Finding Nash Equilibrium using Linear Program with strategy constraints












1












$begingroup$


Finding the Nash Equilibrium p in mixed strategies of a 2-player, symmetric zero-sum game with 3 pure strategies can be done by solving LP:



max $(0, 0, 0, 1)^{T}(p_1, p_2, p_3, epsilon)$



s.t. $A geq b$



Where A is:



begin{bmatrix}
a_{11} & a_{12} & a_{13} & -1 \
a_{21} & a_{22} & a_{23} & -1 \
a_{31} & a_{32} & a_{33} & -1 \
end{bmatrix}



and $b = (0,0,0)$



i.e. We maximize our worse payoff.



Now suppose player 2, who plays $q = (q_1, q_2, q_3)$ is under the constraint $q_1 + q_2 = c$, for some $c$ $0 leq c leq 1$.



Can this be solved? Thanks.










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$endgroup$












  • $begingroup$
    you should provide some background info about how the linear optimization problem solves "We maximize our worse payoff." or how $q$ can appear
    $endgroup$
    – LinAlg
    Dec 22 '18 at 13:16












  • $begingroup$
    @LinAlg q is the opponent's strategy. It is not needed in the original "unconstrained" problem. For more information on the formulation, page 38 of this document might help: math.ucla.edu/~tom/Game_Theory/mat.pdf
    $endgroup$
    – Agrim Pathak
    Dec 23 '18 at 9:04










  • $begingroup$
    Hi! Did you eventually figure out what are conditions for such a constrained optimal mixed strategy to exist?
    $endgroup$
    – Ben Usman
    Mar 5 at 14:35










  • $begingroup$
    @BenUsman Hi, unfortunately I made no progress on this. There are however, iterative algorithms which can converge to NE.
    $endgroup$
    – Agrim Pathak
    Mar 7 at 4:47
















1












$begingroup$


Finding the Nash Equilibrium p in mixed strategies of a 2-player, symmetric zero-sum game with 3 pure strategies can be done by solving LP:



max $(0, 0, 0, 1)^{T}(p_1, p_2, p_3, epsilon)$



s.t. $A geq b$



Where A is:



begin{bmatrix}
a_{11} & a_{12} & a_{13} & -1 \
a_{21} & a_{22} & a_{23} & -1 \
a_{31} & a_{32} & a_{33} & -1 \
end{bmatrix}



and $b = (0,0,0)$



i.e. We maximize our worse payoff.



Now suppose player 2, who plays $q = (q_1, q_2, q_3)$ is under the constraint $q_1 + q_2 = c$, for some $c$ $0 leq c leq 1$.



Can this be solved? Thanks.










share|cite|improve this question











$endgroup$












  • $begingroup$
    you should provide some background info about how the linear optimization problem solves "We maximize our worse payoff." or how $q$ can appear
    $endgroup$
    – LinAlg
    Dec 22 '18 at 13:16












  • $begingroup$
    @LinAlg q is the opponent's strategy. It is not needed in the original "unconstrained" problem. For more information on the formulation, page 38 of this document might help: math.ucla.edu/~tom/Game_Theory/mat.pdf
    $endgroup$
    – Agrim Pathak
    Dec 23 '18 at 9:04










  • $begingroup$
    Hi! Did you eventually figure out what are conditions for such a constrained optimal mixed strategy to exist?
    $endgroup$
    – Ben Usman
    Mar 5 at 14:35










  • $begingroup$
    @BenUsman Hi, unfortunately I made no progress on this. There are however, iterative algorithms which can converge to NE.
    $endgroup$
    – Agrim Pathak
    Mar 7 at 4:47














1












1








1


1



$begingroup$


Finding the Nash Equilibrium p in mixed strategies of a 2-player, symmetric zero-sum game with 3 pure strategies can be done by solving LP:



max $(0, 0, 0, 1)^{T}(p_1, p_2, p_3, epsilon)$



s.t. $A geq b$



Where A is:



begin{bmatrix}
a_{11} & a_{12} & a_{13} & -1 \
a_{21} & a_{22} & a_{23} & -1 \
a_{31} & a_{32} & a_{33} & -1 \
end{bmatrix}



and $b = (0,0,0)$



i.e. We maximize our worse payoff.



Now suppose player 2, who plays $q = (q_1, q_2, q_3)$ is under the constraint $q_1 + q_2 = c$, for some $c$ $0 leq c leq 1$.



