How to optimize this Haskell code summing up the primes in sublinear time? - algorithm

Problem 10 from Project Euler is to find the sum of all the primes below given n.
I solved it simply by summing up the primes generated by the sieve of Eratosthenes. Then I came across much more efficient solution by Lucy_Hedgehog (sub-linear!).
For n = 2⋅10^9:
Python code (from the quote above) runs in 1.2 seconds in Python 2.7.3.
C++ code (mine) runs in about 0.3 seconds (compiled with g++ 4.8.4).
I re-implemented the same algorithm in Haskell, since I'm learning it:
import Data.List
import Data.Map (Map, (!))
import qualified Data.Map as Map
problem10 :: Integer -> Integer
problem10 n = (sieve (Map.fromList [(i, i * (i + 1) `div` 2 - 1) | i <- vs]) 2 r vs) ! n
where vs = [n `div` i | i <- [1..r]] ++ reverse [1..n `div` r - 1]
r = floor (sqrt (fromIntegral n))
sieve :: Map Integer Integer -> Integer -> Integer -> [Integer] -> Map Integer Integer
sieve m p r vs | p > r = m
| otherwise = sieve (if m ! p > m ! (p - 1) then update m vs p else m) (p + 1) r vs
update :: Map Integer Integer -> [Integer] -> Integer -> Map Integer Integer
update m vs p = foldl' decrease m (map (\v -> (v, sumOfSieved m v p)) (takeWhile (>= p*p) vs))
decrease :: Map Integer Integer -> (Integer, Integer) -> Map Integer Integer
decrease m (k, v) = Map.insertWith (flip (-)) k v m
sumOfSieved :: Map Integer Integer -> Integer -> Integer -> Integer
sumOfSieved m v p = p * (m ! (v `div` p) - m ! (p - 1))
main = print $ problem10 $ 2*10^9
I compiled it with ghc -O2 10.hs and run with time ./10.
It gives the correct answer, but takes about 7 seconds.
I compiled it with ghc -prof -fprof-auto -rtsopts 10 and run with ./10 +RTS -p -h.
10.prof shows that decrease takes 52.2% time and 67.5% allocations.
After running hp2ps 10.hp I got such heap profile:
Again looks like decrease takes most of the heap. GHC version 7.6.3.
How would you optimize run time of this Haskell code?
Update 13.06.17:
I tried replacing immutable Data.Map with mutable Data.HashTable.IO.BasicHashTable from the hashtables package, but I'm probably doing something bad, since for tiny n = 30 it already takes too long, about 10 seconds. What's wrong?
Update 18.06.17:
Curious about the HashTable performance issues is a good read. I took Sherh's code using mutable Data.HashTable.ST.Linear, but dropped Data.Judy in instead. It runs in 1.1 seconds, still relatively slow.

I've done some small improvements so it runs in 3.4-3.5 seconds on my machine.
Using IntMap.Strict helped a lot. Other than that I just manually performed some ghc optimizations just to be sure. And make Haskell code more close to Python code from your link. As a next step you could try to use some mutable HashMap. But I'm not sure... IntMap can't be much faster than some mutable container because it's an immutable one. Though I'm still surprised about it's efficiency. I hope this can be implemented faster.
Here is the code:
import Data.List (foldl')
import Data.IntMap.Strict (IntMap, (!))
import qualified Data.IntMap.Strict as IntMap
p :: Int -> Int
p n = (sieve (IntMap.fromList [(i, i * (i + 1) `div` 2 - 1) | i <- vs]) 2 r vs) ! n
where vs = [n `div` i | i <- [1..r]] ++ [n', n' - 1 .. 1]
r = floor (sqrt (fromIntegral n) :: Double)
n' = n `div` r - 1
sieve :: IntMap Int -> Int -> Int -> [Int] -> IntMap Int
sieve m' p' r vs = go m' p'
where
go m p | p > r = m
| m ! p > m ! (p - 1) = go (update m vs p) (p + 1)
| otherwise = go m (p + 1)
update :: IntMap Int -> [Int] -> Int -> IntMap Int
update s vs p = foldl' decrease s (takeWhile (>= p2) vs)
where
sp = s ! (p - 1)
p2 = p * p
sumOfSieved v = p * (s ! (v `div` p) - sp)
decrease m v = IntMap.adjust (subtract $ sumOfSieved v) v m
main :: IO ()
main = print $ p $ 2*10^(9 :: Int)
UPDATE:
Using mutable hashtables I've managed to make performance up to ~5.5sec on Haskell with this implementation.
Also, I used unboxed vectors instead of lists in several places. Linear hashing seems to be the fastest. I think this can be done even faster. I noticed sse42 option in hasthables package. Not sure I've managed to set it correctly but even without it runs that fast.
UPDATE 2 (19.06.2017)
I've managed to make it 3x faster then best solution from #Krom (using my code + his map) by dropping judy hashmap at all. Instead just plain arrays are used. You can come up with the same idea if you notice that keys for S hashmap are either sequence from 1 to n' or n div i for i from 1 to r. So we can represent such HashMap as two arrays making lookups in array depending on searching key.
My code + Judy HashMap
$ time ./judy
95673602693282040
real 0m0.590s
user 0m0.588s
sys 0m0.000s
My code + my sparse map
$ time ./sparse
95673602693282040
real 0m0.203s
user 0m0.196s
sys 0m0.004s
This can be done even faster if instead of IOUArray already generated vectors and Vector library is used and readArray is replaced by unsafeRead. But I don't think this should be done if only you're not really interested in optimizing this as much as possible.
Comparison with this solution is cheating and is not fair. I expect same ideas implemented in Python and C++ will be even faster. But #Krom solution with closed hashmap is already cheating because it uses custom data structure instead of standard one. At least you can see that standard and most popular hash maps in Haskell are not that fast. Using better algorithms and better ad-hoc data structures can be better for such problems.
Here's resulting code.

