Difference between revisions of "Euler problems/21 to 30"
(→Problem 21: Clarify problem and add a solution.) 

(36 intermediate revisions by 17 users not shown)  
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−  == [http://projecteuler.net/index.php?section= 
+  == [http://projecteuler.net/index.php?section=problems&id=21 Problem 21] == 
−  Evaluate the sum of all amicable 
+  Evaluate the sum of all amicable numbers (including those with a pair number over the limit) under 10000. 
+  
+  Solution: 

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

−  Solution: 

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

<haskell> 
<haskell> 

Line 15:  Line 14:  
<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 

Line 29:  Line 28:  
</haskell> 
</haskell> 

−  == [http://projecteuler.net/index.php?section=view&id=22 Problem 22] == 

+  Here is another alternative solution that computes the sumofdivisors for the numbers by iterating over products of their factors (very fast): 

+  
+  <haskell> 

+  import Data.Array 

+  
+  max_ = 100000 

+  
+  gen 100001 = [] 

+  gen n = [(i*n,n)i < [2 .. max_ `div` n]] ++ (gen (n+1)) 

+  
+  arr = accumArray (+) 0 (0,max_) (gen 1) 

+  
+  problem_21_v3 = sum $ filter (\a > let b = (arr!a) in b /= a && (arr!b) == a) [1 .. (10000  1)] 

+  
+  </haskell> 

+  
+  == [http://projecteuler.net/index.php?section=problems&id=22 Problem 22] == 

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

Solution: 
Solution: 

<haskell> 
<haskell> 

−   apply to a list of names 

+  import Data.List 

−  problem_22 :: [String] > Int 

+  import Data.Char 

−  problem_22 = sum . zipWith (*) [ 1 .. ] . map score 

+  problem_22 = 

−  where score = sum . map ( subtract 64 . ord ) 

+  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 

</haskell> 
</haskell> 

−  == [http://projecteuler.net/index.php?section= 
+  == [http://projecteuler.net/index.php?section=problems&id=23 Problem 23] == 
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers. 
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers. 

Solution: 
Solution: 

<haskell> 
<haskell> 

−  import Data.Set hiding (filter, map) 

+  http://www.research.att.com/~njas/sequences/A048242 

−  import Data. 
+  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] 

−  problem_23 :: Integer 

+  rests x = map (x) $ takeWhile (<= x `div` 2) abunds 

−  problem_23 = sum [1..28123]  (fold (+) 0 $ abundant_sums $ abundant 28123) 

+  isSum = any (abunds_array !) . rests 

−  abundant_sums :: [Integer] > Set Integer 

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

−  abundant_sums [] = empty 

−  abundant_sums l@(x:xs) = union (fromList [x + a  a < takeWhile (\y > y <= 28123  x) l]) (abundant_sums xs) 

−  
−  abundant :: Integer > [Integer] 

−  abundant n = [a  a < [1..n], (sum $ factors a)  a > a] 

−  
−  primes :: [Integer] 

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

−  
−  primeFactors :: Integer > [Integer] 

−  primeFactors n = factor n primes 

−  where 

−  factor _ [] = [] 

−  factor m (p:ps)  p*p > m = [m] 

−   m `mod` p == 0 = p : factor (m `div` p) (p:ps) 

−   otherwise = factor m ps 

−  
−  factors :: Integer > [Integer] 

−  factors = perms . map (tail . scanl (*) 1) . group . primeFactors 

−  where 

−  perms :: (Integral a) => [[a]] > [a] 

−  perms [] = [1] 

−  perms (x:xs) = perms xs ++ concatMap (\z > map (*z) $ perms xs) x 

</haskell> 
</haskell> 

−  == [http://projecteuler.net/index.php?section= 
+  == [http://projecteuler.net/index.php?section=problems&id=24 Problem 24] == 
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9? 
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9? 

