 # The localization capture time of a graph

The localization game is a pursuit-evasion game analogous to Cops and Robbers, where the robber is invisible and the cops send distance probes in an attempt to identify the location of the robber. We present a novel graph parameter called the capture time, which measures how long the localization game lasts assuming optimal play. We conjecture that the capture time is linear in the order of the graph, and show that the conjecture holds for graph families such as trees and interval graphs. We study bounds on the capture time for trees and its monotone property on induced subgraphs of trees and more general graphs. We give upper bounds for the capture time on the incidence graphs of projective planes. We finish with new bounds on the localization number and capture time using treewidth.

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## 1. Introduction

Pursuit-evasion games

are combinatorial models for the detection or neutralization of an adversary’s activity on a graph. In such models, agents or cops are attempting to capture an adversary or robber loose on the vertices of a graph. The players move at alternating ticks of the clock, and have restrictions on their movements or relative speed depending on the game played. The most studied such game is Cops and Robbers, where the cops and robber can only move to vertices with which they share an edge. The cop number is the minimum number of cops needed to guarantee the robber’s capture. How the players move and the rules of capture depend on which variant is studied. These variants are motivated by problems in practice or inspired by foundational issues in computer science, discrete mathematics, and artificial intelligence, such as robotics and network security. For surveys of pursuit-evasion games, see

[9, 10], and see  for more background on Cops and Robbers.

The localization game was first introduced for one cop by Seager [27, 28] and was further studied in [5, 6, 7, 11, 15, 18, 19, 20, 22]. In the localization game, two players play on a connected graph, with one player controlling a set of cops, where is a positive integer, and the second controlling a single robber. The robber is invisible to the cops during gameplay. When there is no ambiguity, we identify a player with the vertex it occupies. The game is played over a sequence of discrete time-steps; a round is a cop move and a subsequent robber move.

The robber occupies a vertex of the graph, and when the robber is ready to move during a round, they may move to a neighboring vertex or remain on their current vertex. A move for the cops is a placement of cops on a set of vertices. Note that the cops are not limited to moving to neighboring vertices. At the beginning of the game, the robber chooses a starting vertex. After this, the cops move first, followed by the robber; thereafter, they move in alternate turns. Observe that any subset of cops may move in a given round. In each round, the cops occupy a set of vertices and each cop sends out a cop probe, which gives their distance , from to the robber, where . Hence, in each round, the cops determine a

distance vector

of cop probes. The cops win if they have a strategy to determine, after finitely many rounds, the vertex the robber occupies, at which time we say that the cops capture the robber. The robber wins if they are never captured.

For a connected graph , define the localization number of , written , to be the least integer for which cops have a winning strategy over any possible strategy of the robber; that is, we consider the worst case for the cops in that the robber is omniscient and so knows the entire strategy of the cops. As is at most the parameter is well-defined. Note that where is the metric dimension of

In , it was shown that is bounded above by the pathwidth of and that the localization number is unbounded even on planar graphs obtained by adding a universal vertex to a tree. They also proved that computing is NP-hard for graphs with diameter . Bonato and Kinnersley  studied the localization number for graphs based on their degeneracy. In  , they resolved a conjecture from  relating and the chromatic number; further, they proved that the localization number of outerplanar graphs is at most 2, and they proved an asymptotically tight upper bound on the localization number of the hypercube. The localization number of the incidence graphs of designs was studied in . In particular, they gave exact values for the localization number of the incidence graphs of projective and affine planes, and bounds for the incidence graphs of Steiner systems and transversal designs. The localization number of graph products was considered in , and in diameter 2 graphs such as Kneser, Moore, and polarity graphs in . Localization was studied in random binomial graphs in [20, 19] and in random geometric graphs in .

In the present paper, we focus not only on the number of cops needed to capture the robber in the localization game, but the time or minimum number of rounds it takes to do so. For a graph and an integer the corresponding optimization parameter is which is the minimum number of rounds for the cops to capture the robber. If then we simplify this to . We refer to this graph parameter as the capture time of for the localization game. Note that we assume here that the cops minimize the number of rounds needed for capture, while the robber maximizes the number of rounds for capture. For example, if is an integer, then and .

