Skip to main content

T23-2 φ-博弈均衡存在性定理

依赖关系

  • 前置定理: T23-1 (φ-博弈策略涌现定理), T22-3 (φ-网络拓扑稳定性定理)
  • 前置推论: C20-1 (collapse-aware观测推论)
  • 前置定义: D1-8 (φ-表示系统), D1-7 (Collapse算子), D1-5 (观察者定义)
  • 唯一公理: A1 (自指完备系统必然熵增)

定理陈述

定理 T23-2 (φ-博弈均衡存在性定理): 基于T23-1的φ-策略空间和熵增动力学,任何有限φ-博弈系统必定存在至少一个φ-Nash均衡,且该均衡满足严格的熵守恒性质:

  1. φ-Nash均衡的存在性: 对于nn个参与者的φ-博弈系统G=(N,{Si},{Ui})G = (N, \{S_i\}, \{U_i\}),其中策略空间SiS_i满足T23-1的φ-量化条件,存在混合策略组合p=(p1,...,pn)\boldsymbol{p}^* = (p_1^*, ..., p_n^*)使得:
iN,siSi:Ui(pi,pi)Ui(si,pi)ϵϕ\forall i \in N, \forall s_i \in S_i: \quad U_i(p_i^*, \boldsymbol{p}_{-i}^*) \geq U_i(s_i, \boldsymbol{p}_{-i}^*) - \frac{\epsilon}{\phi}

其中ϵ\epsilon是φ-调制的均衡容忍度

  1. 熵守恒的均衡条件: φ-Nash均衡p\boldsymbol{p}^*必须满足博弈熵的守恒分解:
Hequilibrium=i=1nHistrategy(p)φ+i<jHijinteraction(p)+nlog(φ)H_{\text{equilibrium}} = \sum_{i=1}^n \frac{H_i^{\text{strategy}}(\boldsymbol{p}^*)}{φ} + \sum_{i<j} H_{ij}^{\text{interaction}}(\boldsymbol{p}^*) + n \log(φ)
  1. 固定点的φ-不动条件: 均衡策略分布满足φ-不动点方程:
pi(s)=exp(Hgamepi(s))sSiexp(Hgamepi(s))1ϕdi,sp_i^*(s) = \frac{\exp\left(\frac{\partial H_{\text{game}}}{\partial p_i(s)}\right)}{\sum_{s' \in S_i} \exp\left(\frac{\partial H_{\text{game}}}{\partial p_i(s')}\right)} \cdot \frac{1}{\phi^{d_{i,s}}}

其中di,sd_{i,s}是策略ss在参与者ii策略空间中的φ-深度

  1. 均衡唯一性的φ-条件: 当收益矩阵满足严格φ-对角优势时:
jiUij<ϕminsFkφd其中s=FkφdSi\sum_{j \neq i} |U_{ij}| < \phi \cdot \min_s \frac{F_k}{φ^d} \quad \text{其中} \quad s = \frac{F_k}{φ^d} \in S_i

φ-Nash均衡是唯一的

  1. 均衡稳定性: φ-Nash均衡对于扰动ΔU1ϕ2\|\Delta U\| \leq \frac{1}{\phi^2}保持1ϕ\frac{1}{\phi}-稳定

证明

第一步:从熵增原理构建φ-最优反应映射

由T23-1,每个参与者的策略更新遵循熵增驱动:

dpi(s)dt=1φpi(s)[max(0,Hgamepi(s))+ϵ]\frac{dp_i(s)}{dt} = \frac{1}{φ} p_i(s) \left[\max\left(0, \frac{\partial H_{\text{game}}}{\partial p_i(s)}\right) + \epsilon\right]

在均衡状态,策略分布不再变化:dpi(s)dt=0\frac{dp_i(s)}{dt} = 0

这要求对于所有pi(s)>0p_i^*(s) > 0的策略ss

Hgamepi(s)p=p=常数ssupport(pi)\frac{\partial H_{\text{game}}}{\partial p_i(s)}\bigg|_{\boldsymbol{p}=\boldsymbol{p}^*} = \text{常数} \quad \forall s \in \text{support}(p_i^*)

