# Euler problems/21 to 30

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(→Problem 24: Added another method for Problem 24) |
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Evaluate the sum of all amicable pairs under 10000. | Evaluate the sum of all amicable pairs under 10000. | ||

− | Solution: | + | Solution: |

(http://www.research.att.com/~njas/sequences/A063990) | (http://www.research.att.com/~njas/sequences/A063990) | ||

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<haskell> | <haskell> | ||

problem_21_v2 = sum [n | n <- [2..9999], let m = d n, | problem_21_v2 = sum [n | n <- [2..9999], let m = d n, | ||

− | m > 1, m < 10000, n == d m] | + | m > 1, m < 10000, n == d m, d m /= d (d m)] |

d n = product [(p * product g - 1) `div` (p - 1) | | d n = product [(p * product g - 1) `div` (p - 1) | | ||

g <- group $ primeFactors n, let p = head g | g <- group $ primeFactors n, let p = head g | ||

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let names = sort $ read$"["++ input++"]" | let names = sort $ read$"["++ input++"]" | ||

let scores = zipWith score names [1..] | let scores = zipWith score names [1..] | ||

− | print | + | print . sum $ scores |

where score w i = (i *) . sum . map (\c -> ord c - ord 'A' + 1) $ w | where score w i = (i *) . sum . map (\c -> ord c - ord 'A' + 1) $ w | ||

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

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isSum = any (abunds_array !) . rests | isSum = any (abunds_array !) . rests | ||

− | problem_23 = | + | problem_23 = print . sum . filter (not . isSum) $ [1..n] |

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

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problem_24 = perms "0123456789" 999999 | problem_24 = perms "0123456789" 999999 | ||

+ | </haskell> | ||

+ | |||

+ | Or, using Data.List.permutations, | ||

+ | <haskell> | ||

+ | import Data.List | ||

+ | problem_24 = (!! 999999) . sort $ permutations ['0'..'9'] | ||

+ | </haskell> | ||

+ | |||

+ | Casey Hawthorne | ||

+ | |||

+ | For Project Euler #24 you don't need to generate all the lexicographic permutations by Knuth's method or any other. | ||

+ | |||

+ | You're only looking for the millionth lexicographic permutation of "0123456789" | ||

+ | |||

+ | <haskell> | ||

+ | |||

+ | -- Plan of attack. | ||

+ | |||

+ | -- The "x"s are different numbers | ||

+ | -- 0xxxxxxxxx represents 9! = 362880 permutations/numbers | ||

+ | -- 1xxxxxxxxx represents 9! = 362880 permutations/numbers | ||

+ | -- 2xxxxxxxxx represents 9! = 362880 permutations/numbers | ||

+ | |||

+ | |||

+ | -- 20xxxxxxxx represents 8! = 40320 | ||

+ | -- 21xxxxxxxx represents 8! = 40320 | ||

+ | |||

+ | -- 23xxxxxxxx represents 8! = 40320 | ||

+ | -- 24xxxxxxxx represents 8! = 40320 | ||

+ | -- 25xxxxxxxx represents 8! = 40320 | ||

+ | -- 26xxxxxxxx represents 8! = 40320 | ||

+ | -- 27xxxxxxxx represents 8! = 40320 | ||

+ | |||

+ | |||

+ | module Euler where | ||

+ | |||

+ | import Data.List | ||

+ | |||

+ | factorial n = product [1..n] | ||

+ | |||

+ | -- lexOrder "0123456789" 1000000 "" | ||

+ | |||

+ | lexOrder digits left s | ||

+ | | len == 0 = s ++ digits | ||

+ | | quot > 0 && rem == 0 = lexOrder (digits\\(show (digits!!(quot-1)))) rem (s ++ [(digits!!(quot-1))]) | ||

+ | | quot == 0 && rem == 0 = lexOrder (digits\\(show (digits!!len))) rem (s ++ [(digits!!len)]) | ||

+ | | rem == 0 = lexOrder (digits\\(show (digits!!(quot+1)))) rem (s ++ [(digits!!(quot+1))]) | ||

+ | | otherwise = lexOrder (digits\\(show (digits!!(quot)))) rem (s ++ [(digits!!(quot))]) | ||

+ | where | ||

+ | len = (length digits) - 1 | ||

+ | (quot,rem) = quotRem left (factorial len) | ||

+ | |||

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

Line 87: | Line 139: | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | + | fibs = 0:1:(zipWith (+) fibs (tail fibs)) | |

− | + | t = 10^999 | |

− | problem_25 = | + | |

− | where fibs = | + | problem_25 = length w |

+ | where | ||

+ | w = takeWhile (< t) fibs | ||

+ | </haskell> | ||

+ | |||

+ | |||

+ | Casey Hawthorne | ||

+ | |||

+ | I believe you mean the following: | ||

+ | |||

+ | <haskell> | ||

+ | |||

+ | fibs = 0:1:(zipWith (+) fibs (tail fibs)) | ||

+ | |||

+ | last (takeWhile (<10^1000) fibs) | ||

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

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Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | problem_26 = | + | problem_26 = fst $ maximumBy (comparing snd) |

