Question

我的代码错误答案

n是变量的数量和formula是包含子句的列表

给定一个SAT实例，其中'n'个变量和子句在列表'公式'中编码，如果实例是可满足的，则返回'可满足'，并且'不可满足' 除此以外。 'formula'的每个元素代表一个子句，并且是一个列表整数i表示文字Xi存在于整数中的整数子句和整数-i表示文字~Xi存在于条款。例如，用列表表示子句“X1 v~X11 v X7” [1，-11,7]。

import itertools
n = 4 
formula = [[1, -2, 3], [-1, 3], [-3], [2, 3]]


booleanValues = [True,False] * n

allorderings = set(itertools.permutations(booleanValues, n)) #create possible combinations of variables that can check if formula is satisfiable or not

print(allorderings)

for potential in allorderings:
    l = [] #boolean value for each variable / different combination for each iteration
    for i in potential:
        l.append(i)
    #possible = [False]*n
    aclause = []
    for clause in formula:
        something = []
        #clause is [1,2,3]
        for item in clause:
            if item > 0:
                something.append(l[item-1]) 
            else:
                item = item * -1
                x = l[item-1]
                if x == True:
                    x = False
                else:
                    x = True
                something.append(x)
        counter = 0
        cal = False
        for thingsinclause in something:
            if counter == 0:
                cal = thingsinclause
                counter = counter + 1
            else:
                cal = cal and thingsinclause
                counter = counter + 1

        aclause.append(cal)

    counter2 = 0
    formcheck = False
    for checkformula in aclause:
        if counter2 == 0:
            formcheck = checkformula
            counter2 = counter2 + 1
        else:
            formcheck = formcheck or checkformula
    print("this combination works", checkformula)

Answer 1

以下是更正后的版本：

import itertools
n = 4 
formula = [[1, -2, 3], [-1, 3], [-3], [2, 3]]

allorderings = itertools.product ([False, True], repeat = n)

for potential in allorderings:
    print ("Initial values:", potential)
    allclauses = []
    for clause in formula:
        curclause = []
        for item in clause:
            x = potential[abs (item) - 1]
            curclause.append (x if item > 0 else not x)
        cal = any (curclause)
        allclauses.append (cal)
    print ("Clauses:", allclauses)
    formcheck = all (allclauses)
    print ("This combination works:", formcheck)

要考虑的要点：

您可以使用any和all，而不是引入一些复杂的 - 也是错误的 - 逻辑来查找连接和析取。那更干净，更不容易出错。
要循环的自然对象是itertools.product([False, True], repeat = n)，即可能的布尔值的集[False, True]提升到n的幂。换句话说，n的{{1}}副本Cartesian product。 Here是itertools.product的文档。
我介绍了更多的输出，看看情况如何。这是我用Python3得到的输出（Python2添加括号但打印基本相同）：

[False, True]

boolean sat检查我的代码

1 个答案: