Difference between revisions of "Euler problems/1 to 10"
(Use null.tail or sieve to find primes) 
(Added problem_2_v3) 

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where fibs = 1 : 1 : zipWith (+) fibs (tail fibs) 
where fibs = 1 : 1 : zipWith (+) fibs (tail fibs) 

</haskell> 
</haskell> 

−   

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

+  the Fibonacci sequence themselves form a Fibonaccilike sequence 

+  that satisfies 

⚫  
<haskell> 
<haskell> 

problem_2_v2 = sumEvenFibs $ numEvenFibsLessThan 1000000 
problem_2_v2 = sumEvenFibs $ numEvenFibsLessThan 1000000 

Line 36:  Line 40:  
</haskell> 
</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> 
<haskell> 

−  2 * evenFib (2*n+1) == evenFib n ^2 + evenFib (n+1) ^ 2 

+  problem_2_v3 = 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) = let ac = a*c in (a*d + b*c  4*ac, ac + b*d) 

+  times2E (a, b) = addE (a, b) (a, b) 

</haskell> 
</haskell> 

Revision as of 19:04, 23 September 2007
Contents
Problem 1
Add all the natural numbers below 1000 that are multiples of 3 or 5.
Solution:
problem_1 = sum [ x  x < [1..999], (x `mod` 3 == 0)  (x `mod` 5 == 0)]
problem_1_v2 = sum $ filter (\x > ( x `mod` 3 == 0  x `mod` 5 == 0 ) ) [1..999]
sum1to n = n * (n+1) `div` 2
problem_1_v3 = sumStep 3 999 + sumStep 5 999  sumStep 15 999
where sumStep s n = s * sum1to (n `div` s)
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, x `mod` 2 == 0]
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_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_v3 = 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) = let ac = a*c in (a*d + b*c  4*ac, ac + b*d)
times2E (a, b) = addE (a, b) (a, b)
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)
This can be improved by using
null . tail
instead of
(== 1) . length
.
Problem 4
Find the largest palindrome made from the product of two 3digit numbers.
Solution:
problem_4 = foldr max 0 [ x  y < [100..999], z < [100..999], let x = y * z, let s = show x, s == reverse s]
An alternative to avoid evaluating twice the same pair of numbers:
problem_4' = foldr1 max [ 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 = head [ x  x < [2520,5040..], all (\y > x `mod` y == 0) [1..20]]
An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom:
problem_5' = foldr1 lcm [1..20]
Problem 6
What is the difference between the sum of the squares and the square of the sums?
Solution:
problem_6 = sum [ x^2  x < [1..100]]  (sum [1..100])^2
Problem 7
Find the 10001st prime.
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_7 = head $ drop 10000 primes
As above, this can be improved by using
null . tail
instead of
(== 1) . length
.
Here is an alternative that uses a sieve of Eratosthenes:
primes' = 2 : 3 : sieve (tail primes') [5,7..]
where
sieve (p:ps) x = let (h, _:t) = span (p*p <) x
in h ++ sieve ps (filter (\q > q `mod` p /= 0) t
problem_7_v2 = primes' !! 10000
Problem 8
Discover the largest product of five consecutive digits in the 1000digit number.
Solution:
num = ...  1000 digit number as a string
digits = map digitToInt num
groupsOf _ [] = []
groupsOf n xs = take n xs : groupsOf n ( tail xs )
problem_8 = maximum . map product . groupsOf 5 $ digits
Problem 9
There is only one Pythagorean triplet, {a, b, c}, for which a + b + c = 1000. Find the product abc.
Solution:
problem_9 = head [a*b*c  a < [1..500], b < [a..500], let c = 1000ab, a^2 + b^2 == c^2]
Another solution using Pythagorean Triplets generation:
triplets :: Int > [(Int, Int, Int)]
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]
where limit = floor $ sqrt $ fromIntegral l
tripletWithLength :: Int > [(Int, Int, Int)]
tripletWithLength n = filter ((==n) . f) $ triplets n
where
f (a,b,c) = a+b+c
problem_9 :: Int
problem_9 = prod3 $ head $ tripletWithLength 1000
where
prod3 (a,b,c) = a*b*c
Problem 10
Calculate the sum of all the primes below one million.
Solution:
problem_10 = sum (takeWhile (< 1000000) primes)