There is an analogous temporal parameter defined for the cop number, also referred to as capture time, first introduced in . Since we focus almost exclusively on the localization number in this paper, there will be no confusion between the parameters. Note that as increases, monotonically decreases. In analogy with , we refer to this in temporal speed-up and the number of cops as overprescribed. If then

Capture time was implicitly defined in the first paper of Seager , where only one cop played. The parameter there was called the location number, written which is defined for with as the number of probes (that is, rounds) needed to capture the robber. The location number was studied in several subsequent publications, such as [15, 18, 22, 28]. The present paper is the first place where the capture time is explicitly introduced in the general setting, where there may be more than one cop and possibly more cops than the localization number.

The capture time of a graph is challenging to calculate exactly for general graphs, and as such, we present bounds for general graph families, and a few exact values along the way. We present a conjecture on the capture time in Section 2 claiming that the capture time is linear in the order of the graph. We show the conjecture holds for graph families such as trees and interval graphs. We discuss the monotone property for the capture time on induced subgraphs. In particular, we show that capture time is monotone on induced subgraphs in a general family of graphs including trees. The capture time of trees is considered in Section 3, and we derive upper and lower bounds in Theorems 3.9 and 3.10, respectively. We derive nearly tight values for the capture time on perfect -ary trees in the overprescribed setting. In Section 4, we consider the capture time of the incidence graphs of projective planes of order In , it was shown that such graphs have localization number In Theorem 4.2, we show that if is the incidence graph of a projective plane of order , then for

We present novel bounds on the localization number and capture time in terms of the treewidth of a graph. The final section includes several open problems.

Throughout, all graphs considered are simple, undirected, connected, and finite. For a general reference for graph theory, see . The closed neighborhood of , written consists of a vertex along with neighbors of . The distance between vertices and is denoted by The maximum degree of a graph is denoted by For a graph we denote the subgraph induced by by . A leaf is a vertex of degree 1.

## 2. Well-localizable graphs and the monotone property

A basic question is how large the capture time may be as a function of the order of the graph. One motivation for this question comes from graph searching, where a set of searchers attempts to clear edges of contamination by prescribed rules; for more background on graph searching, see  and the recent survey . In graph searching, it was shown in [2, 24] that for a graph of order there exist winning strategies that require at most rounds. In particular, searchers never need to reclear edges. In contrast, the capture time for the cop number may be superlinear in the number of vertices. For graphs with cop number , it was proved in  that the capture time is . Recently, the bound was proven to be asymptotically tight in [16, 23]. Note that for graphs of order with cop number 1, the capture time is see .

In the setting of the localization game, we do not know of graphs where the is superlinear in the order of . It may be the case that the capture time is at most , where is order of the graph, but we propose the following more modest conjecture.

Localization Capture Time Conjecture (LCTC): If is a graph of order then

 captζ(G)=O(n).

We say that a graph with order is well-localizable if it satisfies The LCTC may be rephrased as saying that all graphs are well-localizable.

One important family where LCTC holds is for trees.

###### Theorem 2.1.

Trees are well-localizable.

###### Proof.

Let be a tree of order . We then have that by . In the case from the cop strategy presented in the proof of Theorem 8 in , the robber is captured in at most rounds, so the LCTC holds.

In the case when consider the following strategy from . Place one cop on a fixed vertex and let the second cop probe each neighbor of looking for a subtree of where the distance to the robber is smallest. Once is discovered, we move to the unique vertex of adjacent to and repeat this procedure on We recursively search in this way, and in at most rounds, identify the robber’s location. Hence, the LCTC holds in this case. ∎

Another family where the LCTC holds is for interval graphs, which are the intersection graphs of intervals on the real line. Interval graphs are precisely those that are chordal and asteroidal-triple-free; for more background on this graph family, see for example, .

To prove that interval graphs are well-localizable, we first need the notions of treewidth and pathwidth. Given a graph , a tree decomposition is a pair , where is a family of subsets of called bags, and is a tree whose vertices are the subsets , satisfying the following three properties.

