PathSpec-ICLR / scripts /opt_tree1.py
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import numpy as np
from scipy.optimize import minimize
# 参数
p = np.array([0.8, 0.5, 0.3, 0.2, 0.12, 0.07])
q = 1 - p # 失效概率
N_total = 80
def objective(n):
"""目标函数:Σ n_i * [1 - q_i^{n_i}]"""
return np.sum(n * (1 - q**n))
# 修正约束条件
constraints = [
{'type': 'eq', 'fun': lambda n: np.sum(n) - N_total}
]
# 设置边界为至少1个节点
bounds = [(1, N_total)] * 6
# 初始猜测(均匀分配)
n0 = np.ones(6) * N_total / 6
n0 = np.maximum(n0, 1) # 确保初始值 >=1
# 优化(使用连续实数)
res = minimize(lambda n: -objective(n), n0,
bounds=bounds,
constraints=constraints,
method='SLSQP',
options={'maxiter': 50,
'ftol': 1e-6,
'disp': True})
n_opt = res.x
max_value = objective(n_opt)
print("最优节点分配(连续解):")
for i in range(6):
print(f"层 {i+1}: {n_opt[i]:.2f} 个节点")
print(f"\n最大值: {max_value:.4f}")
# 检查整数解
from itertools import product
def integer_search(center, radius=3):
best_val = -1
best_n = None
# 生成附近整数组合(确保每个值至少为1)
ranges = []
for i in range(6):
start = max(1, int(center[i]) - radius) # 至少为1
end = min(N_total, int(center[i]) + radius) # 最多N_total
ranges.append(range(start, end + 1))
# 限制搜索范围,避免组合爆炸
count = 0
max_combinations = 100000 # 限制搜索组合数
for combo in product(*ranges):
count += 1
if count > max_combinations:
break
if sum(combo) == N_total and all(x >= 1 for x in combo):
val = objective(np.array(combo))
if val > best_val:
best_val = val
best_n = combo
# 如果没找到合适的解,尝试更简单的启发式搜索
if best_n is None:
print("直接搜索未找到合适解,使用四舍五入法...")
# 四舍五入并调整到总和为80
rounded = np.round(n_opt).astype(int)
diff = N_total - np.sum(rounded)
# 调整差值
if diff > 0:
# 从最小值开始加
sorted_idx = np.argsort(n_opt - rounded)
for i in range(diff):
rounded[sorted_idx[i]] += 1
elif diff < 0:
# 从最大值开始减
sorted_idx = np.argsort(rounded - n_opt)[::-1]
for i in range(-diff):
if rounded[sorted_idx[i]] > 1: # 确保至少为1
rounded[sorted_idx[i]] -= 1
# 确保所有值至少为1
rounded = np.maximum(rounded, 1)
best_n = tuple(rounded)
best_val = objective(np.array(best_n))
return best_n, best_val
int_n, int_val = integer_search(n_opt, radius=3)
print("\n近似最优整数解:")
for i in range(6):
print(f"层 {i+1}: {int_n[i]} 个节点")
print(f"整数值: {int_val:.4f}")
# 验证
print(f"\n验证:")
print(f"总和: {sum(int_n)}")
print(f"所有节点 ≥ 1: {all(x >= 1 for x in int_n)}")
# 附加:使用更智能的整数搜索方法
print("\n\n备选:使用动态规划寻找最优整数解...")
# 由于节点数较少,可以尝试更系统的方法
def find_optimal_integer():
from itertools import combinations_with_replacement
import math
best_val = -1
best_n = None
# 使用星棒法生成所有可能的组合
# C(N_total-1, 6-1) = C(79, 5) ≈ 2.3 million 仍然很大
# 使用更高效的方法:从连续解开始,在附近搜索
# 生成所有6个数字总和为80的组合,每个至少为1
# 这相当于找5个切割点
# 由于组合数仍然较大,我们使用更智能的剪枝
# 简化的搜索:先固定前5层,最后一层由总和决定
candidates = []
# 放宽搜索半径
radius = 4
center = np.round(n_opt).astype(int)
for n1 in range(max(1, center[0]-radius), center[0]+radius+1):
for n2 in range(max(1, center[1]-radius), center[1]+radius+1):
for n3 in range(max(1, center[2]-radius), center[2]+radius+1):
for n4 in range(max(1, center[3]-radius), center[3]+radius+1):
for n5 in range(max(1, center[4]-radius), center[4]+radius+1):
n6 = N_total - (n1+n2+n3+n4+n5)
if n6 >= 1:
combo = (n1, n2, n3, n4, n5, n6)
# 检查是否在合理范围内
if all(abs(combo[i] - center[i]) <= radius+2 for i in range(6)):
val = objective(np.array(combo))
if val > best_val:
best_val = val
best_n = combo
return best_n, best_val
opt_int_n, opt_int_val = find_optimal_integer()
print("\n优化后的整数解:")
for i in range(6):
print(f"层 {i+1}: {opt_int_n[i]} 个节点")
print(f"优化整数值: {opt_int_val:.4f}")