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: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays.

Find an example of a map $M$ with 4 or more hideouts such that $W(M)=3$, or prove that no such map exists.

Solution:
We denote a map as star, that is, the map $\mathcal{S}_{n}$ with one central hideout connected to $n-1$ outer hideouts, none of which is connected to any other hideout.

No such map exists. Let $M$ be a map with at least four hideouts. The proof below shows that either $W(M)>3$ or else $M$ is a star, in which case $W(M)=2$.

First, suppose that there are four distinct hideouts $A_{1}, A_{2}, B_{1}$, and $B_{2}$ such that $A_{1}$ and $A_{2}$ are adjacent, as are $B_{1}$ and $B_{2}$. If the Cops make fewer than four searches, then they cannot search $\left\{A_{1}, A_{2}\right\}$ twice and also search $\left\{B_{1}, B_{2}\right\}$ twice. Without loss of generality, assume the Cops search $\left\{A_{1}, A_{2}\right\}$ at most once. Then the Robber can evade capture by moving from $A_{1}$ to $A_{2}$ and back again, provided that he is lucky enough to start off at the right one. Thus



$W(M)>3$ in this case. For the second case, assume that whenever $A_{1}$ is adjacent to $A_{2}$ and $B_{1}$ is adjacent to $B_{2}$, the four hideouts are not distinct. Start with two adjacent hideouts, and call them $A_{1}$ and $A_{2}$. Consider any hideout, say $B$, that is not one of these two. Then $B$ must be adjacent to some hideout, say $C$. By assumption, $A_{1}, A_{2}, B$, and $C$ are not distinct, so $C=A_{1}$ or $C=A_{2}$. That is, every hideout $B$ is adjacent to $A_{1}$ or to $A_{2}$.

If there are hideouts $B_{1}$ and $B_{2}$, distinct from $A_{1}$ and $A_{2}$ and from each other, such that $B_{1}$ is adjacent to $A_{1}$ and $B_{2}$ is adjacent to $A_{2}$, then it clearly violates the assumption of this case. That is, either every hideout $B$, distinct from $A_{1}$ and $A_{2}$, is adjacent to $A_{1}$, or every such hideout is adjacent to $A_{2}$. Without loss of generality, assume the former.

Because every hideout is adjacent to $A_{1}$ (except for $A_{1}$ itself), the foregoing proves that the map $M$ is a star. It remains to show that there are no "extra" edges in the map. For the sake of contradiction, suppose that $B_{1}$ and $B_{2}$ are adjacent hideouts, both distinct from $A_{1}$. Because the map has at least four hideouts, choose one distinct from these three and call it $A_{2}$. Then these four hideouts violate the assumption of this case.