First as a baseline, the timings of the existing approaches
on my machine:
Original program posted in the question:
time stack exec primorig
95673602693282040
real 0m4.601s
user 0m4.387s
sys 0m0.251s
Second the version using Data.IntMap.Strict from
here
time stack exec primIntMapStrict
95673602693282040
real 0m2.775s
user 0m2.753s
sys 0m0.052s
Shershs code with Data.Judy dropped in here
time stack exec prim-hash2
95673602693282040
real 0m0.945s
user 0m0.955s
sys 0m0.028s
Your python solution.
I compiled it with
python -O -m py_compile problem10.py
and the timing:
time python __pycache__/problem10.cpython-36.opt-1.pyc
95673602693282040
real 0m1.163s
user 0m1.160s
sys 0m0.003s
Your C++ version:
$ g++ -O2 --std=c++11 p10.cpp -o p10
$ time ./p10
sum(2000000000) = 95673602693282040
real 0m0.314s
user 0m0.310s
sys 0m0.003s
I didn't bother to provide a baseline for slow.hs, as I didn't
want to wait for it to complete when run with an argument of
2*10^9.
Subsecond performance
The following program runs in under a second on my machine.
It uses a hand rolled hashmap, which uses closed hashing with
linear probing and uses some variant of knuths hashfunction,
see here.
Certainly it is somewhat tailored to the case, as the lookup
function for example expects the searched keys to be present.
Timings:
time stack exec prim
95673602693282040
real 0m0.725s
user 0m0.714s
sys 0m0.047s
First I implemented my hand rolled hashmap simply to hash
the keys with
key `mod` size
and selected a size multiple times higher than the expected
input, but the program took 22s or more to complete.
Finally it was a matter of choosing a hash function which was
good for the workload.
Here is the program:
import Data.Maybe
import Control.Monad
import Data.Array.IO
import Data.Array.Base (unsafeRead)
type Number = Int
data Map = Map { keys :: IOUArray Int Number
, values :: IOUArray Int Number
, size :: !Int
, factor :: !Int
}
newMap :: Int -> Int -> IO Map
newMap s f = do
k <- newArray (0, s-1) 0
v <- newArray (0, s-1) 0
return $ Map k v s f
storeKey :: IOUArray Int Number -> Int -> Int -> Number -> IO Int
storeKey arr s f key = go ((key * f) `mod` s)
where
go :: Int -> IO Int
go ind = do
v <- readArray arr ind
go2 v ind
go2 v ind
| v == 0 = do { writeArray arr ind key; return ind; }
| v == key = return ind
| otherwise = go ((ind + 1) `mod` s)
loadKey :: IOUArray Int Number -> Int -> Int -> Number -> IO Int
loadKey arr s f key = s `seq` key `seq` go ((key *f) `mod` s)
where
go :: Int -> IO Int
go ix = do
v <- unsafeRead arr ix
if v == key then return ix else go ((ix + 1) `mod` s)
insertIntoMap :: Map -> (Number, Number) -> IO Map
insertIntoMap m#(Map ks vs s f) (k, v) = do
ix <- storeKey ks s f k
writeArray vs ix v
return m
fromList :: Int -> Int -> [(Number, Number)] -> IO Map
fromList s f xs = do
m <- newMap s f
foldM insertIntoMap m xs
(!) :: Map -> Number -> IO Number
(!) (Map ks vs s f) k = do
ix <- loadKey ks s f k
readArray vs ix
mupdate :: Map -> Number -> (Number -> Number) -> IO ()
mupdate (Map ks vs s fac) i f = do
ix <- loadKey ks s fac i
old <- readArray vs ix
let x' = f old
x' `seq` writeArray vs ix x'
r' :: Number -> Number
r' = floor . sqrt . fromIntegral
vs' :: Integral a => a -> a -> [a]
vs' n r = [n `div` i | i <- [1..r]] ++ reverse [1..n `div` r - 1]
vss' n r = r + n `div` r -1
list' :: Int -> Int -> [Number] -> IO Map
list' s f vs = fromList s f [(i, i * (i + 1) `div` 2 - 1) | i <- vs]
problem10 :: Number -> IO Number
problem10 n = do
m <- list' (19*vss) (19*vss+7) vs
nm <- sieve m 2 r vs
nm ! n
where vs = vs' n r
vss = vss' n r
r = r' n
sieve :: Map -> Number -> Number -> [Number] -> IO Map
sieve m p r vs | p > r = return m
| otherwise = do
v1 <- m ! p
v2 <- m ! (p - 1)
nm <- if v1 > v2 then update m vs p else return m
sieve nm (p + 1) r vs
update :: Map -> [Number] -> Number -> IO Map
update m vs p = foldM (decrease p) m $ takeWhile (>= p*p) vs
decrease :: Number -> Map -> Number -> IO Map
decrease p m k = do
v <- sumOfSieved m k p
mupdate m k (subtract v)
return m
sumOfSieved :: Map -> Number -> Number -> IO Number
sumOfSieved m v p = do
v1 <- m ! (v `div` p)
v2 <- m ! (p - 1)
return $ p * (v1 - v2)
main = do { n <- problem10 (2*10^9) ; print n; } -- 2*10^9
I am not a professional with hashing and that sort of stuff, so
this can certainly be improved a lot. Maybe we Haskellers should
improve the of the shelf hash maps or provide some simpler ones.
My hashmap, Shershs code
If I plug my hashmap in Shershs (see answer below) code, see here
we are even down to
time stack exec prim-hash2
95673602693282040
real 0m0.601s
user 0m0.604s
sys 0m0.034s
Why is slow.hs slow?
If you read through the source
for the function insert in Data.HashTable.ST.Basic, you
will see that it deletes the old key value pair and inserts
a new one. It doesn't look up the "place" for the value and
mutate it, as one might imagine, if one reads that it is
a "mutable" hashtable. Here the hashtable itself is mutable,
so you don't need to copy the whole hashtable for insertion
of a new key value pair, but the value places for the pairs
are not. I don't know if that is the whole story of slow.hs
being slow, but my guess is, it is a pretty big part of it.
A few minor improvements
So that's the idea I followed while trying to improve
your program the first time.
See, you don't need a mutable mapping from keys to values.
Your key set is fixed. You want a mapping from keys to mutable
places. (Which is, by the way, what you get from C++ by default.)
And so I tried to come up with that. I used IntMap IORef from
Data.IntMap.Strict and Data.IORef first and got a timing
of
tack exec prim
95673602693282040
real 0m2.134s
user 0m2.141s
sys 0m0.028s
I thought maybe it would help to work with unboxed values
and to get that, I used IOUArray Int Int with 1 element
each instead of IORef and got those timings:
time stack exec prim
95673602693282040
real 0m2.015s
user 0m2.018s
sys 0m0.038s
Not much of a difference and so I tried to get rid of bounds
checking in the 1 element arrays by using unsafeRead and
unsafeWrite and got a timing of
time stack exec prim
95673602693282040
real 0m1.845s
user 0m1.850s
sys 0m0.030s
which was the best I got using Data.IntMap.Strict.
Of course I ran each program multiple times to see if
the times are stable and the differences in run time aren't
just noise.
It looks like these are all just micro-optimizations.
And here is the program that ran fastest for me without using a hand rolled data structure:
import qualified Data.IntMap.Strict as M
import Control.Monad
import Data.Array.IO
import Data.Array.Base (unsafeRead, unsafeWrite)
type Number = Int
type Place = IOUArray Number Number
type Map = M.IntMap Place
tupleToRef :: (Number, Number) -> IO (Number, Place)
tupleToRef = traverse (newArray (0,0))
insertRefs :: [(Number, Number)] -> IO [(Number, Place)]
insertRefs = traverse tupleToRef
fromList :: [(Number, Number)] -> IO Map
fromList xs = M.fromList <$> insertRefs xs
(!) :: Map -> Number -> IO Number
(!) m i = unsafeRead (m M.! i) 0
mupdate :: Map -> Number -> (Number -> Number) -> IO ()
mupdate m i f = do
let place = m M.! i
old <- unsafeRead place 0
let x' = f old
-- make the application of f strict
x' `seq` unsafeWrite place 0 x'
r' :: Number -> Number
r' = floor . sqrt . fromIntegral
vs' :: Integral a => a -> a -> [a]
vs' n r = [n `div` i | i <- [1..r]] ++ reverse [1..n `div` r - 1]
list' :: [Number] -> IO Map
list' vs = fromList [(i, i * (i + 1) `div` 2 - 1) | i <- vs]
problem10 :: Number -> IO Number
problem10 n = do
m <- list' vs
nm <- sieve m 2 r vs
nm ! n
where vs = vs' n r
r = r' n
sieve :: Map -> Number -> Number -> [Number] -> IO Map
sieve m p r vs | p > r = return m
| otherwise = do
v1 <- m ! p
v2 <- m ! (p - 1)
nm <- if v1 > v2 then update m vs p else return m
sieve nm (p + 1) r vs
update :: Map -> [Number] -> Number -> IO Map
update m vs p = foldM (decrease p) m $ takeWhile (>= p*p) vs
decrease :: Number -> Map -> Number -> IO Map
decrease p m k = do
v <- sumOfSieved m k p
mupdate m k (subtract v)
return m
sumOfSieved :: Map -> Number -> Number -> IO Number
sumOfSieved m v p = do
v1 <- m ! (v `div` p)
v2 <- m ! (p - 1)
return $ p * (v1 - v2)
main = do { n <- problem10 (2*10^9) ; print n; } -- 2*10^9
If you profile that, you see that it spends most of the time in the custom lookup function (!),
don't know how to improve that further. Trying to inline (!) with {-# INLINE (!) #-}
didn't yield better results; maybe ghc already did this.

