Section 5.2.5: Mixed strategies for matrix games (LP formulation)

% Boyd & Vandenberghe, "Convex Optimization"
% Joëlle Skaf - 08/24/05
%
% Player 1 wishes to choose u to minimize his expected payoff u'Pv, while
% player 2 wishes to choose v to maximize u'Pv, where P is the payoff
% matrix, u and v are the probability distributions of the choices of each
% player (i.e. u>=0, v>=0, sum(u_i)=1, sum(v_i)=1)
% LP formulation:   minimize    t
%                       s.t.    u >=0 , sum(u) = 1, P'*u <= t*1
%                   maximize    t
%                       s.t.    v >=0 , sum(v) = 1, P*v >= t*1

% Input data
randn('state',0);
n = 12;
m = 12;
P = randn(n,m);

% Optimal strategy for Player 1
fprintf(1,'Computing the optimal strategy for player 1 ... ');

cvx_begin
    variables u(n) t1
    minimize ( t1 )
    u >= 0;
    ones(1,n)*u == 1;
    P'*u <= t1*ones(m,1);
cvx_end

fprintf(1,'Done! \n');

% Optimal strategy for Player 2
fprintf(1,'Computing the optimal strategy for player 2 ... ');

cvx_begin
    variables v(m) t2
    maximize ( t2 )
    v >= 0;
    ones(1,m)*v == 1;
    P*v >= t2*ones(n,1);
cvx_end

fprintf(1,'Done! \n');

% Displaying results
disp('------------------------------------------------------------------------');
disp('The optimal strategies for players 1 and 2 are respectively: ');
disp([u v]);
disp('The expected payoffs for player 1 and player 2 respectively are: ');
[t1 t2]
disp('They are equal as expected!');
Computing the optimal strategy for player 1 ...  
Calling SDPT3: 25 variables, 13 equality constraints
------------------------------------------------------------

 num. of constraints = 13
 dim. of linear var  = 24
 dim. of free   var  =  1 *** convert ublk to lblk
*******************************************************************
   SDPT3: Infeasible path-following algorithms
*******************************************************************
 version  predcorr  gam  expon  scale_data
    NT      1      0.000   1        0    
it pstep dstep pinfeas dinfeas  gap      mean(obj)   cputime
-------------------------------------------------------------------
 0|0.000|0.000|3.6e+01|1.0e+01|6.0e+02| 1.437979e-10| 0:0:00| chol  1  1 
 1|0.917|0.853|3.0e+00|1.6e+00|8.6e+01| 5.803096e-02| 0:0:00| chol  1  1 
 2|1.000|0.986|8.8e-07|3.3e-02|8.7e+00| 2.061331e-01| 0:0:00| chol  1  1 
 3|1.000|0.935|1.7e-06|3.1e-03|2.4e+00|-1.806437e-01| 0:0:00| chol  1  1 
 4|0.870|0.665|4.1e-06|1.1e-03|6.9e-01|-2.077128e-01| 0:0:00| chol  1  1 
 5|1.000|0.088|1.9e-06|1.0e-03|6.3e-01|-2.050085e-01| 0:0:00| chol  1  1 
 6|1.000|0.554|6.9e-07|4.6e-04|3.5e-01|-1.112454e-01| 0:0:00| chol  1  1 
 7|0.985|0.650|3.2e-07|1.6e-04|1.1e-01|-7.743436e-02| 0:0:00| chol  1  1 
 8|0.924|0.351|7.4e-08|1.1e-04|6.6e-02|-7.001485e-02| 0:0:00| chol  1  1 
 9|1.000|0.428|3.1e-08|6.0e-05|3.7e-02|-5.993379e-02| 0:0:00| chol  1  1 
10|1.000|0.497|5.8e-09|3.0e-05|1.9e-02|-5.270502e-02| 0:0:00| chol  1  1 
11|1.000|0.414|1.5e-09|1.8e-05|1.1e-02|-4.968381e-02| 0:0:00| chol  1  1 
12|0.983|0.858|4.5e-10|2.5e-06|1.5e-03|-4.555801e-02| 0:0:00| chol  1  1 
13|0.988|0.985|2.5e-11|1.7e-05|2.5e-05|-4.485120e-02| 0:0:00| chol  1  1 
14|1.000|0.989|3.4e-14|2.8e-07|5.3e-07|-4.484054e-02| 0:0:00| chol  1  1 
15|1.000|0.989|1.1e-14|5.9e-09|9.8e-09|-4.484042e-02| 0:0:00|
  stop: max(relative gap, infeasibilities) < 1.49e-08
-------------------------------------------------------------------
 number of iterations   = 15
 primal objective value = -4.48404196e-02
 dual   objective value = -4.48404289e-02
 gap := trace(XZ)       = 9.85e-09
 relative gap           = 9.04e-09
 actual relative gap    = 8.60e-09
 rel. primal infeas     = 1.14e-14
 rel. dual   infeas     = 5.92e-09
 norm(X), norm(y), norm(Z) = 9.7e-01, 4.0e-01, 7.4e-01
 norm(A), norm(b), norm(C) = 1.4e+01, 2.0e+00, 2.4e+00
 Total CPU time (secs)  = 0.3  
 CPU time per iteration = 0.0  
 termination code       =  0
 DIMACS: 1.1e-14  0.0e+00  7.1e-09  0.0e+00  8.6e-09  9.0e-09
-------------------------------------------------------------------
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): -0.0448404
Done! 
Computing the optimal strategy for player 2 ...  
Calling SDPT3: 25 variables, 13 equality constraints
------------------------------------------------------------

