Re: [Haskell-cafe] New Benchmark Under Review: Magic Squares
Daniel Fischer [EMAIL PROTECTED] wrote: Cool, though the problem of exploding runtime remains, it's only pushed a little further. Now I get a 5x5 magig square in 1 s, a 6x6 in 5.4 s, but 7x7 segfaulted after about 2 1/2 hours - out of memory, I note that your solution uses Arrays. I have recently discovered that the standard array implementations in GHC introduce non-linear performance profiles (wrt to the size of the array). None of the ordinary variations of arrays seemed to make any significant difference, but replacing Array with the new ByteString from fps brought my application's performance back down to the expected linear complexity. Here are some figures, timings all in seconds: dataset size (Mb) Array ByteString -- - -- marschnerlobb0.0690.670.57 silicium 0.1131.371.09 neghip 0.26 2.682.18 hydrogenAtom 2.1031.617.6 lobster 5.46 137 49.3 engine 8.39 286 83.2 statueLeg 10.8420 95.8 BostonTeapot11.8488 107 skull 16.7924 152 Regards, Malcolm ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] New Benchmark Under Review: Magic Squares
Am Mittwoch, 5. Juli 2006 21:28 schrieben Sie:
Hi Daniel,
In the paragraph below it looks like you improved the performance of
5x5 from one and one half hours to one second. Is that a typo or
should I be very, very impressed. :-)
Cheers, David
Err, neither, really. Apparently, I haven't expressed myself immaculately
clearly, so let me try again.
Josh Goldfoot's original code produced a 5x5 magic square on the benchmarking
computer in 8063.01s, on my computer, I hit ctrl-C after about 4 1/2 hours.
My first version produced a 5x5 square in a little over 4 seconds (or was it a
little over 5s, I'm not sure), and a 6x6 square in 86.5s, but since I used
better bounds for the possible moves - e.g., if we regard a 5x5 square with
two entries, 1 at (1,1) and 2 at (1,2), JG's code would give [3 .. 25] as the
list of possible moves for (1,3), whereas I took into account that the sum of
(1,4) and (1,5) is at most 24 + 25 = 49 (and at least 3+4, but that doesn't
help here), thus finding that (1,3) must be at leat 65 - (1+2) - 49 = 13 and
[13 .. 25] as the list of possible moves. So I avoided a lot of dead ends,
but produced a different magic square.
This code I have pushed down to 1s for the 5x5 square and 5.4s for the 6x6
square (simply by replacing intersect [a .. b] with takeWhile (= b) :
dropWhile ( a)).
I have then tuned Josh Goldfoot's code (throwing out the List - Set
conversions, keeping a list of unused numbers and not much else), so that it
produced a 5x5 square in 1 1/2 hours on my computer, giving the same list of
possible moves as the original and hence the same magic square.
That's not bad, but not really awe-inspiring.
However, I've also combined the algorithms, using my better bounds, thus
avoiding many dead ends, but calculating the priorities as if I used the
original bounds, so exploring the branches in the same order and producing
the same square as the original.
This took about 12 minutes for a 5x5 square and impressed me - I expected it
to be significantly slower than the fast code, but a factor of 720 was much
more than I dreamed of.
Cheers,
Daniel
On Jul 4, 2006, at 6:48 AM, Daniel Fischer wrote:
Hi,
I have now tuned Josh Goldfoot's code without changing the order in
which the
magic squares are produced, for a 5x5 magic square, my machine took
about 1
1/2 hours and used 2Mb memory (considering that the original code
did not
finish within 4 1/2 hours here, that should push time on the
benchmarking
machine under 3000s and put us in the lead, I hope).
However, with the improved bounds for the possibilities, I can now
get a 5x5
square in 1s, a 6x6 square in 5.5s (replacing intersect by takeWhile
dropWhile), so it's still sl.
Brent, can I informally submit the code thus, or what formalities
would I have
to perform to submit my code?
