{"id":1937,"date":"2020-02-01T16:20:05","date_gmt":"2020-02-01T15:20:05","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=1937"},"modified":"2021-10-26T16:48:28","modified_gmt":"2021-10-26T14:48:28","slug":"determiner-une-valeur-approche-de-pi-a-laide-des-probabilites-methode-de-monte-carlo-sous-python","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2020\/02\/01\/determiner-une-valeur-approche-de-pi-a-laide-des-probabilites-methode-de-monte-carlo-sous-python\/","title":{"rendered":"D\u00e9terminer une valeur approch\u00e9e de Pi \u00e0 l&rsquo;aide des probabilit\u00e9s (m\u00e9thode de Monte-Carlo sous Python)"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">\\(\\pi\\) est la constante d\u00e9finie comme \u00e9tant le rapport de la circonf\u00e9rence d&rsquo;un cercle et de son diam\u00e8tre. Et on arrive \u00e0 d\u00e9montrer que l&rsquo;aire du disque d\u00e9fini par ce cercle est \u00e9gale \u00e0 : $$\\mathcal{A}=\\pi \\times r^2.$$Nous allons voir dans cet article comme utiliser cette derni\u00e8re \u00e9galit\u00e9 afin de trouver une valeur approch\u00e9e de \\(\\pi\\) en passant par les probabilit\u00e9s.<\/p>\n\n\n\n<!--more-->\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_85 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\/02\/01\/determiner-une-valeur-approche-de-pi-a-laide-des-probabilites-methode-de-monte-carlo-sous-python\/#Approche_theorique\" >Approche th\u00e9orique<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/02\/01\/determiner-une-valeur-approche-de-pi-a-laide-des-probabilites-methode-de-monte-carlo-sous-python\/#Methode_de_Monte-Carlo\" >M\u00e9thode de Monte-Carlo<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/02\/01\/determiner-une-valeur-approche-de-pi-a-laide-des-probabilites-methode-de-monte-carlo-sous-python\/#Pourquoi_cette_methode_est_aussi_pourrie\" >Pourquoi cette m\u00e9thode est aussi pourrie ?<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Approche_theorique\"><\/span>Approche th\u00e9orique<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"204\" height=\"205\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/monte-carlo-1.png\" alt=\"\" class=\"wp-image-1943\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/monte-carlo-1.png 204w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/monte-carlo-1-100x100.png 100w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2020\/02\/monte-carlo-1-150x150.png 150w\" sizes=\"auto, (max-width: 204px) 100vw, 204px\" \/><\/figure><\/div>\n\n\n\n<p class=\"wp-block-paragraph\">Consid\u00e9rons un disque (jaune) de centre O et de rayon 1 inscrit dans un carr\u00e9 (vert). Intuitivement, si l&rsquo;on prend un point au hasard dans le carr\u00e9 vert, la probabilit\u00e9 pour qu&rsquo;il soit dans le disque jaune est \u00e9gale \u00e0 la proportion de ce dernier dans le carr\u00e9, \u00e0 savoir :$$\\frac{\\text{aire du disque}}{\\text{aire du carr\u00e9}}=\\frac{\\pi \\times 1^2}{2 \\times 2}=\\frac{\\pi}{4}.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ainsi, si l&rsquo;on arrive \u00e0 d\u00e9terminer une valeur approch\u00e9e de cette probabilit\u00e9, on arrivera \u00e0 trouver une valeur approch\u00e9e de \\(\\pi\\) en multipliant par 4. C&rsquo;est l&rsquo;objectif de la m\u00e9thode de Monte-Carlo.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Methode_de_Monte-Carlo\"><\/span>M\u00e9thode de Monte-Carlo<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">L&rsquo;id\u00e9e est de consid\u00e9rer l&rsquo;exp\u00e9rience consistant \u00e0 choir un point au hasard dans le carr\u00e9 vert et \u00e0 regarder s&rsquo;il est dans le disque jaune, puis de r\u00e9p\u00e9ter cette exp\u00e9rience un assez grand nombre de fois afin de calculer la proportion de points \u00e0 l&rsquo;int\u00e9rieur du disque par rapport au nombre total de points. Nous allons impl\u00e9menter cela en Python.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">On va imaginer une fonction <strong>is_in() <\/strong>renvoyant 1 si un point choisi au hasard est dans le cercle, et renvoyant 0 sinon. Pour cela, on se place dans un rep\u00e8re (O,I,J), o\u00f9 le disque jaune est de centre O. On choisit au hasard deux nombres <em>a<\/em> et <em>b<\/em>, tous deux compris entre 0 et 1 et on calcule les coordonn\u00e9es d&rsquo;un point M(x;y) tel que \\( x = -1 + 2a \\) et \\(y = -1+ 2b\\); ainsi, M est dans le carr\u00e9 vert. <\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Ensuite, on regarde si M est dans le disque jaune; pour cela, on calcule la distance OM:$$OM=\\sqrt{x^2+y^2}.$$ Si cette distance est inf\u00e9rieure ou \u00e9gale \u00e0 1, cela signifie que M est dans le disque; la fonction doit alors renvoyer \u00ab\u00a01\u00a0\u00bb; sinon, elle renvoie 0.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Cela donne:<\/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=\"\">from random import random\ndef is_in():\n    a = random() # choisit un nombre al\u00e9atoire entre 0 et 1\n    b = random()\n    x = -1 + 2 * a # absisse d'un point M comprise entre -1 et 1\n    y = -1 + 2 * b # ordonn\u00e9e de M comprise entre -1 et 1\n    \n    d = ( x**2 + y**2 )**0.5 # distance OM\n\n    if d &lt;= 1: # si M est dans le disque...\n        return 1\n    else:     # sinon...