计算纳什均衡

最新推荐文章于 2023-05-05 21:13:33 发布

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最新推荐文章于 2023-05-05 21:13:33 发布

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文章标签：算法

本文链接：https://blog.youkuaiyun.com/int_main_Roland/article/details/129848189

EXAMPLE1

[
	[0, 2, -1, 1],
	[-2, 0, 3, 1],
	[1, -3, 0, 1],
]

双人零和对称博弈( $M = -M^T$ )的所有纳什均衡都是对称纳什均衡.
$max⁡pmin⁡qpMqT=min⁡qmax⁡ppMqT=pMpT=−pMpT=0\max\limits_p \min\limits_q pMq^T = \min\limits_q \max\limits_p pMq^T = pMp^T = -pMp^T = 0$

max min 问题如下:

${max⁡p,vvpM≽v1p≽0p1T=1\begin{aligned} & \max\limits_p \min\limits_q pMq^T \\ \iff& \max\limits_p \min\{pM\} \\ \iff& \begin{cases} \max\limits_{p,v} v \\ pM \succcurlyeq v1 \\ p \succcurlyeq 0 \\ p1^T = 1 \\ \end{cases} \\ \end{aligned}$

${min⁡p,v−v2p1−p2+v⩽0−2p0+3p2+v⩽0p0−3p1+v⩽0−p0⩽0−p1⩽0−p2⩽0p0+p1+p2=1\iff \begin{cases} & \min\limits_{p,v} -v \\ & \begin{matrix} & 2p_1 & -p_2 & +v &\leqslant &0 \\ -2p_0 & & +3p_2 & +v &\leqslant &0 \\ p_0 & -3p_1 & & +v &\leqslant &0 \\ -p_0 & & & &\leqslant &0 \\ & -p_1 & & &\leqslant &0 \\ & & -p_2 & &\leqslant &0 \\ p_0 & +p_1 & +p_2 & &= &1 \\ \end{matrix} \\ \end{cases}$
max min 问题解得 $p∗=[12,16,13]p^*=[\frac{1}{2}, \frac{1}{6}, \frac{1}{3}]$ , $v^*=0$ .
min max 问题如下:

${min⁡q,uuqMT≼u1q≽0q1T=1\begin{aligned} & \min\limits_q \max\limits_p qM^Tp^T \\ \iff& \min\limits_q \max\{qM^T\} \\ \iff& \begin{cases} \min\limits_{q, u} u \\ qM^T \preccurlyeq u1 \\ q \succcurlyeq 0 \\ q1^T = 1 \\ \end{cases} \\ \end{aligned}$

${min⁡q,uu2q1−q2−u⩽0−2q0+3q2−u⩽0q0−3q1−u⩽0−q0⩽0−q1⩽0−q2⩽0q0+q1+q2=1\iff \begin{cases} & \min\limits_{q,u} u \\ & \begin{matrix} & 2q_1 & -q_2 & -u &\leqslant &0 \\ -2q_0 & & +3q_2 & -u &\leqslant &0 \\ q_0 & -3q_1 & & -u &\leqslant &0 \\ -q_0 & & & &\leqslant &0 \\ & -q_1 & & &\leqslant &0 \\ & & -q_2 & &\leqslant &0 \\ q_0 & +q_1 & +q_2 & &= &1 \\ \end{matrix} \\ \end{cases}$
min max 问题解得 $q∗=[12,16,13]q^*=[\frac{1}{2}, \frac{1}{6}, \frac{1}{3}]$ , $u^*=0$ .
结论:
计算机求解结果满足 $v^*=u^*$ , 因此 $p^*,q^*)$ 是一个混合策略纳什均衡.
计算机求解结果满足 $p^*=q^*$ , $v^*=u^*=0$ , 与理论分析一致.

具体代码如下:

# https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.linprog.html

import numpy
import scipy

problem_p_v = {
    'c': numpy.array([[0, 0, 0, -1]]),
    'A_ub': numpy.array([
        [0, 2, -1, 1],
        [-2, 0, 3, 1],
        [1, -3, 0, 1],
        [-1, 0, 0, 0],
        [0, -1, 0, 0],
        [0, 0, -1, 0],
    ]),
    'b_ub': numpy.array([
        [0],
        [0],
        [0],
        [0],
        [0],
        [0],
    ]),
    'A_eq': numpy.array([[1, 1, 1, 0]]),
    'b_eq': numpy.array([[1]]),
}
result_p_v = scipy.optimize.linprog(**problem_p_v, bounds=(None, None))
print(result_p_v)

problem_q_u = {
    'c': numpy.array([[0, 0, 0, 1]]),
    'A_ub': numpy.array([
        [0, 2, -1, -1],
        [-2, 0, 3, -1],
        [1, -3, 0, -1],
        [-1, 0, 0, 0],
        [0, -1, 0, 0],
        [0, 0, -1, 0],
    ]),
    'b_ub': numpy.array([
        [0],
        [0],
        [0],
        [0],
        [0],
        [0],
    ]),
    'A_eq': numpy.array([[1, 1, 1, 0]]),
    'b_eq': numpy.array([[1]]),
}
result_q_u = scipy.optimize.linprog(**problem_q_u, bounds=(None, None))
print(result_q_u)