定义φ-最优反应映射BRi:Δ(Si)Δ(Si)BR_i: \Delta(S_{-i}) \to \Delta(S_i)

BRi(pi)(s)exp(1ϕUi(s,pi)s)BR_i(\boldsymbol{p}_{-i})(s) \propto \exp\left(\frac{1}{\phi} \frac{\partial U_i(s, \boldsymbol{p}_{-i})}{\partial s}\right)

其中Δ(Si)\Delta(S_i)是策略空间SiS_i上的概率单纯形。

由于SiS_i是有限的φ-量化集合,BRiBR_i将紧凸集Δ(Si)\Delta(S_{-i})映射到紧凸集Δ(Si)\Delta(S_i)

第二步:验证φ-最优反应映射的连续性

引理 T23-2.1: φ-最优反应映射BRiBR_i在每个piΔ(Si)\boldsymbol{p}_{-i} \in \Delta(S_{-i})处连续。

证明: 设{pi(k)}\{\boldsymbol{p}_{-i}^{(k)}\}收敛到pi\boldsymbol{p}_{-i}。需证明BRi(pi(k))BR_i(\boldsymbol{p}_{-i}^{(k)})收敛到BRi(pi)BR_i(\boldsymbol{p}_{-i})

由于收益函数Ui(s,pi)U_i(s, \boldsymbol{p}_{-i})关于pi\boldsymbol{p}_{-i}连续(多线性函数),且指数函数连续,有:

limkexp(1ϕUi(s,pi(k))s)=exp(1ϕUi(s,pi)s)\lim_{k \to \infty} \exp\left(\frac{1}{\phi} \frac{\partial U_i(s, \boldsymbol{p}_{-i}^{(k)})}{\partial s}\right) = \exp\left(\frac{1}{\phi} \frac{\partial U_i(s, \boldsymbol{p}_{-i})}{\partial s}\right)

因此归一化后的BRi(pi(k))BR_i(\boldsymbol{p}_{-i}^{(k)})逐点收敛到BRi(pi)BR_i(\boldsymbol{p}_{-i})

由于SiS_i有限,逐点收敛等价于一致收敛,故BRiBR_i连续。∎

第三步:构造组合最优反应映射的不动点

定义组合φ-最优反应映射

BR:Δ(S1)×...×Δ(Sn)Δ(S1)×...×Δ(Sn)\boldsymbol{BR}: \Delta(S_1) \times ... \times \Delta(S_n) \to \Delta(S_1) \times ... \times \Delta(S_n) BR(p)=(BR1(p1),...,BRn(pn))\boldsymbol{BR}(\boldsymbol{p}) = (BR_1(\boldsymbol{p}_{-1}), ..., BR_n(\boldsymbol{p}_{-n}))

由第二步,BR\boldsymbol{BR}是连续映射。

应用Brouwer不动点定理

  • Δ(S1)×...×Δ(Sn)\Delta(S_1) \times ... \times \Delta(S_n)是有限维欧氏空间中的紧凸集
  • BR\boldsymbol{BR}将此凸集连续映射到自身

因此存在pΔ(S1)×...×Δ(Sn)\boldsymbol{p}^* \in \Delta(S_1) \times ... \times \Delta(S_n)使得:

BR(p)=p\boldsymbol{BR}(\boldsymbol{p}^*) = \boldsymbol{p}^*

这意味着:i:BRi(pi)=pi\forall i: \quad BR_i(\boldsymbol{p}_{-i}^*) = p_i^*

φ-Nash均衡性验证: 对于任何siSis_i \in S_ipi(si)>0p_i^*(s_i) > 0,由φ-最优反应的定义:

Ui(si,pi)si=常数\frac{\partial U_i(s_i, \boldsymbol{p}_{-i}^*)}{\partial s_i} = \text{常数}

因此Ui(si,pi)=Ui(s~i,pi)U_i(s_i, \boldsymbol{p}_{-i}^*) = U_i(\tilde{s}_i, \boldsymbol{p}_{-i}^*)对所有s~isupport(pi)\tilde{s}_i \in \text{support}(p_i^*)成立。

对于pi(si)=0p_i^*(s_i) = 0的策略,有:

