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(Problem 5)
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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>
sumOnetoN n = n * (n+1) `div` 2
+
import Data.List (union)
problem_1 =
+
problem_1' = sum (union [3,6..999] [5,10..999])
sumStep 3 999 + sumStep 5 999 - sumStep 15 999
+
where
+
problem_1 = sum [x | x <- [1..999], x `mod` 3 == 0 || x `mod` 5 == 0]
  +
</haskell>
  +
  +
Another solution which uses algebraic relationships:
  +
  +
<haskell>
  +
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 16: Line 17:
 
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 26: Line 27:
 
<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 40: Line 41:
 
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 60: Line 61:
 
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 69: Line 70:
 
| otherwise = factor n ps
 
| otherwise = factor n ps
   
problem_3 =
+
problem_3 = last (primeFactors 317584931803)
last (primeFactors 317584931803)
 
 
</haskell>
 
</haskell>
  +
  +
Another solution, not using recursion, is:
  +
<haskell>
  +
problem_3 = (m !! 0) `div` (m !! 1)
  +
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] ==
 
Find the largest palindrome made from the product of two 3-digit numbers.
 
Find the largest palindrome made from the product of two 3-digit numbers.
Line 76: Line 87:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_4 =
+
problem_4 =
foldr1 max [ x |
+
maximum [x | y<-[100..999], z<-[y..999], let x=y*z, let s=show x, s==reverse s]
y <- [100..999],
 
z <- [y..999],
 
let x = y * z,
 
let s = show x,
 
s == reverse s
 
]
 
 
</haskell>
 
</haskell>
   
Line 85: Line 96:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
--http://www.research.att.com/~njas/sequences/A003418
 
 
problem_5 = foldr1 lcm [1..20]
 
problem_5 = foldr1 lcm [1..20]
 
</haskell>
 
</haskell>
Line 93: Line 103:
   
 
Solution:
 
Solution:
  +
<!--
 
<haskell>
 
<haskell>
fun n=
+
fun n = a - b
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
 
where
a=div (n^2 * (n+1)^2) 4
+
a = (sum [1..n])^2
b=div (n * (n+1) * (2*n+1)) 6
+
b = sum (map (^2) [1..n])
problem_6=fun 100
+
  +
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 108: Line 119:
 
<haskell>
 
<haskell>
 
--primes in problem_3
 
--primes in problem_3
problem_7 =
+
problem_7 = primes !! 10000
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] ==
Line 114: Line 125:
   
 
Solution:
 
Solution:
  +
<!--
 
<haskell>
 
<haskell>
 
import Data.Char
 
import Data.Char
Line 120: Line 132:
 
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 133: Line 145:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
triplets l = [[a,b,c]|
+
triplets l = [[a,b,c] | m <- [2..limit],
m <- [2..limit],
+
n <- [1..(m-1)],
n <- [1..(m-1)],
+
let a = m^2 - n^2,
let a = m^2 - n^2,
+
let b = 2*m*n,
let b = 2*m*n,
+
let c = m^2 + n^2,
let c = m^2 + n^2,
+
a+b+c==l]
a+b+c==l
+
where limit = floor . sqrt . fromIntegral $ l
]
+
where limit = floor $ sqrt $ fromIntegral l
+
problem_9 = product . head . triplets $ 1000
problem_9 = product $ head $ triplets 1000
 
 
</haskell>
 
</haskell>
   
Line 149: Line 161:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
--http://www.research.att.com/~njas/sequences/A046731
+
--primes in problem_3
problem_10 =
+
problem_10 = sum (takeWhile (< 1000000) primes)
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)