Sum and product puzzle: Difference between revisions
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Finally, as we expect and need only one solution, Haskell's lazy evaluation strategy will avoid wasted tests if we request only the first item from the possible solution stream.
<lang Haskell>import Data.List (intersect)
------------------ SUM AND PRODUCT PUZZLE ----------------
s1, s2, s3, s4 :: [(Int, Int)]
s1 =
[2 .. 100]
>>= \x ->
[succ x
>>= \y ->
[ (x, y)
| x + y < 100
]
s2 = filter (all ((1 /=) . length . mulEq) . sumEq) s1▼
add, mul :: (Int, Int) -> Int▼
add = uncurry (+)▼
s3 = filter ((1 ==) . length . (`intersect` s2) . mulEq) s2▼
mul = uncurry (*)▼
s4 = filter ((1 ==) . length . (`intersect` s3) . sumEq) s3▼
sumEq, mulEq :: (Int, Int) -> [(Int, Int)]
sumEq p = filter ((add p ==) . add) s1
mulEq p = filter ((mul p ==) . mul) s1
▲add, mul :: (Int, Int) -> Int
▲s2 = filter (all ((1 /=) . length . mulEq) . sumEq) s1
▲add = uncurry (+)
▲mul = uncurry (*)
▲s3 = filter ((1 ==) . length . (`intersect` s2) . mulEq) s2
▲s4 = filter ((1 ==) . length . (`intersect` s3) . sumEq) s3
main :: IO ()
main = print $ take 1 s4</lang>
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