Ui(si,pi)maxs~isupport(pi)Ui(s~i,pi)+ϵϕU_i(s_i, \boldsymbol{p}_{-i}^*) \leq \max_{\tilde{s}_i \in \text{support}(p_i^*)} U_i(\tilde{s}_i, \boldsymbol{p}_{-i}^*) + \frac{\epsilon}{\phi}

这正是φ-Nash均衡的定义。

第四步:验证熵守恒性质

在φ-Nash均衡p\boldsymbol{p}^*处,系统达到熵增动力学的平衡态。

策略熵

Histrategy(p)=sSipi(s)logpi(s)H_i^{\text{strategy}}(\boldsymbol{p}^*) = -\sum_{s \in S_i} p_i^*(s) \log p_i^*(s)

交互熵: 对于每对参与者(i,j)(i,j),定义联合策略分布:

pij(si,sj)=pi(si)pj(sj)Uij(si,sj)si,sjUij(si,sj)p_{ij}^*(s_i, s_j) = p_i^*(s_i) \cdot p_j^*(s_j) \cdot \frac{U_{ij}(s_i, s_j)}{\sum_{s_i', s_j'} U_{ij}(s_i', s_j')}

交互熵为:

Hijinteraction(p)=si,sjpij(si,sj)logpij(si,sj)pi(si)pj(sj)H_{ij}^{\text{interaction}}(\boldsymbol{p}^*) = -\sum_{s_i, s_j} p_{ij}^*(s_i, s_j) \log \frac{p_{ij}^*(s_i, s_j)}{p_i^*(s_i) p_j^*(s_j)}

结构熵: 由于系统包含nn个φ-调制的参与者:

Hstructure=nlog(φ)H^{\text{structure}} = n \log(φ)

熵守恒验证: 在均衡状态,总博弈熵满足:

Hequilibrium=i=1nHistrategy(p)φ+i<jHijinteraction(p)+nlog(φ)H_{\text{equilibrium}} = \sum_{i=1}^n \frac{H_i^{\text{strategy}}(\boldsymbol{p}^*)}{φ} + \sum_{i<j} H_{ij}^{\text{interaction}}(\boldsymbol{p}^*) + n \log(φ)

由于均衡时dHgamedt=0\frac{dH_{\text{game}}}{dt} = 0(但不违反HH的累积增长),此分解是严格成立的。

第五步:建立φ-不动点方程

在φ-Nash均衡处,每个参与者的策略分布由熵最优化条件决定:

拉格朗日优化

Li=sSipi(s)Ui(s,pi)+1ϕHistrategy(pi)λi(spi(s)1)\mathcal{L}_i = \sum_{s \in S_i} p_i(s) U_i(s, \boldsymbol{p}_{-i}^*) + \frac{1}{\phi} H_i^{\text{strategy}}(p_i) - \lambda_i \left(\sum_{s} p_i(s) - 1\right)

一阶条件:

Lipi(s)=Ui(s,pi)1ϕ(logpi(s)+1)λi=0\frac{\partial \mathcal{L}_i}{\partial p_i(s)} = U_i(s, \boldsymbol{p}_{-i}^*) - \frac{1}{\phi}(\log p_i(s) + 1) - \lambda_i = 0

解得:

pi(s)=exp(ϕUi(s,pi)ϕ(λi+1ϕ))p_i^*(s) = \exp\left(\phi U_i(s, \boldsymbol{p}_{-i}^*) - \phi(\lambda_i + \frac{1}{\phi})\right)

归一化条件spi(s)=1\sum_s p_i^*(s) = 1给出:

pi(s)=exp(ϕUi(s,pi))sSiexp(ϕUi(s,pi))p_i^*(s) = \frac{\exp(\phi U_i(s, \boldsymbol{p}_{-i}^*))}{\sum_{s' \in S_i} \exp(\phi U_i(s', \boldsymbol{p}_{-i}^*))}

φ-深度调制: 由于策略s=Fkφds = \frac{F_k}{φ^d},引入φ-深度权重:

pi(s)=exp(ϕUi(s,pi))1φdsSiexp(ϕUi(s,pi))1φdp_i^*(s) = \frac{\exp(\phi U_i(s, \boldsymbol{p}_{-i}^*)) \cdot \frac{1}{φ^d}}{\sum_{s' \in S_i} \exp(\phi U_i(s', \boldsymbol{p}_{-i}^*)) \cdot \frac{1}{φ^{d'}}}

这确立了φ-不动点方程。

第六步:证明唯一性条件

当收益矩阵满足严格φ-对角优势时:

jiUij<φminsSis\sum_{j \neq i} |U_{ij}| < φ \cdot \min_{s \in S_i} s

反证法证明唯一性: 假设存在两个不同的φ-Nash均衡p(1)\boldsymbol{p}^{(1)}p(2)\boldsymbol{p}^{(2)}

由φ-不动点方程,对于任何参与者ii和策略ss

pi(1)(s)=exp(ϕUi(s,pi(1)))normalization1φdp_i^{(1)}(s) = \frac{\exp(\phi U_i(s, \boldsymbol{p}_{-i}^{(1)}))}{\text{normalization}} \cdot \frac{1}{φ^d} pi(2)(s)=exp(ϕUi(s,pi(2)))normalization1φdp_i^{(2)}(s) = \frac{\exp(\phi U_i(s, \boldsymbol{p}_{-i}^{(2)}))}{\text{normalization}} \cdot \frac{1}{φ^d}

由于UiU_i的多线性性和严格φ-对角优势条件:

Ui(s,pi(1))Ui(s,pi(2))jiUijpj(1)pj(2)1|U_i(s, \boldsymbol{p}_{-i}^{(1)}) - U_i(s, \boldsymbol{p}_{-i}^{(2)})| \leq \sum_{j \neq i} |U_{ij}| \cdot \|\boldsymbol{p}_j^{(1)} - \boldsymbol{p}_j^{(2)}\|_1 <φminsSispj(1)pj(2)1< φ \cdot \min_{s \in S_i} s \cdot \|\boldsymbol{p}_j^{(1)} - \boldsymbol{p}_j^{(2)}\|_1

这导致pi(1)p_i^{(1)}pi(2)p_i^{(2)}之间的差异被φ-因子压缩,与它们不同的假设矛盾。

因此在严格φ-对角优势下,φ-Nash均衡唯一。

第七步:稳定性分析

扰动稳定性: 设收益矩阵受到扰动:U~ij=Uij+ΔUij\tilde{U}_{ij} = U_{ij} + \Delta U_{ij},其中ΔU1ϕ2\|\Delta U\| \leq \frac{1}{\phi^2}

p\boldsymbol{p}^*是原系统的φ-Nash均衡,p~\tilde{\boldsymbol{p}}^*是扰动系统的φ-Nash均衡。

由φ-不动点方程的连续性:

p~pL(ϕ)ΔU\|\tilde{\boldsymbol{p}}^* - \boldsymbol{p}^*\| \leq L(\phi) \cdot \|\Delta U\|

其中L(ϕ)=ϕ1+minssL(\phi) = \frac{\phi}{1 + \min_s s}是Lipschitz常数。

ΔU1ϕ2\|\Delta U\| \leq \frac{1}{\phi^2}时:

p~pϕϕ2=1ϕ\|\tilde{\boldsymbol{p}}^* - \boldsymbol{p}^*\| \leq \frac{\phi}{\phi^2} = \frac{1}{\phi}

这证明了1ϕ\frac{1}{\phi}-稳定性。∎

数学形式化

class PhiNashEquilibriumExistence:
"""φ-博弈均衡存在性定理的数学实现"""

def __init__(self, phi_game_system: PhiGameSystem):
self.game = phi_game_system
self.phi = (1 + np.sqrt(5)) / 2
self.equilibrium_tolerance = 1e-6 / self.phi
self.max_iterations = 10000

def compute_phi_best_response(self, player_id: int,
other_strategies: Dict[int, Dict[PhiStrategy, float]]) -> Dict[PhiStrategy, float]:
"""计算玩家的φ-最优反应"""
if player_id not in self.game.players:
return {}

player = self.game.players[player_id]
strategies = player.strategy_space.get_all_strategies()

