A 3x3 array of counters is laid out. Players take turns removing counters. The rule for removing counters is to pick a row or column and take any 1,2 or 3 from it. Whoever removes the last counter wins.
Does the first or second player have a winning strategy?
What is this strategy?
(In reply to tes he can
by Ady TZIDON)
(The problem was posted quite late at night, I overlooked part of Ady's response, and have posted this accordingly. I'm going to leave it here since
a) the next post points out my error,
b) it does suggest some strategies worthy of investigation should A not play as Ady suggested).
1 2 3
A 4 5 6 B
7 8 9
On your premise A takes 5, let B take 6 so A, by mirror takes 4.
B may now take 1,7; 2;8 or 3,9. If A chooses to take 2 then B takes the other 2.
If A takes 1 then B takes the one diagonally opposite in the rectangle/square array left after B's first play. the result inevitably is B wins.
I don't have time to finish my report on this, but I haven't found a scenario for A where B cannot logically win. There are different strategies that B must address, and one of them is to form that quadrilateral. That quadiraleral might be 'compromised' by A's last take in that a triangle remains, B takes from the cell commonly shared. Two cell left, B wins.
There are some senarios that B must avoid playing into but B must consider how to react as to the number that A withdraws.
It's late, I'll relook at my thoughts in the morning.
Edited on June 26, 2009, 1:00 am
Posted by brianjn
on 2009-06-25 12:45:19