# Euler problems/41 to 50

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Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | import Data. | + | -- Assuming isPrime has been implemented |

− | + | import Data.Char (intToDigit) | |

− | + | problem_41 = maximum [ n' | d <- [3..9], n <- permute ['1'..intToDigit d], | |

− | + | let n' = read n, isPrime n'] | |

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

where | where | ||

− | + | permute "" = [""] | |

+ | permute str = [(x:xs)| x <- str, xs <- permute (delete x str)] | ||

</haskell> | </haskell> | ||

Line 26: | Line 21: | ||

trilist = takeWhile (<300) (scanl1 (+) [1..]) | trilist = takeWhile (<300) (scanl1 (+) [1..]) | ||

wordscore xs = sum $ map (subtract 64 . ord) xs | wordscore xs = sum $ map (subtract 64 . ord) xs | ||

− | problem_42 megalist= | + | problem_42 megalist = |

− | length [ wordscore a | | + | length [ wordscore a | a <- megalist, |

− | + | elem (wordscore a) trilist ] | |

− | + | main = do f <- readFile "words.txt" | |

− | + | let words = read $"["++f++"]" | |

− | main=do | + | print $ problem_42 words |

− | + | ||

− | + | ||

− | + | ||

</haskell> | </haskell> | ||

Line 54: | Line 46: | ||

. filter (elem 0) . genSeq [] $ [17,13,11,7,5,3,2] | . filter (elem 0) . genSeq [] $ [17,13,11,7,5,3,2] | ||

− | mults mi ma n = takeWhile (< ma) | + | mults mi ma n = takeWhile (< ma) . dropWhile (<mi) . iterate (+n) $ n |

sequ xs ys = tail xs == init ys | sequ xs ys = tail xs == init ys | ||

Line 60: | Line 52: | ||

addZ n xs = replicate (n - length xs) 0 ++ xs | addZ n xs = replicate (n - length xs) 0 ++ xs | ||

− | genSeq [] (x:xs) = genSeq | + | genSeq [] (x:xs) = genSeq (filter (not . doub) |

− | + | . map (addZ 3 . reverse . explode) | |

− | + | $ mults 9 1000 x) | |

− | + | xs | |

genSeq ys (x:xs) = | genSeq ys (x:xs) = | ||

− | genSeq (do | + | genSeq (do m <- mults 9 1000 x |

− | + | let s = addZ 3 . reverse . explode $ m | |

− | + | y <- filter (sequ s . take 3) $ filter (notElem (head s)) ys | |

− | + | return (head s:y)) | |

− | + | xs | |

− | + | ||

genSeq ys [] = ys | genSeq ys [] = ys | ||

doub xs = nub xs /= xs | doub xs = nub xs /= xs | ||

+ | </haskell> | ||

+ | |||

+ | An arguably cleaner, alternate solution uses nondeterminism + state to create a backtracking monad particularly suited to this problem: | ||

