[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.
Three buckets have capacities of 8, 5 and 3 litres, respectively. At the start of the game, the 8 litre bucket is full and both others are empty. Aiming at reaching exactly 4 litres in one bucket, water is transferred between buckets. However, a fraction 1/k is lost with each transfer. If k=9, it is possible to reach 4 litres in three operations? If k=7, is it at all possible to reach 4 litres?
By sheer random search
k=1/5 z=c(8,5,3) m<-function(s){ i=sample(1:3,2) s[i]=s[i]+ifelse( rep((a<-z[i[1]]-s[i[1]])<(b<-s[i[2]])*(1-k),2), a*c(1,-1-k),b*c(1-k,-1)) s}
I found that most fractions allow to reach 4 litres starting with k=2. (And am unsure the missing ones, like 18 or 21 are not due to a lack of luck… In particular, for k=9, the shortest path is
8.000 0 0 2.375 5 0 0.000 5 2.11 0.000 4 3
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.