\{\begin{array}[]{l}1,\ \ a_{1}a_{2}\cdots a_{n-1}\in P_{p_{1}p_{2}\cdots p_{n-1}},\\ 0,\ \ \mbox{otherwise}.\end{array}\right.

Then the set of ηp1p2pn1=x\eta^{p_{1}p_{2}\cdots p_{n-1}}=x (1p1p2pn1k)(1\!\leq\!p_{1}\!\leq\!p_{2}\!\leq\!\cdots\!\leq\!p_{n-1}\!\leq k) is a group of bases of the solution space 𝒳n1\mathcal{X}_{n-1} of (43). Define ζp1p2pn1q1q2qn1=x\zeta_{p_{1}p_{2}\cdots p_{n-1}}^{q_{1}q_{2}\cdots q_{n-1}}=x (q1q2qn1Pp1p2pn1,q1q2qn1p1p2pn1)(q_{1}q_{2}\cdots q_{n-1}\in P_{p_{1}p_{2}\cdots p_{n-1}},q_{1}q_{2}\cdots q_{n-1}\neq p_{1}p_{2}\cdots p_{n-1}) by

xa1a2an={1,a1a2an=p1p2pn1,1,a1a2an=q1q2qn1,0,otherwise.x_{a_{1}a_{2}\cdots a_{n}}=\left\{\begin{array}[]{l}1,\ \ a_{1}a_{2}\cdots a_{n}=p_{1}p_{2}\cdots p_{n-1},\\ -1,\ \ a_{1}a_{2}\cdots a_{n}=q_{1}q_{2}\cdots q_{n-1},\\ 0,\ \ \mbox{otherwise}.\end{array}\right.

Then M𝒳n1M_{\mathcal{X}_{n-1}} is composed of ηp1p2pn1\eta^{p_{1}p_{2}\cdots p_{n-1}} (1p1p2pn1k)(1\!\leq\!p_{1}\!\leq\!p_{2}\!\leq\!\cdots\!\leq\!p_{n-1}\!\leq k), and M𝒳n1M_{\mathcal{X}_{n-1}^{\perp}} is composed of ζp1p2pn1q1q2qn1\zeta_{p_{1}p_{2}\cdots p_{n-1}}^{q_{1}q_{2}\cdots q_{n-1}} (q1q2qn1Pp1p2pn1,q1q2qn1p1p2pn1)(q_{1}q_{2}\cdots q_{n-1}\in P_{p_{1}p_{2}\cdots p_{n-1}},q_{1}q_{2}\cdots q_{n-1}\neq p_{1}p_{2}\cdots p_{n-1}).

Theorem 4.

The dimension of the symmetric game subspace 𝒮[n;k]\mathcal{S}_{[n;k]} is kCk+n2n1kC_{k+n-2}^{n-1}. A group of bases of 𝒮[n;k]\mathcal{S}_{[n;k]} is composed of the columns of matrix

[W[kn1,k]IkW[kn2,k]Ik2W[kn3,k]Ikn2W[k]Ikn](M𝒳n1Ik),\begin{bmatrix}W_{[k^{n-1},k]}\\ I_{k}\otimes W_{[k^{n-2},k]}\\ I_{k^{2}}\otimes W_{[k^{n-3},k]}\\ \vdots\\ I_{k^{n-2}}\otimes W_{[k]}\\ I_{k^{n}}\end{bmatrix}(M_{\mathcal{X}_{n-1}}\otimes I_{k}), (47)

where M𝒳n1M_{\mathcal{X}_{n-1}} is composed of a bases of linear equations (43). Moreover, the linear equations with the minimum numbers to test symmetric games in 𝒢[n;k]\mathcal{G}_{[n;k]} are

[IknTμ1IknTμ2IknTμn1M𝒳n1TIk]VGT=0.\begin{bmatrix}I_{k^{n}}&-T_{\mu_{1}}&&&&\\ &I_{k^{n}}&-T_{\mu_{2}}&&&\\ &&\ddots&\ddots&&\\ &&&I_{k^{n}}&-T_{\mu_{n-1}}\\ &&&&M_{\mathcal{X}_{n-1}^{\perp}}^{\mathrm{T}}\!\!\!\!\!\otimes I_{k}\end{bmatrix}V_{G}^{\rm T}=0. (48)
{pf}

It is easy to see that

[IknTμ1IknTμ2IknTμn2]=[Ikn1W[k]Ikn3Ikn1IkW[k]Ikn4Ikn1Ikn3W[k]]Ik.\begin{bmatrix}I_{k^{n}}\!-\!T_{\mu_{1}}\\ I_{k^{n}}\!-\!T_{\mu_{2}}\\ \vdots\\ I_{k^{n}}\!-\!T_{\mu_{n-2}}\end{bmatrix}=\begin{bmatrix}I_{k^{n-1}}-W_{[k]}\otimes I_{k^{n-3}}\\ I_{k^{n-1}}-I_{k}\otimes W_{[k]}\otimes I_{k^{n-4}}\\ \vdots\\ I_{k^{n-1}}-I_{k^{n-3}}\otimes W_{[k]}\end{bmatrix}\otimes I_{k}. (49)

