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Line 2: Line 2:
 
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 =
+
import Data.List (union)
sum [ x |
+
problem_1' = sum (union [3,6..999] [5,10..999])
x <- [1..999],
+
(x `mod` 3 == 0) || (x `mod` 5 == 0)
+
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 =
+
problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
sum $ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999]
+
where
</haskell>
 
----
 
<haskell>
 
sumOnetoN n = n * (n+1) `div` 2
 
 
problem_1 =
 
sumStep 3 999 + sumStep 5 999 - sumStep 15 999
 
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 22: Line 24:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_2 =
+
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, even x]
sum [ x |
+
where
x <- takeWhile (<= 1000000) fibs,
 
x `mod` 2 == 0
 
]
 
where
 
 
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
 
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
 
</haskell>
 
</haskell>
Line 32: 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 =
+
problem_2 = sumEvenFibs $ numEvenFibsLessThan 1000000
sumEvenFibs $ numEvenFibsLessThan 1000000
+
where
sumEvenFibs n =
+
sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
(evenFib n + evenFib (n+1) - 2) `div` 4
+
evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
evenFib n =
+
numEvenFibsLessThan 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 46: Line 48:
 
problem_2 = sumEvenFibsLessThan 1000000
 
problem_2 = sumEvenFibsLessThan 1000000
   
sumEvenFibsLessThan n =
+
sumEvenFibsLessThan n = (a + b - 1) `div` 2
(a + b - 1) `div` 2
+
where
where
 
 
n2 = n `div` 2
 
n2 = n `div` 2
(a, b) =
+
(a, b) = foldr f (0,1)
foldr f (0,1) $
+
. takeWhile ((<= n2) . fst)
takeWhile ((<= n2) . fst) $
+
. iterate times2E $ (1, 4)
iterate times2E (1, 4)
+
f x y | fst z <= n2 = z
f x y
+
| otherwise = y
| fst z <= n2 = z
+
where z = x `addE` y
| otherwise = y
+
addE (a, b) (c, d) = (a*d + b*c - 4*ac, ac + b*d)
where z = x `addE` y
+
where ac=a*c
addE (a, b) (c, d) =
+
(a*d + b*c - 4*ac, ac + b*d)
+
times2E (a, b) = addE (a, b) (a, b)
where
 
ac=a*c
 
times2E (a, b) =
 
addE (a, b) (a, b)
 
 
</haskell>
 
</haskell>
   
Line 66: Line 68:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
primes =
+
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
2 : filter ((==1) . length . primeFactors) [3,5..]
+
primeFactors n =
+
primeFactors n = factor n primes
factor n primes
+
where
where
 
 
factor n (p:ps)
 
factor n (p:ps)
 
| p*p > n = [n]
 
| p*p > n = [n]
Line 75: Line 77:
 
| otherwise = factor n ps
 
| otherwise = factor n ps
   
problem_3 =
+
problem_3 = last (primeFactors 317584931803)
last (primeFactors 317584931803)
 
 
</haskell>
 
</haskell>
   
This can be improved by using
+
Another solution, not using recursion, is:
<hask>null . tail</hask>
+
<haskell>
instead of
+
problem_3 = (m !! 0) `div` (m !! 1)
<hask>(== 1) . length</hask>.
+
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] ==
Line 88: Line 90:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_4 =
+
problem_4 =
foldr max 0 [ x |
+
maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]
y <- [100..999],
 
z <- [100..999],
 
let x = y * z,
 
let s = show x,
 
s == reverse s
 
]
 
</haskell>
 
An alternative to avoid evaluating twice the same pair of numbers:
 
<haskell>
 
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 97: Line 99:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_5 =
+
problem_5 = foldr1 lcm [1..20]
head [ x |
 
x <- [2520,5040..],
 
all (\y -> x `mod` y == 0) [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>
   
Line 104: Line 106:
   
 
Solution:
 
Solution:
  +
<!--
 
<haskell>
 
<haskell>
problem_6 =
+
fun n = a - b
sum [ x^2 | x <- [1..100]] - (sum [1..100])^2
+
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 115: Line 118:
 
<haskell>
 
<haskell>
 
--primes in problem_3
 
--primes in problem_3
problem_7 =
+
problem_7 = primes !! 10000
head $ drop 10000 primes
 
 
</haskell>
 
</haskell>
 
As above, this can be improved by using
 
<hask>null . tail</hask>
 
instead of
 
<hask>(== 1) . length</hask>.
 
 
Here is an alternative that uses a
 
[http://www.haskell.org/pipermail/haskell-cafe/2007-February/022854.html sieve of Eratosthenes]:
 
 
<haskell>
 
primes' =
 
2 : 3 : sieve (tail primes') [5,7..]
 
where
 
sieve (p:ps) x =
 
h ++ sieve ps (filter (\q -> q `mod` p /= 0) t
 
where
 
(h, _:t) = span (p*p <) x
 
problem_7_v2 = primes' !! 10000
 
</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 five consecutive digits in the 1000-digit number.
   
 
Solution:
 
Solution:
  +
<!--
 
<haskell>
 
<haskell>
 
import Data.Char
 
import Data.Char
Line 147: Line 131:
 
take n xs : groupsOf n ( tail xs )
 
take n xs : groupsOf n ( tail xs )
 
 
problem_8 x=
+
problem_8 x = maximum . map product . groupsOf 5 $ x
maximum . map product . groupsOf 5 $ x
+
main = do t <- readFile "p8.log"
main=do
+
let digits = map digitToInt $concat $ lines t
t<-readFile "p8.log"
+
print $ problem_8 digits
let digits = map digitToInt $foldl (++) "" $ lines t
+
</haskell>
print $ problem_8 digits
+
-->
  +
<!-- 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 160: Line 144:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_9 =
+
triplets l = [[a,b,c] | m <- [2..limit],
head [a*b*c |
+
n <- [1..(m-1)],
a <- [1..500],
+
let a = m^2 - n^2,
b <- [a..500],
+
let b = 2*m*n,
let c = 1000-a-b,
+
let c = m^2 + n^2,
a^2 + b^2 == c^2
+
a+b+c==l]
]
+
where limit = floor . sqrt . fromIntegral $ l
</haskell>
 
   
Another solution using Pythagorean Triplets generation:
+
problem_9 = product . head . triplets $ 1000
<haskell>
 
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 176: Line 160:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_10 =
+
--primes in problem_3
sum (takeWhile (< 1000000) primes)
+
problem_10 = sum (takeWhile (< 1000000) primes)
 
</haskell>
 
</haskell>

Revision as of 14:17, 22 October 2012

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)