Euler problems/111 to 120
From HaskellWiki
m 

(19 intermediate revisions by 8 users not shown) 
Latest revision as of 08:07, 23 February 2010
Contents 
[edit] 1 Problem 111
Search for 10digit primes containing the maximum number of repeated digits.
Solution:
import Control.Monad (replicateM)  All ways of interspersing n copies of x into a list intr :: Int > a > [a] > [[a]] intr 0 _ y = [y] intr n x (y:ys) = concat [map ((replicate i x ++) . (y :)) $ intr (ni) x ys  i < [0..n]] intr n x _ = [replicate n x]  All 10digit primes containing the maximal number of the digit d maxDigits :: Char > [Integer] maxDigits d = head $ dropWhile null [filter isPrime $ map read $ filter ((/='0') . head) $ concatMap (intr (10n) d) $ replicateM n $ delete d "0123456789"  n < [1..9]] problem_111 = sum $ concatMap maxDigits "0123456789"
[edit] 2 Problem 112
Investigating the density of "bouncy" numbers.
Solution:
import Data.List isIncreasing x = show x == sort (show x) isDecreasing x = reverse (show x) == sort (show x) isBouncy x = not (isIncreasing x) && not (isDecreasing x) findProportion prop = snd . head . filter condition . zip [1..] where condition (a,b) = a >= prop * fromIntegral b problem_112 = findProportion 0.99 $ filter isBouncy [1..]
[edit] 3 Problem 113
How many numbers below a googol (10100) are not "bouncy"?
Solution:
import Array mkArray b f = listArray b $ map f (range b) digits = 100 inc = mkArray ((1, 0), (digits, 9)) ninc dec = mkArray ((1, 0), (digits, 9)) ndec ninc (1, _) = 1 ninc (l, d) = sum [inc ! (l1, i)  i < [d..9]] ndec (1, _) = 1 ndec (l, d) = sum [dec ! (l1, i)  i < [0..d]] problem_113 = sum [inc ! i  i < range ((digits, 0), (digits, 9))] + sum [dec ! i  i < range ((1, 1), (digits, 9))]  digits*9  numbers like 11111 are counted in both inc and dec  1  0 is included in the increasing numbers
Note: inc and dec contain the same data, but it seems clearer to duplicate them.
it is another way to solution this problem:
binomial x y =div (prodxy (y+1) x) (prodxy 1 (xy)) prodxy x y=product[x..y] problem_113=sum[binomial (8+a) a+binomial (9+a) a10a<[1..100]]
[edit] 4 Problem 114
Investigating the number of ways to fill a row with separated blocks that are at least three units long.
Solution:
 fun in p115 problem_114=fun 3 50
[edit] 5 Problem 115
Finding a generalisation for the number of ways to fill a row with separated blocks.
Solution:
binomial x y =div (prodxy (y+1) x) (prodxy 1 (xy)) prodxy x y=product[x..y] fun m n=sum[binomial (k+a) (ka)a<[0..div (n+1) (m+1)],let k=1a*m+n] problem_115 = (+1)$length$takeWhile (<10^6) [fun 50 ii<[1..]]
[edit] 6 Problem 116
Investigating the number of ways of replacing square tiles with one of three coloured tiles.
Solution:
binomial x y =div (prodxy (y+1) x) (prodxy 1 (xy)) prodxy x y=product[x..y] f116 n x=sum[binomial (a+b) aa<[1..div n x],let b=na*x] p116 x=sum[f116 x aa<[2..4]] problem_116 = p116 50
[edit] 7 Problem 117
Investigating the number of ways of tiling a row using differentsized tiles.
Solution:
fibs5 = 0 : 0 :1: 1:zipWith4 (\a b c d>a+b+c+d) fibs5 a1 a2 a3 where a1=tail fibs5 a2=tail a1 a3=tail a2 p117 x=fibs5!!(x+2) problem_117 = p117 50
[edit] 8 Problem 118
Exploring the number of ways in which sets containing prime elements can be made.
Solution:
digits = ['1'..'9']  possible partitions voor prime number sets  leave out patitions with more than 4 1's  because only {2,3,5,7,..} is possible  and the [9]partition because every permutation of all  nine digits is divisable by 3 test xs len>4=False xs==[9]=False otherwise=True where len=length $filter (==1) xs parts = filter test $partitions 9 permutationsOf [] = [[]] permutationsOf xs = [x:xs'  x < xs, xs' < permutationsOf (delete x xs)] combinationsOf 0 _ = [[]] combinationsOf _ [] = [] combinationsOf k (x:xs) = map (x:) (combinationsOf (k1) xs) ++ combinationsOf k xs priemPerms [] = 0 priemPerms ds = fromIntegral . length . filter (isPrime . read) . permutationsOf $ ds setsums [] 0 = [[]] setsums [] _ = [] setsums (x:xs) n  x > n = setsums xs n  otherwise = map (x:) (setsums (x:xs) (nx)) ++ setsums xs n partitions n = setsums (reverse [1..n]) n fc :: [Integer] > [Char] > Integer fc (p:[]) ds = priemPerms ds fc (p:ps) ds = sum [np y * fc ps (ds \\ y)  y < combinationsOf p ds, np y /= 0] where np = priemPerms  here is the 'imperfection' correction method:  make use of duplicate reducing factors for partitions  with repeating factors, f.i. [1,1,1,1,2,3]:  in this case 4 1's > factor = 4!  or for [1,1,1,3,3] : factor = 3! * 2! dupF :: [Integer] > Integer dupF = product . map (product . enumFromTo 1 . fromIntegral . length) . group main = do print . sum . map (\x > fc x digits `div` dupF x) $ parts problem_118 = main
[edit] 9 Problem 119
Investigating the numbers which are equal to sum of their digits raised to some power.
Solution:
import Data.List digits n { 123>[3,2,1] } n<10=[n] otherwise= y:digits x where (x,y)=divMod n 10 problem_119 =sort [(a^b) a<[2..200], b<[2..9], let m=a^b, let n=sum$digits m, n==a]!!29
[edit] 10 Problem 120
Finding the maximum remainder when (a − 1)^{n} + (a + 1)^{n} is divided by a^{2}.
Solution:
fun m=div (m*(8*m^23*m5)) 3 problem_120 = fun 500
I have no idea what the above solution has to do with this
problem, even though it produces the correct answer. I suspect
it is some kind of red herring. Below you will find a more holy
mackerel approach, based on the observation that:
1. (a1)^{n} + (a+1)^{n} = 2 if n is odd, and 2an if n is even (mod a^{2})
2. the maximum of 2an mod a^{2} occurs when n = (a1)/2
I hope this is a little more transparent than the solution proposed above. Henrylaxen Mar 5, 2008
maxRemainder n = 2 * n * ((n1) `div` 2) problem_120 = sum $ map maxRemainder [3..1000]