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Euler problems/171 to 180

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(solution to problem 179)
m
 
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l=length k
 
l=length k
 
p175 x y=
 
p175 x y=
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
 
test=p175 13 17
 
test=p175 13 17
Line 153: Line 153:
 
<haskell>
 
<haskell>
 
{-# OPTIONS -O2 -optc-O #-}
 
{-# OPTIONS -O2 -optc-O #-}
import Data.Array.IO
+
import Data.Array.ST
import Data.Array.Base
+
import Data.Array.Unboxed
 
import Control.Monad
 
import Control.Monad
  +
import Control.Monad.ST
 
import Data.List
 
import Data.List
import qualified Data.Array.IArray as I
 
   
 
r1 = (2,(10^7-1))
 
r1 = (2,(10^7-1))
   
toRange r = [fst r .. snd r]
+
type Sieve s = STUArray s Int Int
   
type Sieve = IOUArray Int Int
+
incN :: Sieve s -> Int -> ST s ()
 
incN :: Sieve -> Int -> IO ()
 
 
incN a n = do
 
incN a n = do
x <- unsafeRead a n
+
x <- readArray a n
unsafeWrite a n (x+1)
+
writeArray a n (x+1)
   
incEveryN :: Sieve -> Int -> IO ()
+
incEveryN :: Sieve s -> Int -> ST s ()
incEveryN a n = do
+
incEveryN a n = mapM_ (incN a) [n,n+n..snd r1]
mapM_ (incN a) [n,n+n..snd r1]
 
   
sieve :: IO Integer
+
sieve :: Int
sieve = do
+
sieve = countConsecutiveInts b
a <- newArray r1 1 :: IO (IOUArray Int Int)
+
where b = runSTUArray $
mapM_ (incEveryN a) [fst r1 .. (snd r1) `div` 2]
+
do a <- newArray r1 1 :: ST s (STUArray s Int Int)
b <- unsafeFreeze a :: IO (UArray Int Int)
+
mapM_ (incEveryN a) [fst r1 .. (snd r1) `div` 2]
return $ countConsecutiveInts b
+
return a
   
countConsecutiveInts :: UArray Int Int -> Integer
+
countConsecutiveInts :: UArray Int Int -> Int
 
countConsecutiveInts a =
 
countConsecutiveInts a =
let r1 = toRange (fst (bounds a), (snd (bounds a)) - 1)
+
let r1 = [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
+
in length $ filter (\y -> a ! y == a ! (y+1)) $ r1
 
main = sieve >>= print
 
   
  +
main = print sieve
 
</haskell>
 
</haskell>
   

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)