+ | |||

+ | <haskell> | ||

+ | import Control.Monad | ||

+ | import Control.Monad.State | ||

+ | import Data.Set | ||

+ | |||

+ | type Select elem a = StateT (Set elem) [] a | ||

+ | |||

+ | select :: (Ord elem) => [elem] -> Select elem elem | ||

+ | select as = do | ||

+ | set <- get | ||

+ | a <- lift as | ||

+ | guard (not (member a set)) | ||

+ | put (insert a set) | ||

+ | return a | ||

+ | |||

+ | runSelect :: Select elem a -> [a] | ||

+ | runSelect m = Prelude.map fst (runStateT m empty) | ||

+ | |||

+ | fromDigits = foldl (\tot d -> 10 * tot + d) 0 | ||

+ | |||

+ | ds = runSelect $ do | ||

+ | d4 <- select [0,2..8] | ||

+ | d3 <- select [0..9] | ||

+ | d5 <- select [0..9] | ||

+ | guard ((d3 + d4 + d5) `mod` 3 == 0) | ||

+ | d6 <- select [0,5] | ||

+ | d7 <- select [0..9] | ||

+ | guard ((100 * d5 + 10 * d6 + d7) `mod` 7 == 0) | ||

+ | d8 <- select [0..9] | ||

+ | guard ((d6 - d7 + d8) `mod` 11 == 0) | ||

+ | d9 <- select [0..9] | ||

+ | guard ((100 * d7 + 10 * d8 + d9) `mod` 13 == 0) | ||

+ | d10 <- select [0..9] | ||

+ | guard ((100 * d8 + 10 * d9 + d10) `mod` 17 == 0) | ||

+ | d2 <- select [0..9] | ||

+ | d1 <- select [0..9] | ||

+ | return (fromDigits [d1, d2, d3, d4, d5, d6, d7, d8, d9, d10]) | ||

+ | |||

+ | answer = sum ds | ||

+ | |||

+ | main = do | ||

+ | print ds | ||

+ | print answer | ||

+ | </haskell> | ||

+ | |||

+ | An almost instant answer can be generated by only creating permutations which fulfil the requirement of particular digits being multiples of certain numbers. | ||

+ | |||

+ | <haskell> | ||

+ | import Data.List ((\\), nub) | ||

+ | |||

+ | main = print q43 | ||

+ | |||

+ | q43 = sum [ read n | (d7d8d9, remDigits) <- permMults digits 17, | ||

+ | (d4d5d6, remDigits') <- permMults remDigits 7, | ||

+ | d4d5d6 !! 1 == '0' || d4d5d6 !! 1 == '5', | ||

+ | (d1d2d3, remDigit) <- permMults remDigits' 2, | ||

+ | let n = remDigit ++ d1d2d3 ++ d4d5d6 ++ d7d8d9, | ||

+ | hasProperty (tail n) primes] | ||

+ | where | ||

+ | digits = "0123456789" | ||

+ | primes = [2,3,5,7,11,13,17] | ||

+ | hasProperty _ [] = True | ||

+ | hasProperty c (p:ps) = (read $ take 3 c) `mod` p == 0 | ||

+ | && hasProperty (tail c) ps | ||

+ | permMults cs p = [ (ds, cs \\ ds) | n <- [p,2*p..987], | ||

+ | let ds = leadingZero n, | ||

+ | ds == nub ds, | ||

+ | all (flip elem cs) ds] | ||

+ | where | ||

+ | leadingZero n | ||

+ | | n < 10 = "00" ++ show n | ||

+ | | n < 100 = "0" ++ show n | ||

+ | | otherwise = show n | ||

</haskell> | </haskell> | ||

Line 82: | Line 150: | ||

<haskell> | <haskell> | ||

import Data.Set | import Data.Set | ||

− | problem_44 = | + | problem_44 = head solutions |

− | + | where solutions = [a-b | a <- penta, | |

− | + | b <- takeWhile (<a) penta, | |

− | + | isPenta (a-b), | |

− | + | isPenta (b+a) ] | |

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

isPenta = (`member` fromList penta) | isPenta = (`member` fromList penta) | ||

penta = [(n * (3*n-1)) `div` 2 | n <- [1..5000]] | penta = [(n * (3*n-1)) `div` 2 | n <- [1..5000]] | ||

</haskell> | </haskell> | ||

+ | |||

+ | The above solution finds the correct answer but searches the pairs in the wrong order. Lengthier and slower but perhaps more correct solution [https://gist.github.com/2079968 here]. | ||

== [http://projecteuler.net/index.php?section=problems&id=45 Problem 45] == | == [http://projecteuler.net/index.php?section=problems&id=45 Problem 45] == | ||

Line 101: | Line 166: | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | isPent n = | + | isPent n = (af == 0) && ai `mod` 6 == 5 |

− | + | where (ai, af) = properFraction . sqrt $ 1 + 24 * (fromInteger n) | |

− | + | ||

− | + | ||

problem_45 = head [x | x <- scanl (+) 1 [5,9..], x > 40755, isPent x] | problem_45 = head [x | x <- scanl (+) 1 [5,9..], x > 40755, isPent x] | ||

