是否可以从此函数中删除递归？

Question

我一直在玩这个，但是看不出明显的解决方案。我想从XinY_Go函数中删除递归。

def XinY_Go(x,y,index,slots):
   if (y - index) == 1:
      slots[index] = x
      print slots
      slots[index] = 0
      return
   for i in range(x+1):
      slots[index] = x-i
      XinY_Go(x-(x-i), y, index + 1, slots)

def XinY(x,y):
   return XinY_Go(x,y,0,[0] * y)

该函数正在计算将X弹珠放入Y个插槽的方法数。以下是一些示例输出：

 >>> xy.XinY(1,2)
 [1, 0]
 [0, 1]
 >>> xy.XinY(2,3)
 [2, 0, 0]
 [1, 1, 0]
 [1, 0, 1]
 [0, 2, 0]
 [0, 1, 1]
 [0, 0, 2]

Answer 1

我们认为作为递归的所有东西也可以被认为是基于堆栈的问题，其中递归函数只使用程序的调用堆栈而不是创建单独的堆栈。这意味着可以使用堆栈重写任何递归函数。

我不太了解python足以为您提供实现，但这应该指向正确的方向。但简而言之，将函数的初始参数推送到堆栈并添加一个循环，只要堆栈的大小大于零就会运行。每次循环迭代弹出一次，每次函数当前调用自身时都按下。

Answer 2

@Joel Coehoorn's suggestion的简单实现如下：

def XinY_Stack(x, y):
    stack = [(x, 0, [0]*y)]
    while stack:
        x, index, slots = stack.pop()
        if (y - index) == 1:
            slots[index] = x
            print slots
            slots[index] = 0
        else:
            for i in range(x + 1):
                slots[index] = x-i
                stack.append((i, index + 1, slots[:]))

示例：

>>> XinY_Stack(2, 3)
[0, 0, 2]
[0, 1, 1]
[0, 2, 0]
[1, 0, 1]
[1, 1, 0]
[2, 0, 0]

基于itertools.product

def XinY_Product(nmarbles, nslots):
    return (slots
            for slots in product(xrange(nmarbles + 1), repeat=nslots)
            if sum(slots) == nmarbles) 

基于嵌套循环

def XinY_Iter(nmarbles, nslots):
    assert 0 < nslots < 22 # 22 -> too many statically nested blocks
    if nslots == 1: return iter([nmarbles])
    # generate code for iter solution
    TAB = "  "
    loopvars   = []
    stmt       = ["def f(n):\n"]
    for i in range(nslots - 1):
        var = "m%d" % i
        stmt += [TAB * (i + 1), "for %s in xrange(n - (%s)):\n"
                 % (var, '+'.join(loopvars) or 0)]
        loopvars.append(var)

    stmt += [TAB * (i + 2), "yield ", ','.join(loopvars),
             ', n - 1 - (', '+'.join(loopvars), ')\n']
    print ''.join(stmt)
    # exec the code within empty namespace
    ns = {}
    exec(''.join(stmt), ns, ns)
    return ns['f'](nmarbles + 1) 

示例：

>>> list(XinY_Product(2, 3))
[(0, 0, 2), (0, 1, 1), (0, 2, 0), (1, 0, 1), (1, 1, 0), (2, 0, 0)]
>>> list(XinY_Iter(2, 3))
def f(n):
  for m0 in xrange(n - (0)):
    for m1 in xrange(n - (m0)):
      yield m0,m1, n - 1 - (m0+m1)

[(0, 0, 2), (0, 1, 1), (0, 2, 0), (1, 0, 1), (1, 1, 0), (2, 0, 0)]

Answer 3

请查看此代码以创建所有排列，我想我可以相对简单地为您的问题实现类似的东西。

How to generate all permutations of a list in python?