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Euler problems/111 to 120

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Solution:
Solution:
<haskell>
<haskell>
-
isPrime x
+
digits = ['1'..'9']
-
|x<100=isPrime' x
+
 
-
|otherwise=foldl (&& )True [millerRabinPrimality x y|y<-[2,7,61]]
+
-- possible partitions voor prime number sets
-
getprimes ""= [[]]
+
-- leave out patitions with more than 4 1's
-
getprimes s1=
+
-- because only {2,3,5,7,..} is possible
-
[n:f|
+
-- and the [9]-partition because every permutation of all
-
let len=length s1,
+
-- nine digits is divisable by 3
-
a<-[1..len],
+
test xs
-
let b=take a s1,
+
|len>4=False
-
let n=read b::Integer,
+
|xs==[9]=False
-
isPrime n,
+
|otherwise=True
-
let k=getprimes $drop a s1,
+
where
-
f<-k,
+
len=length $filter (==1) xs
-
a==len|| n<head f
+
parts = filter test $partitions 9
-
]
+
permutationsOf [] = [[]]
-
perms :: [a] -> [[a]]
+
permutationsOf xs = [x:xs' | x <- xs, xs' <- permutationsOf (delete x xs)]
-
perms [] = [ [] ]
+
combinationsOf 0 _ = [[]]
-
perms (x:xs) =
+
combinationsOf _ [] = []
-
concat (map (between x) (perms xs))
+
combinationsOf k (x:xs) =
-
where
+
map (x:) (combinationsOf (k-1) xs) ++ combinationsOf k xs
-
between e [] = [ [e] ]
+
 
-
between e (y:ys) = (e:y:ys) : map (y:) (between e ys)
+
priemPerms [] = 0
-
fun x=do
+
priemPerms ds =
-
let cs=length$getprimes x
+
fromIntegral . length . filter (isPrime . read) . permutationsOf $ ds
-
if (cs/=0) then
+
setsums [] 0 = [[]]
-
appendFile "p118.log"$(++"\n")$show cs
+
setsums [] _ = []
-
else
+
setsums (x:xs) n
-
return ()
+
| x > n = setsums xs n
-
sToInt =(+0).read
+
| otherwise = map (x:) (setsums (x:xs) (n-x)) ++ setsums xs n
-
problem_118a=do
+
 
-
s<-readFile "p118.log"
+
partitions n = setsums (reverse [1..n]) n
-
print$sum$map sToInt$lines s
+
 
-
main=do
+
fc :: [Integer] -> [Char] -> Integer
-
mapM_ fun $perms ['1'..'9']
+
fc (p:[]) ds = priemPerms ds
-
problem_118a
+
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
problem_118 = main
</haskell>
</haskell>

Revision as of 06:37, 27 January 2008

Contents

1 Problem 111

Search for 10-digit 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 (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"

2 Problem 112

Investigating the density of "bouncy" numbers.

Solution: 

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

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 ! (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

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 (x-y))
prodxy x y=product[x..y]
problem_113=sum[binomial (8+a) a+binomial (9+a) a-10|a<-[1..100]]

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

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 (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..]]

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 (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

7 Problem 117

Investigating the number of ways of tiling a row using different-sized 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

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 (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

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

10 Problem 120

Finding the maximum remainder when (a − 1)n + (a + 1)n is divided by a2.

Solution: 

fun m=div (m*(8*m^2-3*m-5)) 3
problem_120 = fun 500
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