{"id":599,"date":"2014-11-20T07:56:56","date_gmt":"2014-11-20T05:56:56","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=599"},"modified":"2014-12-27T20:32:50","modified_gmt":"2014-12-27T18:32:50","slug":"inversa-a-trozos","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/inversa-a-trozos\/","title":{"rendered":"Inversa a trozos"},"content":{"rendered":"<h4>Enunciado<\/h4>\n<pre lang=\"text\">\n-- Definir la funci\u00f3n \n--    inversa :: Int -> [a] -> [a]\n-- tal que (inversa k xs) es la lista obtenida invirtiendo elementos de\n-- xs, k elementos cada vez. Si el n\u00famero de elementos de xs no es un\n-- m\u00faltiplo de k, entonces los finales elementos de xs se dejen sin\n-- invertir. Por ejemplo,\n--    inversa 3 [1..11]  ==  [3,2,1,6,5,4,9,8,7,10,11]\n--    inversa 4 [1..11]  ==  [4,3,2,1,8,7,6,5,9,10,11]\n--\n-- Comprobar con QuickCheck que la funci\u00f3n inversa es involutiva; es\n-- decir, para todo n\u00famero k>0 y toda lista xs, se tiene que\n-- (inversa k (inversa k xs)) es igual a xs\n<\/pre>\n<h4>Soluciones<\/h4>\n<p>[schedule expon=&#8217;2014-11-27&#8242; expat=\u00bb06:00&#8243;]<\/p>\n<ul>\n<li>Las soluciones se pueden escribir en los comentarios hasta el 27 de noviembre.\n<li>El c\u00f3digo se debe escribir entre una l\u00ednea con &#60;pre lang=\u00bbhaskell\u00bb&#62; y otra con &#60;\/pre&#62;\n<\/ul>\n<p>[\/schedule]<\/p>\n<p>[schedule on=&#8217;2014-11-27&#8242; at=\u00bb06:00&#8243;]<\/p>\n<pre lang=\"haskell\">\r\nimport Test.QuickCheck\r\n\r\n-- 1\u00aa definici\u00f3n\r\ninversa1 :: Int -> [a] -> [a]\r\ninversa1 k xs \r\n    | length xs < k = xs\r\n    | otherwise     = reverse (take k xs) ++ inversa1 k (drop k xs) \r\n\r\n-- 2\u00aa definici\u00f3n \r\ninversa2 :: Int -> [a] -> [a]\r\ninversa2 k xs = aux xs (length xs) where\r\n    aux xs n\r\n        | n < k     = xs\r\n        | otherwise = reverse (take k xs) ++ aux (drop k xs) (n-k)\r\n\r\n\r\n-- La dos definiciones son equivalentes\r\nprop_equivalencia ::Int -> [Int] -> Property\r\nprop_equivalencia k xs =\r\n    k > 0 ==> inversa1 k xs == inversa2 k xs\r\n\r\n-- La comprobaci\u00f3n es\r\n--    ghci> quickCheck prop_equivalencia\r\n--    +++ OK, passed 100 tests.\r\n\r\n-- La segunda es m\u00e1s eficiente\r\n--    ghci> :set +s \r\n--    \r\n--    ghci> last (inversa1 3 [1..100000])\r\n--    100000\r\n--    (16.42 secs, 17576420 bytes)\r\n--     \r\n--    ghci> last (inversa2 3 [1..100000])\r\n--    100000\r\n--    (0.11 secs, 18171356 bytes)\r\n\r\n-- La propiedad es \r\nprop_inversa :: Int -> [Int] -> Property\r\nprop_inversa k xs =\r\n    k > 0 ==> inversa2 k (inversa2 k xs) == xs\r\n\r\n-- La comprobaci\u00f3n es \r\n--    ghci> quickCheck prop_inversa\r\n--    +++ OK, passed 100 tests.\r\n<\/pre>\n<p>[\/schedule]<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Enunciado &#8212; Definir la funci\u00f3n &#8212; inversa :: Int -> [a] -> [a] &#8212; tal que (inversa k xs) es la lista obtenida invirtiendo elementos de &#8212; xs, k elementos cada vez. Si el n\u00famero de elementos de xs no es un &#8212; m\u00faltiplo de k, entonces los finales elementos de xs se dejen sin&#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":[46,28,6,32,47,146],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/599"}],"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=599"}],"version-history":[{"count":4,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/599\/revisions"}],"predecessor-version":[{"id":736,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/599\/revisions\/736"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=599"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=599"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}