{"id":3808,"date":"2020-10-13T16:11:28","date_gmt":"2020-10-13T14:11:28","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=3808"},"modified":"2020-10-13T16:33:47","modified_gmt":"2020-10-13T14:33:47","slug":"equation-de-bezout-en-python","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/13\/equation-de-bezout-en-python\/","title":{"rendered":"\u00c9quation de B\u00e9zout en Python"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">R\u00e9soudre une \u00e9quation de B\u00e9zout en Python n&rsquo;est pas si difficile que ce que l&rsquo;on pourrait imaginer au premier abord.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Nous allons dans un premier temps faire un rappel sur la mani\u00e8re dont on r\u00e9sout math\u00e9matiquement une telle \u00e9quation, puis nous allons voir une impl\u00e9mentation en Python.<\/p>\n\n\n\n<!--more-->\n\n\n\n<div class=\"wp-block-image is-style-rounded\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"280\" height=\"326\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/10\/etienne-bezout.jpg\" alt=\"\u00c9quation de B\u00e9zout en Python\" class=\"wp-image-3813\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/10\/etienne-bezout.jpg 280w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/10\/etienne-bezout-258x300.jpg 258w\" sizes=\"auto, (max-width: 280px) 100vw, 280px\" \/><figcaption>\u00c9tienne B\u00e9zout<\/figcaption><\/figure><\/div>\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_87 counter-hierarchy ez-toc-counter ez-toc-white ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Au menu sur cette page...<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/13\/equation-de-bezout-en-python\/#Equation_de_Bezout_en_Python_rappels_mathematiques\" >\u00c9quation de B\u00e9zout en Python: rappels math\u00e9matiques<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/13\/equation-de-bezout-en-python\/#Equation_de_Bezout_en_Python_definition\" >\u00c9quation de B\u00e9zout en Python: d\u00e9finition<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/13\/equation-de-bezout-en-python\/#Resultat_mathematique_existence_des_solutions\" >R\u00e9sultat math\u00e9matique: existence des solutions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/13\/equation-de-bezout-en-python\/#Resultat_mathematique_valeurs_des_solutions\" >R\u00e9sultat math\u00e9matique: valeurs des solutions<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/13\/equation-de-bezout-en-python\/#Equation_de_Bezout_en_Python_implementation\" >\u00c9quation de B\u00e9zout en Python: impl\u00e9mentation<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/13\/equation-de-bezout-en-python\/#Fonctions_principales\" >Fonctions principales<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/13\/equation-de-bezout-en-python\/#Une_fonction_pgcd_un_peu_speciale\" >Une fonction pgcd un peu sp\u00e9ciale<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/10\/13\/equation-de-bezout-en-python\/#Une_application\" >Une application<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Equation_de_Bezout_en_Python_rappels_mathematiques\"><\/span>\u00c9quation de B\u00e9zout en Python: rappels math\u00e9matiques<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Equation_de_Bezout_en_Python_definition\"><\/span>\u00c9quation de B\u00e9zout en Python: d\u00e9finition<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Une <em>\u00e9quation de B\u00e9zout<\/em> est une \u00e9quation \u00e0 deux variables enti\u00e8res <em>x<\/em> et <em>y<\/em> de la forme:$$ax+by=c$$On la trouve aussi sous le nom d&rsquo;<em><a href=\"https:\/\/fr.wikipedia.org\/wiki\/%C3%89quation_diophantienne\" target=\"_blank\" rel=\"noreferrer noopener\">\u00e9quation diophantienne<\/a><\/em> mais c&rsquo;est tr\u00e8s abusif. En effet, une \u00e9quation de B\u00e9zout est un cas particulier des \u00e9quations diophantiennes.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Resultat_mathematique_existence_des_solutions\"><\/span>R\u00e9sultat math\u00e9matique: existence des solutions<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">L&rsquo;\u00e9quation <em>ax + by = c<\/em> admet des solutions enti\u00e8res uniquement si <em>c<\/em> est divisible par pgcd(<em>a<\/em> ; <em>b<\/em>).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ainsi, l&rsquo;\u00e9quation <em>ax<\/em> + <em>by<\/em> = 1 admet des solutions quand <em>a<\/em> et <em>b<\/em> sont premiers entre eux.