Eating chocolate game on grid Given is a chocolate of size $m\times n$. Anne and Birgitte plays a game, with Anne starting. In each turn, the player has to divide the chocolate into two rectangular parts along the lines, and eat the smaller part (or any part if both are equal.) The first person to eat chocolate of size $1\times 1$ loses.
Who has a winning strategy?
[Source: based on Israeli competition problem]
 A: Okay, if the strategy has to guarentee a win, then it is impossible for one person to win all of the time. 
For example, 3 x 1 forces the first player to lose. The first break would either create a 2 x 1 + 1 x 1 or a 1 x 1 + 2 x 1, and in either case, they would be forced to eat the 1 x 1, making Anna lose.
However, with a 5 x 1, Birgette is forced to lose. Anna's first move would have to be splitting the bar into a 3 x 1 + 2 x 1, forcing her to eat the 2 x 1 piece and leaving Birgette with the 3 x 1 piece, which we have already seen forces a loss.
A: Following cases are the all initial conditions for which the one go the first step will lose the game:
1)there exist k>=2 and u,v>=0, such that n = (2k+1) * 2^u - 1, m = (2k+1) * 2^v - 1;
2)there exist u,v>=0 such that n = 2 * 2^u - 1 and m = 3 * 2^v - 1;
3)there exist u,v>=0 such that n = 3 * 2^u - 1 and m = 2 * 2^v - 1.
If the initial condition is different from the listed above, then one going first will win.   
The complete proof is long and somewhat boring, but I get the feeling through the simple cases when n , m are small. Followings are the main ideas.   
Def:
1)Regard ( n, m ) as the grid point on the place, call it a bad point if the one going fist will lose, and call it a good point if the one going first will win;
2)Call a sequence a[n], n>=0 a "typical sequence", if it satisfies a[n+1] = 2 * a[n] + 1, and the initial term a[0] is either 1 or even, which means that one go backward from a[0] will no longer get a positive integer.   
Some observations:
1)A grid point must be either good or bad;
2)Symmetry: if ( n, m ) is bad/good, then so is ( m, n );
3)If ( n, m ) is a bad point, then so are all points ( k, m ), n<=k<=2n;
4)If ( k, m ) is good for all n/2<=k<=n-1, and ( n, t ) is good for all m/2<=k<=m-1, then ( n, m ) must be a bad point. And if this condition doesn't hold, then ( n, m ) is a good point. 
Main claim:
For every fixed m>=2, all bad points contained in the set {( n, m ) | n is a positive integer} form a typical sequence whose initial term is less or equal to m.  
The proof of this main claim can be done by second mathematical induction. And notice the observations 3) and 4), as well as the symmetry property 2) which may be ignored, all play essential roles within the proof. Nevertheless, there is no trick during the proof, it is just a combination of several almost trivial checks.
Then, when obtain the main claim, one can easily see, maybe a sketch of these points in a coordinate card is needed, all bad points have been found out.
