Reverse Iteration Through Integers

[Mick Krippendorf]

Paul Rubin wrote:
> Yet another way is to use recursion.  I'll leave that as an exercise too.

This time with big numbers:


def trampoline(bouncing, *args, **kwargs):
    while bouncing:
        result, bouncing, args, kwargs = bouncing(*args, **kwargs)
        if result:
            return result()

def bouncy(function):
    return lambda *args, **kwargs:(None, function, args, kwargs)

def land(result=None):
    return lambda:result, None, None, None


def reverse(n):
    @bouncy
    def rev(i=n, j=0, k=0):
        if i:
            return rev(*divmod(i, 10), k=(j+k)*10)
        return land(j + k)
    return trampoline(rev)


print reverse(169883903200298309284038223098439430943092816286 ** 123)


Try it without the @bouncy decoration.

Granted, the code looks like a serious case of Haskell envy, but after
recursion and tail call optimization being cryptic was just the logical
consequence ;-)

Mick.