Le Monde puzzle [#1099]

April 27, 2019
By

(This article was first published on R – Xi'an's Og, and kindly contributed to R-bloggers)

A simple 2×2 Le Monde mathematical puzzle:

Arielle and Brandwein play a game out of two distinct even integers between 1500 and 2500,  and y. Providing one another with either the pair (x/2,y+x/2) or the pair (x+y/2,y/2) until they run out of even possibilities or exceed 6 rounds. When x=2304, what is the value of y that makes Brandwein win?

Which I solved by a recursive function (under the constraint of a maximum of 11 levels of recursion):

nezt=function(x,y,i=1){
  if ((i>11)||((is.odd(x)&is.odd(y)))){ return(-1)
  }else{
    z=-1
    if (is.even(x)) z=-nezt(x/2,y+x/2,i+1)
    if (is.even(y)) z=max(z,-nezt(y/2,x+y/2,i+1))
    return(z)}}

and checking all values of y between 1500 and 2500 when x=2304, which produces y=1792 as the only value when Arielle loses. The reason behind (?) is that both 2304 and 1792 are divisible by 2⁸, which means no strategy avoids reaching stalemate after 8 steps, when it is Arielle’s turn to play.

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 on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more...



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)