[This article was first published on R – Xi'an's Og, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
An express riddle from the Riddler about reopening pools, where lanes are allowed provided there is no swimmer in the lane or in any of the adjacent lanes. If swimmers pick their lane at random (while they can), what is the average number of occupied lanes?
If there are n lanes and E(n) is the expected number of swimmers, E(n) satisfies a recurrence relation determined by the location of the first swimmer:
![E(n)=1+\frac{1}{n}[2E(n-2)+\sum_{i=2}^{n-1}\{E(i-2)+E(n-i-1)\}]](https://s0.wp.com/latex.php?latex=E%28n%29%3D1%2B%5Cfrac%7B1%7D%7Bn%7D%5B2E%28n-2%29%2B%5Csum_%7Bi%3D2%7D%5E%7Bn-1%7D%5C%7BE%28i-2%29%2BE%28n-i-1%29%5C%7D%5D&bg=000000&%23038;fg=B0B0B0&%23038;s=0 "E(n)=1+\frac{1}{n}[2E(n-2)+\sum_{i=2}^{n-1}\{E(i-2)+E(n-i-1)\}]")
with E(0)=0, E(1)=E(2)=1. The above can be checked with a quick R experiment:
en=0
for(t in 1:T){
la=rep(u<-0,N)
while(sum(la)<N){
i=sample(rep((1:N)[!la],2),1)
la[max(1,i-1):min(N,i+1)]=1
u=u+1}
en=en+u}
To leave a comment for the author, please follow the link and comment on their blog: R – Xi'an's Og.
R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.