Euler problems/21 to 30
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(Fixed version 2 to have proper solution. Numbers whose divisor sum equals themselves are not amicable numbers.) 
(→Problem 24: Added another method for Problem 24) 

(16 intermediate revisions by 8 users not shown)  
Line 42:  Line 42:  
let names = sort $ read$"["++ input++"]" 
let names = sort $ read$"["++ input++"]" 

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

−  print . show . sum $ scores 
+  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> 

Line 61:  Line 61:  
isSum = any (abunds_array !) . rests 
isSum = any (abunds_array !) . rests 

−  problem_23 = putStrLn . show . foldl1 (+) . filter (not . isSum) $ [1..n] 
+  problem_23 = print . sum . filter (not . isSum) $ [1..n] 
</haskell> 
</haskell> 

Line 80:  Line 80:  
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!!(quot1)))) rem (s ++ [(digits!!(quot1))]) 

+   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> 

−  valid ( i, n ) = length ( show n ) == 1000 
+  fibs = 0:1:(zipWith (+) fibs (tail fibs)) 
−  +  t = 10^999 

−  problem_25 = fst . head . filter valid . zip [ 1 .. ] $ fibs 
+  
−  where fibs = 1 : 1 : 2 : zipWith (+) fibs ( tail 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> 

Line 98:  Line 150:  
Solution: 
Solution: 

<haskell> 
<haskell> 

−  problem_26 = fst $ maximumBy (\a b > snd a `compare` snd b) 
+  problem_26 = fst $ maximumBy (comparing snd) 
[(n,recurringCycle n)  n < [1..999]] 
[(n,recurringCycle n)  n < [1..999]] 

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

remainders d 0 rs = 0 
remainders d 0 rs = 0 

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

−  in case findIndex (== r') rs of 
+  in case elemIndex r' rs of 
Just i > i + 1 
Just i > i + 1 

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

Line 125:  Line 177:  
<haskell> 
<haskell> 

problem_28 = sum (map (\n > 4*(n2)^2+10*(n1)) [3,5..1001]) + 1 
problem_28 = sum (map (\n > 4*(n2)^2+10*(n1)) [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..(n1)])) 

</haskell> 
</haskell> 

Line 134:  Line 193:  
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> 

Line 141:  Line 223:  
Solution: 
Solution: 

<haskell> 
<haskell> 

−  import Data.Char (ord) 
+  import Data.Char (digitToInt) 
limit :: Integer 
limit :: Integer 

Line 147:  Line 229:  
fifth :: Integer > Integer 
fifth :: Integer > Integer 

−  fifth n = foldr (\a b > (toInteger(ord a)  48)^5 + b) 0 $ show n 
+  fifth = sum . map ((^5) . toInteger . digitToInt) . show 
problem_30 :: Integer 
problem_30 :: Integer 
Latest revision as of 03:52, 14 November 2011
Contents 
[edit] 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..]
[edit] 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
[edit] 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]
[edit] 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!!(quot1)))) rem (s ++ [(digits!!(quot1))])  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)
[edit] 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)
[edit] 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)
[edit] 7 Problem 27
Find a quadratic formula that produces the maximum number of primes for consecutive values of n.
Solution:
problem_27 = (2*a1)*(a^2a+41) where n = 1000 m = head $ filter (\x>x^2x+41>n) [1..] a = m1
[edit] 8 Problem 28
What is the sum of both diagonals in a 1001 by 1001 spiral?
Solution:
problem_28 = sum (map (\n > 4*(n2)^2+10*(n1)) [3,5..1001]) + 1
euler28 n = sum $ scanl (+) 0 (1:(concatMap (replicate 4) [2,4..(n1)]))
[edit] 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]]
[edit] 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]