# Euler problems/171 to 180

### From HaskellWiki

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Finding numbers for which the sum of the squares of the digits is a square. |
Finding numbers for which the sum of the squares of the digits is a square. |
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− | Solution: |
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{{sect-stub}} |
{{sect-stub}} |
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Counting the number of "hollow" square laminae that can form one, two, three, ... distinct arrangements. |
Counting the number of "hollow" square laminae that can form one, two, three, ... distinct arrangements. |
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− | Solution: This was my C++ code, published here without my permission nor any attribution, shame on whoever put it here. [[user:henk263|henk263]] |
+ | {{sect-stub}} |

== [http://projecteuler.net/index.php?section=problems&id=175 Problem 175] == |
== [http://projecteuler.net/index.php?section=problems&id=175 Problem 175] == |
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l=length k |
l=length k |
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p175 x y= |
p175 x y= |
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− | init$foldl (++) "" [a++","| |
+ | concat $ intersperse "," $ |

− | a<-map show $reverse $filter (/=0)$findRat x y] |
+ | map show $reverse $filter (/=0)$findRat x y |

problems_175=p175 123456789 987654321 |
problems_175=p175 123456789 987654321 |
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test=p175 13 17 |
test=p175 13 17 |
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Integer angled Quadrilaterals. |
Integer angled Quadrilaterals. |
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− | Solution: This C++ solution is stolen from balakrishnan. Check out the forum if you want to see his solution to the problem |
+ | {{sect-stub}} |

== [http://projecteuler.net/index.php?section=problems&id=178 Problem 178] == |
== [http://projecteuler.net/index.php?section=problems&id=178 Problem 178] == |
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Step Numbers |
Step Numbers |
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− | Solution: This C++ solution is stolen from balakrishnan. Check out the forum if you want to see his solution to the problem |
+ | Count pandigital step numbers. |

+ | <haskell> |
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+ | import qualified Data.Map as M |
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+ | |||

+ | data StepState a = StepState { minDigit :: a |
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+ | , maxDigit :: a |
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+ | , lastDigit :: a |
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+ | } deriving (Show, Eq, Ord) |
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+ | |||

+ | isSolution (StepState i a _) = i == 0 && a == 9 |
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+ | neighborStates m s@(StepState i a n) = map (\x -> (x, M.findWithDefault 0 s m)) $ |
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+ | [StepState (min i (n - 1)) a (n - 1), StepState i (max a (n + 1)) (n + 1)] |
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+ | |||

+ | allStates = [StepState i a n | (i, a) <- range ((0, 0), (9, 9)), n <- range (i, a)] |
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+ | initialState = M.fromDistinctAscList [(StepState i i i, 1) | i <- [1..9]] |
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+ | stepState m = M.fromListWith (+) $ allStates >>= neighborStates m |
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+ | numSolutionsInMap = sum . map snd . filter (isSolution . fst) . M.toList |
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+ | numSolutionsOfSize n = sum . map numSolutionsInMap . take n $ iterate stepState initialState |
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+ | |||

+ | problem_178 = numSolutionsOfSize 40 |
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+ | </haskell> |
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== [http://projecteuler.net/index.php?section=problems&id=179 Problem 179] == |
== [http://projecteuler.net/index.php?section=problems&id=179 Problem 179] == |
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Consecutive positive divisors. |
Consecutive positive divisors. |
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+ | See http://en.wikipedia.org/wiki/Divisor_function for a simple |
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+ | explanation of calculating the number of divisors of an integer, |
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+ | based on its prime factorization. Then, if you have a lot of |
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+ | time on your hands, run the following program. You need to load |
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+ | the Factoring library from David Amos' wonderful Maths library. |
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+ | See: http://www.polyomino.f2s.com/david/haskell/main.html |
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+ | |||

+ | <haskell> |
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+ | import Factoring |
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+ | import Data.List |
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+ | |||

+ | nFactors :: Integer -> Int |
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+ | nFactors n = |
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+ | let a = primePowerFactorsL n |
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+ | in foldl' (\x y -> x * ((snd y)+1) ) 1 a |
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+ | |||

+ | countConsecutiveInts l = foldl' (\x y -> if y then x+1 else x) 0 a |
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+ | where a = zipWith (==) l (tail l) |
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+ | |||

+ | problem_179 = countConsecutiveInts $ map nFactors [2..(10^7 - 1)] |
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+ | |||

+ | main = print problem_179 |
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+ | |||

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

+ | This is all well and good, but it runs very slowly. (About 4 |
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+ | minutes on my machine) We have to factor every number between 2 |
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+ | and 10^7, which on a non Quantum CPU takes a while. There is |
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+ | another way! |
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+ | |||

