Euler problems/171 to 180
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Latest revision as of 01:49, 13 February 2010
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[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' (p1)) + (choose p 2)*(p172' (p2)) + (choose p 3)*(p172' (p3)) where p172' = p172 (d1) 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 xx<[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 (l1) k++[last k1,1] 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 optcO #} import Data.Array.ST import Data.Array.Unboxed import Control.Monad import Control.Monad.ST import Data.List r1 = (2,(10^71)) 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^(n1) + y^(n1)  z^(n1) ) f3 n x y z = x*y*z*( x^(n2) + y^(n2)  z^(n2) ) 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^nz^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..(b1)] ] 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)