 num. of constraints = 13
 dim. of linear var  = 24
 dim. of free   var  =  1 *** convert ublk to lblk
*******************************************************************
   SDPT3: Infeasible path-following algorithms
*******************************************************************
 version  predcorr  gam  expon  scale_data
    NT      1      0.000   1        0    
it pstep dstep pinfeas dinfeas  gap      mean(obj)   cputime
-------------------------------------------------------------------
 0|0.000|0.000|3.8e+01|1.0e+01|6.0e+02|-1.437979e-10| 0:0:00| chol  1  1 
 1|0.915|0.847|3.2e+00|1.6e+00|8.8e+01| 1.352019e-01| 0:0:00| chol  1  1 
 2|1.000|0.986|9.2e-07|3.3e-02|8.8e+00| 2.751661e-01| 0:0:00| chol  1  1 
 3|1.000|0.899|1.7e-06|4.2e-03|2.5e+00|-1.502898e-01| 0:0:00| chol  1  1 
 4|0.887|0.352|4.3e-06|2.8e-03|1.3e+00|-2.907480e-01| 0:0:00| chol  1  1 
 5|1.000|0.227|4.4e-06|2.2e-03|1.1e+00|-2.153276e-01| 0:0:00| chol  1  1 
 6|1.000|0.596|1.5e-06|8.7e-04|4.1e-01|-9.555833e-02| 0:0:00| chol  1  1 
 7|1.000|0.277|2.2e-07|6.3e-04|3.1e-01|-5.906355e-02| 0:0:00| chol  1  1 
 8|1.000|0.371|1.2e-07|4.0e-04|2.0e-01|-2.566743e-02| 0:0:00| chol  1  1 
 9|1.000|0.499|3.8e-08|2.0e-04|9.9e-02| 5.671373e-03| 0:0:00| chol  1  1 
10|1.000|0.661|7.0e-09|6.7e-05|3.2e-02| 3.014456e-02| 0:0:00| chol  1  1 
11|1.000|0.811|6.9e-10|1.3e-05|6.0e-03| 4.196714e-02| 0:0:00| chol  1  1 
12|0.989|0.970|6.0e-11|3.8e-07|1.8e-04| 4.475275e-02| 0:0:00| chol  1  1 
13|0.989|0.988|3.6e-12|2.0e-06|2.3e-06| 4.483941e-02| 0:0:00| chol  1  1 
14|1.000|0.989|1.9e-14|2.5e-08|4.7e-08| 4.484041e-02| 0:0:00| chol  1  1 
15|1.000|0.989|5.6e-16|5.3e-10|8.9e-10| 4.484042e-02| 0:0:00|
  stop: max(relative gap, infeasibilities) < 1.49e-08
-------------------------------------------------------------------
 number of iterations   = 15
 primal objective value =  4.48404225e-02
 dual   objective value =  4.48404216e-02
 gap := trace(XZ)       = 8.93e-10
 relative gap           = 8.20e-10
 actual relative gap    = 7.80e-10
 rel. primal infeas     = 5.65e-16
 rel. dual   infeas     = 5.35e-10
 norm(X), norm(y), norm(Z) = 7.4e-01, 4.4e-01, 9.7e-01
 norm(A), norm(b), norm(C) = 1.4e+01, 2.0e+00, 2.4e+00
 Total CPU time (secs)  = 0.2  
 CPU time per iteration = 0.0  
 termination code       =  0
 DIMACS: 5.6e-16  0.0e+00  6.5e-10  0.0e+00  7.8e-10  8.2e-10
-------------------------------------------------------------------
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): -0.0448404
Done! 
------------------------------------------------------------------------
The optimal strategies for players 1 and 2 are respectively: 
    0.2695    0.0686
    0.0000    0.1619
    0.0973    0.0000
    0.1573    0.2000
    0.1145    0.0000
    0.0434    0.1545
    0.0000    0.1146
    0.0000    0.0000
    0.2511    0.1030
    0.0670    0.0000
    0.0000    0.0000
    0.0000    0.1974

The expected payoffs for player 1 and player 2 respectively are: 

ans =

   -0.0448   -0.0448

They are equal as expected!