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.
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