Since challenge questions are up in the air, what better time to pose a puzzle-building question that's intrigued me for some time.
Consider an infinite square grid made up of nodes and edges as shown in the following figure:
Each node connects to four edges, which are labeled top, left, right, and bottom as one would intuitively expect. Likewise, every square in the grid has four labeled edges. These labels are depicted in the two figures below.
Every edge in the grid has a parity label, which is either $A$ or $B$. Parity labels obey the following two symmetry rules:
At every node, the multiset of parity labels of the top and right edges must equal the multiset of parity labels on the bottom and left edges.
For example, if both the top and right edges have a parity label of $A$, then both the bottom and left edges must have a parity label of $A$. If the top and right edges have parity labels of $A$ and $B$ respectively, then then parity labels for the bottom and left edges must either be $A,B$ or $B,A$, respectively.
For every grid square, the multiset of parity labels of the top and left edges must equal the multiset of parity labels on the bottom and right edges.
The idea of the puzzle is to find a rule (or rules) for assigning a parity label to each edge in the grid that satisfies rule 1 at every node and rule 2 for every grid square.
There are several trivial ways to do this, such as assigning $A$'s to all edges, assigning $B$'s to all edges, etc., but one additional complication exists. The grid must start in either of the following two states:
Note that rule 2 is satisfied for the fully-assigned grid square in both grids, but rule 1 is not satisfied for the two fully-assigned nodes. This broken symmetry is by design for these particular nodes in these grids.
None of the existing parity labels may be changed, and rule 1 applies everywhere except at the two symmetry-breaking nodes.
The amended goal of the puzzle is therefore to find a rule (or rules) for assigning a parity label to each edge in the grid that satisfies rule 1 at every node except at the two symmetry-breaking nodes, and satisfies rule 2 for every grid square.
It's not intuitively obvious that the puzzle as stated above even has a solution. And indeed, all of my attempts to find a rule to populate the grid have lead to what I'll call "gridlock"—the inability to extend labels further in any direction without violating at least one of the two symmetry rules.
Unfortunately, "COTO gets consistently stuck" is far from a rigorous proof of non-existence of a solution, hence I shall commit the puzzle to the enterprising minds of puzzling.SE.
Does this puzzle have a solution? If so, what is it? If not, is there an elegant proof of this fact?
Answers and intuitions are welcome.