This code of mine evaluates the sum to 2⋅10^9 in 0.3 seconds and the sum to 10^12 (18435588552550705911377) in 19.6 seconds (if given sufficient RAM).
import Control.DeepSeq
import qualified Control.Monad as ControlMonad
import qualified Data.Array as Array
import qualified Data.Array.ST as ArrayST
import qualified Data.Array.Base as ArrayBase
primeLucy :: (Integer -> Integer) -> (Integer -> Integer) -> Integer -> (Integer->Integer)
primeLucy f sf n = g
where
r = fromIntegral $ integerSquareRoot n
ni = fromIntegral n
loop from to c = let go i = ControlMonad.when (to<=i) (c i >> go (i-1)) in go from
k = ArrayST.runSTArray $ do
k <- ArrayST.newListArray (-r,r) $ force $
[sf (div n (toInteger i)) - sf 1|i<-[r,r-1..1]] ++
[0] ++
[sf (toInteger i) - sf 1|i<-[1..r]]
ControlMonad.forM_ (takeWhile (<=r) primes) $ \p -> do
l <- ArrayST.readArray k (p-1)
let q = force $ f (toInteger p)
let adjust = \i j -> do { v <- ArrayBase.unsafeRead k (i+r); w <- ArrayBase.unsafeRead k (j+r); ArrayBase.unsafeWrite k (i+r) $!! v+q*(l-w) }
loop (-1) (-div r p) $ \i -> adjust i (i*p)
loop (-div r p-1) (-min r (div ni (p*p))) $ \i -> adjust i (div (-ni) (i*p))
loop r (p*p) $ \i -> adjust i (div i p)
return k
g :: Integer -> Integer
g m
| m >= 1 && m <= integerSquareRoot n = k Array.! (fromIntegral m)
| m >= integerSquareRoot n && m <= n && div n (div n m)==m = k Array.! (fromIntegral (negate (div n m)))
| otherwise = error $ "Function not precalculated for value " ++ show m
primeSum :: Integer -> Integer
primeSum n = (primeLucy id (\m -> div (m*m+m) 2) n) n
If your integerSquareRoot function is buggy (as reportedly some are), you can replace it here with floor . sqrt . fromIntegral.
Explanation:
As the name suggests it is based upon a generalization of the famous method by "Lucy Hedgehog" eventually discovered by the original poster.
It allows you to calculate many sums of the form (with p prime) without enumerating all the primes up to N and in time O(N^0.75).
Its inputs are the function f (i.e., id if you want the prime sum), its summatory function over all the integers (i.e., in that case the sum of the first m integers or div (m*m+m) 2), and N.
PrimeLucy returns a lookup function (with p prime) restricted to certain values of n: .

Try this and let me know how fast it is:
-- sum of primes
import Control.Monad (forM_, when)
import Control.Monad.ST
import Data.Array.ST
import Data.Array.Unboxed
sieve :: Int -> UArray Int Bool
sieve n = runSTUArray $ do
let m = (n-1) `div` 2
r = floor . sqrt $ fromIntegral n
bits <- newArray (0, m-1) True
forM_ [0 .. r `div` 2 - 1] $ \i -> do
isPrime <- readArray bits i
when isPrime $ do
let a = 2*i*i + 6*i + 3
b = 2*i*i + 8*i + 6
forM_ [a, b .. (m-1)] $ \j -> do
writeArray bits j False
return bits
primes :: Int -> [Int]
primes n = 2 : [2*i+3 | (i, True) <- assocs $ sieve n]
main = do
print $ sum $ primes 1000000
You can run it on ideone. My algorithm is the Sieve of Eratosthenes, and it should be quite fast for small n. For n = 2,000,000,000, the array size may be a problem, in which case you will need to use a segmented sieve. See my blog for more information about the Sieve of Eratosthenes. See this answer for information about a segmented sieve (but not in Haskell, unfortunately).

