何时是PNE(纯策略纳什均衡)?何时是MNE(混合策略纳什均衡)?

题目: 何时是PNE(纯策略纳什均衡)?何时是MNE(混合策略纳什均衡)?

                 +----------------+----------------+|    L    (q)    |    R  (1 - q)  |
+----------------+----------------+----------------+
|    U    (p)    |     a    c     |     e    g     |
+----------------+----------------+----------------+
|    D  (1 - p)  |     b    d     |     f    h     |
+----------------+----------------+----------------+

注B

∧ \wedge ∧ 是逻辑与
∨ \vee ∨ 是逻辑或
¬ \neg ¬ 是逻辑非

注C

( a − b > 0 ∧ f − e > 0 ) ∨ ( a − b < 0 ∧ f − e < 0 ) (a-b > 0 \wedge f-e > 0) \vee (a-b < 0 \wedge f-e < 0) (a−b>0∧f−e>0)∨(a−b<0∧f−e<0) 时, 一定有 f − e a − b + f − e ∈ ( 0 , 1 ) \frac{f-e}{a-b+f-e} \in (0,1) a−b+f−ef−e∈(0,1).
( c − g > 0 ∧ h − d > 0 ) ∨ ( c − g < 0 ∧ h − d < 0 ) (c-g > 0 \wedge h-d > 0) \vee (c-g < 0 \wedge h-d < 0) (c−g>0∧h−d>0)∨(c−g<0∧h−d<0) 时, 一定有 h − d c − g + h − d ∈ ( 0 , 1 ) \frac{h-d}{c-g+h-d} \in (0,1) c−g+h−dh−d∈(0,1).

收益函数和最优反应如下:

U 1 ( p , q ) = p [ q a + ( 1 − q ) e ] + ( 1 − p ) [ q b + ( 1 − q ) f ] = p [ ( a − b + f − e ) q − ( f − e ) ] + [ q ( b − f ) + f ] \begin{aligned} U_1(p, q) &= p \left[ qa + (1-q)e \right] + (1-p) \left[ qb + (1-q)f \right] \\ &= p \left[ (a-b+f-e)q - (f-e) \right] + \left[ q(b-f) + f \right] \\ \end{aligned} U1(p,q)=p[qa+(1−q)e]+(1−p)[qb+(1−q)f]=p[(a−b+f−e)q−(f−e)]+[q(b−f)+f]

{ p = { 1 , q > [ 0 , 1 ] , q = f − e a − b + f − e 0 , q < , a − b > 0 ∧ f − e > 0 p ≡ 1 , a − b > 0 ∧ f − e < 0 p = { 0 , q > [ 0 , 1 ] , q = f − e a − b + f − e 1 , q < , a − b < 0 ∧ f − e < 0 p ≡ 0 , a − b < 0 ∧ f − e > 0 p = { 0 , q < 1 [ 0 , 1 ] , q = 1 , a − b = 0 ∧ f − e > 0 p = { 1 , q < 1 [ 0 , 1 ] , q = 1 , a − b = 0 ∧ f − e < 0 p = { [ 0 , 1 ] , q = 0 1 , q > 0 , a − b > 0 ∧ f − e = 0 p = { [ 0 , 1 ] , q = 0 0 , q > 0 , a − b < 0 ∧ f − e = 0 p ≡ [ 0 , 1 ] , a − b = 0 ∧ f − e = 0 \begin{cases} &p = \begin{cases} 1, & \phantom{q} > \\ [0,1], & q = \frac{f-e}{a-b+f-e} \\ 0, & \phantom{q} < \\ \end{cases}, && a-b > 0 \wedge f-e > 0 \\ &p \equiv 1, && a-b > 0 \wedge f-e < 0 \\ &p = \begin{cases} 0, & \phantom{q} > \\ [0,1], & q = \frac{f-e}{a-b+f-e} \\ 1, & \phantom{q} < \\ \end{cases}, && a-b < 0 \wedge f-e < 0 \\ &p \equiv 0, && a-b < 0 \wedge f-e > 0 \\ &p = \begin{cases} 0, & q < 1 \\ [0,1], & q = 1 \\ \end{cases}, && a-b = 0 \wedge f-e > 0 \\ &p = \begin{cases} 1, & q < 1 \\ [0,1], & q = 1 \\ \end{cases}, && a-b = 0 \wedge f-e < 0 \\ &p = \begin{cases} [0,1], & q = 0 \\ 1, & q > 0 \\ \end{cases}, && a-b > 0 \wedge f-e = 0 \\ &p = \begin{cases} [0,1], & q = 0 \\ 0, & q > 0 \\ \end{cases}, && a-b < 0 \wedge f-e = 0 \\ &p \equiv [0, 1], && a-b = 0 \wedge f-e = 0 \\ \end{cases} ⎩ ⎨ ⎧p=⎩ ⎨ ⎧1,[0,1],0,q>q=a−b+f−ef−eq<,p≡1,p=⎩ ⎨ ⎧0,[0,1],1,q>q=a−b+f−ef−eq<,p≡0,p={0,[0,1],q<1q=1,p={1,[0,1],q<1q=1,p={[0,1],1,q=0q>0,p={[0,1],0,q=0q>0,p≡[0,1],a−b>0∧f−e>0a−b>0∧f−e<0a−b<0∧f−e<0a−b<0∧f−e>0a−b=0∧f−e>0a−b=0∧f−e<0a−b>0∧f−e=0a−b<0∧f−e=0a−b=0∧f−e=0

