Alice and Bob play a game on a Cartesian Coordinate Plane. At the beginning, Alice chooses a lattice point $ \left(x_{0}, y_{0}\right) $ and places a pudding. Then they plays by turns (B goes first) according to the rules a. If $ A $ places a pudding on $ \left(x,y\right) $ in the last round, then $ B $ can only place a pudding on one of $ \left(x+2, y+1\right), \left(x+2, y-1\right), \left(x-2, y+1\right), \left(x-2, y-1\right) $ b. If $ B $ places a pudding on $ \left(x,y\right) $ in the last round, then $ A $ can only place a pudding on one of $ \left(x+1, y+2\right), \left(x+1, y-2\right), \left(x-1, y+2\right), \left(x-1, y-2\right) $ Furthermore, if there is already a pudding on $ \left(a,b\right) $, then no one can place a pudding on $ \left(c,d\right) $ where $ c \equiv a \pmod{n}, d \equiv b \pmod{n} $. 1. Who has a winning strategy when $ n = 2018 $ 1. Who has a winning strategy when $ n = 2019 $
Problem
Source: 2019 Taiwan TST Round 1
Tags: analytic geometry, combinatorics
15.06.2020 19:12
Any Solutions?
16.06.2020 11:39
For n=2018, Bob wins. WLOG bob always does $(x+2,y+1)$, and if we were to color the grid like a checkerboard, there’s a 1-1 correspondence between a white square $(x,y)$ and a black square $(x+2\mod 2018, y+1\mod 2018)$. Each turn, alice must fill a new white square, while bob can just fill the corresponding black square, which is guaranteed to be empty.
16.06.2020 12:29
dantx5 wrote: For n=2018, Bob wins. WLOG bob always does $(x+2,y+1)$, and if we were to color the grid like a checkerboard, there’s a 1-1 correspondence between a white square $(x,y)$ and a black square $(x+2\mod 2018, y+1\mod 2018)$. Each turn, alice must fill a new white square, while bob can just fill the corresponding black square, which is guaranteed to be empty. Can Bob always fill a corresponding black square? I mean if the corresponding square that Bob is supposed to fill is $(x,y)$ but if a square $(a,b)$ is already occupied such that $a \equiv x \pmod{2018}$ and $b \equiv y \pmod{2018}$ then Bob cannot place a pudding on $(x,y)$
16.06.2020 20:50
Pretend the black square (x,y) is already occupied by Bob. In order for that to happen, Alice must go to white square (x-2, y-1) the turn before, and then bob goes (x,y). In order for (x,y) to be visited again by Bob, Alice must go to (x-2,y-1) again, but she can’t, because it’s already occupied.