\n        return 0<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">[Ajout du 14\/06\/2021 suite \u00e0 un commentaire] Comme le rayon de notre disque est \u00e9gal \u00e0 1, on peut aussi comparer le carr\u00e9 de la distance, ce qui permet de gagner en terme de complexit\u00e9 algorithmique, et donc gagner du temps :<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">from random import random\ndef is_in():\n    a = random() # choisit un nombre al\u00e9atoire entre 0 et 1\n    b = random()\n    x = -1 + 2 * a # absisse d'un point M comprise entre -1 et 1\n    y = -1 + 2 * b # ordonn\u00e9e de M comprise entre -1 et 1\n    \n    d = x**2 + y**2  # carr\u00e9 de la distance OM\n\n    if d &lt;= 1: # si M est dans le disque...\n        return 1\n    else:     # sinon...\n        return 0<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Maintenant, nous devons simuler un grand nombre de fois cette exp\u00e9rience, disons 10 000 fois (par exemple).<\/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=\"\">N = 0 # nombre de points \u00e0 l'int\u00e9rieur du disque\nf = 5000000\n\nfor n in range(f):\n    N += is_in()\n\nprint(\"PI est \u00e0 peu pr\u00e8s \u00e9gal \u00e0 {}\".format(4 * N \/ f))<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Prendre 10 000 points peut para\u00eetre \u00e9norme, mais il n&rsquo;en est rien&#8230; En effet, la valeur retourn\u00e9e est assez \u00e9loiogn\u00e9e de \\(\\pi\\) (j&rsquo;ai obtenu 3.14296). Bon, vous allez me dire, c&rsquo;est pas si \u00e9loign\u00e9 que \u00e7a cr\u00e9vindiou ! Mouais&#8230; M&rsquo;enfin&#8230; Y&rsquo;a mieux ! En prenant  10 000 000 points, on arrive \u00e0 3.1417784, et pour n =  100 000 000 (avec un peut de patience car il faut tout de m\u00eame ex\u00e9cuter pas mal d&rsquo;op\u00e9rations et cela ne se fait pas en quelques secondes uniquement), j&rsquo;obtiens 3.14124084&#8230; Pas s\u00fbr que \u00e7a vallait l&rsquo;attente de plusieurs minutes&#8230;<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Pourquoi_cette_methode_est_aussi_pourrie\"><\/span>Pourquoi cette m\u00e9thode est aussi pourrie ?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">En fait, ce n&rsquo;est pas la m\u00e9thode qui est nulle, mais les r\u00e9sultats trouv\u00e9s par Python. Il ne faut pas oublier d\u00e9j\u00e0 que le hasard n&rsquo;existe pas en informatique. Ainsi, la simulation du hasard n&rsquo;est qu&rsquo;un pseudo-hasard, ce qui peut expliquer en partie les r\u00e9sultats d\u00e9cevants que nous avons obtenus. De plus, les points pris al\u00e9atoirement peuvent \u00eatre trop proches, voire confondus, ce qui fait qu&rsquo;on peut compter deux fois un \u00ab\u00a0m\u00eame\u00a0\u00bb point et on peut imaginer que tous les points pris al\u00e9atoirement \u00ab\u00a0ne couvrent pas enti\u00e8rement le disque\u00a0\u00bb (oui, c&rsquo;est assez abusif de dire \u00e7a, mais bon&#8230; j&rsquo;essaie de simplifier). En effet, imaginez au pire des cas que tous les points choisis au hasard tombent au m\u00eame endroit dans le disque : la probabilit\u00e9 obtenue sera alors \u00e9gale \u00e0 1, soit une valeur approch\u00e9e de \\(\\pi\\) \u00e9gale \u00e0 4&#8230; ! <\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Alors bien s\u00fbr, l&rsquo;algorithme de \u00ab\u00a0random\u00a0\u00bb est performant et il n&rsquo;y a aucune chance qu&rsquo;un grand nombre de points tombent presque au m\u00eame endroit, mais cet algorithme ne refl\u00e8te pas non plus la r\u00e9alit\u00e9&#8230;<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Donc en th\u00e9orie, la m\u00e9thode de Monte-Carlo est int\u00e9ressante, mais dans la pratique, elle ne vaut pas grand-chose&#8230;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\\(\\pi\\) est la constante d\u00e9finie comme \u00e9tant le rapport de la circonf\u00e9rence d&rsquo;un cercle et de son diam\u00e8tre. Et on arrive \u00e0 d\u00e9montrer que l&rsquo;aire du disque d\u00e9fini par ce cercle est \u00e9gale \u00e0 : $$\\mathcal{A}=\\pi \\times r^2.$$Nous allons voir dans cet article comme utiliser cette derni\u00e8re \u00e9galit\u00e9 afin de [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[21,6,5],"tags":[164],"class_list":["post-1937","post","type-post","status-publish","format-standard","hentry","category-enseignement","category-mathematiques","category-python","tag-monte-carlo"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.1 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>D\u00e9terminer une valeur approch\u00e9e de Pi \u00e0 l&#039;aide des probabilit\u00e9s (m\u00e9thode de Monte-Carlo sous Python) - Mathweb.fr<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.mathweb.fr\/euclide\/2020\/02\/01\/determiner-une-valeur-approche-de-pi-a-laide-des-probabilites-methode-de-monte-carlo-sous-python\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"D\u00e9terminer une valeur approch\u00e9e de Pi \u00e0 l&#039;aide des probabilit\u00e9s (m\u00e9thode de Monte-Carlo sous Python) - Mathweb.fr\" \/>\n<meta property=\"og:description\" content=\"(pi) est la constante d\u00e9finie comme \u00e9tant le rapport de la circonf\u00e9rence d&rsquo;un cercle et de son diam\u00e8tre. 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