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[[Category:Programming exercise spoilers]]
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== [http://projecteuler.net/index.php?section=problems&id=1 Problem 1] ==
-
== [http://projecteuler.net/index.php?section=view&id=1 Problem 1] ==
+
Add all the natural numbers below 1000 that are multiples of 3 or 5.
Add all the natural numbers below 1000 that are multiples of 3 or 5.
-
Solution:
+
Two solutions using <hask>sum</hask>:
<haskell>
<haskell>
-
problem_1 = sum [ x | x <- [1..999], (x `mod` 3 == 0) || (x `mod` 5 == 0)]
+
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>
</haskell>
 +
 +
Another solution which uses algebraic relationships:
<haskell>
<haskell>
-
problem_1_v2 = sum $ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999]
+
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
</haskell>
</haskell>
-
----
 
-
<haskell>
 
-
sum1to n = n * (n+1) `div` 2
 
-
problem_1_v3 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
+
== [http://projecteuler.net/index.php?section=problems&id=2 Problem 2] ==
-
where sumStep s n = s * sum1to (n `div` s)
+
-
</haskell>
+
-
 
+
-
== [http://projecteuler.net/index.php?section=view&id=2 Problem 2] ==
+
Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million.
Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million.
Solution:
Solution:
<haskell>
<haskell>
-
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, x `mod` 2 == 0]
+
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x]
-
where fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
+
where
 +
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
</haskell>
</haskell>
Line 33: Line 34:
<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>.
<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>.
<haskell>
<haskell>
-
problem_2_v2 = sumEvenFibs $ numEvenFibsLessThan 1000000
+
problem_2 = sumEvenFibs $ numEvenFibsLessThan 1000000
-
sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
+
where
-
evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
+
sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
-
numEvenFibsLessThan n =
+
evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
-
floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)
+
numEvenFibsLessThan n =
 +
floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)
</haskell>
</haskell>
Line 44: Line 46:
(up to at least 10^1000000 on my computer):
(up to at least 10^1000000 on my computer):
<haskell>
<haskell>
-
problem_2_v3 = sumEvenFibsLessThan 1000000
+
problem_2 = sumEvenFibsLessThan 1000000
 +
 
sumEvenFibsLessThan n = (a + b - 1) `div` 2
sumEvenFibsLessThan n = (a + b - 1) `div` 2
where
where
n2 = n `div` 2
n2 = n `div` 2
-
(a, b) = foldr f (0,1) $ takeWhile ((<= n2) . fst) $ iterate times2E (1, 4)
+
(a, b) = foldr f (0,1)
 +
. takeWhile ((<= n2) . fst)
 +
. iterate times2E $ (1, 4)
f x y | fst z <= n2 = z
f x y | fst z <= n2 = z
| otherwise = y
| otherwise = y
where z = x `addE` y
where z = x `addE` y
-
addE (a, b) (c, d) = let ac = a*c in (a*d + b*c - 4*ac, ac + b*d)
+
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)
times2E (a, b) = addE (a, b) (a, b)
</haskell>
</haskell>
-
== [http://projecteuler.net/index.php?section=view&id=3 Problem 3] ==
+
== [http://projecteuler.net/index.php?section=problems&id=3 Problem 3] ==
Find the largest prime factor of 317584931803.
Find the largest prime factor of 317584931803.
Line 62: Line 69:
<haskell>
<haskell>
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
 +
primeFactors n = factor n primes
primeFactors n = factor n primes
-
where factor n (p:ps) | p*p > n = [n]
+
where
-
| n `mod` p == 0 = p : factor (n `div` p) (p:ps)
+
factor n (p:ps)
-
| otherwise = factor n 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)
problem_3 = last (primeFactors 317584931803)
</haskell>
</haskell>
-
This can be improved by using
+
Another solution, not using recursion, is:
-
<hask>null . tail</hask>
+
<haskell>
-
instead of
+
problem_3 = (m !! 0) `div` (m !! 1)
-
<hask>(== 1) . length</hask>.
+
where
 +
m = reverse $
 +
takeWhile (<=n) (scanl1 (*) [ x | x <- 2:[3,5..], (n `mod` x) == 0 ])
 +
n = 600851475143
 +
</haskell>
-
== [http://projecteuler.net/index.php?section=view&id=4 Problem 4] ==
+
== [http://projecteuler.net/index.php?section=problems&id=4 Problem 4] ==
Find the largest palindrome made from the product of two 3-digit numbers.
Find the largest palindrome made from the product of two 3-digit numbers.
Solution:
Solution:
<haskell>
<haskell>
-
problem_4 = foldr max 0 [ x | y <- [100..999], z <- [100..999], let x = y * z, let s = show x, s == reverse s]
+
problem_4 =
-
</haskell>
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maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]
-
An alternative to avoid evaluating twice the same pair of numbers:
+
-
<haskell>
+
-
problem_4' = foldr1 max [ x | y <- [100..999], z <- [y..999], let x = y * z, let s = show x, s == reverse s]
+
</haskell>
</haskell>
-
== [http://projecteuler.net/index.php?section=view&id=5 Problem 5] ==
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== [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?
What is the smallest number divisible by each of the numbers 1 to 20?
Solution:
Solution:
<haskell>
<haskell>
-
problem_5 = head [ x | x <- [2520,5040..], all (\y -> x `mod` y == 0) [1..20]]
+
problem_5 = foldr1 lcm [1..20]
-
</haskell>
+
-
An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom:
+
-
<haskell>
+
-
problem_5' = foldr1 lcm [1..20]
+
</haskell>
</haskell>
-
== [http://projecteuler.net/index.php?section=view&id=6 Problem 6] ==
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== [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?
What is the difference between the sum of the squares and the square of the sums?
Solution:
Solution:
 +
<!--
<haskell>
<haskell>
-
problem_6 = sum [ x^2 | x <- [1..100]] - (sum [1..100])^2
+
fun n = a - b
 +
where
 +
a=(n^2 * (n+1)^2) `div` 4
 +
b=(n * (n+1) * (2*n+1)) `div` 6
 +
 
