Difference between revisions of "Euler problems/1 to 10"

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[[Category:Programming exercise spoilers]]
 
 
== [http://projecteuler.net/index.php?section=problems&id=1 Problem 1] ==
 
== [http://projecteuler.net/index.php?section=problems&id=1 Problem 1] ==
 
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 = sum [ x | x <- [1..999], (x `mod` 3 == 0) || (x `mod` 5 == 0)]
 
  +
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>
 
</haskell>
  +
  +
Another solution which uses algebraic relationships:
   
 
<haskell>
 
<haskell>
problem_1_v2 = sum $ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999]
 
  +
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
 
</haskell>
 
</haskell>
   
Line 17: Line 24:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, x `mod` 2 == 0]
+
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x]
where fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
+
where
  +
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
  +
</haskell>
  +
  +
The following two solutions use the fact that the even-valued terms in
  +
the Fibonacci sequence themselves form a Fibonacci-like sequence
  +
that satisfies
  +
<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>.
  +
<haskell>
  +
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)
  +
</haskell>
  +
  +
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):
  +
<haskell>
  +
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)
  +
  +
</haskell>
  +
  +
  +
Another elegant, quick solution, based on some background mathematics as in comments:
  +
  +
<haskell>
  +
-- Every third term is even, and every third term beautifully follows:
  +
-- fib n = 4*fib n-3 + fib n-6
  +
evenFibs = 2 : 8 : zipWith (+) (map (4*) (tail evenFibs)) evenFibs
  +
  +
-- So, evenFibs are: e(n) = 4*e(n-1) + e(n-2)
  +
-- [there4]:4e(n) = e(n+1) - e(n-1)
  +
-- 4e(n-1) = e(n) - e(n-2)
  +
-- 4e(n-2) = e(n-1) - e(n-3)
  +
-- ...
  +
-- 4e(3) = e(4) - e(2)
  +
-- 4e(2) = e(3) - e(1)
  +
-- 4e(1) = e(2) - e(0)
  +
-- -------------------------------
  +
-- Total: 4([sum] e(k) - e(0)) = e(n+1) + e(n) - e(1) - e(0)
  +
-- => [sum] e(k) = (e(n+1) + e(n) - e(1) + 3e(0))/4 = 1089154 for
  +
-- first 10 terms
  +
  +
sumEvenFibsBelow :: Int -> Int
  +
sumEvenFibsBelow n = ((last $ take (x+1) evenFibs) +
  +
(last $ take x evenFibs) -
  +
8 + 6) `div` 4
  +
where x = length (takeWhile (<= n) evenFibs)
  +
 
</haskell>
 
</haskell>
   
 
== [http://projecteuler.net/index.php?section=problems&id=3 Problem 3] ==
 
== [http://projecteuler.net/index.php?section=problems&id=3 Problem 3] ==
Find the largest prime factor of 317584931803.
+
Find the largest prime factor of 600851475143.
   
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
+
primes = 2 : filter (null . tail . primeFactors) [3,5..]
  +
 
primeFactors n = factor n primes
 
primeFactors n = factor n primes
where factor n (p:ps) | p*p > n = [n]
 
  +
where
| n `mod` p == 0 = p : factor (n `div` p) (p:ps)
+
factor n (p:ps)
| otherwise = factor n 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)
+
problem_3 = last (primeFactors 600851475143)
 
</haskell>
 
</haskell>
   
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Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_4 = foldr max 0 [ x | y <- [100..999], z <- [100..999], let x = y * z, let s = show x, s == reverse s]
 
  +
problem_4 =
  +
maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]
 
</haskell>
 
</haskell>
   
Line 48: Line 56:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_5 = head [ x | x <- [2520,5040..], all (\y -> x `mod` y == 0) [1..20]]
+
problem_5 = foldr1 lcm [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>
  +
  +
Another solution: <code>16*9*5*7*11*13*17*19</code>. Product of maximal powers of primes in the range.
   
