运筹学 Operations Research H1(a1,B)=1H1(1,B2)=-1H1(a2,B1) H1(a2,B2)=1 H2(12B1)=-1H2(x12B2)=1H2(a2,B1)=1H2(a2,B2)=-1 二人有限零和对策(2- player finite zero- sum game) ={12} t1,2,… S2={B1,B2;…,Bn}mn<+∞ H1+H,=0 设H1(a,B1)=an,则H2(a,B)=-an (局中人1的)收益矩阵:A=(a1)mn,其中an=H1(a,B,) 2021/2/202021/2/20 5 运 筹 学 Operations Research H1 (1 ,1 ) =1 H1 (1 , 2 ) = −1 H1 ( 2 ,1 ) = −1 H1 ( 2 , 2 ) =1 H2 (1 ,1 ) = −1 H2 (1 , 2 ) =1 H2 ( 2 ,1 ) =1 H2 (2 , 2 ) = −1 二人有限零和对策（2-player finite zero-sum game）:…… I = {1,2} { , , , } S1 = 1  2   m { , , , } S2 = 1  2   n H1 + H2 = 0 m,n  + ( , ) ( , ) . 设H1  i  j = ai j，则H2  i  j =－ai j （局中人1的）收益矩阵： ( ) , A = aij mn ( , ). 其中aij = H1  i  j