# Difference between revisions of "Euler problems/171 to 180"

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Henrylaxen (talk | contribs) (solution to problem 179) |
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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.IO |
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+ | import Data.Array.Base |
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+ | import Control.Monad |
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+ | import Data.List |
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+ | import qualified Data.Array.IArray as I |
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+ | |||

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

+ | toRange r = [fst r .. snd r] |
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+ | |||

+ | type Sieve = IOUArray Int Int |
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+ | |||

+ | incN :: Sieve -> Int -> IO () |
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+ | incN a n = do |
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+ | x <- unsafeRead a n |
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+ | unsafeWrite a n (x+1) |
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+ | |||

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

+ | sieve :: IO Integer |
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+ | sieve = do |
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+ | a <- newArray r1 1 :: IO (IOUArray Int Int) |
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+ | mapM_ (incEveryN a) [fst r1 .. (snd r1) `div` 2] |
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+ | b <- unsafeFreeze a :: IO (UArray Int Int) |
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+ | return $ countConsecutiveInts b |
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+ | |||

+ | countConsecutiveInts :: UArray Int Int -> Integer |
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+ | countConsecutiveInts a = |
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+ | let r1 = toRange (fst (bounds a), (snd (bounds a)) - 1) |
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+ | in foldl' (\x y -> if a I.! y == a I.! (y+1) then x+1 else x) 0 r1 |
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+ | |||

+ | main = sieve >>= print |
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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] == |

## Revision as of 18:10, 14 May 2008

## Contents

## Problem 171

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

## 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
```

## 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)
```

## Problem 174

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

## 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=
init$foldl (++) "" [a++","|
a<-map show $reverse $filter (/=0)$findRat x y]
problems_175=p175 123456789 987654321
test=p175 13 17
```

## 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]
```

## Problem 177

Integer angled Quadrilaterals.

## 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
```

## 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.IO
import Data.Array.Base
import Control.Monad
import Data.List
import qualified Data.Array.IArray as I
r1 = (2,(10^7-1))
toRange r = [fst r .. snd r]
type Sieve = IOUArray Int Int
incN :: Sieve -> Int -> IO ()
incN a n = do
x <- unsafeRead a n
unsafeWrite a n (x+1)
incEveryN :: Sieve -> Int -> IO ()
incEveryN a n = do
mapM_ (incN a) [n,n+n..snd r1]
sieve :: IO Integer
sieve = do
a <- newArray r1 1 :: IO (IOUArray Int Int)
mapM_ (incEveryN a) [fst r1 .. (snd r1) `div` 2]
b <- unsafeFreeze a :: IO (UArray Int Int)
return $ countConsecutiveInts b
countConsecutiveInts :: UArray Int Int -> Integer
countConsecutiveInts a =
let r1 = toRange (fst (bounds a), (snd (bounds a)) - 1)
in foldl' (\x y -> if a I.! y == a I.! (y+1) then x+1 else x) 0 r1
main = sieve >>= print
```

## 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)
```