Suma ordenada de listas infinitas ordenadas

3 Comentarios

1. Haskell

   sumaOrdenada :: [Integer] -> [Integer] -> [Integer] sumaOrdenada (x:xs) (y:ys) = x+y : map (+x) ys `mezcla` map (+y) xs `mezcla` sumaOrdenada ys xs mezcla :: [Integer] -> [Integer] -> [Integer] mezcla (x:xs) (y:ys) | x == y = x : mezcla xs ys | x < y = x : mezcla xs (y:ys) | otherwise = y : mezcla (x:xs) ys

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   sumaOrdenada :: [Integer] -> [Integer] -> [Integer] sumaOrdenada (x:xs) (y:ys) =     x+y : map (+x) ys `mezcla` map (+y) xs `mezcla` sumaOrdenada ys xs   mezcla :: [Integer] -> [Integer] -> [Integer] mezcla (x:xs) (y:ys) | x == y    = x : mezcla xs ys                      | x < y     = x : mezcla xs (y:ys)                      | otherwise = y : mezcla (x:xs) ys

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2. Haskell

   sumaOrdenada :: [Integer] -> [Integer] -> [Integer] sumaOrdenada xs ys = [a | a <- [0..], isSum a xs (takeWhile (a>=) ys)] where isSum a (x:xs) ys | x > a = False | otherwise = elem (a-x) ys || isSum a xs ys

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   sumaOrdenada :: [Integer] -> [Integer] -> [Integer] sumaOrdenada xs ys = [a | a <- [0..], isSum a xs (takeWhile (a>=) ys)]         where isSum a (x:xs) ys | x > a = False                                 | otherwise = elem (a-x) ys || isSum a xs ys

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   1. Al comprobar uno a uno los números naturales, esta solución tiene problemas con las series compuestas de números grandes con grandes espacios. Por ejemplo:

      > let xs = iterate (^2) 2 > let ys = iterate (^2) 3 > (sumaOrdenada xs ys)!!30 1853020188851843

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      > let xs = iterate (^2) 2 > let ys = iterate (^2) 3 > (sumaOrdenada xs ys)!!30 1853020188851843

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