{"id":8117,"date":"2023-05-10T06:00:41","date_gmt":"2023-05-10T04:00:41","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=8117"},"modified":"2023-05-02T13:20:37","modified_gmt":"2023-05-02T11:20:37","slug":"10-may-23","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/10-may-23\/","title":{"rendered":"TAD de los polinomios: Divisi\u00f3n de polinomios"},"content":{"rendered":"<p>Usando el <a href=\"https:\/\/bit.ly\/3KwqXYu\">tipo abstracto de los polinomios<\/a>, definir las funciones<\/p>\n<pre lang=\"text\">\n   cociente :: (Fractional a, Eq a) =>\n               Polinomio a -> Polinomio a -> Polinomio a\n   resto    :: (Fractional a, Eq a) =>\n               Polinomio a -> Polinomio a -> Polinomio a\n<\/pre>\n<p>tales que<\/p>\n<ul>\n<li><code>cociente p q<\/code> es el cociente de la divisi\u00f3n de <code>p<\/code> entre <code>q<\/code>. Por ejemplo,<\/li>\n<\/ul>\n<pre lang=\"text\">\n     \u03bb> pol1 = consPol 3 2 (consPol 2 9 (consPol 1 10 (consPol 0 4 polCero)))\n     \u03bb> pol1\n     2*x^3 + 9*x^2 + 10*x + 4\n     \u03bb> pol2 = consPol 2 1 (consPol 1 3 polCero)\n     \u03bb> pol2\n     x^2 + 3*x\n     \u03bb> cociente pol1 pol2\n     2.0*x + 3.0\n<\/pre>\n<ul>\n<li><code>resto p q<\/code> es el resto de la divisi\u00f3n de <code>p<\/code> entre <code>q<\/code>. Por ejemplo,<\/li>\n<\/ul>\n<pre lang=\"text\">\n     \u03bb> resto pol1 pol2\n     1.0*x + 4.0\n<\/pre>\n<p><b>Soluciones<\/b><\/p>\n<p>A continuaci\u00f3n se muestran las <a href=\"#haskell\">soluciones en Haskell<\/a> y las <a href=\"#python\">soluciones en Python<\/a>.<\/p>\n<p><a name=\"haskell\"><\/a><br \/>\n<b>Soluciones en Haskell<\/b><\/p>\n<pre lang=\"haskell\">\nimport TAD.Polinomio (Polinomio, polCero, consPol, grado, coefLider)\nimport Pol_Crea_termino (creaTermino)\nimport Pol_Producto_polinomios (multPol, multPorTerm)\nimport Pol_Resta_de_polinomios (restaPol)\nimport Pol_Multiplicacion_de_un_polinomio_por_un_numero (multEscalar)\n\ncociente :: (Fractional a, Eq a) =>\n            Polinomio a -> Polinomio a -> Polinomio a\ncociente p q\n  | n2 == 0   = multEscalar (1\/a2) p\n  | n1 < n2   = polCero\n  | otherwise = consPol n3 a3 (cociente p3 q)\n  where n1 = grado p\n        a1 = coefLider p\n        n2 = grado q\n        a2 = coefLider q\n        n3 = n1-n2\n        a3 = a1\/a2\n        p3 = restaPol p (multPorTerm (creaTermino n3 a3) q)\n\nresto :: (Fractional a, Eq a) =>\n         Polinomio a -> Polinomio a -> Polinomio a\nresto p q = restaPol p (multPol (cociente p q) q)\n<\/pre>\n<p><a name=\"python\"><\/a><br \/>\n<b>Soluciones en Python<\/b><\/p>\n<pre lang=\"python\">\nfrom src.Pol_Crea_termino import creaTermino\nfrom src.Pol_Multiplicacion_de_un_polinomio_por_un_numero import multEscalar\nfrom src.Pol_Producto_polinomios import multPol, multPorTerm\nfrom src.Pol_Resta_de_polinomios import restaPol\nfrom src.TAD.Polinomio import Polinomio, coefLider, consPol, grado, polCero\n\n\ndef cociente(p: Polinomio[float], q: Polinomio[float]) -> Polinomio[float]:\n    n1 = grado(p)\n    a1 = coefLider(p)\n    n2 = grado(q)\n    a2 = coefLider(q)\n    n3 = n1 - n2\n    a3 = a1 \/ a2\n    p3 = restaPol(p, multPorTerm(creaTermino(n3, a3), q))\n    if n2 == 0:\n        return multEscalar(1 \/ a2, p)\n    if n1 < n2:\n        return polCero()\n    return consPol(n3, a3, cociente(p3, q))\n\ndef resto(p: Polinomio[float], q: Polinomio[float]) -> Polinomio[float]:\n    return restaPol(p, multPol(cociente(p, q), q))\n<\/pre>\n","protected":false},"excerpt":{"rendered":"<p>Usando el tipo abstracto de los polinomios, definir las funciones cociente :: (Fractional a, Eq a) => Polinomio a -> Polinomio a -> Polinomio a resto :: (Fractional a, Eq a) => Polinomio a -> Polinomio a -> Polinomio a tales que cociente p q es el cociente de la divisi\u00f3n de p entre q&#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":[581],"tags":[265],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/8117"}],"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=8117"}],"version-history":[{"count":1,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/8117\/revisions"}],"predecessor-version":[{"id":8118,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/8117\/revisions\/8118"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=8117"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=8117"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=8117"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}