I wrote:
merge' (x:xs) ys = x : merge xs ys
hammingx = 1 : foldr1 merge' [map (h x) hammingx | x - hammingx]
Sorry. 'foldr1' is still slightly too strict. (Why doesn't the Haskell
report define foldr1 in terms of foldr?)
The code that works is
marge' [] ys = ys
merge'
original-
De: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] En nombre de Bertram Felgenhauer
Enviado el: viernes, 18 de enero de 2008 8:36
Para: haskell-cafe@haskell.org
Asunto: Re: [Haskell-cafe] Hamming's Problem
Jose Luis Reyes F. wrote:
Hi,
In exercise 2 of http://www.walenz.org/Dijkstra
Jose Luis Reyes F. wrote:
Thanks for your help, I tried to solve the problem this weekend but I have
some dudes.
My reasoning is:
The sequence is [1,a1,a2,a3,..] where a1 = h 1 1, a2 = h 1 a1, and so on
So, I want to construct the list of lists
[[1,a1,a2,a3],
[h 1 1, h 1 a1, h 1
Jose Luis Reyes F. wrote:
Hi,
In exercise 2 of http://www.walenz.org/Dijkstra/page0145.html we need to
write a function that holds
(1)The value 1 is in the sequence
(2)If x and y are in the sequence, so is f(x,y), where f has the
properties
a. f(x,y) x
b. (y1 y2) =
On 18/01/2008, Bertram Felgenhauer [EMAIL PROTECTED] wrote:
I have some ideas, but maybe you want to work on a correct solution
first?
There is an implementation on the wikipedia article for Haskell:
hamming = 1 : map (2*) hamming `merge` map (3*) hamming `merge` map (5*) hamming
where
Hi,
In exercise 2 of http://www.walenz.org/Dijkstra/page0145.html we need to
write a function that holds
(1)The value 1 is in the sequence
(2)If x and y are in the sequence, so is f(x,y), where f has the
properties
a. f(x,y) x
b. (y1 y2) = (f(x,y1)f(x,y2))
This
The author of Pugs (Perl6 implemented in Haskell) gives a nice solution
to the problem of generating the Hamming numbers in the following interview:
http://www.perl.com/pub/a/2005/09/08/autrijus-tang.html?page=last
___
Haskell-Cafe mailing list
