[Haskell-cafe] Re: Need some help with an infinite list
tphyahoo at gmail.com
Sat Jun 20 14:16:34 EDT 2009
could someone explain sharing?
In the code below, allstrings2 is 6X as fast as allstrings. I assume
because of sharing, but I don't intuitively see a reason why.
can someone give me some pointers, perhaps using debug.trace or other
tools (profiling?) to show where the first version is being
letters = ['a'..'z']
strings 0 = [""]
strings n = [ c : s | c <- letters, s <- strings (n-1) x ]
allstrings = concat $ map strings [1..]
allstrings2 = let sss = [""] : [ [ c:s | c <- letters, s <- ss ] | ss <- sss ]
in concat $ tail sss
t = allstrings !! wanted
t2 = allstrings2 !! wanted
wanted = (10^2)
2009/6/18 Lee Duhem <lee.duhem at gmail.com>:
> On Fri, Jun 19, 2009 at 6:17 AM, Matthew Brecknell<haskell at brecknell.org> wrote:
>> On Thu, 2009-06-18 at 23:57 +0800, Lee Duhem wrote:
>>> [...] I have prepared a blog post for how
>>> I worked out some of these answers, here is the draft of it, I hope it
>>> can help you too.
>> Nice post! Certainly, pen-and-paper reasoning like this is a very good
>> way to develop deeper intuitions.
>>> Answer 1 (by Matthew Brecknell):
>>> concat $ tail $ iterate (map (:) ['a' .. 'z'] <*>) []
>> I actually said "tail $ concat $ iterate ...", because I think the
>> initial empty string is logically part of the sequence. Tacking "tail"
>> on the front then produces the subsequence requested by the OP.
> Yes, I changed your solution from "tail $ concat $ iterate ..." to
> "concat $ tail $ iterate ...", because I think cut useless part out early
> is good idea, forgot to mention that, sorry.
>> I should have given more credit to Reid for this solution. I'm always
>> delighted to see people using monadic combinators (like replicateM) in
>> the list monad, because I so rarely think to use them this way. Sadly,
>> my understanding of these combinators is still somewhat stuck in IO,
>> where I first learned them. I never would have thought to use <*> this
>> way if I had not seen Reid's solution first.
> Actually, I first figure out how Reid's solution works, then figure out yours.
> After that, I found, for me, your solution's logic is easier to understand,
> so I take it as my first example. As I said at the end, or as I'll
> said at the end,
> Reid' solution and yours are the same (except effective)
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