Xi'an's Og

A recursive programming  Le Monde mathematical puzzle:

Given n tokens with 10≤n≤25, Alice and Bob play the following game: the first player draws an integer1≤m≤6 at random. This player can then take 1≤r≤min(2m,n) tokens. The next player is then free to take 1≤s≤min(2r,n-r) tokens. The player taking the last tokens is the winner. There is a winning strategy for Alice if she starts with m=3 and if Bob starts with m=2. Deduce the value of n.

Although I first wrote a brute force version of the following code, a moderate amount of thinking leads to conclude that the person given n remaining token and an adversary choice of m tokens such that 2m≥n always win by taking the n remaining tokens:

[sourcecode language=”r” gutter=”false”]
optim=function(n,m){

outcome=(n<2*m+1)
if (n>2*m){
for (i in 1:(2*m))
outcome=max(outcome,1-optim(n-i,i))
}
return(outcome)
}
[/sourcecode]

eliminating solutions which dividers are not solutions themselves:

[sourcecode language=”r” gutter=”false”]
sol=lowa=plura[plura<100]
for (i in 3:6){
sli=plura[(plura>10^(i-1))&(plura<10^i)]
ace=sli-10^(i-1)*(sli%/%10^(i-1))
lowa=sli[apply(outer(ace,lowa,FUN="=="),
1,max)==1]
lowa=sort(unique(lowa))
sol=c(sol,lowa)}
[/sourcecode]

which leads to the output

[sourcecode language=”r” gutter=”false”]
> subs=rep(0,16)
> for (n in 10:25) subs[n-9]=optim(n,3)
> for (n in 10:25) if (subs[n-9]==1) subs[n-9]=1-optim(n,2)
> subs
[1] 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
> (10:25)[subs==1]
[1] 18
[/sourcecode]

Ergo, the number of tokens is 18!

This entry was posted on April 1, 2015 at 12:15 am and is filed under Books, Kids, R, Statistics, University life with tags Le Monde, mathematical puzzle, R, recursive function. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.