TITLE: Breaking chocolate bars game
QUESTION [13 upvotes]: About two weeks ago, a friend of mine taught me the following game without his knowing the answer. It may be famous, but I haven't known it.
There are $N\ (\in\mathbb N)$ chocolate bars composed of $a_i\times b_i\ (i=1,2,\cdots, N)$ squares of chocolate. Here, suppose that $a_i$ is the length of the vertical edge of each square, and that $b_i$ is the length of the horizontal edge of each square. Also, let $a_i,b_i$ be natural numbers. (see below. $N=4$ case.) 
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Two players $A, B$ take turns picking up a piece of chocolate and cutting it along the grid lines. A player cuts through and creates a new pieces. The first player $A$ always has to cut it horizontally. The second player $B$ always has to cut it vertically. One must not turn around the squares. A player that has no place to cut loses. (see below. $N=3$ case in which the game will be continued.) 
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Examples : If there is a chocolate bar composed of $3\times 2$ squares of chocolate, then the player $B$ wins. If there is a chocolate bar composed of $4\times 2$ squares of chocolate, then the player $A$ wins. 

Question : In what cases does the player $A$ win? Why? 

I have a conjecture. 
My conjecture : The player $A$ wins $\iff \sum_{i=1}^{N}\lfloor \log_2 a_i\rfloor\gt \sum_{i=1}^{N}\lfloor\log_2 b_i\rfloor$ holds.
This seems true for $N=1$ case and some other smaller cases, but I don't think I've succeeded in proving this strictly. Of course it may not be true. Can anyone help?
Edit : I edited the notation. I hope this time is not confusing.

REPLY [2 votes]: This game is called Cutcake, and it is explicitly analyzed in both "On Numbers and Games" and "Winning Ways". (For some reason the spam-blocker thinks this answer is spam; not sure why.)