Op 2006-04-04, Tomi Lindberg schreef <[EMAIL PROTECTED]>: > First, thanks to Antoon and Alexander for replying. > > Antoon Pardon wrote: > >> It would be better to construct distributions for one >> die and make a function that can 'add' two distributions >> together. > > As both replies pointed to this direction, I tried to take > that route. Here's the unpolished code I came up with. Does > it look even remotely sane way to accomplish my goal?

> -- code begins -- > > # A die with n faces > D = lambda n: [x+1 for x in range(n)] > > # A new die with 6 faces > d6 = D(6) > > # Adds another die to results. > def add_dice(sums, die): > # If first die, all values appear once > if not sums: > for face in die: > sums[face] = 1 > # Calculating the number of appearances for additional > # dice > else: > new_sums = {} > for k in sums.keys(): > for f in die: > if new_sums.has_key(k+f): > new_sums[k+f] += sums[k] > else: > new_sums[k+f] = sums[k] > sums = new_sums > return sums > > sums = add_dice({}, d6) > sums = add_dice(sums, d6) > sums = add_dice(sums, d6) > > -- code ends -- IMO you are making things too complicated and not general enough. Here is my proposal. ----- import operator class Distribution(dict): '''A distribution is a dictionary where the keys are dice totals and the values are the number of possible ways this total can come up ''' def __add__(self, term): '''Take two distributions and combine them into one.''' result = Distribution() for k1, v1 in self.iteritems(): for k2, v2 in term.iteritems(): k3 = k1 + k2 v3 = v1 * v2 try: result[k3] += v3 except KeyError: result[k3] = v3 return result def __rmul__(self, num): tp = num * [self] return reduce(operator.add, tp) def D(n): ''' One die has a distribution where each result has one possible way of coming up ''' return Distribution((i,1) for i in xrange(1,n+1)) sum3d6 = 3 * D(6) sum2d6p2d4 = 2 * D(6) + 2 * D(4) ----- -- Antoon Pardon -- http://mail.python.org/mailman/listinfo/python-list