Permutación de elementos consecutivos

José A. Alonso, 18-marzo-2019

Definir la función

   permutaConsecutivos :: [a] -> [a]

tal que (permutaConsecutivos xs) es la lista obtenida permutando los elementos consecutivos de xs. Por ejemplo,

   permutaConsecutivos [1..8]         ==  [2,1,4,3,6,5,8,7]
   permutaConsecutivos [1..9]         ==  [2,1,4,3,6,5,8,7,9]
   permutaConsecutivos "simplemente"  ==  "ispmelemtne"

Soluciones

import Data.Array
 
-- 1ª solución
-- ===========
 
permutaConsecutivos :: [a] -> [a]
permutaConsecutivos (x:y:zs) = y : x : permutaConsecutivos zs
permutaConsecutivos xs       = xs
 
-- 2ª solución
-- ===========
 
permutaConsecutivos2 :: [a] -> [a]
permutaConsecutivos2 xs 
  | even n    = elems (array (1,n) [(i,f i) | i <- [1..n]])
  | otherwise = elems (array (1,n) ((n,v!n) : [(i,f i) | i <- [1..n-1]]))
  where
    n = length xs
    v = listArray (1,n) xs
    f i | even i =    v ! (i - 1)
        | otherwise = v ! (i + 1)
 
-- Comparación de eficiencia
-- =========================
 
-- La comparación es
--    λ> length (permutaConsecutivos [1..(3*10^6)])
--    3000000
--    (2.21 secs, 504,102,648 bytes)
--    λ> length (permutaConsecutivos2 [1..(3*10^6)])
--    3000000
--    (6.18 secs, 1,248,127,688 bytes)

Pensamiento

Entre el vivir y el soñar
hay una tercera cosa.
Adivínala.

Antonio Machado

Inicial

Recursión

8 soluciones de “Permutación de elementos consecutivos”

Responder
frahidzam
18-marzo-2019
permutaConsecutivos :: [a] -> [a] permutaConsecutivos [] = [] permutaConsecutivos [x] = [x] permutaConsecutivos (x:y:xs) = y:x:(permutaConsecutivos xs)
permutaConsecutivos :: [a] -> [a] permutaConsecutivos [] = [] permutaConsecutivos [x] = [x] permutaConsecutivos (x:y:xs) = y:x:(permutaConsecutivos xs)
Responder
adogargon
18-marzo-2019
permutaConsecutivos :: [a] -> [a] permutaConsecutivos = concatMap reverse. divide divide :: [a] -> [[a]] divide [] = [] divide [x] = [[x]] divide (x:y:xs) = [x,y]:divide xs
permutaConsecutivos :: [a] -> [a] permutaConsecutivos = concatMap reverse. divide divide :: [a] -> [[a]] divide [] = [] divide [x] = [[x]] divide (x:y:xs) = [x,y]:divide xs
Responder
luipromor
18-marzo-2019
permutaConsecutivos :: [a] -> [a] permutaConsecutivos [] = [] permutaConsecutivos [x] = [x] permutaConsecutivos (x:y:xs) = y:x: permutaConsecutivos xs
permutaConsecutivos :: [a] -> [a] permutaConsecutivos [] = [] permutaConsecutivos [x] = [x] permutaConsecutivos (x:y:xs) = y:x: permutaConsecutivos xs
Responder
javmarcha1
18-marzo-2019
permutaConsecutivos :: [a] -> [a] permutaConsecutivos xs | length xs < 2 = xs permutaConsecutivos (x:y:xs) = y:[x] ++ permutaConsecutivos xs
permutaConsecutivos :: [a] -> [a] permutaConsecutivos xs | length xs < 2 = xs permutaConsecutivos (x:y:xs) = y:[x] ++ permutaConsecutivos xs
Responder
ireprirod
18-marzo-2019
permutaConsecutivos :: [a] -> [a] permutaConsecutivos [] = [] permutaConsecutivos [x] = [x] permutaConsecutivos (x:y:xs) = y:x:permutaConsecutivos xs
permutaConsecutivos :: [a] -> [a] permutaConsecutivos [] = [] permutaConsecutivos [x] = [x] permutaConsecutivos (x:y:xs) = y:x:permutaConsecutivos xs
Responder
frahidzam
20-marzo-2019
Solución en Maxima
permutaConsecutivos (xs) := if emptyp (xs) then [] elseif emptyp (rest(xs,1)) then [first (xs)] else append ([second (xs), first (xs)],permutaConsecutivos(rest (xs,2)))$
permutaConsecutivos (xs) := if emptyp (xs) then [] elseif emptyp (rest(xs,1)) then [first (xs)] else append ([second (xs), first (xs)],permutaConsecutivos(rest (xs,2)))$
Responder
alebarmor1
21-marzo-2019
```
consec [] = []
consec [x] = [(x,x)]
consec (x:y:xs) = [(x,y)] ++ (consec xs)
twist p = (snd p, fst p)
giroLista xs = map (twist) xs
parALista p = [fst p, snd p]
listaParLista xs = concat (map (parALista) xs)
permutaConsecutivos xs = take (length xs) (listaParLista (giroLista (consec xs)))
```
Responder

ireprirod
28-abril-2019

Definición en Maxima:
permutaConsecutivos (xs):= if (emptyp (xs) or length (xs)=1) then xs
else cons(second(xs),cons(first(xs), permutaConsecutivos(rest(xs,2))))$