Talk:Cop-win graph

There is a problem with the paragraph about hereditarily cop-win graphs. If they are defined as the graphs in which every induced subgraph is cop-win (as it is now), then they do not correspond to bridged graphs (consider a cycle of length 6 plus a universal vertex). In the abstract of the paper of Anstee and Farber, it is written that "a connected graph is bridged if and only if every isometric subgraph is a cop-win graph".

So one need either to change the definition of hereditarily cop-win graphs by replacing induced by isometric, or to rephrase/remove the subsequent sentences.
 * Looking at ISGCI it seems you are right, it should be isometric not just induced. —David Eppstein (talk) 20:20, 4 April 2018 (UTC)

The article states that the king's graph is cop-win, but that can't be true. The 2 by 2 version of the king's graph is the complete graph of order 4, which isn't cop win. 100.1.249.103 (talk) 02:10, 14 June 2018 (UTC)
 * Huh? Every complete graph is an immediate win for the cop. —David Eppstein (talk) 05:44, 14 June 2018 (UTC)

Something wrong...
In section Recognition algorithms and the family of all dismantling orders read:

Repeatedly find a vertex v that is an endpoint of an edge participating in a number of triangles equal to the degree of v, delete v, and decrement the triangles per edge of each remaining edge that formed a triangle with v.

Diamond graph has 2 triangles. 2 vertex has degree 2, but number of triangles equal 1. 2 vertex has degree 3, but number of triangles equal 2. So number of triangles equal to the degree of v never holds...

Another case. Take any path. There are no triangles at all. So number of triangles equal to the degree of v again does not held. Jumpow (talk) 09:54, 3 April 2019 (UTC)
 * There are two issues here, one with your article but the other with your misreading of it. (1) There's an off-by-one error; it should be degree minus one. (2) It's not the number of triangles v belongs to, and it's not the number of triangles in the whole graph; it's whether there's an edge vw that participates in deg(v)-1 triangles. —David Eppstein (talk) 17:01, 3 April 2019 (UTC)


 * Ok, 1. Agree 2. Not agree. What about missing triangles in graph? For example - lattice. So number of triangles equal to the degree of v minus one again does not held (triangles = 0, min degree = 2). It looks like I something missed... Every dismantling graph has triangles? And why exist vertex with property deg(v)-1 = # triangles? Jumpow (talk) 18:06, 3 April 2019 (UTC)
 * The square lattice is not a cop-win graph. The example in the article that looks like a lattice is actually a King's graph, a different graph with many triangles. And in a cop-win graph, if v is a dominated vertex, it must have a neighbor w that dominates it, in which case edge vw participates in exactly deg(v)-1 triangles, one with each other neighbor of v. —David Eppstein (talk) 19:23, 3 April 2019 (UTC)