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[[Category:Programming exercise spoilers]]
 
 
== [http://projecteuler.net/index.php?section=problems&id=11 Problem 11] ==
 
== [http://projecteuler.net/index.php?section=problems&id=11 Problem 11] ==
What is the greatest product of four numbers on the same straight line in the [http://projecteuler.net/index.php?section=view&id=11 20 by 20 grid]?
+
What is the greatest product of four numbers on the same straight line in the [http://projecteuler.net/index.php?section=problems&id=11 20 by 20 grid]?
   
 
Solution:
 
Solution:
  +
using Array and Arrows, for fun :
 
<haskell>
 
<haskell>
problem_11 = undefined
+
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 = print . maximum . prods . input =<< getContents
 
</haskell>
 
</haskell>
   
Line 13: Line 13:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
  +
--primeFactors in problem_3
 
problem_12 = head $ filter ((> 500) . nDivisors) triangleNumbers
 
problem_12 = head $ filter ((> 500) . nDivisors) triangleNumbers
where triangleNumbers = scanl1 (+) [1..]
+
where nDivisors n = product $ map ((+1) . length) (group (primeFactors n))
nDivisors n = product $ map ((+1) . length) (group (primeFactors n))
+
triangleNumbers = scanl1 (+) [1..]
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
 
 
</haskell>
 
</haskell>
   
Line 23: Line 24:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
nums = ... -- put the numbers in a list
+
problem_13 = take 10 . show . sum $ nums
+
main = do xs <- fmap (map read . lines) (readFile "p13.log")
  +
print . take 10 . show . sum $ xs
 
</haskell>
 
</haskell>
   
Line 31: Line 32:
   
 
Solution:
 
Solution:
  +
<haskell>
  +
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]
  +
</haskell>
  +
  +
Faster solution, using unboxed types and parallel computation:
 
<haskell>
 
<haskell>
p14s :: Integer -> [Integer]
+
import Control.Parallel
p14s n = n : p14s' n
+
import Data.Word
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
+
collatzLen :: Int -> Word32 -> Int
  +
collatzLen c 1 = c
  +
collatzLen c n = collatzLen (c+1) $ if n `mod` 2 == 0 then n `div` 2 else 3*n+1
  +
  +
pmax x n = x `max` (collatzLen 1 n, n)
   
problem_14 = fst $ head $ sortBy (\(_,x) (_,y) -> compare y x) [(x, length $ p14s x) | x <- [1 .. 999999]]
+
solve xs = foldl pmax (1,1) xs
   
  +
main = print soln
  +
where
  +
s1 = solve [2..500000]
  +
s2 = solve [500001..1000000]
  +
soln = s2 `par` (s1 `pseq` max s1 s2)
 
</haskell>
 
</haskell>
   
Alternate solution, illustrating use of strict folding:
+
Even faster solution, using an Array to memoize length of sequences :
  +
<haskell>
  +
import Data.Array
  +
import Data.List
  +
import Data.Ord (comparing)
  +
  +
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 $ maximumBy (comparing snd) $ assocs $ syrs 1000000
  +
</haskell>
  +
  +
<!--
  +
This is a trivial solution without any memoization, right?
  +
  +
Using a list to memoize the lengths
   
 
<haskell>
 
<haskell>
 
import Data.List
 
import Data.List
   
problem_14 = j 1000000 where
+
-- computes the sequence for a given n
f :: Int -> Integer -> Int
+
l n = n:unfoldr f n where
f k 1 = k
+
f 1 = Nothing -- we're done here
f k n = f (k+1) $ if even n then div n 2 else 3*n + 1
+
-- for reasons of speed we do div and mod in one go
g x y = if snd x < snd y then y else x
+
f n = let (d,m)=divMod n 2 in case m of
h x n = g x (n, f 1 n)
+
0 -> Just (d,d) -- n was even
j n = fst $ foldl' h (1,1) [2..n-1]
+
otherwise -> let k = 3*n+1 in Just (k,k) -- n was odd
  +
  +
  +
answer = foldl1' f $ -- computes the maximum of a list of tuples
  +
-- save the length of the sequence and the number generating it in a tuple
  +
[(length $! l x, x) | x <- [1..1000000]] where
  +
f (a,c) (b,d) -- one tuple is greater than other if the first component (=sequence-length) is greater
  +
| a > b = (a,c)
  +
| otherwise = (b,d)
  +
  +
main = print answer
 
</haskell>
 
</haskell>
  +
-->
   
 
== [http://projecteuler.net/index.php?section=problems&id=15 Problem 15] ==
 
== [http://projecteuler.net/index.php?section=problems&id=15 Problem 15] ==
Line 59: Line 79:
   
 
Solution:
 
Solution:
  +
A direct computation:
  +
<haskell>
  +
problem_15 = iterate (scanl1 (+)) (repeat 1) !! 20 !! 20
  +
</haskell>
  +
  +
Thinking about it as a problem in combinatorics:
  +
  +
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:
  +
 
<haskell>
 
<haskell>
problem_15 = iterate (scanl1 (+)) (repeat 1) !! 20 !! 20
+
problem_15 = product [21..40] `div` product [2..20]
 