 
== [http://projecteuler.net/index.php?section=problems&id=6 Problem 6] ==
 
== [http://projecteuler.net/index.php?section=problems&id=6 Problem 6] ==
Line 55: Line 65:
   
 
Solution:
 
Solution:
  +
<!--
 
<haskell>
 
<haskell>
problem_6 = sum [ x^2 | x <- [1..100]] - (sum [1..100])^2
 
  +
fun n = a - b
  +
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 64: Line 95:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
 
  +
--primes in problem_3
primeFactors n = factor n primes
 
  +
problem_7 = primes !! 10000
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_7 = head $ drop 10000 primes
 
 
</haskell>
 
</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 thirteen consecutive digits in the 1000-digit number.
   
 
Solution:
 
Solution:
  +
<!--
  +
<haskell>
  +
import Data.Char
  +
groupsOf _ [] = [] -- incorrect, overall: last
  +
groupsOf n xs = -- subsequences will be shorter than n!!
  +
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
  +
</haskell>
  +
-->
 
<haskell>
 
<haskell>
num = ... -- 1000 digit number as a string
 
  +
import Data.Char
digits = map digitToInt num
 
  +
import Data.List
   
groupsOf _ [] = []
 
  +
euler_8 = do
groupsOf n xs = take n xs : groupsOf n ( tail xs )
 
  +
str <- readFile "number.txt"
 
  +
print . maximum . map product
problem_8 = maximum . map product . groupsOf 5 $ digits
 
  +
. foldr (zipWith (:)) (repeat [])
  +
. take 13 . tails . map (fromIntegral . digitToInt)
  +
. concat . lines $ str
 
</haskell>
 
</haskell>
   
Line 91: Line 132:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_9 = head [a*b*c | a <- [1..500], b <- [a..500], let c = 1000-a-b, a^2 + b^2 == c^2]
 
  +
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 99: Line 148:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
  +
--primes in problem_3
 
problem_10 = sum (takeWhile (< 1000000) primes)
 
problem_10 = sum (takeWhile (< 1000000) primes)
 
</haskell>
 
</haskell>
 
 
[[Category:Tutorials]]
 
[[Category:Code]]
 

Latest revision as of 02:31, 8 May 2016

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

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)


Another elegant, quick solution, based on some background mathematics as in comments:

-- Every third term is even, and every third term beautifully follows:
-- fib n = 4*fib n-3 + fib n-6
evenFibs = 2 : 8 : zipWith (+) (map (4*) (tail evenFibs)) evenFibs

-- So, evenFibs are: e(n) = 4*e(n-1) + e(n-2)
-- [there4]:4e(n)   = e(n+1) - e(n-1)
--          4e(n-1) = e(n)   - e(n-2)
--          4e(n-2) = e(n-1) - e(n-3)
--          ...
--          4e(3)   = e(4)   - e(2)
--          4e(2)   = e(3)   - e(1)
--          4e(1)   = e(2)   - e(0)
--         -------------------------------
-- Total: 4([sum] e(k) - e(0)) = e(n+1) + e(n) - e(1) - e(0)
-- => [sum] e(k) = (e(n+1) + e(n) - e(1) + 3e(0))/4 = 1089154 for
-- first 10 terms

sumEvenFibsBelow :: Int -> Int
sumEvenFibsBelow n = ((last $ take (x+1) evenFibs) +
                      (last $ take x evenFibs) -
                      8 + 6) `div` 4
  where x = length (takeWhile (<= n) evenFibs)

Problem 3

Find the largest prime factor of 600851475143.

Solution:

primes = 2 : filter (null . tail . 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 600851475143)

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]

Problem 5

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

Solution:

problem_5 = foldr1 lcm [1..20]

Another solution: 16*9*5*7*11*13*17*19. Product of maximal powers of primes in the range.

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])

Problem 7

Find the 10001st prime.

Solution:

--primes in problem_3
problem_7 = primes !! 10000

Problem 8

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

Solution:

import Data.Char 
import Data.List 

euler_8 = do
   str <- readFile "number.txt"
   print . maximum . map product
         . foldr (zipWith (:)) (repeat [])
         . take 13 . tails . map (fromIntegral . digitToInt)
         . concat . lines $ str

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

Problem 10

Calculate the sum of all the primes below one million.

Solution:

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