--
-
{- The Computer Language Shootout
http://shootout.alioth.debian.org/
benchmark implementation
contributed by Josh Goldfoot
modified by Daniel Fischer to improve performance -}
{- An implementation using Data.Graph would be much faster. This
implementation
is designed to demonstrate the benchmark algorithm. -}
import Data.Array
import Data.List
import System (getArgs)
main = do
n - getArgs = return . read . head
let mn = (n * (1 + n * n)) `div` 2 -- the magic number
initialNode = makeSquare n mn (listArray ((1,1),(n,n))
(repeat 0), [1
.. n^2])
allSquares = bestFirst (successorNodes n mn) (initialNode:[])
putStrLn $ printMatrix n $ grid $ head allSquares
where
printMatrix n grid = unlines [ (rowlist grid n y) | y -
[1..n]]
where
rowlist grid n y = unwords [show $ grid ! (x,y) | x
- [1..n]]
data Square = Square { grid :: Array (Int,Int) Int,
ffm :: ([Int], Int, Int),
unused :: [Int],
priority :: !Int }
{- bestFirst: Given a queue with one initial node and a function,
successors,
that takes a node and returns a list of nodes that are created
by making
all possible moves in a single cell, implements the Best-First
algorithm,
and returns a list of all nodes that end up with priority
zero. In this
implementation we only ever use the first node.
-}
bestFirst _ [] = []
bestFirst successors (frontnode:priorityq)
| priority frontnode == 0 = frontnode:bestFirst successors
priorityq
| otherwise = bestFirst successors $ foldr (insertBy
compSquare) priorityq
(successors frontnode)
where
{- The priority queue is sorted first by
the node's calculated priority; then, if the priorities
are equal, by whichever node has the lowest numbers
in the top-left of the array (or
Re: [Haskell-cafe] New Benchmark Under Review: Magic Squares
Hi,
I have now tuned Josh Goldfoot's code without changing the order in which the
magic squares are produced, for a 5x5 magic square, my machine took about 1
1/2 hours and used 2Mb memory (considering that the original code did not
finish within 4 1/2 hours here, that should push time on the benchmarking
machine under 3000s and put us in the lead, I hope).
However, with the improved bounds for the possibilities, I can now get a 5x5
square in 1s, a 6x6 square in 5.5s (replacing intersect by takeWhile
dropWhile), so it's still sl.
Brent, can I informally submit the code thus, or what formalities would I have
to perform to submit my code?
---
{- The Computer Language Shootout
http://shootout.alioth.debian.org/
benchmark implementation
contributed by Josh Goldfoot
modified by Daniel Fischer to improve performance -}
{- An implementation using Data.Graph would be much faster. This
implementation
is designed to demonstrate the benchmark algorithm. -}
import Data.Array
import Data.List
import System (getArgs)
main = do
n - getArgs = return . read . head
let mn = (n * (1 + n * n)) `div` 2 -- the magic number
initialNode = makeSquare n mn (listArray ((1,1),(n,n)) (repeat 0), [1
.. n^2])
allSquares = bestFirst (successorNodes n mn) (initialNode:[])
putStrLn $ printMatrix n $ grid $ head allSquares
where
printMatrix n grid = unlines [ (rowlist grid n y) | y - [1..n]]
where
rowlist grid n y = unwords [show $ grid ! (x,y) | x - [1..n]]
data Square = Square { grid :: Array (Int,Int) Int,
ffm :: ([Int], Int, Int),
unused :: [Int],
priority :: !Int }
{- bestFirst: Given a queue with one initial node and a function, successors,
that takes a node and returns a list of nodes that are created by making
all possible moves in a single cell, implements the Best-First algorithm,
and returns a list of all nodes that end up with priority zero. In this
implementation we only ever use the first node.
-}
bestFirst _ [] = []
bestFirst successors (frontnode:priorityq)
| priority frontnode == 0 = frontnode:bestFirst successors priorityq
| otherwise = bestFirst successors $ foldr (insertBy compSquare) priorityq
(successors frontnode)
where
{- The priority queue is sorted first by
the node's calculated priority; then, if the priorities
are equal, by whichever node has the lowest numbers
in the top-left of the array (or the next cell over,
and so on). -}
compSquare a b = case compare (priority a) (priority b) of
EQ - compare (grid a) (grid b)
ot - ot
{- successorNodes: Find the cell with the fewest
possible moves left, and then creates a new node for each possible move
in that cell.
