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Euler problems/21 to 30

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(→Problem 25: restore old solution to problem 25 to avoid gratuitous complexity)
Current revision (03:52, 14 November 2011) (edit) (undo)
(→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)
 +
 
 +
This is a little slow because of the naive method used to compute the divisors.
 +
<haskell>
 +
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]
 +
</haskell>
 +
 
 +
Here is an alternative using a faster way of computing the sum of divisors.
<haskell>
<haskell>
-
--http://www.research.att.com/~njas/sequences/A063990
+
problem_21_v2 = sum [n | n <- [2..9999], let m = d n,
-
problem_21 = sum [220, 284, 1184, 1210, 2620, 2924, 5020, 5564, 6232, 6368]
+
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..]
</haskell>
</haskell>
Line 19: 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 38: 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 57: 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!!(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 64: 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 75: Line 164:
Solution:
Solution:
<haskell>
<haskell>
-
problem_26 = head [a | a<-[999,997..], and [isPrime a, isPrime $ a `div` 2]]
+
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>
Line 95: Line 191:
<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>
Line 104: Line 207:
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 111: Line 237:
Solution:
Solution:
<haskell>
<haskell>
-
--http://www.research.att.com/~njas/sequences/A052464
+
import Data.Char (digitToInt)
-
problem_30 = sum [4150, 4151, 54748, 92727, 93084, 194979]
+
 
 +
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]
</haskell>
</haskell>

Current revision

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
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..(n-1)]))

9 Problem 29

How many distinct terms are in the sequence generated by ab 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]
Retrieved from "http://www.haskell.org/haskellwiki/Euler_problems/21_to_30"