Fix solve_system function

[A-Lounsbury]

Test it on games that have verifiable mixed strategy equilibria.

It currently seems to always return empty sets for some reason, even though I’ve printed the polynomials to verify that they’re correct.

Maybe implement this?

[A-Lounsbury]

We can get it to solve for some variables (how it decides which ones to solve for, I do not know). This is the output for the following game

game1 = [
    [
        [[-3, -3, 2], [0, 5, 4]],
        [[-5, 0, 1], [-1, -1, 1]]
    ],
    [
        [[0, 0, 1], [5, 1, 3]],
        [[1, 5, 0], [6, 6, 2]]
    ]
]

[{b_0: 3/2 - 1/c_0}]
[{c_0: 2}]
[{a_0: 3*(b_0 - 1)/(3*b_0 - 4)}]

I just don't know how to tell sympy to do the whole computation, plugging in $c = 2$ into the first equation, and then $b = 1$ into the third equation to give $a = 0$, or if that's even possible.

[A-Lounsbury]

This issue is misnamed now that we're working on the solve_system function. Renaming.

[A-Lounsbury]

Where is {c_0:2} coming from? These are supposed to be probabilities...

[A-Lounsbury]

These are our polynomials for game2:

1, 2, 1   7, 4, 2 
5, 6, 3   3, 8, 1     

2, 4, 0   6, 8, 1     
10, 12, 3   14, 16, 4 
b_0*c_0 + 2*b_0*(1 - c_0) + c_0*(7 - 7*b_0) + (1 - c_0)*(6 - 6*b_0)
5*b_0*c_0 + 10*b_0*(1 - c_0) + c_0*(3 - 3*b_0) + (1 - c_0)*(14 - 14*b_0)

2*a_0*c_0 + 4*a_0*(1 - c_0) + c_0*(6 - 6*a_0) + (1 - c_0)*(12 - 12*a_0) 
4*a_0*c_0 + 8*a_0*(1 - c_0) + c_0*(8 - 8*a_0) + (1 - c_0)*(16 - 16*a_0) 

a_0*b_0 + 2*a_0*(1 - b_0) + b_0*(3 - 3*a_0) + (1 - a_0)*(1 - b_0)       
a_0*(1 - b_0) + b_0*(3 - 3*a_0) + (1 - b_0)*(4 - 4*a_0)

They appear to be correct.

[A-Lounsbury]

Unsimplified differences:

-4*b_0*c_0 - 8*b_0*(1 - c_0) - c_0*(3 - 3*b_0) + c_0*(7 - 7*b_0) + (1 - c_0)*(6 - 6*b_0) - (1 - c_0)*(14 - 14*b_0)
-2*a_0*c_0 - 4*a_0*(1 - c_0) + c_0*(6 - 6*a_0) - c_0*(8 - 8*a_0) + (1 - c_0)*(12 - 12*a_0) - (1 - c_0)*(16 - 16*a_0)
a_0*b_0 + a_0*(1 - b_0) + (1 - a_0)*(1 - b_0) - (1 - b_0)*(4 - 4*a_0)

Simplified differences:

-8*b_0*c_0 + 12*c_0 - 8
2*c_0 - 4
-3*a_0*b_0 + 4*a_0 + 3*b_0 - 3

I really have no idea.

[A-Lounsbury]

That's game2. Here are the differences for game1:

-3, -3, 2   0, 5, 4 
-5, 0, 1   -1, -1, 1 

0, 0, 1   5, 1, 3    
1, 5, 0   6, 6, 2 

b_0*c_0 + 2*c_0 - 1
-9*a_0*c_0 + 2*c_0 - 1
-2*a_0*b_0 + 2*a_0 + 2*b_0 - 1

[A-Lounsbury]

The coefficients for the first two polynomials for players 1 and 2 in game1 happen to be the same

-3, -3, 2   0, 5, 4 
-5, 0, 1   -1, -1, 1 

0, 0, 1   5, 1, 3    
1, 5, 0   6, 6, 2    
PLAYER 1:
                        COEF: -3
                        COEF: 0 
                        COEF: 0 
                        COEF: 5 

                        COEF: -5
                        COEF: -1
                        COEF: 1 
                        COEF: 6 

PLAYER 2:
                        COEF: -3
                        COEF: 0 
                        COEF: 0 
                        COEF: 5 

                        COEF: 5
                        COEF: -1
                        COEF: 1
                        COEF: 6

POLYNOMIALS:
-3*b_0*c_0 + (1 - c_0)*(5 - 5*b_0)
-5*b_0*c_0 + b_0*(1 - c_0) + c_0*(b_0 - 1) + (1 - c_0)*(6 - 6*b_0)

-3*a_0*c_0 + (1 - c_0)*(5 - 5*a_0)
5*a_0*c_0 + a_0*(1 - c_0) + c_0*(a_0 - 1) + (1 - c_0)*(6 - 6*a_0)

2*a_0*b_0 + 4*a_0*(1 - b_0) + b_0*(1 - a_0) + (1 - a_0)*(1 - b_0)
a_0*b_0 + 3*a_0*(1 - b_0) + (1 - b_0)*(2 - 2*a_0)