Related

Gradient Descent algorithm not converging in Haskell

I am trying to implement the gradient descent algorithm in Andrew Ng's ML course. After reading in the data, I try to implement the following, updating my list of theta values 1000 times, with the expectation of some convergence.
The algorithm in question is gradientDescent. I know that typically a cause of this problem is that alpha can be too large, but when I change alpha by a factor of n for example, my results change by a factor of n. The same happens when I change iterations by a factor of n. I want to say this could be to do with haskell's laziness, but I'm completely unsure. Any help would be appreciated.
module LR1V where
import qualified Data.Matrix as M
import System.IO
import Data.List.Split
import qualified Data.Vector as V
main :: IO ()
main = do
contents <- getContents
let lns = lines contents :: [String]
entries = map (splitOn ",") lns :: [[String]]
mbPoints = mapM readPoints entries :: Maybe [[Double]]
case mbPoints of
Just points -> runData points
_ -> putStrLn "Error: it is possible the file is incorrectly formatted"
readPoints :: [String] -> Maybe [Double]
readPoints dat#(x:y:_) = return $ map read dat
readPoints _ = Nothing
runData :: [[Double]] -> IO ()
runData pts = do
let (mxs,ys) = runPoints pts
c = M.ncols mxs
m = M.nrows mxs
thetas = M.zero 1 (M.ncols mxs)
alpha = 0.01
iterations = 1000
results = gradientDescent mxs ys thetas alpha m c iterations
print results
runPoints :: [[Double]] -> (M.Matrix Double, [Double])
runPoints pts = (xs, ys) where
xs = M.fromLists $ addX0 $ map init pts
ys = map last pts
-- X0 will always be 1
addX0 :: [[Double]] -> [[Double]]
addX0 = map (1.0 :)
-- theta is 1xn and x is nx1, where n is the amount of features
-- so it is safe to assume a scalar results from the multiplication
hypothesis :: M.Matrix Double -> M.Matrix Double -> Double
hypothesis thetas x =
M.getElem 1 1 (M.multStd thetas x)
gradientDescent :: M.Matrix Double
-> [Double]
-> M.Matrix Double
-> Double
-> Int
-> Int
-> Int
-> [Double]
gradientDescent mxs ys thetas alpha m n it =
let x i = M.colVector $ M.getRow i mxs
y i = ys !! (i-1)
h i = hypothesis thetas (x i)
thL = zip [1..] $ M.toList thetas :: [(Int, Double)]
z i j = ((h i) - (y i))*(M.getElem i j $ mxs)
sumSquares j = sum [z i j | i <- [1..m]]
thetaJ t j = t - ((alpha * (1/ (fromIntegral m))) * (sumSquares j))
result = map snd $ foldl (\ts _ -> [(j,thetaJ t j) | (j,t) <- ts]) thL [1..it] in
result
and the data...
6.1101,17.592
5.5277,9.1302
8.5186,13.662
7.0032,11.854
5.8598,6.8233
8.3829,11.886
7.4764,4.3483
8.5781,12
6.4862,6.5987
5.0546,3.8166
5.7107,3.2522
14.164,15.505
5.734,3.1551
8.4084,7.2258
5.6407,0.71618
5.3794,3.5129
6.3654,5.3048
5.1301,0.56077
6.4296,3.6518
7.0708,5.3893
6.1891,3.1386
20.27,21.767
5.4901,4.263
6.3261,5.1875
5.5649,3.0825
18.945,22.638
12.828,13.501
10.957,7.0467
13.176,14.692
22.203,24.147
5.2524,-1.22
6.5894,5.9966
9.2482,12.134
5.8918,1.8495
8.2111,6.5426
7.9334,4.5623
8.0959,4.1164
5.6063,3.3928
12.836,10.117
6.3534,5.4974
5.4069,0.55657
6.8825,3.9115
11.708,5.3854
5.7737,2.4406
7.8247,6.7318
7.0931,1.0463
5.0702,5.1337
5.8014,1.844
11.7,8.0043
5.5416,1.0179
7.5402,6.7504
5.3077,1.8396
7.4239,4.2885
7.6031,4.9981
6.3328,1.4233
6.3589,-1.4211
6.2742,2.4756
5.6397,4.6042
9.3102,3.9624
9.4536,5.4141
8.8254,5.1694
5.1793,-0.74279
21.279,17.929
14.908,12.054
18.959,17.054
7.2182,4.8852
8.2951,5.7442
10.236,7.7754
5.4994,1.0173
20.341,20.992
10.136,6.6799
7.3345,4.0259
6.0062,1.2784
7.2259,3.3411
5.0269,-2.6807
6.5479,0.29678
7.5386,3.8845
5.0365,5.7014
10.274,6.7526
5.1077,2.0576
5.7292,0.47953
5.1884,0.20421
6.3557,0.67861
9.7687,7.5435
6.5159,5.3436
8.5172,4.2415
9.1802,6.7981
6.002,0.92695
5.5204,0.152
5.0594,2.8214
5.7077,1.8451
7.6366,4.2959
5.8707,7.2029
5.3054,1.9869
8.2934,0.14454
13.394,9.0551
5.4369,0.61705
When alpha is 0.01, my thetas evaluate to [58.39135051546406,653.2884974555699]. When alpha is 0.001 my values become [5.839135051546473,65.32884974555617]. When iterations is changed to 10,000 my values return to what they were before.
It appears that with each run of updating theta values, the approximation function h(x) was using the initial theta vector each time, rather than the updated vector. Now, I get an alright approximation of my theta values. However, increasing the number of iterations by a large factor changes my results in an odd way.

