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)
A path on $n$ hideouts is a map with $n$ hideouts, connected in one long string. (More formally, a map is a path if and only if two hideouts are adjacent to exactly one hideout each and all other hideouts are adjacent to exactly two hideouts each.) It is denoted by $\mathcal{P}_{n}$. The maps $\mathcal{P}_{3}$ through $\mathcal{P}_{6}$ are shown below.

![](https://cdn.mathpix.com/cropped/2023_12_21_a4ab1664c3b96ebcfbd1g-1.jpg?height=174&width=1548&top_left_y=951&top_left_x=337)

Show that $D\left(\mathcal{P}_{n}, 1\right) \leq 2 n$ for $n \geq 3$.

Solution:
The following argument shows that $C\left(\mathcal{P}_{n}\right)=1$, and that capture occurs in at most $2 n-4$ days. It helps to draw the hideout map as in the following diagram, so that odd-numbered hideouts are all on one level and even-numbered hideouts are all on another level; the case where $n$ is odd is shown below. (A similar argument applies where $n$ is even.)

![](https://cdn.mathpix.com/cropped/2023_12_21_fbf8df629daaa704cef9g-1.jpg?height=203&width=631&top_left_y=2240&top_left_x=801)



Each night, the Robber has the choice of moving diagonally left or diagonally right on this map, but he is required to move from top to bottom or vice-versa.

For the first $n-2$ days, the Cop should search hideouts $A_{2}, A_{3}, \ldots, A_{n-1}$, in that order. For the next $n-2$ days, the Cop should search hideouts $A_{n-1}, A_{n-2}, \ldots, A_{2}$, in that order.

The Cop always moves from top to bottom, or vice-versa, except that he searches $A_{n-1}$ two days in a row. If the Cop is lucky, the Robber chose an even-numbered hideout on the first day. In this case, the Cop and the Robber are always on the same level (top or bottom) for the first half ( $n-2$ days) of the search. The Cop is searching from left to right, and the Robber started out to the right of the Cop (or at $A_{2}$, the first hideout searched), so eventually the Robber is caught (at $A_{n-1}$, if not earlier).

If the Robber chose an odd-numbered hideout on the first day, then the Cop and Robber will be on different levels for the first half of the search, but on the same level for the second half. For the second half of the search, the Robber is to the left of the Cop (or possibly at $A_{n-1}$, which would happen if the Robber moved unwisely or if he spent day $n-2$ at hideout $A_{n}$ ). But now the Cop is searching from right to left, so again, the Cop will eventually catch the Robber.

The same strategy can be justified without using the zig-zag diagram above. Suppose that the Robber is at hideout $A_{R}$ on a given day, and the Cop searches hideout $A_{C}$. Whenever the Cop moves to the next hideout or the preceding hideout, $C$ changes by \pm 1 , and the Robber's constraint forces $R$ to change by \pm 1 . Thus if the Cop uses the strategy above, on each of the first $n-2$ days, either the difference $R-C$ stays the same or it decreases by 2 . If on the first day $R-C$ is even, either $R-C$ is 0 (the Robber was at $A_{2}$ and is caught immediately) or is positive (because $C=2$ and $R \geq 4$ ). Because the difference $R-C$ is even and decreases (if at all) by 2 each day, it cannot go from positive to negative without being zero. If on the first day $R-C$ is odd, then the Robber avoids capture through day $n-2$ (because $R-C$ is still odd), but then on day $n-1, R$ changes (by \pm 1 ) while $C$ does not. So $R-C$ is now even (and is either 0 or negative), and henceforth either remains the same or increases by 2 each day, so again, $R-C$ must be zero at some point between day $n-1$ and day $2 n-4$, inclusive.

Final answer: 证明题，略