Line 120: | Line 183: | ||

<haskell> | <haskell> | ||

import Data.List | import Data.List | ||

− | isPrime x | + | isPrime x | x==3 = True |

− | + | | otherwise = millerRabinPrimality x 2 | |

− | + | problem_46 = find (\x -> not (isPrime x) && check x) [3,5..] | |

− | problem_46 = | + | where |

− | + | check x = not . any isPrime | |

+ | . takeWhile (>0) | ||

+ | . map (\y -> x - 2 * y * y) $ [1..] | ||

+ | </haskell> | ||

+ | |||

+ | Alternate Solution: | ||

+ | |||

+ | Considering that the answer is less than 6000, there's no need for fancy solutions. The following is as fast as most C++ solutions. | ||

+ | |||

+ | <haskell> | ||

+ | primes :: [Int] | ||

+ | primes = 2 : filter isPrime [3, 5..] | ||

+ | |||

+ | isPrime :: Int -> Bool | ||

+ | isPrime n = all (not . divides n) $ takeWhile (\p -> p^2 <= n) primes | ||

where | where | ||

− | + | divides n p = n `mod` p == 0 | |

− | + | ||

+ | compOdds :: [Int] | ||

+ | compOdds = filter (not . isPrime) [3, 5..] | ||

+ | |||

+ | verifConj :: Int -> Bool | ||

+ | verifConj n = any isPrime (takeWhile (>0) $ map (\i -> n - 2*i*i) [1..]) | ||

+ | |||

+ | problem_46 :: Int | ||

+ | problem_46 = head $ filter (not . verifConj) compOdds | ||

</haskell> | </haskell> | ||

Line 138: | Line 223: | ||

problem_47 = find (all ((==4).snd)) . map (take 4) . tails | problem_47 = find (all ((==4).snd)) . map (take 4) . tails | ||

. zip [1..] . map (length . factors) $ [1..] | . zip [1..] . map (length . factors) $ [1..] | ||

− | fstfac x = [(head a ,length a)|a<-group$primeFactors x] | + | fstfac x = [(head a ,length a) | a <- group $ primeFactors x] |

− | fac [(x,y)]=[x^a|a<-[0..y]] | + | fac [(x,y)] = [x^a | a <- [0..y]] |

− | fac (x:xs)=[a*b|a<-fac [x],b<-fac xs] | + | fac (x:xs) = [a*b | a <- fac [x], b <- fac xs] |

− | factors x=fac$fstfac x | + | factors x = fac $ fstfac x |

primes = 2 : filter ((==1) . length . primeFactors) [3,5..] | primes = 2 : filter ((==1) . length . primeFactors) [3,5..] | ||

primeFactors n = factor n primes | primeFactors n = factor n primes | ||

− | + | where factor _ [] = [] | |

− | + | ||

factor m (p:ps) | p*p > m = [m] | factor m (p:ps) | p*p > m = [m] | ||

− | | m `mod` p == 0 = | + | | m `mod` p == 0 = [p, m `div` p] |

| otherwise = factor m ps | | otherwise = factor m ps | ||

+ | </haskell> | ||

+ | |||

+ | |||

+ | Alternate Solution: | ||

+ | The previous solution actually didn't give the correct answer for me. The following method did. | ||

