# Euler problems/1 to 10

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− | + | == [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. | ||

+ | |||

+ | Solution: | ||

+ | <haskell> | ||

+ | problem_1 = | ||

+ | sum [ x | | ||

+ | x <- [1..999], | ||

+ | (x `mod` 3 == 0) || (x `mod` 5 == 0) | ||

+ | ] | ||

+ | </haskell> | ||

+ | |||

+ | <haskell> | ||

+ | problem_1_v2 = | ||

+ | sum $ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999] | ||

+ | </haskell> | ||

+ | ---- | ||

+ | <haskell> | ||

+ | sumOnetoN n = n * (n+1) `div` 2 | ||

+ | |||

+ | problem_1 = | ||

+ | sumStep 3 999 + sumStep 5 999 - sumStep 15 999 | ||

+ | where | ||

+ | sumStep s n = s * sumOnetoN (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. | ||

+ | |||

+ | Solution: | ||

+ | <haskell> | ||

+ | problem_2 = | ||

+ | sum [ x | | ||

+ | x <- takeWhile (<= 1000000) fibs, | ||

+ | x `mod` 2 == 0 | ||

+ | ] | ||

+ | 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_v2 = | ||

+ | sumEvenFibs $ numEvenFibsLessThan 1000000 | ||

+ | 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) | ||

+ | </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) $ | ||

+ | 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) | ||

+ | </haskell> | ||

+ | |||

+ | == [http://projecteuler.net/index.php?section=view&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 | ||

+ | 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) | ||

+ | </haskell> | ||

+ | |||

+ | This can be improved by using | ||

+ | <hask>null . tail</hask> | ||

+ | instead of | ||

+ | <hask>(== 1) . length</hask>. | ||

+ | |||

+ | == [http://projecteuler.net/index.php?section=view&id=4 Problem 4] == | ||

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

+ | |||

+ | Solution: | ||

+ | <haskell> | ||

+ | problem_4 = | ||

+ | foldr max 0 [ x | | ||

+ | y <- [100..999], | ||

+ | z <- [100..999], | ||

+ | let x = y * z, | ||

+ | let s = show x, | ||

+ | s == reverse s | ||

+ | ] | ||

+ | </haskell> | ||

+ | 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> | ||

+ | |||

+ | == [http://projecteuler.net/index.php?section=view&id=5 Problem 5] == | ||

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

+ | |||

+ | Solution: | ||

+ | <haskell> | ||

+ | problem_5 = | ||

+ | head [ x | | ||

+ | x <- [2520,5040..], | ||

+ | all (\y -> x `mod` y == 0) [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> | ||

+ | |||

+ | == [http://projecteuler.net/index.php?section=view&id=6 Problem 6] == | ||

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

+ | |||

+ | Solution: | ||

+ | <haskell> | ||

+ | problem_6 = | ||

+ | sum [ x^2 | x <- [1..100]] - (sum [1..100])^2 | ||

+ | </haskell> | ||

+ | |||

+ | == [http://projecteuler.net/index.php?section=view&id=7 Problem 7] == | ||

+ | Find the 10001st prime. | ||

+ | |||

+ | Solution: | ||

+ | <haskell> | ||

+ | --primes in problem_3 | ||

+ | problem_7 = | ||

+ | head $ drop 10000 primes | ||

+ | </haskell> | ||

+ | |||

+ | As above, this can be improved by using | ||

+ | <hask>null . tail</hask> | ||

+ | 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]: | ||

+ | |||

+ | <haskell> | ||

+ | primes' = | ||

+ | 2 : 3 : sieve (tail primes') [5,7..] | ||

+ | where | ||

+ | sieve (p:ps) x = | ||

+ | h ++ sieve ps (filter (\q -> q `mod` p /= 0) t | ||

+ | where | ||

+ | (h, _:t) = span (p*p <) x | ||

+ | problem_7_v2 = primes' !! 10000 | ||

+ | </haskell> | ||

+ | |||

+ | == [http://projecteuler.net/index.php?section=view&id=8 Problem 8] == | ||

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

+ | |||

+ | Solution: | ||

+ | <haskell> | ||

+ | import Data.Char | ||

+ | groupsOf _ [] = [] | ||

+ | groupsOf n xs = | ||

+ | take n xs : groupsOf n ( tail xs ) | ||

+ | |||

+ | problem_8 x= | ||

+ | maximum . map product . groupsOf 5 $ x | ||

+ | main=do | ||

+ | t<-readFile "p8.log" | ||

+ | let digits = map digitToInt $foldl (++) "" $ lines t | ||

+ | print $ problem_8 digits | ||

+ | </haskell> | ||

+ | |||

+ | == [http://projecteuler.net/index.php?section=view&id=9 Problem 9] == | ||

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

+ | |||

+ | Solution: | ||

+ | <haskell> | ||

+ | problem_9 = | ||

+ | head [a*b*c | | ||

+ | a <- [1..500], | ||

+ | b <- [a..500], | ||

+ | let c = 1000-a-b, | ||

+ | a^2 + b^2 == c^2 | ||

+ | ] | ||

+ | </haskell> | ||

+ | |||

+ | Another solution using Pythagorean Triplets generation: | ||

+ | <haskell> | ||

+ | 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 | ||

+ | </haskell> | ||

+ | |||

+ | == [http://projecteuler.net/index.php?section=view&id=10 Problem 10] == | ||

+ | Calculate the sum of all the primes below one million. | ||

+ | |||

+ | Solution: | ||

+ | <haskell> | ||

+ | problem_10 = | ||

+ | sum (takeWhile (< 1000000) primes) | ||

+ | </haskell> |

## Revision as of 04:53, 30 January 2008

## Contents |

## 1 Problem 1

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

Solution:

problem_1 = sum [ x | x <- [1..999], (x `mod` 3 == 0) || (x `mod` 5 == 0) ]

problem_1_v2 = sum $ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999]

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

## 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, x `mod` 2 == 0 ] 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

problem_2_v2 = sumEvenFibs $ numEvenFibsLessThan 1000000 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)

This can be improved by using

instead of

## 4 Problem 4

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

Solution:

problem_4 = foldr max 0 [ x | y <- [100..999], z <- [100..999], let x = y * z, let s = show x, s == reverse s ]

An alternative to avoid evaluating twice the same pair of numbers:

problem_4' = foldr1 max [ 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 = head [ x | x <- [2520,5040..], all (\y -> x `mod` y == 0) [1..20] ]

An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom:

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 [ x^2 | x <- [1..100]] - (sum [1..100])^2

## 7 Problem 7

Find the 10001st prime.

Solution:

--primes in problem_3 problem_7 = head $ drop 10000 primes

As above, this can be improved by using

instead of

Here is an alternative that uses a sieve of Eratosthenes:

primes' = 2 : 3 : sieve (tail primes') [5,7..] where sieve (p:ps) x = h ++ sieve ps (filter (\q -> q `mod` p /= 0) t where (h, _:t) = span (p*p <) x problem_7_v2 = primes' !! 10000

## 8 Problem 8

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

Solution:

import Data.Char groupsOf _ [] = [] groupsOf n xs = take n xs : groupsOf n ( tail xs ) problem_8 x= maximum . map product . groupsOf 5 $ x main=do t<-readFile "p8.log" let digits = map digitToInt $foldl (++) "" $ lines t print $ problem_8 digits

## 9 Problem 9

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

Solution:

problem_9 = head [a*b*c | a <- [1..500], b <- [a..500], let c = 1000-a-b, a^2 + b^2 == c^2 ]

Another solution using Pythagorean Triplets generation:

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:

problem_10 = sum (takeWhile (< 1000000) primes)