U 2 ( q , p ) = q [ p c + ( 1 − p ) d ] + ( 1 − q ) [ p g + ( 1 − p ) h ] = q [ ( c − g + h − d ) p − ( h − d ) ] + [ p ( g − h ) + h ] \begin{aligned} U_2(q, p) &= q \left[ pc + (1-p)d \right] + (1-q) \left[ pg + (1-p)h \right] \\ &= q \left[ (c-g+h-d)p - (h-d) \right] + \left[ p(g-h) + h \right] \\ \end{aligned} U2(q,p)=q[pc+(1−p)d]+(1−q)[pg+(1−p)h]=q[(c−g+h−d)p−(h−d)]+[p(g−h)+h]

{ q = { 1 , p > [ 0 , 1 ] , p = h − d c − g + h − d 0 , p < , c − g > 0 ∧ h − d > 0 q ≡ 1 , c − g > 0 ∧ h − d < 0 q = { 0 , p > [ 0 , 1 ] , p = h − d c − g + h − d 1 , p < , c − g < 0 ∧ h − d < 0 q ≡ 0 , c − g < 0 ∧ h − d > 0 q = { 0 , p < 1 [ 0 , 1 ] , p = 1 , c − g = 0 ∧ h − d > 0 q = { 1 , p < 1 [ 0 , 1 ] , p = 1 , c − g = 0 ∧ h − d < 0 q = { [ 0 , 1 ] , p = 0 1 , p > 0 , c − g > 0 ∧ h − d = 0 q = { [ 0 , 1 ] , p = 0 0 , p > 0 , c − g < 0 ∧ h − d = 0 q ≡ [ 0 , 1 ] , c − g = 0 ∧ h − d = 0 \begin{cases} &q = \begin{cases} 1, & \phantom{p} > \\ [0,1], & p = \frac{h-d}{c-g+h-d} \\ 0, & \phantom{p} < \\ \end{cases}, && c-g > 0 \wedge h-d > 0 \\ &q \equiv 1, && c-g > 0 \wedge h-d < 0 \\ &q = \begin{cases} 0, & \phantom{p} > \\ [0,1], & p = \frac{h-d}{c-g+h-d} \\ 1, & \phantom{p} < \\ \end{cases}, && c-g < 0 \wedge h-d < 0 \\ &q \equiv 0, && c-g < 0 \wedge h-d > 0 \\ &q = \begin{cases} 0, & p < 1 \\ [0,1], & p = 1 \\ \end{cases}, && c-g = 0 \wedge h-d > 0 \\ &q = \begin{cases} 1, & p < 1 \\ [0,1], & p = 1 \\ \end{cases}, && c-g = 0 \wedge h-d < 0 \\ &q = \begin{cases} [0,1], & p = 0 \\ 1, & p > 0 \\ \end{cases}, && c-g > 0 \wedge h-d = 0 \\ &q = \begin{cases} [0,1], & p = 0 \\ 0, & p > 0 \\ \end{cases}, && c-g < 0 \wedge h-d = 0 \\ &q \equiv [0, 1], && c-g = 0 \wedge h-d = 0 \\ \end{cases} ⎩ ⎨ ⎧q=⎩ ⎨ ⎧1,[0,1],0,p>p=c−g+h−dh−dp<,q≡1,q=⎩ ⎨ ⎧0,[0,1],1,p>p=c−g+h−dh−dp<,q≡0,q={0,[0,1],p<1p=1,q={1,[0,1],p<1p=1,q={[0,1],1,p=0p>0,q={[0,1],0,p=0p>0,q≡[0,1],c−g>0∧h−d>0c−g>0∧h−d<0c−g<0∧h−d<0c−g<0∧h−d>0c−g=0∧h−d>0c−g=0∧h−d<0c−g>0∧h−d=0c−g<0∧h−d=0c−g=0∧h−d=0