Can this be solved? Thanks.










share|cite|improve this question











$endgroup$




Finding the Nash Equilibrium p in mixed strategies of a 2-player, symmetric zero-sum game with 3 pure strategies can be done by solving LP:



max $(0, 0, 0, 1)^{T}(p_1, p_2, p_3, epsilon)$



s.t. $A geq b$



Where A is:



begin{bmatrix}
a_{11} & a_{12} & a_{13} & -1 \
a_{21} & a_{22} & a_{23} & -1 \
a_{31} & a_{32} & a_{33} & -1 \
end{bmatrix}



and $b = (0,0,0)$



i.e. We maximize our worse payoff.



Now suppose player 2, who plays $q = (q_1, q_2, q_3)$ is under the constraint $q_1 + q_2 = c$, for some $c$ $0 leq c leq 1$.



Can this be solved? Thanks.







optimization linear-programming game-theory nash-equilibrium






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 0:54







Agrim Pathak

















asked Dec 22 '18 at 0:48









Agrim PathakAgrim Pathak

1113




1113












  • $begingroup$
    you should provide some background info about how the linear optimization problem solves "We maximize our worse payoff." or how $q$ can appear
    $endgroup$
    – LinAlg
    Dec 22 '18 at 13:16












  • $begingroup$
    @LinAlg q is the opponent's strategy. It is not needed in the original "unconstrained" problem. For more information on the formulation, page 38 of this document might help: math.ucla.edu/~tom/Game_Theory/mat.pdf
    $endgroup$
    – Agrim Pathak
    Dec 23 '18 at 9:04










  • $begingroup$
    Hi! Did you eventually figure out what are conditions for such a constrained optimal mixed strategy to exist?
    $endgroup$
    – Ben Usman
    Mar 5 at 14:35










  • $begingroup$
    @BenUsman Hi, unfortunately I made no progress on this. There are however, iterative algorithms which can converge to NE.
    $endgroup$
    – Agrim Pathak
    Mar 7 at 4:47


















  • $begingroup$
    you should provide some background info about how the linear optimization problem solves "We maximize our worse payoff." or how $q$ can appear
    $endgroup$
    – LinAlg
    Dec 22 '18 at 13:16












  • $begingroup$
    @LinAlg q is the opponent's strategy. It is not needed in the original "unconstrained" problem. For more information on the formulation, page 38 of this document might help: math.ucla.edu/~tom/Game_Theory/mat.pdf
    $endgroup$
    – Agrim Pathak
    Dec 23 '18 at 9:04










  • $begingroup$
    Hi! Did you eventually figure out what are conditions for such a constrained optimal mixed strategy to exist?
    $endgroup$
    – Ben Usman
    Mar 5 at 14:35










  • $begingroup$
    @BenUsman Hi, unfortunately I made no progress on this. There are however, iterative algorithms which can converge to NE.
    $endgroup$
    – Agrim Pathak
    Mar 7 at 4:47
















$begingroup$
you should provide some background info about how the linear optimization problem solves "We maximize our worse payoff." or how $q$ can appear
$endgroup$
– LinAlg
Dec 22 '18 at 13:16






$begingroup$
you should provide some background info about how the linear optimization problem solves "We maximize our worse payoff." or how $q$ can appear
$endgroup$
– LinAlg
Dec 22 '18 at 13:16














$begingroup$
@LinAlg q is the opponent's strategy. It is not needed in the original "unconstrained" problem. For more information on the formulation, page 38 of this document might help: math.ucla.edu/~tom/Game_Theory/mat.pdf
$endgroup$
– Agrim Pathak
Dec 23 '18 at 9:04




$begingroup$
@LinAlg q is the opponent's strategy. It is not needed in the original "unconstrained" problem. For more information on the formulation, page 38 of this document might help: math.ucla.edu/~tom/Game_Theory/mat.pdf
$endgroup$
– Agrim Pathak
Dec 23 '18 at 9:04












$begingroup$
Hi! Did you eventually figure out what are conditions for such a constrained optimal mixed strategy to exist?
$endgroup$
– Ben Usman
Mar 5 at 14:35




$begingroup$
Hi! Did you eventually figure out what are conditions for such a constrained optimal mixed strategy to exist?
$endgroup$
– Ben Usman
Mar 5 at 14:35












$begingroup$
@BenUsman Hi, unfortunately I made no progress on this. There are however, iterative algorithms which can converge to NE.
$endgroup$
– Agrim Pathak
Mar 7 at 4:47




$begingroup$
@BenUsman Hi, unfortunately I made no progress on this. There are however, iterative algorithms which can converge to NE.
$endgroup$
– Agrim Pathak
Mar 7 at 4:47










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