-}
successorNodes n mn squarenode =
map (makeSquare n mn) [(thegrid//[((x, y), i)], delete i un) | i -
possibilities]
where
thegrid = grid squarenode
un = unused squarenode
(possibilities, x, y) = ffm squarenode
{- makeSquare: Creates a node for the priority queue. In the process, this
calculates the cell with the fewest possible moves, and also calculates
this node's priority. The priority function is:
(number of zeros in the grid)
plus
(number of possible moves in the cell with the fewest possible moves)
the lower the priority, the sooner the node will be popped from the queue.
-}
makeSquare n mn (thegrid,un) =
Square { grid = thegrid, ffm = moveChoices, unused = un, priority =
calcPriority }
where
moveChoices@(poss,_,_) = findFewestMoves n mn thegrid un
calcPriority = length un + length poss
{- findFewestMoves: Go through the grid (starting at the top-left, and moving
right and down), checking all 0 cells to find the cell with the fewest
possible moves.
-}
findFewestMoves n mn grid un
| null un = ([],0,0)
| otherwise = (movelist, mx, my)
where
openSquares = [ (x,y) | y - [1..n], x - [1..n], (grid ! (x,y)) == 0]
pm = possibleMoves n mn grid un
openMap = map (\(x,y) - (pm (x,y), (x,y))) openSquares
mycompare f g = compare ((length . fst) f) ((length . fst) g)
(movelist, (mx, my)) = minimumBy mycompare openMap
{- possibleMoves: Return all moves that can go in the cell x,y for a given
grid. A move is possible if the move (number) is not already
in the grid, and if, after making that move, it is still possible to
satisfy the magic square conditions (all rows, columns, diagonals adding
up to mn, the magic number)
-}
possibleMoves n mn grid un (x,y)
| grid ! (x,y) /= 0 = []
| null oneZeroGroups = takeWhile (= highest) un -- [1 .. highest]
`intersect` un
Re: [Haskell-cafe] New Benchmark Under Review: Magic Squares
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Daniel, I have now tuned Josh Goldfoot's code without changing the order in which the magic squares are produced, for a 5x5 magic square, my machine took about 1 1/2 hours and used 2Mb memory (considering that the original code did not finish within 4 1/2 hours here, that should push time on the benchmarking machine under 3000s and put us in the lead, I hope). Thanks for your efforts on this project. I'm actually more interested in using your earlier solution, since it is so much faster. Right now, the magic square code rises in runtime from 1.5 seconds to 4 hours with an increase of 1 in the square's dimension. I would much rather use a technique that had a more linear (or even exponential) increase! I would propose modifying the other entries (since there are only a handful) to match the output of your original solution. What do you think? - -Brent -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.2.2 (Darwin) iD8DBQFEqpVmzGDdrzfvUpURAkPpAJ9oKTwzmUyTAoA6yQdOo7APKnXCqACghJEV id5EqEyVKrvSlJlLH9JZTN0= =jNXB -END PGP SIGNATURE- ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] New Benchmark Under Review: Magic Squares
Am Dienstag, 4. Juli 2006 18:20 schrieben Sie:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Daniel,
I have now tuned Josh Goldfoot's code without changing the order in
which the
magic squares are produced, for a 5x5 magic square, my machine took
about 1
1/2 hours and used 2Mb memory (considering that the original code
did not
finish within 4 1/2 hours here, that should push time on the
benchmarking
machine under 3000s and put us in the lead, I hope).
Thanks for your efforts on this project. I'm actually more
interested in using your earlier solution, since it is so much
faster. Right now, the magic square code rises in runtime from 1.5
seconds to 4 hours with an increase of 1 in the square's dimension.
I would much rather use a technique that had a more linear (or even
exponential) increase!
I would propose modifying the other entries (since there are only a
handful) to match the output of your original solution.