特别地 ¬ [ ( a − b = 0 ∧ f − e = 0 ) ∨ ( c − g = 0 ∧ h − d = 0 ) ] \neg[(a-b = 0 \wedge f-e = 0) \vee (c-g = 0 \wedge h-d = 0)] ¬[(a−b=0∧f−e=0)∨(c−g=0∧h−d=0)] 时, 可以画图分析如下:

统计PNE个数(即?P)和MNE个数(即?M)如下:

注: ~M意为∞M(无穷个MNE)

        +-------+-------+-------+-------+-------+-------+-------+-------+-------+| a-b>0 | a-b>0 | a-b<0 | a-b<0 | a-b=0 | a-b=0 | a-b>0 | a-b<0 | a-b=0 ||  and  |  and  |  and  |  and  |  and  |  and  |  and  |  and  |  and  || f-e>0 | f-e<0 | f-e<0 | f-e>0 | f-e>0 | f-e<0 | f-e=0 | f-e=0 | f-e=0 |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| c-g>0 |       |       |       |       |       |       |       |       |       |
|  and  | 2P 1M | 1P 0M | 0P 1M | 1P 0M | 2P ~M | 1P ~M | 2P ~M | 1P ~M | 2P ~M |
| h-d>0 |       |       |       |       |       |       |       |       |       |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| c-g>0 |       |       |       |       |       |       |       |       |       |
|  and  | 1P 0M | 1P 0M | 1P 0M | 1P 0M | 2P ~M | 2P ~M | 1P 0M | 1P 0M | 2P ~M |
| h-d<0 |       |       |       |       |       |       |       |       |       |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| c-g<0 |       |       |       |       |       |       |       |       |       |
|  and  | 0P 1M | 1P 0M | 2P 1M | 1P 0M | 1P ~M | 2P ~M | 1P ~M | 2P ~M | 2P ~M |
| h-d<0 |       |       |       |       |       |       |       |       |       |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| c-g<0 |       |       |       |       |       |       |       |       |       |
|  and  | 1P 0M | 1P 0M | 1P 0M | 1P 0M | 1P 0M | 1P 0M | 2P ~M | 2P ~M | 2P ~M |
| h-d>0 |       |       |       |       |       |       |       |       |       |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| c-g=0 |       |       |       |       |       |       |       |       |       |
|  and  | 2P ~M | 2P ~M | 1P ~M | 1P 0M | 2P 0M | 2P ~M | 3P ~M | 2P ~M | 3P ~M |
| h-d>0 |       |       |       |       |       |       |       |       |       |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| c-g=0 |       |       |       |       |       |       |       |       |       |
|  and  | 1P ~M | 2P ~M | 2P ~M | 1P 0M | 2P ~M | 3P ~M | 2P ~M | 2P 0M | 3P ~M |
| h-d<0 |       |       |       |       |       |       |       |       |       |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| c-g>0 |       |       |       |       |       |       |       |       |       |
|  and  | 2P ~M | 1P 0M | 1P ~M | 2P ~M | 3P ~M | 2P ~M | 2P 0M | 2P ~M | 3P ~M |
| h-d=0 |       |       |       |       |       |       |       |       |       |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| c-g<0 |       |       |       |       |       |       |       |       |       |
|  and  | 1P ~M | 1P 0M | 2P ~M | 2P ~M | 2P ~M | 2P 0M | 2P ~M | 3P ~M | 3P ~M |
| h-d=0 |       |       |       |       |       |       |       |       |       |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+
| c-g=0 |       |       |       |       |       |       |       |       |       |
|  and  | 2P ~M | 2P ~M | 2P ~M | 2P ~M | 3P ~M | 3P ~M | 3P ~M | 3P ~M | 4P ~M |
| h-d=0 |       |       |       |       |       |       |       |       |       |
+-------+-------+-------+-------+-------+-------+-------+-------+-------+-------+

分析如下:

不存在PNE

不存在PNE(即0P)仅限于两种情形, 而这两种情形恰好是 “纯策略最优反应” 完全不重合的两种情形. 因此我们得出结论: 只要 “纯策略最优反应” 有重合, 那么就一定存在PNE; 如果 “纯策略最优反应” 不重合, 那么就不存在PNE.

