5. $A$ and $B$ play alternating turns on a $2012 \times 2013$ board with enough pieces of the following types: Type $1$: Piece like Type $2$ but with one square at the right of the bottom square. Type $2$: Piece of $2$ consecutive squares, one over another. Type $3$: Piece of $1$ square. At his turn, $A$ must put a piece of the type $1$ on available squares of the board. $B$, at his turn, must put exactly one piece of each type on available squares of the board. The player that cannot do more movements loses. If $A$ starts playing, decide who has a winning strategy. Note: The pieces can be rotated but cannot overlap; they cannot be out of the board. The pieces of the types $1$, $2$ and $3$ can be put on exactly $3$, $2$ and $1$ squares of the board respectively.
Problem
Source: Problem 5
Tags: rotation, combinatorics proposed, combinatorics
20.04.2019 04:23
Anyone??
20.04.2019 04:29
um sorry im stupid so idk this sorry i'll see what i can help with though i just need to understand the problem
04.10.2019 12:25
For a contradiction, assume A wins no matter how B plays. In the first two moves, B will make (in 2 corners) 2 structures that i don't know how to draw here . They allow him 2 play 2 dominoes or two squares at any time, and A can't put a type 1 over them. Let's call them $banks$ Next, B will play a type 1, and assemble another type one from a type 2 and 3, so let's assume he plays two type ones. (He can play however) Now let's look at the final position. A places a type 1, and B has nowhere to play a type 1 (because he can put the other two in the structures he made). Now we look at a position before A and B made there last moves, and lets call positions of the type ones they put after, $pos 1$ and $pos 2$ for B and $pos 3$ for A. So, B doesn't have to put a type 1 on $pos 2$, instead he just plays $pos 1$ . If A plays either $pos 2$ or $pos 3$, B just plays the other one and puts the rest in the other $bank$ and wins. So by our assumption, A can play some $pos 4$. Now let's rewind again, and let B play $pos 1$ and $pos 4$, but because we assume A wins, there is a $pos 5$ (possibly coinciding with the previous few) he can play. That can't make B lose, because if it did, B could play that in the position where he has $pos 1$ and A has $pos 4$ and make A lose. So i hope this means whenever B loses, he can either alter his moves and make A lose, or alter his moves and not lose without using any of the $banks$ PS. can someone help me to actually draw these $banks$