下面大家将会看到的是一个极其简单而又极其复杂的“迷宫”，这无疑是我在本年度见到的最变态的谜题：从左边入口处的 2011 进去，在迷宫里转悠，最后变成 2012 从右边出来。你可以在迷宫里转圈，可以重复之前走过的路，但不能往回退着走。

Start at 2011. By moving through the maze and doing any arithmetic operations you encounter, exit the maze with a result of 2012. You may pass through an operation several times, but not twice in a row.

原题和答案在这里：http://www2.stetson.edu/~efriedma/holiday/2011/index.html

在网上找了两个python程序。

程序一：

plus7,div2,mul3,min5=(‘+ 7′,’/ 2′,’* 3′,’- 5′)
left,middle,right=range(3)

class node:
    pre_node=None
    pre_opr=None
    pre_pos=None
    this_node=None
    def __init__(self,value,opr,pos):
        self.value=value
        self.opr=opr
        self.pos=pos
    def hash(self):
        return self.value*157+ord(self.opr[0])*3+self.pos
    def next(self):
        list=self._next()
        for child in list:
            child.pre_node=self.this_node
            child.pre_opr=self.opr
            child.pre_pos=self.pos
        return list
    def _next(self):
        if self.opr==plus7:
            child_var=self.value+7
            if self.pos==left:
                return node(child_var,div2,middle),node(child_var,mul3,middle),node(child_var,min5,middle)
            elif self.pos==middle:
                return node(child_var,div2,left),
        elif self.opr==div2:
            if self.value&1:
                return tuple()
            child_var=int(self.value)//2
            if self.pos==left:
                return node(child_var,plus7,middle),node(child_var,mul3,middle),node(child_var,min5,middle)
            elif self.pos==middle:
                return node(child_var,plus7,left),
        elif self.opr==mul3:
            child_var=int(self.value*3)
            if self.pos==middle:
                return node(child_var,min5,right),
            elif self.pos==right:
                return node(child_var,plus7,middle),node(child_var,div2,middle),node(child_var,min5,middle)
        elif self.opr==min5:
            child_var=self.value-5
            if self.pos==middle:
                return node(child_var,mul3,right),
            elif self.pos==right:
                return node(child_var,plus7,middle),node(child_var,div2,middle),node(child_var,mul3,middle)

node1=node(2011,plus7,left)
tree=[node1,]
hashlist=set([node1.hash(),])
i=0
while not (tree[i].pos==right and tree[i].value==2012):
    tree[i].this_node=i
    result=tree[i].next()
    for item in result:
        hashitem=item.hash()
        if hashitem in hashlist:
            continue
        else:
            tree.append(item)
            hashlist.add(hashitem)
    i+=1
    #if i%100==0:
    #    print(i,tree[i].pos,tree[i].value)
print(‘Tried’,i,’times!’)
list=[]
while tree[i].pre_node:
    list.append(i)
    i=tree[i].pre_node
list.append(0)
list.reverse()
for i in range(len(list)-1):

    print(tree[list[i]].value,tree[list[i]].opr,’=',tree[list[i+1]].value)

程序二：

#!/usr/bin/env python
import sys

todo=[(2011,1,'+7',[])]
state=0

while(todo):
    newtodo=[]
    for num, state, cmd, history in todo:
#        print num, state, cmd, history
        history=list(history)
        if cmd==’+7′:
            history.append(‘+7′)
            num+=7
        elif cmd==’/2′:
            history.append(‘/2′)
            num/=2
        elif cmd==’-5′:
            history.append(‘-5′)
            num-=5
        else:
            history.append(‘*3′)
            num*=3
        if num==2012:
            print history
            sys.exit(0)
        if state==0:
            if cmd==’+7′:
                if(num%2==0):
                    newtodo.append((num,1,’/2′,history))
            else:
                newtodo.append((num,1,’+7′, history))
        elif state==1:
            if cmd==’+7′:
                if(num%2==0):
                    newtodo.append((num,0,’/2′,history))
                newtodo.append((num,2,’*3′,history))
                newtodo.append((num,2,’-5′,history))
            elif cmd==’/2′:
                newtodo.append((num,0,’+7′,history))
                newtodo.append((num,2,’*3′,history))
                newtodo.append((num,2,’-5′,history))
            elif cmd==’*3′:
                newtodo.append((num,0,’+7′,history))
                if(num%2==0):
                    newtodo.append((num,0,’/2′,history))
                newtodo.append((num,2,’-5′,history))
            else:
                newtodo.append((num,0,’+7′,history))
                if(num%2==0):
                    newtodo.append((num,0,’/2′,history))
                newtodo.append((num,2,’*3′,history))
        elif state==2:
            if cmd==’*3′:
                newtodo.append((num,1,’-5′,history))
            else:
                newtodo.append((num,1,’*3′,history))
    todo=newtodo

我自己也写了个Python版本，用递归来实现：

#encoding=utf-8
table = ['+', '*', '/', '-']
hasmethod = 0
def calc(method, cursum, curmethods, leftright):
    global hasmethod

    desmethod = []
    if hasmethod:
        return
    if method == ‘+’:
        cursum += 7
        curmethods += ‘+’
        if cursum % 2 == 0:
            desmethod.append(‘/’)
        if leftright == ‘r’:
            desmethod.append(‘*’)
            desmethod.append(‘-’)
    elif method == ‘*’:
        cursum *= 3
        curmethods += ‘*’
        if leftright == ‘l’:
            if cursum % 2 == 0:
                desmethod.append(‘/’)
            desmethod.append(‘+’)
        desmethod.append(‘-’)
    elif method == ‘-’:
        cursum -= 5
        curmethods += ‘-’
        if leftright == ‘l’:
            if cursum % 2 == 0:
                desmethod.append(‘/’)
            desmethod.append(‘+’)
        desmethod.append(‘*’)
    elif method == ‘/’:
        cursum /= 2
        curmethods += ‘/’
        desmethod.append(‘+’)
        if leftright == ‘r’:
            desmethod.append(‘*’)
            desmethod.append(‘-’)
    if cursum == 2012:
        ensuremethod(curmethods)
        curmethods = curmethods.replace(‘+’, ‘+7 ‘)
        curmethods = curmethods.replace(‘-’, ‘-5 ‘)
        curmethods = curmethods.replace(‘*’, ‘*3 ‘)
        curmethods = curmethods.replace(‘/’, ‘/2 ‘)
        print curmethods
        hasmethod = 1
    elif len(curmethods) < 30:  #防止递归调用层次过深
        for m in desmethod:
            curleftright = leftright
            #控制方向，从左进还是从右进
            if (m in ['+', '/'] and method in ['+', '/']) or (m in ['*', '-'] and method in ['*', '-']):
                curleftright = ‘r’ if leftright == ‘l’ else ‘l’
            calc(m, cursum, curmethods, curleftright)

def ensuremethod(methodstr):
    print methodstr
    print len(methodstr)
    s = 2011
    str0 = ”
    for i in methodstr:
        strs = str(s)
        if i == ‘+’:
            s += 7
            str0 = ‘+7′
        elif i == ‘-’:
            s -= 5
            str0 = ‘-5′
        elif i == ‘*’:
            s *= 3
            str0 = ‘*3′
        elif i == ‘/’:
            s /= 2
            str0 = ‘/2′
        print ‘%s%s=%d’ % (strs, str0, s)
    print s

hasmethod = 0
calc(‘+’, 2011, ”, ‘r’)