By Theorem 3 and Lemma 6, we can easily get the dimension of the symmetric game subspace 𝒮[n;k]\mathcal{S}_{[n;k]} is kCk+n2n1kC_{k+n-2}^{n-1}. Using M𝒳n1M_{\mathcal{X}_{n-1}} composed of the bases of (43), we get a group of bases of the solution space of (34):

[Tμ1Tμ2Tμn1(M𝒳n1Ik)Tμ2Tμn1(M𝒳n1Ik)Tμn1(M𝒳n1Ik)M𝒳n1Ik].\begin{bmatrix}T_{\mu_{1}}T_{\mu_{2}}\cdots T_{\mu_{n-1}}(M_{\mathcal{X}_{n-1}}\otimes I_{k})\\ T_{\mu_{2}}T_{\mu_{n-1}}(M_{\mathcal{X}_{n-1}}\otimes I_{k})\\ \vdots\\ T_{\mu_{n-1}}(M_{\mathcal{X}_{n-1}}\otimes I_{k})\\ M_{\mathcal{X}_{n-1}}\otimes I_{k}\end{bmatrix}. (50)

By the properties of swap matrices, we have that

TμsTμs+1Tμn1=Iks1W[kns,k]T_{\mu_{s}}T_{\mu_{s+1}}\cdots T_{\mu_{n-1}}=I_{k}^{s-1}\otimes W_{[k^{n-s},k]}

for each 1sn11\leq s\leq n-1. So, (50) is just (47). Moreover, from (49), it follows that (34) is equivalent to (48). Since the coefficient matrix of (48) has a full row rank, the testing equations have the minimum number. \blacksquare

Remark 1.

In Theorem 19.2.1 of [5], the dimension of the symmetric game subspace is obtained by using the concept of strategy multiplicity vector of [4]. However, Theorem 4 gives the dimension of a symmetric game subspace via a purely algebraic approach without using the strategy multiplicity vector, which makes the proof greatly simplified.

Remark 2.

In [15] and [20], construction algorithms of a group of bases for the symmetric game subspace are given by using the classification of strategy profiles. Different from [15] and [20], Theorem 4 directly constructs a group of bases of a symmetric game subspace from a group of bases of the solution space of linear equations (43). In addition, the testing equations for symmetric games with the minimum number is easily obtained.

Example 1.

Consider G𝒢[3;2]G\in\mathcal{G}_{[3;2]}.

M𝒳2=[100010010001],M𝒳2=[0110](11)(12)(21)(22).M_{\mathcal{X}_{2}}\!\!=\!\!\left[\begin{array}[]{cccc}1&~{}0&~{}0\\ 0&~{}1&~{}0\\ 0&~{}1&~{}0\\ 0&~{}0&~{}1\\ \end{array}\right]\!\!,\ M_{\mathcal{X}_{2}^{\perp}}\!\!=\!\!\left[\begin{array}[]{cccc}0\\ 1\\ -1\\ 0\\ \end{array}\right]\ \ \begin{array}[]{c}(11)\\ (12)\\ (21)\\ (22)\end{array}.
Tμ1=W[2]I2=δ8[12563478];T_{\mu_{1}}=W_{[2]}\otimes I_{2}=\delta_{8}[1~{}2~{}5~{}6~{}3~{}4~{}7~{}8];
Tμ2=I2W[2]=δ8[13245768].T_{\mu_{2}}=I_{2}\otimes W_{[2]}=\delta_{8}[1~{}3~{}2~{}4~{}5~{}7~{}6~{}8].

GG is a symmetric game if and only if

[I8Tμ1I8Tμ2M𝒳2TI2]VGT=0.\begin{bmatrix}I_{8}&-T_{\mu_{1}}\\ &I_{8}&-T_{\mu_{2}}\\ &&M_{\mathcal{X}_{2}^{\perp}}^{\mathrm{T}}\otimes I_{2}\end{bmatrix}V_{G}^{\rm T}=0. (51)

Let

V3c=[x1x2x3x4x5x6x7x8].V_{3}^{c}=[x_{1}~{}x_{2}~{}x_{3}~{}x_{4}~{}x_{5}~{}x_{6}~{}x_{7}~{}x_{8}].

According to (M𝒳2TI2)(V3c)T=0(M_{\mathcal{X}_{2}^{\perp}}^{\mathrm{T}}\otimes I_{2})(V_{3}^{c})^{\mathrm{T}}=0, we have

x3=x5,x4=x6.x_{3}=x_{5},~{}x_{4}=x_{6}.

