Personal tools

Euler problems/1 to 10

From HaskellWiki

< Euler problems(Difference between revisions)
Jump to: navigation, search
(change view to problems)
(Problem 5)
(36 intermediate revisions by 10 users not shown)
Line 2: Line 2:
 
Add all the natural numbers below 1000 that are multiples of 3 or 5.
 
Add all the natural numbers below 1000 that are multiples of 3 or 5.
  
Solution:
+
Two solutions using <hask>sum</hask>:
 
<haskell>
 
<haskell>
problem_1 =  
+
import Data.List (union)
    sum [ x |
+
problem_1' = sum (union [3,6..999] [5,10..999])
    x <- [1..999],
+
 
    (x `mod` 3 == 0) || (x `mod` 5 == 0)
+
problem_1  = sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]
    ]
+
 
</haskell>
 
</haskell>
 +
 +
Another solution which uses algebraic relationships:
  
 
<haskell>
 
<haskell>
problem_1_v2 =
+
problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
    sum $ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999]
+
  where
</haskell>
+
----
+
<haskell>
+
sumOnetoN n = n * (n+1) `div` 2
+
+
problem_1 =  
+
    sumStep 3 999 + sumStep 5 999 - sumStep 15 999
+
    where
+
 
     sumStep s n = s * sumOnetoN (n `div` s)
 
     sumStep s n = s * sumOnetoN (n `div` s)
      
+
     sumOnetoN n = n * (n+1) `div` 2
 
</haskell>
 
</haskell>
  
Line 31: Line 24:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_2 =  
+
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x]
    sum [ x |
+
  where
    x <- takeWhile (<= 1000000) fibs,
+
    x `mod` 2 == 0
+
    ]
+
    where
+
 
     fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
 
     fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
 
</haskell>
 
</haskell>
Line 45: Line 34:
 
<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>.
 
<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>.
 
<haskell>
 
<haskell>
problem_2_v2 =  
+
problem_2 = sumEvenFibs $ numEvenFibsLessThan 1000000
    sumEvenFibs $ numEvenFibsLessThan 1000000
+
  where
sumEvenFibs n =
+
    sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
    (evenFib n + evenFib (n+1) - 2) `div` 4
+
    evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
evenFib n =  
+
    numEvenFibsLessThan n =
    round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
+
              floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)
numEvenFibsLessThan n =
+
    floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)
+
 
</haskell>
 
</haskell>
  
Line 61: Line 48:
 
problem_2 = sumEvenFibsLessThan 1000000
 
problem_2 = sumEvenFibsLessThan 1000000
  
sumEvenFibsLessThan n =  
+
sumEvenFibsLessThan n = (a + b - 1) `div` 2
    (a + b - 1) `div` 2
+
  where
    where
+
 
     n2 = n `div` 2
 
     n2 = n `div` 2
     (a, b) =  
+
     (a, b) = foldr f (0,1)
        foldr f (0,1) $
+
            . takeWhile ((<= n2) . fst)
        takeWhile ((<= n2) . fst) $
+
            . iterate times2E $ (1, 4)
        iterate times2E (1, 4)
+
     f x y | fst z <= n2 = z
     f x y  
+
          | otherwise  = y
        | fst z <= n2 = z
+
      where z = x `addE` y
        | otherwise  = y
+
addE (a, b) (c, d) = (a*d + b*c - 4*ac, ac + b*d)
        where z = x `addE` y
+
  where ac=a*c
addE (a, b) (c, d) =  
+
 
    (a*d + b*c - 4*ac, ac + b*d)
+
times2E (a, b) = addE (a, b) (a, b)
    where
+
    ac=a*c
+
times2E (a, b) =
+
    addE (a, b) (a, b)
+
 
</haskell>
 
</haskell>
  
Line 86: Line 68:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
primes =  
+
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
    2 : filter ((==1) . length . primeFactors) [3,5..]
+
 
primeFactors n =
+
primeFactors n = factor n primes
    factor n primes
+
  where
    where
+
 
     factor n (p:ps)  
 
     factor n (p:ps)  
 
         | p*p > n        = [n]
 
         | p*p > n        = [n]
Line 96: Line 77:
 
         | otherwise      = factor n ps
 
         | otherwise      = factor n ps
  
problem_3 =  
+
problem_3 = last (primeFactors 317584931803)
    last (primeFactors 317584931803)
+
 
</haskell>
 
</haskell>
  
This can be improved by using
+
Another solution, not using recursion, is:
<hask>null . tail</hask>
+
<haskell>
instead of
+
problem_3 = (m !! 0) `div` (m !! 1)
<hask>(== 1) . length</hask>.
+
  where
 +
    m = reverse $
 +
        takeWhile (<=n) (scanl1 (*) [ x | x <- 2:[3,5..], (n `mod` x) == 0 ])
 +
    n = 600851475143
 +
</haskell>
  
 
== [http://projecteuler.net/index.php?section=problems&id=4 Problem 4] ==
 
== [http://projecteuler.net/index.php?section=problems&id=4 Problem 4] ==
Line 110: Line 94:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_4 =  
+
problem_4 =
    foldr max 0 [ x |
+
  maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]
    y <- [100..999],
+
    z <- [100..999],
+
    let x = y * z,
+
    let s = show x,
+
    s == reverse s
+
    ]
+
</haskell>
+
An alternative to avoid evaluating twice the same pair of numbers:
+
<haskell>
+
problem_4' =
+
    foldr1 max [ x |
+
    y <- [100..999],
+
    z <- [y..999],
+
    let x = y * z,
+
    let s = show x,
+
    s == reverse s
+
    ]
+
 
