Difference between revisions of "Euler problems/1 to 10"
m (Completed definition of primes in problem 7) 
(Added solution to problem 2) 

(55 intermediate revisions by 17 users not shown)  
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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 
+  problem_2 = sum [ x  x < takeWhile (<= 1000000) fibs, even x] 
−  where 
+  where 
+  fibs = 1 : 1 : zipWith (+) fibs (tail fibs) 

+  </haskell> 

+  
+  The following two solutions use the fact that the evenvalued terms in 

+  the Fibonacci sequence themselves form a Fibonaccilike 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 n3 + fib n6 

+  evenFibs = 2 : 8 : zipWith (+) (map (4*) (tail evenFibs)) evenFibs 

+  
+   So, evenFibs are: e(n) = 4*e(n1) + e(n2) 

+   [there4]:4e(n) = e(n+1)  e(n1) 

+   4e(n1) = e(n)  e(n2) 

+   4e(n2) = e(n1)  e(n3) 

+   ... 

+   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 
+  Find the largest prime factor of 600851475143. 
Solution: 
Solution: 

<haskell> 
<haskell> 

−  primes = 2 : filter ( 
+  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 

−  +  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 
+  problem_3 = last (primeFactors 600851475143) 
</haskell> 
</haskell> 

Line 40:  Line 47:  
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 = 
+  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 
+  Discover the largest product of thirteen consecutive digits in the 1000digit 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> 

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Solution: 
Solution: 

<haskell> 
<haskell> 

−  problem_9 = head [a*b*c  a < [1..500], b < [a..500], let c = 1000ab, a^2 + b^2 == c^2] 

+  triplets l = [[a,b,c]  m < [2..limit], 

+  n < [1..(m1)], 

+  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
Contents
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 evenvalued 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 evenvalued terms in
the Fibonacci sequence themselves form a Fibonaccilike 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 n3 + fib n6
evenFibs = 2 : 8 : zipWith (+) (map (4*) (tail evenFibs)) evenFibs
 So, evenFibs are: e(n) = 4*e(n1) + e(n2)
 [there4]:4e(n) = e(n+1)  e(n1)
 4e(n1) = e(n)  e(n2)
 4e(n2) = e(n1)  e(n3)
 ...
 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 3digit 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 1000digit 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..(m1)],
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)