python的约束库constraint解决《2018刑侦科题目》

Github地址：https://github.com/ZJW9633/hello-word/blob/master/Xingzhenke

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题目分析：

10道题互相关联，耦合性强，暴力求解需4^10种可能，本人使用python的约束库constraint设置约束进行解决。

python代码：

#-*- coding:utf-8 -*-
# 《2018刑侦科推理试题》非穷举的 Python 解法
# 需要先安装约束解决库 `pip install python-constraint`
# author:jinwei
# date:2018/04/1

from constraint import *
from time import time
problem = Problem()

# a1 - a10 表示第一题到第十题的答案变量，答案使用“1”表示“A”, “2”表示“B”，以此类推
vars = ["a1", "a2", "a3", "a4", "a5", "a6", "a7", "a8", "a9", "a10"]
problem.addVariables(vars, [1, 2, 3, 4])

#第 1 题
#略
#第 2 题
def a2_func(a2, a5):
    return (a2 == 1 and a5 == 3) or (a2 == 2 and a5 == 4) or (a2 == 3 or a5 == 1) or (a2 == 4 or a5 == 2)
problem.addConstraint(a2_func, ["a2", "a5"])

#第 3 题
def a3_func(a2, a3, a4,a6):
    return (a3 == 1 and a6 == a2 == a4 != a3) or (a3 == 2 and a3 == a2 == a4 != a6) 
        or (a3 == 3 and a3 == a6 == a4 != a2) or (a3 == 4 and a3 == a6 == a2 != a4)
problem.addConstraint(a3_func, ['a2','a3','a4','a6'])


#第 4 题
def a4_func(a1, a2, a4, a5, a6, a7, a9, a10):
    return (a4 == 1 and a1 == a5) or (a4 == 2 and a2 == a7) or (a4 == 3 and a1 == a9) or (a4 == 4 and a6 == a10)
problem.addConstraint(a4_func, ['a1','a2','a4','a5','a6','a7','a9','a10'])

#第 5 题
def a5_func( a4, a5, a7, a8, a9):
    return (a5 == a8 == 1) or (a5 == a4 == 2) or (a5 == a9 == 3) or (a5 == a7 == 4)
problem.addConstraint(a5_func, ['a4','a5','a7','a8','a9'])

#第 6 题
def a6_func(a1, a2, a3, a4, a5, a6 ,a8, a9, a10):
    return (a6 == 1 and a2 == a4 == a8) or (a6 == 2 and a1 == a6 == a8) 
           or (a6 == 3 and a3 == a10 == a8) or (a6 == 4 and a5 == a9 == a8)
problem.addConstraint(a6_func, ['a1','a2','a3','a4','a5','a6','a8','a9','a10'])

#第 7 题
def a7_func(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10):
    all_answers = [a1, a2, a3, a4, a5, a6, a7, a8, a9, a10]
    counter = [0, 0, 0, 0]
    for a in all_answers:
        counter[a - 1] += 1
    imin = counter.index(min(counter)) + 1
    n = len(set(counter))
    # 这题为“此10题种被选中的选项最少的为，ABCD选择，说明 ABCD 的数量个不相同”
    return n == 4 and ((a7 == 1 and imin == 3) or (a7 == 2 and imin == 2) or (a7 == 3 and imin == 1) or (a7 == 4 and imin == 4))
problem.addConstraint(a7_func, vars)

#第 8 题
def a8_func(a8, a7, a5, a2, a10, a1):
    adj = lambda x: abs(a1 - x) != 1
    return (a8 == 1 and adj(a7)) or (a8 == 2 and adj(a5)) or (a8 == 3 and adj(a2)) or (a8 == 4 and adj(a10))
problem.addConstraint(a8_func, ["a8", "a7", "a5", "a2", "a10", "a1"])

#第 9 题
def a9_func(a1, a2, a5, a6, a9, a10):
    cond1 = a1 == a6
    cond = lambda x: cond1 != (x == a5)
    return (a9 == 1 and cond(a6)) or (a9 == 2 and cond(a10)) or (a9 == 3 and cond(a2)) or (a9 == 4 and cond(a9))
problem.addConstraint(a9_func, ["a1", "a2", "a5", "a6", "a9", "a10"])

#第 10 题
def a10_func(a1, a2, a3, a4, a5, a6, a7, a8, a9, a10):
    all_answers = [a1, a2, a3, a4, a5, a6, a7, a8, a9, a10]
    counter = [0, 0, 0, 0]
    for a in all_answers:
        counter[a - 1] += 1
    v = abs(max(counter) - min(counter))
    return (a10 == 1 and v == 3) or (a10 == 2 and v == 2) or (a10 == 3 and v == 4) or (a10 == 4 and v == 1)
problem.addConstraint(a10_func, vars)


# 约束设置完毕，开始求解
start_time=time()
solutions = problem.getSolutions()
elsap_time=time()-start_time
print("解决时间:{0}s".format(elsap_time))
print("答案组数：{0}".format(len(solutions)))

chars = ['A', 'B', 'C', 'D']

for si in range(0, len(solutions)):
    print("
---------- 第 {0} 组结果 ----------".format(si + 1))
    s = solutions[si]
    for i in range(1, 11):
        key = "a" + str(i)
        answer = chars[s[key] - 1]
        print("{0}".format( answer),end='')

代码结果：

解决时间:0.5780596733093262s
答案组数：1

---------- 第 1 组结果 ----------
BCACACDABA
Process finished with exit code 0

总结：

出现的问题：在进行约束条件的时，将传入所有参数a1~a10，并且将vars传入constraint中，增加了不必要的计算，大大降低了效率。

改进前运行所需时间：

改进：重新编写约束函数，只传入必要的参数，提高效率

改进后的运行时间：0.5s