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Euler problems/181 to 190

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(another solution to problem 185)
(solution to problem 187)
Line 248: Line 248:
   
 
problem_185 = head . solve $ initial
 
problem_185 = head . solve $ initial
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=187 Problem 187] ==
  +
Semiprimes
  +
  +
Solution:
  +
  +
The solution to this problem isn't terribly difficult, once you
  +
know that the numbers the problem is referring to are called
  +
semiprimes. In fact there is an excellent write-up at:
  +
http://mathworld.wolfram.com/Semiprime.html which provides an
  +
explicit formula for the number of semiprimes less than a given
  +
number.
  +
  +
The problem with this formula is the use of pi(n) where pi is
  +
the number of primes less than n. For Mathematica users this is
  +
no problem since the function is built in, but for us poor
  +
Haskeller's we have to build it ourselves. This is what took
  +
the most time for me, to come up with a pi(n) function that can
  +
compute pi(50000000) in less than the lifetime of the universe.
  +
Thus I embarked on a long and circuitous journey that eventually
  +
led me to the PI module below, which does little more than read
  +
a 26MB file of prime numbers into memory and computes a map from
  +
which we can calculate pi(n). I am sure there must be a better
  +
way of doing this, and I look forward to this entry being
  +
amended (or replaced) with a more reasonable solution.
  +
HenryLaxen Mar 24, 2008
  +
  +
  +
<haskell>
  +
  +
module PI (prime,primes,pI) where
  +
import Control.Monad.ST
  +
import System.IO.Unsafe
  +
import qualified Data.Map as M
  +
import Data.ByteString.Char8 (readFile,lines,readInt)
  +
import Data.Maybe
  +
import Data.Array.ST
  +
import Data.Array.IArray
  +
  +
r = runSTUArray
  +
(do a <- newArray (0,3001134) 0 :: ST s (STUArray s Int Int)
  +
writeArray a 0 1
  +
mapM_ (\(x,y) -> writeArray a x y) (zip [0..] primes)
  +
return a)
  +
  +
{-# NOINLINE s #-}
  +
s = M.fromDistinctAscList $ map (\(x,y) -> (y,x)) (assocs r)
  +
  +
prime n = r!n
  +
  +
pI :: Int -> Int
  +
pI n =
  +
case M.splitLookup n s of
  +
(_,Just x,_) -> (x+1)
  +
(_,_,b) -> snd (head (M.assocs b))
  +
  +
{-# NOINLINE primes #-}
  +
primes :: [Int]
  +
primes = unsafePerformIO $ do
  +
l <- Data.ByteString.Char8.readFile "primes_to_5M.txt"
  +
let p = Data.ByteString.Char8.lines l
  +
r = map (fst . fromJust . readInt) p::[Int]
  +
return r
  +
  +
  +
import PI
  +
import Data.List
  +
  +
s n k = ( n `div` (prime (k-1))) - k + 1
  +
  +
semiPrimeCount n =
  +
let last = pI $ floor $ sqrt (fromIntegral n)
  +
s k = pI ( n `div` (prime (k-1))) - k + 1
  +
pI2 = foldl' (\x y -> s y + x) 0 [1..last]
  +
in pI2
  +
  +
main = print (semiPrimeCount 100000000)
  +
 
</haskell>
 
</haskell>

Revision as of 19:53, 24 March 2008

Contents

1 Problem 181

Investigating in how many ways objects of two different colours can be grouped.

2 Problem 182

RSA encryption.

Solution:

fun a1 b1 = sum [ e | e <- [2..a*b-1],
                      gcd e (a*b) == 1,
                      gcd (e-1) a == 2,
                      gcd (e-1) b == 2 ]
  where a = a1-1
        b = b1-1
 
problem_182 = fun 1009 3643

3 Problem 183

Maximum product of parts.

Solution:

-- Does the decimal expansion of p/q terminate?
terminating p q = 1 == reduce [2,5] (q `div` gcd p q)
        where reduce   []   n = n
              reduce (x:xs) n | n `mod` x == 0 = reduce (x:xs) (n `div` x)
                              | otherwise      = reduce xs n
 
-- The expression (round $ fromIntegral n / e) computes the integer k
-- for which (n/k)^k is at a maximum. Also note that, given a rational number
-- r and a natural number k, the decimal expansion of r^k terminates if
-- and only if the decimal expansion of r does.
answer = sum [if terminating n (round $ fromIntegral n / e) then -n else n
                | n <- [5 .. 10^4]]
        where e = exp 1
 
main = print answer

4 Problem 184

Triangles containing the origin.

