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Do them on your own!
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== [http://projecteuler.net/index.php?section=problems&id=1 Problem 1] ==
  +
Add all the natural numbers below 1000 that are multiples of 3 or 5.
  +
  +
Two solutions using <hask>sum</hask>:
  +
<haskell>
  +
import Data.List (union)
  +
problem_1' = sum (union [3,6..999] [5,10..999])
  +
  +
problem_1 = sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]
  +
</haskell>
  +
  +
Another solution which uses algebraic relationships:
  +
  +
<haskell>
  +
problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
  +
where
  +
sumStep s n = s * sumOnetoN (n `div` s)
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sumOnetoN n = n * (n+1) `div` 2
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=2 Problem 2] ==
  +
Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million.
  +
  +
Solution:
  +
<haskell>
  +
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x]
  +
where
  +
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
  +
</haskell>
  +
  +
The following two solutions use the fact that the even-valued terms in
  +
the Fibonacci sequence themselves form a Fibonacci-like sequence
  +
that satisfies
  +
<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>.
  +
<haskell>
  +
problem_2 = sumEvenFibs $ numEvenFibsLessThan 1000000
  +
where
  +
sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
  +
evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
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numEvenFibsLessThan n =
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floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)
  +
</haskell>
  +
  +
The first two solutions work because 10^6 is small.
  +
The following solution also works for much larger numbers
  +
(up to at least 10^1000000 on my computer):
  +
<haskell>
  +
problem_2 = sumEvenFibsLessThan 1000000
  +
  +
sumEvenFibsLessThan n = (a + b - 1) `div` 2
  +
where
  +
n2 = n `div` 2
  +
(a, b) = foldr f (0,1)
  +
. takeWhile ((<= n2) . fst)
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. iterate times2E $ (1, 4)
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f x y | fst z <= n2 = z
  +
| otherwise = y
  +
where z = x `addE` y
  +
addE (a, b) (c, d) = (a*d + b*c - 4*ac, ac + b*d)
  +
where ac=a*c
  +
  +
times2E (a, b) = addE (a, b) (a, b)
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=3 Problem 3] ==
  +
Find the largest prime factor of 317584931803.
  +
  +
Solution:
  +
<haskell>
  +
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
  +
  +
primeFactors n = factor n primes
  +
where
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factor n (p:ps)
  +
| p*p > n = [n]
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| n `mod` p == 0 = p : factor (n `div` p) (p:ps)
  +
| otherwise = factor n ps
  +
  +
problem_3 = last (primeFactors 317584931803)
  +
</haskell>
  +
  +
Another solution, not using recursion, is:
  +
<haskell>
  +
problem_3 = (m !! 0) `div` (m !! 1)
  +
where
  +
m = reverse $
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takeWhile (<=n) (scanl1 (*) [ x | x <- 2:[3,5..], (n `mod` x) == 0 ])
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n = 600851475143
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=4 Problem 4] ==
  +
Find the largest palindrome made from the product of two 3-digit numbers.
  +
  +
Solution:
  +
<haskell>
  +
problem_4 =
  +
maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=5 Problem 5] ==
  +
What is the smallest number divisible by each of the numbers 1 to 20?
  +
  +
Solution:
  +
<haskell>
  +
problem_5 = foldr1 lcm [1..20]
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=6 Problem 6] ==
  +
What is the difference between the sum of the squares and the square of the sums?
  +
  +
Solution:
  +
<!--
  +
<haskell>
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fun n = a - b
  +
where
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a=(n^2 * (n+1)^2) `div` 4
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b=(n * (n+1) * (2*n+1)) `div` 6
  +
  +
problem_6 = fun 100
  +
</haskell>
  +
-->
  +
<!-- Might just be me, but I find this a LOT easier to read. Perhaps not as good mathematically, but it runs in an instant, even for n = 25000.
  +
<haskell>
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fun n = a - b
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where
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a = (sum [1..n])^2
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b = sum (map (^2) [1..n])
  +
  +
problem_6 = fun 100
  +
</haskell>
  +
-->
  +
<!-- I just made it a oneliner... -->
  +
<haskell>
  +
problem_6 = (sum [1..100])^2 - sum (map (^2) [1..100])
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=7 Problem 7] ==
  +
Find the 10001st prime.
  +
  +
Solution:
  +
<haskell>
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--primes in problem_3
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problem_7 = primes !! 10000
  +
</haskell>
  +
== [http://projecteuler.net/index.php?section=problems&id=8 Problem 8] ==
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Discover the largest product of five consecutive digits in the 1000-digit number.
  +
  +
Solution:
  +
<!--
  +
<haskell>
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import Data.Char
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groupsOf _ [] = []
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groupsOf n xs =
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take n xs : groupsOf n ( tail xs )
  +
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problem_8 x = maximum . map product . groupsOf 5 $ x
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main = do t <- readFile "p8.log"
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let digits = map digitToInt $concat $ lines t
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print $ problem_8 digits
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</haskell>
  +
-->
  +
<!-- I just cleaned up a little. -->
  +
<haskell>
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import Data.Char (digitToInt)
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import Data.List (tails)
  +
  +
problem_8 = do str <- readFile "number.txt"
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-- This line just converts our str(ing) to a list of 1000 Ints
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let number = map digitToInt (concat $ lines str)
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print $ maximum $ map (product . take 5) (tails number)
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=9 Problem 9] ==
  +
There is only one Pythagorean triplet, {''a'', ''b'', ''c''}, for which ''a'' + ''b'' + ''c'' = 1000. Find the product ''abc''.
  +
  +
Solution:
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<haskell>
  +
triplets l = [[a,b,c] | m <- [2..limit],
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n <- [1..(m-1)],
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let a = m^2 - n^2,
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let b = 2*m*n,
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let c = m^2 + n^2,
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a+b+c==l]
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where limit = floor . sqrt . fromIntegral $ l
  +
  +
problem_9 = product . head . triplets $ 1000
  +
</haskell>
  +
  +
== [http://projecteuler.net/index.php?section=problems&id=10 Problem 10] ==
  +
Calculate the sum of all the primes below one million.
  +
  +
Solution:
  +
<haskell>
  +
--primes in problem_3
  +
problem_10 = sum (takeWhile (< 1000000) primes)
  +
</haskell>