Haskell performance using dynamic programming

I am attempting to calculate the Levenshtein distance between two strings using dynamic programming. This is being done through Hackerrank, so I have timing constraints. I used a techenique I saw in: How are Dynamic Programming algorithms implemented in idiomatic Haskell? and it seems to be working. Unfortunaly, it is timing out in one test case. I do not have access to the specific test case, so I don't know the exact size of the input.
import Control.Monad
import Data.Array.IArray
import Data.Array.Unboxed
main = do
n <- readLn
replicateM_ n $ do
s1 <- getLine
s2 <- getLine
print $ editDistance s1 s2
editDistance :: String -> String -> Int
editDistance s1 s2 = dynamic editDistance' (length s1, length s2)
where
s1' :: UArray Int Char
s1' = listArray (1,length s1) s1
s2' :: UArray Int Char
s2' = listArray (1,length s2) s2
editDistance' table (i,j)
| min i j == 0 = max i j
| otherwise = min' (table!((i-1),j) + 1) (table!(i,(j-1)) + 1) (table!((i-1),(j-1)) + cost)
where
cost = if s1'!i == s2'!j then 0 else 1
min' a b = min (min a b)
dynamic :: (Array (Int,Int) Int -> (Int,Int) -> Int) -> (Int,Int) -> Int
dynamic compute (xBnd, yBnd) = table!(xBnd,yBnd)
where
table = newTable $ map (\coord -> (coord, compute table coord)) [(x,y) | x<-[0..xBnd], y<-[0..yBnd]]
newTable xs = array ((0,0),fst (last xs)) xs
I've switched to using arrays, but that speed up was insufficient. I cannot use Unboxed arrays, because this code relies on laziness. Are there any glaring performance mistakes I have made? Or how else can I speed it up?
The backward equations for edit distance calculations are:
f(i, j) = minimum [
1 + f(i + 1, j), -- delete from the 1st string
1 + f(i, j + 1), -- delete from the 2nd string
f(i + 1, j + 1) + if a(i) == b(j) then 0 else 1 -- substitute or match
]
So within each dimension, you need nothing more than the very next index: + 1. This is a sequential access pattern, not random access to require arrays; and can be implemented using lists and nested right folds:
editDistance :: Eq a => [a] -> [a] -> Int
editDistance a b = head . foldr loop [n, n - 1..0] $ zip a [m, m - 1..]
where
(m, n) = (length a, length b)
loop (s, l) lst = foldr go [l] $ zip3 b lst (tail lst)
where
go (t, i, j) acc#(k:_) = inc `seq` inc:acc
where inc = minimum [i + 1, k + 1, if s == t then j else j + 1]
You may test this code in Hackerrank Edit Distance Problem as in:
import Control.Applicative ((<$>))
import Control.Monad (replicateM_)
import Text.Read (readMaybe)
editDistance :: Eq a => [a] -> [a] -> Int
editDistance a b = ... -- as implemented above
main :: IO ()
main = do
Just n <- readMaybe <$> getLine
replicateM_ n $ do
a <- getLine
b <- getLine
print $ editDistance a b
which passes all tests with a decent performance.

Randomized algorithm not behaving as expected

I am implementing an approximate counting algorithm where we:
Maintain a counter X using log (log n) bits
Initialize X to 0
When an item arrives, increase X by 1 with probability (½)X
When the stream is over, output 2X − 1 so that E[2X]= n + 1
My implementation is as follows:
import System.Random
type Prob = Double
type Tosses = Int
-- * for sake of simplicity we assume 0 <= p <= 1
tos :: Prob -> StdGen -> (Bool,StdGen)
tos p s = (q <= 100*p, s')
where (q,s') = randomR (1,100) s
toses :: Prob -> Tosses -> StdGen -> [(Bool,StdGen)]
toses _ 0 _ = []
toses p n s = let t#(b,s') = tos p s in t : toses p (pred n) s'
toses' :: Prob -> Tosses -> StdGen -> [Bool]
toses' p n = fmap fst . toses p n
morris :: StdGen -> [a] -> Int
morris s xs = go s xs 0 where
go _ [] n = n
go s (_:xs) n = go s' xs n' where
(h,s') = tos (0.5^n) s
n' = if h then succ n else n
main :: IO Int
main = do
s <- newStdGen
return $ morris s [1..10000]
The problem is that my X is always incorrect for any |stream| > 2, and it seems like for all StdGen and |stream| > 1000, X = 7
I tested the same algorithm in Matlab and it works there, so I assume it's either
an issue with my random number generator, or
raising 1/2 to a large n in Double
Please suggest a path forward?
The problem is actually very simple: with randomR (1,100) you preclude values within the first percent, so you have a complete cutoff at high powers of 1/2 (which all lie in that small interval). Actually a general thing: ranges should start at zero, not at one†, unless there's a specific reason.
But why even use a range of 100 in the first place? I'd just make it
tos :: Prob -> StdGen -> (Bool,StdGen)
tos p s = (q <= p, s')
where (q,s') = randomR (0,1) s
†I know, Matlab gets this wrong all over the place. Just one of the many horrible things about that language.
Unrelated to your problem: as chi remarked this kind of code looks a lot nicer if you use a suitable random monad, instead of manually passing around StdGens.
import Data.Random
import Data.Random.Source.Std
type Prob = Double
tos :: Prob -> RVar Bool
tos p = do
q <- uniform 0 1
return $ q <= p
morris :: [a] -> RVar Int
morris xs = go xs 0 where
go [] n = return n
go (_:xs) n = do
h <- tos (0.5^n)
go xs $ if h then succ n else n
morrisTest :: Int -> IO Int
morrisTest n = do
runRVar (morris [1..n]) StdRandom