+ | |||

+ | <haskell> | ||

+ | |||

+ | import Data.List | ||

+ | import Data.Numbers | ||

+ | import Data.Numbers.Primes | ||

+ | import qualified Data.Set as Set | ||

+ | |||

+ | dPrimeFactors n = Set.fromList $ primeFactors n | ||

+ | |||

+ | dPFList n = [(k, dPrimeFactors k) | ||

+ | | k <- filter (\z -> (not $ isPrime z)) [1..n]] | ||

+ | |||

+ | nConsec n s = | ||

+ | let dpf = dPFList s | ||

+ | fltrd = filter (\z -> Set.size (snd z) == n) dpf | ||

+ | gps = [take (fromIntegral n) (drop (fromIntegral k) fltrd) | ||

+ | | k <- [0..(length fltrd - n)] ] | ||

+ | gps2 = filter (\z -> isConsec (map fst z)) gps | ||

+ | in filter (\zz -> Set.empty == | ||

+ | foldl (\acc z -> Set.intersection acc (snd z)) | ||

+ | (snd (head zz)) | ||

+ | zz) gps2 | ||

+ | |||

+ | isConsec xs = (sort xs) == [(minimum xs)..(maximum xs)] | ||

+ | |||

+ | problem_47 = (fst . head . head) $ nConsec 4 20000 | ||

+ | |||

</haskell> | </haskell> | ||

Line 161: | Line 278: | ||

<haskell> | <haskell> | ||

− | problem_48 = | + | problem_48 = (`mod` limit) $ sum [powMod limit n n | n <- [1..1000]] |

− | where | + | where limit=10^10 |

− | + | </haskell> | |

+ | |||

+ | Another one-liner for this problem, with no use of other functions is the following: | ||

+ | <haskell> | ||

+ | problem_48 = reverse $ take 10 | ||

+ | $ reverse $ show $ sum $ map (\x -> x^x) [1..1000] | ||

</haskell> | </haskell> | ||

Line 173: | Line 295: | ||

<haskell> | <haskell> | ||

− | |||

import Data.List | import Data.List | ||

+ | |||

isPrime x | isPrime x | ||

− | |x==3=True | + | | x==3 = True |

− | |otherwise=millerRabinPrimality x 2 | + | | otherwise = millerRabinPrimality x 2 |

− | + | ||

− | + | ||

− | problem_49 = | + | primes4 = filter isPrime [1000..9999] |

− | + | ||

− | + | problem_49 = [ (a,b,c) | a <- primes4, | |

− | + | b <- dropWhile (<= a) primes4, | |

− | + | sort (show a) == sort (show b), | |

− | + | let c = 2 * b - a, | |

− | + | c `elem` primes4, | |

− | + | sort (show a) == sort (show c) ] | |

− | + | ||

− | + | ||

− | + | ||

</haskell> | </haskell> | ||

Line 209: | Line 326: | ||

problem_50 = findPrimeSum $ take 546 primes | problem_50 = findPrimeSum $ take 546 primes | ||

+ | </haskell> | ||

+ | |||

+ | * This code is wrong: if you switch ''init'' and ''tail'', it produces wrong result, so it is by sheer luck that it produces the right answer at all. As written, it will produce the longest chain of primes ending at 546-th prime, summing up to a prime - using 546 as it is the longest prefix of primes summing up to less than a million. That's not what the problem asks for. | ||

+ | |||

+ | :What's to guarantee us there's no longer chain ending at 545-th prime? 544-th? For instance, for 1,100,000 the longest sequence ends at 568-th instead of 571-st prime which is what the above code would use. | ||

+ | |||

+ | :Moreover, cutting the search short at just first 546 primes is wrong too. What if the longest chain was really short, like 10 or 20 primes? Then we'd have to go much higher into the primes. We have no way of knowing that length in advance. | ||

+ | |||

+ | * Here's my solution, it's not the fastest but is correct, feel free to criticise (isPrime and primes not included): | ||

+ | <haskell> | ||

+ | import Euler.Helpers | ||

+ | import qualified Data.List as L | ||

+ | |||

+ | prime n = takeWhileSum n primes | ||

+ | takeWhileSum n = takeWhileArr (\x -> sum x <= n) | ||

+ | takeWhileArr f xs = takeWhileF f [] xs | ||

+ | where | ||

+ | takeWhileF f rs [] = reverse rs | ||

+ | takeWhileF f rs (x:xs) | ||

+ | | f (x:rs) = takeWhileF f (x:rs) xs | ||

+ | | otherwise = reverse rs | ||

+ | |||

+ | primeSums n = map (map (\x -> (isPrime x,x) ) . takeWhile (<n) . scanl1 (+)) (L.tails (prime n)) | ||