+ | We can sieve for the answer. Every number has 1 for a factor. |
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+ | Every other number has 2 as a factor, and every third number has |
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+ | 3 as a factor. So we run a sieve that counts (increments) |
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+ | itself for every integer. When we are done, we run through the |
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+ | resulting array and look at neighboring elements. If they are |
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+ | equal, we increment a counter. This version runs in about 9 |
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+ | seconds on my machine. HenryLaxen May 14, 2008 |
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+ | |||

+ | <haskell> |
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+ | {-# OPTIONS -O2 -optc-O #-} |
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+ | import Data.Array.ST |
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+ | import Data.Array.Unboxed |
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+ | import Control.Monad |
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+ | import Control.Monad.ST |
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+ | import Data.List |
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+ | |||

+ | r1 = (2,(10^7-1)) |
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+ | |||

+ | type Sieve s = STUArray s Int Int |
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+ | |||

+ | incN :: Sieve s -> Int -> ST s () |
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+ | incN a n = do |
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+ | x <- readArray a n |
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+ | writeArray a n (x+1) |
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+ | |||

+ | incEveryN :: Sieve s -> Int -> ST s () |
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+ | incEveryN a n = mapM_ (incN a) [n,n+n..snd r1] |
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+ | |||

+ | sieve :: Int |
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+ | sieve = countConsecutiveInts b |
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+ | where b = runSTUArray $ |
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+ | do a <- newArray r1 1 :: ST s (STUArray s Int Int) |
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+ | mapM_ (incEveryN a) [fst r1 .. (snd r1) `div` 2] |
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+ | return a |
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+ | |||

+ | countConsecutiveInts :: UArray Int Int -> Int |
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+ | countConsecutiveInts a = |
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+ | let r1 = [fst (bounds a) .. snd (bounds a) - 1] |
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+ | in length $ filter (\y -> a ! y == a ! (y+1)) $ r1 |
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+ | |||

+ | main = print sieve |
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+ | </haskell> |
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− | {{sect-stub}} |
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== [http://projecteuler.net/index.php?section=problems&id=180 Problem 180] == |
== [http://projecteuler.net/index.php?section=problems&id=180 Problem 180] == |

## Latest revision as of 01:49, 13 February 2010

## Contents |

## [edit] 1 Problem 171

Finding numbers for which the sum of the squares of the digits is a square.

## [edit] 2 Problem 172

Investigating numbers with few repeated digits.

Solution:

factorial n = product [1..toInteger n] fallingFactorial x n = product [x - fromInteger i | i <- [0..toInteger n - 1] ] choose n k = fallingFactorial n k `div` factorial k -- how many numbers can we get having d digits and p positions p172 0 _ = 0 p172 d p | p < 4 = d^p | otherwise = (p172' p) + p*(p172' (p-1)) + (choose p 2)*(p172' (p-2)) + (choose p 3)*(p172' (p-3)) where p172' = p172 (d-1) problem_172= (p172 10 18) * 9 `div` 10

## [edit] 3 Problem 173

Using up to one million tiles how many different "hollow" square laminae can be formed? Solution:

problem_173= let c=div (10^6) 4 xm=floor$sqrt $fromIntegral c k=[div c x|x<-[1..xm]] in sum k-(div (xm*(xm+1)) 2)

## [edit] 4 Problem 174

Counting the number of "hollow" square laminae that can form one, two, three, ... distinct arrangements.

## [edit] 5 Problem 175

Fractions involving the number of different ways a number can be expressed as a sum of powers of 2. Solution:

sternTree x 0=[] sternTree x y= m:sternTree y n where (m,n)=divMod x y findRat x y |odd l=take (l-1) k++[last k-1,1] |otherwise=k where k=sternTree x y l=length k p175 x y= concat $ intersperse "," $ map show $reverse $filter (/=0)$findRat x y problems_175=p175 123456789 987654321 test=p175 13 17

## [edit] 6 Problem 176

Rectangular triangles that share a cathetus. Solution:

--k=47547 --2*k+1=95095 = 5*7*11*13*19 lst=[5,7,11,13,19] primes=[2,3,5,7,11] problem_176 = product[a^b|(a,b)<-zip primes (reverse n)] where la=div (last lst+1) 2 m=map (\x->div x 2)$init lst n=m++[la]

## [edit] 7 Problem 177

Integer angled Quadrilaterals.

## [edit] 8 Problem 178

Step Numbers

Count pandigital step numbers.

import qualified Data.Map as M data StepState a = StepState { minDigit :: a , maxDigit :: a , lastDigit :: a } deriving (Show, Eq, Ord) isSolution (StepState i a _) = i == 0 && a == 9 neighborStates m s@(StepState i a n) = map (\x -> (x, M.findWithDefault 0 s m)) $ [StepState (min i (n - 1)) a (n - 1), StepState i (max a (n + 1)) (n + 1)] allStates = [StepState i a n | (i, a) <- range ((0, 0), (9, 9)), n <- range (i, a)] initialState = M.fromDistinctAscList [(StepState i i i, 1) | i <- [1..9]] stepState m = M.fromListWith (+) $ allStates >>= neighborStates m numSolutionsInMap = sum . map snd . filter (isSolution . fst) . M.toList numSolutionsOfSize n = sum . map numSolutionsInMap . take n $ iterate stepState initialState problem_178 = numSolutionsOfSize 40

## [edit] 9 Problem 179

Consecutive positive divisors. See http://en.wikipedia.org/wiki/Divisor_function for a simple explanation of calculating the number of divisors of an integer, based on its prime factorization. Then, if you have a lot of time on your hands, run the following program. You need to load the Factoring library from David Amos' wonderful Maths library. See: http://www.polyomino.f2s.com/david/haskell/main.html

import Factoring import Data.List nFactors :: Integer -> Int nFactors n = let a = primePowerFactorsL n in foldl' (\x y -> x * ((snd y)+1) ) 1 a countConsecutiveInts l = foldl' (\x y -> if y then x+1 else x) 0 a where a = zipWith (==) l (tail l) problem_179 = countConsecutiveInts $ map nFactors [2..(10^7 - 1)] main = print problem_179

This is all well and good, but it runs very slowly. (About 4 minutes on my machine) We have to factor every number between 2 and 10^7, which on a non Quantum CPU takes a while. There is another way!

We can sieve for the answer. Every number has 1 for a factor. Every other number has 2 as a factor, and every third number has 3 as a factor. So we run a sieve that counts (increments) itself for every integer. When we are done, we run through the resulting array and look at neighboring elements. If they are equal, we increment a counter. This version runs in about 9 seconds on my machine. HenryLaxen May 14, 2008

{-# OPTIONS -O2 -optc-O #-} import Data.Array.ST import Data.Array.Unboxed import Control.Monad import Control.Monad.ST import Data.List r1 = (2,(10^7-1)) type Sieve s = STUArray s Int Int incN :: Sieve s -> Int -> ST s () incN a n = do x <- readArray a n writeArray a n (x+1) incEveryN :: Sieve s -> Int -> ST s () incEveryN a n = mapM_ (incN a) [n,n+n..snd r1] sieve :: Int sieve = countConsecutiveInts b where b = runSTUArray $ do a <- newArray r1 1 :: ST s (STUArray s Int Int) mapM_ (incEveryN a) [fst r1 .. (snd r1) `div` 2] return a countConsecutiveInts :: UArray Int Int -> Int countConsecutiveInts a = let r1 = [fst (bounds a) .. snd (bounds a) - 1] in length $ filter (\y -> a ! y == a ! (y+1)) $ r1 main = print sieve

## [edit] 10 Problem 180

Rational zeros of a function of three variables. Solution:

import Data.Ratio {- After some algebra, we find: f1 n x y z = x^(n+1) + y^(n+1) - z^(n+1) f2 n x y z = (x*y + y*z + z*x) * ( x^(n-1) + y^(n-1) - z^(n-1) ) f3 n x y z = x*y*z*( x^(n-2) + y^(n-2) - z^(n-2) ) f n x y z = f1 n x y z + f2 n x y z - f3 n x y z f n x y z = (x+y+z) * (x^n+y^n-z^n) Now the hard part comes in realizing that n can be negative. Thanks to Fermat, we only need examine the cases n = [-2, -1, 1, 2] Which leads to: f(-2) z = xy/sqrt(x^2 + y^2) f(-1) z = xy/(x+y) f(1) z = x+y f(2) z = sqrt(x^2 + y^2) -} unique :: Eq(a) => [a] -> [a] unique [] = [] unique (x:xs) | elem x xs = unique xs | otherwise = x : unique xs -- Not quite correct, but I don't care about the zeros ratSqrt :: Rational -> Rational ratSqrt x = let a = floor $ sqrt $ fromIntegral $ numerator x b = floor $ sqrt $ fromIntegral $ denominator x c = (a%b) * (a%b) in if x == c then (a%b) else 0 -- Not quite correct, but I don't care about the zeros reciprocal :: Rational -> Rational reciprocal x | x == 0 = 0 | otherwise = denominator x % numerator x problem_180 = let order = 35 range :: [Rational] range = unique [ (a%b) | b <- [1..order], a <- [1..(b-1)] ] fm2,fm1,f1,f2 :: [[Rational]] fm2 = [[x,y,z] | x<-range, y<-range, let z = x*y * reciprocal (ratSqrt(x*x+y*y)), elem z range] fm1 = [[x,y,z] | x<-range, y<-range, let z = x*y * reciprocal (x+y), elem z range] f1 = [[x,y,z] | x<-range, y<-range, let z = (x+y), elem z range] f2 = [[x,y,z] | x<-range, y<-range, let z = ratSqrt(x*x+y*y), elem z range] result = sum $ unique $ map (\x -> sum x) (fm2++fm1++f1++f2) in (numerator result + denominator result)