Andrey Yankin
Greetings to all!
anew, which means that delayed arrays are inefficient when the data is needed multiple times"(http://www.haskell.org/haskellwiki/Numeric_Haskell:_A_Repa_Tutorial).I started diving into Guiding Parallel Array Fusion with Indexed Types mentioned
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 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 processors! 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)) and replace g <- computeP $ accumulate n grid :: IO (Array U DIM2 Int) by 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. Dominic.