Project Euler No. 14 Haskell

I'm trying to resolve problem 14 of Project Euler (http://projecteuler.net/problem=14) and I hit a dead end using Haskell.
Now, I know that the numbers may be small enough and I could do a brute force, but that isn't the purpose of my exercise.
I am trying to memorize the intermediate results in a Map of type Map Integer (Bool, Integer) with the meaning of:
- the first Integer (the key) holds the number
- the Tuple (Bool, Interger) holds either (True, Length) or (False, Number)
where Length = length of the chain
Number = the number before him
Ex:
for 13: the chain is 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
My map should contain :
13 - (True, 10)
40 - (False, 13)
20 - (False, 40)
10 - (False, 20)
5 - (False, 10)
16 - (False, 5)
8 - (False, 16)
4 - (False, 8)
2 - (False, 4)
1 - (False, 2)
Now when I search for another number like 40 i know that the chain has (10 - 1) length and so on.
I want now, if I search for 10, not only to tell me that length of 10 is (10 - 3) length and update the map, but also I want to update 20, 40 in case they are still (False, _)
My code:
import Data.Map as Map
solve :: [Integer] -> Map Integer (Bool, Integer)
solve xs = solve' xs Map.empty
where
solve' :: [Integer] -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
solve' [] table = table
solve' (x:xs) table =
case Map.lookup x table of
Nothing -> countF x 1 (x:xs) table
Just (b, _) ->
case b of
True -> solve' xs table
False -> {-WRONG-} solve' xs table
f :: Integer -> Integer
f x
| x `mod` 2 == 0 = x `quot` 2
| otherwise = 3 * x + 1
countF :: Integer -> Integer -> [Integer] -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
countF n cnt (x:xs) table
| n == 1 = solve' xs (Map.insert x (True, cnt) table)
| otherwise = countF (f n) (cnt + 1) (x:xs) $ checkMap (f n) n table
checkMap :: Integer -> Integer -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
checkMap n rez table =
case Map.lookup n table of
Nothing -> Map.insert n (False, rez) table
Just _ -> table
At the {-WRONG-} part we should update all the values like in the following example:
--We are looking for 10:
10 - (False, 20)
|
V {-finally-} update 10 => (True, 10 - 1 - 1 - 1)
20 - (False, 40) ^
| |
V update 20 => 20 - (True, 10 - 1 - 1)
40 - (False, 13) ^
| |
V update 40 => 40 - (True, 10 - 1)
13 - (True, 10) ^
| |
---------------------------
The problem is that I don't know if its possible to do 2 things in a function like updating a number and continue the recurence. In a C like language I may do something like (pseudocode):
void f(int n, tuple(b,nr), int &length, table)
{
if(b == False) f (nr, (table lookup nr), 0, table);
// the bool is true so we got a length
else
{
length = nr;
return;
}
// Since this is a recurence it would work as a stack, producing the right output
table update(n, --cnt);
}
The last instruction would work since we are sending cnt by reference. Also we always know that it will finish at some point and cnt should not be < 1.
The easiest optimization (as you have identified) is memoization. You have attempted create a memoization system yourself, however have come across issues on how to store the memoized values. There are solutions to doing this in a maintainable way, such as using a State monad or a STArray. However, there is a much simpler solution to your problem - use haskell's existing memoization. Haskell by default remembers constant values, so if you create a value that stores the collatz values, it will be automatically memoized!
A simple example of this is the following fibonacci definition:
fib :: Int -> Integer
fib n = fibValues !! n where
fibValues = 1 : 1 : zipWith (+) fibValues (tail fibValues)
The fibValues is a [Integer], and as it is just a constant value, it is memoized. However, that doesn't mean it is all memoized at once, since as it is an infinte list, this would never finish. Instead, the values are only calculated when needed, as haskell is lazy.
So if you do something similar with your problem, you will get memoization without a lot of the work. However, using a list like above won't work well in your solution. This is because the collatz algorithm uses many different values to get the result for a given number, so the container used will require random access to be efficient. The obvious choice is an array.
collatzMemoized :: Array Integer Int
Next, we need to fill up the array with the correct values. I'll write this function pretending a collatz function exists that calculates the collatz value for any n. Also, note that arrays are fixed size, so a value needs to be used to determine the maximum number to memoize. I'll use a million, but any value can be used (it is a memory/speed tradeoff).
collatzMemoized = listArray (1, maxNumberToMemoize) $ map collatz [1..maxNumberToMemoize] where
maxNumberToMemroize = 1000000
That is pretty straightforward, the listArray is given bounds, and the a list of all the collatz values in that range is given to it. Remember that this won't calculate all the collatz values straight away, as the values are lazy.
Now, the collatz function can be written. The most important part is to only check the collatzMemoized array if the number being checked is within its bounds:
collatz :: Integer -> Int
collatz 1 = 1
collatz n
| inRange (bounds collatzMemoized) nextValue = 1 + collatzMemoized ! nextValue
| otherwise = 1 + collatz nextValue
where
nextValue = case n of
1 -> 1
n | even n -> n `div` 2
| otherwise -> 3 * n + 1
In ghci, you can now see the effectiveness of the memoization. Try collatz 200000. It will take about 2 seconds to finish. However, if you run it again, it will complete instantly.
Finally, the solution can be found:
maxCollatzUpTo :: Integer -> (Integer, Int)
maxCollatzUpTo n = maximumBy (compare `on` snd) $ zip [1..n] (map collatz [1..n]) where
and then printed:
main = print $ maxCollatzUpTo 1000000
If you run main, the result will be printed in about 10 seconds.
Now, a small problem with this approach is it uses a lot of stack space. It will work fine in ghci (which seems to use be more flexible with regards to stack space). However, if you compile it and try to run the executable, it will crash (with a stack space overflow). So to run the program, you have to specify more when you compile it. This can be done by adding -with-rtsopts='K64m' to the compile options. This increases the stack to 64mb.
Now the program can be compiled and ran:
> ghc -O3 --make -with-rtsopts='-K6m' problem.hs
Running ./problem will give the result in less than a second.
You are going about memoization the hard way, trying to write an imperative program in Haskell. Borrowing from David Eisenstat's solution, we'll solve it as j_random_hacker suggested:
collatzLength :: Integer -> Integer
collatzLength n
| n == 1 = 1
| even n = 1 + collatzLength (n `div` 2)
| otherwise = 1 + collatzLength (3*n + 1)
The dynamic programming solution for this is to replace the recursion with looking things up in a table. Let's make a function where we can replace the recursive call:
collatzLengthDef :: (Integer -> Integer) -> Integer -> Integer
collatzLengthDef r n
| n == 1 = 1
| even n = 1 + r (n `div` 2)
| otherwise = 1 + r (3*n + 1)
Now we could define the recursive algorithm as
collatzLength :: Integer -> Integer
collatzLength = collatzLengthDef collatzLength
