Difference between revisions of "Euler problems/1 to 10"
Jump to navigation
Jump to search
Marypoppins (talk | contribs) |
|||
| Line 1: | Line 1: | ||
| + | Do them on your own! |
||
| − | == [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 21:39, 29 January 2008
Do them on your own!