Euler problems/11 to 20

Problem 11

What is the greatest product of four numbers on the same straight line in the 20 by 20 grid?

Solution:

import System.Process
import IO
import List

slurpURL url = do
    (_,out,_,_) <- runInteractiveCommand $ "curl " ++ url
    hGetContents out
    
parse_11 src =
    let npre p = or.(zipWith (/=) p)
        clip p q xs = takeWhile (npre q) $ dropWhile (npre p) xs
        trim s =
            let (x,y) = break (== '<') s
                (_,z) = break (== '>') y
            in  if null z then x else x ++ trim (tail z)
    in  map ((map read).words.trim) $ clip "08" "</p>" $ lines src

solve_11 xss =
    let mult w x y z = w*x*y*z
        zipf f (w,x,y,z) = zipWith4 f w x y z
        zifm = zipf mult
        zifz = zipf (zipWith4 mult)
        tupl = zipf (\w x y z -> (w,x,y,z)) 
        skew (w,x,y,z) = (w, drop 1 x, drop 2 y, drop 3 z)
        sker (w,x,y,z) = skew (z,y,x,w)
        skex x = skew (x,x,x,x)
        maxl = foldr1 max
        maxf f g = maxl $ map (maxl.f) $ g xss
    in  maxl
            [ maxf (zifm.skex) id
            , maxf id          (zifz.skex)
            , maxf (zifm.skew) (tupl.skex)
            , maxf (zifm.sker) (tupl.skex) ] 

problem_11 = do
    src <- slurpURL "http://projecteuler.net/print.php?id=11"
    print $ solve_11 $ parse_11 src

Alternative, slightly easier to comprehend:

import Data.List 
 
diag b = [b !! n !! n | 
    n <- [0 .. length b - 1], 
    n < (length $transpose b)
    ]
getAllDiags f g = map f 
    [drop n . take (length g) $ g |
    n <- [1.. (length g - 1)]
    ]
problem_11 num= maximumBy (\(x, _) (y, _) -> compare x y) 
    $zip (map product allfours) allfours
    where 
    rows = num
    cols = transpose rows
    diagLs = 
        diag rows : diagup ++ diagdown
        where
        diagup = getAllDiags diag rows
        diagdown = getAllDiags diag cols
    diagRs = 
        diag (reverse rows) : diagup ++ diagdown
        where
        diagup = getAllDiags diag (reverse num)
        diagdown = getAllDiags diag (transpose $ reverse num)
    allposs = rows ++ cols ++ diagLs ++ diagRs
    allfours = [x | 
        xss <- allposs, 
        xs <- inits xss,
        x <- tails xs,
        length x == 4
        ]
 
split :: Char -> String -> [String]
split = unfoldr . split'
 
split' :: Char -> String -> Maybe (String, String)
split' c l
  | null l = Nothing
  | otherwise = Just (h, drop 1 t)
  where (h, t) = span (/=c) l
sToInt x=map ((+0).read) $split ' ' x
main=do
    a<-readFile "p11.log"
    let b=map sToInt $lines a
    print $problem_11 b

Second alternative, using Array and Arrows, for fun :

import Control.Arrow
import Data.Array

input :: String -> Array (Int,Int) Int
input = listArray ((1,1),(20,20)) . map read . words

senses = [(+1) *** id,(+1) *** (+1), id *** (+1), (+1) *** (\n -> n - 1)]

inArray a i = inRange (bounds a) i

prods :: Array (Int, Int) Int -> [Int]
prods a = [product xs | 
           i <- range $ bounds a
          , s <- senses
          , let is = take 4 $ iterate s i
          , all (inArray a) is
          , let xs = map (a!) is
          ]

main = getContents >>= print . maximum . prods . input

Problem 12

What is the first triangle number to have over five-hundred divisors?

Solution:

problem_12 = head $ filter ((> 500) . nDivisors) triangleNumbers
    where triangleNumbers = scanl1 (+) [1..]
          nDivisors n     = product $ map ((+1) . length) (group (primeFactors n))
          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 13

Find the first ten digits of the sum of one-hundred 50-digit numbers.