由此我们得出结论: 不存在PNE(只存在MNE)当且仅当 “纯策略最优反应” 完全不重合. 这一命题总是成立.

以下是不存在PNE(即0P)的两种情形, 已经用中括号框出 “纯策略最优反应”.
- ( a − b > 0 ∧ f − e > 0 ) ∧ ( c − g < 0 ∧ h − d < 0 ) (a-b > 0 \wedge f-e > 0) \wedge (c-g < 0 \wedge h-d < 0) (a−b>0∧f−e>0)∧(c−g<0∧h−d<0)
```
                 +----------------+----------------+|    L    (q)    |    R  (1 - q)  |
+----------------+----------------+----------------+
|    U    (p)    |    [a]   c     |     e   [g]    |
+----------------+----------------+----------------+
|    D  (1 - p)  |     b   [d]    |    [f]   h     |
+----------------+----------------+----------------+
```
- ( a − b < 0 ∧ f − e < 0 ) ∧ ( c − g > 0 ∧ h − d > 0 ) (a-b < 0 \wedge f-e < 0) \wedge (c-g > 0 \wedge h-d > 0) (a−b<0∧f−e<0)∧(c−g>0∧h−d>0)
```
                 +----------------+----------------+|    L    (q)    |    R  (1 - q)  |
+----------------+----------------+----------------+
|    U    (p)    |     a   [c]    |    [e]   g     |
+----------------+----------------+----------------+
|    D  (1 - p)  |    [b]   d     |     f   [h]    |
+----------------+----------------+----------------+
```
不存在MNE

不存在MNE(即0M)没有 “覆盖整个概率测度空间” 的规律, 但是可以归纳出一些 “几乎完全覆盖整个概率测度空间” 的规律. 如果我们忽视所有 ? − ? = 0 ?-?=0 ?−?=0 的情形, 所有不存在MNE的情形当且仅当 ( a − b > 0 ∧ f − e < 0 ) ∨ ( a − b < 0 ∧ f − e > 0 ) ∨ ( c − g > 0 ∧ h − d < 0 ) ∨ ( c − g < 0 ∧ h − d > 0 ) (a-b > 0 \wedge f-e < 0) \vee (a-b < 0 \wedge f-e > 0) \vee (c-g > 0 \wedge h-d < 0) \vee (c-g < 0 \wedge h-d > 0) (a−b>0∧f−e<0)∨(a−b<0∧f−e>0)∨(c−g>0∧h−d<0)∨(c−g<0∧h−d>0). 注意到:
- 整个概率测度空间 Δ = Δ p × Δ q \Delta = \Delta_p \times \Delta_q Δ=Δp×Δq 是二维的, 所有 ? − ? = 0 ?-?=0 ?−?=0 的情形相应的概率测度子空间由于等式约束的存在都是一维甚至零维的, 而在二维测度空间中可数个一维或者零维子空间测度之和为零.
- ( a − b > 0 ∧ f − e < 0 ) ∨ ( a − b < 0 ∧ f − e > 0 ) ∨ ( c − g > 0 ∧ h − d < 0 ) ∨ ( c − g < 0 ∧ h − d > 0 ) (a-b > 0 \wedge f-e < 0) \vee (a-b < 0 \wedge f-e > 0) \vee (c-g > 0 \wedge h-d < 0) \vee (c-g < 0 \wedge h-d > 0) (a−b>0∧f−e<0)∨(a−b<0∧f−e>0)∨(c−g>0∧h−d<0)∨(c−g<0∧h−d>0) 的实际含义依次是: U严格占优D; D严格占优U; L严格占优R; R严格占优L.
由此我们得出结论: 不存在MNE(只存在PNE) 当且仅当存在严格占优纯策略. 这一命题几乎总是成立. (这一命题成立的概率子空间的测度 ∣ Δ 成立 ∣ |\Delta_{\text{成立}}| ∣Δ成立∣ 等于整个概率测度空间的测度 ∣ Δ ∣ |\Delta| ∣Δ∣)

