Classical FP problem in python : Hamming problem
Francis Girard
francis.girard at free.fr
Wed Jan 26 23:41:07 CET 2005
Le mardi 25 Janvier 2005 09:01, Michael Spencer a écrit :
> Francis Girard wrote:
> > The following implementation is even more speaking as it makes
> > self-evident and almost mechanical how to translate algorithms that run
> > after their tail from recursion to "tee" usage :
>
> Thanks, Francis and Jeff for raising a fascinating topic. I've enjoyed
> trying to get my head around both the algorithm and your non-recursive
> implementation.
>
Yes, it's been fun.
> Here's a version of your implementation that uses a helper class to make
> the algorithm itself prettier.
>
> from itertools import tee, imap
>
> def hamming():
> def _hamming():
> yield 1
> for n in imerge(2 * hamming, imerge(3 * hamming, 5 * hamming)):
> yield n
>
> hamming = Tee(_hamming())
> return iter(hamming)
>
>
> class Tee(object):
> """Provides an indepent iterator (using tee) on every iteration
> request Also implements lazy iterator arithmetic"""
> def __init__(self, iterator):
> self.iter = tee(iterator,1)[0]
> def __iter__(self):
> return self.iter.__copy__()
> def __mul__(self, number):
> return imap(lambda x: x * number,self.__iter__())
>
> def imerge(xs, ys):
> x = xs.next()
> y = ys.next()
> while True:
> if x == y:
> yield x
> x = xs.next()
> y = ys.next()
> elif x < y:
> yield x
> x = xs.next()
> else: # if y < x:
> yield y
> y = ys.next()
>
> >>> hg = hamming()
> >>> for i in range(10000):
>
> ... n = hg.next()
> ... if i % 1000 == 0: print i, n
> ...
> 0 1
> 1000 51840000
> 2000 8100000000
> 3000 279936000000
> 4000 4707158941350
> 5000 50960793600000
> 6000 409600000000000
> 7000 2638827906662400
> 8000 14332723200000000
> 9000 68024448000000000
>
>
Interesting idea.
> Regards
>
> Michael
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