河内塔-Python中的中途求解算法

有可能中途解决河内塔问题吗?我已经做了大量的研究,寻找能够解决一半用户配置问题的代码,但我还没有找到。这是一个任务,我要求代码从用户停止解算的地方接管,并继续为用户解算,而不必将谜题重置为方块。

我知道有递归算法随时可用,但这不是我要搜索的。 我正在搜索一些算法,这些算法可以从用户解决问题的地方一直到,然后从那里继续解决问题。 有什么想法吗?

到目前为止,我已经提出了一种算法,它将优化后的算法(通过递归实现)存储到一个数组中,然后检查用户的输入是否等于数组中的任何输入,然后从那里继续求解。然而,问题在于在优化算法数组中找不到用户配置。

以下是迄今为止我的代码(我排除了stack.py代码):

def solveHalfway(n, start, end, middle, count):
    gameInstance.stackA = [3,2]
    gameInstance.stackB = []
    gameInstance.stackC = [1]
    loopCounter = 0 # initialise loopCounter as 0
    moveCounter = 0 # initialise the move index the user is stuck at
    indicator = 0 # to indicate whether the user's config equals the solution's config
    while loopCounter < arrayOfStacks.size(): # while loopCounter size has not reached the end of arrayOfStacks
        if loopCounter != 0 and loopCounter % 3 == 0:  # if 3 stacks have been dequeued
            moveCounter += 1
            if gameInstance.getUserConfig() == tempStack.data:  #check whether user's config is equal to the solution's config
                indicator += 1
                print "User is stuck at move: ", moveCounter  #this will be the current move the user is at
                while arrayOfStacks.size() != 0: # while not the end of arrayOfStacks
                    correctMovesStack.push(arrayOfStacks.dequeue())  # add the moves to correctMovesStack
                    if correctMovesStack.size() == 3: # if 3 stacks have been dequeued
                        print "Step:", moveCounter , correctMovesStack.data # display the step number plus the correct move to take
                        moveCounter+=1 # increase move by 1
                        while correctMovesStack.size() != 0: # if correct moves stack isn't empty
                            correctMovesStack.pop() # empty the stack
                return
            else:
                while tempStack.size() != 0: # check if tempStack is empty
                    tempStack.pop()  # empty tempStack so that it can be used for the next loop
            tempStack.push(arrayOfStacks.dequeue()) #dequeue from arrayOfStacks for a total of 3 times and push it to tempStack
        else:
            tempStack.push(arrayOfStacks.dequeue()) #dequeue from arrayOfStacks for a total of 3 times and push it to tempStack
        loopCounter +=1 # increase loop counter by 1
    if indicator == 0:
        moveWith3Towers(noOfDisks, stackA, stackC, stackB, count)
    print indicator