I don't mean to be blunt, but have you guys taken a course in linear algebra?

On Mon, Dec 3, 2012 at 9:21 PM, Trevor L. McDonell <tmcdonell@cse.unsw.edu.au> wrote:

As far as I am aware, the only description is in the Repa paper. I you are right, it really should be explained properly somewhere… 

At a simpler example, here is the outer product of two vectors [1].

vvProd :: (IsNum e, Elt e) => Acc (Vector e) -> Acc (Vector e) -> Acc (Matrix e)

vvProd xs ys = A.zipWith (*) xsRepl ysRepl

  where

    n           = A.size xs

    m           = A.size ys

    xsRepl      = A.replicate (lift (Z :. All :. m  )) xs

    ysRepl      = A.replicate (lift (Z :. n   :. All)) ys

If we then `A.fold (+) 0` the matrix, it would reduce along each row producing a vector. So the first element of that vector is going to be calculated as (xs[0] * ys[0] + xs[0] * ys[1] +  … xs[0] * ys[m-1]). That's the idea we want for our matrix multiplication … but I agree, it is difficult for me to visualise as well.

I do the same sort of trick with the n-body demo to get all n^2 particle interactions.

-Trev

 [1]: http://en.wikipedia.org/wiki/Outer_product#Vector_multiplication

On 04/12/2012, at 3:41 AM, Clark Gaebel <cgaebel@uwaterloo.ca> wrote:

Ah. I see now. Silly Haskell making inefficient algorithms hard to write and efficient ones easy. It's actually kind of annoying when learning, but probably for the best.

Is there a good write-up of the algorithm you're using somewhere? The Repa paper was very brief in its explaination, and I'm having trouble visualizing the mapping of the 2D matricies into 3 dimensions.

  - Clark

On Mon, Dec 3, 2012 at 2:06 AM, Trevor L. McDonell <tmcdonell@cse.unsw.edu.au> wrote:

Hi Clark,

The trick is that most accelerate operations work over multidimensional arrays, so you can still get around the fact that we are limited to flat data-parallelism only.

Here is matrix multiplication in Accelerate, lifted from the first Repa paper [1].

import Data.Array.Accelerate as A

type Matrix a = Array DIM2 a

matMul :: (IsNum e, Elt e) => Acc (Matrix e) -> Acc (Matrix e) -> Acc (Matrix e)

matMul arr brr

  = A.fold (+) 0

  $ A.zipWith (*) arrRepl brrRepl

  where

    Z :. rowsA :. _     = unlift (shape arr)    :: Z :. Exp Int :. Exp Int

    Z :. _     :. colsB = unlift (shape brr)    :: Z :. Exp Int :. Exp Int

    arrRepl             = A.replicate (lift $ Z :. All   :. colsB :. All) arr

    brrRepl             = A.replicate (lift $ Z :. rowsA :. All   :. All) (A.transpose brr)

If you use github sources rather than the hackage package, those intermediate replicates will get fused away.

Cheers,

-Trev

 [1] http://www.cse.unsw.edu.au/~chak/papers/KCLPL10.html

On 03/12/2012, at 5:07 PM, Clark Gaebel <cgaebel@uwaterloo.ca> wrote:

Hello cafe,

I've recently started learning about cuda and hetrogenous programming, and have been using accelerate [1] to help me out. Right now, I'm running into trouble in that I can't call parallel code from sequential code. Turns out GPUs aren't exactly like Repa =P.

Here's what I have so far:

import qualified Data.Array.Accelerate as A
import Data.Array.Accelerate ( (:.)(..)
                             , Acc
                             , Vector
                             , Scalar
                             , Elt
                             , fold
                             , slice
                             , constant
                             , Array
                             , Z(..), DIM1, DIM2
                             , fromList
                             , All(..)
                             , generate
                             , lift, unlift
                             , shape
                             )
import Data.Array.Accelerate.Interpreter ( run )

dotP :: (Num a, Elt a) => Acc (Vector a) -> Acc (Vector a) -> Acc (Scalar a)
dotP xs ys = fold (+) 0 $ A.zipWith (*) xs ys

type Matrix a = Array DIM2 a

getRow :: Elt a => Int -> Acc (Matrix a) -> Acc (Vector a)
getRow n mat = slice mat . constant $ Z :. n :. All

-- Naive matrix multiplication:
--
-- index (i, j) is equal to the ith row of 'a' `dot` the jth row of 'b'
matMul :: A.Acc (Matrix Double) -> A.Acc (Matrix Double) -> A.Acc (Matrix Double)
matMul a b' = A.generate (constant $ Z :. nrows :. ncols) $
                \ix ->
                  let (Z :. i :. j) = unlift ix
                   in getRow i a `dotP` getRow j b
    where
        b = A.transpose b' -- I assume row indexing is faster than column indexing...
        (Z :. nrows :.   _  ) = unlift $ shape a
        (Z :.   _   :. ncols) = unlift $ shape b


This, of course, gives me errors right now because I'm calling getRow and dotP from within the generation function, which expects Exp[ression]s, not Acc[elerated computation]s.

So maybe I need to replace that line with an inner for loop? Is there an easy way to do that with Accelerate?

Thanks for your help,
  - Clark

[1] http://hackage.haskell.org/package/accelerate

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