{"id":6007,"date":"2021-01-26T06:00:45","date_gmt":"2021-01-26T04:00:45","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=6007"},"modified":"2021-02-02T08:57:50","modified_gmt":"2021-02-02T06:57:50","slug":"limitacion-del-numero-de-repeticiones","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/limitacion-del-numero-de-repeticiones\/","title":{"rendered":"Limitaci\u00f3n del n\u00famero de repeticiones"},"content":{"rendered":"<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   conRepeticionesAcotadas :: Eq a => [a] -> Int -> [a]\n<\/pre>\n<p>tal que (conRepeticionesAcotadas xs n) es una lista que contiene cada elemento de xs como m\u00e1ximo n veces sin reordenar (se supone que n es un n\u00famero positivo).. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   conRepeticionesAcotadas [1,2,3,1,2,1,3,2,3,5] 1  ==  [1,2,3,5]\n   conRepeticionesAcotadas [1,2,3,1,2,1,3,2,3,5] 2  ==  [1,2,3,1,2,3,5]\n   conRepeticionesAcotadas [1,2,3,1,2,1,3,2,3,5] 3  ==  [1,2,3,1,2,1,3,2,3,5]\n   conRepeticionesAcotadas [1,2,3,1,2,1,3,2,3,5] 4  ==  [1,2,3,1,2,1,3,2,3,5]\n<\/pre>\n<h4>Soluciones<\/h4>\n<pre lang=\"haskell\">\nimport Data.List (foldl')\nimport Data.Maybe (fromJust, isNothing)\nimport Test.QuickCheck (Property, (==>),quickCheck)\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\nconRepeticionesAcotadas :: Eq a => [a] -> Int -> [a]\nconRepeticionesAcotadas xs n = reverse (aux [] xs)\n  where aux zs []     = zs\n        aux zs (y:ys) | m < n     = aux (y:zs) ys\n                      | otherwise = aux zs ys\n          where m = nOcurrencias y zs\n\n-- (nOcurrencias x ys) es el n\u00famero de ocurrencias de x en ys. Por\n-- ejemplo,\n--    nOcurrencias 7 [7,2,7,7,5]  ==  3\nnOcurrencias :: Eq a => a -> [a] -> Int\nnOcurrencias x ys = length (filter (== x) ys)\n\n-- Se puede simplificar la definici\u00f3n de nOcurrencias:\nnOcurrencias2 :: Eq a => a -> [a] -> Int\nnOcurrencias2 x = length . filter (== x)\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\nconRepeticionesAcotadas2 :: Eq a => [a] -> Int -> [a]\nconRepeticionesAcotadas2 xs n = reverse (foldl' aux [] xs)\n  where aux zs y | m < n     = y:zs\n                 | otherwise = zs\n          where m = nOcurrencias y zs\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\nconRepeticionesAcotadas3 :: Eq a => [a] -> Int -> [a]\nconRepeticionesAcotadas3 xs n = reverse (aux [] [] xs)\n  where aux as _ []      = as\n        aux as bs (y:ys) | y `elem` bs = aux as bs ys\n                         | m < n       = aux (y:as) bs ys\n                         | otherwise   = aux as (y:bs) ys\n          where m = nOcurrencias y as\n\n-- 4\u00aa soluci\u00f3n\n-- ===========\n\nconRepeticionesAcotadas4 :: Eq a => [a] -> Int -> [a]\nconRepeticionesAcotadas4 xs n = aux xs []\n  where aux [] _      = []\n        aux (y:ys) ps | r == Nothing = y : aux ys ((y,1) : ps)\n                      | m < n        = y : aux ys ((y,m+1) : ps)\n                      | otherwise    = aux ys ps\n                      where r = busca y ps\n                            Just m = r\n\n-- (busca x ps) es justamente la segunda componente del primer par de ps\n-- cuya primera componente es xs, si ps tiene alg\u00fan par cuya primera\n-- componente es x; y Nothing en caso contrario. Por ejemplo,\n--    busca 'a' [('b',2),('a',3),('a',1)]  ==  Just 3\n--    busca 'c' [('b',2),('a',3),('a',1)]  ==  Nothing\nbusca :: Eq a => a -> [(a,b)] -> Maybe b\nbusca x ps\n  | null ys   = Nothing\n  | otherwise = Just (head ys)\n  where ys = [n | (y,n) <- ps, y == x]\n\n-- 5\u00aa soluci\u00f3n\n-- ===========\n\nconRepeticionesAcotadas5 :: Eq a => [a] -> Int -> [a]\nconRepeticionesAcotadas5 xs n = aux xs []\n  where aux [] _      = []\n        aux (y:ys) ps | isNothing r = y : aux ys ((y,1) : ps)\n                      | m < n       = y : aux ys ((y,m+1) : ps)\n                      | otherwise   = aux ys ps\n                      where r = lookup y ps\n                            m = fromJust r\n\n-- Equivalencia de las definiciones\n-- ================================\n\n-- La propiedad es\nprop_conRepeticionesAcotadas :: [Int] -> Int -> Property\nprop_conRepeticionesAcotadas xs n =\n  n > 0 ==>\n  all (==(conRepeticionesAcotadas xs n))\n      [ conRepeticionesAcotadas2 xs n\n      , conRepeticionesAcotadas3 xs n\n      , conRepeticionesAcotadas4 xs n\n      , conRepeticionesAcotadas5 xs n]\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_conRepeticionesAcotadas\n--    +++ OK, passed 100 tests.\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> length (conRepeticionesAcotadas (concat [[1..n] | n <- [1..500]]) 2)\n--    999\n--    (5.14 secs, 64,372,768 bytes)\n--    \u03bb> length (conRepeticionesAcotadas2 (concat [[1..n] | n <- [1..500]]) 2)\n--    999\n--    (4.95 secs, 62,322,880 bytes)\n--    \u03bb> length (conRepeticionesAcotadas3 (concat [[1..n] | n <- [1..500]]) 2)\n--    999\n--    (0.38 secs, 38,764,952 bytes)\n--    \u03bb> length (conRepeticionesAcotadas4 (concat [[1..n] | n <- [1..500]]) 2)\n--    999\n--    (5.66 secs, 2,429,904,144 bytes)\n--    \u03bb> length (conRepeticionesAcotadas5 (concat [[1..n] | n <- [1..500]]) 2)\n--    999\n--    (0.68 secs, 48,536,872 bytes)\n<\/pre>\n<h4>Nuevas soluciones<\/h4>\n<ul>\n<li>En los comentarios se pueden escribir nuevas soluciones.\n<li>El c\u00f3digo se debe escribir entre una l\u00ednea con &#60;pre lang=&quot;haskell&quot;&#62; y otra con &#60;\/pre&#62;\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Definir la funci\u00f3n conRepeticionesAcotadas :: Eq a => [a] -> Int -> [a] tal que (conRepeticionesAcotadas xs n) es una lista que contiene cada elemento de xs como m\u00e1ximo n veces sin reordenar (se supone que n es un n\u00famero positivo).. Por ejemplo, conRepeticionesAcotadas [1,2,3,1,2,1,3,2,3,5] 1 == [1,2,3,5] conRepeticionesAcotadas [1,2,3,1,2,1,3,2,3,5] 2 == [1,2,3,1,2,3,5] conRepeticionesAcotadas [1,2,3,1,2,1,3,2,3,5]&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"footnotes":"","_jetpack_memberships_contains_paid_content":false},"categories":[4],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6007"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/comments?post=6007"}],"version-history":[{"count":2,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6007\/revisions"}],"predecessor-version":[{"id":6041,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6007\/revisions\/6041"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=6007"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=6007"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=6007"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}