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

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− | == [http://projecteuler.net/index.php?section= | + | == [http://projecteuler.net/index.php?section=problems&id=111 Problem 111] == |

Search for 10-digit primes containing the maximum number of repeated digits. | Search for 10-digit primes containing the maximum number of repeated digits. | ||

Line 25: | Line 25: | ||

</haskell> | </haskell> | ||

− | == [http://projecteuler.net/index.php?section= | + | == [http://projecteuler.net/index.php?section=problems&id=112 Problem 112] == |

Investigating the density of "bouncy" numbers. | Investigating the density of "bouncy" numbers. | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | problem_112 = | + | 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..] | ||

</haskell> | </haskell> | ||

− | == [http://projecteuler.net/index.php?section= | + | == [http://projecteuler.net/index.php?section=problems&id=113 Problem 113] == |

How many numbers below a googol (10100) are not "bouncy"? | How many numbers below a googol (10100) are not "bouncy"? | ||

Line 60: | Line 69: | ||

Note: inc and dec contain the same data, but it seems clearer to duplicate them. | Note: inc and dec contain the same data, but it seems clearer to duplicate them. | ||

− | == [http://projecteuler.net/index.php?section= | + | 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. | Investigating the number of ways to fill a row with separated blocks that are at least three units long. | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | problem_114 = | + | -- fun in p115 |

+ | problem_114=fun 3 50 | ||

</haskell> | </haskell> | ||

− | == [http://projecteuler.net/index.php?section= | + | == [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. | Finding a generalisation for the number of ways to fill a row with separated blocks. | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | problem_115 = | + | 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> | </haskell> | ||

− | == [http://projecteuler.net/index.php?section= | + | == [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. | Investigating the number of ways of replacing square tiles with one of three coloured tiles. | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | problem_116 = | + | 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> | </haskell> | ||

− | == [http://projecteuler.net/index.php?section= | + | == [http://projecteuler.net/index.php?section=problems&id=117 Problem 117] == |

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

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | problem_117 = | + | 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> | </haskell> | ||

− | == [http://projecteuler.net/index.php?section= | + | == [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. | Exploring the number of ways in which sets containing prime elements can be made. | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | problem_118 = | + | 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 = | ||

+ | 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 | ||

</haskell> | </haskell> | ||

− | == [http://projecteuler.net/index.php?section= | + | == [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. | Investigating the numbers which are equal to sum of their digits raised to some power. | ||

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | problem_119 = | + | 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> | </haskell> | ||

− | == [http://projecteuler.net/index.php?section= | + | == [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 | + | Finding the maximum remainder when (a − 1)<sup>n</sup> + (a + 1)<sup>n</sup> is divided by a<sup>2</sup>. |

Solution: | Solution: | ||

<haskell> | <haskell> | ||

− | problem_120 = | + | fun m=div (m*(8*m^2-3*m-5)) 3 |

+ | problem_120 = fun 500 | ||

+ | </haskell> | ||

+ | |||

+ | |||

+ | 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. (a-1)<sup>n</sup> + (a+1)<sup>n</sup> = 2 if n is odd, and 2an if n is even (mod a<sup>2</sup>) | ||

+ | |||

+ | 2. the maximum of 2an mod a<sup>2</sup> occurs when n = (a-1)/2 | ||

+ | |||

+ | I hope this is a little more transparent than the solution | ||

+ | proposed above. Henrylaxen Mar 5, 2008 | ||

+ | |||

+ | <haskell> | ||

+ | maxRemainder n = 2 * n * ((n-1) `div` 2) | ||

+ | problem_120 = sum $ map maxRemainder [3..1000] | ||

</haskell> | </haskell> |

## Latest revision as of 08:07, 23 February 2010

## Contents

## 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"
```

## 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 ! (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]]
```

## 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 (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..]]
```

## 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
```

## 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
```

## 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 =
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^2-3*m-5)) 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. (a-1)^{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 = (a-1)/2

I hope this is a little more transparent than the solution proposed above. Henrylaxen Mar 5, 2008

```
maxRemainder n = 2 * n * ((n-1) `div` 2)
problem_120 = sum $ map maxRemainder [3..1000]
```