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Resultat_mathematique_valeurs_des_solutions\"><\/span>R\u00e9sultat math\u00e9matique: valeurs des solutions<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Pour trouver les solutions de l&rsquo;\u00e9quation <em>ax<\/em> + <em>by<\/em> = 1 quand <em>a<\/em> et <em>b<\/em> sont premiers entre eux, il faut avant tout trouver une solution particuli\u00e8re.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Pour cela, on utilise l&rsquo;algorithme d&rsquo;Euclide. Prenons un exemple: on souhaite r\u00e9soudre l&rsquo;\u00e9quation 17<em>x<\/em> + 13<em>y<\/em> = 1. L&rsquo;algorithme d&rsquo;Euclide donne:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">17 = 1 \u00d7 13 + 4\n13 = 3 \u00d7 4 + 1\n4 = 4 \u00d7 1 + 0<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">On \u00ab\u00a0remonte\u00a0\u00bb l&rsquo;algorithme \u00e0 partir de \u00ab\u00a01\u00a0\u00bb:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">1 = 13 - 3 \u00d7 4\n1 = 13 - 3 \u00d7 (17 - 1 \u00d7 13)\n1 = 4 \u00d7 13 - 3 \u00d7 17<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Une solution particuli\u00e8re est donc <em>x<\/em> = -3 et <em>y<\/em> = 4. On a alors:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">17  x  + 13 y  = 1\n17(-3) + 13(4) = 1<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Donc, en soustrayant la seconde ligne \u00e0 la premi\u00e8re, on a:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">17(x+3) + 13(y-4)=0\n17(x+3) = 13(4-y)<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">D&rsquo;apr\u00e8s le <a href=\"https:\/\/fr.wikipedia.org\/wiki\/Lemme_d%27Euclide\" target=\"_blank\" rel=\"noreferrer noopener\">th\u00e9or\u00e8me de Gauss<\/a>, comme 13 et 17 sont premiers entre eux, 13 divise <em>x<\/em>+3. Il existe donc un entier <em>k<\/em> tel que:$$x+3=13k\\quad\\text{soit}\\quad x=-3+13k.$$En injectant dans l&rsquo;\u00e9quation \\(17(x+3)=13(4-y)\\) cette valeur de <em>x<\/em>, on obtient:$$17\\times13k=13(4-y)$$soit, en divisant par 13:$$17k=4-y$$et donc:$$y=4-17k.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ainsi, toutes les solutions de l&rsquo;\u00e9quation sont de la forme:$$(-3+13k;4-17k)\\quad,\\quad k\\in\\mathbb{Z}.$$<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Equation_de_Bezout_en_Python_implementation\"><\/span>\u00c9quation de B\u00e9zout en Python: impl\u00e9mentation<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Fonctions_principales\"><\/span>Fonctions principales<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Nous allons consid\u00e9rer que les \u00e9quations \u00e0 r\u00e9soudre sont de la forme <em>ax<\/em> + <em>by<\/em> = 1, avec pgcd(<em>a<\/em> ; <em>b<\/em>) = 1.<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"false\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">def bezout_fct(a,b):\n    if b == 0:\n        return 1,0\n    else:\n        u , v = bezout_fct(b , a % b)\n        return v , u - (a\/\/b)*v\n\ndef resoud_equation(a,b,c):\n    u,v = bezout_fct(a,b)\n    return \"Les solutions de l'\u00e9quation {}x + {}y = {} sont:\\n({} + {}k , {} - {}k)\".format(a,b,c,u,b,v,a)<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">La premi\u00e8re fonction retourne les deux valeurs particuli\u00e8res de l&rsquo;\u00e9quation.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">La seconde retourne l&rsquo;ensemble des solutions.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">&gt;&gt;&gt; resoud_equation(17,13,1)\nLes solutions de l'\u00e9quation 17x + 13y = 1 sont:\n(-3 + 13k , 4 - 17k)<\/pre>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Une_fonction_pgcd_un_peu_speciale\"><\/span>Une fonction pgcd un peu sp\u00e9ciale<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"false\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">from math import log\n\ndef pgcd(a,b,la=0,lb=0,lq=0,lr=0,olda=0,oldb=0,affiche=False,bezout=False,nombre=False):\n    if a &lt; b:\n        a,b = b,a\n    r = a % b\n    \n    if affiche == True:\n        q = a \/\/ b\n    \n        if la == 0:\n            la = str( int( log( a , 10 ) ) + 1 )\n            lb = str( int( log( b , 10 ) ) + 1 )\n            lq = str( int( log( q , 10 ) ) + 2 )\n            lr = str( int( log( r , 10 ) ) + 1 )\n    \n        s = '{:'+la+'} = {:'+lq+'} \u00d7 {:'+lb+'} + {:'+lr+'}'\n        print(s.format(a,q,b,r))\n    \n    if r == 0:\n        if affiche == True:\n            response = '\\npgcd({};{}) = {}'.format(olda,oldb,b)\n        else:\n            if nombre == False:\n                response = 'pgcd({};{}) = {}'.format(olda,oldb,b)\n            else:\n                response = b\n        \n        if bezout == True:\n            u , v = bezout_fct(olda,oldb)\n            response += \"\\n\\nUne solution particuli\u00e8re de l'\u00e9quation {}x + {}y = 1 est : x = {}, y = {}.