[Haskell-cafe] Repa: traverse delayed array multiple times
dominic at steinitz.org
Mon Jan 14 09:08:24 CET 2013
Andrey Yankin <yankin013 <at> gmail.com> writes:
> Greetings to all!
repa?I wrote this rough sketch that shows what I am into. Apparently, program
is severely slow. I think reason is:"Every time an element is requested from a
delayed array it is calculated
> anew, which means that delayed arrays are inefficient when the data is
> needed multiple
started diving into Guiding Parallel Array Fusion with Indexed Types mentioned
there. Is it a right place to find any answer to my question?
It's certainly possible to do this in repa. As I understand it, that is pretty
much what repa is for. I'm not sure about your explanation for the slowdown
although it sounds plausible.
I amended your code slightly and it runs pretty quickly for me using my 2
updater :: Source r Int => Array r DIM2 Int -> Array D DIM2 Int
updater a = traverse a id step
updaterM :: Monad m => Int -> Array U DIM2 Int -> m (Array U DIM2 Int)
updaterM n = foldr (>=>) return (replicate n (computeP . updater))
g <- computeP $ accumulate n grid :: IO (Array U DIM2 Int)
g <- updaterM n grid
BTW I think you can use stencils for what you are doing (I assume it is some
sort of relaxation method) as the coefficients are constant. I think this
should speed things up further as you won't be testing the boundary conditions
on every loop.
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