+ | [(n,recurringCycle n) | n <- [1..999]] | ||

+ | where recurringCycle d = remainders d 10 [] | ||

+ | remainders d 0 rs = 0 | ||

+ | remainders d r rs = let r' = r `mod` d | ||

+ | in case elemIndex r' rs of | ||

+ | Just i -> i + 1 | ||

+ | Nothing -> remainders d (10*r') (r':rs) | ||

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

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<haskell> | <haskell> | ||

problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1 | problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1 | ||

+ | </haskell> | ||

+ | |||

+ | Alternatively, one can use the fact that the distance between the diagonal numbers increases by 2 in every concentric square. Each square contains four gaps, so the following <hask>scanl</hask> does the trick: | ||

+ | |||

+ | <haskell> | ||

+ | euler28 n = sum $ scanl (+) 0 | ||

+ | (1:(concatMap (replicate 4) [2,4..(n-1)])) | ||

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

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import Control.Monad | import Control.Monad | ||

problem_29 = length . group . sort $ liftM2 (^) [2..100] [2..100] | problem_29 = length . group . sort $ liftM2 (^) [2..100] [2..100] | ||

+ | </haskell> | ||

+ | |||

+ | We can also solve it in a more naive way, without using Monads, like this: | ||

+ | <haskell> | ||

+ | import List | ||

+ | problem_29 = length $ nub pr29_help | ||

+ | where pr29_help = [z | y <- [2..100], | ||

+ | z <- lift y] | ||

+ | lift y = map (\x -> x^y) [2..100] | ||

+ | </haskell> | ||

+ | |||

+ | Simpler: | ||

+ | |||

+ | <haskell> | ||

+ | import List | ||

+ | problem_29 = length $ nub [x^y | x <- [2..100], y <- [2..100]] | ||

+ | </haskell> | ||

+ | |||

+ | Instead of using lists, the Set data structure can be used for a significant speed increase: | ||

+ | |||

+ | <haskell> | ||

+ | import Set | ||

+ | problem_29 = size $ fromList [x^y | x <- [2..100], y <- [2..100]] | ||

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

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Solution: | Solution: | ||

<haskell> | <haskell> | ||

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

− | + | ||

− | + | ||

− | + | limit :: Integer | |

+ | limit = snd $ head $ dropWhile (\(a,b) -> a > b) $ zip (map (9^5*) [1..]) (map (10^) [1..]) | ||

+ | fifth :: Integer -> Integer | ||

+ | fifth = sum . map ((^5) . toInteger . digitToInt) . show | ||

− | + | problem_30 :: Integer | |

− | problem_30 | + | problem_30 = sum $ filter (\n -> n == fifth n) [2..limit] |

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

− | + | ||

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− | + | ||

</haskell> | </haskell> |

## Revision as of 03:52, 14 November 2011

## Contents |

## 1 Problem 21

Evaluate the sum of all amicable pairs under 10000.

Solution: (http://www.research.att.com/~njas/sequences/A063990)

This is a little slow because of the naive method used to compute the divisors.

problem_21 = sum [m+n | m <- [2..9999], let n = divisorsSum ! m, amicable m n] where amicable m n = m < n && n < 10000 && divisorsSum ! n == m divisorsSum = array (1,9999) [(i, sum (divisors i)) | i <- [1..9999]] divisors n = [j | j <- [1..n `div` 2], n `mod` j == 0]

Here is an alternative using a faster way of computing the sum of divisors.

problem_21_v2 = sum [n | n <- [2..9999], let m = d n, m > 1, m < 10000, n == d m, d m /= d (d m)] d n = product [(p * product g - 1) `div` (p - 1) | g <- group $ primeFactors n, let p = head g ] - n primeFactors = pf primes where pf ps@(p:ps') n | p * p > n = [n] | r == 0 = p : pf ps q | otherwise = pf ps' n where (q, r) = n `divMod` p primes = 2 : filter (null . tail . primeFactors) [3,5..]

## 2 Problem 22

What is the total of all the name scores in the file of first names?

