Even shorter:
c=mapM(\k->[0..k])
- Lyle
Fritz Ruehr wrote:
Well, as far as that goes, we can shave off a little bit (around 7%)
this way:
combs = mapM (\k->[0..k])
(As a bonus, it's even a bit more cryptic/symbolic, in the fine
tradition of APL one-liner character-shavings.)
But who's countin
Well, as far as that goes, we can shave off a little bit (around 7%)
this way:
combs = mapM (\k->[0..k])
(As a bonus, it's even a bit more cryptic/symbolic, in the fine
tradition of APL one-liner character-shavings.)
But who's counting? :) :) :)
-- Fritz Ruehr
On Aug 11, 2004, at 3:22 PM,
I bow to your superior ingenuity. :)
- Lyle
>
> Why so long-winded :-)?
>
> combs = mapM (enumFromTo 0)
>
> mike
>
> Lyle Kopnicky <[EMAIL PROTECTED]> writes:
> ...
>> Here is the
>> improved version:
>>
>> combs [] = [[]]
>> combs (n:r) = [i:cr | i <- [0..n], cr <- combs r]
> ...
>
___
Hi Cale,
On Wed, Aug 11, 2004 at 02:38:51PM -0400, Cale Gibbard wrote:
> I would write it as one of the following,
>
> keys [] = [[]]
> keys (x:xs) = [0..x] >>= (\k -> map (k:) (keys xs))
> -- or,
> keys' [] = [[]]
> keys' (x:xs) = do { k <- [0..x] ; map (k:) (keys' xs) }
> -- or,
> keys'' []
Why so long-winded :-)?
combs = mapM (enumFromTo 0)
mike
Lyle Kopnicky <[EMAIL PROTECTED]> writes:
...
> Here is the
> improved version:
>
> combs [] = [[]]
> combs (n:r) = [i:cr | i <- [0..n], cr <- combs r]
...
___
Haskell-Cafe mailing lis
Henning Thielemann wrote:
On Wed, 11 Aug 2004, Lyle Kopnicky wrote:
Here's my version:
combs [] = []
combs [n] = [[i] | i <- [0..n]]
combs (n:r) = let combsr = combs r in [i:cr | i <- [0..n], cr <- combsr]
Since there is one zero combination, it should be
combs [] = [[]]
Ah, yes. I
Oops, just realised that I didn't include the mailing list in my
original reply to Florian -- here was my reply:
-
Hi,
I would write it as one of the following,
keys [] = [[]]
keys (x:xs) = [0..x] >>= (\k -> map (k:) (keys xs))
-- or,
keys' [] = [[]]
keys' (x:xs) = do { k <- [0..x] ;
On Wed, 11 Aug 2004, Lyle Kopnicky wrote:
> Here's my version:
>
> combs [] = []
> combs [n] = [[i] | i <- [0..n]]
> combs (n:r) = let combsr = combs r in [i:cr | i <- [0..n], cr <- combsr]
Since there is one zero combination, it should be
> combs [] = [[]]
Then you can also remove the defini
Here's my version:
combs [] = []
combs [n] = [[i] | i <- [0..n]]
combs (n:r) = let combsr = combs r in [i:cr | i <- [0..n], cr <- combsr]
- Lyle
Florian Boehl wrote:
Hi,
I'ld like to generate a list (of lists) that contains all combinations
of natural numbers stored in another list. It should work
On Tue, 10 Aug 2004, Florian Boehl wrote:
> I'ld like to generate a list (of lists) that contains all combinations
> of natural numbers stored in another list. It should work like a
> combination-lock. E.g.:
>
> [2,2] -> [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]]
I have written a s
G'day all.
At 06:01 10/08/04 +0200, Florian Boehl wrote:
>If I know the length 'l' of the 'locklist', I can solve the
>problem via generators. E.g.:
>
>l = 2: [[a,b] | a <- [0..locklist!!0], b <- [0..locklist!!1]]
>
>But if the length is unknown (because it's dynamic) this solutions (of
>course)
At 06:01 10/08/04 +0200, Florian Boehl wrote:
Hi,
I'ld like to generate a list (of lists) that contains all combinations
of natural numbers stored in another list. It should work like a
combination-lock. E.g.:
[2,2] -> [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]]
If I know the length 'l'
Just make the function recursive.
There is a simple relation between,
l [a,b,c] and l [b,c]
Tom
On Tue, 10 Aug 2004 14:01, Florian Boehl wrote:
> Hi,
>
> I'ld like to generate a list (of lists) that contains all combinations
> of natural numbers stored in another list. It should work like a
> co
Hi,
I'ld like to generate a list (of lists) that contains all combinations
of natural numbers stored in another list. It should work like a
combination-lock. E.g.:
[2,2] -> [[0,0],[0,1],[0,2],[1,0],[1,1],[1,2],[2,0],[2,1],[2,2]]
If I know the length 'l' of the 'locklist', I can solve the
problem
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