以下是忽视所有 ? − ? = 0 ?-?=0 ?−?=0 的情形的子表, 已经用中括号框出不存在MNE(即0M)的情形.
```
        +-------+-------+-------+-------+| a-b>0 | a-b>0 | a-b<0 | a-b<0 ||  and  |  and  |  and  |  and  || f-e>0 | f-e<0 | f-e<0 | f-e>0 |
+-------+-------+-------+-------+-------+
| c-g>0 |       |   ++++|       |   ++++|
|  and  | 2P 1M | 1P[0M]| 0P 1M | 1P[0M]|
| h-d>0 |       |   ++++|       |   ++++|
+-------+-------+-------+-------+-------+
| c-g>0 |   ++++|   ++++|   ++++|   ++++|
|  and  | 1P[0M]| 1P[0M]| 1P[0M]| 1P[0M]|
| h-d<0 |   ++++|   ++++|   ++++|   ++++|
+-------+-------+-------+-------+-------+
| c-g<0 |       |   ++++|       |   ++++|
|  and  | 0P 1M | 1P[0M]| 2P 1M | 1P[0M]|
| h-d<0 |       |   ++++|       |   ++++|
+-------+-------+-------+-------+-------+
| c-g<0 |   ++++|   ++++|   ++++|   ++++|
|  and  | 1P[0M]| 1P[0M]| 1P[0M]| 1P[0M]|
| h-d>0 |   ++++|   ++++|   ++++|   ++++|
+-------+-------+-------+-------+-------+
```
总结如下:
1. 不存在PNE(只存在MNE) 当且仅当 “纯策略最优反应” 完全不重合. 这一命题总是成立.
2. 不存在MNE(只存在PNE) 当且仅当存在严格占优纯策略. 这一命题"几乎总是成立.
3. (推论) “纯策略最优反应” 完全不重合时, 不存在严格占优纯策略; 存在严格占优纯策略时, “纯策略最优反应” 有重合. 这一命题"几乎总是成立.
4. (推论) “纯策略最优反应” 有重合而且不存在严格占优纯策略时, 既存在PNE又存在MNE. 这一命题"几乎总是成立.

附录 - csv格式表格

,a−b>0∧f−e>0,a−b>0∧f−e<0,a−b<0∧f−e<0,a−b<0∧f−e>0,a−b=0∧f−e>0,a−b=0∧f−e<0,a−b>0∧f−e=0,a−b<0∧f−e=0,a−b=0∧f−e=0
c−g>0∧h−d>0,2P1M,1P0M,0P1M,1P0M,2P∞M,1P∞M,2P∞M,1P∞M,2P∞M
c−g>0∧h−d<0,1P0M,1P0M,1P0M,1P0M,2P∞M,2P∞M,1P0M,1P0M,2P∞M
c−g<0∧h−d<0,0P1M,1P0M,2P1M,1P0M,1P∞M,2P∞M,1P∞M,2P∞M,2P∞M
c−g<0∧h−d>0,1P0M,1P0M,1P0M,1P0M,1P0M,1P0M,2P∞M,2P∞M,2P∞M
c−g=0∧h−d>0,2P∞M,2P∞M,1P∞M,1P0M,2P0M,2P∞M,3P∞M,2P∞M,3P∞M
c−g=0∧h−d<0,1P∞M,2P∞M,2P∞M,1P0M,2P∞M,3P∞M,2P∞M,2P0M,3P∞M
c−g>0∧h−d=0,2P∞M,1P0M,1P∞M,2P∞M,3P∞M,2P∞M,2P0M,2P∞M,3P∞M
c−g<0∧h−d=0,1P∞M,1P0M,2P∞M,2P∞M,2P∞M,2P0M,2P∞M,3P∞M,3P∞M
c−g=0∧h−d=0,2P∞M,2P∞M,2P∞M,2P∞M,3P∞M,3P∞M,3P∞M,3P∞M,4P∞M

