Euler problems/21 to 30
(Removing category tags. See Talk:Euler_problems)
Revision as of 08:46, 1 December 2007
Evaluate the sum of all amicable pairs under 10000.
Solution: This is a little slow because of the naive method used to compute the divisors.
problem_21 = sum [m+n | m <- [2..9999], let n = divisorsSum ! m, amicable m n] where amicable m n = m < n && n < 10000 && divisorsSum ! n == m divisorsSum = array (1,9999) [(i, sum (divisors i)) | i <- [1..9999]] divisors n = [j | j <- [1..n `div` 2], n `mod` j == 0]
Here is an alternative using a faster way of computing the sum of divisors.
problem_21_v2 = sum [n | n <- [2..9999], let m = d n, m > 1, m < 10000, n == d m] d n = product [(p * product g - 1) `div` (p - 1) | g <- group $ primeFactors n, let p = head g ] - n primeFactors = pf primes where pf ps@(p:ps') n | p * p > n = [n] | r == 0 = p : pf ps q | otherwise = pf ps' n where (q, r) = n `divMod` p primes = 2 : filter (null . tail . primeFactors) [3,5..]
What is the total of all the name scores in the file of first names?
-- apply to a list of names problem_22 :: [String] -> Int problem_22 = sum . zipWith (*) [ 1 .. ] . map score where score = sum . map ( subtract 64 . ord )
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
import Data.Set hiding (filter, map) import Data.List (scanl, group) problem_23 :: Integer problem_23 = sum [1..28123] - (fold (+) 0 $ abundant_sums $ abundant 28123) abundant_sums :: [Integer] -> Set Integer abundant_sums  = empty abundant_sums l@(x:xs) = union (fromList [x + a | a <- takeWhile (\y -> y <= 28123 - x) l]) (abundant_sums xs) abundant :: Integer -> [Integer] abundant n = [a | a <- [1..n], (sum $ factors a) - a > a] primes :: [Integer] primes = 2 : filter ((==1) . length . primeFactors) [3,5..] primeFactors :: Integer -> [Integer] primeFactors n = factor n primes where factor _  =  factor m (p:ps) | p*p > m = [m] | m `mod` p == 0 = p : factor (m `div` p) (p:ps) | otherwise = factor m ps factors :: Integer -> [Integer] factors = perms . map (tail . scanl (*) 1) . group . primeFactors where perms :: (Integral a) => [[a]] -> [a] perms  =  perms (x:xs) = perms xs ++ concatMap (\z -> map (*z) $ perms xs) x
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
perms  _=  perms xs n= do let m=fac$(length(xs) -1) let y=div n m let x = xs!!y x:( perms ( delete x $ xs ) (mod n m)) problem_24 = perms "0123456789" 999999
What is the first term in the Fibonacci sequence to contain 1000 digits?
valid ( i, n ) = length ( show n ) == 1000 problem_25 = fst . head . filter valid . zip [ 1 .. ] $ fibs where fibs = 1 : 1 : 2 : zipWith (+) fibs ( tail fibs )
Find the value of d < 1000 for which 1/d contains the longest recurring cycle.
problem_26 = fst $ maximumBy (\a b -> snd a `compare` snd b) [(n,recurringCycle n) | n <- [1..999]] where recurringCycle d = remainders d 10  remainders d 0 rs = 0 remainders d r rs = let r' = r `mod` d in case findIndex (== r') rs of Just i -> i + 1 Nothing -> remainders d (10*r') (r':rs)
Find a quadratic formula that produces the maximum number of primes for consecutive values of n.
The following is written in literate Haskell:
> import Data.List To be sure we get the maximum type checking of the compiler, we switch off the default type > default () Generate a list of primes. It works by filtering out numbers that are divisable by a previously found prime > primes :: [Int] > primes = sieve (2 : [3, 5..]) > where > sieve (p:xs) = p : sieve (filter (\x -> x `mod` p > 0) xs) > isPrime :: Int -> Bool > isPrime x = x `elem` (takeWhile (<= x) primes) The lists of values we are going to try for a and b; b must be a prime, as n² + an + b is equal to b when n = 0 > testRangeA :: [Int] > testRangeA = [-1000 .. 1000] > testRangeB :: [Int] > testRangeB = takeWhile (< 1000) primes The search > bestCoefficients :: (Int, Int, Int) > bestCoefficients = > maximumBy (\(x, _, _) (y, _, _) -> compare x y) $ > [f a b | a <- testRangeA, b <- testRangeB] > where Generate a list of results of the quadratic formula (only the contiguous primes) wrap the result in a triple, together with a and b > f :: Int -> Int -> (Int, Int, Int) > f a b = ( length $ contiguousPrimes a b > , a > , b > ) > contiguousPrimes :: Int -> Int -> [Int] > contiguousPrimes a b = takeWhile isPrime (map (quadratic a b) [0..]) The quadratic formula > quadratic :: Int -> Int -> Int -> Int > quadratic a b n = n * n + a * n + b > problem_27 = > do > let (l, a, b) = bestCoefficients > > putStrLn $ "" > putStrLn $ "Problem Euler 27" > putStrLn $ "" > putStrLn $ "The best quadratic formula found is:" > putStrLn $ " n * n + " ++ show a ++ " * n + " ++ show b > putStrLn $ "" > putStrLn $ "The number of primes is: " ++ (show l) > putStrLn $ "" > putStrLn $ "The primes are:" > print $ take l $ contiguousPrimes a b > putStrLn $ ""
What is the sum of both diagonals in a 1001 by 1001 spiral?
corners :: Int -> (Int, Int, Int, Int) corners i = (n*n, 1+(n*(2*m)), 2+(n*(2*m-1)), 3+(n*(2*m-2))) where m = (i-1) `div` 2 n = 2*m+1 sumcorners :: Int -> Int sumcorners i = a+b+c+d where (a, b, c, d) = corners i sumdiags :: Int -> Int sumdiags i | even i = error "not a spiral" | i == 3 = s + 1 | otherwise = s + sumdiags (i-2) where s = sumcorners i problem_28 = sumdiags 1001
You can note that from 1 to 3 there's (+2), and such too for 5, 7 and 9, it then goes up to (+4) 4 times, and so on, adding 2 to the number to add for each level of the spiral. You can so avoid all need for multiplications and just do additions with the following code :
problem_28 = sum . scanl (+) 1 . concatMap (replicate 4) $ [2,4..1000]
How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
problem_29 = length . group . sort $ [a^b | a <- [2..100], b <- [2..100]]
10 Problem 30
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
import Data.Char (ord) limit :: Integer limit = snd $ head $ dropWhile (\(a,b) -> a > b) $ zip (map (9^5*) [1..]) (map (10^) [1..]) fifth :: Integer -> Integer fifth n = foldr (\a b -> (toInteger(ord a) - 48)^5 + b) 0 $ show n problem_30 :: Integer problem_30 = sum $ filter (\n -> n == fifth n) [2..limit]