# Breaking of pieces from a 10 x 15 candy bar - winning strategy?

Two friends are playing a game where they break of pieces from a rectangular 10 x 15 candy bar. The game continues until either player gets a 1 x 1 piece. In the first move player A breaks the bar along a line. The second move is that player B chooses one of the two pieces and breaks it along a line (creating 3 total pieces). Third move is that player A can choose any of the three pieces (not just the latest two), and breaks it along a line. The game continues until either player breaks a piece leaving a 1 x 1 piece.

The question is if either of the players have a winning strategy if i) the loser is the player that creates a 1 x 1 piece ii) the winner is the player that creates a 1 x 1 piece

My reasoning is that after move $$n$$ there will be $$n + 1$$ pieces. Assuming optimal play there will be only 1 x 2 pieces left when either player is forced into losing in (i). That's exactly 75 pieces, so that's after move 74 which player B makes. That means that player A loses?

I don't really have a good idea of coming up with a strategy for (ii) though, any help of clue would be very much helpful!

• I don't see how "assuming optimal play" would lead to there never being any 1×3 pieces. Aug 12, 2019 at 10:37
• In (i) note that both 1x2 and 1x3 pieces are "dead," so the strategies will be driven by creating the right number of 1x3 pieces. For instance, if we started with a 1x6 piece and I went first, I would split it in half and make two 1x3s in order to win. All other moves would make me lose. Aug 12, 2019 at 10:39
• For (ii) the Sprague-Grundy theorem still applies if we declare that any move that creates a $1\times n$ piece is forbidden (it would be an immediately losing move anyway). So for a complete analysis there would be less that $9\times 14$ positions (i.e. piece sizes) to consider, and then combine them as nimbers. Aug 12, 2019 at 10:50

For both variants, $$A$$ has the winning strategy:
$$A$$'s first move will always be to break the bar in half, into two $$5 \times 15$$ pieces. Then $$A$$ will mirror $$B$$'s moves, so that at the end of $$A$$'s turn, there will always be an even number of pieces of size $$m \times n,$$ for every applicable $$m,n,$$ until $$B$$ makes a losing move.
For variant $$(i),$$ we define a losing move to be when $$B$$ makes a $$1 \times 1$$ piece. Obviously, at this point, $$B$$ will have lost with no more input from $$A$$.
For variant $$(ii),$$ we define a losing move to be when $$B$$ makes a $$n \times 1$$ (or $$1\times n$$) piece. It's easy to argue that the first time this occurs will happen on $$B$$'s turn and also that $$n > 1$$. Therefore, $$A$$ can break off the end and be the first to create a $$1 \times 1$$ piece, winning the game.