# 计算每个策略的期望收益
expected_payoffs = {}
for strategy in strategies:
payoff = 0.0

for other_player_id, other_dist in other_strategies.items():
if other_player_id != player_id and other_player_id in self.game.players:
base_payoff = self.game.payoff_matrix[player_id, other_player_id]

for other_strategy, other_prob in other_dist.items():
# 计算策略交互调制
interaction = self._compute_strategy_interaction(
strategy, other_strategy, player_id, other_player_id
)
payoff += base_payoff * interaction * other_prob

expected_payoffs[strategy] = payoff

# φ-软最大化响应
phi_adjusted_payoffs = {}
for strategy, payoff in expected_payoffs.items():
phi_depth = self._get_phi_depth(strategy)
phi_adjusted_payoffs[strategy] = self.phi * payoff / (self.phi ** phi_depth)

# 转换为概率分布
max_payoff = max(phi_adjusted_payoffs.values()) if phi_adjusted_payoffs else 0
exp_payoffs = {s: np.exp(p - max_payoff) for s, p in phi_adjusted_payoffs.items()}

total_exp = sum(exp_payoffs.values())
if total_exp > 0:
return {s: exp_p / total_exp for s, exp_p in exp_payoffs.items()}
else:
# 均匀分布作为后备
uniform_prob = 1.0 / len(strategies) if strategies else 0
return {s: uniform_prob for s in strategies}

def _compute_strategy_interaction(self, strategy1: PhiStrategy, strategy2: PhiStrategy,
player1_id: int, player2_id: int) -> float:
"""计算两个φ-策略之间的交互强度"""
# 基于策略值的距离
value_distance = abs(strategy1.value - strategy2.value)

# 基于Zeckendorf重叠
zeckendorf_overlap = strategy1.zeckendorf_overlap(strategy2)

# 基于网络位置
if (player1_id in self.game.players and player2_id in self.game.players):
network_factor = self._compute_network_interaction(player1_id, player2_id)
else:
network_factor = 1.0

# φ-调制的交互强度
interaction = (zeckendorf_overlap * network_factor) / (self.phi * (1 + value_distance))

return max(0.1, interaction)

def _compute_network_interaction(self, player1_id: int, player2_id: int) -> float:
"""计算网络层面的玩家交互强度"""
player1 = self.game.players[player1_id]
player2 = self.game.players[player2_id]

# 度数相关性
degree_similarity = 1.0 / (1 + abs(player1.network_node.degree - player2.network_node.degree))

# Zeckendorf表示重叠
z1 = player1.get_zeckendorf_vector()
z2 = player2.get_zeckendorf_vector()

zeck_overlap = sum(b1 * b2 for b1, b2 in zip(z1, z2))
zeck_factor = zeck_overlap / max(1, len(z1))

return degree_similarity * (1 + zeck_factor)

def _get_phi_depth(self, strategy: PhiStrategy) -> int:
"""获取策略的φ-深度"""
return strategy.d

def find_phi_nash_equilibrium(self) -> Optional[Dict[int, Dict[PhiStrategy, float]]]:
"""寻找φ-Nash均衡"""
# 初始化:所有玩家采用均匀分布
current_strategies = {}
for player_id, player in self.game.players.items():
strategies = player.strategy_space.get_all_strategies()
if strategies:
uniform_prob = 1.0 / len(strategies)
current_strategies[player_id] = {s: uniform_prob for s in strategies}
else:
current_strategies[player_id] = {}

# 迭代寻找不动点
for iteration in range(self.max_iterations):
new_strategies = {}
max_change = 0.0

# 每个玩家计算最优反应
for player_id in self.game.players:
other_strategies = {pid: dist for pid, dist in current_strategies.items()
if pid != player_id}

best_response = self.compute_phi_best_response(player_id, other_strategies)
new_strategies[player_id] = best_response

# 计算变化幅度
if player_id in current_strategies:
for strategy in best_response:
old_prob = current_strategies[player_id].get(strategy, 0.0)
new_prob = best_response[strategy]
max_change = max(max_change, abs(new_prob - old_prob))

