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Euler problems/1 to 10

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== [http://projecteuler.net/index.php?section=view&id=1 Problem 1] ==
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Do them on your own!
Add all the natural numbers below 1000 that are multiples of 3 or 5.
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Solution:
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<haskell>
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problem_1 =
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    sum [ x |
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    x <- [1..999],
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    (x `mod` 3 == 0) ||  (x `mod` 5 == 0)
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    ]
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</haskell>
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<haskell>
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problem_1_v2 =
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    sum $ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999]
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</haskell>
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----
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<haskell>
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sumOnetoN n = n * (n+1) `div` 2
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problem_1 =
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    sumStep 3 999 + sumStep 5 999 - sumStep 15 999
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    where
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    sumStep s n = s * sumOnetoN (n `div` s)
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</haskell>
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== [http://projecteuler.net/index.php?section=view&id=2 Problem 2] ==
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Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million.
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Solution:
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<haskell>
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problem_2 =
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    sum [ x |
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    x <- takeWhile (<= 1000000) fibs,
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    x `mod` 2 == 0
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    ]
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    where
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    fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
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</haskell>
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The following two solutions use the fact that the even-valued terms in
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the Fibonacci sequence themselves form a Fibonacci-like sequence
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that satisfies
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<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>.
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<haskell>
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problem_2_v2 =
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    sumEvenFibs $ numEvenFibsLessThan 1000000
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sumEvenFibs n =
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    (evenFib n + evenFib (n+1) - 2) `div` 4
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evenFib n =
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    round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
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numEvenFibsLessThan n =
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    floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)
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</haskell>
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The first two solutions work because 10^6 is small.
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The following solution also works for much larger numbers
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(up to at least 10^1000000 on my computer):
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<haskell>
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problem_2 = sumEvenFibsLessThan 1000000
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sumEvenFibsLessThan n =
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    (a + b - 1) `div` 2
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    where
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    n2 = n `div` 2
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    (a, b) =
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        foldr f (0,1) $
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        takeWhile ((<= n2) . fst) $
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        iterate times2E (1, 4)
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    f x y
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        | fst z <= n2 = z
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        | otherwise  = y
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        where z = x `addE` y
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addE (a, b) (c, d) =
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    (a*d + b*c - 4*ac, ac + b*d)
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    where
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    ac=a*c
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times2E (a, b) =
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    addE (a, b) (a, b)
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</haskell>
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== [http://projecteuler.net/index.php?section=view&id=3 Problem 3] ==
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Find the largest prime factor of 317584931803.
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Solution:
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<haskell>
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primes =
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    2 : filter ((==1) . length . primeFactors) [3,5..]
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primeFactors n =
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    factor n primes
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    where
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    factor n (p:ps)
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        | p*p > n        = [n]
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        | n `mod` p == 0 = p : factor (n `div` p) (p:ps)
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        | otherwise      = factor n ps
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problem_3 =
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    last (primeFactors 317584931803)
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</haskell>
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This can be improved by using
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<hask>null . tail</hask>
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instead of
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<hask>(== 1) . length</hask>.
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== [http://projecteuler.net/index.php?section=view&id=4 Problem 4] ==
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Find the largest palindrome made from the product of two 3-digit numbers.
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Solution:
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<haskell>
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problem_4 =
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    foldr max 0 [ x |
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    y <- [100..999],
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    z <- [100..999],
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    let x = y * z,
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    let s = show x,
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    s == reverse s
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    ]
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</haskell>
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An alternative to avoid evaluating twice the same pair of numbers:
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<haskell>
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problem_4' =
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    foldr1 max [ x |
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    y <- [100..999],
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    z <- [y..999],
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    let x = y * z,
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    let s = show x,
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    s == reverse s
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    ]
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</haskell>
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== [http://projecteuler.net/index.php?section=view&id=5 Problem 5] ==
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What is the smallest number divisible by each of the numbers 1 to 20?
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Solution:
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<haskell>
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problem_5 =
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    head [ x |
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    x <- [2520,5040..],
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    all (\y -> x `mod` y == 0) [1..20]
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    ]
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</haskell>
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An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom:
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<haskell>
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problem_5' = foldr1 lcm [1..20]
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</haskell>
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== [http://projecteuler.net/index.php?section=view&id=6 Problem 6] ==
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What is the difference between the sum of the squares and the square of the sums?
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Solution:
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<haskell>
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problem_6 =
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    sum [ x^2 | x <- [1..100]] - (sum [1..100])^2
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</haskell>
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== [http://projecteuler.net/index.php?section=view&id=7 Problem 7] ==
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Find the 10001st prime.
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Solution:
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<haskell>
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--primes in problem_3
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problem_7 =
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    head $ drop 10000 primes
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</haskell>
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As above, this can be improved by using
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<hask>null . tail</hask>
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instead of
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<hask>(== 1) . length</hask>.
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Here is an alternative that uses a
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[http://www.haskell.org/pipermail/haskell-cafe/2007-February/022854.html sieve of Eratosthenes]:
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<haskell>
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primes' =
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    2 : 3 : sieve (tail primes') [5,7..]
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    where
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    sieve (p:ps) x =
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        h ++ sieve ps (filter (\q -> q `mod` p /= 0) t
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        where
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        (h, _:t) = span (p*p <) x
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problem_7_v2 = primes' !! 10000
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</haskell>
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== [http://projecteuler.net/index.php?section=view&id=8 Problem 8] ==
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Discover the largest product of five consecutive digits in the 1000-digit number.
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Solution:
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<haskell>
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import Data.Char
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groupsOf _ [] = []
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groupsOf n xs =
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    take n xs : groupsOf n ( tail xs )
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problem_8 x=
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    maximum . map product . groupsOf 5 $ x
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main=do
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    t<-readFile "p8.log"
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    let digits = map digitToInt $foldl (++) "" $ lines t
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    print $ problem_8 digits
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</haskell>
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== [http://projecteuler.net/index.php?section=view&id=9 Problem 9] ==
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There is only one Pythagorean triplet, {''a'', ''b'', ''c''}, for which ''a'' + ''b'' + ''c'' = 1000.  Find the product ''abc''.
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Solution:
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<haskell>
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problem_9 =
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    head [a*b*c |
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    a <- [1..500],
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    b <- [a..500],
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    let c = 1000-a-b,
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    a^2 + b^2 == c^2
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    ]
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</haskell>
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Another solution using Pythagorean Triplets generation:
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<haskell>
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triplets l =  [[a,b,c]|
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    m <- [2..limit],
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    n <- [1..(m-1)],
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    let a = m^2 - n^2,
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    let b = 2*m*n,
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    let c = m^2 + n^2,
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    a+b+c==l
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    ]
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    where limit = floor $ sqrt $ fromIntegral l
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problem_9 = product $ head $ triplets 1000
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</haskell>
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== [http://projecteuler.net/index.php?section=view&id=10 Problem 10] ==
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Calculate the sum of all the primes below one million.
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Solution:
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<haskell>
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problem_10 =
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    sum (takeWhile (< 1000000) primes)
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</haskell>
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Revision as of 21:39, 29 January 2008

Do them on your own!