Question

我正在尝试使用给定的x，y和z查找可能的排列总数。 x是初始数字，y是最终数字，z是按下的按钮总数。我应该只是像国际象棋中的骑士一样从一个数字到一个数字，在一个＆＃39; L＆＃39;形状。例如，如果您刚刚拨打1，则您拨打的下一个号码必须是6或8.如果您刚刚拨打了6，则下一个号码必须是1或7。目前，我的实现输出了我给出的所有数字的正确答案。然而，由于计算时间是指数的，所以上帝可怕的慢。我想知道的是我如何在线性时间内或多或少地计算出来。 z将始终介于1和100之间。

##Computes the number of phone numbers one
##can dial that start with the digit x, end
##in the digit y, and consist of z digits in
##total. Output this number as a
##string representing the number in base-10.
##Must follow "knights rule" moves, like chess
##########_________##########
##########|1||2||3|##########
##########|_||_||_|##########
##########|4||5||6|##########
##########|_||_||_|##########
##########|7||8||9|##########
##########|_||_||_|##########
##########|_||0||_|##########
##########^^^^^^^^^##########
dict = {0: [4, 6], 1: [6, 8], 2: [7, 9], 3: [4, 8],
    4: [0, 3, 9], 5: [], 6: [0, 1, 7], 7: [2, 6], 8: [1, 3],
    9: [2, 4]}


def recAnswer(current, y, z, count, total):
    if count == z and current == y:
            total += 1
            return total
    count+=1
    if count > z:
            return total
    for i in dict.get(current):
            total = recAnswer(i, y, z, count, total)
    return total

def answer(x, y, z):
    if x == y:
            if z%2 == 0:
                    return '0'
    elif x == 5 or y == 5:
            if z == 1 and x == y:
                    return '1'
            else:
                    return '0'
    elif x%2 != 0 and y%2 == 0:
            if z%2 != 0:
                    return '0'
    elif x%2 == 0 and y%2 != 0:
            if z%2 != 0:
                    return '0'
    elif x%2 == 0 and y%2 ==0:
            if z%2 == 0:
                    return '0'
    elif x%2 != 0 and y%2 != 0:
            if z%2 == 0:
                    return '0'

    total = recAnswer(x,y,z,1,0)
    return str(total)

def test():
    for i in xrange(1,15,1):
            print i,":",answer(1,3,i)

    print answer(6, 2, 5)
    print answer(1, 6, 3)
    print answer(1, 1, 99)


test()

Answer 1

您的代码速度慢的原因是您最终反复访问（并重新计算）相同的组合。您可以使用称为memoization的技术缩短重新计算部分。

记忆很容易添加，但让我们重新设计你的递归函数，以便调用函数进行累积，而不是函数本身。换句话说，不要传递总数并仅返回此子路径的组合：

def recAnswer(current, y, z, count):
    if count == z and current == y:
        return 1

    count += 1
    if count > z:
        return 0

    total = 0
    for i in dict.get(current):
        total += recAnswer(memo, i, y, z, count)

    return total

此更改不会更改计算本身;结果仍然相同。

现在让我们将所有重复的调用切断为相同的参数。我们将字典memo传递给函数。这个dict的关键是你的函数参数的元组。作为递归函数的第一步，检查计算是否已完成。作为初始计算的最后一步，将解决方案添加到dict：

def recAnswer(memo, current, y, z, count):
    # dict key is the tuple of arguments
    key = (current, y, z, count)

    # Have we been here before? If so, return memoized answer
    if key in memo:
        return memo[key]

    if count == z and current == y:
        return 1

    count += 1
    if count > z:
        return 0

    total = 0
    for i in dict.get(current):
        total += recAnswer(memo, i, y, z, count)

    # Store answer for later use
    memo[key] = total

    return total

然后用空字典进行计算：

total = recAnswer({}, x, y, z, 1)

附录：现在我已经了解了@decorator s，我将使用memoization来装饰该函数，以便不更改原始函数。好吧，我将再做一次更改，正如Janne在评论中提到的那样：我将目标cound和当前计数合并为一个变量，计数从目标值开始并倒数到零而是达到目标值。

首先，memoization装饰器，它将是一个包含装饰函数的类，func和memoization字典。该类必须使用所需数量的参数实现功能__call__：

class memoized(object):
    """Decorator function that adds the memoization"""

    def __init__(self, func):
        self.func = func
        self.memo = {}

    def __call__(self, current, target, count): 
        key = (current, target, count)
        if key not in self.memo:
            self.memo[key] = self.func(current, target, count)
        return self.memo[key]

现在是在def inition之前使用装饰器的简化函数：

@memoized
def recAnswer(current, target, count):
    """Unmemoized original function"""

    if count == 0:
        return int(current == target)       # False: 0, True: 1

    total = 0
    for next in dict[current]:
        total += recAnswer(next, target, count - 1)

    return total

@memoized装饰器通过recAnswer调用函数memoized.__call__来处理memoization。像这样调用递归函数：

total = recAnswer(x, y, z - 1)

（这里的-1考虑到在原始代码中，计数从1开始。）

可能还有改进的余地。例如，您可以使用splat语法为memoized装饰器类变量创建参数数量，以便您可以将memoizer重新用于其他函数：

def __call__(self, *args): 
    if args not in self.memo:
        self.memo[args] = self.func(*args)
    return self.memo[args]

所有这一切的结果是，如果你有一个问题，你一次又一次地重新评估同一组参数，只需跟踪以前计算的结果可以给你一个巨大的加速，而不必摆弄基本实施。

Answer 2

来自codereview

"""Solve the phone/chess paths problem from this question:
https://codereview.stackexchange.com/questions/71988
"""

# Move dictionary
MOVES = {0: [4, 6],
         1: [6, 8],
         2: [7, 9],
         3: [4, 8],
         4: [0, 3, 9],
         5: [],
         6: [0, 1, 7],
         7: [2, 6],
         8: [1, 3],
         9: [2, 4]}

# Cache implementation
def cache(func):
    """Standard cache decorator implementation."""
    cache_dct = {}
    def wrapper(*args):
        if args not in cache_dct:
            cache_dct[args] = func(*args)
        return cache_dct[args]
    return wrapper

# Recusive function 
@cache
def rec(x, y, z):
    """Recursively count the number of path
    to go from x to y in z moves.
    """
    if not z:
        return int(x == y)
    return sum(rec(x, y, z-1) for x in MOVES[x])

# Paths function
def paths(x, y, z):
    """Count the number of paths to go from x to y
    with z key strokes.
    """
    if not z:
        return 0
    return rec(x, y, z-1)

# Main execution
if __name__ == "__main__":
    example = "paths(1, 1, 99)"
    print example + " = " + str(eval(example))
    # Output: paths(1, 1, 99) = 30810672576979228350316940764381184

电话号码的排列从按钮x开始，到按钮y结束，z号码长

2 个答案: