# My Weblog

## Proving the correctness of code using SMT sovler.

SMT solvers are great tool in computer science. One use of these solvers to prove the correctness of code. I wrote Haskell code to compute the inverse of polynomial using extended euclidean algorithm in Galois Field GF ( 2n ). If we try to analyze the code of inversePolynomial then every thing is normal except passing the depth parameter. It is unnecessary because we know that recursion is going to terminate. Initially I wrote the code without depth and the program was not terminating so I wrote mail to Levent Erkok and according to him

Your code suffers from what’s known as the “symbolic termination”. See Section 7.4 of: http://goo.gl/0E7wkm for a discussion of this issue in the context of Cryptol; but the same comments apply here as well. Briefly, when you recurse in inversePoly, SBV does not know when to stop the recursion of both branches in the ite: Thus it keeps expanding both branches ad infinitum.

Luckily, symbolic termination is usually easy to deal with once you identify what the problem is. Typically, one needs to introduce a recursion depth counter, which would stop recursion when it reaches a pre-determined depth that is determined to be safe. In fact, SBV comes with the regular GCD example that talks about this issue in detail:

I’d recommend reading the source code of the above; it should clarify the problem. You’ll need to find out a recursion-depth that’ll work for your example (try running it on concrete values and coming up with a safe number); and then prove that the bound is actually correct as explained in the above. The recursion depth bound just needs to be an overapproximation: If the algorithm needs 20 iterations but you use 30 as your recursion depth, nothing will go wrong; it’ll just produce bigger code and thus be less efficient and maybe more complicated for the backend prover. Of course, if you put a depth of 18 when 20 is needed, then it will be wrong and you won’t be able to establish correctness.

My verification condition is inverse of inverse of polynomial is same as polynomial. Right now I have no idea about proving the upper bound on depth so I just took 20. If you have any suggestion/comments then please let me know.


import Data.SBV.Bridge.Z3
import Data.Maybe

inversePolynomial ::  [ Int ] -> [ Int ] ->  SWord16
inversePolynomial poly reducer =  inversePoly  reducer' rem' first' second' depth'  where
poly' =  polynomial poly :: SWord16
reducer' =  polynomial reducer :: SWord16
first' =  0 :: SWord16
second' = 1 :: SWord16
depth' = 20 :: SWord16
( quot', rem' ) = pDivMod poly' reducer'

inversePoly :: SWord16 -> SWord16 -> SWord16 -> SWord16  ->  SWord16 -> SWord16
inversePoly  reducer'' poly'' first'' second'' depth'' =
ite ( depth'' .== 0 ||| rem'' .== ( 0 :: SWord16 ) ) (  second'' )  ( inversePoly  poly'' rem'' second'' ret ( depth'' - 1 ) )  where
( quot'', rem'' ) = pDivMod reducer'' poly''
ret =  pAdd ( pMult ( quot'', second'', [] ) ) first''

isInversePolyCorrect :: SWord16 -> SWord16 -> SBool
isInversePolyCorrect poly reducer = inversePoly reducer ( inversePoly reducer rem first second depth ) first second depth .== rem where
( quot, rem ) = pDivMod  poly reducer
first = 0 :: SWord16
second = 1 :: SWord16
depth = 20 :: SWord16

*Main> showPoly ( inversePolynomial  [ 6,4,1,0] [ 8,4,3,1,0] )
"x^7 + x^6 + x^3 + x"
*Main> proveWith z3 { verbose=False }  $forAll ["x"]$ \x ->  pMod x ( 283 :: SWord16 ) ./=0 ==> isInversePolyCorrect x ( 283 :: SWord16 )
Q.E.D.
*Main> proveWith cvc4 { verbose=False }  $forAll ["x"]$ \x ->  pMod x ( 283 :: SWord16 ) ./=0 ==> isInversePolyCorrect x ( 283 :: SWord16 )
*** Exception: fd:10: hFlush: resource vanished (Broken pipe)
*Main> proveWith yices { verbose=False }  $forAll ["x"]$ \x ->  pMod x ( 283 :: SWord16 ) ./=0 ==> isInversePolyCorrect x ( 283 :: SWord16 )
Q.E.D.


September 21, 2013

## Counting Inversions

First week of algorithm course programming assignment is to compute the number of inversion count of given array. We have a given array A[0...n - 1] of n distinct positive integers and we have to count all the pairs ( i , j ) such that i < j and A[i] > A[j].
1. Brute Force
Brute force idea is very simple. Run two loops and keep counting the respective pairs for which i < j and A[i] > A[j]. The complexity of brute force is O( n2 ).

bruteForce :: [ Int ] -> Int
bruteForce [] = 0
bruteForce [_] = 0
bruteForce ( x : xs ) = ( length . filter ( x > ) $xs ) + bruteForce xs *Main> bruteForce [ 3 , 1 , 2] Loading package array-0.4.0.1 ... linking ... done. Loading package deepseq-1.3.0.1 ... linking ... done. Loading package bytestring-0.10.0.0 ... linking ... done. 2 *Main> bruteForce [ 2,3,8,6,1] 5  2. Divide and Conquer We can modify the merge sort to compute the inversion count. The idea is to compute the inversion count of during the merge phase. Inversion count of array C resulting from merging two sorted array A and B is sum of inversion count of A, inversion count of B and their cross inversion. The inversion count of single element is 0 ( base case ). The only trick is compute the cross inversion. What will be the inversion count of following array if we merge them ? A = [ 1 , 2 , 3 ] B = [ 4 , 5 , 6 ] Clearly 0 because all element of A is less than the first element of B. What is the inversion count of C = [ 4 , 5 , 6 ] and D = [ 1 , 2 , 3 ] ? First element of D ( 1 ) is less than first element of C ( 4 ) so we have total three pairs ( 4 , 1 ) ( 5 , 1 ) ( 6 , 1). Similarly for 2 and 3 ( total 9 inversion count ). It reveals that if we have any element y in D is less than certain element x in C then all the element of A after x will also be greater than y and we will have those many cross inversion counts. More clear explanation. Wrote a quick Haskell code and got the answer. import Data.List import Data.Maybe ( fromJust ) import qualified Data.ByteString.Lazy.Char8 as BS inversionCnt :: [ Int ] -> [ Int ] -> [ Int ] -> Int -> ( Int , [ Int ] ) inversionCnt [] ys ret cnt = ( cnt , reverse ret ++ ys ) inversionCnt xs [] ret cnt = ( cnt , reverse ret ++ xs ) inversionCnt u@( x : xs ) v@( y : ys ) ret cnt | x <= y = inversionCnt xs v ( x : ret ) cnt | otherwise = inversionCnt u ys ( y : ret ) ( cnt + length u ) merge :: ( Int , [ Int ] ) -> ( Int , [ Int ] ) -> ( Int , [ Int ] ) merge ( cnt_1 , xs ) ( cnt_2 , ys ) = ( cnt_1 + cnt_2 + cnt , ret ) where ( cnt , ret ) = inversionCnt xs ys [] 0 mergeSort :: [ Int ] -> ( Int , [ Int ] ) mergeSort [ x ] = ( 0 , [ x ] ) mergeSort xs = merge ( mergeSort xs' ) ( mergeSort ys' ) where ( xs' , ys' ) = splitAt ( div ( length xs ) 2 ) xs main = BS.interact$ ( \x -> BS.pack $show x ++ "\n" ) . fst . mergeSort . map ( fst . fromJust . BS.readInt :: BS.ByteString -> Int ) . BS.lines Mukeshs-MacBook-Pro:Puzzles mukeshtiwari$ ghc  -fforce-recomp -O2  Inversion.hs
[1 of 1] Compiling Main             ( Inversion.hs, Inversion.o )
Mukeshs-MacBook-Pro:Puzzles mukeshtiwari$time ./Inversion < IntegerArray.txt 2407905288 real 0m8.002s user 0m7.881s sys 0m0.038s  I tried to compare this solution with GeeksforGeeks‘s solution written in C and it was a complete surprise to me. Mukeshs-MacBook-Pro:Puzzles mukeshtiwari$ time ./a.out <  IntegerArray.txt
Number of inversions are 2407905288

real	0m0.039s
user	0m0.036s
sys	0m0.002s


Almost 200 times faster! My Haskell intuition told me that I am doing certainly some wrong. I started analyzing the code and got the point. In inversionCnt function, I am doing extra work for computing the length of list u ( O ( n ) ) so rather than doing this extra work, I can pass the length.

import Data.List
import Data.Maybe ( fromJust )
import qualified Data.ByteString.Lazy.Char8 as BS

inversionCnt :: [ Int ] -> [ Int ] -> [ Int ] -> Int -> Int -> ( Int , [ Int ] )
inversionCnt [] ys  ret _ cnt = ( cnt , reverse ret ++ ys )
inversionCnt xs []  ret _ cnt = ( cnt , reverse ret ++ xs )
inversionCnt u@( x : xs ) v@( y : ys )  ret n cnt
| x <= y = inversionCnt xs v  (  x : ret   )  ( pred n ) cnt
| otherwise = inversionCnt u ys  ( y : ret ) n ( cnt + n )

merge ::  ( Int , [ Int ] ) -> ( Int , [ Int ] ) -> ( Int , [ Int ] )
merge ( cnt_1 , xs ) ( cnt_2 , ys ) = ( cnt_1 + cnt_2 + cnt , ret ) where
( cnt , ret ) = inversionCnt xs ys [] ( length xs )  0

mergeSort ::  [ Int ] -> ( Int , [ Int ] )
mergeSort [ x ] = ( 0 , [ x ] )
mergeSort  xs   = merge ( mergeSort xs' ) ( mergeSort ys' ) where
( xs' , ys' ) = splitAt ( div ( length xs ) 2 ) xs

main =  BS.interact $( \x -> BS.pack$ show x ++ "\n" ) . fst
.  mergeSort . map ( fst . fromJust .  BS.readInt :: BS.ByteString -> Int  )
. BS.lines

Mukeshs-MacBook-Pro:Puzzles mukeshtiwari$time ./Inversion < IntegerArray.txt 2407905288 real 0m0.539s user 0m0.428s sys 0m0.031s  Now 10 times slower and still lot of room for improvement but for now I am happy with this :). You can also try to solve INVCNT. Run this code on Ideone. July 4, 2013 ## ZeroMQ for distributed computing. This post is influenced by ØMQ – The Guide By Pieter Hintjens and translation of codes in Haskell. I suggest you to read The Guide By Pieter Hintjens and if you are interested in Haskell code then you see these codes. ### Request-Reply The client sends “Accept hello from Client. ” to the server, which replies with “I got you.”. This Haskell server opens a ØMQ socket on port 5555, reads requests on it, and replies with “I got you” to each request.The REQ-REP socket pair is in lockstep. The client issues send and then receive, in a loop (or once if that’s all it needs). Doing any other sequence (e.g., sending two messages in a row) will result in a return code of -1 from the send or receive call.  import System.ZMQ3 import Control.Monad import qualified Data.ByteString.Char8 as BS import Data.ByteString.Lazy.Internal as BL import Data.IORef import Control.Concurrent ( threadDelay ) main = do c <- context s <- socket c Rep bind s "tcp://127.0.0.1:5555" counter <- newIORef 0 forever$ do
print res
send' s [] ( BL.packChars $"I got you. " ++ show t ) modifyIORef counter ( +1 ) threadDelay 10000 close s destroy c return ()  Client import System.ZMQ3 import Control.Monad import qualified Data.ByteString.Char8 as BS import qualified Data.ByteString.Lazy.Internal as BL import Data.IORef import Control.Concurrent ( threadDelay ) main = do c <- context s <- socket c Req connect s "tcp://127.0.0.1:5555" counter <- newIORef 0 forever$ do
send' s [] ( BL.packChars $"Accept hello from Client. " ++ show t ) msg <- receive s print msg modifyIORef counter ( +1 ) threadDelay 10000 return ()  Running a server with two clients.  Mukeshs-MacBook-Pro:ZMQ mukeshtiwari$ ./ZeroMqServer
"Accept hello from Client. 0"
"Accept hello from Client. 1"
"Accept hello from Client. 2"
"Accept hello from Client. 3"
"Accept hello from Client. 4"
"Accept hello from Client. 5"
"Accept hello from Client. 6"
"Accept hello from Client. 7"
"Accept hello from Client. 8"
"Accept hello from Client. 9"
"Accept hello from Client. 10"
"Accept hello from Client. 11"
"Accept hello from Client. 12"
"Accept hello from Client. 13"
"Accept hello from Client. 14"
"Accept hello from Client. 15"

"Accept hello from Client. 0"
"Accept hello from Client. 228"
"Accept hello from Client. 1"
"Accept hello from Client. 229"
"Accept hello from Client. 2"
"Accept hello from Client. 230"
"Accept hello from Client. 3"
"Accept hello from Client. 231"
"Accept hello from Client. 4"
"Accept hello from Client. 232"

First client.
Mukeshs-MacBook-Pro:ZMQ mukeshtiwari$./ZeroMqClient "I got you. 0" "I got you. 1" "I got you. 2" "I got you. 3" "I got you. 4" "I got you. 5" "I got you. 6" "I got you. 7" "I got you. 8" "I got you. 9" "I got you. 10" "I got you. 11" "I got you. 12" Second client Mukeshs-MacBook-Pro:ZMQ mukeshtiwari$ ./ZeroMqClient
"I got you. 228"
"I got you. 230"
"I got you. 232"
"I got you. 234"
"I got you. 236"
"I got you. 238"
"I got you. 240"
"I got you. 242"
"I got you. 244"
"I got you. 246"
"I got you. 248"
"I got you. 250"
"I got you. 252"
"I got you. 254"
"I got you. 256"
"I got you. 258"
"I got you. 260"
"I got you. 262"
"I got you. 264"
"I got you. 266"


### Publish-Subscribe

Data publishing server which publishes weather data for zip codes in range 500 and 2000.

import System.ZMQ3
import qualified Data.ByteString.Char8 as BS hiding ( putStrLn )
import qualified Data.ByteString.Lazy.Internal as BL
import System.Random

main = do
c <- context
publisher <- socket c Pub
bind publisher "tcp://127.0.0.1:5556"
bind publisher "ipc://weather.ipc"
forever $do zipcode <- randomRIO ( ( 500 , 2000 ) :: ( Int , Int ) ) temp <- randomRIO ( 10 , 45 ) :: IO Int relhum <- randomRIO ( 0 , 100 ) :: IO Int putStrLn$ show zipcode ++ " " ++ show temp ++ " " ++ show relhum
send' publisher [] ( BL.packChars $show zipcode ++ " " ++ show temp ++ " " ++ show relhum ) close publisher destroy c return ()  Client who is only interested in two zip codes. import System.ZMQ3 import Control.Monad import qualified Data.ByteString.Char8 as BS import qualified Data.ByteString.Lazy.Internal as BL import Control.Concurrent ( threadDelay ) import System.Random main = do c <- context subscriber <- socket c Sub connect subscriber "tcp://127.0.0.1:5556" subscribe subscriber ( BS.pack "1000" ) subscribe subscriber ( BS.pack "1010" ) forever$ do
print update

close subscriber
destroy c



Our server is publishing lot of data but client is only interested in two zip codes.



Mukeshs-MacBook-Pro:ZMQ mukeshtiwari$./ZeroMqWeatherPubServer 1568 27 85 924 41 46 1461 15 46 1867 44 28 1013 23 100 1052 13 6 1720 45 6 1480 12 94 1852 33 4 1295 20 6 925 18 77 935 37 94 1670 11 6 1285 39 38 1613 44 99 1888 26 62 1011 21 45 993 45 45 1402 26 86 1639 13 65 1285 18 40 1960 38 18 1160 27 39 1374 16 59 665 25 22 Mukeshs-MacBook-Pro:ZMQ mukeshtiwari$ ./ZeroMqWeatherClient
"1000 34 89"
"1010 19 70"
"1010 15 86"
"1000 38 57"
"1010 12 42"
"1000 25 1"
"1000 28 78"
"1000 25 16"
"1000 28 82"
"1010 28 98"
"1000 12 77"
"1010 11 16"
"1010 44 14"
"1010 12 89"
"1000 32 8"
"1010 37 55"
"1010 17 21"
"1000 13 96"
"1000 18 51"
"1010 16 38"
"1000 18 21"
"1010 37 60"
"1010 17 25"
"1000 33 43"
"1000 34 44"
"1010 33 78"
"1010 35 63"
"1000 39 50"
"1000 45 70"
"1000 26 3"
"1010 34 12"
"1010 26 3"
"1000 32 75"
"1010 14 68"
"1000 44 75"
"1010 27 54"
"1000 21 39"
"1010 12 65"
"1010 43 29"
"1010 25 60"


### Divide and Conquer

For this problem we will calculate the number of primes less than $10^{7}$. Our ventilator will push the task to workers and they will perform the task. After finishing the job, they will send the result back to sink.

Ventilator

import System.ZMQ3
import qualified Data.ByteString.Char8 as BS
import Data.ByteString.Lazy.Internal as BL
import Data.IORef

main = do
c <- context
sender <- socket c Push
bind sender "tcp://127.0.0.1:5557"

sink <- socket c Push
connect sink "tcp://127.0.0.1:5558"

putStrLn "Press enter when workers are ready."
_ <- getChar
putStrLn "Sending the task to workers.\n"

send' sink [] ( BL.packChars . show $0 ) forM_ [ 0..9999 ]$ \ i -> do
putStrLn $"Sending the range [ " ++ show ( 1000 * i + 1 ) ++ " .. " ++ show ( 1000 * i + 1 + 999 ) ++ "] to worker for primality testing." send' sender [] ( BL.packChars . show$ 1000 * i + 1 )

close sink
close sender
destroy c



Worker for computing the prime number.

import System.ZMQ3
import qualified Data.ByteString.Char8 as BS
import qualified Data.ByteString.Lazy.Internal as BL
import Data.IORef
import Data.Bits

powM :: Integer -> Integer -> Integer ->  Integer
powM a d n = powM' a d n where
powM' a d n
| d == 0 = 1
| d == 1 = mod a n
| otherwise = mod q n  where
p = powM'  ( mod ( a^2 ) n ) ( shiftR d 1 ) n
q = if (.&.) d 1 == 1 then mod ( a * p ) n else p

calSd :: Integer -> ( Integer , Integer )
calSd n = ( s , d ) where
s = until ( \x -> testBit ( n - 1) ( fromIntegral x ) )  ( +1 ) 0
d = div ( n - 1 ) (  shiftL 1 ( fromIntegral s )  )

rabinMiller::Integer->Integer->Integer->Integer-> Bool
rabinMiller  n s d a
| n == a = True
| otherwise =  case x == 1 of
True -> True
_ ->   any ( == pred n ) . take ( fromIntegral s )
. iterate (\e -> mod ( e^2 ) n ) $x where x = powM a d n isPrime::Integer-> Bool isPrime n | n <= 1 = False | n == 2 = True | even n = False | otherwise = all ( == True ) . map ( rabinMiller n s d )$  [ 2 , 3 , 5 , 7 , 11 , 13 , 17 ] where
( s , d ) = calSd n

primeRange :: Integer -> Integer -> [ Bool ]
primeRange m n = map isPrime [ m .. n ]

main = do
c <- context

sender <- socket c Push
connect sender "tcp://127.0.0.1:5558"

forever $do num <- receive receiver let n = read . BS.unpack$ num  :: Integer
let len = length . filter ( == True ) . primeRange  n  $n + 999 putStrLn$ "Received range [ " ++  show n ++ " .. " ++ show ( n + 999 ) ++ "  and number of primes in this range is " ++ show len
send' sender [] ( BL.packChars . show $len ) close receiver close sender destroy c  Sink for collecting the results. import System.ZMQ3 import Control.Monad import qualified Data.ByteString.Char8 as BS import qualified Data.ByteString.Lazy.Internal as BL import Data.IORef import Control.Concurrent ( threadDelay ) main = do c <- context receiver <- socket c Pull bind receiver "tcp://127.0.0.1:5558" m <- receive receiver print m sum <- newIORef 0 forM_ [ 1.. 10000 ]$ \i -> do
let n = read . BS.unpack $num :: Integer putStrLn$ "Received a number " ++ show n ++ " from one of worker"
modifyIORef sum ( + n )

putStrLn $"Total number of primes in the range is " ++ show t close receiver destroy c  Running Ventilator, 3 workers and sink Mukeshs-MacBook-Pro:ZMQ mukeshtiwari$ ./TaskVent
Sending the range [ 9992001 .. 9993000] to worker for primality testing.
Sending the range [ 9993001 .. 9994000] to worker for primality testing.
Sending the range [ 9994001 .. 9995000] to worker for primality testing.
Sending the range [ 9995001 .. 9996000] to worker for primality testing.
Sending the range [ 9996001 .. 9997000] to worker for primality testing.
Sending the range [ 9997001 .. 9998000] to worker for primality testing.
Sending the range [ 9998001 .. 9999000] to worker for primality testing.
Sending the range [ 9999001 .. 10000000] to worker for primality testing.

Mukeshs-MacBook-Pro:ZMQ mukeshtiwari$./TaskWorkder Received range [ 9978001 .. 9979000 and number of primes in this range is 60 Received range [ 9981001 .. 9982000 and number of primes in this range is 49 Received range [ 9984001 .. 9985000 and number of primes in this range is 69 Received range [ 9987001 .. 9988000 and number of primes in this range is 57 Received range [ 9990001 .. 9991000 and number of primes in this range is 60 Received range [ 9993001 .. 9994000 and number of primes in this range is 71 Received range [ 9996001 .. 9997000 and number of primes in this range is 58 Received range [ 9999001 .. 10000000 and number of primes in this range is 53 Task worker - 2 Mukeshs-MacBook-Pro:ZMQ mukeshtiwari$ ./TaskWorkder
Received range [ 9982001 .. 9983000  and number of primes in this range is 69
Received range [ 9985001 .. 9986000  and number of primes in this range is 62
Received range [ 9988001 .. 9989000  and number of primes in this range is 58
Received range [ 9991001 .. 9992000  and number of primes in this range is 57
Received range [ 9994001 .. 9995000  and number of primes in this range is 64
Received range [ 9997001 .. 9998000  and number of primes in this range is 67
Received range [ 9986001 .. 9987000  and number of primes in this range is 59
Received range [ 9989001 .. 9990000  and number of primes in this range is 58
Received range [ 9992001 .. 9993000  and number of primes in this range is 58
Received range [ 9995001 .. 9996000  and number of primes in this range is 62
Received range [ 9998001 .. 9999000  and number of primes in this range is 64

Sink
Received a number 53 from one of worker
Received a number 67 from one of worker
Received a number 64 from one of worker
Received a number 52 from one of worker
Received a number 59 from one of worker
Received a number 58 from one of worker
Received a number 58 from one of worker
Received a number 62 from one of worker
Received a number 64 from one of worker
Total number of primes in the range is 664579


First start all the workers and sink and then run ventilator. See more about distributed computing in Haskell Eden and distributed-process. If you have any suggestion then please let me know. Some of the contents and images are taken from Pieter Hintjens‘s tutorial by his permission.

May 13, 2013

## Missionaries and cannibals problem

In the missionaries and cannibals problem, three missionaries and three cannibals must cross a river using a boat which can carry at most two people, under the constraint that, for both banks, if there are missionaries present on the bank, they cannot be outnumbered by cannibals (if they were, the cannibals would eat the missionaries.) The boat cannot cross the river by itself with no people on board. We can solve this problem using depth first search. Representing the states as number of missionaries, cannibals and boat location in following way.


data Side = Side { missionaries :: Int , cannibals :: Int } deriving ( Show , Eq  )
data Loc = L | R deriving ( Show , Eq  )
data State = State { left :: Side , right :: Side , bloc :: Loc } deriving ( Show , Eq  )



Target is to move every missionaries and cannibals from left to right so our initial state is initialState and final state is finalState.


initialState = State { left =  Side 3 3  , right =  Side 0 0  , bloc = L }
finalState = State  { left =  Side 0 0  , right = Side 3 3  , bloc = R }



Possible movements
1. move ( 2 , 0 )
2. move ( 1 , 0 )
3. move ( 1 , 1 )
4. move ( 0 , 1 )
5. move ( 0 , 2 )
where move ( M , C ) is movement of M missionaries and C cannibals from one side to other side depending on boat location. If boat location is left then move ( M , C ) is moving M missionaries and C cannibals from left to right and vice versa. Using depth first search, we can explore every possible movement

path :: [ State ] -> [ State ] -> [ State ]
path []   vis = reverse vis
path  ( x : xs )  vis
| x  == finalState  = reverse ( x : vis )
| elem x vis = path xs vis
| otherwise = path xs'  vis'  where
vis'  = x : vis
xs' =   ( filter  isLegal  ( move x ) ) ++ xs


Encoding the moves and testing the legality


move :: State -> [ State ]
move ( State ( Side ml cl ) ( Side mr cr ) L ) =
[ State ( Side ( ml - 2 ) cl ) ( Side ( mr + 2 ) cr ) R
, State ( Side ( ml - 1 ) cl ) ( Side ( mr + 1 ) cr ) R
, State ( Side ( ml - 1 ) ( cl - 1 ) ) ( Side ( mr + 1 ) ( cr + 1 ) ) R
, State ( Side ml ( cl - 2 ) ) ( Side mr ( cr + 2 ) ) R
, State ( Side ml ( cl - 1 ) ) ( Side mr ( cr + 1 ) ) R
]
move  ( State ( Side ml cl ) ( Side mr cr ) R ) =
[ State ( Side ( ml + 2 ) cl ) ( Side ( mr - 2 ) cr ) L
, State ( Side ( ml + 1 ) cl ) ( Side ( mr - 1 ) cr ) L
, State ( Side ( ml + 1 ) ( cl + 1 ) ) ( Side ( mr - 1 ) ( cr - 1 ) ) L
, State ( Side ml ( cl + 2 ) ) ( Side mr ( cr - 2 ) ) L
, State ( Side ml ( cl + 1 ) ) ( Side mr ( cr - 1 ) ) L
]

isLegal :: State -> Bool
isLegal ( State ( Side ml cl ) ( Side mr cr ) y )
| ml == 0 = mr >= cr  && cr >= 0
| mr == 0 = ml >= cl  && cl >= 0
| otherwise = ml >= cl && mr >= cr && cl >= 0 && cr >= 0



Complete source code ( In case if you are lazy )

import Data.List

data Side = Side { missionaries :: Int , cannibals :: Int } deriving ( Show , Eq  )
data Loc = L | R deriving ( Show , Eq  )
data State = State { left :: Side , right :: Side , bloc :: Loc } deriving ( Show , Eq  )

initialState = State { left =  Side 3 3  , right =  Side 0 0  , bloc = L }
finalState = State  { left =  Side 0 0  , right = Side 3 3  , bloc = R }

path :: [ State ] -> [ State ] -> [ State ]
path []   vis = reverse vis
path  ( x : xs )  vis
| x  == finalState  = reverse ( x : vis )
| elem x vis = path xs vis
| otherwise = path xs'  vis'  where
vis'  = x : vis
xs' =   ( filter  isLegal  ( move x ) ) ++ xs

move :: State -> [ State ]
move ( State ( Side ml cl ) ( Side mr cr ) L ) =
[ State ( Side ( ml - 2 ) cl ) ( Side ( mr + 2 ) cr ) R
, State ( Side ( ml - 1 ) cl ) ( Side ( mr + 1 ) cr ) R
, State ( Side ( ml - 1 ) ( cl - 1 ) ) ( Side ( mr + 1 ) ( cr + 1 ) ) R
, State ( Side ml ( cl - 2 ) ) ( Side mr ( cr + 2 ) ) R
, State ( Side ml ( cl - 1 ) ) ( Side mr ( cr + 1 ) ) R
]
move  ( State ( Side ml cl ) ( Side mr cr ) R ) =
[ State ( Side ( ml + 2 ) cl ) ( Side ( mr - 2 ) cr ) L
, State ( Side ( ml + 1 ) cl ) ( Side ( mr - 1 ) cr ) L
, State ( Side ( ml + 1 ) ( cl + 1 ) ) ( Side ( mr - 1 ) ( cr - 1 ) ) L
, State ( Side ml ( cl + 2 ) ) ( Side mr ( cr - 2 ) ) L
, State ( Side ml ( cl + 1 ) ) ( Side mr ( cr - 1 ) ) L
]

isLegal :: State -> Bool
isLegal ( State ( Side ml cl ) ( Side mr cr ) y )
| ml == 0 = mr >= cr  && cr >= 0
| mr == 0 = ml >= cl  && cl >= 0
| otherwise = ml >= cl && mr >= cr && cl >= 0 && cr >= 0



 Main> path [ initialState ] [] [State {left = Side {missionaries = 3, cannibals = 3}, right = Side {missionaries = 0, cannibals = 0}, bloc = L},State {left = Side {missionaries = 2, cannibals = 2}, right = Side {missionaries = 1, cannibals = 1}, bloc = R},State {left = Side {missionaries = 3, cannibals = 2}, right = Side {missionaries = 0, cannibals = 1}, bloc = L},State {left = Side {missionaries = 3, cannibals = 0}, right = Side {missionaries = 0, cannibals = 3}, bloc = R},State {left = Side {missionaries = 3, cannibals = 1}, right = Side {missionaries = 0, cannibals = 2}, bloc = L},State {left = Side {missionaries = 1, cannibals = 1}, right = Side {missionaries = 2, cannibals = 2}, bloc = R},State {left = Side {missionaries = 2, cannibals = 2}, right = Side {missionaries = 1, cannibals = 1}, bloc = L},State {left = Side {missionaries = 0, cannibals = 2}, right = Side {missionaries = 3, cannibals = 1}, bloc = R},State {left = Side {missionaries = 0, cannibals = 3}, right = Side {missionaries = 3, cannibals = 0}, bloc = L},State {left = Side {missionaries = 0, cannibals = 1}, right = Side {missionaries = 3, cannibals = 2}, bloc = R},State {left = Side {missionaries = 1, cannibals = 1}, right = Side {missionaries = 2, cannibals = 2}, bloc = L},State {left = Side {missionaries = 0, cannibals = 0}, right = Side {missionaries = 3, cannibals = 3}, bloc = R}] *Main> map bloc . path [ initialState ] $[] [L,R,L,R,L,R,L,R,L,R,L,R]  If you have any suggestion or comment then please let me know. April 19, 2013 ## SPOJ Pell (Mid pelling) Pell (Mid pelling) is related to Pell’s equation. It is similar to Project Euler 66 and SPOJ EQU2. I just wrote the quick solution from mathworld but I found Chakravala method very interesting. Accepted Haskell code. import qualified Data.ByteString.Char8 as BS import Data.List import Data.Maybe ( fromJust ) continuedFraction :: Integer -> [ Integer ] continuedFraction n = map ( \ ( a , _ , _ ) -> a ) . iterate fun$ ( d , 0 , 1 )
where
d = truncate . sqrt . fromIntegral $n fun ( a0 , p0 , q0 ) = ( a1 , p1 , q1 ) where p1 = a0 * q0 - p0 q1 = div ( n - p1 ^ 2 ) q0 a1 = div ( d + p1 ) q1 pellSolver :: Integer -> BS.ByteString pellSolver n | perfectSqr n = BS.pack. show$ ( -1 )
| otherwise =  ( BS.pack . show $p ) BS.append ( BS.pack " " ) BS.append ( BS.pack.show$ q ) where
d = truncate . sqrt . fromIntegral $n lst = takeWhile ( /= 2 * d ) . continuedFraction$ n
len = length lst
r@( x : y : xs ) = take ( if even len then len else 2 * len ) .
continuedFraction $n ( p0 , p1 , q0 , q1 ) = ( x , x * y + 1 , 1 , y ) ( p , _ ) = foldl' compute ( p1 , p0 )$ xs
( q , _ ) = foldl' compute ( q1 , q0 ) $xs compute :: ( Integer , Integer ) -> Integer -> ( Integer , Integer ) compute ( p1 , p0 ) a = ( a * p1 + p0 , p1 ) perfectSqr :: Integer -> Bool perfectSqr n = d * d == n where d = truncate . sqrt . fromIntegral$ n

main = BS.interact $BS.unlines . map ( pellSolver . readI ) . tail . BS.lines  March 23, 2013 ## Parsing Email ID Now a days I am exploring one of the most awesome feature of Haskell, parsing. There are lot of parsing libraries but Parsec is the popular one. This parsing code is written for SPOJ EMAIL ID problem ( Unfortunately it’s getting time limit exceed. I have seen couple of python solution accepted so I am hopeful that there must be another algorithm probably using regular expressions to solve the problem ) but you can build more sophisticated email-id parser by adding more functionality. Tony Morris excellent parsing tutorial is must read. import Data.List import qualified Text.Parsec.ByteString as PB import Text.Parsec.Prim import Text.Parsec.Char import Text.Parsec.Combinator import qualified Data.ByteString.Char8 as BS import Control.Applicative hiding ( ( <|> ) , many ) validChars :: PB.Parser Char validChars = alphaNum <|> oneOf "._" dontCare :: PB.Parser Char dontCare = oneOf "~!@#$%^&*()<>?,."

{--
_ <- many dontCare
fi <- alphaNum
se <- validChars
th <- validChars
fo <- validChars
ft <- validChars
let addr = fi : se : th : fo : ft : restAddr
char '@'
dom <- many1 alphaNum
rest <- try ( string ".com" <|> string ".org"
<|>  string ".edu" ) <|> try ( string ".co.in" )
_ <- many dontCare
return $addr ++ ( '@': dom ++ rest ) --} emailAddress :: PB.Parser String emailAddress = conCatfun <$> ( many dontCare *> alphaNum ) <*> validChars <*>
validChars <*> validChars <*> validChars <*> many alphaNum  <*>
( char '@' *> many1 alphaNum ) <*> ( try ( string ".com" <|>
string ".org" <|>  string ".edu" ) <|> try ( string ".co.in" )
<* many dontCare ) where
conCatfun a b c d e f dom rest =
( a : b : c : d : e : f ) ++ ( '@' : dom ) ++ rest

collectEmail :: BS.ByteString -> String
collectEmail email = case  parse emailAddress "" email of
Left err ->  ""

process :: ( Int , [ String ] ) -> BS.ByteString
process ( k , xs ) = ( BS.pack "Case " ) BS.append ( BS.pack . show $k ) BS.append ( BS.pack ": " ) BS.append ( BS.pack . show . length$ xs )
BS.append ( BS.pack "\n" ) BS.append ( BS.pack
(  unlines . filter ( not . null ) $xs ) ) main = BS.interact$ BS.concat .  map process . zip [ 1.. ] . map ( map collectEmail .
BS.words ) . tail . BS.lines


Mukeshs-MacBook-Pro:Haskell mukeshtiwari$cat t.txt 2 svm11@gmail.com svm11@gmail.com svm12@gmail.co.in ~!@#$%^&*()<>?svm12@gmail.co.in~!@#$%^&*() Mukeshs-MacBook-Pro:Haskell mukeshtiwari$ ./Spoj_11105 < t.txt
Case 1: 1
svm11@gmail.com
Case 2: 2
svm11@gmail.com
svm12@gmail.co.in
svm12@gmail.co.in


#### Update

I tried again to solve this problem using regular expression. Following the tutorial, I wrote this code but I got compiler error because of old version of ghc on SPOJ ( GHC-6.10.4 ). It’s working fine on my system but I still have to test if it is correct and fast enough to get accepted.

import Data.List
import Text.Regex.Posix
import qualified Data.ByteString.Char8 as BS

pat :: BS.ByteString
pat = BS.pack "[^~!@#$%^&*()<>?,.]*[a-zA-Z0-9][a-zA-Z0-9._][a-zA-Z0-9._][a-zA-Z0-9._][a-zA-Z0-9._][a-zA-Z0-9._]*@[a-zA-Z0-9]+.(com|edu|org|co.in)[^~!@#$%^&*()<>?,.a-zA-Z0-9]*"

collectEmail :: BS.ByteString -> BS.ByteString
collectEmail email = ( =~ ) email pat

process :: ( Int , [ BS.ByteString ] ) -> BS.ByteString
process ( k , xs ) = ( BS.pack "Case " ) BS.append ( BS.pack . show $k ) BS.append ( BS.pack ": " ) BS.append ( BS.pack . show . length$ xs )
BS.append ( BS.pack "\n" ) BS.append ( BS.unlines xs )

main = BS.interact $BS.concat . map process . zip [ 1 .. ] . map ( filter ( not . BS.null ) . map collectEmail . BS.words ) . tail . BS.lines  Mukeshs-MacBook-Pro:SPOJ mukeshtiwari$ cat t.txt
2
svm11@gmail.com
svm11@gmail.com svm12@gmail.co.in %&^%&%&%&%&^%&^%&^%&^mukeshtiwari.iiitm@gmail.com%&%^&^%&%&%&%&^%&%&^%&^%&^% %$%$#%#%#%#%$#%&%&%&tiwa@gmail.com Mukeshs-MacBook-Pro:SPOJ mukeshtiwari$ ./Spoj_11105 < t.txt
Case 1: 1
svm11@gmail.com
Case 2: 3
svm11@gmail.com
svm12@gmail.co.in
mukeshtiwari.iiitm@gmail.com


March 8, 2013

## Generating primes in parallel

We can use Miller-Rabin primality testing to test a number is prime or not with certain probability. We can use this method on chunk of data to get the primes in parallel. I wrote this code to see how it goes and it’s quite interesting. My other post about data parallelism and Don Stewart answer regarding different options to improve parallelism. The crux of this code is

primeRange :: Integer -> Integer -> [ Bool ]
primeRange m n = ( map isPrime [ m .. n ] ) using parListChunk 10000 rdeepseq


import Data.Bits
import Control.Parallel.Strategies

powM :: Integer -> Integer -> Integer ->  Integer
powM a d n = powM' a d n where
powM' a d n
| d == 0 = 1
| d == 1 = mod a n
| otherwise = mod q n  where
p = powM'  ( mod ( a^2 ) n ) ( shiftR d 1 ) n
q = if (.&.) d 1 == 1 then mod ( a * p ) n else p

calSd :: Integer -> ( Integer , Integer )
calSd n = ( s , d ) where
s = until ( \x -> testBit ( n - 1) ( fromIntegral x ) )  ( +1 ) 0
d = div ( n - 1 ) (  shiftL 1 ( fromIntegral s )  )

rabinMiller::Integer->Integer->Integer->Integer-> Bool
rabinMiller  n s d a
| n == a = True
| otherwise =  case x == 1 of
True -> True
_ ->   any ( == pred n ) . take ( fromIntegral s )
. iterate (\e -> mod ( e^2 ) n ) $x where x = powM a d n isPrime::Integer-> Bool isPrime n | n <= 1 = False | n == 2 = True | even n = False | otherwise = all ( == True ) . map ( rabinMiller n s d )$  [ 2 , 3 , 5 , 7 , 11 , 13 , 17 ] where
( s , d ) = calSd n

primeRange :: Integer -> Integer -> [ Bool ]
primeRange m n = ( map isPrime [ m .. n ] ) using parListChunk 10000 rdeepseq

main = do
let t = filter ( == True ) . primeRange 2 $10 ^ 6 print . length$ t

Mukeshs-MacBook-Pro:Haskell mukeshtiwari$ghc -O2 -rtsopts --make RabinMiller.hs -threaded -fforce-recomp [1 of 1] Compiling Main ( RabinMiller.hs, RabinMiller.o ) Linking RabinMiller ... Mukeshs-MacBook-Pro:Haskell mukeshtiwari$ time ./RabinMiller +RTS -N1
78498

real	0m4.841s
user	0m4.695s
sys	0m0.075s
Mukeshs-MacBook-Pro:Haskell mukeshtiwari$time ./RabinMiller +RTS -N2 78498 real 0m3.115s user 0m5.969s sys 0m0.215s Mukeshs-MacBook-Pro:Haskell mukeshtiwari$ time ./RabinMiller +RTS -N3
78498

real	0m2.687s
user	0m7.556s
sys	0m0.415s


I am no expert in parallelism or Haskell so if you have any comment or suggestion then please let me know.

February 5, 2013

## Parsing with Applicative Functors

While reading Write Yourself a Scheme in 48 Hours, I thought of converting the monadic codes of parsing into applicative style. Learn you a Haskell for Great Good has excellent explanation about functors and applicative functors. Real World Haskell has excellent chapter on using applicative functors for parsing. Below is my attempt to convert the monadic codes of parsing into applicative style for Scheme parsing. On parsing note, see the excellent tutorial of Tony Morris and Albert Y. C. Lie.

import Control.Applicative hiding ( many , ( <|> )  )
import Text.ParserCombinators.Parsec hiding ( spaces )

symbol :: Parser Char
symbol =  oneOf "!$%&|+-*/:<=?>@^_~#" spaces :: Parser () spaces = skipMany space parseString :: Parser LispVal parseString = String <$> ( char '"' *>  x <*  char '"' ) where
x = many ( noneOf "\"" )
--This is not completely applicative style because I have to take the result for testing conditions.
parseAtom :: Parser LispVal
parseAtom = do
atom <-  ( : ) <$> ( letter <|> symbol ) <*> many ( letter <|> digit <|> symbol ) return$ case atom of
"#t" -> Bool True
"#f" -> Bool False
_ -> Atom atom

parseNumber :: Parser LispVal
parseNumber = Number . read <$> many1 digit parseList :: Parser LispVal parseList = List <$> sepBy parseExpr space

parseDottedList :: Parser LispVal
parseDottedList = DottedList <$> ( endBy parseExpr space ) <*> ( char '.' *> spaces *> parseExpr ) parseQuoted :: Parser LispVal parseQuoted = f <$>   ( char '\'' *> parseExpr ) where
f x =   List [ Atom "quote" , x ]

parseExpr :: Parser LispVal
parseExpr =  parseAtom
<|> parseString
<|> parseNumber
<|> parseQuoted
<|> ( char '(' *> spaces *> ( ( try parseList ) <|> parseDottedList ) <* spaces  <* char ')' )



Complete code on github ( Apart from applicative parsing every other code is from Scheme parsing tutorial ). If you have any comments or suggestion then please let me know.

January 18, 2013

## Newton–Raphson method to solve Ax+Bsin(x)=C

I was reading Newton–Raphson method and realized that you can encode this whole algorithm using until function in Haskell.

Type signature of until
ghci>:t until
until :: (a -> Bool) -> (a -> a) -> a -> a


To encode Newton-Raphson algorithm in until

until ( you testing condition ) ( function )  ( initial value of x )


TRIGALGE is related to Newton-Raphson . $\sin x$ will be in between [ -1.0 , 1.0 ] so $Ax - B \leq C \leq Ax + B$. My initial guess was $\frac{C-B}{A}$. See this stackoverflow discussion for initial approximation. Accepted Haskell code.

import Data.ByteString.Lazy.Char8 as BS hiding  ( map , tail , filter  , null )
import Data.Maybe ( fromJust )
import Text.Printf ( printf )

diffEQ :: Double -> Double -> Double -> Double
diffEQ a b x = a + b * cos x

valEQ :: Double -> Double -> Double -> Double -> Double
valEQ a b c x = a * x   +  b * sin x - c

evalFun :: [ Int ]  -> Double
evalFun [ a' , b' , c' ]  = ret where
( a , b , c  ) = ( fromIntegral a' , fromIntegral b' , fromIntegral c' )
ret = fst . until ( \ ( _ , cnt ) -> cnt  >=  500 )  ( \( x , cnt ) -> ( x - (  valEQ a b c x  /  diffEQ a b x ) , succ cnt ) ) $( ( c - b ) / a , 0 ) readD :: BS.ByteString -> Int readD = fst . fromJust . BS.readInt main = BS.interact$ BS.unlines . map (  BS.pack . ( printf "%.6f" :: Double -> String ) . evalFun .  map readD ) . filter ( not.null ) . map BS.words . tail . BS.lines



Changing the initial approximation of $\frac{C}{A}$ and 50 iteration improves the time by 0.60 seconds ( from 0.80 to 0.22 ).

January 4, 2013

#include<stdio.h>

//comment all the functions related to lock to see the race condition.

void* coun_incr(void* arg )
{
int i;
//going to lock this critical code

int* count = ( int* ) arg;
for( i = 0 ; i < 1000000 ; i++ )
*count += 10;

//release the lock because we have finished the whole iteration
return NULL;
}

int main()
{
//initialize the lock
pthread_mutex_init ( &lock , NULL );
int count = 0 ;
pthread_create( &tid[0] , NULL , coun_incr , &count ) ;
pthread_create( &tid[1] , NULL , coun_incr , &count ) ;
pthread_create( &tid[2] , NULL , coun_incr , &count ) ;
//wait for thread to finish otherwise main will finish and terminate all the three threads.
pthread_join ( tid[0] , NULL );
pthread_join ( tid[1] , NULL );
pthread_join ( tid[2] , NULL );
printf("%d\n", count );

//destroy the lock
}

Some results in inconsistent state.
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$gcc -lpthread Thread_1.c Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./a.out
13482740
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./a.out 21974930 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./a.out
24638680
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./a.out 17794120 Using locks on critical code. Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ gcc -lpthread Thread_1.c
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./a.out 30000000 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./a.out
30000000
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./a.out 30000000 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./a.out
30000000


The almost same code in Haskell using mvar.

import Control.Concurrent
import GHC.IORef

incr_count :: MVar () -> MVar Int -> IO ()
incr_count m n = ( forM_ [ 1..10000 ] $\_ -> modifyMVar_ n ( return . ( + 10 ) ) ) >> putMVar m () main :: IO() main = do count <- newMVar 0 list <- forM [1..10]$ \_ -> newEmptyMVar
forM_ list $\var -> forkIO . incr_count var$ count
forM_ list $\var -> takeMVar var val <- takeMVar count print val Ok, modules loaded: Main. ghci>main 1000000 ghci>main 1000000 ghci>main 1000000  We can create race condition using IORef. We are creating a mutable variable count of type IORef and 10 parallel threads are modifying it. import Control.Concurrent import Control.Monad import Data.IORef incr_count :: MVar () -> IORef Int -> IO () incr_count m n = ( forM_ [ 1 .. 10000 ]$ \_ -> modifyIORef' n ( + 10   ))  >> putMVar m ()

main :: IO ()
main = do
count <- newIORef 0
list <- forM [1..10] $\_ -> newEmptyMVar forM_ list$ \var -> forkIO . incr_count var $count forM_ list$ \var ->  takeMVar var
print val

Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./Thread_1 1000000 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1
1000000
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./Thread_1 +RTS -N2 619360 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1  +RTS -N2
614090
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./Thread_1 +RTS -N2 521930 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1
1000000
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./Thread_1 1000000 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1
1000000
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./Thread_1 1000000 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1
1000000
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./Thread_1 1000000 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1
500010
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./Thread_1 1000000 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1  +RTS -N2
597710


import Control.Concurrent
import Control.Concurrent.STM
import Data.IORef

incr_count :: TVar Int -> STM ()
incr_count m = forM_ [ 1 .. 10000 ] $\_ -> modifyTVar' m ( + 10 ) main = do count <- newTVarIO 0 list <- forM [1..10]$ \_ -> newEmptyMVar
forM_ list $\var -> forkIO ( ( atomically . incr_count$ count ) >> putMVar var () )
forM_ list $\var -> takeMVar var --threadDelay 1000000 val <- readTVarIO count print val Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1
1000000
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./Thread_1 +RTS -N2 1000000 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1 +RTS -N2
1000000
Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$./Thread_1 +RTS -N4 1000000 Mukeshs-MacBook-Pro:GitRepo mukeshtiwari$ ./Thread_1 +RTS -N4
1000000


I am no expert in C or Haskell and still learning. If you have feedback or comment then please let me know.

December 18, 2012