1. That is, each graph vertex is associated with at least one tree vertex.

2. For every edge in the graph, there is a subset that contains both and .

3. If , and are vertices, and is on the path from to then .

The width of a tree decomposition is the cardinality of its largest set minus one. The treewidth of a graph written is the minimum width among all possible tree decompositions of . We refer to a tree decomposition with width equaling the treewidth as optimal. If we restrict to be a path, then the resulting parameter is called the pathwidth of , written Note that

We have the following.

###### Theorem 2.2.

If satisfies then is well-localizable. In particular, interval graphs are well-localizable.

###### Proof.

As was shown in , for all graphs , we have that In the proof of this bound, the bags are linearly ordered as where The cops begin by occupying every vertex except one of which ensures the robber will not start in , or they are captured in one round. The cops then move to occupying all but one vertex, and then and so on. In this fashion, the robber can never enter a bag presently or one previously occupied by cops. The robber will be eventually be captured in (given that they are omniscient and will maximize the length of the game).

The cop strategy described in the previous paragraph takes at most rounds to capture the robber. Hence, if then

Interval graphs satisfy and so the final statement of the theorem holds. ∎

We next prove that the LCTC holds for complete -partite graphs. We let denote the chromatic number of

###### Theorem 2.3.

For a positive integer, let be a complete -partite graph of order with parts , for , such that for . Let be the number of parts of cardinality . If then the following statements hold.

1. If , then we have that

 ζ(G)=n−χ(G)−|Xk|+ρ+1 and captζ(G)≤|Xk|−1;
2. If , then we have that

 ζ(G)=n−χ(G)−|Xk|+2 and captζ(G)≤|Xk|−1.

In particular,

###### Proof.

Let if , and if . If there are fewer than cops, then this implies that either: (i) there is a part not containing the robber that will contain at least two cop-free vertices on the next cop turn; or (ii) there will be two parts of cardinality one that are cop-free on the next turn (although this case does not happen if ).

In (i), the robber may move to one of the two vertices of that will be cop-free, and since the cops cannot distinguish these two vertices on the same part, the robber avoids capture. In (ii), the robber may move to either of the two parts of cardinality one that are cop-free on the next turn, as the cops will not be able to distinguish these two. Therefore, we have that .

To show that , we play with cops and show that the robber can be captured. We place cops on each independent set of cardinality if . There are cops left to place. We place cops on each non-singleton set where . This consists of cops, so there is one cop left to place. Each cop placed so far remains stationary for the rest of play. Finally, we place the last remaining cop on a vertex in , which will move during play.

If the robber moves to one of the vertices that contain a cop on each turn, then the robber is captured. If the robber moves to the only part of cardinality that does not contain a cop, then all cops probe a distance of , which results in a unique candidate and the robber is captured. If the robber moves to a cop-free vertex of a non-singleton part , where , then all cops on that part probe a distance of , and all other cops probe a distance of , which results in a unique candidate and so the robber is captured.

Therefore, the robber will be captured if they ever move from the part . As is an independent set, the robber cannot move between vertices of . On each successive turn, the cop player plays the unique moving cop on a new vertex of . After at most rounds, the cop player will either occupy the vertex of the robber, or the robber will be known to be on the unique vertex of that the cop has not yet visited. ∎

### 2.1. The monotone property on induced subgraphs

When studying capture time, it is useful to know how induced subgraphs affect it. We consider whether the capture time is monotone on induced subgraphs: that is, if a graph has an induced subgraph , then . We show that capture time is monotone on induced subgraphs for a general family of graphs that includes all trees. The latter fact will be useful in the next section.

In general, capture time may fail to be monotone on induced subgraphs. We define as in Figure 1.

By  we know that . Furthermore, it is straightforward to check that . The graph is an induced subgraph of the Cartesian grid . Label the vertices of by Placing cops on and results in a capture in one round. Hence, and

We consider certain subgraphs where the capture time is monotone on induced subgraphs in a strong sense, where it holds for all greater than the localization number. We say that an induced subgraph of is special if is connected and the only paths in between distinct vertices in are entirely contained in Note that we assume here that paths do not repeat vertices or edges.

###### Theorem 2.4.

Let be an integer. If is a connected graph and is a special induced subgraph of then

 captζ,k(H)≤captζ,k(G).

