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%
% Example of computation of minimax (and Nash) strategies for
% bimatrix games. Note that it only computes one minimax
% solution. How would you modify the program to have it compute
% all the minimax solutions?
%
% Author: K. Passino
% Version: 2/5/02
%
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clear all
% Set the number of different possible values of the decision variables
m=5; % If change will need to modify specific J1 value chosen below
n=3;
% Set the payoff matrices J1(i,j) and J2(i,j):
% One set of payoff matrices that gives two Nash equilibria (1,3) and (4,1) is as follows.
% Note that for these payoff matrices the minimax strategy is not a Nash solution.
J1 =[-1 5 -3;
-2 5 1;
4 3 -2;
-5 -1 5;
3 0 2]
J2 =[-1 2 -3;
2 -3 1;
-2 3 1;
-1 1 -1;
4 -4 1]
% For generating random bimatrix games:
J1=round(10*rand(m,n)-5*ones(m,n)) % Make it random integers between -5 and +5
J2=round(10*rand(m,n)-5*ones(m,n)) % Make it random integers between -5 and +5
% These can be used to find other equilibria. For instance, if you use the following two
% payoff matrices then you will get one (admissible) Nash equilibrium which is then the
% minimax strategy also. This shows that for some games the minimax strategy
% is quite good.
J1 =[4 1 -4;
-2 5 3;
-3 2 -1;
4 4 4;
-3 -5 2]
J2 =[2 -1 0;
-2 4 0;
-3 0 -1;
-3 3 4;
-3 0 -5]
% Compute the minimax strategy (the name "minimax" clearly arises from the computations involved in
% each player trying to minimize its maximum loss):
% First, compute the security strategy for player 1
[maxvals1,indexmax1]=max(J1'); % This is the max value for each row (note transpose)
[minimaxvalue1,secstratP1]=min(maxvals1) % Display the security value of P1 and its security strategy
% (P1 loses no more than minimaxvalue1, no matter what strategy
% P2 uses)
% Second, compute the security strategy for player 2 (note difference from computation of security
% strategies for matrix games since both payoff matrices are "loss" matrices, and note that computation of the
% minimax strategies is completely independent)
[maxvals2,indexmax2]=max(J2); % This is the min value for each column (note no transpose)
[minimaxvalue2,secstratP2]=min(maxvals2) % Display the security value of P2 and its security strategy
% The outcome of the game will be (with a minimax strategy)
outcome1=J1(secstratP1,secstratP2)
outcome2=J2(secstratP1,secstratP2)
% Compute the Nash equilibria:
flag=0; % Flag for saying if there is no Nash equilibria
for i=1:m
for j=1:n
if J1(i,j)<=min(J1(:,j)) & J2(i,j)<=min(J2(i,:)), % Conduct two inequality tests
display('Nash equilibrium and outcome:') % If satisfied, then diplay solution
i
j
J1(i,j)
J2(i,j)
flag=1; % Indicates that there was one Nash equilibrium (or more)
end
end
end
if flag==0
display('There were no Nash equilibria')
end
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% Do another example:
clear all
% Define payoff (cost) functions for each player
theta1=-4:0.05:4; % Ok, while we think of it as an infinite game computationally
% we of course only study a finite number of points.
m=length(theta1);
theta2=-5:0.05:5;
n=length(theta2);
% A set of cost functions, J1 for player 1 and J2 for player 2
for ii=1:length(theta1)
for jj=1:length(theta2)
J1(ii,jj)=-2*(exp( (-(theta1(ii)-2)^2)/5 +(-4*(theta1(ii)*theta2(jj))/20) + ((-(theta2(jj)-3)^2)/2)));
J2(ii,jj)=-1*(exp( (-(theta1(ii)-1)^2)/4 + (5*(theta1(ii)*theta2(jj))/10) + ((-(theta2(jj)+1)^2)/2)));
end
end
% Next, compute the reaction curves for each player and will plot on top of J1 and J2
for j=1:length(theta2)
[temp,t1]=min(J1(:,j)); % Find the min point on J1 with theta2 fixed, for all theta2
R1(j)=theta1(t1); % Compute the theta1 value that is the best reaction to each theta2
end
for i=1:length(theta1)
[temp,t2]=min(J2(i,:)); % Find the min point on J2 with theta1 fixed
R2(i)=theta2(t2); % Compute the theta2 value that is the best reaction to each theta1
end
% Compute the minimax strategy (the name "minimax" clearly arises from the computations involved in
% each player trying to minimize its maximum loss):
% First, compute the security strategy for player 1
[maxvals1,indexmax1]=max(J1'); % This is the max value for each row (note transpose)
[minimaxvalue1,secstratP1]=min(maxvals1) % Display the security value of P1 and its security strategy
% (P1 loses no more than minimaxvalue1, no matter what strategy
% P2 uses)
theta1(secstratP1)
% Second, compute the security strategy for player 2 (note difference from computation of security
% strategies for matrix games since both payoff matrices are "loss" matrices, and note that computation of the
% minimax strategies is completely independent)
[maxvals2,indexmax2]=max(J2); % This is the min value for each column (note no transpose)
[minimaxvalue2,secstratP2]=min(maxvals2) % Display the security value of P2 and its security strategy
theta2(secstratP2)
% The outcome of the game will be (with a minimax strategy)
outcome1=J1(secstratP1,secstratP2)
outcome2=J2(secstratP1,secstratP2)
figure(1)
clf
contour(theta2,theta1,J1,10)
hold on
contour(theta2,theta1,J2,10)
hold on
plot(theta2,R1,'k-')
hold on
plot(R2,theta1,'k--')
hold on
plot(theta2(secstratP2),theta1(secstratP1),'x')
xlabel('\theta_2')
ylabel('\theta_1')
title('J_1, J_2, R_1 (-), R_2 (--), "x" marks a minimax solution')
hold off
%-------------------------------------
% End of program
%-------------------------------------