Solution:

problem_184 = undefined

5 Problem 185

Number Mind

Solution:

This approach does NOT solve the problem in under a minute, unless of course you are extremely lucky. The best time I've seen so far has been about 76 seconds. Before I came up with this code, I tried to search for the solution by generating a list of all possible solutions based on the information given in the guesses. This was feasible with the 5 digit problem, but completely intractable with the 16 digit problem. The approach here, simple yet effective, is to make a random guess, and then vary each digit in the guess from [0..9], generating a score of how well the guess matched the given numbermind clues. You then improve the guess by selecting those digits that had a unique low score. It turns out this approach converges rather quickly, but can often be stuck in cycles, so we test for this and try a differenct random first guess if a cycle is detected. Once you run the program, you might have time for a cup of coffee, or maybe even a dinner. HenryLaxen 2008-03-12

import Data.List
import Control.Monad
import Data.Char
import System.Random
 
type Mind = [([Char],Int)]
 
values :: [Char]
values = "0123456789"
 
 
score :: [Char] -> [Char] -> Int
score guess answer = foldr (\(a,b) y -> if a == b then y+1 else y) 0 
      (zip guess answer)
 
scores :: Mind -> [Char] -> [Int]
scores m g = map (\x -> abs ((snd x) - score (fst x) g)) m
 
scoreMind :: Mind -> [Char] -> Int
scoreMind m g = sum $ scores m g
 
 
ex1 :: Mind
ex1 = 
  [("90342",2),
   ("39458",2),
   ("51545",2),
   ("34109",1),
   ("12531",1),
   ("70794",0)]
 
ex2 :: Mind   
ex2 = 
  [
   ("5616185650518293",2),
   ("3847439647293047",1),
   ("5855462940810587",3),
   ("9742855507068353",3),
   ("4296849643607543",3),
   ("3174248439465858",1),
   ("4513559094146117",2),
   ("7890971548908067",3),
   ("8157356344118483",1),
   ("2615250744386899",2),
   ("8690095851526254",3),
   ("6375711915077050",1),
   ("6913859173121360",1),
   ("6442889055042768",2),
   ("2321386104303845",0),
   ("2326509471271448",2),
   ("5251583379644322",2),
   ("1748270476758276",3),
   ("4895722652190306",1),
   ("3041631117224635",3),
   ("1841236454324589",3),
   ("2659862637316867",2)]
 
 
guesses :: [Char] -> Int -> [[Char]]
guesses str pos = [ left ++ n:(tail right) | n<-values]
  where (left,right) = splitAt pos str
 
bestGuess :: Mind -> [[Char]] -> [Int]
bestGuess mind guesses =
  let scores = map (scoreMind mind) guesses
      bestScore = minimum scores
      bestGuesses = findIndices (==bestScore) scores
  in bestGuesses
 
iterateGuesses :: Mind -> [Char] -> [Char]
iterateGuesses mind value = 
   let allguesses = map (guesses value) [0..(length value)-1]
       mins = map (bestGuess mind) allguesses
   in nextguess value mins
 
nextguess :: [Char] -> [[Int]] -> [Char]
nextguess prev mins = 
  let choose x = if length (snd x) == 1 then intToDigit ((snd x)!!0) else fst x
      both = zip prev mins
  in  foldr (\x y -> (choose x) : y) "" both
 
 
iterateMind :: Mind -> [Char] -> [([Char], Int)]
iterateMind mind n =
  let a = drop 2 $ inits $ iterate (iterateGuesses mind) n
      b = last $ takeWhile (\x -> (last x) `notElem` (init x)) a
      c = map (scoreMind mind) b
  in zip b c      
 
 
randomStart :: (Num a, Enum a) => a -> IO [Char]
randomStart n = mapM (\_ -> getStdRandom (randomR ('0','9'))) [1..n]
 
main :: IO ()
main = do
  let ex = ex1
  x <- randomStart (length (fst (head ex)))
  let y = iterateMind ex x
  let done = (snd (last  y) == 0)
  when done (putStrLn $ (fst.last) y)
  unless done main

Here's another solution, and this one squeaks by in just under a minute on my machine. The basic idea is to just do a back-tracking search, but with some semi-smart pruning and guess ordering. The code is in pretty much the order I wrote it, so most prefixes of this code should also compile. This also means you should be able to figure out what each function does one at a time.

import Control.Monad
import Data.List
import qualified Data.Set as S
 
ensure p x = guard (p x) >> return x
selectDistinct 0 _  = [[]]
selectDistinct n [] = []
selectDistinct n (x:xs) = map (x:) (selectDistinct (n - 1) xs) ++ selectDistinct n xs
 
data Environment a = Environment { guesses      :: [(Int, [a])]
                                 , restrictions :: [S.Set a]
                                 , assignmentsLeft :: Int
                                 } deriving (Eq, Ord, Show)
 
reorder e = e { guesses = sort . guesses $ e }
domain  = S.fromList "0123456789"
initial = Environment gs (replicate a S.empty) a where
    a   = length . snd . head $ gs
    gs  = [(2, "5616185650518293"), (1, "3847439647293047"), (3, "5855462940810587"), (3, "9742855507068353"), (3, "4296849643607543"), (1, "3174248439465858"), (2, "4513559094146117"), (3, "7890971548908067"), (1, "8157356344118483"), (2, "2615250744386899"), (3, "8690095851526254"), (1, "6375711915077050"), (1, "6913859173121360"), (2, "6442889055042768"), (0, "2321386104303845"), (2, "2326509471271448"), (2, "5251583379644322"), (3, "1748270476758276"), (1, "4895722652190306"), (3, "3041631117224635"), (3, "1841236454324589"), (2, "2659862637316867")]
 
