Thanks! I'll read through "Matricies and Linear Algebra" over the next few
days.
- Clark
On Tue, Dec 4, 2012 at 7:43 PM, Alexander Solla <alex.so...@gmail.com>wrote:
> Sorry, I didn't realize that course was offered next year. I read through
> "Matrices and Linear Algebra" when I was in high school. And used
> Friedberg, Insel, Spence's "Linear Algebra" in college.
>
>
> On Tue, Dec 4, 2012 at 4:37 PM, Alexander Solla <alex.so...@gmail.com>wrote:
>
>> Well, an m x n matrix corresponds to a linear transformation in at most
>> min{m,n} dimensions. In particular, this means that a 2x2 matrix
>> corresponds to a plane, line, or the origin of 3-space, as a linear
>> subspace. Which of those the matrix corresponds to depends on the matrix's
>> "rank", which is the number of linearly independent columns (or rows) in
>> the matrix.
>>
>> Do you really need to know /which/ plane or line a matrix corresponds to?
>> If so, reduce it using Gaussian elimination and, if appropriate, compute
>> its eigenvectors or span. Otherwise, think of it as a generic
>> plane/line/0-point.
>>
>> Outer products represent more of these simple facts about linear algebra.
>> The product of an mx1 and 1xn matrices is an mxn matrix with rank at most
>> 1. Trouble visualizing this means you are missing the essential facts (for
>> the general picture as the product as a line or origin), or requires some
>> computational details -- reducing the matrix using Gaussian elimination and
>> determining its span.
>>
>> As I said, I don't mean to be harsh, but playing with a vector algebra
>> package without understanding vectors is like playing with a calculator
>> without understanding multiplication. You're better off learning what
>> multiplication represents first, before using a machine to do it fast. So,
>> I can humbly recommend taking a course on the subject. For example,
>> https://www.coursera.org/course/matrix
>>
>>
>> On Tue, Dec 4, 2012 at 4:13 PM, Clark Gaebel <cgae...@uwaterloo.ca>wrote:
>>
>>> 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.so...@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 <
>>>> tmcdon...@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 <cgae...@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 <
>>>>> tmcdon...@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 <cgae...@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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>>>>>>
>>>>>>
>>>>>
>>>>>
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>>>>
>>>
>>
>
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