附录 - 绘图代码

import matplotlib as mpl
import matplotlib.pyplot as plt# Debian bullseye KDE
plt.rcParams['font.sans-serif'] = 'Noto Sans CJK JP'  # SC+TC+HK + JP + KR
# for font in mpl.font_manager.fontManager.ttflist:
#     if font.name.startswith('Noto Sans CJK'):
#         print(font.name,font.fname)A0 = 0
A1 = 1
Ax = 0.5
caseUD = ['a−b>0∧f−e>0','a−b>0∧f−e<0','a−b<0∧f−e<0','a−b<0∧f−e>0','a−b=0∧f−e>0','a−b=0∧f−e<0','a−b>0∧f−e=0','a−b<0∧f−e=0',
]
respUD = {'a−b>0∧f−e>0': ((A0,A0),(Ax,A0),(Ax,A1), (A1,A1)),'a−b>0∧f−e<0': ((A0,A1),(A1,A1)),'a−b<0∧f−e<0': ((A0,A1),(Ax,A1),(Ax,A0), (A1,A0)),'a−b<0∧f−e>0': ((A0,A0),(A1,A0)),'a−b=0∧f−e>0': ((A0,A0),(A1,A0),(A1,A1)),'a−b=0∧f−e<0': ((A0,A1),(A1,A1),(A1,A0)),'a−b>0∧f−e=0': ((A0,A0),(A0,A1),(A1,A1)),'a−b<0∧f−e=0': ((A0,A1),(A0,A0),(A1,A0)),
}
caseLR = ['c−g>0∧h−d>0','c−g>0∧h−d<0','c−g<0∧h−d<0','c−g<0∧h−d>0','c−g=0∧h−d>0','c−g=0∧h−d<0','c−g>0∧h−d=0','c−g<0∧h−d=0',
]
respLR = {'c−g>0∧h−d>0': ((A0,A0),(Ax,A0),(Ax,A1), (A1,A1)),'c−g>0∧h−d<0': ((A0,A1),(A1,A1)),'c−g<0∧h−d<0': ((A0,A1),(Ax,A1),(Ax,A0), (A1,A0)),'c−g<0∧h−d>0': ((A0,A0),(A1,A0)),'c−g=0∧h−d>0': ((A0,A0),(A1,A0),(A1,A1)),'c−g=0∧h−d<0': ((A0,A1),(A1,A1),(A1,A0)),'c−g>0∧h−d=0': ((A0,A0),(A0,A1),(A1,A1)),'c−g<0∧h−d=0': ((A0,A1),(A0,A0),(A1,A0)),
}fig, axes = plt.subplots(9, 9, figsize=(16, 16), dpi=256, layout='constrained')
for i in range(9):for j in range(9):if i == 0 and j == 0:axes[i][j].set_axis_off()else:# matplotlib.axes.Axesaxes[i][j].set_aspect('equal')# matplotlib.axis.XAxisaxes[i][j].set_xlim((-0.1, 1.1))axes[i][j].set_xticks((0, 1))axes[i][j].set_xlabel('q', rotation='horizontal')axes[i][j].xaxis.set_ticks_position('top')axes[i][j].xaxis.set_label_position('top')# matplotlib.axis.YAxisaxes[i][j].set_ylim((-0.1, 1.1))axes[i][j].set_yticks((0, 1))axes[i][j].set_ylabel('p', rotation='horizontal')axes[i][j].yaxis.set_ticks_position('left')axes[i][j].yaxis.set_label_position('left')
for j in range(1, 9):case = caseUD[j-1]resp = respUD[case]for i in range(0, 9):if i == 0:axes[i][j].set_title(case)for (q0, p0), (q1, p1) in zip(resp[:-1], resp[1:]):if (q0, p0) < (q1, p1):(q0, p0), (q1, p1) = (q1, p1), (q0, p0)axes[i][j].plot((q0, q1), (p0, p1), color='orange', alpha=0.6, linewidth=5)
for i in range(1, 9):case = caseLR[i-1]resp = respLR[case]for j in range(0, 9):if j == 0:axes[i][j].set_title(case, rotation='vertical', x=-0.4, y=0.1)for (p0, q0), (p1, q1) in zip(resp[:-1], resp[1:]):if (p0, q0) < (p1, q1):(p0, q0), (p1, q1) = (p1, q1), (p0, q0)axes[i][j].plot((q0, q1), (p0, p1), color='green', alpha=0.4, linewidth=5)
fig.savefig('./a.png')