Solution:

sToInt =(+0).read
main=do
    a<-readFile "p13.log" 
    let b=map sToInt $lines a
    let c=take 10 $ show $ sum b
    print c

Problem 14

Find the longest sequence using a starting number under one million.

Solution:

p14s :: Integer -> [Integer]
p14s n = n : p14s' n
  where p14s' n = if n' == 1 then [1] else n' : p14s' n'
          where n' = if even n then n `div` 2 else (3*n)+1

problem_14 = fst $ head $ sortBy (\(_,x) (_,y) -> compare y x) [(x, length $ p14s x) | x <- [1 .. 999999]]

Alternate solution, illustrating use of strict folding:

import Data.List

problem_14 = j 1000000 where
    f :: Int -> Integer -> Int
    f k 1 = k
    f k n = f (k+1) $ if even n then div n 2 else 3*n + 1
    g x y = if snd x < snd y then y else x
    h x n = g x (n, f 1 n)
    j n   = fst $ foldl' h (1,1) [2..n-1]

Faster solution, using an Array to memoize length of sequences :

import Data.Array
import Data.List

syrs n = a
    where a = listArray (1,n) $ 0:[1 + syr n x | x <- [2..n]]
          syr n x = if x' <= n then a ! x' else 1 + syr n x'
              where x' = if even x then x `div` 2 else 3 * x + 1

main = print $ foldl' maxBySnd (0,0) $ assocs $ syrs 1000000
    where maxBySnd x@(_,a) y@(_,b) = if a > b then x else y

Problem 15

Starting in the top left corner in a 20 by 20 grid, how many routes are there to the bottom right corner?

Solution:

problem_15 = iterate (scanl1 (+)) (repeat 1) !! 20 !! 20

Here is a bit of explanation, and a few more solutions:

Each route has exactly 40 steps, with 20 of them horizontal and 20 of them vertical. We need to count how many different ways there are of choosing which steps are horizontal and which are vertical. So we have:

problem_15_v2 = product [21..40] `div` product [2..20]

The first solution calculates this using the clever trick of contructing Pascal's triangle along its diagonals.

Here is another solution that constructs Pascal's triangle in the usual way, row by row:

problem_15_v3 = iterate (\r -> zipWith (+) (0:r) (r++[0])) [1] !! 40 !! 20

Problem 16

What is the sum of the digits of the number 2¹⁰⁰⁰?

Solution:

import Data.Char
problem_16 = sum k
    where
    s=show $2^1000
    k=map digitToInt s

Problem 17

How many letters would be needed to write all the numbers in words from 1 to 1000?

Solution:

-- not a very concise or beautiful solution, but food for improvements :)

names = concat $
  [zip  [(0, n) | n <- [0..19]]
        ["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight"
        ,"Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen"
        ,"Sixteen", "Seventeen", "Eighteen", "Nineteen"]
  ,zip  [(1, n) | n <- [0..9]]
        ["", "Ten", "Twenty", "Thirty", "Fourty", "Fifty", "Sixty", "Seventy"
        ,"Eighty", "Ninety"]
  ,[((2,0), "")]
  ,[((2, n), look (0,n) ++ " Hundred and") | n <- [1..9]]
  ,[((3,0), "")]
  ,[((3, n), look (0,n) ++ " Thousand") | n <- [1..9]]]

look n = fromJust . lookup n $ names

spell n = unwords $ if last s == "and" then init s else s
  where
    s     = words . unwords $ map look digs'
    digs  = reverse . zip [0..] . reverse . map digitToInt . show $ n
    digs' = case lookup 1 digs of
                Just 1  ->
                  let [ten,one] = filter (\(a,_) -> a<=1) digs in
                      (digs \\ [ten,one]) ++ [(0,(snd ten)*10+(snd one))]
                otherwise -> digs

problem_17 xs = sum . map (length . filter (`notElem` " -") . spell) $ xs

This is another solution. I think it is much cleaner than the one above.