Solution: 
Solution: 

<haskell> 
<haskell> 

+  import Data.List 

+  
+  fac 0 = 1 

+  fac n = n * fac (n  1) 

perms [] _= [] 
perms [] _= [] 

−  perms xs n= 
+  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 

−  problem_24 = perms "0123456789" 999999 

</haskell> 
</haskell> 

−  == [http://projecteuler.net/index.php?section=view&id=25 Problem 25] == 

+  Or, using Data.List.permutations, 

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

−  
−  Solution: 

<haskell> 
<haskell> 

−  valid ( i, n ) = length ( show n ) == 1000 

+  import Data.List 

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

−  problem_25 = fst . head . filter valid . zip [ 1 .. ] $ fibs 

−  where fibs = 1 : 1 : 2 : zipWith (+) fibs ( tail fibs ) 

</haskell> 
</haskell> 

−  == [http://projecteuler.net/index.php?section=view&id=26 Problem 26] == 

+  Casey Hawthorne 

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

+  
+  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" 

−  Solution: 

<haskell> 
<haskell> 

−  problem_26 = fst $ maximumBy (\a b > snd a `compare` snd b) 

−  [(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 findIndex (== r') rs of 

−  Just i > i + 1 

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

−  </haskell> 

−  == [http://projecteuler.net/index.php?section=view&id=27 Problem 27] == 

+   Plan of attack. 

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

−  Solution: 

+   The "x"s are different numbers 

+   0xxxxxxxxx represents 9! = 362880 permutations/numbers 

+   1xxxxxxxxx represents 9! = 362880 permutations/numbers 

+   2xxxxxxxxx represents 9! = 362880 permutations/numbers 

−  The following is written in [http://haskell.org/haskellwiki/Literate_programming#Haskell_and_literate_programming literate Haskell]: 

−  <haskell> 

−  > import Data.List 

−  To be sure we get the maximum type checking of the compiler, 

+   20xxxxxxxx represents 8! = 40320 

−  we switch off the default type 

+   21xxxxxxxx represents 8! = 40320 

−  > default () 

+   23xxxxxxxx represents 8! = 40320 

+   24xxxxxxxx represents 8! = 40320 

+   25xxxxxxxx represents 8! = 40320 

+   26xxxxxxxx represents 8! = 40320 

+   27xxxxxxxx represents 8! = 40320 

−  Generate a list of primes. 

−  It works by filtering out numbers that are 

−  divisable by a previously found prime 

−  > primes :: [Int] 

+  module Euler where 

−  > primes = sieve (2 : [3, 5..]) 

−  > where 

−  > sieve (p:xs) = p : sieve (filter (\x > x `mod` p > 0) xs) 

−  > isPrime :: Int > Bool 

+  import Data.List 

−  > isPrime x = x `elem` (takeWhile (<= x) primes) 

+  factorial n = product [1..n] 

−  The lists of values we are going to try for a and b; 

+   lexOrder "0123456789" 1000000 "" 

−  b must be a prime, as n² + an + b is equal to b when n = 0 

−  > testRangeA :: [Int] 

+  lexOrder digits left s 

−  > testRangeA = [1000 .. 1000] 

+   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) 

−  > testRangeB :: [Int] 

+  </haskell> 

−  > testRangeB = takeWhile (< 1000) primes 

+  == [http://projecteuler.net/index.php?section=problems&id=25 Problem 25] == 

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

−  The search 

+  Solution: 

+  <haskell> 

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

+  t = 10^999 

−  > bestCoefficients :: (Int, Int, Int) 

+  problem_25 = length w 

−  > bestCoefficients = 

+  where 

−  > maximumBy (\(x, _, _) (y, _, _) > compare x y) $ 

+  w = takeWhile (< t) fibs 

−  > [f a b  a < testRangeA, b < testRangeB] 

+  </haskell> 

−  > where 

−  Generate a list of results of the quadratic formula 

−  (only the contiguous primes) 

−  wrap the result in a triple, together with a and b 

−  > f :: Int > Int > (Int, Int, Int) 

+  Casey Hawthorne 

−  > f a b = ( length $ contiguousPrimes a b 

−  > , a 

−  > , b 

−  > ) 

−  > contiguousPrimes :: Int > Int > [Int] 

+  I believe you mean the following: 

−  > contiguousPrimes a b = takeWhile isPrime (map (quadratic a b) [0..]) 

