[Haskell-cafe] Re: Fair diagonals
Louis Wasserman
wasserman.louis at gmail.com
Thu Nov 5 16:04:35 EST 2009
I figured out an inductive approach as follows, which lets you derive
stripeN from stripe(N-1). This could be TemplateHaskell'd if you have a
bound on N; I'm still trying to figure out a type-magical alternative.
Suppose stripe(N-1) :: x -> [b] -> [c]
Then
stripeN :: (a -> [b]) -> x -> [a] -> [[c]]
stripeN f x [] = []
stripeN f x (a:as) = case stripe(N-1) x (f a) of
[] -> stripeN f x as
(b:bs) -> [b]:zipCons bs (stripeN f x as)
and then diagN is obtained by applying concat an appropriate number of
times. It's fair. The real-question is how to type-magically work it for
arbitrarily many coordinates...
Louis Wasserman
wasserman.louis at gmail.com
http://profiles.google.com/wasserman.louis
On Wed, Nov 4, 2009 at 4:38 AM, Martijn van Steenbergen <
martijn at van.steenbergen.nl> wrote:
> Louis Wasserman wrote:
>
>> +1 on Control.Monad.Omega. In point of fact, your diagN function is
>> simply
>>
>> diagN = runOmega . mapM Omega
>>
>> You'll find it an interesting exercise to grok the source of
>> Control.Monad.Omega, obviously, but essentially, you're replacing concatMap
>> with a fair (diagonal) traversal order version.
>>
>
> Thanks for the replies!
>
> I've looked at Omega but it's not fair enough. The sums of the indices are
> not non-decreasing:
>
> map sum $ runOmega . mapM each $ [[1..], [1..], [1..]]
> [3,4,4,4,5,5,5,5,6,6,5,6,6,7,7,5,6,7,7,8,8,6,6,7,8,8,9,9,6,7,...
>
> Is there another way to use Omega that meets this (very important)
> criterion or is Omega not the right tool here?
>
> Thanks,
>
> Martijn.
>
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