What do you think?
Cool, though the problem of exploding runtime remains, it's only pushed a
little further. Now I get a 5x5 magig square in 1 s, a 6x6 in 5.4 s, but 7x7
segfaulted after about 2 1/2 hours - out of memory, I believe.
And, as mentioned in passing, using 'intersect' in the first version is
slowing things down, so here is my currently fastest (undoubtedly, the
experts could still make it faster by clever unboxing):
import Data.Array.Unboxed
import Data.List
import System.Environment (getArgs)
main :: IO ()
main = getArgs = return . read . head = msquare
msquare :: Int - IO ()
msquare n = let mn = (n*(n*n+1)) `quot` 2
grd = listArray ((1,1),(n,n)) (repeat 0)
unus = [1 .. n*n]
ff = findFewestMoves n mn grd unus
ini = Square grd unus ff (2*n*n)
allSquares = bestFirst (successorNodes n mn) [ini]
in putStrLn $ showGrid n . grid $ head allSquares
data Square = Square { grid :: UArray (Int,Int) Int
, unused :: [Int]
, ffm :: ([Int], Int, Int, Int)
, priority :: !Int
} deriving Eq
instance Ord Square where
compare (Square g1 _ _ p1) (Square g2 _ _ p2)
= case compare p1 p2 of
EQ - compare g1 g2
ot - ot
showMat :: [[Int]] - ShowS
showMat lns = foldr1 ((.) . (. showChar '\n')) $ showLns
where
showLns =
map (foldr1 ((.) . (. showChar ' ')) . map shows) lns
showGrid :: Int - UArray (Int,Int) Int - String
showGrid n g = showMat [[g ! (r,c) | c - [1 .. n]] | r - [1 .. n]]
bestFirst :: (Square - [Square]) - [Square] - [Square]
bestFirst _ [] = []
bestFirst successors (front:queue)
| priority front == 0 = front : bestFirst successors queue
| otherwise = bestFirst successors $ foldr insert queue (successors front)
successorNodes n mn sq
= map (place sq n mn (r,c)) possibilities
where
(possibilities,_,r,c) = ffm sq
place :: Square - Int - Int - (Int,Int) - Int - Square
place (Square grd unus _ _) n mn (r,c) k
= Square grd' uns moveChoices p
where
grd' = grd//[((r,c),k)]
moveChoices@(_,len,_,_) = findFewestMoves n mn grd' uns
uns = delete k unus
p = length uns + len
findFewestMoves :: Int - Int - UArray (Int,Int) Int - [Int] -
([Int],Int,Int,Int)
findFewestMoves n mn grid unus
| null unus = ([],0,0,0)
| otherwise = (movelist, length movelist, mr, mc)
where
openSquares = [(r,c) | r - [1 .. n], c - [1 .. n], grid ! (r,c) ==
0]
pm = possibleMoves n mn grid unus
openMap = map (\(x,y) - (pm x y,x,y)) openSquares
mycompare (a,_,_) (b,_,_) = compare (length a) (length b)
(movelist,mr,mc) = minimumBy mycompare openMap
possibleMoves :: Int - Int - UArray (Int,Int) Int - [Int] - Int - Int -
[Int]
possibleMoves n mn grid unus r c
| grid ! (r,c) /= 0 = []
| otherwise = takeWhile (= ma) $ dropWhile ( mi) unus
where
cellGroups
| r == c r + c == n + 1 = [d1, d2, theRow, theCol]
| r == c = [d1, theRow, theCol]
| r + c == n + 1 = [d2, theRow, theCol]
| otherwise = [theRow, theCol]
d1 = diag1 grid n
d2 = diag2 grid n
theRow = gridRow grid n r
theCol = gridCol grid n c
lows = scanl (+) 0 unus
higs = scanl (+) 0 $ reverse unus
rge :: [Int] - (Int,Int)
rge cg = let k = count0s cg - 1
lft = mn - sum cg
