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July 19, 2014 23:29
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Ejercicio de Parcial Haskell
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| encontrarMinimo :: Perm a -> a -> Integer | |
| encontrarMinimo permutacion x0 = foldr (\x rec -> if ((aplicarNVeces permutacion x0 x)== x0) then rec else rec + 1) 0 [0..] | |
| ciclo :: Eq a => Perm a -> a -> a -> [a] | |
| ciclo p inicial = [aplicarNVeces p inicial x| x <- [0..(encontrarMinimo p inicial)]] | |
| aplicarNVeces :: Perm a -> a -> Int -> a | |
| aplicarNVeces p x0 x = foldr (\x rec -> (p rec)) x0 [0..x] |
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donde uso [0..] es porque el enunciado decia que se sabia que existia el valor