Therefore,

V3c=[a1a2a3a4a3a4a5a6]V_{3}^{c}=[a_{1}~{}a_{2}~{}a_{3}~{}a_{4}~{}a_{3}~{}a_{4}~{}a_{5}~{}a_{6}]

where a1,a2,a3,a4,a5,a6a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}\in\mathbb{R}. According to (V2c)T=Tμ2(V3c)T(V_{2}^{c})^{\mathrm{T}}=T_{\mu_{2}}(V_{3}^{c})^{\mathrm{T}}, (V1c)T=Tμ1(V2c)T(V_{1}^{c})^{\mathrm{T}}=T_{\mu_{1}}(V_{2}^{c})^{\mathrm{T}},

V2c=[a1a3a2a4a3a5a4a6],V_{2}^{c}=[a_{1}~{}a_{3}~{}a_{2}~{}a_{4}~{}a_{3}~{}a_{5}~{}a_{4}~{}a_{6}],
V1c=[a1a3a3a5a2a4a4a6].V_{1}^{c}=[a_{1}~{}a_{3}~{}a_{3}~{}a_{5}~{}a_{2}~{}a_{4}~{}a_{4}~{}a_{6}].

Therefore, the game is a symmetric game if and only if its payoff matrix is in the form

[a1a3a3a5a2a4a4a6a1a3a2a4a3a5a4a6a1a2a3a4a3a4a5a6].\displaystyle\begin{bmatrix}a_{1}&a_{3}&a_{3}&a_{5}&a_{2}&a_{4}&a_{4}&a_{6}\\ a_{1}&a_{3}&a_{2}&a_{4}&a_{3}&a_{5}&a_{4}&a_{6}\\ a_{1}&a_{2}&a_{3}&a_{4}&a_{3}&a_{4}&a_{5}&a_{6}\\ \end{bmatrix}.
Example 2.

For any G𝒢[4;2]G\in\mathcal{G}_{[4;2]}, we let

Tμ1=W[2]I4=δ16[12349101112567813141516],\displaystyle\begin{split}T_{\mu_{1}}=&W_{[2]}\otimes I_{4}\\ =&\delta_{16}[1~{}2~{}3~{}4~{}9~{}10~{}11~{}12~{}5~{}6~{}7~{}8~{}13~{}14~{}15~{}16],\end{split}
Tμ2=I2W[2]I2=δ16[12563478910131411121516],\displaystyle\begin{split}T_{\mu_{2}}=&I_{2}\otimes W_{[2]}\otimes I_{2}\\ =&\delta_{16}[1~{}2~{}5~{}6~{}3~{}4~{}7~{}8~{}9~{}10~{}13~{}14~{}11~{}12~{}15~{}16],\end{split}
Tμ3=I4W[2]=δ16[13245768911101213151416].\displaystyle\begin{split}T_{\mu_{3}}=&I_{4}\otimes W_{[2]}\\ =&\delta_{16}[1~{}3~{}2~{}4~{}5~{}7~{}6~{}8~{}9~{}11~{}10~{}12~{}13~{}15~{}14~{}16].\end{split}

By (31), we have

(M𝒳3TI2)(V4c)T=0,(M_{\mathcal{X}_{3}^{\bot}}^{\mathrm{T}}\otimes I_{2})(V_{4}^{c})^{\mathrm{T}}=0,
(V3c)T=Tμ3(V4c)T,(V_{3}^{c})^{\mathrm{T}}=T_{\mu_{3}}(V_{4}^{c})^{\mathrm{T}},
(V2c)T=Tμ2(V3c)T,(V_{2}^{c})^{\mathrm{T}}=T_{\mu_{2}}(V_{3}^{c})^{\mathrm{T}},
(V1c)T=Tμ1(V2c)T.(V_{1}^{c})^{\mathrm{T}}=T_{\mu_{1}}(V_{2}^{c})^{\mathrm{T}}.

Let V4c=(x1,x2,,x16)V_{4}^{c}=(x_{1},x_{2},\cdots,x_{16}). According to (M𝒳3TI2)(V4c)T=0(M_{\mathcal{X}_{3}^{\bot}}^{\mathrm{T}}\otimes I_{2})(V_{4}^{c})^{\mathrm{T}}=0, we have

x3\displaystyle x_{3} =x5=x9,\displaystyle=x_{5}=x_{9},
x4\displaystyle x_{4} =x6=x10,\displaystyle=x_{6}=x_{10},
x7\displaystyle x_{7} =x11=x13,\displaystyle=x_{11}=x_{13},
x8\displaystyle x_{8} =x12=x14.\displaystyle=x_{12}=x_{14}.