in (lft - (higs!!k),lft - (lows!!k))
(mi,ma) = foldr1 mima $ map rge cellGroups
mima (a,b) (c,d) = (max a c, min b d)
gridRow, gridCol :: UArray (Int,Int) Int - Int - Int - [Int]
diag1, diag2 :: UArray (Int,Int) Int - Int - [Int]
gridRow grid n r = [grid ! (r,i) | i - [1 .. n]]
gridCol grid n c = [grid ! (i,c) | i - [1 .. n]]
diag1 grid n = [grid ! (i,i) | i - [1 .. n]]
diag2 grid n = [grid
Re: [Haskell-cafe] New Benchmark Under Review: Magic Squares
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 On Jul 4, 2006, at 5:20 PM, Daniel Fischer wrote: I would propose modifying the other entries (since there are only a handful) to match the output of your original solution. What do you think? Cool, though the problem of exploding runtime remains, it's only pushed a little further. Now I get a 5x5 magig square in 1 s, a 6x6 in 5.4 s, but 7x7 segfaulted after about 2 1/2 hours - out of memory, I believe. Hrm. Well, I still prefer the growth of search space in your version over the original, since it *was* going from 0.01 s (3x3) to 0.10 (4x4) to 4 hours (5x5). Going 1s-5.4s -x hours is at least a bit more controlled. I wonder if anyone can propose a slightly smaller problem, or a better algorithm? Thanks, - -Brent -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.2.2 (Darwin) iD8DBQFEq0b7zGDdrzfvUpURAumWAJ4itR4eayB3mj5hYEtxbK630mF4IgCeO3PA qFF7cLTW4xk36J/nQOON+F4= =C1xL -END PGP SIGNATURE- ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] New Benchmark Under Review: Magic Squares
Perhaps you could post a new entry page on our shootout wiki?
http://www.haskell.org/hawiki/ShootoutEntry
This makes it easier for people to keep contributing.
Cheers,
Don
daniel.is.fischer:
Am Sonntag, 2. Juli 2006 01:58 schrieb Brent Fulgham:
We recently began considering another benchmark for the shootout,
namely a Magic Square via best-first search. This is fairly
inefficient, and we may need to shift to another approach due to the
extremely large times required to find a solution for larger squares.
A slightly less naive approach to determining the possible moves dramatically
reduces the effort, while Josh Goldfoot's code did not finish within 4 1/2
hours on my machine, a simple modification (see below) reduced runtime for
N = 5 to 4.3 s, for N = 6 to 86.5 s.
Unfortunately, the squares are now delivered in a different order, so my
programme would probably be rejected :-(
I thought the Haskell community might be interested in the
performance we have measured so far (see http://
shootout.alioth.debian.org/sandbox/fulldata.php?
test=magicsquaresp1=java-0p2=javaclient-0p3=ghc-0p4=psyco-0
Interestingly, Java actually beats the tar out of GHC and Python for
N=5x5 (and I assume higher, though this already takes on the order of
2 hours to solve on the benchmark machine). Memory use in GHC stays
nice and low, but the time to find the result rapidly grows.
I was hoping for an order of magnitude increase with each increase in
N, but discovered that it is more like an exponential...