+ | main = print . maximum $ map index (primeSums 100000) | ||

+ | where index x = if null $ ind x | ||

+ | then (0,0) | ||

+ | else (last $ ind x, snd (x !! (last $ ind x))) | ||

+ | ind = L.findIndices (fst) | ||

</haskell> | </haskell> |

## Latest revision as of 10:22, 25 August 2012

## Contents |

## [edit] 1 Problem 41

What is the largest n-digit pandigital prime that exists?

Solution:

-- Assuming isPrime has been implemented import Data.Char (intToDigit) problem_41 = maximum [ n' | d <- [3..9], n <- permute ['1'..intToDigit d], let n' = read n, isPrime n'] where permute "" = [""] permute str = [(x:xs)| x <- str, xs <- permute (delete x str)]

## [edit] 2 Problem 42

How many triangle words can you make using the list of common English words?

Solution:

import Data.Char trilist = takeWhile (<300) (scanl1 (+) [1..]) wordscore xs = sum $ map (subtract 64 . ord) xs problem_42 megalist = length [ wordscore a | a <- megalist, elem (wordscore a) trilist ] main = do f <- readFile "words.txt" let words = read $"["++f++"]" print $ problem_42 words

## [edit] 3 Problem 43

Find the sum of all pandigital numbers with an unusual sub-string divisibility property.

Solution:

import Data.List l2n :: (Integral a) => [a] -> a l2n = foldl' (\a b -> 10*a+b) 0 swap (a,b) = (b,a) explode :: (Integral a) => a -> [a] explode = unfoldr (\a -> if a==0 then Nothing else Just $ swap $ quotRem a 10) problem_43 = sum . map l2n . map (\s -> head ([0..9] \\ s):s) . filter (elem 0) . genSeq [] $ [17,13,11,7,5,3,2] mults mi ma n = takeWhile (< ma) . dropWhile (<mi) . iterate (+n) $ n sequ xs ys = tail xs == init ys addZ n xs = replicate (n - length xs) 0 ++ xs genSeq [] (x:xs) = genSeq (filter (not . doub) . map (addZ 3 . reverse . explode) $ mults 9 1000 x) xs genSeq ys (x:xs) = genSeq (do m <- mults 9 1000 x let s = addZ 3 . reverse . explode $ m y <- filter (sequ s . take 3) $ filter (notElem (head s)) ys return (head s:y)) xs genSeq ys [] = ys doub xs = nub xs /= xs

An arguably cleaner, alternate solution uses nondeterminism + state to create a backtracking monad particularly suited to this problem:

import Control.Monad import Control.Monad.State import Data.Set type Select elem a = StateT (Set elem) [] a select :: (Ord elem) => [elem] -> Select elem elem select as = do set <- get a <- lift as guard (not (member a set)) put (insert a set) return a runSelect :: Select elem a -> [a] runSelect m = Prelude.map fst (runStateT m empty) fromDigits = foldl (\tot d -> 10 * tot + d) 0 ds = runSelect $ do d4 <- select [0,2..8] d3 <- select [0..9] d5 <- select [0..9] guard ((d3 + d4 + d5) `mod` 3 == 0) d6 <- select [0,5] d7 <- select [0..9] guard ((100 * d5 + 10 * d6 + d7) `mod` 7 == 0) d8 <- select [0..9] guard ((d6 - d7 + d8) `mod` 11 == 0) d9 <- select [0..9] guard ((100 * d7 + 10 * d8 + d9) `mod` 13 == 0) d10 <- select [0..9] guard ((100 * d8 + 10 * d9 + d10) `mod` 17 == 0) d2 <- select [0..9] d1 <- select [0..9] return (fromDigits [d1, d2, d3, d4, d5, d6, d7, d8, d9, d10]) answer = sum ds main = do print ds print answer