 +
problem_6 = fun 100
</haskell>
</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>
 +
fun n = a - b
 +
where
 +
a = (sum [1..n])^2
 +
b = sum (map (^2) [1..n])
-
== [http://projecteuler.net/index.php?section=view&id=7 Problem 7] ==
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problem_6 = fun 100
-
Find the 10001st prime.
+
</haskell>
-
 
+
-->
-
Solution:
+
<!-- I just made it a oneliner... -->
<haskell>
<haskell>
-
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
+
problem_6 = (sum [1..100])^2 - sum (map (^2) [1..100])
-
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_7 = head $ drop 10000 primes
+
</haskell>
</haskell>
-
As above, this can be improved by using
+
== [http://projecteuler.net/index.php?section=problems&id=7 Problem 7] ==
-
<hask>null . tail</hask>
+
Find the 10001st prime.
-
instead of
+
-
<hask>(== 1) . length</hask>.
+
-
 
+
-
Here is an alternative that uses a
+
-
[http://www.haskell.org/pipermail/haskell-cafe/2007-February/022854.html sieve of Eratosthenes]:
+
 +
Solution:
<haskell>
<haskell>
-
primes' = 2 : 3 : sieve (tail primes') [5,7..]
+
--primes in problem_3
-
where
+
problem_7 = primes !! 10000
-
sieve (p:ps) x = let (h, _:t) = span (p*p <) x
+
-
in h ++ sieve ps (filter (\q -> q `mod` p /= 0) t
+
-
problem_7_v2 = primes' !! 10000
+
</haskell>
</haskell>
-
 
+
== [http://projecteuler.net/index.php?section=problems&id=8 Problem 8] ==
-
== [http://projecteuler.net/index.php?section=view&id=8 Problem 8] ==
+
Discover the largest product of five consecutive digits in the 1000-digit number.
Discover the largest product of five consecutive digits in the 1000-digit number.
Solution:
Solution:
 +
<!--
<haskell>
<haskell>
-
num = ... -- 1000 digit number as a string
+
import Data.Char
-
digits = map digitToInt num
+
-
 
+
groupsOf _ [] = []
groupsOf _ [] = []
-
groupsOf n xs = take n xs : groupsOf n ( tail xs )
+
groupsOf n xs =
-
 
+
take n xs : groupsOf n ( tail xs )
-
problem_8 = maximum . map product . groupsOf 5 $ digits
+
 +
problem_8 x = maximum . map product . groupsOf 5 $ x
 +
main = do t <- readFile "p8.log"
 +
let digits = map digitToInt $concat $ lines t
 +
print $ problem_8 digits
 +
</haskell>
 +
-->
 +
<!-- I just cleaned up a little. -->
 +
<haskell>
 +
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)
</haskell>
</haskell>
-
== [http://projecteuler.net/index.php?section=view&id=9 Problem 9] ==
+
== [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''.
There is only one Pythagorean triplet, {''a'', ''b'', ''c''}, for which ''a'' + ''b'' + ''c'' = 1000. Find the product ''abc''.
Solution:
Solution:
<haskell>
<haskell>
-
problem_9 = head [a*b*c | a <- [1..500], b <- [a..500], let c = 1000-a-b, a^2 + b^2 == c^2]
+
triplets l = [[a,b,c] | m <- [2..limit],
-
</haskell>
+
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
-
Another solution using Pythagorean Triplets generation:
+
problem_9 = product . head . triplets $ 1000
-
<haskell>
+
-
triplets :: Int -> [(Int, Int, Int)]
+
-
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]
+
-
where limit = floor $ sqrt $ fromIntegral l
+
-
 
+
-
tripletWithLength :: Int -> [(Int, Int, Int)]
+
-
tripletWithLength n = filter ((==n) . f) $ triplets n
+
-
where
+
-
f (a,b,c) = a+b+c
+
-
 
+
-
problem_9 :: Int
+
-
problem_9 = prod3 $ head $ tripletWithLength 1000
+
-
where
+
-
prod3 (a,b,c) = a*b*c
+
</haskell>
</haskell>
-
== [http://projecteuler.net/index.php?section=view&id=10 Problem 10] ==
+
== [http://projecteuler.net/index.php?section=problems&id=10 Problem 10] ==
Calculate the sum of all the primes below one million.
Calculate the sum of all the primes below one million.
Solution:
Solution:
<haskell>
<haskell>
 +
--primes in problem_3
problem_10 = sum (takeWhile (< 1000000) primes)
problem_10 = sum (takeWhile (< 1000000) primes)
</haskell>
</haskell>
-
 
-
 
-
[[Category:Tutorials]]
 
-
[[Category:Code]]
 

Current revision

Contents

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

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)

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

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]

5 Problem 5

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

Solution: 

problem_5 = foldr1 lcm [1..20]

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])

7 Problem 7

Find the 10001st prime.

Solution: 

--primes in problem_3
problem_7 = primes !! 10000

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)

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

10 Problem 10

Calculate the sum of all the primes below one million.

Solution: 

--primes in problem_3
problem_10 = sum (takeWhile (< 1000000) primes)
Retrieved from "http://www.haskell.org/haskellwiki/Euler_problems/1_to_10"