{"id":6313,"date":"2021-04-27T06:00:47","date_gmt":"2021-04-27T04:00:47","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=6313"},"modified":"2021-05-04T08:21:43","modified_gmt":"2021-05-04T06:21:43","slug":"maximos-de-una-funcion-recursiva-ome2002-p3","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/maximos-de-una-funcion-recursiva-ome2002-p3\/","title":{"rendered":"M\u00e1ximos de una funci\u00f3n recursiva (OME2002 P3)"},"content":{"rendered":"<p>El enunciado del <a href=\"https:\/\/bit.ly\/3tueZVv\">problema 5 de la OME (Olimpiada Matem\u00e1tica Espa\u00f1ola) del 2002<\/a> es<\/p>\n<blockquote><p>\n  La funci\u00f3n g se define sobre los n\u00fameros naturales y satisface las condiciones:<\/p>\n<ul>\n<li>g(1) = 1<\/li>\n<li>g(2n) = g(n)<\/li>\n<li>g(2n + 1) = g(2n) + 1<\/li>\n<\/ul>\n<p>  Sea n un n\u00famero natural tal que 1 \u2264 n \u2264 2002. Calcula el valor m\u00e1ximo M de g(n). Calcula tambi\u00e9n cu\u00e1ntos valores de n satisfacen g(n) = M.\n<\/p><\/blockquote>\n<p>Los valores de la funci\u00f3n g para n de 1 a 30 son<\/p>\n<pre lang=\"text\">\n   1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4\n<\/pre>\n<p>Definir la funci\u00f3n<\/p>\n<pre lang=\"text\">\n   maximoG :: Integer -> Integer\n<\/pre>\n<p>tal que (maximoG m) es el m\u00e1ximo de los valores de g(n) para n en {1, 2,&#8230;, m}. Por ejemplo,<\/p>\n<pre lang=\"text\">\n   maximoG 30           ==  4\n   maximoG (10^(10^5))  ==  332192\n<\/pre>\n<p>Usando la funci\u00f3n maximoG, calcular los valores pedidos en el problema.<\/p>\n<h4>Soluciones<\/h4>\n<pre lang=\"haskell\">\nimport Data.List (genericLength, genericTake, group)\nimport Test.QuickCheck (Property, (==>), quickCheck)\n\n-- 1\u00aa soluci\u00f3n\n-- ===========\n\nmaximoG :: Integer -> Integer\nmaximoG m = maximum (genericTake m sucesionG)\n\n-- sucesionG es la lista de los valores de g. Por ejemplo,\n--    \u03bb> take 30 sucesionG\n--    [1,1,2,1,2,2,3,1,2,2,3,2,3,3,4,1,2,2,3,2,3,3,4,2,3,3,4,3,4,4]\nsucesionG :: [Integer]\nsucesionG = map g [1..]\n\n-- (g n) es el valor de g(n).\ng :: Integer -> Integer\ng 1 = 1\ng n | even n    = g (n `div` 2)\n    | otherwise = g (n `div` 2) + 1\n\n-- 2\u00aa soluci\u00f3n\n-- ===========\n\n-- Observando los siguientes c\u00e1lculos\n--   \u03bb> map maximoG [1..40]\n--   [1,1,2,2,2,2,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5]\n--   \u03bb> take 10 (map length (group (map maximoG [1..])))\n--   [2,4,8,16,32,64,128,256,512,1024]\n\nmaximoG2 :: Integer -> Integer\nmaximoG2 m = head [x | x <- [1..], m + 1 < 2^x] - 1\n\n-- 3\u00aa soluci\u00f3n\n-- ===========\n\nmaximoG3 :: Integer -> Integer\nmaximoG3 m = genericLength (takeWhile (<m+2) potenciasDeDos) - 1\n\n-- potenciasDeDos es la lista de las potencias de dos. Por ejemplo,\n--    take 10 potenciasDeDos  ==  [1,2,4,8,16,32,64,128,256,512]\npotenciasDeDos :: [Integer]\npotenciasDeDos = iterate (*2) 1\n\n-- 4\u00aa soluci\u00f3n\n-- ===========\n\nmaximoG4 :: Integer -> Integer\nmaximoG4 m = floor (logBase 2 (fromIntegral (m+1)))\n\n-- Comprobaci\u00f3n de equivalencia\n-- ============================\n\n-- La propiedad es\nprop_maximoG :: Integer -> Property\nprop_maximoG m =\n  m > 0 ==>\n  all (== (maximoG m))\n      [ maximoG2 m,\n        maximoG3 m,\n        maximoG4 m]\n\n-- La comprobaci\u00f3n es\n--    \u03bb> quickCheck prop_maximoG\n--    +++ OK, passed 100 tests.\n\n-- Nota: Aunque QuickCheck no ha encontrado ning\u00fan contraejemplo, la\n-- definici\u00f3n maximoG4 que usa n\u00fameros decimales falla para n\u00fameros muy\n-- grandes. Por ejemplo,\n--    \u03bb> maximoG4 (10^308)\n--    1023\n--    \u03bb> maximoG4 (10^309)\n--    1797693134862315907729305190789024733617976978942306572734300...\n--    \u03bb> maximoG3 (10^308)\n--    1023\n--    \u03bb> maximoG3 (10^309)\n--    1026\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> maximoG (10^6)\n--    19\n--    (8.80 secs, 6,015,772,760 bytes)\n--    \u03bb> maximoG2 (10^6)\n--    19\n--    (0.02 secs, 106,752 bytes)\n--    \u03bb> maximoG3 (10^6)\n--    19\n--    (0.02 secs, 102,592 bytes)\n--    \u03bb> maximoG4 (10^6)\n--    19\n--    (0.01 secs, 98,464 bytes)\n--\n--    \u03bb> maximoG2 (10^308)\n--    1023\n--    (0.03 secs, 3,266,096 bytes)\n--    \u03bb> maximoG3 (10^308)\n--    1023\n--    (0.02 secs, 266,856 bytes)\n--    \u03bb> maximoG4 (10^308)\n--    1023\n--    (0.02 secs, 103,176 bytes)\n--\n--    \u03bb> maximoG2 (10^16789)\n--    55771\n--    (1.70 secs, 1,201,022,600 bytes)\n--    \u03bb> maximoG3 (10^16789)\n--    55771\n--    (0.24 secs, 214,872,864 bytes)\n\n-- C\u00e1lculos del problema\n-- =====================\n\n-- El m\u00e1ximo se calcula como sigue:\n--    \u03bb> maximum (take 2002 sucesionG)\n--    10\n-- Por tanto, el m\u00e1ximo es 10.\n\n-- Los valores de n tales que g(n) es el m\u00e1ximo se calcula como sigue:\n--    \u03bb> [n | n <- [1..2002], g n == 10]\n--    [1023,1535,1791,1919,1983]\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>El enunciado del problema 5 de la OME (Olimpiada Matem\u00e1tica Espa\u00f1ola) del 2002 es La funci\u00f3n g se define sobre los n\u00fameros naturales y satisface las condiciones: g(1) = 1 g(2n) = g(n) g(2n + 1) = g(2n) + 1 Sea n un n\u00famero natural tal que 1 \u2264 n \u2264 2002. Calcula el valor&#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":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6313"}],"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=6313"}],"version-history":[{"count":3,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6313\/revisions"}],"predecessor-version":[{"id":6397,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6313\/revisions\/6397"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=6313"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=6313"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=6313"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}