# Euler problems/1 to 10

### From HaskellWiki

(Added problem2_v2) |
(Use null.tail or sieve to find primes) |
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problem_3 = last (primeFactors 317584931803) | problem_3 = last (primeFactors 317584931803) | ||

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

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| otherwise = factor n ps | | otherwise = factor n ps | ||

problem_7 = head $ drop 10000 primes | 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 = 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> | ||

## Revision as of 17:25, 23 September 2007

## 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]

sum1to n = n * (n+1) `div` 2 problem_1_v3 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999 where sumStep s n = s * sum1to (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)

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)

This works because 10^6 is small. To work with large numbers, where using Double is not possible, you can define evenFib and numEvenFibsLessThan by using the identities:

2 * evenFib (2*n+1) == evenFib n ^2 + evenFib (n+1) ^ 2 2 * evenFib (2*n) == evenFib n * evenFib (n+1) - 2 * evenFib n ^ 2

## 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 = 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_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 = let (h, _:t) = span (p*p <) x in h ++ sieve ps (filter (\q -> q `mod` p /= 0) t problem_7_v2 = primes' !! 10000

## 8 Problem 8

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

Solution:

num = ... -- 1000 digit number as a string digits = map digitToInt num groupsOf _ [] = [] groupsOf n xs = take n xs : groupsOf n ( tail xs ) problem_8 = maximum . map product . groupsOf 5 $ 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 :: 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

## 10 Problem 10

Calculate the sum of all the primes below one million.

Solution:

problem_10 = sum (takeWhile (< 1000000) primes)