Suppose that you have a stack of identical, frictionless, uniform-density $1 \times 2$ bricks arranged in the ordinary configuration (each row is offset by $1$ relative to the row below.)
Question
Is there a combinatorial rule that captures whether or not a given stack of bricks would be stable? In particular, that no small vertical force will cause the stack to move.
In other words, how can one determine whether or not a configuration is stable without doing a static force analysis.
(Admittedly, this question is a little hand-wavey, so let me know if I can clarify anything.)
Examples
Clearly, a stack of bricks that does not have any "overhangs" should be stable:
But we can also allow a cantilever if there's a brick above:
Non-examples
However, having a brick above is not sufficient, because a stack like this should not be stable.
Similarly, having a brick above is necessary because a configuration like this is in unstable equilibrium; in particular the upper-right brick in this example would fall if an arbitrarily small force were applied to the right side.
Although probably not a combinatorial rule per se, here is an algorithm that could be used to find unstable equilibria or configurations that would not be in equilibria:
1) Identify all bricks with only 1 supporting brick.
2) For each of these bricks consider the bricks above it connected to it by an upward facing V. (These are the only bricks that put any weight onto the brick, and therefore determine its stability.)
3) If this group of bricks has a center of gravity that is directly over the supporting brick, then the configuration is stable. If the center of gravity is directly over the edge of the supporting brick then it is unstable. If the center of gravity is over the unsupported side then it is not in equilibrium and will fall due to gravity.
4) If all singly supported bricks are in stable equilibrium, then the entire configuration is also. Else, if all singly supported bricks are either stable or unstable, then the configuration is unstable. Else, the configuration is not in an equilibrium.