a chain of collapses

June 19, 2018
By

[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.

A quick riddler resolution during a committee meeting (!) of a short riddle: 36 houses stand in a row and collapse at times t=1,2,..,36. In addition, once a house collapses, the neighbours if still standing collapse at the next time unit. What are the shortest and longest lifespans of this row?

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.

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.



If you got this far, why not subscribe for updates from the site? Choose your flavor: e-mail, twitter, RSS, or facebook...

Comments are closed.

Search R-bloggers

Sponsors

Never miss an update!
Subscribe to R-bloggers to receive
e-mails with the latest R posts.
(You will not see this message again.)

Click here to close (This popup will not appear again)