Note that each -connected component (or block) of is a special induced subgraph. Hence, by Theorem 2.4, is bounded below by the maximum capture time of any -connected component.

Before we begin the proof of Theorem 2.4, we recall some notation. An induced subgraph is a retract of if there is a homomorphism such that for that is, is the identity on The map is called a retraction.

###### Proof of Theorem 2.4..

If we consider the graph formed by removing the edges of from , then the graph is composed of a number of components, say , each of which contains exactly one vertex in . We note that any cannot contain two vertices in , or else a path exists between the two vertices that is not entirely contained within . Further, there cannot be a with no vertices in , or else must not have been connected. We let denote the unique vertex of that is contained in . See Figure 2.

For each and , any path connecting to must contain . Further, a minimal path between and must be composed of a minimal path from to and a minimal path from to . Hence,

 d(u,v)=d(u,vi)+d(vi,v).

For , those vertices in are exactly those vertices in , where denotes the set of vertices of distance from . Define the function by if and by if . Note that this mapping is a retraction onto if we consider to be reflexive (that is, each vertex has a loop). Making the assumption that is reflexive has no impact on distances in the graph, and so we assume it without loss of generality for the remainder of the proof.

Suppose the robber restricts to playing in , the cop knows the robber is in and the cops may probe any vertex in In this variant of the localization game played on let be the resulting capture time. Note that We show that , which will be enough to complete the proof.

We employ a shadow strategy, where the primary game is in , and the secondary game is in . The images of a cop onto are referred to as shadow cops and written Suppose that in round , after the cop has moved, the cops knew the robber was on the vertex set . At time , the cops play on vertices in the primary game and as a result, the cop receives a distance vector . The cop finds that the robber is on as a result of its probe, since it is known that the robber is on . Therefore, the cop exactly knows that the robber is on the vertex set

 Vt=(N[Vt−1]∩V(H))∩m⋂i=1(Ndi(Ci)∩V(H)),

where is any vertex in or a neighbor of a vertex in . We previously showed in that , so therefore, we may rewrite this as

 Vt=(N[Vt−1]∩V(H))∩m⋂i=1(Ndi−d(Ci,f(Ci))(f(Ci))∩V(H)).

Suppose for the sake of contradiction that on the primary game, the cop can capture the robber in rounds. Suppose in round , the cops probe vertices . The robber translates this to the secondary game by playing shadow cops on vertices in , yielding a distance vector . In round , the shadow cops know that the robber is on the set . The shadow cop finds that the robber is on as a result of its probe. Therefore, the shadow cop exactly knows that the robber is on the vertex set

 V′t=(N[V′t−1])∩m⋂i=1(Nd′i(f(Ci)))=(N[V′t−1]∩V(H))∩m⋂i=1(Nd′i(f(Ci))∩V(H)).

By noting that , we have that . Consequentially, as only when , we have that the robber is captured in the secondary game in . This is a contradiction, as we require at least rounds to capture the robber with cops on . Therefore, in the primary game, the cop requires at least rounds to capture the robber. We then have that , as required. ∎

The connected induced subgraphs of trees are always special, and so Theorem 2.4 gives the following.

###### Corollary 2.5.

Capture time is monotone on trees. In particular, if is a tree with subtree then for all integers ,

Note that for larger than the localization number, we may find graphs where the capture time when playing with cops is 1, but is unbounded for induced subgraphs. Consider the hypercube of dimension , written . As proved in , If we label the vertices of by binary -tuples, we can place a cop on each of the vertices whose label contains at most one to see that . However, is bipartite and so has an independent set of order . If , then

## 3. Trees

Trees have localization number either 1 or 2, as proved in . As we proved in Theorem 1, trees are well-localizable. We refine bounds on the capture time of trees in the overprescribed setting (that is, where the number of cops is more than the localization number), presenting bounds that are functions of the maximum degree.

When studying the capture time of trees, leaves play a special role.

###### Lemma 3.1.

For an integer , if is a tree with leaves, then we have that .

###### Proof.

The cops play on the leaves. Each vertex on a tree is contained in a shortest path connecting two leaves. This implies that the robber, if not on a leaf and captured immediately, is on the shortest path connecting two cops, say and . Suppose that cop has distance to the robber, for . These two cops can identify that the robber is on the shortest path connecting and as . The robber is identified to be on the unique vertex on the shortest path connecting and that is distance from . ∎

For the proof of the next lemma, and other upcoming results, we need the following definition. Let be a rooted tree with root . We call a vertex a descendant of vertex in if is a neighbor of , and sits on the unique path between and the root .

When there are two or more cops we may assume without loss of generality that they play exclusively on the leaves.

###### Lemma 3.2.

If there are at least two cops playing the localization game on a tree , then it never hinders the cops to be on leaves.

###### Proof.

Suppose that in some round, there is a cop probing a vertex that is not a leaf of . Let the other cops probe vertices respectively. Say that the robber is on vertex .

Root the tree at , and let be a leaf that is a descendant of . Suppose is instead placed on (and all other cops remain where they were). Let be the shortest path between and and let be the shortest path between and . Observe that . Between them, cops and can determine the vertex on that is closest to the robber. To see this, note that

 d(R,x)=(d(u′1,R)+d(u2,R)−d(u′1,u2))/2.

From this, it follows that

 d(u′1,x)=d(u′1,R)−d(R,x)=(d(u′1,R)−d(u2,R)+d(u′1,u2))/2

and this, together with knowing it lies on , is enough to determine .

Supposing probes and not , we show that the cops know . Whether or , we have . Since the cops know and , the cops can still determine . Thus, playing cop at leaf is at least as good as playing the cop at . We may repeat this argument until all cops are on leaves. ∎

Throughout the remainder of the section, we let and be positive integers. The perfect -ary tree of height has levels labeled . The th level contains one vertex, the root, which has neighbors. For , every vertex in the th level has exactly one neighbor in level and neighbors in level . In particular, the th level contains vertices all at distance from the root. These are naturally arising structures; for example, there is a natural bijection between the vertices of and the set of sequences of length at most with entries in , where two vertices are adjacent if one of the corresponding sequences is equal to the other sequence with one entry appended at the end.

Note that the perfect -ary tree has leaves. Also, for any the induced subgraph on all vertices at distance or more from the root consists of copies of .

We will prove upper and lower bounds on the capture time for perfect -ary trees. Our analysis splits into two cases, depending on whether the number of cops is less than or greater than .

We begin with a result that is effective when the number of cops is less than .

###### Lemma 3.3.

For an integer, we have that

 captζ,k(Thm)≥h⌊m−1k⌋.
###### Proof.

We will show that the robber can avoid capture if less than rounds are played. We assume that the robber picks a leaf and remains on that leaf for the remaining rounds, and the cops know that the robber is employing this strategy. The robber has other options but if the cops cannot capture the robber in fewer rounds in this case, they will not be able to do so when the robber has more freedom.

Let denote the rooted subtrees formed by deleting the root vertex , with the root of each such subtree being the vertex that was adjacent to . Each cop probes at most one of these subtrees in the first round. During the first rounds, at most of these subtrees was visited by a cop. The robber may therefore have chosen to start on a leaf of one of the subtrees, say , that was not visited by a cop in these first rounds. There may have been multiple subtrees that were not visited by the cops, so suppose that the robber tells the cop that they are on after these rounds. This only serves to reduce the capture time.

Note that the distance from a cop on vertex to any of the leaf vertices of an unplayed subtree is , so the cops cannot distinguish the leaves of an unplayed subtree. We may now proceed inductively, since the unplayed subtree is a copy of . We repeat the argument over iterations, each taking rounds, yielding a capture time of at least . ∎

See Figure 3 for an example of Lemma 3.3. We derive an upper bound that differs from the lower bound by an additive factor of at most .

###### Lemma 3.4.

For an integer we have that

 captζ,k(Thm)≤h⌈m−1k⌉.
###### Proof.

We modify the proof of Lemma 3.3 to instead show an upper bound. Let denote the rooted subtrees formed by deleting the root vertex , with the root of each such subtree being the vertex that was connected to .

