Add Haskell solutions for Problems 1, 2, 3
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Haskell/p001.hs
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Haskell/p001.hs
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-- If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
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--
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-- Find the sum of all the multiples of 3 or 5 below 1000.
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sumMultiples :: (Integral n) => n
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sumMultiples = sum(filter p [ n | n <- [1..999] ])
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where p n = n `mod` 3 == 0 || n `mod` 5 == 0
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main :: IO ()
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main = do
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let result = sumMultiples
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putStrLn $ "Project Euler, Problem 1\n"
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++ "Answer: " ++ (show result)
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Haskell/p002.hs
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Haskell/p002.hs
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-- Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
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--
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-- 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
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--
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-- By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
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fib :: (Integral n) => n -> n
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fib 0 = 0
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fib 1 = 1
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fib n = fib (n-1) + fib(n-2)
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sumEvenFib :: (Integral n) => n
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sumEvenFib = sum $ filter even $ takeWhile (<=4000000) (map fib [0..])
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main :: IO ()
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main = do
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let result = sumEvenFib
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putStrLn $ "Project Euler, Problem 2\n"
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++ "Answer: " ++ (show result)
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23
Haskell/p003.hs
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23
Haskell/p003.hs
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-- The prime factors of 13195 are 5, 7, 13 and 29. --
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-- What is the largest prime factor of the number 600851475143?
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isPrime :: (Integral n) => n -> Bool
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isPrime 1 = False
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isPrime 2 = True
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isPrime 3 = True
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isPrime n =
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n `mod` 2 /= 0 && n `mod` 3 /= 0 && null [ x | x <- candidates, n `mod` x == 0 || n `mod` (x+2) == 0 ]
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where candidates = [5,11..limit]
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limit = floor(sqrt(fromIntegral n)) + 1
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maxPrimeFactor :: (Integral n) => n -> n
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maxPrimeFactor n
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| isPrime n = n
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| n `mod` 2 == 0 = maxPrimeFactor $ fromIntegral n `div` 2
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| otherwise = maxPrimeFactor $ fromIntegral n `div` head [i | i <- [3,5..], n `mod` i == 0 && isPrime i]
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main :: IO ()
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main = do
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let result = maxPrimeFactor 600851475143
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putStrLn $ "Project Euler, Problem 3\n"
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++ "Answer: " ++ (show result)
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