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Euler problems/1 to 10

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(Switched out problem 6 for a much easier-to-understand solution. Hope nobody minds)
Current revision (14:17, 22 October 2012) (edit) (undo)
(→Problem 5)
 
(28 intermediate revisions not shown.)
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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>
 +
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]
 +
</haskell>
 +
 
 +
Another solution which uses algebraic relationships:
 +
 
<haskell>
<haskell>
-
sumOnetoN n = n * (n+1) `div` 2
 
problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
where
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 15: Line 24:
Solution:
Solution:
<haskell>
<haskell>
-
problem_2 =
+
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x]
-
sum [ x | x <- takeWhile (<= 1000000) fibs,
+
where
-
x `mod` 2 == 0]
+
-
where
+
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
</haskell>
</haskell>
Line 27: 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 = sumEvenFibs $ numEvenFibsLessThan 1000000
+
problem_2 = sumEvenFibs $ numEvenFibsLessThan 1000000
-
sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
+
where
-
evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
+
sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
-
numEvenFibsLessThan n =
+
evenFib 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 71: Line 79:
problem_3 = last (primeFactors 317584931803)
problem_3 = last (primeFactors 317584931803)
</haskell>
</haskell>
 +
 +
Another solution, not using recursion, is:
 +
<haskell>
 +
problem_3 = (m !! 0) `div` (m !! 1)
 +
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] ==
Find the largest palindrome made from the product of two 3-digit numbers.
Find the largest palindrome made from the product of two 3-digit numbers.
Line 76: Line 94:
Solution:
Solution:
<haskell>
<haskell>
-
problem_4 = maximum [ x | y <- [100..999],
+
problem_4 =
-
z <- [y..999],
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maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]
-
let x = y * z,
+
-
let s = show x,
+
-
s == reverse s ]
+
</haskell>
</haskell>
Line 88: Line 103:
Solution:
Solution:
<haskell>
<haskell>
-
--http://www.research.att.com/~njas/sequences/A003418
 
problem_5 = foldr1 lcm [1..20]
problem_5 = foldr1 lcm [1..20]
</haskell>
</haskell>
Line 100: Line 114:
fun n = a - b
fun n = a - b
where
where
-
a=div (n^2 * (n+1)^2) 4
+
a=(n^2 * (n+1)^2) `div` 4
-
b=div (n * (n+1) * (2*n+1)) 6
+
b=(n * (n+1) * (2*n+1)) `div` 6
problem_6 = fun 100
problem_6 = fun 100
</haskell>
</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.
-
<!-- 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>
<haskell>
fun n = a - b
fun n = a - b
Line 116: Line 129:
problem_6 = fun 100
problem_6 = fun 100
</haskell>
</haskell>
 +
-->
 +
<!-- I just made it a oneliner... -->
 +
<haskell>
 +
problem_6 = (sum [1..100])^2 - sum (map (^2) [1..100])
 +
</haskell>
 +
== [http://projecteuler.net/index.php?section=problems&id=7 Problem 7] ==
== [http://projecteuler.net/index.php?section=problems&id=7 Problem 7] ==
Find the 10001st prime.
Find the 10001st prime.
Line 122: Line 141:
<haskell>
<haskell>
--primes in problem_3
--primes in problem_3
-
problem_7 = primes !! 10001
+
problem_7 = primes !! 10000
</haskell>
</haskell>
== [http://projecteuler.net/index.php?section=problems&id=8 Problem 8] ==
== [http://projecteuler.net/index.php?section=problems&id=8 Problem 8] ==
Line 128: Line 147:
Solution:
Solution:
 +
<!--
<haskell>
<haskell>
import Data.Char
import Data.Char
Line 136: Line 156:
problem_8 x = maximum . map product . groupsOf 5 $ x
problem_8 x = maximum . map product . groupsOf 5 $ x
main = do t <- readFile "p8.log"
main = do t <- readFile "p8.log"
-
let digits = map digitToInt $foldl (++) "" $ lines t
+
let digits = map digitToInt $concat $ lines t
print $ problem_8 digits
print $ problem_8 digits
 +
</haskell>
 +
-->
 +
<!-- 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 161: Line 192:
Solution:
Solution:
<haskell>
<haskell>
-
--http://www.research.att.com/~njas/sequences/A046731
+
--primes in problem_3
problem_10 = sum (takeWhile (< 1000000) primes)
problem_10 = sum (takeWhile (< 1000000) primes)
</haskell>
</haskell>

Current revision

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)
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