le Monde puzzle [#745]
The puzzle in Le Monde this weekend is not that clear (for a change!), so I may be confused in the following exposition:
Three card players are betting with a certain (and different) number of chips each, between 4 and 9. After each game, the looser doubles the number of chips of the winner (while the second keeps her chips). The game stops if the looser cannot pay the winner. Find the initial configuration such that, at some point in the party, all players have the same number of chips.
So, if (x1,x2,x3) is the chip configuration at time t for the ordered players, (2x1,x2,x3-x1) is the chip configuration at time (t+1). Rather than running an exhaustive search, given the limited number of possibilities, I decided to search at random, ending up with the R function
[sourcecode language=”r” gutter=”false”]
lemonde=function(chip,chap){
x=rep(-1,3)
while (min(x)<0){
start=x=sample(chip:chap,3)
while (length(unique(start))==1)
start=x=sample(chip:chap,3)
while ((min(x)>-1)&&(length(unique(x))>1)){
x=sample(x) #random winner
x=c(2*x[1],x[2],x[3]-x[1])
}}
list(start=sort(start),finish=x)
}
[/sourcecode]
leading to
[sourcecode language=”r” gutter=”false”]
> lemonde(4,9)
$start
[1] 7 8 9
$finish
[1] 8 8 8
[/sourcecode]
with a unique starting point. More interestingly, other configurations may have several starting points. Of course, a mathematical analysis of the problem would bring more light on the difference. Maybe the issue of Le Monde next weekend (i.e., tonight!) will be enough.
[sourcecode language=”r” gutter=”false”]
> lemonde(4,10)
$start
[1] 5 9 10
$finish
[1] 8 8 8
> lemonde(4,10)
$start
[1] 6 8 10
$finish
[1] 8 8 8
> lemonde(4,10)
$start
[1] 7 8 9
$finish
[1] 8 8 8
[/sourcecode]
December 9, 2011 at 12:11 am
[…] a loooong break, here is one Le Monde mathematical puzzle I had time to look at, prior to going to Dauphine for a […]
October 21, 2011 at 10:30 am
When the game ends, the player who just won has an even number of chips. Hence the only possible end distributions are (6,6,6) and (8,8,8).
It is easy to check that the only distributions leading to (6,6,6) are (3,6,9) and (3,3,12), neither of which is admissible. Hence the end distribution is (8,8,8).
If there are 24 chips in total, the initial distribution must be (7,8,9) or (6,9,9). (6,9,9) cannot work: all these numbers are multiples of 3, hence this will still be true at all subsequent rounds.
Starting at (7,8,9) there is the path
(7,8,9)->(14,8,2)->(12,8,4)->(8,8,8).
October 21, 2011 at 11:42 am
Great, you beat le Monde!