acceptableCounts e = small >= 0 && big <= assignmentsLeft e where
    ns    = (0:) . map fst . guesses $ e
    small = minimum ns
    big   = maximum ns
 
positions s = map fst . filter (not . snd) . zip [0..] . zipWith S.member s
acceptableRestriction  r (n, s) = length (positions s r) >= n
acceptableRestrictions e = all (acceptableRestriction (restrictions e)) (guesses e)
 
firstGuess = head . guesses
sanityCheck e = acceptableRestrictions e && acceptableCounts e
 
solve e@(Environment _  _ 0) = [[]]
solve e@(Environment [] r _) = sequence $ map (S.toList . (domain S.\\)) r
solve e' = do
    is   <- m
    newE <- f is
    rest <- solve newE
    return $ interleaveAscIndices is (l is) rest
        where
    f = ensure sanityCheck . update e
    m = selectDistinct n (positions g (restrictions e)) 
    e = reorder e'
    l = fst . flip splitAscIndices g
    (n, g) = firstGuess e
 
splitAscIndices = indices 0 where
    indices _ [] xs = ([], xs)
    indices n (i:is) (x:xs)
        | i == n = let (b, e) = indices (n + 1) is     xs in (x:b, e)
        | True   = let (b, e) = indices (n + 1) (i:is) xs in (b, x:e)
 
interleaveAscIndices = indices 0 where
    indices _ [] [] ys = ys
    indices n (i:is) (x:xs) ys
        | i == n = x : indices (n + 1) is xs ys
        | True   = head ys : indices (n + 1) (i:is) (x:xs) (tail ys)
 
update (Environment ((_, a):gs) r l) is = Environment newGs restriction (l - length is) where
    (assignment, newRestriction)    = splitAscIndices is a
    (_, oldRestriction)             = splitAscIndices is r
    restriction                     = zipWith S.insert newRestriction oldRestriction
    newGs                           = map updateEntry gs
    updateEntry (n', a')            = (newN, newA) where
        (dropped, newA) = splitAscIndices is a'
        newN            = n' - length (filter id $ zipWith (==) assignment dropped)
 
problem_185 = head . solve $ initial

6 Problem 187

Semiprimes

Solution:

The solution to this problem isn't terribly difficult, once you know that the numbers the problem is referring to are called semiprimes. In fact there is an excellent write-up at: http://mathworld.wolfram.com/Semiprime.html which provides an explicit formula for the number of semiprimes less than a given number.

The problem with this formula is the use of pi(n) where pi is the number of primes less than n. For Mathematica users this is no problem since the function is built in, but for us poor Haskeller's we have to build it ourselves. This is what took the most time for me, to come up with a pi(n) function that can compute pi(50000000) in less than the lifetime of the universe. Thus I embarked on a long and circuitous journey that eventually led me to the PI module below, which does little more than read a 26MB file of prime numbers into memory and computes a map from which we can calculate pi(n). I am sure there must be a better way of doing this, and I look forward to this entry being amended (or replaced) with a more reasonable solution. HenryLaxen Mar 24, 2008


module PI (prime,primes,pI) where 
import Control.Monad.ST
import System.IO.Unsafe
import qualified Data.Map as M
import Data.ByteString.Char8 (readFile,lines,readInt) 
import Data.Maybe
import Data.Array.ST
import Data.Array.IArray
 
r = runSTUArray 
            (do a <- newArray (0,3001134) 0 :: ST s (STUArray s Int Int)
                writeArray a 0 1
                mapM_ (\(x,y) -> writeArray a x y) (zip [0..] primes)
                return a) 
 
{-# NOINLINE s #-}
s = M.fromDistinctAscList $ map (\(x,y) -> (y,x)) (assocs r)
 
prime n = r!n
 
pI :: Int -> Int
pI n = 
    case M.splitLookup n s of
       (_,Just x,_) -> (x+1)
       (_,_,b) -> snd (head (M.assocs b))
 
{-# NOINLINE primes #-}
primes :: [Int]
primes = unsafePerformIO $ do
  l <- Data.ByteString.Char8.readFile "primes_to_5M.txt"
  let p = Data.ByteString.Char8.lines l
      r = map (fst . fromJust . readInt) p::[Int]
  return r  
 
 
import PI
import Data.List
 
s n k =       ( n `div` (prime (k-1))) - k + 1
 
semiPrimeCount n =
  let last = pI $ floor $ sqrt (fromIntegral n)
      s k = pI ( n `div` (prime (k-1))) - k + 1
      pI2 = foldl' (\x y -> s y + x) 0 [1..last]
  in pI2
 
main = print (semiPrimeCount 100000000)