Le Monde puzzle [#1049]

An algorithmic Le Monde mathematical puzzle with a direct

Alice and Bob play a game by picking alternatively one of the remaining digits between 1 and 10 and putting it in either one of two available stacks, 1 or 2. Their respective gains are the products of the piles (1 for Alice and 2 for Bob).

The problem is manageable by a recursive function

facten=factorial(10)
pick=function(play=1,remz=matrix(0,2,5)){
 if ((min(remz[1,])>0)||(min(remz[2,])>0)){#finale
  remz[remz==0]=(1:10)[!(1:10)%in%remz]
  return(prod(remz[play,]))
  }else{
   gainz=0
   for (i in (1:10)[!(1:10)%in%remz]){
     propz=rbind(c(remz[1,remz[1,]>0],i,
     rep(0,sum(remz[1,]==0)-1)),remz[2,])
     gainz=max(gainz,facten/pick(3-play,remz=propz))}
   for (i in (1:10)[!(1:10)%in%remz]){
     propz=rbind(remz[1,],c(remz[2,remz[2,]>0],i,
     rep(0,sum(remz[2,]==0)-1)))
     gainz=max(gainz,facten/pick(3-play,remz=propz))}
return(gainz)}}

that shows the optimal gain for Alice is 3360=2x5x6x7x 8, versus Bob getting 1080=1x3x4x9x10. The moves ensuring the gain are 2-10-…

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

w

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.