{"id":1918,"date":"2015-12-28T06:00:58","date_gmt":"2015-12-28T04:00:58","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=1918"},"modified":"2016-01-04T08:22:10","modified_gmt":"2016-01-04T06:22:10","slug":"operacion-sobre-todos-los-pares","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/operacion-sobre-todos-los-pares\/","title":{"rendered":"Operaci\u00f3n sobre todos los pares"},"content":{"rendered":"

Definir la funci\u00f3n<\/p>\n

\n   todosPares :: (a -> b -> c) -> [a] -> [b] -> [c]\n<\/pre>\n
tal que (todosPares f xs ys) es el resultado de aplicar la operaci\u00f3n f a todos los pares de xs e ys. Por ejemplo,<\/p>\n
\n   todosPares (*) [2,3,5] [7,11]            == [14,22,21,33,35,55]\n   todosPares (\\x y -> x:show y) \"ab\" [7,5] == [\"a7\",\"a5\",\"b7\",\"b5\"]\n<\/pre>\n
Soluciones<\/h4>\n
\n-- 1\u00aa definici\u00f3n (por comprensi\u00f3n)\ntodosPares1 :: (a -> b -> c) -> [a] -> [b] -> [c]\ntodosPares1 f xs ys = [f x y | x <- xs, y <- ys] \n\n-- 2\u00aa definici\u00f3n (por recursi\u00f3n)\ntodosPares2 :: (a -> b -> c) -> [a] -> [b] -> [c]\ntodosPares2 _ []     _  = []\ntodosPares2 f (x:xs) ys = map (f x) ys ++ todosPares2 f xs ys\n\n-- 3\u00aa definici\u00f3n (recursi\u00f3n con auxiliar)\ntodosPares3 :: (a -> b -> c) -> [a] -> [b] -> [c]\ntodosPares3 f xs ys = aux xs where\n   aux []     = []\n   aux (x:xs) = map (f x) ys ++ aux xs\n\n-- 4\u00aa definici\u00f3n (recursi\u00f3n con auxiliar)\ntodosPares4 :: (a -> b -> c) -> [a] -> [b] -> [c]\ntodosPares4 f xs ys = \n    (foldr (\\x zs -> map (f x) ys ++ zs) []) xs\n<\/pre>\n","protected":false},"excerpt":{"rendered":"
Definir la funci\u00f3n todosPares :: (a -> b -> c) -> [a] -> [b] -> [c] tal que (todosPares f xs ys) es el resultado de aplicar la operaci\u00f3n f a todos los pares de xs e ys. Por ejemplo, todosPares (*) [2,3,5] [7,11] == [14,22,21,33,35,55] todosPares (\\x y -> x:show y) \u00abab\u00bb [7,5] ==…<\/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":[5],"tags":[8,94,10,11,6],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/1918"}],"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=1918"}],"version-history":[{"count":4,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/1918\/revisions"}],"predecessor-version":[{"id":1954,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/1918\/revisions\/1954"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=1918"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=1918"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=1918"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}