An almost instant answer can be generated by only creating permutations which fulfil the requirement of particular digits being multiples of certain numbers.

import Data.List ((\\), nub) main = print q43 q43 = sum [ read n | (d7d8d9, remDigits) <- permMults digits 17, (d4d5d6, remDigits') <- permMults remDigits 7, d4d5d6 !! 1 == '0' || d4d5d6 !! 1 == '5', (d1d2d3, remDigit) <- permMults remDigits' 2, let n = remDigit ++ d1d2d3 ++ d4d5d6 ++ d7d8d9, hasProperty (tail n) primes] where digits = "0123456789" primes = [2,3,5,7,11,13,17] hasProperty _ [] = True hasProperty c (p:ps) = (read $ take 3 c) `mod` p == 0 && hasProperty (tail c) ps permMults cs p = [ (ds, cs \\ ds) | n <- [p,2*p..987], let ds = leadingZero n, ds == nub ds, all (flip elem cs) ds] where leadingZero n | n < 10 = "00" ++ show n | n < 100 = "0" ++ show n | otherwise = show n

## [edit] 4 Problem 44

Find the smallest pair of pentagonal numbers whose sum and difference is pentagonal.

Solution:

import Data.Set problem_44 = head solutions where solutions = [a-b | a <- penta, b <- takeWhile (<a) penta, isPenta (a-b), isPenta (b+a) ] isPenta = (`member` fromList penta) penta = [(n * (3*n-1)) `div` 2 | n <- [1..5000]]

The above solution finds the correct answer but searches the pairs in the wrong order. Lengthier and slower but perhaps more correct solution here.

## [edit] 5 Problem 45

After 40755, what is the next triangle number that is also pentagonal and hexagonal?

Solution:

isPent n = (af == 0) && ai `mod` 6 == 5 where (ai, af) = properFraction . sqrt $ 1 + 24 * (fromInteger n) problem_45 = head [x | x <- scanl (+) 1 [5,9..], x > 40755, isPent x]

## [edit] 6 Problem 46

What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?

Solution:

This solution is inspired by exercise 3.70 in *Structure and Interpretation of Computer Programs*, (2nd ed.).

millerRabinPrimality on the Prime_numbers page

import Data.List isPrime x | x==3 = True | otherwise = millerRabinPrimality x 2 problem_46 = find (\x -> not (isPrime x) && check x) [3,5..] where check x = not . any isPrime . takeWhile (>0) . map (\y -> x - 2 * y * y) $ [1..]

Alternate Solution:

Considering that the answer is less than 6000, there's no need for fancy solutions. The following is as fast as most C++ solutions.

primes :: [Int] primes = 2 : filter isPrime [3, 5..] isPrime :: Int -> Bool isPrime n = all (not . divides n) $ takeWhile (\p -> p^2 <= n) primes where divides n p = n `mod` p == 0 compOdds :: [Int] compOdds = filter (not . isPrime) [3, 5..] verifConj :: Int -> Bool verifConj n = any isPrime (takeWhile (>0) $ map (\i -> n - 2*i*i) [1..]) problem_46 :: Int problem_46 = head $ filter (not . verifConj) compOdds

## [edit] 7 Problem 47

Find the first four consecutive integers to have four distinct primes factors.

Solution:

import Data.List problem_47 = find (all ((==4).snd)) . map (take 4) . tails . zip [1..] . map (length . factors) $ [1..] fstfac x = [(head a ,length a) | a <- group $ primeFactors x] fac [(x,y)] = [x^a | a <- [0..y]] fac (x:xs) = [a*b | a <- fac [x], b <- fac xs] factors x = fac $ fstfac x primes = 2 : filter ((==1) . length . primeFactors) [3,5..] primeFactors n = factor n primes where factor _ [] = [] factor m (p:ps) | p*p > m = [m] | m `mod` p == 0 = [p, m `div` p] | otherwise = factor m ps