Solution:

import Data.List import Data.Char problem_22 = do input <- readFile "names.txt" let names = sort $ read$"["++ input++"]" let scores = zipWith score names [1..] print . sum $ scores where score w i = (i *) . sum . map (\c -> ord c - ord 'A' + 1) $ w

## 3 Problem 23

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Solution:

--http://www.research.att.com/~njas/sequences/A048242 import Data.Array n = 28124 abundant n = eulerTotient n - n > n abunds_array = listArray (1,n) $ map abundant [1..n] abunds = filter (abunds_array !) [1..n] rests x = map (x-) $ takeWhile (<= x `div` 2) abunds isSum = any (abunds_array !) . rests problem_23 = print . sum . filter (not . isSum) $ [1..n]

## 4 Problem 24

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

Solution:

import Data.List fac 0 = 1 fac n = n * fac (n - 1) perms [] _= [] perms xs n= x : perms (delete x xs) (mod n m) where m = fac $ length xs - 1 y = div n m x = xs!!y problem_24 = perms "0123456789" 999999

Or, using Data.List.permutations,

import Data.List problem_24 = (!! 999999) . sort $ permutations ['0'..'9']

Casey Hawthorne

For Project Euler #24 you don't need to generate all the lexicographic permutations by Knuth's method or any other.

You're only looking for the millionth lexicographic permutation of "0123456789"

-- Plan of attack. -- The "x"s are different numbers -- 0xxxxxxxxx represents 9! = 362880 permutations/numbers -- 1xxxxxxxxx represents 9! = 362880 permutations/numbers -- 2xxxxxxxxx represents 9! = 362880 permutations/numbers -- 20xxxxxxxx represents 8! = 40320 -- 21xxxxxxxx represents 8! = 40320 -- 23xxxxxxxx represents 8! = 40320 -- 24xxxxxxxx represents 8! = 40320 -- 25xxxxxxxx represents 8! = 40320 -- 26xxxxxxxx represents 8! = 40320 -- 27xxxxxxxx represents 8! = 40320 module Euler where import Data.List factorial n = product [1..n] -- lexOrder "0123456789" 1000000 "" lexOrder digits left s | len == 0 = s ++ digits | quot > 0 && rem == 0 = lexOrder (digits\\(show (digits!!(quot-1)))) rem (s ++ [(digits!!(quot-1))]) | quot == 0 && rem == 0 = lexOrder (digits\\(show (digits!!len))) rem (s ++ [(digits!!len)]) | rem == 0 = lexOrder (digits\\(show (digits!!(quot+1)))) rem (s ++ [(digits!!(quot+1))]) | otherwise = lexOrder (digits\\(show (digits!!(quot)))) rem (s ++ [(digits!!(quot))]) where len = (length digits) - 1 (quot,rem) = quotRem left (factorial len)

## 5 Problem 25

What is the first term in the Fibonacci sequence to contain 1000 digits?

Solution:

fibs = 0:1:(zipWith (+) fibs (tail fibs)) t = 10^999 problem_25 = length w where w = takeWhile (< t) fibs

Casey Hawthorne

I believe you mean the following:

fibs = 0:1:(zipWith (+) fibs (tail fibs)) last (takeWhile (<10^1000) fibs)

## 6 Problem 26

Find the value of d < 1000 for which 1/d contains the longest recurring cycle.

Solution:

problem_26 = fst $ maximumBy (comparing snd) [(n,recurringCycle n) | n <- [1..999]] where recurringCycle d = remainders d 10 [] remainders d 0 rs = 0 remainders d r rs = let r' = r `mod` d in case elemIndex r' rs of Just i -> i + 1 Nothing -> remainders d (10*r') (r':rs)

## 7 Problem 27

Find a quadratic formula that produces the maximum number of primes for consecutive values of n.

Solution:

problem_27 = -(2*a-1)*(a^2-a+41) where n = 1000 m = head $ filter (\x->x^2-x+41>n) [1..] a = m-1

## 8 Problem 28

What is the sum of both diagonals in a 1001 by 1001 spiral?

Solution:

problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1

euler28 n = sum $ scanl (+) 0 (1:(concatMap (replicate 4) [2,4..(n-1)]))

## 9 Problem 29

How many distinct terms are in the sequence generated by a^{b} for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

Solution:

import Control.Monad problem_29 = length . group . sort $ liftM2 (^) [2..100] [2..100]

We can also solve it in a more naive way, without using Monads, like this:

import List problem_29 = length $ nub pr29_help where pr29_help = [z | y <- [2..100], z <- lift y] lift y = map (\x -> x^y) [2..100]

Simpler:

import List problem_29 = length $ nub [x^y | x <- [2..100], y <- [2..100]]

Instead of using lists, the Set data structure can be used for a significant speed increase:

import Set problem_29 = size $ fromList [x^y | x <- [2..100], y <- [2..100]]

## 10 Problem 30

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

Solution:

import Data.Char (digitToInt) limit :: Integer limit = snd $ head $ dropWhile (\(a,b) -> a > b) $ zip (map (9^5*) [1..]) (map (10^) [1..]) fifth :: Integer -> Integer fifth = sum . map ((^5) . toInteger . digitToInt) . show problem_30 :: Integer problem_30 = sum $ filter (\n -> n == fifth n) [2..limit]