Here is the Region Select game based on the knot theory.

Question. Is it possible to find an invariant that divides the initial set of lamps knots into a solvable and an unsolvable game?

My attempt is:

I have found the papers Dror Bar-Natan, 1992, Ted Stanford, 1992.


If $P$ is a knot projection (rather than a link of multiple components), then any starting configuration of lamps can be turned all off or all on. In Region crossing change is an unknotting operation, Ayaka Shimizu proves that for any fixed lamp, there is a sequence of region flips that can change the given lamp but leave all others fixed.

The basic idea of the proof follows. Think of the lamps as crossing information, so that the game board is now a knot diagram. Then flipping all the lamps in a region corresponds to changing all crossings around a region. The procedure to change one crossing is:

  1. Orient the knot, and resolve the chosen crossing according to the orientation:enter image description here
  2. The resulting link diagram has two components. Apply a checkerboard coloring for one of the components (pretending the other component is not there).
  3. Perform a region crossing change for each of the shaded regions (in the original knot diagram).

Here is an example of the algorithm in action (taken from an expository article by Shimizu):enter image description here

Since it's possible to change a single crossing (or flip a single lamp in the original formulation) through a sequence of region flips, that means any configuration of on/off lamps (including all-on and all-off) is obtainable from any initial configuration.

  • $\begingroup$ I will add as a comment that the case for diagrams of links of multiple components has recently been solved by Heather Russell and Oliver Dasbach. When the diagram has more than one component, there can be more than one equivalence class (think of the standard diagram of the Hopf link). If I remember, I will update this answer once they post their paper to the arxiv. $\endgroup$ – Adam Lowrance Nov 19 '17 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.