\".format(olda,oldb,u,v)\n            \n        return response\n    else:\n        if olda == 0:\n            olda, oldb = a, b\n        return pgcd(b,r,la,lb,lq,lr,olda,oldb,affiche,bezout,nombre)<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Cette fonction est l\u00e9g\u00e8rement plus compliqu\u00e9e qu&rsquo;une simple fonction retournant le PGCD de deux nombres car je voulais qu&rsquo;elle fasse bien plus \u00e0 l&rsquo;origine.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">En effet, je voulais que cette fonction puisse afficher l&rsquo;algorithme d&rsquo;Euclide. Avec cette fonction, on a:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">&gt;&gt;&gt; print( pgcd(155,35,affiche=True) )\n155 = 4 \u00d7 35 + 15\n35 = 2 \u00d7 15 + 5\n15 = 3 \u00d7 5 + 0\n\npgcd(155;35) = 5<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Il n&rsquo;\u00e9tait pas utile d&rsquo;avoir une fonction si compliqu\u00e9e pour r\u00e9soudre une \u00e9quation de B\u00e9zout, mais j&rsquo;avais envie d&rsquo;un petit \u00ab\u00a0plus\u00a0\u00bb.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Une_application\"><\/span>Une application<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<figure class=\"wp-block-pullquote\"><blockquote><p>Au 8e si\u00e8cle, un groupe compos\u00e9 d\u2019hommes et de femmes a d\u00e9pens\u00e9 100 pi\u00e8ces de monnaie dans une<\/p><p>auberge. Les hommes ont d\u00e9pens\u00e9 8 pi\u00e8ces chacun et les femmes 5 pi\u00e8ces chacune.<\/p><p>Combien pouvait-il y avoir d\u2019hommes et de femmes dans le groupe ?<\/p><\/blockquote><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Si <em>x<\/em> repr\u00e9sente le nombre d&rsquo;hommes et <em>y<\/em> celui des femmes, on a l&rsquo;\u00e9quation suivante:$$8x+5y=100.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Le programme pr\u00e9c\u00e9dent ne donne pas les solutions particuli\u00e8res de cette \u00e9quation car <em>c<\/em> est diff\u00e9rent de 1. Je dois donc le modifier un peu \u00e0 l&rsquo;arrache:<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"false\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">def resoud_equation(a,b,c):\n    d = pgcd(a,b,nombre=True)\n    if c == 1 and d == 1:\n        u,v = bezout_fct(a,b)\n        return \"Les solutions de l'\u00e9quation {}x + {}y = {} sont: ({} + {}k , {} - {}k)\".format(a,b,c,u,b,v,a)\n    elif d == 1:\n        y = 0\n        while (c - b*y)%a != 0:\n            y += 1\n        return \"Les solutions de l'\u00e9quation {}x + {}y = {} sont: ({} + {}k , {} - {}k)\".format(a,b,c,(c-b*y)\/\/a,b,y,a)\n    else:\n        return \"L'\u00e9quation {}x + {}y = {} n'a pas de solutions enti\u00e8res.\".format(a,b,c)<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Je vous l&rsquo;accorde, c&rsquo;est bien moche&#8230; Mais \u00e7a fait le job (en s&rsquo;en contentera pour le moment). On obtient alors:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">>>> resoud_equation(8,5,100)\nLes solutions de l'\u00e9quation 8x + 5y = 100 sont: (10 + 5k , 4 - 8k)<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Comme les valeurs de <em>x<\/em> et <em>y<\/em> sont strictement positives, il n&rsquo;y a que deux solutions:<\/p>\n\n\n\n<ul class=\"wp-block-list\"><li>si <em>k<\/em> = 0 alors <em>x<\/em> = 10 et <em>y<\/em> = 4;<\/li><li>si <em>k<\/em> = -1 alors <em>x<\/em> = 5 et <em>y<\/em> = 12.<\/li><\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Et voil\u00e0 ! Notre probl\u00e8me est r\u00e9solu !<\/p>\n","protected":false},"excerpt":{"rendered":"<p>R\u00e9soudre une \u00e9quation de B\u00e9zout en Python n&rsquo;est pas si difficile que ce que l&rsquo;on pourrait imaginer au premier abord. Nous allons dans un premier temps faire un rappel sur la mani\u00e8re dont on r\u00e9sout math\u00e9matiquement une telle \u00e9quation, puis nous allons voir une impl\u00e9mentation en Python.<\/p>\n","protected":false},"author":1,"featured_media":3814,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6,5],"tags":[51,244,246,216],"class_list":["post-3808","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques","category-python","tag-algorithme","tag-bezout","tag-equations-diophantiennes","tag-euclide"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>\u00c9quation de B\u00e9zout en Python - Mathweb.fr - Impl\u00e9mentation concr\u00e8te<\/title>\n<meta name=\"description\" content=\"Nous allons voir dans cet article une mani\u00e8re d&#039;impl\u00e9menter une \u00e9quation de B\u00e9zout en Python. 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