[Haskell-cafe] Project Euler Problem 357 in Haskell

[Lyndon Maydwell]

Could Int be overflowing?

On Tue, Nov 8, 2011 at 7:21 PM, mukesh tiwari
<mukeshtiwari.iiitm at gmail.com> wrote:
> Hello all
> Being a Haskell enthusiastic , first I tried to solve this problem in
> Haskell but it running for almost 10 minutes on my computer but not getting
> the answer. A similar C++ program outputs the answer almost instant so could
> some one please tell me how to improve this Haskell program.
>
> import Control.Monad.ST
> import Data.Array.ST
> import Data.Array.Unboxed
> import Control.Monad
>
> prime :: Int -> UArray Int Bool
> prime n = runSTUArray $ do
>     arr <- newArray ( 2 , n ) True :: ST s ( STUArray s Int Bool )
>     forM_ ( takeWhile ( \x -> x*x <= n ) [ 2 .. n ] ) $ \i -> do
>         ai <- readArray arr i
>         when ( ai  ) $ forM_ [ i^2 , i^2 + i .. n ] $ \j -> do
>             writeArray arr j False
>
>     return arr
>
> pList :: UArray Int Bool
> pList = prime $  10 ^ 8
>
> divPrime :: Int -> Bool
> divPrime n = all ( \d -> if mod n d == 0 then pList ! ( d + div  n  d ) else
> True )  $  [ 1 .. truncate . sqrt . fromIntegral  $ n ]
>
>
> main = putStrLn . show . sum  $ [ if and [ pList ! i , divPrime . pred $ i ]
> then pred  i else 0 | i <- [ 2 .. 10 ^ 8 ] ]
>
>
> C++ program which outputs the answer almost instant.
>
> #include<cstdio>
> #include<iostream>
> #include<vector>
> #define Lim 100000001
> using namespace std;
>
> bool prime [Lim];
> vector<int> v ;
>
> void isPrime ()
>      {
> 		for( int i = 2 ; i * i <= Lim ; i++)
> 		 if ( !prime [i]) for ( int j = i * i ; j <= Lim ; j += i ) prime [j] = 1
> ;
>
> 		for( int i = 2 ; i <= Lim ; i++) if ( ! prime[i] ) v.push_back( i ) ;
> 		//cout<<v.size()<<endl;
> 		//for(int i=0;i<10;i++) cout<<v[i]<<" ";cout<<endl;
>
>      }
>
> int main()
> 	{
> 		isPrime();
> 		int n = v.size();
> 		long long sum = 0;
> 		for(int i = 0 ; i < n ; i ++)
> 		 {
> 			int k = v[i]-1;
> 			bool f = 0;
> 			for(int i = 1 ; i*i<= k ; i++)
> 				if ( k % i == 0 && prime[ i + ( k / i ) ] )  { f=1 ; break ; }
>
> 			if ( !f ) sum += k;
> 		 }
> 		cout<<sum<<endl;
> 	}
>
> Regards
> Mukesh Tiwari
>
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