In round with the cops play on . Note that if the robber ever plays on the root they are captured immediately, as the root is the unique vertex at distance from all the cops (this requires at least two cops). Thus, the robber must remain on the same for all these rounds. We may assume without loss of generality that if for some and , then in round the cops can determine that the robber was in , and that the cops could not know this before round . This follows since for all and for all , where is the vertex occupied by the robber that lies on the subtree rooted at . Hence, will be the unique vertex with a shorter distance to the robber than the others. If is not of this form, then and after rounds we deduce the robber is in as it was not in any .

After at most rounds the cops know which subtree the robber is on and the robber can never move to the root of the tree without being captured. Now we move to playing on and apply induction. Since the robber can never move to the root , the robber can never escape the subtree . This induction lasts for rounds. ∎

We therefore have the following.

###### Corollary 3.5.

For all integers we have that

 h⌊m−1k⌋≤captζ,k(Thm)≤h⌈m−1k⌉.

We note that this corollary is most useful when the number of cops is less than . If , then these bounds only show and we can find much improved bounds.

We next present a result that is effective in the case when the number of cops is greater than or equal .

###### Lemma 3.6.

For an integer , we have that

 captζ,k(Thm)≥h1+⌊logmk⌋.
###### Proof.

Since we are only proving a lower bound, we may assume that the robber picks a leaf and remains on that leaf throughout the game and the cops know that the robber is employing this strategy; the robber has other options but if the cops cannot capture the robber in fewer rounds in this case, they will not be able to do so when the robber has more freedom.

Let so that . We claim that in round there is some subtree isomorphic to such that the set of vertices where the robber could be is exactly the leaves of . In round , the initial robber placement could be on any leaf of so clearly this holds in round . We proceed by induction on .

Suppose that the claim holds in round for the subtree isomorphic to . Consider the induced subgraph on all vertices of at distance or more from the root of . We know this consists of copies of . The cops pick vertices, so there is one of these subtrees isomorphic to with no cops on it. Call this and suppose that the robber chose a leaf of this subtree; this is possible since the robber knew in advance which vertices the cops would pick. For any vertex outside the distance from to a leaf of is the same for every leaf of , so the cops cannot distinguish between the leaves of . In particular, the set of vertices the robber could be on contains all leaves of , and the claim holds in round .

Now, it is evident that the cops cannot capture the robber in rounds for . Hence, , and the tree has at least leaves, and so there are at least vertices that the robber could be on in round . ∎

We can also find an analogous upper bound.

###### Lemma 3.7.

If , then we have that

 captζ,k(Thm)≤h⌊logmk⌋.
###### Proof.

Let and note that . It is enough to show that the theorem holds with cops owing to temporal speed-up.

We will prove the lemma by induction on . When , by Lemma 3.1 the cops can win in one round by playing on all of the leaves of , and so we are done.

Suppose . In the first round, the cops play on the vertices at distance from the root. Call this set . If the robber was on a vertex at distance from the root, then the maximum distance probed from is exactly , so the cops can determine .

If , then the vertices in that are descendants of are at distance from and all other leaves are at distance from . Thus, given the set of distances, we can determine all leaves that are descendants of the vertex the robber is on and taking their unique common ancestor gives .

Otherwise, and there is a unique vertex in with the minimum distance to . The cops know that the robber is on the subtree rooted at so on future rounds play passes to this subtree. This subtree is isomorphic to and so by induction the cops can find the robber in a further rounds. Thus, the cops can capture the robber in rounds. ∎

We therefore have the following.

###### Corollary 3.8.

For

 h1+⌊logm(k)⌋≤captζ,k(Thm)≤h⌊logm(k)⌋

The upper and lower bounds differ by a multiplicative factor of , so if is large compared to this difference is relatively small.

Every tree of maximum degree and radius is a subtree of . By Corollary 2.5 and Lemmas 3.7 and 3.4, we derive the following.

###### Theorem 3.9.

For a tree , we have that:

As for a lower bound, we know that a tree of maximum degree contains a star , so by the monotone property of capture time on trees we have the following.

###### Theorem 3.10.

For a tree , if then we have that

 captζ,k(T)≥⌈Δ(T)−1k⌉.

## 4. Incidence Graphs of Projective Planes

A projective plane consists of a set of points and lines satisfying the following axioms.