EXAMPLE2

[
    [1, -2, 6, -4],
    [2, -7, 2, 4],
    [-3, 4, -4, -3],
    [-8, 3, -2, 3],
]

max min 问题如下:

${min⁡p,v−v[1−MT⋮1−10⋱⋮−10][pTv]≼0[1⋯10][pTv]=[1]\begin{aligned} & \max\limits_p \min\limits_q pMq^T \\ \iff& \max\limits_p \min\{pM\} \\ \iff& \begin{cases} \max\limits_{p,v} v \\ pM \succcurlyeq v1 \\ p \succcurlyeq 0 \\ p1^T = 1 \\ \end{cases} \\ \iff& \begin{cases} \min\limits_{p,v} -v \\ \begin{bmatrix} & & &1 \\ &-M^T & &\vdots \\ & & &1 \\ -1 & & &0 \\ &\ddots & &\vdots \\ & &-1 &0 \\ \end{bmatrix} \begin{bmatrix} \\ p^T \\ \\ v \end{bmatrix} \preccurlyeq 0 \\ \begin{bmatrix} 1 &\cdots &1 &0 \end{bmatrix} \begin{bmatrix} \\ p^T \\ \\ v \end{bmatrix} = [1] \\ \end{cases} \\ \end{aligned}$
max min 问题解得 $p^*=[0.14,0.35,0.51,0]$ , $v^*=-0.69$ .
min max 问题如下:

${min⁡q,uu[−1M⋮−1−10⋱⋮−10][qTu]≼0[1⋯10][qTu]=[1]\begin{aligned} & \min\limits_q \max\limits_p qM^Tp^T \\ \iff& \min\limits_q \max\{qM^T\} \\ \iff& \begin{cases} \min\limits_{q, u} u \\ qM^T \preccurlyeq u1 \\ q \succcurlyeq 0 \\ q1^T = 1 \\ \end{cases} \\ \iff& \begin{cases} \min\limits_{q,u} u \\ \begin{bmatrix} & & &-1 \\ &M & &\vdots \\ & & &-1 \\ -1 & & &0 \\ &\ddots & &\vdots \\ & &-1 &0 \\ \end{bmatrix} \begin{bmatrix} \\ q^T \\ \\ u \end{bmatrix} \preccurlyeq 0 \\ \begin{bmatrix} 1 &\cdots &1 &0 \end{bmatrix} \begin{bmatrix} \\ q^T \\ \\ u \end{bmatrix} = [1] \\ \end{cases} \\ \end{aligned}$
min max 问题解得 $q^*=[0.53,0.33,0,0.14]$ , $u^*=-0.69$ .
结论:
计算机求解结果满足 $v^*=u^*$ , 因此 $p^*,q^*)$ 是一个混合策略纳什均衡.

具体代码如下:

# https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.linprog.html

import numpy
import scipy

M = numpy.array([
    [1, -2, 6, -4],
    [2, -7, 2, 4],
    [-3, 4, -4, -3],
    [-8, 3, -2, 3],
])

problem_p_v = {
    'c': numpy.concatenate((numpy.zeros((1, 4)), numpy.array([[-1]])), axis=1),
    'A_ub': numpy.concatenate(
        (
            numpy.concatenate((-M.T, numpy.ones((4, 1))), axis=1),
            numpy.concatenate((-numpy.identity(4), numpy.zeros((4, 1))), axis=1),
        ),
        axis=0,
    ),
    'b_ub': numpy.zeros((8, 1)),
    'A_eq': numpy.concatenate((numpy.ones((1, 4)), numpy.array([[0]])), axis=1),
    'b_eq': numpy.array([[1]]),
}
result_p_v = scipy.optimize.linprog(**problem_p_v, bounds=(None, None))
print(result_p_v)

problem_q_u = {
    'c': numpy.concatenate((numpy.zeros((1, 4)), numpy.array([[1]])), axis=1),
    'A_ub': numpy.concatenate(
        (
            numpy.concatenate((M, -numpy.ones((4, 1))), axis=1),
            numpy.concatenate((-numpy.identity(4), numpy.zeros((4, 1))), axis=1),
        ),
        axis=0,
    ),
    'b_ub': numpy.zeros((8, 1)),
    'A_eq': numpy.concatenate((numpy.ones((1, 4)), numpy.array([[0]])), axis=1),
    'b_eq': numpy.array([[1]]),
}
result_q_u = scipy.optimize.linprog(**problem_q_u, bounds=(None, None))
print(result_q_u)