Euler problems/21 to 30

Problem 21

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]

Problem 22

What is the total of all the name scores in the file of first names?

Solution:

-- apply to a list of names
problem_22 :: [String] -> Int
problem_22 = sum . zipWith (*) [ 1 .. ] . map score
    where score = sum . map ( subtract 64 . ord )

Problem 23

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Solution:

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 [] = [1]
        perms (x:xs) = perms xs ++ concatMap (\z -> map (*z) $ perms xs) x

Problem 24

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

Solution:

perms [] = [[]]
perms xs = do
    x <- xs
    map ( x: ) ( perms . delete x $ xs )

problem_24 = ( perms "0123456789" ) !! 999999

Problem 25

What is the first term in the Fibonacci sequence to contain 1000 digits?

Solution:

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 )

Problem 26

Find the value of d < 1000 for which 1/d contains the longest recurring cycle.

Solution:

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)

Problem 27

Find a quadratic formula that produces the maximum number of primes for consecutive values of n.

Solution:

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 $ ""

Problem 28

What is the sum of both diagonals in a 1001 by 1001 spiral?

Solution:

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..n-1]

Problem 29

How many distinct terms are in the sequence generated by a^b for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

Solution:

problem_29 = length . group . sort $ [a^b | a <- [2..100], b <- [2..100]]

Problem 30

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

Solution:

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]