Difference between revisions of "Euler problems/111 to 120"
m 
m 

Line 161:  Line 161:  
fc (p:[]) ds = priemPerms ds 
fc (p:[]) ds = priemPerms ds 

fc (p:ps) ds = 
fc (p:ps) ds = 

−  +  sum [np y * fc ps (ds \\ y)  y < combinationsOf p ds, np y /= 0] 

−  where 

−  fcmul x y 

−   np y == 0 = x 

−   otherwise = x + np y * fc ps (ds \\ y) 

where 
where 

np = priemPerms 
np = priemPerms 
Latest revision as of 08:07, 23 February 2010
Contents
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"
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..]
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]]
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
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..]]
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
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
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
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
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]