</haskell>
 
</haskell>
  
Line 136: Line 103:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_5 =
+
problem_5 = foldr1 lcm [1..20]
    head [ x |
+
    x <- [2520,5040..],
+
    all (\y -> x `mod` y == 0) [1..20]
+
    ]
+
</haskell>
+
An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom:
+
<haskell>
+
problem_5' = foldr1 lcm [1..20]
+
 
</haskell>
 
</haskell>
  
Line 151: Line 110:
  
 
Solution:
 
Solution:
 +
<!--
 
<haskell>
 
<haskell>
problem_6 =  
+
fun n = a - b
     sum [ x^2 | x <- [1..100]] - (sum [1..100])^2
+
  where
 +
    a=(n^2 * (n+1)^2) `div` 4
 +
    b=(n * (n+1) * (2*n+1)) `div` 6
 +
 
 +
problem_6 = fun 100
 +
</haskell>
 +
-->
 +
<!-- Might just be me, but I find this a LOT easier to read. Perhaps not as good mathematically, but it runs in an instant, even for n = 25000.
 +
<haskell>
 +
fun n = a - b
 +
     where
 +
        a = (sum [1..n])^2
 +
        b = sum (map (^2) [1..n])
 +
 
 +
problem_6 = fun 100
 +
</haskell>
 +
-->
 +
<!-- I just made it a oneliner... -->
 +
<haskell>
 +
problem_6 = (sum [1..100])^2 - sum (map (^2) [1..100])
 
</haskell>
 
</haskell>
  
Line 162: Line 141:
 
<haskell>
 
<haskell>
 
--primes in problem_3
 
--primes in problem_3
problem_7 =  
+
problem_7 = primes !! 10000
    head $ drop 10000 primes
+
 
</haskell>
 
</haskell>
 
As above, this can be improved by using
 
<hask>null . tail</hask>
 
instead of
 
<hask>(== 1) . length</hask>.
 
 
Here is an alternative that uses a
 
[http://www.haskell.org/pipermail/haskell-cafe/2007-February/022854.html sieve of Eratosthenes]:
 
 
<haskell>
 
primes' =
 
    2 : 3 : sieve (tail primes') [5,7..]
 
    where
 
    sieve (p:ps) x =
 
        h ++ sieve ps (filter (\q -> q `mod` p /= 0) t
 
        where
 
        (h, _:t) = span (p*p <) x
 
problem_7_v2 = primes' !! 10000
 
</haskell>
 
 
 
== [http://projecteuler.net/index.php?section=problems&id=8 Problem 8] ==
 
== [http://projecteuler.net/index.php?section=problems&id=8 Problem 8] ==
 
Discover the largest product of five consecutive digits in the 1000-digit number.
 
Discover the largest product of five consecutive digits in the 1000-digit number.
  
 
Solution:
 
Solution:
 +
<!--
 
<haskell>
 
<haskell>
 
import Data.Char
 
import Data.Char
Line 195: Line 154:
 
     take n xs : groupsOf n ( tail xs )
 
     take n xs : groupsOf n ( tail xs )
 
   
 
   
problem_8 x=  
+
problem_8 x = maximum . map product . groupsOf 5 $ x
    maximum . map product . groupsOf 5 $ x
+
main = do t <- readFile "p8.log"  
main=do
+
          let digits = map digitToInt $concat $ lines t
    t<-readFile "p8.log"  
+
          print $ problem_8 digits
    let digits = map digitToInt $foldl (++) "" $ lines t
+
</haskell>
    print $ problem_8 digits
+
-->
 +
<!-- I just cleaned up a little. -->
 +
<haskell>
 +
import Data.Char (digitToInt)
 +
import Data.List (tails)
 +
         
 +
problem_8 = do str <- readFile "number.txt"
 +
              -- This line just converts our str(ing) to a list of 1000 Ints
 +
              let number = map digitToInt (concat $ lines str)
 +
              print $ maximum $ map (product . take 5) (tails number)
 
</haskell>
 
</haskell>
  
Line 208: Line 176:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_9 =  
+
triplets l = [[a,b,c] | m <- [2..limit],
    head [a*b*c |
+
                        n <- [1..(m-1)],  
    a <- [1..500],  
+
                        let a = m^2 - n^2,
    b <- [a..500],  
+
                        let b = 2*m*n,  
    let c = 1000-a-b,  
+
                        let c = m^2 + n^2,
    a^2 + b^2 == c^2
+
                        a+b+c==l]
     ]
+
     where limit = floor . sqrt . fromIntegral $ l
</haskell>
+
  
Another solution using Pythagorean Triplets generation:
+
problem_9 = product . head . triplets $ 1000
<haskell>
+
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
+
 
</haskell>
 
</haskell>
  
Line 236: Line 192:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_10 =  
+
--primes in problem_3
    sum (takeWhile (< 1000000) primes)
+
problem_10 = sum (takeWhile (< 1000000) primes)
 
</haskell>
 
</haskell>

Revision as of 14:17, 22 October 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 = sumEvenFibs $ numEvenFibsLessThan 1000000
  where
    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:

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:

problem_6 = (sum [1..100])^2 - sum (map (^2) [1..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 (digitToInt)
import Data.List (tails)
 
problem_8 = do str <- readFile "number.txt"
               -- This line just converts our str(ing) to a list of 1000 Ints
               let number = map digitToInt (concat $ lines str)
               print $ maximum $ map (product . take 5) (tails number)

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:

--primes in problem_3
problem_10 = sum (takeWhile (< 1000000) primes)