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Euler problems/1 to 10

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(Problem 1)
(Problem 2)
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Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_2 =
+
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x]
sum [ x | x <- takeWhile (<= 1000000) fibs,
+
where
even x]
 
where
 
 
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
 
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
 
</haskell>
 
</haskell>

Revision as of 16:22, 4 September 2012

Contents

1 Problem 1

Add all the natural numbers below 1000 that are multiples of 3 or 5.

Two solutions using sum:

import Data.List (union)
problem_1' = sum (union [3,6..999] [5,10..999])
 
problem_1  = sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]

Another solution which uses algebraic relationships:

problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
  where
    sumStep s n = s * sumOnetoN (n `div` s)
    sumOnetoN n = n * (n+1) `div` 2

2 Problem 2

Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million.

Solution:

problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x]
  where
    fibs = 1 : 1 : zipWith (+) fibs (tail fibs)

The following two solutions use the fact that the even-valued terms in the Fibonacci sequence themselves form a Fibonacci-like sequence that satisfies

evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)
.
problem_2_v2 = sumEvenFibs $ numEvenFibsLessThan 1000000
sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
numEvenFibsLessThan n =
   floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)

The first two solutions work because 10^6 is small. The following solution also works for much larger numbers (up to at least 10^1000000 on my computer):

problem_2 = sumEvenFibsLessThan 1000000
 
sumEvenFibsLessThan n = (a + b - 1) `div` 2
  where
    n2 = n `div` 2
    (a, b) = foldr f (0,1)
             . takeWhile ((<= n2) . fst)
             . iterate times2E $ (1, 4)
    f x y | fst z <= n2 = z
          | otherwise   = y
      where z = x `addE` y
addE (a, b) (c, d) = (a*d + b*c - 4*ac, ac + b*d)
  where ac=a*c
 
times2E (a, b) = addE (a, b) (a, b)

3 Problem 3

Find the largest prime factor of 317584931803.

Solution:

primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
 
primeFactors n = factor n primes
  where
    factor n (p:ps) 
        | p*p > n        = [n]
        | n `mod` p == 0 = p : factor (n `div` p) (p:ps)
        | otherwise      = factor n ps
 
problem_3 = last (primeFactors 317584931803)

Another solution, not using recursion, is:

problem_3 = (m !! 0) `div` (m !! 1)
    where
    m = reverse $
        takeWhile (<=n) (scanl1 (*) [ x | x <- 2:[3,5..], (n `mod` x) == 0 ])
    n = 600851475143

4 Problem 4

Find the largest palindrome made from the product of two 3-digit numbers.

Solution:

problem_4 = maximum [ x | y <- [100..999],
                          z <- [y..999],
                          let x = y * z,
                          let s = show x,
                          s == reverse s ]

5 Problem 5

What is the smallest number divisible by each of the numbers 1 to 20?

Solution:

--http://www.research.att.com/~njas/sequences/A003418
problem_5 = foldr1 lcm [1..20]

6 Problem 6

What is the difference between the sum of the squares and the square of the sums?

Solution:

fun n = a - b
    where
        a = (sum [1..n])^2
        b = sum (map (^2) [1..n])
 
problem_6 = fun 100

7 Problem 7

Find the 10001st prime.

Solution:

--primes in problem_3
problem_7 = primes !! 10000

8 Problem 8

Discover the largest product of five consecutive digits in the 1000-digit number.

Solution:

import Data.Char
groupsOf _ [] = []
groupsOf n xs = 
    take n xs : groupsOf n ( tail xs )
 
problem_8 x = maximum . map product . groupsOf 5 $ x
main = do t <- readFile "p8.log" 
          let digits = map digitToInt $concat $ lines t
          print $ problem_8 digits

9 Problem 9

There is only one Pythagorean triplet, {a, b, c}, for which a + b + c = 1000. Find the product abc.

Solution:

triplets l = [[a,b,c] | m <- [2..limit],
                        n <- [1..(m-1)], 
                        let a = m^2 - n^2, 
                        let b = 2*m*n, 
                        let c = m^2 + n^2,
                        a+b+c==l]
    where limit = floor . sqrt . fromIntegral $ l
 
problem_9 = product . head . triplets $ 1000

10 Problem 10

Calculate the sum of all the primes below one million.

Solution:

--http://www.research.att.com/~njas/sequences/A046731
problem_10 = sum (takeWhile (< 1000000) primes)