{"id":6967,"date":"2022-04-22T10:58:03","date_gmt":"2022-04-22T08:58:03","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=6967"},"modified":"2022-04-30T16:21:34","modified_gmt":"2022-04-30T14:21:34","slug":"ordenada-ciclicamente","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/ordenada-ciclicamente\/","title":{"rendered":"Ordenada c\u00edclicamente"},"content":{"rendered":"<p>Se dice que una sucesi\u00f3n x(1), &#8230;, x(n) est\u00e1 ordenada c\u00edclicamente si existe un \u00edndice i tal que la sucesi\u00f3n<\/p>\n<pre lang=\"text\">\n   x(i), x(i+1), ..., x(n), x(1), ..., x(i-1)\n<\/pre>\n<p>est\u00e1 ordenada crecientemente de forma estricta.<\/p>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   ordenadaCiclicamente :: Ord a => [a] -> Maybe Int\n<\/pre>\n<p>tal que <code>(ordenadaCiclicamente xs)<\/code> es el \u00edndice a partir del cual est\u00e1 ordenada, si la lista est\u00e1 ordenado c\u00edclicamente y <code>Nothing<\/code> en caso contrario. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   ordenadaCiclicamente [1,2,3,4]      ==  Just 0\n   ordenadaCiclicamente [5,8,1,3]      ==  Just 2\n   ordenadaCiclicamente [4,6,7,5,1,3]  ==  Nothing\n   ordenadaCiclicamente [1,0,3,2]      ==  Nothing\n   ordenadaCiclicamente [1,2,0]        ==  Just 2\n   ordenadaCiclicamente \"cdeab\"        ==  Just 3\n<\/pre>\n<p>Nota: Se supone que el argumento es una lista no vac\u00eda sin elementos repetidos.<\/p>\n<h4>Soluciones<\/h4>\n<pre lang=\"haskell\">\nmodule Ordenada_ciclicamente where\n\nimport Test.QuickCheck (Arbitrary, Gen, NonEmptyList (NonEmpty), Property,\n                        arbitrary, chooseInt, collect, quickCheck)\nimport Data.List       (nub, sort)\nimport Data.Maybe      (isJust, listToMaybe)\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\nordenadaCiclicamente1 :: Ord a => [a] -> Maybe Int\nordenadaCiclicamente1 xs = aux 0 xs\n  where n = length xs\n        aux i zs\n          | i == n      = Nothing\n          | ordenada zs = Just i\n          | otherwise   = aux (i+1) (siguienteCiclo zs)\n\n-- (ordenada xs) se verifica si la lista xs est\u00e1 ordenada\n-- crecientemente. Por ejemplo,\n--   ordenada \"acd\"   ==  True\n--   ordenada \"acdb\"  ==  False\nordenada :: Ord a => [a] -> Bool\nordenada []     = True\nordenada (x:xs) = all (x <) xs &#038;&#038; ordenada xs\n\n-- (siguienteCiclo xs) es la lista obtenida a\u00f1adiendo el primer elemento\n-- de xs al final del resto de xs. Por ejemplo,\n--   siguienteCiclo [3,2,5]  =>  [2,5,3]\nsiguienteCiclo :: [a] -> [a]\nsiguienteCiclo []     = []\nsiguienteCiclo (x:xs) = xs ++ [x]\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\nordenadaCiclicamente2 :: Ord a => [a] -> Maybe Int\nordenadaCiclicamente2 xs =\n  listToMaybe [n | n <- [0..length xs-1],\n                   ordenada (drop n xs ++ take n xs)]\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\nordenadaCiclicamente3 :: Ord a => [a] -> Maybe Int\nordenadaCiclicamente3 xs\n  | ordenada (bs ++ as) = Just k\n  | otherwise           = Nothing\n  where (_,k)   = minimum (zip xs [0..])\n        (as,bs) = splitAt k xs\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_ordenadaCiclicamente1 :: NonEmptyList Int -> Bool\nprop_ordenadaCiclicamente1 (NonEmpty xs) =\n  ordenadaCiclicamente1 xs == ordenadaCiclicamente2 xs\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_ordenadaCiclicamente1\n--    +++ OK, passed 100 tests.\n\n-- La propiedad para analizar los casos de prueba\nprop_ordenadaCiclicamente2 :: NonEmptyList Int -> Property\nprop_ordenadaCiclicamente2 (NonEmpty xs) =\n  collect (isJust (ordenadaCiclicamente1 xs)) $\n  ordenadaCiclicamente1 xs == ordenadaCiclicamente2 xs\n\n-- El an\u00e1lisis es\n--    \u03bb> quickCheck prop_ordenadaCiclicamente2\n--    +++ OK, passed 100 tests:\n--    89% False\n--    11% True\n\n-- Tipo para generar listas\nnewtype Lista = L [Int]\n  deriving Show\n\n-- Generador de listas.