[Haskell-cafe] Naive matrix multiplication with Accelerate
Clark Gaebel
cgaebel at uwaterloo.ca
Wed Dec 5 01:13:00 CET 2012
No. But that doesn't stop me from being curious with Accelerate. Might you
have a better explaination for what's happening here than Trevor's?
- Clark
On Tue, Dec 4, 2012 at 7:08 PM, Alexander Solla <alex.solla at gmail.com>wrote:
> 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 at 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 at 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 at 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 at 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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>>
>>
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