# Euler problems/11 to 20

### From HaskellWiki

CaleGibbard (Talk | contribs) (Fix layout.) |
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prods :: Array (Int, Int) Int -> [Int] |
prods :: Array (Int, Int) Int -> [Int] |
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− | prods a = [product xs | |
+ | prods a = [product xs | i <- range $ bounds a, |

− | i <- range $ bounds a |
+ | s <- senses, |

− | , s <- senses |
+ | let is = take 4 $ iterate s i, |

− | , let is = take 4 $ iterate s i |
+ | all (inArray a) is, |

− | , all (inArray a) is |
+ | let xs = map (a!) is] |

− | , let xs = map (a!) is |
+ | main = print . maximum . prods . input =<< getContents |

− | ] |
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− | main = getContents >>= print . maximum . prods . input |
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</haskell> |
</haskell> |
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76576500, 236215980,7534947420] |
76576500, 236215980,7534947420] |
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--primeFactors in problem_3 |
--primeFactors in problem_3 |
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− | problem_12 = |
+ | problem_12 = head $ filter ((> 500) . nDivisors) triangleNumbers |

− | head $ filter ((> 500) . nDivisors) triangleNumbers |
+ | where nDivisors n = product $ map ((+1) . length) (group (primeFactors n)) |

− | where |
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− | nDivisors n = |
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− | product $ map ((+1) . length) (group (primeFactors n)) |
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</haskell> |
</haskell> |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | sToInt =(+0).read |
+ | |

− | main=do |
+ | main = do xs <- fmap (map read . lines) (readFile "p13.log") |

− | a<-readFile "p13.log" |
+ | print . take 10 . show . sum $ xs |

− | let b=map sToInt $lines a |
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− | let c=take 10 $ show $ sum b |
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− | print c |
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</haskell> |
</haskell> |
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<haskell> |
<haskell> |
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− | problem_15 = |
+ | problem_15 = product [21..40] `div` product [2..20] |

− | product [21..40] `div` product [2..20] |
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</haskell> |
</haskell> |
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import Data.Char |
import Data.Char |
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problem_16 = sum k |
problem_16 = sum k |
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− | where |
+ | where s = show (2^1000) |

− | s=show $2^1000 |
+ | k = map digitToInt s |

− | k=map digitToInt s |
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</haskell> |
</haskell> |
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| x == 1000 = "onethousand" |
| x == 1000 = "onethousand" |
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− | where |
+ | where firstDigit x = digitToInt . head . show $ x |

− | firstDigit x = digitToInt$head (show x) |
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− | problem_17 = |
+ | problem_17 = length . concatMap decompose $ [1..1000] |

− | length$concat (map decompose [1..1000]) |
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</haskell> |
</haskell> |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | problem_18 = |
+ | problem_18 = head $ foldr1 g tri |

− | head $ foldr1 g tri |
+ | where |

− | where |
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f x y z = x + max y z |
f x y z = x + max y z |
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g xs ys = zipWith3 f xs ys $ tail ys |
g xs ys = zipWith3 f xs ys $ tail ys |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | problem_19 = |
+ | problem_19 = length . filter (== sunday) . drop 12 . take 1212 $ since1900 |

− | length $ filter (== sunday) $ drop 12 $ take 1212 since1900 |
+ | since1900 = scanl nextMonth monday . concat $ |

− | since1900 = |
+ | replicate 4 nonLeap ++ cycle (leap : replicate 3 nonLeap) |

− | scanl nextMonth monday $ concat $ |
+ | |

− | replicate 4 nonLeap ++ cycle (leap : replicate 3 nonLeap) |
+ | nonLeap = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] |

− | nonLeap = |
+ | |

− | [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] |
+ | leap = 31 : 29 : drop 2 nonLeap |

− | leap = |
+ | |

− | 31 : 29 : drop 2 nonLeap |
+ | nextMonth x y = (x + y) `mod` 7 |

− | nextMonth x y = |
+ | |

− | (x + y) `mod` 7 |
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sunday = 0 |
sunday = 0 |
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monday = 1 |
monday = 1 |
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import Data.Time.Calendar.WeekDate |
import Data.Time.Calendar.WeekDate |
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− | problem_19_v2 = |
+ | problem_19_v2 = length [() | y <- [1901..2000], |

− | length [() | |
+ | m <- [1..12], |

− | y <- [1901..2000], |
+ | let (_, _, d) = toWeekDate $ fromGregorian y m 1, |

− | m <- [1..12], |
+ | d == 7] |

− | let (_, _, d) = toWeekDate $ fromGregorian y m 1, |
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− | d == 7 |
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− | ] |
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</haskell> |
</haskell> |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | numPrime x p=takeWhile(>0) [div x (p^a)|a<-[1..]] |
+ | numPrime x p = takeWhile(>0) [div x (p^a) | a<-[1..]] |

− | fastFactorial n= |
+ | |

− | product[a^x| |
+ | fastFactorial n = |

− | a<-takeWhile(<n) primes, |
+ | product[a^x | a <- takeWhile(<n) primes, |

− | let x=sum$numPrime n a |
+ | let x = sum $ numPrime n a ] |

− | ] |
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digits n |
digits n |
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− | |n<10=[n] |
+ | | n<10 = [n] |

− | |otherwise= y:digits x |
+ | | otherwise = y:digits x |

− | where |
+ | where (x,y) = divMod n 10 |

− | (x,y)=divMod n 10 |
+ | |

− | problem_20= sum $ digits $fastFactorial 100 |
+ | problem_20 = sum . digits . fastFactorial $ 100 |

</haskell> |
</haskell> |

## Revision as of 19:19, 19 February 2008

## Contents |

## 1 Problem 11

What is the greatest product of four numbers on the same straight line in the 20 by 20 grid?

