## Le Monde puzzle [#860]

**A** Le Monde mathematical puzzle that connects to my awalé post of last year:

For N≤18, N balls are placed in N consecutive holes. Two players, Alice and Bob, consecutively take two balls at a time provided those balls are in contiguous holes. The loser is left with orphaned balls. What is the values of N such that Bob can win, no matter what is Alice’s strategy?

**I** solved this puzzle by the following R code that works recursively on N by eliminating all possible adjacent pairs of balls and checking whether or not there is a winning strategy for the other player.

topA=function(awale){ # return 1 if current player can win, 0 otherwise best=0 if (max(awale[-1]*awale[-N])==1){ #there are adjacent balls remaining for (i in (1:(N-1))[awale[1:(N-1)]==1]){ if (awale[i+1]==1){ bwale=awale bwale[c(i,i+1)]=0 best=max(best,1-topA(bwale)) } }} return(best) } for (N in 2:18) print(topA(rep(1,N)))

which returns the solution

[1] 1 [1] 1 [1] 1 [1] 0 [1] 1 [1] 1 [1] 1 [1] 0 [1] 1 [1] 1 [1] 1 [1] 1 [1] 1 [1] 0 [1] 1 [1] 1 [1] 1 <pre>

(brute-force) answering the question that N=5,9,15 are the values where Alice has no winning strategy if Bob plays in an optimal manner**.** (The case N=5 is obvious as there always remains two adjacent 1′s once Alice removed any adjacent pair. The case N=9 can also be shown to be a lost cause by enumeration of Alice’s options.)

April 6, 2014 at 9:33 pm

This assuming Alice goes first?

April 7, 2014 at 5:40 am

Yes indeed.