Just for fun: Countdown numbers game solver

[Paul Rubin]

dg.google.groups at thesamovar.net writes:
> Unfortunately I realise I stated the problem imprecisely. You're only
> allowed to use each number once (otherwise there's a trivial solution
> for every problem, i.e. x/x + x/x + x/x + ... + x/x repeated y times
> for target y given any source number x). Trying your program on 234
> from [100,9,7,6,3,1] gives you 9*9*3-9 using the 9 three times.

Here's an inefficient solution, that doesn't find 234 but finds 253.
If you see a double post, it's because I posted something similar a
little while ago but cancelled it since it had a bug.  I'm not sure
this one is correct either ;-).

    from operator import *

    def countdown(nums, trace='', ops=[add,mul,sub,div]):
        n0,n1s = nums[0], nums[1:]
        if not n1s:
            yield n0, str(n0)
            return
        for f in ops:
            for r,t in countdown(n1s, trace, [add, mul, sub]):
                if f != div or r != 0 and n0 % r == 0:
                    yield f(n0, r), '%s(%s,%s)'% (f.__name__, n0, t)

    def find_repr(target, nums):
        # print all representations of target from nums
        for x,t in countdown(nums):
            if x == target:
                print x,t

    find_repr(253, [100,9,7,6,3,1])