Solution: using Array and Arrows, for fun :

import Control.Arrow import Data.Array input :: String -> Array (Int,Int) Int input = listArray ((1,1),(20,20)) . map read . words senses = [(+1) *** id,(+1) *** (+1), id *** (+1), (+1) *** (\n -> n - 1)] inArray a i = inRange (bounds a) i prods :: Array (Int, Int) Int -> [Int] prods a = [product xs | i <- range $ bounds a, s <- senses, let is = take 4 $ iterate s i, all (inArray a) is, let xs = map (a!) is] main = print . maximum . prods . input =<< getContents

## 2 Problem 12

What is the first triangle number to have over five-hundred divisors?

Solution:

--http://www.research.att.com/~njas/sequences/A084260 triangleNumbers = [630, 5460, 25200, 73920, 97020, 157080, 1185030, 2031120, 2162160, 17907120, 76576500, 236215980,7534947420] --primeFactors in problem_3 problem_12 = head $ filter ((> 500) . nDivisors) triangleNumbers where nDivisors n = product $ map ((+1) . length) (group (primeFactors n))

## 3 Problem 13

Find the first ten digits of the sum of one-hundred 50-digit numbers.

Solution:

main = do xs <- fmap (map read . lines) (readFile "p13.log") print . take 10 . show . sum $ xs

## 4 Problem 14

Find the longest sequence using a starting number under one million.

Solution: Faster solution, using an Array to memoize length of sequences :

--http://www.research.att.com/~njas/sequences/A033958 problem_14=837799

## 5 Problem 15

Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?

Solution: Here is a bit of explanation, and a few more solutions:

Each route has exactly 40 steps, with 20 of them horizontal and 20 of them vertical. We need to count how many different ways there are of choosing which steps are horizontal and which are vertical. So we have:

problem_15 = product [21..40] `div` product [2..20]

## 6 Problem 16

What is the sum of the digits of the number 2^{1000}?

Solution:

import Data.Char problem_16 = sum k where s = show (2^1000) k = map digitToInt s

## 7 Problem 17

How many letters would be needed to write all the numbers in words from 1 to 1000?

Solution:

import Char one = ["one","two","three","four","five","six","seven","eight", "nine","ten","eleven","twelve","thirteen","fourteen","fifteen", "sixteen","seventeen","eighteen", "nineteen"] ty = ["twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"] decompose x | x == 0 = [] | x < 20 = one !! (x-1) | x >= 20 && x < 100 = ty !! (firstDigit (x) - 2) ++ decompose ( x - firstDigit (x) * 10) | x < 1000 && x `mod` 100 ==0 = one !! (firstDigit (x)-1) ++ "hundred" | x > 100 && x <= 999 = one !! (firstDigit (x)-1) ++ "hundredand" ++decompose ( x - firstDigit (x) * 100) | x == 1000 = "onethousand" where firstDigit x = digitToInt . head . show $ x problem_17 = length . concatMap decompose $ [1..1000]

## 8 Problem 18

Find the maximum sum travelling from the top of the triangle to the base.

Solution:

problem_18 = head $ foldr1 g tri where f x y z = x + max y z g xs ys = zipWith3 f xs ys $ tail ys tri = [ [75], [95,64], [17,47,82], [18,35,87,10], [20,04,82,47,65], [19,01,23,75,03,34], [88,02,77,73,07,63,67], [99,65,04,28,06,16,70,92], [41,41,26,56,83,40,80,70,33], [41,48,72,33,47,32,37,16,94,29], [53,71,44,65,25,43,91,52,97,51,14], [70,11,33,28,77,73,17,78,39,68,17,57], [91,71,52,38,17,14,91,43,58,50,27,29,48], [63,66,04,68,89,53,67,30,73,16,69,87,40,31], [04,62,98,27,23,09,70,98,73,93,38,53,60,04,23]]

## 9 Problem 19

You are given the following information, but you may prefer to do some research for yourself.

- 1 Jan 1900 was a Monday.
- Thirty days has September,
- April, June and November.
- All the rest have thirty-one,
- Saving February alone,

Which has twenty-eight, rain or shine. And on leap years, twenty-nine.

- A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

Solution:

problem_19 = length . filter (== sunday) . drop 12 . take 1212 $ since1900 since1900 = scanl nextMonth monday . concat $ replicate 4 nonLeap ++ cycle (leap : replicate 3 nonLeap) nonLeap = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31] leap = 31 : 29 : drop 2 nonLeap nextMonth x y = (x + y) `mod` 7 sunday = 0 monday = 1

Here is an alternative that is simpler, but it is cheating a bit:

import Data.Time.Calendar import Data.Time.Calendar.WeekDate problem_19_v2 = length [() | y <- [1901..2000], m <- [1..12], let (_, _, d) = toWeekDate $ fromGregorian y m 1, d == 7]

## 10 Problem 20

Find the sum of digits in 100!

Solution:

numPrime x p = takeWhile(>0) [div x (p^a) | a<-[1..]] fastFactorial n = product[a^x | a <- takeWhile(<n) primes, let x = sum $ numPrime n a ] digits n | n<10 = [n] | otherwise = y:digits x where (x,y) = divMod n 10 problem_20 = sum . digits . fastFactorial $ 100