最近很火的一道题目，用笔画了很久都找不到其中的联系，无奈使用最笨的穷举法，题目如下：

Python运算：

import time

moc = ["", "", "", "", "", "", "", "", "", ""]


def mach_question():
    times = 0
    start = time.time()
    for i in ["A", "B", "C", "D"]:
        # 推测第一道
        moc[0] = i
        # 推测第二道
        for j in ["A", "B", "C", "D"]:
            moc[1] = j
            # 推测第三道题
            for k in ["A", "B", "C", "D"]:
                moc[2] = k
                # 推测第四道
                for l in ["A", "B", "C", "D"]:
                    moc[3] = l
                    # 推测第五到
                    for m in ["A", "B", "C", "D"]:
                        moc[4] = m
                        # 推测第六道
                        for n in ["A", "B", "C", "D"]:
                            moc[5] = n
                            ## 推测第七道
                            for q in ["A", "B", "C", "D"]:
                                moc[6] = q
                                if m_3(moc):
                                    ## 推测第八道
                                    for p in ["A", "B", "C", "D"]:
                                        moc[7] = p
                                        ## 推测第九道
                                        for b in ["A", "B", "C", "D"]:
                                            moc[8] = b
                                            ## 推算第十项
                                            for c in ["A", "B", "C", "D"]:
                                                moc[9] = c
                                                times = times + 1
                                                if m_5(moc):
                                                    if m_4(moc):
                                                        if m_6(moc) and m_7(moc) and m_9(moc):
                                                            print('匹配:%d次' % times)
                                                            print(moc)
                                                        break
                                                    else:
                                                        break
                                                else:
                                                    break
                                else:
                                    break
    end = time.time()
    print('运行时间：%ss' % (end - start))


## 运算第三道的方法
def m_3(item):
    k = item[2]
    if k == 'A':
        if item[5] != 'A' and item[5] == item[1] == item[3]:
            return True
        ##如果第三道题为B
    if k == 'B':
        if item[5] != 'B' and item[1] == item[3] == 'B':
            return True
        ##如果第三道题为C
    if k == 'C':
        if item[1] != 'C' and item[5] == item[3] == 'C':
            return True
        ##如果第三道题为D
    if k == 'D':
        if item[3] != 'D' and item[5] == item[1] == 'D':
            return True
    return False


def m_4(item):
    n = item[3]
    if n == 'A':
        if item[0] == item[4] and item[1] != item[6] and item[0] != item[8] and item[5] != item[9]:
            return True
    if n == 'B':
        if item[1] == item[6] and item[0] != item[4] and item[0] != item[8] and item[5] != item[9]:
            return True
    if n == 'C':
        if item[0] == item[8] and item[1] != item[6] and item[0] != item[4] and item[5] != item[9]:
            return True
    if n == 'D':
        if item[5] == item[9] and item[1] != item[6] and item[0] != item[8] and item[0] != item[4]:
            return True
    return False


## 运算符合第五道的方法
def m_5(item):
    m = item[4]
    if m == 'A' and item[1] == 'C' and item[7] == 'A' and item[3] != 'A' and item[8] != 'A' and item[6] != 'A':
        return True
    if m == 'B' and item[1] == 'D' and item[3] == 'B' and item[7] != 'B' and item[8] != 'B' and item[6] != 'B':
        return True
    if m == 'C' and item[1] == 'A' and item[8] == 'C' and item[7] != 'C' and item[3] != 'C' and item[6] != 'C':
        return True
    if m == 'D' and item[1] == 'B' and item[6] == 'D' and item[7] != 'D' and item[3] != 'D' and item[8] != 'D':
        return True
    return False


## 运算第六道
def m_6(item):
    j = item[5]
    if j == 'A':
        if item[1] == item[3] == item[7]:
            return True
    if j == 'B':
        if item[0] == item[5] == item[7]:
            return True
    if j == 'C':
        if item[2] == item[9] == item[7]:
            return True
    if j == 'D':
        if item[4] == item[8] == item[7]:
            return True
    return False


## 匹配第七道题
def m_7(item):
    n = item[7]
    one = ord(item[0])
    if n == 'A':
        return abs(one - ord(item[6])) > 1
    if n == 'B':
        return abs(one - ord(item[4])) > 1
    if n == 'C':
        return abs(one - ord(item[1])) > 1
    if n == 'D':
        return abs(one - ord(item[9])) > 1


##匹配第九道题

def m_9(item):
    q = item[8]
    if item[0] == item[6]:
        if q == 'A':
            return item[4] != item[5]
        if q == 'B':
            return item[4] != item[9]
        if q == 'C':
            return item[4] != item[1]
        if q == 'D':
            return item[4] != item[8]
    else:
        if q == 'A':
            return item[4] == item[5]
        if q == 'B':
            return item[4] == item[9]
        if q == 'C':
            return item[4] == item[1]
        if q == 'D':
            return item[4] == item[8]


mach_question()

运算结果如下：

匹配:4658次
['B', 'C', 'A', 'C', 'A', 'C', 'D', 'A', 'B', 'A']
运行时间：0.011355876922607422s