Cops and Robbers is a Pursuit-evasion game played on graphs with two players, one controls the cops and the other one controls the robber. The game begins by the cops selecting some vertices as their initial positions. Then the robber, knowing the positions of cops, selects his initial vertex. From now on, first the cops move and then the robber moves, where moving means going to a neighboring vertex or staying at the same position. The goal of the cops is to capture the robber, which means having a cop at the same vertex as the robber, and the goal of the robber is to prevent this from happening. The minimum number of cops that guarantee the robber’s capture in a graph is called the cop number of and is denoted by .
One of the most important open problems in this area is Meyniel’s Conjecture.
If , then .
This conjecture has received lots of attentions but it is still far away from being proved. In fact the following conjecture, called Weak Meyniel’s Conjecture, is still widely open.
There is an such that every graph of order has .
Suppose that Conjecture 2 holds for every such that . Then we say that we have Weak Meyniel Conjecture with exponent .
2 Degree reduction
In this section, we show that from every graph we can construct a graph with the following properties:
has smaller maximum degree than , and
is not much larger than .
In the construction, we will replace each vertex of , whose degree is , with a copy of the following graph , where . To form the graph , we start with a set of mutually nonadjacent vertices that are partitioned into almost equal parts (so that for ). Take additional vertices for and join each to all vertices in . The resulting bipartite graph has vertices, each has degree about , and each has degree . See Figure 1 for an example.
In the following theorem we replace each vertex of a graph with a subgraph of the form , where is the degree of the vertex, and gives a construction of a graph with desired properties (a)–(c). Recall that denotes the maximum vertex degree in .
For every graph , there is a graph with the following properties:
Let be a vertex of degree in and let be its neighbors. Replace by a copy of the gadget described above and join to for . Apply this change to all vertices of to get . After doing this, each edge of has been replaced by an edge joining two -vertices in the corresponding gadgets and corresponding to and .
(a) Let . Then cops cannot capture the robber in . Consider the corresponding escape strategy. We will use it to show that the robber can escape from cops in , thus confirming that .
At the beginning of the game each cop will choose a vertex (in ) as their initial position. Each vertex is in a unit , which corresponds to a vertex in . The vertex will be called the shadow of the vertices in this copy of . The robber will assume that each cop is in the corresponding shadow vertex in and will play the escape strategy in . Initially, he has a vertex to pick in , and he will pick the vertex in . From now on, the robber will move based on the movements of (the shadows of) cops in . Whenever the shadows of cops have moved in , the robber has an escaping move in which can be translated into a series of at most 3 moves in . Since the shadow of the robber won’t be captured in the next move of cops in , the robber won’t get captured in in the next 3 moves. During these three moves of the robber, the cops will also make 3 moves, but the shadow of each cop in will either stay the same or move to an adjacent vertex in . Thus, the robber can interpret the change of the shadows as the moves of the cops in . It will be his turn to move, so he can continue using the escaping strategy for . Therefore the robber in can copy the strategy of the robber in and escape from cops in .
(b) In order to prove the stated bound we take . For this choice of , the degrees of vertices and in are all approximately equal to each other. The degree of each is equal to and the degree of any is at most
(c) Let . First note that if , then will have the desired properties itself. So we may assume that . If , then we are replacing each vertex with at most 7 vertices, thus , as desired.
For vertices of degree , we consider to be as in part (b). Then has vertices. It is easy to see that for , and a direct computation shows that the same bound holds for . Thus, . ∎
By repeatedly using Theorem 3, we obtain a new proof of the following result of Andreae [bounded_degree].
Corollary 4 (Andreae [bounded_degree]).
For any constant and there exist graphs of degree at most whose cop number is at least .
These results are also applicable to digraphs, see Section 4.
3 Reducing degree further
Using Theorem 3 repeatedly, we can decrease the degree of each vertex down to 3. To analyse the number of steps needed, let us assume that is a graph on vertices with and . We also let . Then we have:
to 0.8 — X[c] — X[c] — X[c] — X[c]— Graph & vertices & & cop number
& & &
& & &
By repeating the strategy to get from we will have:
to 0.8 — X[c] — X[c] — X[c] — X[c]— & & &
If , then in steps the maximum degree of the graph will become 3. In other words, by repeating the argument times, degrees of all vertices will be at most 3 and the number of vertices of the graph will be at most
This yields the following corollary.
Let be a graph on vertices with maximum degree , then there exist a subcubic graph on vertices such that .
Let be any function such that and let for . It has been proved in [random, random2] that as long as , we have a.a.s.,
In this graph the maximum degree is at most a.a.s. Applying the above approach we will get , a graph on vertices with . Therefore, by a change of variable it is easy to check that if is a (subcubic) graph on vertices then .
For every value of and large enough , there are subcubic graphs on vertices with cop number .
4 Digraphs of bounded degree
In this section we will use the same technique as above to find Eulerian digraphs of bounded degree and with arbitrarily large cop number. The gadget that we are going to get is different but will have the same properties.
Let the maximum out-degree of the digraph be and the maximum in-degree . Consider and . Now find the smallest such that and make a complete binary tree where are the leaves of the tree and direct all edges towards the root. Also find the smallest such that and make a complete binary tree whose leaves are and direct all the edges away from the root. Now merge the roots of these directed trees. Note that this gadget has less than vertices and the distance from any to any is fixed and equal to (for and ).
Now for any vertex where and are in and out-neighbors of , connect (with a directed edge) to () and similarly connect to (). See Figure 3. Note that we can delete the unnecessary vertices.
If we replace all vertices of a digraph with this gadget to get , it is easy to see that in and out-degree of vertices of is bounded by 2, number of vertices of is at most times the number of vertices of and by a similar lemma as Lemma 5 we have .
5 Connection to Meyniel’s Conjecture
We say that a family of graphs satisfies weak Meyniel’s Conjecture with exponent if for every graph in the family, where and the asymptotics is with respect to .
Meyniel’s Conjecture for subcubic graphs implies weak Meyniel’s Conjecture with exponent for the general case.
Assume that Meyniel’s conjecture is true for subcubic graphs. For a fixed , if has a vertex of degree at least , then we will put a cop on it to cover the vertex and its neighbourhood. So by using at most cops we can get a graph of maximum degree at most . Note that one of the components of this new graph is the territory of the robber. From this graph we can get , a subcubic graph on and by Meyniel’s conjecture (for subcubic graphs) . Therefore we have
Considering , we get . ∎
The same proof yields the following more general relationship.
Weak Meyniel’s conjecture for subcubic graphs with exponent implies weak Meyniel’s conjecture with exponent for the general case.