Now we could also make a tabled version of this (it takes a number for the table size, and returns a collatzLength function that is calculated using a table of that size):
-- A utility function that makes memoizing things easier
buildTable :: (Ix i) => (i, i) -> (i -> e) -> Array i e
buildTable bounds f = array $ map (\x -> (x, f x)) $ range bounds
collatzLengthTabled :: Integer -> Integer -> Integer
collatzLengthTabled n = collatzLengthTableLookup
where
bounds = (1, n)
table = buildTable bounds (collatzLengthDef collatzLengthTableLookup)
collatzLengthTableLookup =
\x -> Case inRange bounds x of
True -> table ! x
_ -> (collatzLengthDef collatzLengthTableLookup) x
This works by defining the collatzLength to be a table lookup, with the table being the definition of the function, but with recursive calls replaced by table lookup. The table lookup function checks to see if the argument to the function is in the range that is tabled, and falls back on the definition of the function. We can even make this work for tabling any function like this:
tableRange :: (Ix a) => (a, a) -> ((a -> b) -> a -> b) -> a -> b
tableRange bounds definition = tableLookup
where
table = buildTable bounds (definition tableLookup)
tableLookup =
\x -> Case inRange bounds x of
True -> table ! x
_ -> (definition tableLookup) x
collatzLengthTabled n = tableRange (1, n) collatzLengthDef
You just need to make sure that you
let memoized = collatzLengthTabled 10000000
... memoized ...
So that only one table is built in memory.
I remember finding memoisation of dynamic programming algorithms very counterintuitive in Haskell, and it's been a while since I've done it, but hopefully the following trick works for you.
But first, I don't quite understand your current DP scheme, though I suspect it may be quite inefficient as it seems like it will need to update many entries for each answer. (a) I don't know how to do this in Haskell, and (b) you don't need to do this to solve the problem efficiently ;-)
I suggest the following approach instead: first build an ordinary recursive function that computes the right answer for an input number. (Hint: it will have a signature like collatzLength :: Int -> Int.) When you have this function working, just replace its definition with the definition of an array whose elements are defined lazily with the array function using an association list, and replace all recursive calls to the function to array lookups (e.g. collatzLength 42 would become collatzLength ! 42). This will automagically populate the array in the necessary order! So your "top-level" collatzLength object will now actually be an array, rather than a function.
As I suggested above, I would use an array instead of a map datatype to hold the DP table, since you will need to store values for all integer indices from 1 up to 1,000,000.
I don't have a Haskell compiler handy, so I apologize for any broken code.
Without memoization, there's a function
collatzLength :: Integer -> Integer
collatzLength n
| n == 1 = 1
| even n = 1 + collatzLength (n `div` 2)
| otherwise = 1 + collatzLength (3*n + 1)
With memoization, the type signature is
memoCL :: Map Integer Integer -> Integer -> (Map Integer Integer, Integer)
since memoCL receives a table as input and gives the updated table as output. What memoCL needs to do is intercept the return of the recursive call with a let form and insert the new result.
-- table must have an initial entry for 1
memoCL table n = case Map.lookup n table of
Just m -> (table, m)
Nothing -> let (table', m) = memoCL table (collatzStep n) in (Map.insert n (1 + m) table', 1 + m)
collatzStep :: Integer -> Integer
collatzStep n = if even n then n `div` 2 else 3*n + 1
At some point you'll get sick of the above idiom. Then it's time for monads.
I eventually modify the {-WRONG-} part to do what it should with a call to mark x (b, n) [] xs table where
mark :: Integer -> (Bool, Integer) -> [Integer] -> [Integer] -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
mark crtElem (b, n) list xs table
| b == False = mark n (findElem n table) (crtElem:list) xs table
| otherwise = continueWith n list xs table
continueWith :: Integer -> [Integer] -> [Integer] -> Map Integer (Bool, Integer) -> Map Integer (Bool, Integer)
continueWith _ [] xs table = solve' xs table
continueWith cnt (y:ys) xs table = continueWith (cnt - 1) ys xs (Map.insert y (True, cnt - 1) table)
findElem :: Integer -> Map Integer (Bool, Integer) -> (Bool, Integer)
findElem n table =
case Map.lookup n table of
Nothing -> (False, 0)
Just (b, nr) -> (b, nr)
But it seams that there are better (and far less verbose) answers than this 1
Maybe you might find interesting how I solved the problem. Its is pretty functional though it might be not the most efficient thing on earth :)
You can find the code here: https://github.com/fmancinelli/project-euler/blob/master/haskell/project-euler/Problem014.hs
P.S.: Disclaimer: I was doing Project Euler exercises in order to learn Haskell, so the quality of the solution could be debatable.
Since we are studying recursion schemes, here's one for you.
Let's consider functor N(A,B,X)=A+B*X, which is a stream of Bs with the last element being A.
{-# LANGUAGE DeriveFunctor
, TypeFamilies
, TupleSections #-}
import Data.Functor.Foldable
import qualified Data.Map as M
import Data.List
import Data.Function
import Data.Int
data N a b x = Z a | S b x deriving (Functor)
This stream is handy for several kinds of iterations. For one, we can use it to represent a chain of Ints in a Collatz sequence:
type instance Base Int64 = N Int Int64
instance Foldable Int64 where
project 1 = Z 1
project x | odd x = S x $ 3*x+1
project x = S x $ x `div` 2
This is just a algebra, not a initial one, because the transformation is not a isomorphism (same chain of Ints is part of a chain for 2*x and (x-1)/3), but this is sufficient to represent the fixpoint Base Int64 Int64.
With this definition, cata is going to feed the chain to the algebra given to it, and you can use it to construct a memo Map of integers to the chain length. Finally, anamorphism can use it to generate a stream of solutions to the problem of different sizes:
problems = ana (uncurry $ cata . phi) (M.empty, 1) where
phi :: M.Map Int64 Int ->
Base Int64 (Prim [(Int64, Int)] (M.Map Int64 Int, Int64)) ->
Prim [(Int64, Int)] (M.Map Int64 Int, Int64)
phi m (Z v) = found m 1 v
phi m (S x ~(Cons (_, v') (m', _))) = maybe (notFound m' x v') (found m x) $
M.lookup x m
The ~ before (Cons ...) means lazy pattern matching. We don't touch the pattern until the values are needed. If not for lazy pattern matching, it would always construct the whole chain, and using the map would be useless. With lazy pattern matching we only construct the values v' and m' if the chain length for x was not in the map.
Helper functions construct the stream of (Int, chain length) pairs:
found m x v = Cons (x, v) (m, x+1)
notFound m x v = Cons (x, 1+v) (M.insert x (1+v) m, x+1)
Now just take the first 999999 problems, and figure out the one that has the longest chain:
main = print $ maximumBy (compare `on` snd) $ take 999999 problems
This works slower than array-based solution, because Map lookup is logarithmic of map size, but this solution is not fixed size. Still, it finishes in about 5 seconds.

