[Haskell-cafe] list comprehansion performance has hug different

[Junior White]

Thinks! I think compiler should do this for us, isn't it?


On Wed, Jan 30, 2013 at 7:54 PM, Adrian Keet <arkeet at gmail.com> wrote:

>  The whole point here is to evaluate both lists inside the list
> comprehension only once. There is a very simple way to accomplish this:
>
> [q:qs | let qss = queens' (k-1), q <- [1..n], qs <- qss]
>
> Here, queens' (k-1) is only evaluated once, and is shared for all q.
>
> (Note: If queens' (k-1) is polymorphic (which it is) and you use
> -XNoMonomorphismRestriction, then you better add a type annotation to qss
> to ensure sharing.)
>
> Adrian
>
>
> On 2013/01/30 1:51, Doaitse Swierstra wrote:
>
> From the conclusion that both programs compute the same result it can be
> concluded that  the fact that you have made use of a list comprehension has
> forced  you to make a choice which should not matter, i.e. the order in
> which to place the generators. This should be apparent from your code.
>
>  My approach is such a situation is to "define your own generator"
> (assuming here that isSafe needs both its parameters):
>
>  pl `x` ql = [ (p,q) | p <-pl, q <- ql]
>
>  queens3 n =  map reverse $ queens' n
>     where queens' 0       = [[]]
>
>           queens' k       = [q:qs | (qs, q) <- queens' (k-1) `x` [1..n],
> isSafe q qs]
>           isSafe   try qs = not (try `elem` qs || sameDiag try qs)
>
>           sameDiag try qs = any (\(colDist,q) -> abs (try - q) == colDist)
> $ zip [1..] qs
>
>  Of course you can make more refined versions of `x`, which perform all
> kinds of fair enumeration, but that is not the main point here. It is the
> fact that the parameters to `x` are only evaluated once which matters here.
>
>   Doaitse
>
>  On Jan 29, 2013, at 10:25 , Junior White <efiish at gmail.com> wrote:
>
>  Hi Cafe,
>    I have two programs for the same problem "Eight queens problem",
> the link is http://www.haskell.org/haskellwiki/99_questions/90_to_94.
>    My two grograms only has little difference, but the performance, this
> is my solution:
>
>  -- solution 1------------------------------------------------------------
>  queens1 :: Int -> [[Int]]
>
> queens1 n = map reverse $ queens' n
>
>     where queens' 0       = [[]]
>
>           queens' k       = [q:qs | q <- [1..n], qs <- queens' (k-1),
> isSafe q qs]
>           isSafe   try qs = not (try `elem` qs || sameDiag try qs)
>
>           sameDiag try qs = any (λ(colDist, q) -> abs (try - q) ==
> colDist) $ zip [1..] qs
>
>  -- solution
> 2--------------------------------------------------------------
>  queens2 :: Int -> [[Int]]
>
> queens2 n = map reverse $ queens' n
>
>     where queens' 0       = [[]]
>
>           queens' k       = [q:qs | qs <- queens' (k-1), q <- [1..n],
> isSafe q qs]
>           isSafe   try qs = not (try `elem` qs || sameDiag try qs)
>
>           sameDiag try qs = any (λ(colDist,q) -> abs (try - q) == colDist)
> $ zip [1..] qs
>
>  the performance difference is: (set :set +s in ghci)
>  *Main> length (queens1 8)
> 92
> (287.85 secs, 66177031160 bytes)
> *Main> length (queens2 8)
> 92
> (0.07 secs, 17047968 bytes)
> *Main>
>
>  The only different in the two program is in the first is "q <- [1..n],
> qs <- queens' (k-1)," and the second is "qs <- queens' (k-1), q <- [1..n]".
>
>  Does sequence in list comprehansion matter? And why?
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