</haskell>
 
</haskell>
   
Line 68: Line 99:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_16 = sum.(map (read.(:[]))).show $ 2^1000
+
import Data.Char
  +
problem_16 = sum k
  +
where s = show (2^1000)
  +
k = map digitToInt s
 
</haskell>
 
</haskell>
   
Line 76: Line 107:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
-- not a very concise or beautiful solution, but food for improvements :)
+
import Char
   
names = concat $
+
one = ["one","two","three","four","five","six","seven","eight",
[zip [(0, n) | n <- [0..19]]
+
"nine","ten","eleven","twelve","thirteen","fourteen","fifteen",
["", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight"
+
"sixteen","seventeen","eighteen", "nineteen"]
,"Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen"
+
ty = ["twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"]
,"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
+
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"
   
spell n = unwords $ if last s == "and" then init s else s
+
where firstDigit x = digitToInt . head . show $ x
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
+
problem_17 = length . concatMap decompose $ [1..1000]
 
</haskell>
 
</haskell>
   
Line 95: Line 126:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_18 = head $ foldr1 g tri where
+
problem_18 = head $ foldr1 g tri
  +
where
 
f x y z = x + max y z
 
f x y z = x + max y z
 
g xs ys = zipWith3 f xs ys $ tail ys
 
g xs ys = zipWith3 f xs ys $ tail ys
Line 117: Line 148:
   
 
== [http://projecteuler.net/index.php?section=problems&id=19 Problem 19] ==
 
== [http://projecteuler.net/index.php?section=problems&id=19 Problem 19] ==
How many Sundays fell on the first of the month during the twentieth century?
+
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:
 
Solution:
 
<haskell>
 
<haskell>
problem_19 = undefined
+
problem_19 = length . filter (== sunday) . drop 12 . take 1212 $ since1900
</haskell>
+
since1900 = scanl nextMonth monday . concat $
  +
replicate 4 nonLeap ++ cycle (leap : replicate 3 nonLeap)
   
== [http://projecteuler.net/index.php?section=problems&id=20 Problem 20] ==
+
nonLeap = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
Find the sum of digits in 100!
 
   
Solution:
+
leap = 31 : 29 : drop 2 nonLeap
<haskell>
+
problem_20 = let fac n = product [1..n] in
+
nextMonth x y = (x + y) `mod` 7
foldr ((+) . Data.Char.digitToInt) 0 $ show $ fac 100
+
  +
sunday = 0
  +
monday = 1
 
</haskell>
 
</haskell>
   
Alternate solution, summing digits directly, which is faster than the show, digitToInt route.
+
Here is an alternative that is simpler, but it is cheating a bit:
   
 
<haskell>
 
<haskell>
dsum 0 = 0
+
import Data.Time.Calendar
dsum n = let ( d, m ) = n `divMod` 10 in m + ( dsum d )
+
import Data.Time.Calendar.WeekDate
   
problem_20' = dsum . product $ [ 1 .. 100 ]
+
problem_19_v2 = length [() | y <- [1901..2000],
  +
m <- [1..12],
  +
let (_, _, d) = toWeekDate $ fromGregorian y m 1,
  +
d == 7]
 
</haskell>
 
</haskell>
   
[[Category:Tutorials]]
+
== [http://projecteuler.net/index.php?section=problems&id=20 Problem 20] ==
[[Category:Code]]
+
Find the sum of digits in 100!
  +
  +
Solution:
  +
<haskell>
  +
problem_20 = sum $ map Char.digitToInt $ show $ product [1..100]
  +
</haskell>

Latest revision as of 14:07, 2 December 2011

Contents

[edit] 1 Problem 11

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

Solution: 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 = print . maximum . prods . input =<< getContents

[edit] 2 Problem 12

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

Solution:

--primeFactors in problem_3
problem_12 = head $ filter ((> 500) . nDivisors) triangleNumbers
  where nDivisors n = product $ map ((+1) . length) (group (primeFactors n))    
        triangleNumbers = scanl1 (+) [1..]

[edit] 3 Problem 13

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

Solution:

main = do xs <- fmap (map read . lines) (readFile "p13.log")
          print . take 10 . show . sum $ xs

[edit] 4 Problem 14

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

Solution:

 
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 unboxed types and parallel computation:

import Control.Parallel
import Data.Word
 
collatzLen :: Int -> Word32 -> Int
collatzLen c 1 = c
collatzLen c n = collatzLen (c+1) $ if n `mod` 2 == 0 then n `div` 2 else 3*n+1
 
pmax x n = x `max` (collatzLen 1 n, n)
 
solve xs = foldl pmax (1,1) xs
 
main = print soln
    where
        s1 = solve [2..500000]
        s2 = solve [500001..1000000]
        soln = s2 `par` (s1 `pseq` max s1 s2)

Even faster solution, using an Array to memoize length of sequences :

import Data.Array
import Data.List
import Data.Ord (comparing)
 
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 $ maximumBy (comparing snd) $ assocs $ syrs 1000000


[edit] 5 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: A direct computation:

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

Thinking about it as a problem in combinatorics:

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

[edit] 6 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

[edit] 7 Problem 17

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

Solution:

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 . concatMap decompose $ [1..1000]

[edit] 8 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]]

[edit] 9 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]

[edit] 10 Problem 20

Find the sum of digits in 100!

Solution:

problem_20 = sum $ map Char.digitToInt $ show $ product [1..100]