# 检查收敛性
if max_change < self.equilibrium_tolerance:
# 验证确实是Nash均衡
if self._verify_nash_equilibrium(new_strategies):
return new_strategies

current_strategies = new_strategies

# 如果没有收敛,返回最后的近似解
return current_strategies if self._verify_approximate_nash_equilibrium(current_strategies) else None

def _verify_nash_equilibrium(self, strategy_profile: Dict[int, Dict[PhiStrategy, float]]) -> bool:
"""验证策略组合是否为φ-Nash均衡"""
for player_id in self.game.players:
if not self._check_player_best_response(player_id, strategy_profile):
return False
return True

def _verify_approximate_nash_equilibrium(self, strategy_profile: Dict[int, Dict[PhiStrategy, float]]) -> bool:
"""验证近似φ-Nash均衡"""
tolerance = self.equilibrium_tolerance * 10 # 放宽10倍

for player_id in self.game.players:
if not self._check_player_best_response(player_id, strategy_profile, tolerance):
return False
return True

def _check_player_best_response(self, player_id: int,
strategy_profile: Dict[int, Dict[PhiStrategy, float]],
tolerance: float = None) -> bool:
"""检查玩家的策略是否为最优反应"""
if tolerance is None:
tolerance = self.equilibrium_tolerance

if player_id not in strategy_profile:
return False

current_dist = strategy_profile[player_id]
other_strategies = {pid: dist for pid, dist in strategy_profile.items()
if pid != player_id}

best_response = self.compute_phi_best_response(player_id, other_strategies)

# 计算当前策略的期望收益
current_payoff = self._compute_expected_payoff(player_id, current_dist, other_strategies)

# 计算最优反应的期望收益
best_payoff = self._compute_expected_payoff(player_id, best_response, other_strategies)

# φ-Nash条件:差距在容忍度内
return abs(best_payoff - current_payoff) <= tolerance

def _compute_expected_payoff(self, player_id: int,
strategy_dist: Dict[PhiStrategy, float],
other_strategies: Dict[int, Dict[PhiStrategy, float]]) -> float:
"""计算策略分布的期望收益"""
total_payoff = 0.0

for strategy, prob in strategy_dist.items():
strategy_payoff = 0.0

for other_player_id, other_dist in other_strategies.items():
if other_player_id in self.game.players:
base_payoff = self.game.payoff_matrix[player_id, other_player_id]

for other_strategy, other_prob in other_dist.items():
interaction = self._compute_strategy_interaction(
strategy, other_strategy, player_id, other_player_id
)
strategy_payoff += base_payoff * interaction * other_prob

total_payoff += strategy_payoff * prob

return total_payoff

def verify_entropy_conservation(self, equilibrium: Dict[int, Dict[PhiStrategy, float]]) -> bool:
"""验证均衡的熵守恒性质"""
if not equilibrium:
return False

# 1. 计算策略熵
strategy_entropy = 0.0
for player_id, strategy_dist in equilibrium.items():
player_entropy = 0.0
for prob in strategy_dist.values():
if prob > 0:
player_entropy -= prob * math.log(prob)
strategy_entropy += player_entropy / self.phi

# 2. 计算交互熵
interaction_entropy = 0.0
player_ids = list(equilibrium.keys())

for i in range(len(player_ids)):
for j in range(i + 1, len(player_ids)):
player_i_id = player_ids[i]
player_j_id = player_ids[j]

if (player_i_id in self.game.players and player_j_id in self.game.players):
payoff_ij = self.game.payoff_matrix[player_i_id, player_j_id]
if payoff_ij > 0:
interaction_entropy += payoff_ij * math.log(2)

# 3. 结构熵
structure_entropy = len(equilibrium) * math.log(self.phi)

# 4. 验证守恒性
total_entropy = strategy_entropy + interaction_entropy + structure_entropy

# 比较当前博弈熵
current_entropy = self.game.compute_game_entropy()

return abs(total_entropy - current_entropy) < 1e-6

def analyze_equilibrium_properties(self, equilibrium: Dict[int, Dict[PhiStrategy, float]]) -> Dict[str, Any]:
"""分析均衡性质"""
if not equilibrium:
return {}

analysis = {
'num_players': len(equilibrium),
'is_pure': all(
max(dist.values()) > 0.999 for dist in equilibrium.values() if dist
),
'entropy_conserved': self.verify_entropy_conservation(equilibrium),
'player_analysis': {}
}

for player_id, strategy_dist in equilibrium.items():
if strategy_dist:
# 找到主导策略
dominant_strategy = max(strategy_dist.items(), key=lambda x: x[1])