Latest revision as of 14:17, 22 October 2012

Contents

[edit] 1 Problem 1

Add all the natural numbers below 1000 that are multiples of 3 or 5.

Two solutions using
sum
:
import Data.List (union)
problem_1' = sum (union [3,6..999] [5,10..999])
 
problem_1  = sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]

Another solution which uses algebraic relationships:

problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
  where
    sumStep s n = s * sumOnetoN (n `div` s)
    sumOnetoN n = n * (n+1) `div` 2

[edit] 2 Problem 2

Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million.

Solution:

problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x]
  where
    fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

The following two solutions use the fact that the even-valued terms in the Fibonacci sequence themselves form a Fibonacci-like sequence that satisfies

evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)
.
problem_2 = sumEvenFibs $ numEvenFibsLessThan 1000000
  where
    sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
    evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
    numEvenFibsLessThan n =
              floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)

The first two solutions work because 10^6 is small. The following solution also works for much larger numbers (up to at least 10^1000000 on my computer):

problem_2 = sumEvenFibsLessThan 1000000
 
sumEvenFibsLessThan n = (a + b - 1) `div` 2
  where
    n2 = n `div` 2
    (a, b) = foldr f (0,1)
             . takeWhile ((<= n2) . fst)
             . iterate times2E $ (1, 4)
    f x y | fst z <= n2 = z
          | otherwise   = y
      where z = x `addE` y
addE (a, b) (c, d) = (a*d + b*c - 4*ac, ac + b*d)
  where ac=a*c
 
times2E (a, b) = addE (a, b) (a, b)

[edit] 3 Problem 3

Find the largest prime factor of 317584931803.

Solution:

primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
 
primeFactors n = factor n primes
  where
    factor n (p:ps) 
        | p*p > n        = [n]
        | n `mod` p == 0 = p : factor (n `div` p) (p:ps)
        | otherwise      = factor n ps
 
problem_3 = last (primeFactors 317584931803)

Another solution, not using recursion, is:

problem_3 = (m !! 0) `div` (m !! 1)
  where
    m = reverse $
        takeWhile (<=n) (scanl1 (*) [ x | x <- 2:[3,5..], (n `mod` x) == 0 ])
    n = 600851475143

[edit] 4 Problem 4

Find the largest palindrome made from the product of two 3-digit numbers.

Solution:

problem_4 =
  maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]

[edit] 5 Problem 5

What is the smallest number divisible by each of the numbers 1 to 20?

Solution:

problem_5 = foldr1 lcm [1..20]

[edit] 6 Problem 6

What is the difference between the sum of the squares and the square of the sums?

Solution:

problem_6 = (sum [1..100])^2 - sum (map (^2) [1..100])

[edit] 7 Problem 7

Find the 10001st prime.

Solution:

--primes in problem_3
problem_7 = primes !! 10000

[edit] 8 Problem 8

Discover the largest product of five consecutive digits in the 1000-digit number.

Solution:

import Data.Char (digitToInt)
import Data.List (tails)
 
problem_8 = do str <- readFile "number.txt"
               -- This line just converts our str(ing) to a list of 1000 Ints
               let number = map digitToInt (concat $ lines str)
               print $ maximum $ map (product . take 5) (tails number)

[edit] 9 Problem 9

There is only one Pythagorean triplet, {a, b, c}, for which a + b + c = 1000. Find the product abc.

Solution:

triplets l = [[a,b,c] | m <- [2..limit],
                        n <- [1..(m-1)], 
                        let a = m^2 - n^2, 
                        let b = 2*m*n, 
                        let c = m^2 + n^2,
                        a+b+c==l]
    where limit = floor . sqrt . fromIntegral $ l
 
problem_9 = product . head . triplets $ 1000

[edit] 10 Problem 10

Calculate the sum of all the primes below one million.

Solution:

--primes in problem_3
problem_10 = sum (takeWhile (< 1000000) primes)