# Euler problems/11 to 20

### From HaskellWiki

(→Problem 14: fourth solution does not memoize) |
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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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<haskell> |
<haskell> |
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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 |
+ | triangleNumbers = scanl1 (+) [1..] |

− | triangleNumbers = scanl1 (+) [1..] |
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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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Solution: |
Solution: |
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− | Faster solution, using an Array to memoize length of sequences : |
+ | <haskell> |

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

+ | problem_14 = j 1000000 where |
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+ | f :: Int -> Integer -> Int |
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+ | f k 1 = k |
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+ | f k n = f (k+1) $ if even n then div n 2 else 3*n + 1 |
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+ | g x y = if snd x < snd y then y else x |
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+ | h x n = g x (n, f 1 n) |
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+ | j n = fst $ foldl' h (1,1) [2..n-1] |
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+ | </haskell> |
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+ | |||

+ | Faster solution, using unboxed types and parallel computation: |
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+ | <haskell> |
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+ | import Control.Parallel |
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+ | import Data.Word |
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+ | |||

+ | collatzLen :: Int -> Word32 -> Int |
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+ | collatzLen c 1 = c |
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+ | collatzLen c n = collatzLen (c+1) $ if n `mod` 2 == 0 then n `div` 2 else 3*n+1 |
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+ | |||

+ | pmax x n = x `max` (collatzLen 1 n, n) |
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+ | |||

+ | solve xs = foldl pmax (1,1) xs |
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+ | |||

+ | main = print soln |
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+ | where |
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+ | s1 = solve [2..500000] |
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+ | s2 = solve [500001..1000000] |
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+ | soln = s2 `par` (s1 `pseq` max s1 s2) |
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+ | </haskell> |
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+ | |||

+ | Even faster solution, using an Array to memoize length of sequences : |
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<haskell> |
<haskell> |
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import Data.Array |
import Data.Array |
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import Data.List |
import Data.List |
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+ | import Data.Ord (comparing) |
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syrs n = |
syrs n = |
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main = |
main = |
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− | print $ foldl' maxBySnd (0,0) $ assocs $ syrs 1000000 |
+ | print $ maximumBy (comparing snd) $ assocs $ syrs 1000000 |

− | where |
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− | maxBySnd x@(_,a) y@(_,b) = if a > b then x else y |
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</haskell> |
</haskell> |
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+ | |||

+ | <!-- |
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+ | This is a trivial solution without any memoization, right? |
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+ | |||

+ | Using a list to memoize the lengths |
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+ | |||

+ | <haskell> |
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+ | import Data.List |
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+ | |||

+ | -- computes the sequence for a given n |
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+ | l n = n:unfoldr f n where |
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+ | f 1 = Nothing -- we're done here |
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+ | -- for reasons of speed we do div and mod in one go |
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+ | f n = let (d,m)=divMod n 2 in case m of |
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+ | 0 -> Just (d,d) -- n was even |
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+ | otherwise -> let k = 3*n+1 in Just (k,k) -- n was odd |
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+ | |||

+ | |||

+ | answer = foldl1' f $ -- computes the maximum of a list of tuples |
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+ | -- save the length of the sequence and the number generating it in a tuple |
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+ | [(length $! l x, x) | x <- [1..1000000]] where |
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+ | f (a,c) (b,d) -- one tuple is greater than other if the first component (=sequence-length) is greater |
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+ | | a > b = (a,c) |
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+ | | otherwise = (b,d) |
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+ | |||

+ | main = print answer |
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+ | </haskell> |
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+ | --> |
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== [http://projecteuler.net/index.php?section=problems&id=15 Problem 15] == |
== [http://projecteuler.net/index.php?section=problems&id=15 Problem 15] == |
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Solution: |
Solution: |
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− | Here is a bit of explanation, and a few more solutions: |
+ | A direct computation: |

+ | <haskell> |
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+ | problem_15 = iterate (scanl1 (+)) (repeat 1) !! 20 !! 20 |
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+ | </haskell> |
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+ | |||

+ | Thinking about it as a problem in combinatorics: |
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Each route has exactly 40 steps, with 20 of them horizontal and 20 of |
Each route has exactly 40 steps, with 20 of them horizontal and 20 of |
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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> |
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− | |||

− | The first solution calculates this using the clever trick of contructing |
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− | [http://en.wikipedia.org/wiki/Pascal's_triangle Pascal's triangle] |
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− | along its diagonals. |
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− | |||

− | Here is another solution that constructs Pascal's triangle in the usual way, |
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− | row by row: |
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− | |||

− | <haskell> |
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− | problem_15_v2 = |
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− | iterate (\r -> zipWith (+) (0:r) (r++[0])) [1] !! 40 !! 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..]] |
+ | problem_20 = sum $ map Char.digitToInt $ show $ product [1..100] |

− | fastFactorial n= |
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− | product[a^x| |
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− | a<-takeWhile(<n) primes, |
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− | let x=sum$numPrime n a |
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− | ] |
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− | digits n |
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− | |n<10=[n] |
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− | |otherwise= y:digits x |
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− | where |
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− | (x,y)=divMod n 10 |
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− | problem_20= sum $ digits $fastFactorial 100 |
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</haskell> |
</haskell> |

## Latest revision as of 14:07, 2 December 2011

## Contents |

## [edit] 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

## [edit] 2 Problem 12

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

Solution:

--primeFactors in problem_3 problem_12 = head $ filter ((> 500) . nDivisors) triangleNumbers where nDivisors n = product $ map ((+1) . length) (group (primeFactors n)) triangleNumbers = scanl1 (+) [1..]

## [edit] 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

## [edit] 4 Problem 14

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

Solution:

import Data.List problem_14 = j 1000000 where f :: Int -> Integer -> Int f k 1 = k f k n = f (k+1) $ if even n then div n 2 else 3*n + 1 g x y = if snd x < snd y then y else x h x n = g x (n, f 1 n) j n = fst $ foldl' h (1,1) [2..n-1]

Faster solution, using unboxed types and parallel computation:

import Control.Parallel import Data.Word collatzLen :: Int -> Word32 -> Int collatzLen c 1 = c collatzLen c n = collatzLen (c+1) $ if n `mod` 2 == 0 then n `div` 2 else 3*n+1 pmax x n = x `max` (collatzLen 1 n, n) solve xs = foldl pmax (1,1) xs main = print soln where s1 = solve [2..500000] s2 = solve [500001..1000000] soln = s2 `par` (s1 `pseq` max s1 s2)

Even faster solution, using an Array to memoize length of sequences :

import Data.Array import Data.List import Data.Ord (comparing) syrs n = a where a = listArray (1,n) $ 0:[1 + syr n x | x <- [2..n]] syr n x = if x' <= n then a ! x' else 1 + syr n x' where x' = if even x then x `div` 2 else 3 * x + 1 main = print $ maximumBy (comparing snd) $ assocs $ syrs 1000000

## [edit] 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: A direct computation:

problem_15 = iterate (scanl1 (+)) (repeat 1) !! 20 !! 20

Thinking about it as a problem in combinatorics:

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]

## [edit] 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

## [edit] 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]

## [edit] 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]]

## [edit] 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]

## [edit] 10 Problem 20

Find the sum of digits in 100!

Solution:

problem_20 = sum $ map Char.digitToInt $ show $ product [1..100]