Performance of looping over an Unboxed array in Haskell

First of all, it's great. However, I came across a situation where my benchmarks turned up weird results. I am new to Haskell, and this is first time I've gotten my hands dirty with mutable arrays and Monads. The code below is based on this example.
I wrote a generic monadic for function that takes numbers and a step function rather than a range (like forM_ does). I compared using my generic for function (Loop A) against embedding an equivalent recursive function (Loop B). Having Loop A is noticeably faster than having Loop B. Weirder, having both Loop A and B together is faster than having Loop B by itself (but slightly slower than Loop A by itself).
Some possible explanations I can think of for the discrepancies. Note that these are just guesses:
Something I haven't learned yet about how Haskell extracts results from monadic functions.
Loop B faults the array in a less cache efficient manner than Loop A. Why?
I made a dumb mistake; Loop A and Loop B are actually different.
Note that in all 3 cases of having either or both Loop A and Loop B, the program produces the same output.
Here is the code. I tested it with ghc -O2 for.hs using GHC version 6.10.4 .
import Control.Monad
import Control.Monad.ST
import Data.Array.IArray
import Data.Array.MArray
import Data.Array.ST
import Data.Array.Unboxed
for :: (Num a, Ord a, Monad m) => a -> a -> (a -> a) -> (a -> m b) -> m ()
for start end step f = loop start where
loop i
| i <= end = do
f i
loop (step i)
| otherwise = return ()
primesToNA :: Int -> UArray Int Bool
primesToNA n = runSTUArray $ do
a <- newArray (2,n) True :: ST s (STUArray s Int Bool)
let sr = floor . (sqrt::Double->Double) . fromIntegral $ n+1
-- Loop A
for 4 n (+ 2) $ \j -> writeArray a j False
-- Loop B
let f i
| i <= n = do
writeArray a i False
f (i+2)
| otherwise = return ()
in f 4
forM_ [3,5..sr] $ \i -> do
si <- readArray a i
when si $
forM_ [i*i,i*i+i+i..n] $ \j -> writeArray a j False
return a
primesTo :: Int -> [Int]
primesTo n = [i | (i,p) <- assocs . primesToNA $ n, p]
main = print $ primesTo 30000000
I just tried benchmarking this with Criterion and GHC 6.12.1, and Loop A looks only slightly faster for me. I definitely don't get the weird "both together are faster than B alone" effect.
Also, if your step function really is just a step and doesn't do anything wacky with its argument, the following version of for seems a bit faster, especially for smaller arrays:
for' :: (Enum a, Num a, Ord a, Monad m) => a -> a -> (a -> a) -> (a -> m b) -> m ()
for' start end step = forM_ $ enumFromThenTo start (step start) end
Here are the results from Criterion, where loopA' is your loop A using my for', and where loopC is both A and B together:
benchmarking loopA...
mean: 2.372893 s, lb 2.370982 s, ub 2.374914 s, ci 0.950
std dev: 10.06753 ms, lb 8.820194 ms, ub 11.66965 ms, ci 0.950
benchmarking loopA'...
mean: 2.368167 s, lb 2.354312 s, ub 2.381413 s, ci 0.950
std dev: 69.50334 ms, lb 65.94236 ms, ub 73.17173 ms, ci 0.950
benchmarking loopB...
mean: 2.423160 s, lb 2.419131 s, ub 2.427260 s, ci 0.950
std dev: 20.78412 ms, lb 18.06613 ms, ub 24.99021 ms, ci 0.950
benchmarking loopC...
mean: 4.308503 s, lb 4.304875 s, ub 4.312110 s, ci 0.950
std dev: 18.48732 ms, lb 16.19325 ms, ub 21.32299 ms, ci 0.950<
And here's the code:
module Main where
import Control.Monad
import Control.Monad.ST
import Data.Array.ST
import Data.Array.Unboxed
import Criterion.Main
for :: (Num a, Ord a, Monad m) => a -> a -> (a -> a) -> (a -> m b) -> m ()
for start end step f = loop start where
loop i
| i <= end = do
f i
loop (step i)
| otherwise = return ()
for' :: (Enum a, Num a, Ord a, Monad m) => a -> a -> (a -> a) -> (a -> m b) -> m ()
for' start end step = forM_ $ enumFromThenTo start (step start) end
loopA arr n = for 4 n (+ 2) $ flip (writeArray arr) False
loopA' arr n = for' 4 n (+ 2) $ flip (writeArray arr) False
loopB arr n =
let f i | i <= n = do writeArray arr i False
f (i+2)
| otherwise = return ()
in f 4
loopC arr n = do
loopA arr n
loopB arr n
runPrimes loop n = do
let sr = floor . (sqrt::Double->Double) . fromIntegral $ n+1
a <- newArray (2,n) True :: (ST s (STUArray s Int Bool))
loop a n
forM_ [3,5..sr] $ \i -> do
si <- readArray a i
when si $
forM_ [i*i,i*i+i+i..n] $ \j -> writeArray a j False
return a
primesA n = [i | (i,p) <- assocs $ runSTUArray $ runPrimes loopA n, p]
primesA' n = [i | (i,p) <- assocs $ runSTUArray $ runPrimes loopA' n, p]
primesB n = [i | (i,p) <- assocs $ runSTUArray $ runPrimes loopB n, p]
primesC n = [i | (i,p) <- assocs $ runSTUArray $ runPrimes loopC n, p]
main = let n = 10000000 in
defaultMain [ bench "loopA" $ nf primesA n
, bench "loopA'" $ nf primesA' n
, bench "loopB" $ nf primesB n
, bench "loopC" $ nf primesC n ]
Perhaps compare and contrast with the Shootout nsieve program? in any case, the only way to know what really is happening is to look at the core (e.g. with the ghc-core tool).
{-# OPTIONS -O2 -optc-O -fbang-patterns -fglasgow-exts -optc-march=pentium4 #-}
--
-- The Computer Language Shootout
-- http://shootout.alioth.debian.org/
--
-- Contributed by Don Stewart 2005
-- nsieve over an ST monad Bool array
--
import Control.Monad.ST
import Data.Array.ST
import Data.Array.Base
import System
import Control.Monad
import Data.Bits
import Text.Printf
main = do
n <- getArgs >>= readIO . head :: IO Int
mapM_ (\i -> sieve (10000 `shiftL` (n-i))) [0, 1, 2]
sieve n = do
let r = runST (do a <- newArray (2,n) True :: ST s (STUArray s Int Bool)
go a n 2 0)
printf "Primes up to %8d %8d\n" (n::Int) (r::Int) :: IO ()
go !a !m !n !c
| n == m = return c
| otherwise = do
e <- unsafeRead a n
if e then let loop j
| j < m = do
x <- unsafeRead a j
when x $ unsafeWrite a j False
loop (j+n)
| otherwise = go a m (n+1) (c+1)
in loop (n `shiftL` 1)
else go a m (n+1) c

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