3.3 Matlab模拟 我们用Matlab软件模拟出可能的对阵情况，并根据获胜概率模拟出单站比赛的比赛结 果。由于所有的参赛选手等级分都在2400分以上，K的值可以统一取10。我们把等级分差 距x和W的值做一下线性拟合后便得到了它们之间的关系式 H=10×(80+0.5) L=10×(0.5-650 2 D=H-L=52 当棋手获胜时，W=0,反之W=1。利用等级分计算公式 R1=Ro+K(W-We) 我们可以获得每站比赛结束后各位选手的等级分排名。Matlab源程序如下： A=[0.000,0.333,0.531.0.656,0.611,0.674,0.559.0.580: 0.667,0.000,0.542,0.607,0.633,0.552,0.650,0.582: 0.469,0.458,0.000,0.484,0.591,0.442,0.688,0.500; 0.344,0.393,0.516,0.000,0.500,0.545,0.500,0.548: 0.389,0.367,0.409,0.500,0.000,0.429,0.500,0.528: 0.326,0.448,0.558,0.455,0.571,0.000,0.563,0.518: 0.441,0.350,0.313,0.500,0.500,0.438,0.000,0.467: 0.420,0.418.0.500,0.452,0.472,0.482,0.533,0.000]: x=randperm(8) y=[4]: for gj=1:8 op(gj)=0 end forg=1:10000 for i=1:4 m=rand; if(m=A(x(2*i-1),x(2*i))) y(i)=x(2*i-1); else y(i)=x(2*i): end end y z=[2]; for i=1:2 n=rand; if(n<=A(y(2*i-1),y(2*i))) z(i)=y(2*i-1); else z(i)=y(2*i); end 73.3 Matlab[ ·Ç^Matlab^á[—åUÈ ú¹ßø䂺ëV«[—¸’'m'm( J"du§kÎm ¿Ã?©—32400©±˛ßKäå±⁄ò10"·Çr?© Âx⁄WeäâòeÇ5[‹ ￾B ßÇÉm'X™ H = 10 × ( x 650 + 0.5) L = 10 × (0.5 − x 650 ) D = H − L = 2 65 x ⁄úëûßW = 0ßáÉW = 1"|^?©Oé˙™ R1 = R0 + K(W − We) ·Ç屺z’'m(Â￾à†¿Ã?©¸¶"Matlab ßSXeµ A= [ 0 . 0 0 0 , 0 . 3 3 3 , 0 . 5 3 1 , 0 . 6 5 6 , 0 . 6 1 1 , 0 . 6 7 4 , 0 . 5 5 9 , 0 . 5 8 0 ; 0 . 6 6 7 , 0 . 0 0 0 , 0 . 5 4 2 , 0 . 6 0 7 , 0 . 6 3 3 , 0 . 5 5 2 , 0 . 6 5 0 , 0 . 5 8 2 ; 0 . 4 6 9 , 0 . 4 5 8 , 0 . 0 0 0 , 0 . 4 8 4 , 0 . 5 9 1 , 0 . 4 4 2 , 0 . 6 8 8 , 0 . 5 0 0 ; 0 . 3 4 4 , 0 . 3 9 3 , 0 . 5 1 6 , 0 . 0 0 0 , 0 . 5 0 0 , 0 . 5 4 5 , 0 . 5 0 0 , 0 . 5 4 8 ; 0 . 3 8 9 , 0 . 3 6 7 , 0 . 4 0 9 , 0 . 5 0 0 , 0 . 0 0 0 , 0 . 4 2 9 , 0 . 5 0 0 , 0 . 5 2 8 ; 0 . 3 2 6 , 0 . 4 4 8 , 0 . 5 5 8 , 0 . 4 5 5 , 0 . 5 7 1 , 0 . 0 0 0 , 0 . 5 6 3 , 0 . 5 1 8 ; 0 . 4 4 1 , 0 . 3 5 0 , 0 . 3 1 3 , 0 . 5 0 0 , 0 . 5 0 0 , 0 . 4 3 8 , 0 . 0 0 0 , 0 . 4 6 7 ; 0 . 4 2 0 , 0 . 4 1 8 , 0 . 5 0 0 , 0 . 4 5 2 , 0 . 4 7 2 , 0 . 4 8 2 , 0 . 5 3 3 , 0 . 0 0 0 ] ; x=randperm ( 8 ) y = [ 4 ] ; f o r g j = 1: 8 op ( g j ) =0 end f o r g = 1: 1 0 0 0 0 f o r i = 1: 4 m=rand ; i f (m<=A( x ( 2∗ i −1) , x ( 2∗ i ) ) ) y ( i ) =x ( 2∗ i −1) ; e l s e y ( i ) =x ( 2∗ i ) ; end end y z = [ 2 ] ; f o r i = 1: 2 n=rand ; i f ( n<=A( y ( 2∗ i −1) , y ( 2∗ i ) ) ) z ( i ) =y ( 2∗ i −1) ; e l s e z ( i ) =y ( 2∗ i ) ; end 7