+  <haskell> 

−  The quadratic formula 

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

−  > quadratic :: Int > Int > Int > Int 

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

−  > quadratic a b n = n * n + a * n + b 

+  </haskell> 

+  == [http://projecteuler.net/index.php?section=problems&id=26 Problem 26] == 

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

−  > problem_27 = 

+  Solution: 

−  > do 

+  <haskell> 

−  > let (l, a, b) = bestCoefficients 

+  problem_26 = fst $ maximumBy (comparing snd) 

−  > 

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

−  > putStrLn $ "" 

+  where recurringCycle d = remainders d 10 [] 

−  > putStrLn $ "Problem Euler 27" 

+  remainders d 0 rs = 0 

−  > putStrLn $ "" 

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

−  > putStrLn $ "The best quadratic formula found is:" 

+  in case elemIndex r' rs of 

−  +  Just i > i + 1 

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

−  > 
+  </haskell> 
−  > putStrLn $ "" 

−  > putStrLn $ "The primes are:" 

−  > print $ take l $ contiguousPrimes a b 

−  > putStrLn $ "" 

+  == [http://projecteuler.net/index.php?section=problems&id=27 Problem 27] == 

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

+  Solution: 

+  <haskell> 

+  problem_27 = (2*a1)*(a^2a+41) 

+  where n = 1000 

+  m = head $ filter (\x>x^2x+41>n) [1..] 

+  a = m1 

</haskell> 
</haskell> 

−  == [http://projecteuler.net/index.php?section= 
+  == [http://projecteuler.net/index.php?section=problems&id=28 Problem 28] == 
What is the sum of both diagonals in a 1001 by 1001 spiral? 
What is the sum of both diagonals in a 1001 by 1001 spiral? 

Solution: 
Solution: 

<haskell> 
<haskell> 

−  corners :: Int > (Int, Int, Int, Int) 

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

−  corners i = (n*n, 1+(n*(2*m)), 2+(n*(2*m1)), 3+(n*(2*m2))) 

+  </haskell> 

−  where m = (i1) `div` 2 

−  n = 2*m+1 

−  sumcorners :: Int > Int 

+  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: 

−  sumcorners i = a+b+c+d where (a, b, c, d) = corners i 

−  sumdiags :: Int > Int 

+  <haskell> 

−  sumdiags i  even i = error "not a spiral" 

+  euler28 n = sum $ scanl (+) 0 

−  +  (1:(concatMap (replicate 4) [2,4..(n1)])) 

−   otherwise = s + sumdiags (i2) 

−  where s = sumcorners i 

−  
−  problem_28 = sumdiags 1001 

</haskell> 
</haskell> 

−  You can note that from 1 to 3 there's (+2), and such too for 5, 7 and 9, it then goes up to (+4) 4 times, and so on, adding 2 to the number to add for each level of the spiral. You can so avoid all need for multiplications and just do additions with the following code : 

+  == [http://projecteuler.net/index.php?section=problems&id=29 Problem 29] == 

−  <haskell>problem_28 = sum . scanl (+) 1 . concatMap (replicate 4) $ [2,4..1000]</haskell> 

−  
−  == [http://projecteuler.net/index.php?section=view&id=29 Problem 29] == 

How many distinct terms are in the sequence generated by a<sup>b</sup> for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100? 
How many distinct terms are in the sequence generated by a<sup>b</sup> for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100? 

Solution: 
Solution: 

<haskell> 
<haskell> 

−  problem_29 = length . group . sort $ [a^b  a < [2..100], b < [2..100]] 

+  import Control.Monad 

+  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> 

−  == [http://projecteuler.net/index.php?section= 
+  == [http://projecteuler.net/index.php?section=problems&id=30 Problem 30] == 
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits. 
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits. 

Solution: 
Solution: 

<haskell> 
<haskell> 

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

Line 237:  Line 252:  
fifth :: Integer > Integer 
fifth :: Integer > Integer 

−  fifth 
+  fifth = sum . map ((^5) . toInteger . digitToInt) . show 
problem_30 :: Integer 
problem_30 :: Integer 
Latest revision as of 15:53, 11 October 2015
Contents
Problem 21
Evaluate the sum of all amicable numbers (including those with a pair number over the limit) 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..]
Here is another alternative solution that computes the sumofdivisors for the numbers by iterating over products of their factors (very fast):
import Data.Array
max_ = 100000
gen 100001 = []
gen n = [(i*n,n)i < [2 .. max_ `div` n]] ++ (gen (n+1))
arr = accumArray (+) 0 (0,max_) (gen 1)
problem_21_v3 = sum $ filter (\a > let b = (arr!a) in b /= a && (arr!b) == a) [1 .. (10000  1)]
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
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]
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)
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)
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)
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
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
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 scanl
does the trick:
euler28 n = sum $ scanl (+) 0
(1:(concatMap (replicate 4) [2,4..(n1)]))
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]]
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]