import Char

one = ["one","two","three","four","five","six","seven","eight","nine", "ten",
  "eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen", "nineteen"]
ty = ["twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"]

decompose x | x == 0                       = []
            | x < 20                       = one !! (x-1)
            | x >= 20 && x < 100           = ty !! (firstDigit (x) - 2) ++
                                             decompose ( x - firstDigit (x) * 10)
            | x < 1000 && x `mod` 100 ==0  = one !! (firstDigit (x)-1) ++ "hundred"
            | x > 100 && x <= 999          = one !! (firstDigit (x)-1) ++ "hundredand" ++ 
                                             decompose ( x - firstDigit (x) * 100)
            | x == 1000                    = "onethousand"

  where
    firstDigit x = digitToInt$head (show x)

problem_17 = length$concat (map decompose [1..1000])

Problem 18

Find the maximum sum travelling from the top of the triangle to the base.

Solution:

problem_18 = head $ foldr1 g tri where
    f x y z = x + max y z
    g xs ys = zipWith3 f xs ys $ tail ys
    tri = [
        [75],
        [95,64],
        [17,47,82],
        [18,35,87,10],
        [20,04,82,47,65],
        [19,01,23,75,03,34],
        [88,02,77,73,07,63,67],
        [99,65,04,28,06,16,70,92],
        [41,41,26,56,83,40,80,70,33],
        [41,48,72,33,47,32,37,16,94,29],
        [53,71,44,65,25,43,91,52,97,51,14],
        [70,11,33,28,77,73,17,78,39,68,17,57],
        [91,71,52,38,17,14,91,43,58,50,27,29,48],
        [63,66,04,68,89,53,67,30,73,16,69,87,40,31],
        [04,62,98,27,23,09,70,98,73,93,38,53,60,04,23]]

Problem 19

You are given the following information, but you may prefer to do some research for yourself.

• 1 Jan 1900 was a Monday.
• Thirty days has September,
• April, June and November.
• All the rest have thirty-one,
• Saving February alone,

Which has twenty-eight, rain or shine. And on leap years, twenty-nine.

• A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

Solution:

problem_19 = 
    length $ filter (== sunday) $ drop 12 $ take 1212 since1900
since1900 = scanl nextMonth monday $ concat $
            replicate 4 nonLeap ++ cycle (leap : replicate 3 nonLeap)
nonLeap = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
leap = 31 : 29 : drop 2 nonLeap
nextMonth x y = (x + y) `mod` 7
sunday = 0
monday = 1

Here is an alternative that is simpler, but it is cheating a bit:

import Data.Time.Calendar
import Data.Time.Calendar.WeekDate

problem_19_v2 = 
    length [() | 
    y <- [1901..2000], 
    m <- [1..12],
    let (_, _, d) = toWeekDate $ fromGregorian y m 1,
    d == 7
    ]

Problem 20

Find the sum of digits in 100!

Solution:

problem_20 = let fac n = product [1..n] in
             foldr ((+) . Data.Char.digitToInt) 0 $ show $ fac 100

Alternate solution, summing digits directly, which is faster than the show, digitToInt route.

dsum 0 = 0
dsum n = let ( d, m ) = n `divMod` 10 in m + ( dsum d )

problem_20' = dsum . product $ [ 1 .. 100 ]

Alternate solution, fast Factorial, which is faster than the another two.

numPrime x p=takeWhile(>0) [div x (p^a)|a<-[1..]]
merge xs@(x:xt) ys@(y:yt) = case compare x y of
    LT -> x : (merge xt ys)
    EQ -> x : (merge xt yt)
    GT -> y : (merge xs yt)
    
diff  xs@(x:xt) ys@(y:yt) = case compare x y of
    LT -> x : (diff xt ys)
    EQ -> diff xt yt
    GT -> diff xs yt
 
primes    = [2,3,5] ++ (diff [7,9..] nonprimes) 
nonprimes = foldr1 f . map g $ tail primes
    where f (x:xt) ys = x : (merge xt ys)
          g p = [ n*p | n <- [p,p+2..]]
fastFactorial n=
    product[a^x|
    a<-takeWhile(<n) primes,
    let x=sum$numPrime n a
    ]
digits n 
{-  change 123 to [3,2,1]
 -}
    |n<10=[n]
    |otherwise= y:digits x 
    where
    (x,y)=divMod n 10
problem_20= sum $ digits $fastFactorial 100