Then,

V4c=[b1b2b3b4b3b4b5b6b3b4b5b6b5b6b7b8],\displaystyle V_{4}^{c}=[b_{1}~{}b_{2}~{}b_{3}~{}b_{4}~{}b_{3}~{}b_{4}~{}b_{5}~{}b_{6}~{}b_{3}~{}b_{4}~{}b_{5}~{}b_{6}~{}b_{5}~{}b_{6}~{}b_{7}~{}b_{8}],

where b1,b2,,b8b_{1},b_{2},\cdots,b_{8}\in\mathbb{R}.

V3c\displaystyle V_{3}^{c} =V4cTμ3\displaystyle=V_{4}^{c}T_{\mu_{3}}
=[b1b3b2b4b3b5b4b6b3b5b4b6b5b7b6b8];\displaystyle=[b_{1}~{}b_{3}~{}b_{2}~{}b_{4}~{}b_{3}~{}b_{5}~{}b_{4}~{}b_{6}~{}b_{3}~{}b_{5}~{}b_{4}~{}b_{6}~{}b_{5}~{}b_{7}~{}b_{6}~{}b_{8}];
V2c\displaystyle V_{2}^{c} =V3cTμ2\displaystyle=V_{3}^{c}T_{\mu_{2}}
=[b1b3b3b5b2b4b4b6b3b5b5b7b4b6b6b8];\displaystyle=[b_{1}~{}b_{3}~{}b_{3}~{}b_{5}~{}b_{2}~{}b_{4}~{}b_{4}~{}b_{6}~{}b_{3}~{}b_{5}~{}b_{5}~{}b_{7}~{}b_{4}~{}b_{6}~{}b_{6}~{}b_{8}];
V1c\displaystyle V_{1}^{c} =V2cTμ1\displaystyle=V_{2}^{c}T_{\mu_{1}}
=[b1b3b3b5b3b5b5b7b2b4b4b6b4b6b6b8].\displaystyle=[b_{1}~{}b_{3}~{}b_{3}~{}b_{5}~{}b_{3}~{}b_{5}~{}b_{5}~{}b_{7}~{}b_{2}~{}b_{4}~{}b_{4}~{}b_{6}~{}b_{4}~{}b_{6}~{}b_{6}~{}b_{8}].

Therefore, the game is a symmetric game if and only if its payoff matrix is in the form

[b1b3b3b5b3b5b5b7b2b4b4b6b4b6b6b8b1b3b3b5b2b4b4b6b3b5b5b7b4b6b6b8b1b3b2b4b3b5b4b6b3b5b4b6b5b7b6b8b1b2b3b4b3b4b5b6b3b4b5b6b5b6b7b8].\displaystyle\begin{bmatrix}b_{1}~{}b_{3}~{}b_{3}~{}b_{5}~{}b_{3}~{}b_{5}~{}b_{5}~{}b_{7}~{}b_{2}~{}b_{4}~{}b_{4}~{}b_{6}~{}b_{4}~{}b_{6}~{}b_{6}~{}b_{8}\\ b_{1}~{}b_{3}~{}b_{3}~{}b_{5}~{}b_{2}~{}b_{4}~{}b_{4}~{}b_{6}~{}b_{3}~{}b_{5}~{}b_{5}~{}b_{7}~{}b_{4}~{}b_{6}~{}b_{6}~{}b_{8}\\ b_{1}~{}b_{3}~{}b_{2}~{}b_{4}~{}b_{3}~{}b_{5}~{}b_{4}~{}b_{6}~{}b_{3}~{}b_{5}~{}b_{4}~{}b_{6}~{}b_{5}~{}b_{7}~{}b_{6}~{}b_{8}\\ b_{1}~{}b_{2}~{}b_{3}~{}b_{4}~{}b_{3}~{}b_{4}~{}b_{5}~{}b_{6}~{}b_{3}~{}b_{4}~{}b_{5}~{}b_{6}~{}b_{5}~{}b_{6}~{}b_{7}~{}b_{8}\\ \end{bmatrix}.

4 Conclusion

In this paper, new equivalent conditions for testing symmetric games are obtained via a matrix semi-tensor product method with adjacent transpositions. Compared with the existing literature, this paper provides simple method with purely algebraic operations. Based on the new equivalent conditions for symmetric games, the dimension of a symmetric game subspace has been directly calculated, and a group of bases of the symmetric game subspace has been easily constructed. In addition, the testing equations with the minimum number have been derived. {ack} The research work of X. Liu is supprted in part by Natural Science Foundation of Shandong Province of China under Grant ZR2020QA028, and in part by Doctoral Research Foundation of Weifang University under Grant 2019BS03. The research work of T. Li and J. Zhu is supported in part by National Natural Science Foundation (NNSF) of China under Grants 62103194 and 61673012, respectively.

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