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Euler problems/111 to 120

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== [http://projecteuler.net/index.php?section=problems&id=111 Problem 111] ==
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Search for 10-digit primes containing the maximum number of repeated digits.
 
 
Solution:
 
<haskell>
 
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 (n-i) x ys
 
| i <- [0..n]]
 
intr n x _ = [replicate n x]
 
 
-- All 10-digit 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 (10-n) d) $
 
replicateM n $ delete d "0123456789"
 
| n <- [1..9]]
 
 
problem_111 = sum $ concatMap maxDigits "0123456789"
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=problems&id=112 Problem 112] ==
 
Investigating the density of "bouncy" numbers.
 
 
Solution:
 
<haskell>
 
isIncreasing' n p
 
| n == 0 = True
 
| p >= p1 = isIncreasing' (n `div` 10) p1
 
| otherwise = False
 
where
 
p1 = n `mod` 10
 
 
isIncreasing :: Int -> Bool
 
isIncreasing n = isIncreasing' (n `div` 10) (n `mod` 10)
 
 
isDecreasing' n p
 
| n == 0 = True
 
| p <= p1 = isDecreasing' (n `div` 10) p1
 
| otherwise = False
 
where
 
p1 = n `mod` 10
 
 
isDecreasing :: Int -> Bool
 
isDecreasing n = isDecreasing' (n `div` 10) (n `mod` 10)
 
 
isBouncy n = not (isIncreasing n) && not (isDecreasing n)
 
nnn=1500000
 
num150 =length [x|x<-[1..nnn],isBouncy x]
 
p112 n nb
 
| fromIntegral nnb / fromIntegral n >= 0.99 = n
 
| otherwise = prob112' (n+1) nnb
 
where
 
nnb = if isBouncy n then nb + 1 else nb
 
 
problem_112=p112 (nnn+1) num150
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=problems&id=113 Problem 113] ==
 
How many numbers below a googol (10100) are not "bouncy"?
 
 
Solution:
 
<haskell>
 
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 ! (l-1, i) | i <- [d..9]]
 
 
ndec (1, _) = 1
 
ndec (l, d) = sum [dec ! (l-1, 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
 
</haskell>
 
Note: inc and dec contain the same data, but it seems clearer to duplicate them.
 
 
it is another way to solution this problem:
 
<haskell>
 
binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
 
prodxy x y=product[x..y]
 
problem_113=sum[binomial (8+a) a+binomial (9+a) a-10|a<-[1..100]]
 
</haskell>
 
== [http://projecteuler.net/index.php?section=problems&id=114 Problem 114] ==
 
Investigating the number of ways to fill a row with separated blocks that are at least three units long.
 
 
Solution:
 
<haskell>
 
-- fun in p115
 
problem_114=fun 3 50
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=problems&id=115 Problem 115] ==
 
Finding a generalisation for the number of ways to fill a row with separated blocks.
 
 
Solution:
 
<haskell>
 
binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
 
prodxy x y=product[x..y]
 
fun m n=sum[binomial (k+a) (k-a)|a<-[0..div (n+1) (m+1)],let k=1-a*m+n]
 
problem_115 = (+1)$length$takeWhile (<10^6) [fun 50 i|i<-[1..]]
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=problems&id=116 Problem 116] ==
 
Investigating the number of ways of replacing square tiles with one of three coloured tiles.
 
 
Solution:
 
<haskell>
 
binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
 
prodxy x y=product[x..y]
 
f116 n x=sum[binomial (a+b) a|a<-[1..div n x],let b=n-a*x]
 
p116 x=sum[f116 x a|a<-[2..4]]
 
problem_116 = p116 50
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=problems&id=117 Problem 117] ==
 
Investigating the number of ways of tiling a row using different-sized tiles.
 
 
Solution:
 
<haskell>
 
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
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=problems&id=118 Problem 118] ==
 
Exploring the number of ways in which sets containing prime elements can be made.
 
 
Solution:
 
<haskell>
 
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 (k-1) 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) (n-x)) ++ setsums xs n
 
 
partitions n = setsums (reverse [1..n]) n
 
 
fc :: [Integer] -> [Char] -> Integer
 
fc (p:[]) ds = priemPerms ds
 
fc (p:ps) ds =
 
foldl fcmul 0 . combinationsOf p $ ds
 
where
 
fcmul x y
 
| np y == 0 = x
 
| otherwise = x + np y * fc ps (ds \\ y)
 
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 = foldl (\ x y -> x * product [1..y]) 1 . map (fromIntegral . length) . group
 
 
main = do
 
print . sum . map (\x -> fc x digits `div` dupF x) $ parts
 
problem_118 = main
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=problems&id=119 Problem 119] ==
 
Investigating the numbers which are equal to sum of their digits raised to some power.
 
 
Solution:
 
<haskell>
 
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
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=problems&id=120 Problem 120] ==
 
Finding the maximum remainder when (a − 1)n + (a + 1)n is divided by a2.
 
 
Solution:
 
<haskell>
 
fun m=div (m*(8*m^2-3*m-5)) 3
 
problem_120 = fun 500
 
</haskell>
 

Revision as of 21:45, 29 January 2008

Do them on your own!