\nlistaArbitraria :: Gen Lista\nlistaArbitraria = do\n  x <- arbitrary\n  xs <- arbitrary\n  let ys = x : xs\n  k <- chooseInt (0, length ys)\n  let (as,bs) = splitAt k (sort (nub ys))\n  return (L (bs ++ as))\n\n-- Lista es una subclase de Arbitrary.\ninstance Arbitrary Lista where\n  arbitrary = listaArbitraria\n\n-- La propiedad para analizar los casos de prueba\nprop_ordenadaCiclicamente3 :: Lista -> Property\nprop_ordenadaCiclicamente3 (L xs) =\n  collect (isJust (ordenadaCiclicamente1 xs)) $\n  ordenadaCiclicamente1 xs == ordenadaCiclicamente2 xs\n\n-- El an\u00e1lisis es\n--    \u03bb> quickCheck prop_ordenadaCiclicamente3\n--    +++ OK, passed 100 tests (100% True).\n\n-- Tipo para generar\nnewtype Lista2 = L2 [Int]\n  deriving Show\n\n-- Generador de listas\nlistaArbitraria2 :: Gen Lista2\nlistaArbitraria2 = do\n  x' <- arbitrary\n  xs <- arbitrary\n  let ys = x' : xs\n  k <- chooseInt (0, length ys)\n  let (as,bs) = splitAt k (sort (nub ys))\n  n <- chooseInt (0,1)\n  return (if even n\n          then L2 (bs ++ as)\n          else L2 ys)\n\n-- Lista es una subclase de Arbitrary.\ninstance Arbitrary Lista2 where\n  arbitrary = listaArbitraria2\n\n-- La propiedad para analizar los casos de prueba\nprop_ordenadaCiclicamente4 :: Lista2 -> Property\nprop_ordenadaCiclicamente4 (L2 xs) =\n  collect (isJust (ordenadaCiclicamente1 xs)) $\n  ordenadaCiclicamente1 xs == ordenadaCiclicamente2 xs\n\n-- El an\u00e1lisis es\n--    \u03bb> quickCheck prop_ordenadaCiclicamente4\n--    +++ OK, passed 100 tests:\n--    51% True\n--    49% False\n\n-- La propiedad es\nprop_ordenadaCiclicamente :: Lista2 -> Bool\nprop_ordenadaCiclicamente (L2 xs) =\n  all (== ordenadaCiclicamente1 xs)\n      [ordenadaCiclicamente2 xs,\n       ordenadaCiclicamente3 xs]\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_ordenadaCiclicamente\n--    +++ OK, passed 100 tests.\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> ordenadaCiclicamente1 ([100..4000] ++ [1..99])\n--    Just 3901\n--    (3.27 secs, 2,138,864,568 bytes)\n--    \u03bb> ordenadaCiclicamente2 ([100..4000] ++ [1..99])\n--    Just 3901\n--    (2.44 secs, 1,430,040,008 bytes)\n--    \u03bb> ordenadaCiclicamente3 ([100..4000] ++ [1..99])\n--    Just 3901\n--    (1.18 secs, 515,549,200 bytes)\n<\/pre>\n<p>El c\u00f3digo se encuentra en <a href=\"https:\/\/github.com\/jaalonso\/Exercitium\/blob\/main\/src\/Ordenada_ciclicamente.hs\">GitHub<\/a>.<\/p>\n<p>La elaboraci\u00f3n de las soluciones se describe en el siguiente v\u00eddeo<\/p>\n<p><iframe loading=\"lazy\" width=\"560\" height=\"315\" src=\"https:\/\/www.youtube.com\/embed\/CI090GISHUc\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\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>Se dice que una sucesi\u00f3n x(1), &#8230;, x(n) est\u00e1 ordenada c\u00edclicamente si existe un \u00edndice i tal que la sucesi\u00f3n x(i), x(i+1), &#8230;, x(n), x(1), &#8230;, x(i-1) est\u00e1 ordenada crecientemente de forma estricta. Definir la funci\u00f3n ordenadaCiclicamente :: Ord a => [a] -> Maybe Int tal que (ordenadaCiclicamente xs) es el \u00edndice a partir del&#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":[2],"tags":[41,483,533,554,8,498,500,46,91,550,555,556,28,246,340,552,551,411,11,553,6,371,14,73,47,146],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6967"}],"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=6967"}],"version-history":[{"count":3,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6967\/revisions"}],"predecessor-version":[{"id":6987,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6967\/revisions\/6987"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=6967"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=6967"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=6967"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}