{"id":4555,"date":"2019-01-18T06:00:48","date_gmt":"2019-01-18T04:00:48","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=4555"},"modified":"2022-03-25T20:37:10","modified_gmt":"2022-03-25T18:37:10","slug":"minimo-numero-de-operaciones-para-transformar-un-numero-en-otro","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/minimo-numero-de-operaciones-para-transformar-un-numero-en-otro\/","title":{"rendered":"M\u00ednimo n\u00famero de operaciones para transformar un n\u00famero en otro"},"content":{"rendered":"<p>Se considera el siguiente par de operaciones sobre los n\u00fameros:<\/p>\n<ul>\n<li>multiplicar por dos<\/li>\n<li>restar uno.<\/li>\n<\/ul>\n<p>Dados dos n\u00fameros x e y se desea calcular el menor n\u00famero de operaciones para transformar x en y. Por ejemplo, el menor n\u00famero de operaciones para transformar el 4 en 7 es 2:<\/p>\n<pre lang=\"text\"> \n   4 ------> 8 ------> 7\n      (*2)      (-1)\n<\/pre>\n<p>y el menor n\u00famero de operaciones para transformar 2 en 5 es 4<\/p>\n<pre lang=\"text\"> \n   2 ------> 4 ------> 3 ------> 6 ------> 5\n      (*2)      (-1)      (*2)      (-1)\n<\/pre>\n<p>Definir las siguientes funciones<\/p>\n<pre lang=\"text\"> \n   arbolOp :: Int -> Int -> Arbol\n   minNOp  :: Int -> Int -> Int\n<\/pre>\n<p>tales que<\/p>\n<ul>\n<li>(arbolOp x n) es el \u00e1rbol de profundidad n obtenido aplic\u00e1ndole a x las dos operaciones. Por ejemplo, <\/li>\n<\/ul>\n<pre lang=\"text\">   \n    \u03bb> arbolOp 4 1\n    N 4 (H 8) (H 3)\n    \u03bb> arbolOp 4 2\n    N 4 (N 8 (H 16) (H 7))\n        (N 3 (H 6) (H 2))\n    \u03bb> arbolOp 2 3\n    N 2 (N 4\n           (N 8 (H 16) (H 7))\n           (N 3 (H 6) (H 2)))\n        (N 1\n           (N 2 (H 4) (H 1))\n           (H 0))\n    \u03bb> arbolOp 2 4\n    N 2 (N 4 (N 8\n                (N 16 (H 32) (H 15))\n                (N 7 (H 14) (H 6)))\n             (N 3\n                (N 6 (H 12) (H 5))\n                (N 2 (H 4) (H 1))))\n        (N 1 (N 2\n                (N 4 (H 8) (H 3))\n                (N 1 (H 2) (H 0)))\n             (H 0))\n<\/pre>\n<ul>\n<li>(minNOp x y) es el menor n\u00famero de operaciones necesarias para transformar x en y. Por ejemplo,<\/li>\n<\/ul>\n<pre lang=\"text\">   \n     minNOp 4 7  ==  2\n     minNOp 2 5  ==  4\n<\/pre>\n<h4>Soluciones<\/h4>\n<pre lang=\"haskell\">\ndata Arbol = H Int\n           | N Int Arbol Arbol\n  deriving (Show, Eq)\n\narbolOp :: Int -> Int -> Arbol\narbolOp 0 _ = H 0\narbolOp x 0 = H x\narbolOp x n = N x (arbolOp (2 * x) (n - 1)) (arbolOp (x - 1) (n - 1))\n\nocurre :: Int -> Arbol -> Bool\nocurre x (H y)     = x == y\nocurre x (N y i d) = x == y || ocurre x i || ocurre x d\n\nminNOp :: Int -> Int -> Int\nminNOp x y =\n  head [n | n <- [0..]\n          , ocurre y (arbolOp x n)]\n<\/pre>\n<h4>Pensamiento<\/h4>\n<blockquote><p>\n\u00bfDijiste media verdad?<br \/>\nDir\u00e1n que mientes dos veces<br \/>\nsi dices la otra mitad. <\/p>\n<p>Antonio Machado\n<\/p><\/blockquote>\n","protected":false},"excerpt":{"rendered":"<p>Se considera el siguiente par de operaciones sobre los n\u00fameros: multiplicar por dos restar uno. Dados dos n\u00fameros x e y se desea calcular el menor n\u00famero de operaciones para transformar x en y. Por ejemplo, el menor n\u00famero de operaciones para transformar el 4 en 7 es 2: 4 &#8212;&#8212;> 8 &#8212;&#8212;> 7 (*2)&#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":[269,8,71,6,133],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/4555"}],"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=4555"}],"version-history":[{"count":3,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/4555\/revisions"}],"predecessor-version":[{"id":4637,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/4555\/revisions\/4637"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=4555"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=4555"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=4555"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}