# [Haskell-cafe] Hamming's Problem

Jose Luis Reyes F. jlareyes at cybercia.com
Mon Jan 21 07:42:08 EST 2008

```Bertram

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 a2,..],
[h a1 1, h a1 a1, h a1 a2,...]
...
[h an 1, h an a2,h an a3,...]
...
]

The function is

hammingx = 1 : foldl1 merge [ map (h x) hammingx | x <- hammingx]

However this is not a solution because the list is a infinite list of
infinite lists.

Please some advices to resolve this problem.

Thanks
Jose Luis

-----Mensaje original-----
De: haskell-cafe-bounces at haskell.org
[mailto:haskell-cafe-bounces at haskell.org] En nombre de Bertram Felgenhauer
Enviado el: viernes, 18 de enero de 2008 8:36
Para: haskell-cafe at haskell.org
Asunto: Re: [Haskell-cafe] Hamming's Problem

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) => (f(x,y1)>f(x,y2))
>
> This is a solution for this problem, but an inefficient one
>
> hammingff :: [Integer]
> hammingff = 1 : merge [ h x y | x <- hammingff, y <- hammingff ]
>                       [ h x y | y <- hammingff, x <- hammingff ]

That's not sufficient. In fact, because hammingff is an infinite sequence,
this is equivalent to

hammingff = 1 : merge [ h (head hammingff) y | y <- hammingff ]
[ h x (head hammingff) | x <- hammingff ]

> h x y = 2*x+3*y
[snip merge function]

Indeed, the first few terms of hammingff are [1,5,13,17,29], and the
value h 5 5 = 25 is missing.

> anybody  has a better solution?.

I have some ideas, but maybe you want to work on a correct solution
first?

Btw, with your choice of h, the result list will contain all natural
numbers that are not divisible by 3 and are = 1 (mod 4). Evaluating the
first n elements of that list will require O(n^2) evaluations of h.

Bertram
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