Alternate Solution:
The previous solution actually didn't give the correct answer for me. The following method did.

import Data.List import Data.Numbers import Data.Numbers.Primes import qualified Data.Set as Set dPrimeFactors n = Set.fromList $ primeFactors n dPFList n = [(k, dPrimeFactors k) | k <- filter (\z -> (not $ isPrime z)) [1..n]] nConsec n s = let dpf = dPFList s fltrd = filter (\z -> Set.size (snd z) == n) dpf gps = [take (fromIntegral n) (drop (fromIntegral k) fltrd) | k <- [0..(length fltrd - n)] ] gps2 = filter (\z -> isConsec (map fst z)) gps in filter (\zz -> Set.empty == foldl (\acc z -> Set.intersection acc (snd z)) (snd (head zz)) zz) gps2 isConsec xs = (sort xs) == [(minimum xs)..(maximum xs)] problem_47 = (fst . head . head) $ nConsec 4 20000

## [edit] 8 Problem 48

Find the last ten digits of 1^{1} + 2^{2} + ... + 1000^{1000}.

Solution: If the problem were more computationally intensive, modular exponentiation might be appropriate. With this problem size the naive approach is sufficient.

powMod on the Prime_numbers page

problem_48 = (`mod` limit) $ sum [powMod limit n n | n <- [1..1000]] where limit=10^10

Another one-liner for this problem, with no use of other functions is the following:

problem_48 = reverse $ take 10 $ reverse $ show $ sum $ map (\x -> x^x) [1..1000]

## [edit] 9 Problem 49

Find arithmetic sequences, made of prime terms, whose four digits are permutations of each other.

Solution: millerRabinPrimality on the Prime_numbers page

import Data.List isPrime x | x==3 = True | otherwise = millerRabinPrimality x 2 primes4 = filter isPrime [1000..9999] problem_49 = [ (a,b,c) | a <- primes4, b <- dropWhile (<= a) primes4, sort (show a) == sort (show b), let c = 2 * b - a, c `elem` primes4, sort (show a) == sort (show c) ]

## [edit] 10 Problem 50

Which prime, below one-million, can be written as the sum of the most consecutive primes?

Solution: (prime and isPrime not included)

import Control.Monad findPrimeSum ps | isPrime sumps = Just sumps | otherwise = findPrimeSum (tail ps) `mplus` findPrimeSum (init ps) where sumps = sum ps problem_50 = findPrimeSum $ take 546 primes

- This code is wrong: if you switch
*init*and*tail*, it produces wrong result, so it is by sheer luck that it produces the right answer at all. As written, it will produce the longest chain of primes ending at 546-th prime, summing up to a prime - using 546 as it is the longest prefix of primes summing up to less than a million. That's not what the problem asks for.

- What's to guarantee us there's no longer chain ending at 545-th prime? 544-th? For instance, for 1,100,000 the longest sequence ends at 568-th instead of 571-st prime which is what the above code would use.

- Moreover, cutting the search short at just first 546 primes is wrong too. What if the longest chain was really short, like 10 or 20 primes? Then we'd have to go much higher into the primes. We have no way of knowing that length in advance.

- Here's my solution, it's not the fastest but is correct, feel free to criticise (isPrime and primes not included):

import Euler.Helpers import qualified Data.List as L prime n = takeWhileSum n primes takeWhileSum n = takeWhileArr (\x -> sum x <= n) takeWhileArr f xs = takeWhileF f [] xs where takeWhileF f rs [] = reverse rs takeWhileF f rs (x:xs) | f (x:rs) = takeWhileF f (x:rs) xs | otherwise = reverse rs primeSums n = map (map (\x -> (isPrime x,x) ) . takeWhile (<n) . scanl1 (+)) (L.tails (prime n)) main = print . maximum $ map index (primeSums 100000) where index x = if null $ ind x then (0,0) else (last $ ind x, snd (x !! (last $ ind x))) ind = L.findIndices (fst)