La sucesi\u00f3n de Fibonacci es<\/p>\n
\n 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,...\n<\/pre>\ncuyos dos primeros t\u00e9rminos son 0 y 1 y los restantentes se obtienen sumando los dos anteriores.<\/p>\n
El \u00e1rbol de computaci\u00f3n de su 5\u00ba t\u00e9rmino es<\/p>\n
\n 5\n \/ \\\n \/ \\\n \/ \\\n \/ \\\n \/ \\\n 3 2 \n \/ \\ \/ \\ \n \/ \\ 1 1\n 2 1 \/ \\ \n \/ \\ \/ \\ 1 0 \n 1 1 1 0\n \/ \\ \n 1 0 \n<\/pre>\nque, usando los \u00e1rboles definidos por<\/p>\n
\n data Arbol = H Int\n | N Int Arbol Arbol\n deriving (Eq, Show)\n<\/pre>\nse puede representar por<\/p>\n
\n N 5 \n (N 3 \n (N 2 \n (N 1 (H 1) (H 0))\n (H 1)) \n (N 1 (H 1) (H 0))) \n (N 2 \n (N 1 (H 1) (H 0)) \n (H 1)) \n<\/pre>\nDefinir las funciones<\/p>\n
\n arbolFib :: Int -> Arbol\n nElementosArbolFib :: Int -> Int\n<\/pre>\ntales que<\/p>\n
\n \u03bb> arbolFib 5\n N 5 \n (N 3 \n (N 2 \n (N 1 (H 1) (H 0))\n (H 1)) \n (N 1 (H 1) (H 0))) \n (N 2 \n (N 1 (H 1) (H 0)) \n (H 1))\n \u03bb> arbolFib 6\n N 8\n (N 5\n (N 3\n (N 2\n (N 1 (H 1) (H 0))\n (H 1))\n (N 1 (H 1) (H 0)))\n (N 2\n (N 1 (H 1) (H 0))\n (H 1)))\n (N 3\n (N 2\n (N 1 (H 1) (H 0)) (H 1))\n (N 1 (H 1) (H 0)))\n<\/pre>\n\n nElementosArbolFib 5 == 15\n nElementosArbolFib 6 == 25\n nElementosArbolFib 30 == 2692537\n<\/pre>\nSoluciones<\/h4>\n\ndata Arbol = H Int\n | N Int Arbol Arbol\n deriving (Eq, Show)\n\n-- 1\u00aa definici\u00f3n\n-- =============\n\narbolFib :: Int -> Arbol\narbolFib 0 = H 0\narbolFib 1 = H 1\narbolFib n = N (fib n) (arbolFib (n-1)) (arbolFib (n-2))\n\n-- (fib n) es el n-\u00e9simo elemento de la sucesi\u00f3n de Fibonacci. Por\n-- ejemplo,\n-- fib 5 == 5\n-- fib 6 == 8\nfib :: Int -> Int\nfib 0 = 0\nfib 1 = 1\nfib n = fib (n-1) + fib (n-2)\n\n-- 2\u00aa definici\u00f3n\n-- =============\n\narbolFib2 :: Int -> Arbol\narbolFib2 0 = H 0\narbolFib2 1 = H 1\narbolFib2 2 = N 1 (H 1) (H 0)\narbolFib2 3 = N 2 (N 1 (H 1) (H 0)) (H 1)\narbolFib2 n = N (a1 + a2) (N a1 i1 d1) (N a2 i2 d2)\n where (N a1 i1 d1) = arbolFib2 (n-1)\n (N a2 i2 d2) = arbolFib2 (n-2)\n\n-- 3\u00aa definici\u00f3n\n-- =============\n\narbolFib3 :: Int -> Arbol\narbolFib3 0 = H 0\narbolFib3 1 = H 1\narbolFib3 2 = N 1 (H 1) (H 0)\narbolFib3 3 = N 2 (N 1 (H 1) (H 0)) (H 1)\narbolFib3 n = N (a + b) i d\n where i@(N a _ _) = arbolFib3 (n-1)\n d@(N b _ _) = arbolFib3 (n-2)\n\n-- 1\u00aa definici\u00f3n de nElementosArbolFib\n-- ===================================\n\nnElementosArbolFib :: Int -> Int\nnElementosArbolFib = length . elementos . arbolFib3\n\n-- (elementos a) es la lista de elementos del \u00e1rbol a. Por ejemplo,\n-- \u03bb> elementos (arbolFib 5)\n-- [5,3,2,1,1,0,1,1,1,0,2,1,1,0,1]\n-- \u03bb> elementos (arbolFib 6)\n-- [8,5,3,2,1,1,0,1,1,1,0,2,1,1,0,1,3,2,1,1,0,1,1,1,0]\nelementos :: Arbol -> [Int]\nelementos (H x) = [x]\nelementos (N x i d) = x : elementos i ++ elementos d\n\n-- 2\u00aa definici\u00f3n de nElementosArbolFib\n-- ===================================\n\nnElementosArbolFib2 :: Int -> Int\nnElementosArbolFib2 0 = 1\nnElementosArbolFib2 1 = 1\nnElementosArbolFib2 n = 1 + nElementosArbolFib2 (n-1)\n + nElementosArbolFib2 (n-2)\n<\/pre>\nPensamiento<\/h4>\n\nToda visi\u00f3n requiere distancia.<\/p>\n
Antonio Machado\n<\/p><\/blockquote>\n","protected":false},"excerpt":{"rendered":"
La sucesi\u00f3n de Fibonacci es 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,… cuyos dos primeros t\u00e9rminos son 0 y 1 y los restantentes se obtienen sumando los dos anteriores. El \u00e1rbol de computaci\u00f3n de su 5\u00ba t\u00e9rmino es 5 \/ \\ \/ \\ \/ \\ \/ \\ \/ \\ 3 2 \/ \\ \/ \\ \/ \\ 1 1 2 1…<\/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":[269,28,6],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/4395"}],"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=4395"}],"version-history":[{"count":3,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/4395\/revisions"}],"predecessor-version":[{"id":4578,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/4395\/revisions\/4578"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=4395"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=4395"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=4395"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}