1. There is exactly one line incident with every pair of distinct points.

2. There is exactly one point incident with every pair of distinct lines.

3. There are four points such that no line is incident with more than two of them.

We only consider finite projective planes. It can be shown that projective planes have many points and lines for an integer , each point is on -many lines, and each line contains -many points. We let be the set of points and be the set of lines. The only orders where projective planes are known to exist are prime powers; indeed, this is a deep conjecture in finite geometry. For these items and further background on projective planes, see .

For a given projective plane , define the bipartite graph with red vertices the points of , and the blue vertices represent the lines. Vertices of different colors are adjacent if they are incident. We call this the incidence graph of . See Figure 5 for the incidence graph of the Fano plane, the projective plane of order . Figure 5. The Fano plane and its incidence graph, the Heawood graph. Lines are represented by triples.

As was proved in , the incidence graph of a projective plane of order satisfies We present an upper bound on the capture time of these graphs in the present section.

The following is a warm-up for larger values of

###### Lemma 4.1.

The capture time of the Heawood graph is .

###### Proof.

We identify the Heawood graph as the incidence graph of the Fano plane with and . For the first round, the cops probe vertices 1, 4 and 6 on Each vertex in is uniquely identifiable so the robber must reside on Next, move the cops to 2, 3, and 5. All of the vertices of are uniquely identifiable so the robber could not have moved over there. If the robber was on 1,2,3,4,5 or 6, then they were immediately identified when the vertex was probed. Otherwise, the robber is on 7, the unique vertex that was not ever visited by cops and is of distance 2 from all the cops. Observe that the first round reduced the possible location of the robber in to exclude while the second step reduced to exclude leaving only as a possibility to probe distance without previously being identified.

We show that the robber cannot be captured in one round. We identify the ways in which the cops can be distributed among the vertices. There are four cases to consider: (1) all three are on , (2) all three are on , (3) two are on and one is on , or (4) one is on and two are on . As the design is symmetric, checking cases (1) and (3) suffice. We leave the remaining details to the reader. ∎

The following bounds the capture time of the incidence graphs of projective planes of order.

###### Theorem 4.2.

Let be the incidence graph of a projective plane of order . For an integer, we have that

In the case Theorem 4.2 gives that for the incidence graph of a projective plane of order

 captζ(G)≤q−1+⌈q2⌉.

Hence, the incidence graphs of projective planes are well-localizable and so satisfy the LCTC.

###### Proof.

As we wish to show an upper bound on the capture time, we simply need a cop strategy where cops can capture the robber in rounds. We do this over two phases. In the first phase, the cops take rounds to find that the robber is on a set of vertices for any vertex in . In the second phase, the cops take additional rounds to capture the robber. We bound both and .

We start by providing the strategy for the first phase. Let and . The cops play on the vertices in , and without loss of generality, play on of the vertices in , labeled as . Note that we may make this assumption since capture time monotonically decreases with , and for larger the ceilings give the same upper bound.

If the robber is on , then there is a unique path of length two from to the robber, say . If there is a cop on , then this cop will probe a distance of , so the cops know that the robber is on , and the first phase would end. If there is not a cop on , then , and each cop in will probe a distance of , which can only happen if the robber is on a neighbor of , and the first phase would end. The cops on will all probe a distance of if and only if the robber is on , so the first phase ends if the robber is on .

We may therefore assume the robber is on a vertex in . If the first phase does not end on the first round, then the cops know that the robber is not on or on a neighbor of a vertex in . This means that the robber must reside on the vertex or on a neighbor of a vertex in . The cops on each probe in the case the robber is on , so the first phase would immediately be over if this were the case.

On the th round, the cops play on and , where is a set of vertices chosen from . By similar reasoning to the above, the first phase ends unless the robber is on a neighbor of . After rounds, either some cop probes a distance of and so the first phase is over, or there is one vertex in , and the robber is known by the cops to be residing on the neighborhood of this vertex. Therefore, .

We now give the strategy for the second phase. Suppose that the robber was found to reside on , for some vertex , where .

