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.

![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836)

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 $D\left(\mathcal{C}_{6}, 3\right)=2$.

![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=339&width=399&top_left_y=1050&top_left_x=912)
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.

Determine $W(M)$ for each of the maps in figure a and figure b.
![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=350&width=355&top_left_y=1603&top_left_x=926)
figure a

![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=353&width=353&top_left_y=2095&top_left_x=932)
figure b

Solution:
or the map $M$ from figure a, $W(M)=7$. The most efficient strategy is to use 7 Cops to blanket all the hideouts on the first day. Any strategy using fewer than 7 Cops would require 6 Cops on each of two consecutive days: given that any hideout can be reached from any other hideout, leaving more than one hideout unsearched on one day makes it is impossible to eliminate any hideouts the following day. So any other strategy would require a minimum of 12 Cop workdays.

For the map $M$ from figure b, $W(M)=8$. The strategy outlined in $2 \mathrm{~b}$ used three Cops for a maximum of four workdays, yielding 12 Cop workdays. The most efficient strategy is to use 4 Cops, positioned at $\{B, E, H, K\}$ for 2 days each. The following argument demonstrates that 8 Cop workdays is in fact minimal. First, notice that there is no advantage to searching one of the hideouts between vertices of the square (for example, $C$ ) without searching the other hideout between the same vertices (for example, $D$ ). The Robber can reach $C$ on day $n$ if and only if he is at either $B$ or $E$ on day $n-1$, and in either case he could just as well go to $D$ instead of $C$. So there is no situation in which the Robber is certain to be caught at $C$ rather than at $D$. Additionally, the Robber's possible locations on day $n+1$ are the same whether he is at $C$ or $D$ on day $n$, so searching one rather than the other fails to rule out any locations for future days. So any successful strategy that involves searching $C$ should also involve searching $D$ on the same day, and similarly for $F$ and $G, I$ and $J$, and $L$ and $A$. On the other hand, if the Robber must be at one of $C$ and $D$ on day $n$, then he must be at either $B$ or $E$ on day $n+1$, because those are the only adjacent hideouts. So any strategy that involves searching both hideouts of one of the off-the-square pairs on day $n$ is equivalent to a strategy that searches the adjacent on-the-square hideouts on day $n+1$; the two strategies use the same number of Cops for the same number of workdays. Thus the optimal number of Cop workdays can be achieved using strategies that only search the "corner" hideouts $B, E, H, K$.

Restricting the search to only those strategies searching corner hideouts $B, E, H, K$, a total of 8 workdays can be achieved by searching all four hideouts on two consecutive days: if the Robber is at one of the other eight hideouts the first day, he must move to one of the two adjacent corner hideouts the second day. But each of these corner hideouts is adjacent to two other corner hideouts. So if only one hideout is searched, for no matter how many consecutive days, the following day, the Robber could either be back at the previously-searched hideout or be at any other hideout: no possibilities are ruled out. If two adjacent corners are searched, the Cops do no better, as the following argument shows. Suppose that $B$ and $E$ are both searched for two consecutive days. Then the Cops can rule out $B, E, C$, and $D$ as possible locations, but if the Cops then switch to searching either $H$ or $K$ instead of $B$ or $E$, the Robber can go back to $C$ or $D$ within two days. So searching two adjacent corner hideouts for two days is fruitless and costs four Cop workdays. Searching diagonally opposite corner hideouts is even less fruitful, because doing so rules out none of the other hideouts as possible Robber locations. Using three Cops each day, it is easy to imagine scenarios in which the Robber evades capture for three days before being caught: for example, if $B, E, H$ are searched for two consecutive days, the Robber goes from $I$ or $J$ to $K$ to $L$. Therefore if three Cops are used, four days are required for a total of 12 Cop workdays.

If there are more than four Cops, the preceding arguments show that the number of Cops must be even to produce optimal results (because there is no advantage to searching one hideout between vertices of the square without searching the other). Using six Cops with four at corner hideouts yields no improvement, because the following day, the Robber could get to any of the four corner hideouts, requiring at least four Cops the second day, for ten Cop workdays. If two or fewer Cops are at corner hideouts, the situation is even worse, because if the Robber is not caught that day, he has at least nine possible hideouts the following day (depending on whether the unsearched corners are adjacent or diagonally opposite to each other). Using eight Cops (with four at corner vertices) could eliminate one corner vertex as a possible location for the second day (if the non-corner hideouts searched are on adjacent sides of the square), but eight Cop workdays have already been used on the first day. So 8 Cop workdays is minimal.

Final answer: 12, 8