Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Since a house with index i would collapse on its own by time i, the longest lifespan is 36, which can be achieved with the extra rule when the collapsing times are perfectly ordered. For the shortest lifespan, I ran a short R code implementing the rules and monitoring its minimum. Which found 7 as the minimal number for 10⁵ draws. However, with an optimal ordering, one house plus one or two neighbours of the most recently collapsed, leading to a maximal number of collapsed houses after k time units being
1+2(k-1)+1+2(k-2)+….=k+k(k-1)=k²
which happens to be equal to 36 for k=6. (Which was also obtained in 10⁶ draws!) This also gives the solution for any value of k.
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.