some OT: how to solve this kind of problem in our program?

[bearophileHUGS at lycos.com]

Using Psyco this version is much faster, you can test it on your PC
compared to the other one (the whole running time, Psyco compilation
too):
Psyco is unable to speed up generator functions, so you have to return
true lists.
Giving the func to the permutation function, you can avoid lot of list
copying and unpacking.


try:
    import psyco
    psyco.full()
except ImportError:
    pass

d0, d1 = 1, 2


def func(p):
    a0,a1,a2,b0,b1,b2,c0,c1,c2 = p
    # do application evaluation here
    b1b2 = 10*b1+b2
    a1a2 = 10*a1+a2
    c1c2 = 10*c1+c2
    if d1*a0*b1b2*c1c2 + d1*b0*a1a2*c1c2 + d1*c0*a1a2*b1b2 \
       == d0*a1a2*b1b2*c1c2:
        return sorted( [[a0, a1, a2], [b0, b1, b2], [c0, c1, c2]] )
    else:
        return None


def accepted_permutations(alist, func):
    # func must return None for the unacceptable results
    # Algoritm from Phillip Paul Fuchs, modified
    result = []
    items = alist[:]
    n = len(alist)
    p = range(n+1)
    i = 1
    r = func(alist)
    if r is not None: result.append(r)
    while i < n:
        p[i] -= 1
        if i & 1:
            j = p[i]
        else:
            j = 0
        alist[j], alist[i] = alist[i], alist[j]
        r = func(alist)
        if r is not None: result.append(r)
        i = 1
        while p[i] == 0:
            p[i] = i
            i += 1
    return result


def main():
  result = []
  for aresult in accepted_permutations(range(1, 10), func):
    if aresult not in result:
      result.append(aresult)
      [[a0, a1, a2], [b0, b1, b2], [c0, c1, c2]] = aresult
      print '  %0d     %0d     %0d     %0d' % (a0, b0, c0, d0)
      print '--- + --- + --- = ---'
      print ' %0d%0d    %0d%0d    %0d%0d
%0d'%(a1,a2,b1,b2,c1,c2,d1)
      print

main()

Bye,
bearophile