On Mon, Jan 4, 2010 at 10:05 AM, Daniel Fischer
Am Montag 04 Januar 2010 02:17:06 schrieb Patrick LeBoutillier:
Hi,
This question didn't get any replies on the beginners list, I thought
I'd try it here...
Sorry, been occupied with other things. I already took a look, but hadn't anything conclusive enough to reply yet.
No sweat... I didn't mean to be pushy :) Thanks a lot for all the pointers, they have speeded up my code a lot. Patrick
I've written (and improved using other solutions I've found on the
net) a simple sudoku solver which I'm trying to profile. Here is the
code:
import Array
Better
import Data.Array.Unboxed
*much* faster
import List (transpose, nub, (\\))
import Data.List
data Sudoku = Sudoku { unit :: Int, cells :: Array (Int, Int) Int,
cells :: UArray (Int,Int) Int
holes :: [(Int, Int)] }
cell :: Sudoku -> (Int, Int) -> Int
cell s i = (cells s) ! i
instance Read Sudoku where
readsPrec _ s = [(Sudoku unit cells holes, "")]
where unit = length . words . head . lines $ s
cells = listArray ((1, 1), (unit, unit)) (map read . words $ s)
holes = [ c | c <- indices cells, (cells ! c) == 0]
instance Show Sudoku where
show s = unlines [unwords [show $ cell s (x,y) | x <- [1 .. unit s]]
| y <- [1 .. unit s]]
genNums :: Sudoku -> (Int, Int) -> [Int]
genNums s c@(i,j) = ([1 .. u] \\) . nub $ used
where
nub isn't nice. It's quadratic in the length of the list. Use e.g.
map head . group . sort
or
Data.[Int]Set.toList . Data.[Int]Set.fromList
if the type is in Ord (and you don't need the distinct elements in the order they come in). That gives an O(n*log n) nub with a sorted result.
And (\\) isn't particularly fast either (O(m*n), where m and n are the lengths of the lists). If you use one of the above instead of nub, you can use the O(min m n) 'minus' for sorted lists:
xxs@(x:xs) `minus` yys@(y:ys)
| x < y = x : xs `minus` yys
| x == y = xs `minus` ys
| otherwise = xxs `minus` ys
xs `minus` _ = xs
Here, you can do better:
genNums s c@(i,j) = nums
where
nums = [n | n <- [1 .. u], arr!n]
arr :: [U]Array Int Bool
arr = accumArray (\_ _ -> False) True (0,u) used
used = (row s u i j) ++ (col s u i j) ++ (square s sq u i j)
u = unit s
Not good to calculate sq here. You'll use it many times, calculate once and store it in s.
sq = truncate . sqrt . fromIntegral $ u
row s u i j = [cell s (i, y) | y <- [1 .. u]]
col s u i j = [cell s (x, j) | x <- [1 .. u]]
square s sq u i j = [cell s (x, y) | y <- [1 .. u], x <- [1 .. u], f x i, f
y j] where f a b = div (a-1) sq == div (b-1) sq
Test for f y j before you generate x to skip early.
square s sq u i j = [cell s (ni+x,nj+y) | x <- [1 .. sq], y <- [1 .. sq]]
where
qi = (i-1) `div` sq
qj = (j-1) `div` sq
ni = qi*sq
nj = qj*sq
solve :: Sudoku -> [Sudoku]
solve s =
case holes s of
[] -> [s]
(h:hs) -> do
n <- genNums s h
let s' = Sudoku (unit s) ((cells s) // [(h, n)]) hs
solve s'
main = print . head . solve . read =<< getContents
When I compile as such:
$ ghc -O2 --make Sudoku.hs -prof -auto-all -caf-all -fforce-recomp
and run it on the following puzzle:
0 2 3 4
3 4 1 0
2 1 4 0
0 3 2 1
I get the following profiling report:
Fri Jan 1 10:34 2010 Time and Allocation Profiling Report (Final)
Sudoku +RTS -p -RTS
total time = 0.00 secs (0 ticks @ 20 ms)
That means the report is basically useless. Not entirely, because the allocation figures may already contain useful information. Run on a 9x9 puzzle (a not too hard one, but not trivial either).
Also, run the profiling with -P instead of -p, you'll get more info about time and allocation then.
total alloc = 165,728 bytes (excludes profiling overheads)
COST CENTRE MODULE %time %alloc
CAF GHC.Handle 0.0 10.7
CAF Text.Read.Lex 0.0 2.1
CAF GHC.Read 0.0 1.2
square Main 0.0 2.8
solve Main 0.0 1.3
show_aVx Main 0.0 3.7
readsPrec_aYF Main 0.0 60.6
main Main 0.0 9.6
genNums Main 0.0 5.0
cell Main 0.0 1.2
individual inherited
COST CENTRE MODULE
no. entries %time %alloc %time %alloc
MAIN MAIN
1 0 0.0 0.3 0.0 100.0
main Main
186 1 0.0 9.6 0.0 85.6
show_aVx Main
196 2 0.0 3.7 0.0 3.7
cell Main
197 16 0.0 0.0 0.0 0.0
solve Main
188 5 0.0 1.3 0.0 11.8
genNums Main
189 8 0.0 5.0 0.0 10.4
square Main
194 88 0.0 2.8 0.0 3.2
cell Main
195 16 0.0 0.4 0.0 0.4
col Main
192 4 0.0 0.7 0.0 1.1
cell Main
193 16 0.0 0.4 0.0 0.4
row Main
190 4 0.0 0.7 0.0 1.1
cell Main
191 16 0.0 0.4 0.0 0.4
readsPrec_aYF Main
187 3 0.0 60.6 0.0 60.6
CAF GHC.Read
151 1 0.0 1.2 0.0 1.2
CAF Text.Read.Lex
144 8 0.0 2.1 0.0 2.1
CAF GHC.Handle
128 4 0.0 10.7 0.0 10.7
CAF GHC.Conc
127 1 0.0 0.0 0.0 0.0
Does the column 'entries' represent the number of times the function
was called?
Number of times it was 'entered', not quite the same as the number of times it was called.
I think (Warning: speculation ahead, I don't *know* how the profiles are generated) it's thus:
Say you call a function returning a list. One call, first entry. It finds the beginning of the list, the first k elements and hands them to the caller. Caller processes these, asks "can I have more, or was that it?". Same call, second entry: f looks for more, finds the next m elements, hands them to caller. Caller processes. Repeat until whatever happens first, caller doesn't ask whether there's more or callee finds there's nothing more (or hits bottom).
If so, I don't understand how the 'square' function could
be called 88 times when it's caller is only called 8 times. Same thing
with 'genNums' (called 8 times, and solve called 5 times)
What am I missing here?
Patrick
-- ===================== Patrick LeBoutillier Rosemère, Québec, Canada