We first suppose that . Let be a vertex on the same part as . The cops play vertices on , and vertices on , which we label as . If some cop on probes a distance of , then the robber is captured. If all cops on probe a distance of , then the robber is on the unique vertex of that does not contain a cop. Therefore, the robber cannot stay on a vertex of without being captured, so we assume the robber moves. Suppose the robber moved from vertex to a neighboring vertex . If the robber moved to , then they would be immediately captured by the cops. The cop knows that the robber is on . Each cop on has exactly one neighbor on , and no two cops on share the same neighbor on . The robber is captured if they are on the neighbor of a cop on . We may therefore assume that the robber is on one of the other vertices of , which implies that the cops have found that the robber is residing on , for some vertex , where .

Now if , the cops instead play vertices on , and vertices on , which we label as . If some cop on probes a distance of , then the robber if found. Otherwise the robber has moved during its last turn. If both cops on probe a distance of , then the robber is on . Otherwise, one of the cops on , say on vertex , probes a distance of , and the cops know the robber is on . Each of the cops in is adjacent to one vertex of , and vice versa. Thus, if one of the cops in probes a distance of , the cops know the exact location of the robber. If all of the cops in probe a distance of , then the robber must be on the unique vertex of without a neighbor in .

We repeat the argument above inductively for rounds, in which case there can be at most one vertex that the robber could be residing on, and the robber is captured.∎

## 5. Bounds using treewidth

We consider new bounds on the localization number and capture time using treewidth. Define the tree radius of a graph , written , as the minimum radius of all tree decompositions of that have width . We have the following result.

###### Theorem 5.1.

For a graph , we have that

 ζ(G)≤(tw(G)+1)tr(G).

If an optimal tree decomposition has leaves, then we have that

 captζ,(tw(G)+1)tr(G)≤L.
###### Proof.

Let be the bags of a tree decomposition of with width and radius . Label each bag as unvisited. We begin by selecting a vertex that is in the center of the tree decomposition. For the unique path from to some leaf of distance from , say , place a cop on each vertex of in the vertex set . Label each bag in as occupied. This set of vertices contains at most vertices. If the robber is not on the set of vertices , then the robber is not captured and so moves.

The cop now chooses a new path that intersects the last path in as many vertices as possible. For example, if there is a leaf in the tree decomposition that is also adjacent to , then the cop may choose the path . We label the bags that are no longer in the current path as visited. Note that by the cop strategy,

1. no visited bag will be revisited by the cops;

2. no unvisited bag is adjacent to a visited bag in the tree decomposition; and

3. if a bag is unvisited in the current round, then each of the neighbors of in the next round will be either unvisited or occupied.

In particular, note that if some vertex of the original graph is not contained in an occupied bag, then may be contained in occupied bags or unoccupied bags, but not both.

We may assume that the robber was previously on a vertex with the property that all bags that contain are unvisited. If the robber moved to a vertex that is now contained in an occupied bag, then the robber is captured as each such vertex contains a cop. Supposing then that is not in an occupied bag, then by (3) and since was only in unvisited bags, can only be in unvisited bags, unless the robber is captured. The cop continues this process until all paths starting at and ending at a leaf (of distance at most from ) have been probed. Eventually there will be no unvisited bags, by which time the robber has been captured. As such, the cop must have one round for each leaf, and so the number of rounds needed for capture with this strategy is at most

We also have the following upper bounds.

###### Theorem 5.2.

For a graph we have that

 ζ(G)≤(tw(G)+1)(Δ(G)+1),

and

 captζ,(tw(G)+1)(Δ(G)+1)≤tr(G)+1.
###### Proof.

Let be bags of a tree decomposition of with width and radius . We begin by selecting a vertex that is in the center of the tree decomposition. Using at most cops, we place a cop on each vertex of , and a cop on each neighbor of a vertex in . If we suppose that the robber is not on the same vertex as a cop, then the cop with the smallest distance to the robber must lie on a neighboring vertex of , and not in itself.

Suppose that a closest cop to the robber was on vertex . Note that is not in . If we delete the bag from the tree decomposition, then a forest is formed where there is exactly one subtree with a bag containing . We can then take to be the unique bag that is in this subtree and which was connected to in the tree decomposition. This procedure can then be recursively applied, where on each round a new bag is selected that is further away from than the bag in the previous round. We note that by this procedure, if the robber is on vertex during turn and has avoided capture, then every bag that contains has the property that the unique path from to in the tree decomposition contains the path