From ad6da5f0c25327eedc85e4608f25a2d360ad183b Mon Sep 17 00:00:00 2001 From: af867c30e46f4b444ec768838323b91d Date: Sat, 6 Jun 2020 10:37:33 +0000 Subject: [PATCH] Update toy_document_orgmode_R_en.org --- module2/exo1/toy_document_orgmode_R_en.org | 60 +++++++++++----------- 1 file changed, 30 insertions(+), 30 deletions(-) diff --git a/module2/exo1/toy_document_orgmode_R_en.org b/module2/exo1/toy_document_orgmode_R_en.org index 6f86037..a782a5f 100644 --- a/module2/exo1/toy_document_orgmode_R_en.org +++ b/module2/exo1/toy_document_orgmode_R_en.org @@ -18,34 +18,34 @@ My computer tells me that $\pi$ is /approximatively/ pi #+end_src -#+RESULTS: -: [1] 3.141593 -* Buffon's needle -Applying the method of [[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]], we get the *approximation* -#+begin_src R :results output :session *R* :exports both -set.seed(42) -N = 100000 -x = runif(N) -theta = pi/2*runif(N) -2/(mean(x+sin(theta)>1)) -#+end_src -#+RESULTS: -: [1] 3.14327 -* Using a surface fraction argument -A method that is easier to understand and does not make use of the $\sin$ function is based on the fact that if $X\sim U(0,1)$ and $Y\sim U(0,1)$, then $P[X^2+Y^2\leq 1] = \pi/4$ (see [[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo method" on Wikipedia]]). The following code uses this approach: -#+begin_src R :results output graphics :file figure_pi_mc1.png :exports both :width 600 :height 400 :session *R* -set.seed(42) -N = 1000 -df = data.frame(X = runif(N), Y = runif(N)) -df$Accept = (df$X**2 + df$Y**2 <=1) -library(ggplot2) -ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw() -#+end_src -#+RESULTS: -[[file:figure_pi_mc1.png]] -It is then straightforward to obtain a (not really good) approximation to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1: -#+begin_src R :results output :session *R* :exports both -4*mean(df$Accept) -#+end_src -#+RESULTS: +#+RESULTS: +: [1] 3.141593 +* Buffon's needle +Applying the method of [[https://en.wikipedia.org/wiki/Buffon%2527s_needle_problem][Buffon's needle]], we get the *approximation* +#+begin_src R :results output :session *R* :exports both +set.seed(42) +N = 100000 +x = runif(N) +theta = pi/2*runif(N) +2/(mean(x+sin(theta)>1)) +#+end_src +#+RESULTS: +: [1] 3.14327 +* Using a surface fraction argument +A method that is easier to understand and does not make use of the $\sin$ function is based on the fact that if $X\sim U(0,1)$ and $Y\sim U(0,1)$, then $P[X^2+Y^2\leq 1] = \pi/4$ (see [[https://en.wikipedia.org/wiki/Monte_Carlo_method]["Monte Carlo method" on Wikipedia]]). The following code uses this approach: +#+begin_src R :results output graphics :file figure_pi_mc1.png :exports both :width 600 :height 400 :session *R* +set.seed(42) +N = 1000 +df = data.frame(X = runif(N), Y = runif(N)) +df$Accept = (df$X**2 + df$Y**2 <=1) +library(ggplot2) +ggplot(df, aes(x=X,y=Y,color=Accept)) + geom_point(alpha=.2) + coord_fixed() + theme_bw() +#+end_src +#+RESULTS: +[[file:figure_pi_mc1.png]] +It is then straightforward to obtain a (not really good) approximation to $\pi$ by counting how many times, on average, $X^2 + Y^2$ is smaller than 1: +#+begin_src R :results output :session *R* :exports both +4*mean(df$Accept) +#+end_src +#+RESULTS: : [1] 3.156 \ No newline at end of file -- 2.18.1