Thanks,
-Brent
Modified code, still best-first search:
import Data.Array.Unboxed
import Data.List
import System.Environment (getArgs)
main :: IO ()
main = getArgs = return . read . head = msquare
msquare :: Int - IO ()
msquare n = let mn = (n*(n*n+1)) `quot` 2
grd = listArray ((1,1),(n,n)) (repeat 0)
unus = [1 .. n*n]
ff = findFewestMoves n mn grd unus
ini = Square grd unus ff (2*n*n)
allSquares = bestFirst (successorNodes n mn) [ini]
in putStrLn $ showGrid n . grid $ head allSquares
data Square = Square { grid :: UArray (Int,Int) Int
, unused :: [Int]
, ffm :: ([Int], Int, Int, Int)
, priority :: !Int
} deriving Eq
instance Ord Square where
compare (Square g1 _ _ p1) (Square g2 _ _ p2)
= case compare p1 p2 of
EQ - compare g1 g2
ot - ot
showMat :: [[Int]] - ShowS
showMat lns = foldr1 ((.) . (. showChar '\n')) $ showLns
where
showLns = map (foldr1 ((.) . (. showChar ' ')) . map shows)
lns
showGrid :: Int - UArray (Int,Int) Int - String
showGrid n g = showMat [[g ! (r,c) | c - [1 .. n]] | r - [1 .. n]]
bestFirst :: (Square - [Square]) - [Square] - [Square]
bestFirst _ [] = []
bestFirst successors (front:queue)
| priority front == 0 = front : bestFirst successors queue
| otherwise = bestFirst successors $ foldr insert queue (successors front)
successorNodes n mn sq
= map (place sq n mn (r,c)) possibilities
where
(possibilities,_,r,c) = ffm sq
place :: Square - Int - Int - (Int,Int) - Int - Square
place (Square grd unus _ _) n mn (r,c) k
= Square grd' uns moveChoices p
where
grd' = grd//[((r,c),k)]
moveChoices@(_,len,_,_) = findFewestMoves n mn grd' uns
uns = delete k unus
p = length uns + len
findFewestMoves n mn grid unus
| null unus = ([],0,0,0)
| otherwise = (movelist, length movelist, mr, mc)
where
openSquares = [(r,c) | r - [1 .. n], c - [1 .. n], grid ! (r,c) ==
0]
pm = possibleMoves n mn grid unus
openMap = map (\(x,y) - (pm x y,x,y)) openSquares
mycompare (a,_,_) (b,_,_) = compare (length a) (length b)
(movelist,mr,mc) = minimumBy mycompare openMap
possibleMoves n mn grid unus r c
| grid ! (r,c) /= 0 = []
| otherwise = intersect [mi .. ma] unus -- this is the difference that
-- does it: better bounds
where
cellGroups
| r == c r + c == n + 1 = [d1, d2, theRow, theCol]
| r == c = [d1, theRow, theCol]
| r + c == n + 1 = [d2, theRow, theCol]
| otherwise = [theRow, theCol]
d1 = diag1 grid n
d2 = diag2 grid n
theRow = gridRow grid n r
theCol = gridCol grid n c
lows = scanl (+) 0 unus
higs = scanl (+) 0 $ reverse unus
rge cg = let k = count0s cg - 1
lft = mn - sum cg
in (lft - (higs!!k),lft - (lows!!k))
(mi,ma) = foldr1 mima $ map rge cellGroups
mima (a,b) (c,d) = (max a c, min b d)
gridRow grid n r = [grid ! (r,i) | i - [1 .. n]]
gridCol grid n c = [grid ! (i,c) | i - [1 .. n]]
diag1 grid n = [grid ! (i,i) |
Re: [Haskell-cafe] New Benchmark Under Review: Magic Squares
Hello Brent, Sunday, July 2, 2006, 3:58:11 AM, you wrote: We recently began considering another benchmark for the shootout, namely a Magic Square via best-first search. This is fairly i've slightly beautified your printMatrix code: . where showMatrix n grid = join \n [ showRow y | y-[1..n] ] where showRow y = join [ show $ grid!(x,y) | x-[1..n] ] join filler pss = concat (intersperse filler pss) inefficient, and we may need to shift to another approach due to the extremely large times required to find a solution for larger squares. it's interesting to see one more compiler-dependent (as opposite to libraries-dependent) benchmark in shootout. It seems that the devil hides in the last function, possibleMoves. i tried to replace using of Data.Set with Data.Set.Enum by David F. Place, but got only 5% improvement. This procedure heavily uses lists and that is not the fastest data structure, especially in Haskell where lists are lazy. One possible solution may be to use lists of strict (and automatically unboxed) elements and/or lists that are strict in their links. Another possible solution will be to use unboxed arrays and implement all the required List routines for them. About the overall algorithm - it tends to recompute data that is almost not changed, such as list of already used numbers. It resembles me sudoku solvers that was discussed here several months ago - its highly possible that optimization tricks developed for this task will be appropriate to speed up magic squares too. -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
