Difference between revisions of "Euler problems/1 to 10"
(Removing category tags. See Talk:Euler_problems) 

Line 4:  Line 4:  
Solution: 
Solution: 

<haskell> 
<haskell> 

−  problem_1 = sum [ x  
+  problem_1 = sum [ x  
+  x < [1..999], 

+  (x `mod` 3 == 0)  (x `mod` 5 == 0) 

+  ] 

</haskell> 
</haskell> 

Line 23:  Line 23:  
Solution: 
Solution: 

<haskell> 
<haskell> 

−  problem_2 = sum [ x  
+  problem_2 = sum [ x  
−  +  x < takeWhile (<= 1000000) fibs, 

+  x `mod` 2 == 0 

+  ] 

+  where 

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

</haskell> 
</haskell> 

Line 79:  Line 79:  
Solution: 
Solution: 

<haskell> 
<haskell> 

−  problem_4 = foldr max 0 [ x  
+  problem_4 = foldr max 0 [ x  
+  y < [100..999], 

+  z < [100..999], 

+  let x = y * z, 

+  let s = show x, 

+  s == reverse s 

+  ] 

</haskell> 
</haskell> 

An alternative to avoid evaluating twice the same pair of numbers: 
An alternative to avoid evaluating twice the same pair of numbers: 

<haskell> 
<haskell> 

−  problem_4' = foldr1 max [ x  
+  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 91:  Line 91:  
Solution: 
Solution: 

<haskell> 
<haskell> 

−  problem_5 = head [ x  
+  problem_5 = head [ x  
+  x < [2520,5040..], 

+  all (\y > x `mod` y == 0) [1..20] 

+  ] 

</haskell> 
</haskell> 

An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom: 
An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom: 

Line 140:  Line 140:  
Solution: 
Solution: 

<haskell> 
<haskell> 

−  num = ...  1000 digit number as a string 

+  import Data.Char 

−  digits = map digitToInt num 

−  
groupsOf _ [] = [] 
groupsOf _ [] = [] 

groupsOf n xs = take n xs : groupsOf n ( tail xs ) 
groupsOf n xs = take n xs : groupsOf n ( tail xs ) 

−  
+  
−  problem_8 = maximum . map product . groupsOf 5 $ 
+  problem_8 x= maximum . map product . groupsOf 5 $ x 
+  main=do 

+  t<readFile "p8.log" 

+  let digits=map digitToInt $foldl (++) ""$lines t 

+  print $problem_8 digits 

</haskell> 
</haskell> 

Line 159:  Line 157:  
Another solution using Pythagorean Triplets generation: 
Another solution using Pythagorean Triplets generation: 

<haskell> 
<haskell> 

−  triplets 
+  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 
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 

⚫  
−  where 

−  prod3 (a,b,c) = a*b*c 

</haskell> 
</haskell> 

Revision as of 10:57, 5 January 2008
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:
import Data.Char
groupsOf _ [] = []
groupsOf n xs = 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 $foldl (++) ""$lines t
print $problem_8 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 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:
problem_10 = sum (takeWhile (< 1000000) primes)