Let me be a bit more precise here (at risk of being pedantic) to make sure we're on the same page. (Also, if you have a concrete idea of how to reformulate the text so this is clearer, I'm definitely interested!)
Here are two well-defined conjectures:
Conjecture 1. A packing-coloring of the infinite square grid exists using colors {1, ..., k} if, and only if, there exists one with a chessboard pattern (meaning that wlog all vertices (a, b) such that a+b is odd get color 1).
Conjecture 2. Let $D_{r, k, c}$ be the instance consisting of whether $D_r$ can be packing-colored with $k$ colors assuming it gets color $c$ in the center. Then enforcing that all vertices (a, b) such a+b is odd get color 1 does not change the satisfiability of the instance.
Conjecture 1 is true, based on our paper: if k <= 14, then no packing-coloring exists anyway, so the conjecture is vacuously true. If k >= 15, then we know of a packing-coloring that respects the chessboard pattern (the one in Figure 12), and so the conjecture holds.
Conjecture 2 is false, this is where the smallest counterexample in the post kicks in.
Note that given that 13 colors are not enough, we knew that if there was a solution with 14 colors, it must use color 6 somewhere, and by restricting ourselves to $D_{14}$ around such a vertex, we run into the issue!
About your last comment, we use diamond graphs to prove lower bounds (i.e., that a certain number of colors is not enough), while finite square grids are used to prove upper bounds (as in Figure 12).
About aperiodic colorings, see the other comment in this thread were we discuss about it a bit :)
Here are two well-defined conjectures:
Conjecture 1. A packing-coloring of the infinite square grid exists using colors {1, ..., k} if, and only if, there exists one with a chessboard pattern (meaning that wlog all vertices (a, b) such that a+b is odd get color 1).
Conjecture 2. Let $D_{r, k, c}$ be the instance consisting of whether $D_r$ can be packing-colored with $k$ colors assuming it gets color $c$ in the center. Then enforcing that all vertices (a, b) such a+b is odd get color 1 does not change the satisfiability of the instance.
Conjecture 1 is true, based on our paper: if k <= 14, then no packing-coloring exists anyway, so the conjecture is vacuously true. If k >= 15, then we know of a packing-coloring that respects the chessboard pattern (the one in Figure 12), and so the conjecture holds.
Conjecture 2 is false, this is where the smallest counterexample in the post kicks in.
Note that given that 13 colors are not enough, we knew that if there was a solution with 14 colors, it must use color 6 somewhere, and by restricting ourselves to $D_{14}$ around such a vertex, we run into the issue!
About your last comment, we use diamond graphs to prove lower bounds (i.e., that a certain number of colors is not enough), while finite square grids are used to prove upper bounds (as in Figure 12).
About aperiodic colorings, see the other comment in this thread were we discuss about it a bit :)