Subfield: Combinatorics


Question: This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

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Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.


The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.

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Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$.

Prove that $C(M) \leq n / 2$ for any map $M$ with the property that, for all hideouts $H_{1}$ and $H_{2}$, either all paths from $H_{1}$ to $H_{2}$ contain an odd number of edges, or all paths from $H_{1}$ to $H_{2}$ contain an even number of edges.

Solution:
The given condition actually implies that the graph is bipartite. Let $A_{1}$ be a hideout in $M$, and let $\mathcal{A}$ be the set of all hideouts $V$ such that all paths from $A_{1}$ to $V$ have an even number of edges, as well as $A_{1}$ itself; let $\mathcal{B}$ be the set of all other hideouts in $M$. Notice that there are no edges from $A_{1}$ to any other element $A_{i}$ in $\mathcal{A}$, because if there were, that edge would create a path from $A_{1}$ to $A_{i}$ with an odd number of edges (namely 1). So $A_{1}$ has edges only to hideouts in $\mathcal{B}$. In fact, there can be no edge from any hideout $A_{i}$ in $\mathcal{A}$ to any other hideout $A_{j}$ in $\mathcal{A}$, because if there were, there would be a path with an odd number of edges from $A_{1}$ to $A_{j}$ via $A_{i}$. Similarly, there can be no edge from any hideout $B_{i}$ in $\mathcal{B}$ to any other hideout $B_{j}$ in $\mathcal{B}$, because if there were such an edge, there would be a path from $A_{1}$ to $B_{j}$ with an even number of edges via $B_{i}$. Thus hideouts in $\mathcal{B}$ are connected only to hideouts in $\mathcal{A}$ and vice versa; hence $M$ is bipartite.