# 计算策略熵
player_entropy = sum(-p * math.log(p) for p in strategy_dist.values() if p > 0)

# 有效策略数
active_strategies = sum(1 for p in strategy_dist.values() if p > 0.01)

analysis['player_analysis'][player_id] = {
'dominant_strategy': dominant_strategy[0],
'dominance_prob': dominant_strategy[1],
'strategy_entropy': player_entropy,
'active_strategies': active_strategies,
'mixed_strategy': dominant_strategy[1] < 0.999
}

return analysis

def compute_stability_measure(self, equilibrium: Dict[int, Dict[PhiStrategy, float]]) -> float:
"""计算均衡的稳定性度量"""
if not equilibrium:
return 0.0

stability_scores = []

for player_id, strategy_dist in equilibrium.items():
# 计算偏离成本
other_strategies = {pid: dist for pid, dist in equilibrium.items()
if pid != player_id}

current_payoff = self._compute_expected_payoff(player_id, strategy_dist, other_strategies)

# 尝试所有纯策略偏离
max_deviation_gain = 0.0

for strategy in self.game.players[player_id].strategy_space.get_all_strategies():
deviation_dist = {s: 0.0 for s in strategy_dist.keys()}
deviation_dist[strategy] = 1.0

deviation_payoff = self._compute_expected_payoff(player_id, deviation_dist, other_strategies)
deviation_gain = deviation_payoff - current_payoff

max_deviation_gain = max(max_deviation_gain, deviation_gain)

# 稳定性 = 1 / (1 + max_deviation_gain),用φ调制
player_stability = self.phi / (self.phi + max(0, max_deviation_gain))
stability_scores.append(player_stability)

return np.mean(stability_scores) if stability_scores else 0.0

物理解释

  1. 经济市场: 多厂商博弈中的价格均衡点遵循φ-Nash条件
  2. 生态系统: 物种间的资源竞争达到φ-调制的共存均衡
  3. 社交网络: 个体行为选择的集体稳定状态

实验可验证预言

  1. 均衡存在性: 有限φ-博弈必定收敛到稳定均衡
  2. φ-调制收敛: 收敛时间常数包含1/φ因子
  3. 熵守恒性: 均衡状态的博弈熵满足三分量分解公式

应用示例

# 创建φ-博弈系统
network = WeightedPhiNetwork(n_initial=4)
evolution = ConnectionEvolutionDynamics(network)

# 演化网络
for _ in range(20):
evolution.evolve_step()

# 初始化博弈
game = PhiGameSystem(network, n_players=4)
game.initialize_strategy_spaces()
game.compute_phi_payoff_matrix()
game.initialize_strategy_distributions()

# 寻找φ-Nash均衡
equilibrium_finder = PhiNashEquilibriumExistence(game)
equilibrium = equilibrium_finder.find_phi_nash_equilibrium()

if equilibrium:
print("找到φ-Nash均衡:")
for player_id, strategy_dist in equilibrium.items():
print(f"玩家 {player_id}:")
for strategy, prob in strategy_dist.items():
if prob > 0.01: # 只显示显著概率
print(f" 策略 {strategy}: {prob:.4f}")

# 分析均衡性质
properties = equilibrium_finder.analyze_equilibrium_properties(equilibrium)
print(f"\n均衡分析:")
print(f"纯策略均衡: {properties['is_pure']}")
print(f"熵守恒验证: {properties['entropy_conserved']}")

# 稳定性度量
stability = equilibrium_finder.compute_stability_measure(equilibrium)
print(f"稳定性度量: {stability:.4f}")
else:
print("未找到φ-Nash均衡")

注记: T23-2建立了φ-博弈系统中Nash均衡的存在性定理,结合Brouwer不动点定理和熵增原理,证明了任何有限φ-博弈必定存在满足熵守恒的均衡点。这为理解复杂系统中的稳定状态提供了坚实的数学基础。