Small Connected Planar Graph with 1-Cop-Move Number 4

12/15/2019
by   Wei Quan Lim, et al.
0

This paper describes a 720-vertex connected planar graph G such that cop1(G), denoting the minimum number of cops needed to catch the robber in the 1-cop-move game on G, is at least 4 and at most 7. Furthermore, G has a connected subgraph H such that cop1(H) is exactly 4, meaning that 4 cops are barely sufficient to catch the robber in the 1-cop-move game on H. This is a significant improvement over the graph given by Gao and Yang in 2017.

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Abstract

This paper describes a -vertex connected planar graph such that , denoting the minimum number of cops needed to catch the robber in the 1-cop-move game on , is at least and at most . Furthermore, has a connected subgraph such that is exactly , meaning that cops are barely sufficient to catch the robber in the 1-cop-move game on . This is a significant improvement over the graph given by Gao and Yang in 2017 [8].

Acknowledgements

I would like to thank Ziyuan Gao for introducing me to this interesting problem in the first place, as well as for giving me very helpful feedback on drafts of this paper. This research was partly supported by Singapore MOE AcRF Tier 2 project MOE2018-T2-1-160.

1 Introduction

The abstract game of Cops and Robbers is a perfect-information 2-player game on a graph with two other parameters and . Player (for Cops), has cops, and first places each of them at a vertex of . Player (for Robbers) then places the robber at a (different) vertex of . After that, and take turns to make a move. On ’s turn, may move the robber by step, namely from its current vertex along an edge to a neighbouring vertex. On ’s turn, may move up to cops, each by step. The classical variant where was introduced decades ago [1], whereas the variant where has been the subject of mathematical study only in the past few years [10, 3, 12, 2]. In general, this game is called the -cop-move game with cops on . If eventually some cop moves to the same vertex as the robber, then wins, otherwise wins. (As defined here, neither cops nor robbers are not forced to move on each turn. For some other variants see [5, 7, 10].)

A natural question is, how many cops are needed to catch the robber on a given graph? Specifically, the classical cop number for , denoted by , is the minimum such that player (Cops) wins (i.e. has a winning strategy for) the -cop-move game with cops on . And the -cop-move number for , denoted by , is the minimum such that wins the -cop-move game with cops on . The class of graphs with cop number has been characterized for by Nowakowski and Winkler [9] and for general by Clarke and MacGillivray [6]. It is also natural to ask whether the cop number is bounded for the class of finite connected planar graphs, since the edge connections in a planar graphs are in some sense local. Indeed, Aigner and Frommel showed that for every graph in . In contrast, much less is known about the -cop-move game for  [4]. Although Bal et al. [2] did show that for every graph in with vertices, it is conjectured that there is in fact a fixed upper bound on for every graph in , but this remains unproven.

Recently, Gao and Yang constructed a graph with  [8], settling the question of whether there is even such a graph, which was posed as an open problem by Sullivan et al. [11]. However, they were unable to prove their conjecture that , nor were they able to find a simpler construction. is constructed by replacing each face of a dodecahedron with a number of nested pentagonal layers, where the -th layer from the centre has vertices. In their paper, they used layers (in each face), resulting in more than vertices in . It seems that, although the number of layers can be reduced without essentially affecting their solution, the resulting graph is likely to still have more than vertices.

This paper provides an improved answer to that problem, namely a much smaller graph with merely vertices and a proof that , as well as a connected subgraph of with . It is hoped that the techniques used here, while somewhat ad-hoc, will be helpful in figuring out the answer to the (still-open) question of whether there is a graph with -cop-move number or even larger.

2 The Construction

To build the desired graph , we first start from the truncated icosahedron (a.k.a. the soccer ball graph) with pentagonal faces and hexagonal faces, and retain its vertices but replace its faces as depicted in the diagram on the right for one pentagonal face and three of its neighbouring hexagonal faces.

The blue vertices (each with degree in ) are the vertices of , and the black vertices are added vertices. Note that has the same symmetries (i.e. automorphism group) as . There are black vertices added to each pentagonal face, and black vertices added to each hexagonal faces, and blue vertices in total, and so has vertices in all.

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3 The Robber Evades 3 Cops

We shall now establish that , by explaining the winning strategy for player (Robber). The general idea is for the robber to stick to the key vertices, defined as the vertices of (blue in the diagram), and move safely to another key vertex whenever a cop gets too close, where moving safely to a vertex means to move to in such a way that the cops cannot catch the robber along the way and no cop is next to when the robber reaches . We shall use the following easy lemma throughout the analysis.

Lemma 1 (Nearness Lemma).

On the robber’s turn, if the robber is nearer to a key vertex than any cop, then the robber can move safely to by following any shortest path to without stopping.

Proof ().

Take any shortest path from the robber’s starting vertex to . For each vertex on , just after the robber reaches , no cop can reach immediately after that, since its starting vertex is further from than and so .       

Specifically, after the cops are placed, places the robber at a key vertex that has no cop at or next to it (which is always possible since each cop can be at or next to at most one key vertex), and then over subsequent turns repeats the following indefinitely:

  1. Stay phase: Stay at the key vertex (i.e. do not move the robber) until a cop moves to a vertex adjacent to . By symmetry, there are essentially possible positions for relative to as depicted in Figure 1, and no cop is at any other neighbour of .

  2. Travel phase: Let be the red-dotted regions in Figure 1 on the lower-left, upper-left, and right respectively. Each encloses vertices within steps from some key vertex that is steps away from , except and some neighbours of . There are two possible situations:

    • There is exactly one cop in each of .

    • There is no cop in some of .

  3. In either situation, it is possible to move safely to some key vertex, as we shall show subsequently.

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Figure 1: At the start of the travel phase, the robber (solid green circle) is at a key vertex and there is a cop at exactly one neighbour of , and there are possible positions of (red-circled vertices).

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3.1 One cop in each region

We first deal with the situation where there is exactly one cop in each of . As noted earlier, there are essentially cases for (and no cop is at any other neighbour of ):

  1. is in and step away from .

  2. is in and step away from .

  3. is in and step away from .

3.1.1 Case 1

If the cop in is not at the X-marked vertex in the diagram on the right, then the robber can use the green path to move safely to one of the green-circled key vertices ( is the one on the left) or back to .

More precisely, after the robber takes the first step along the green path, if the cop at immediately starts moving along the red path, then the robber can move back to . Otherwise, the robber can continue along the green path, and by the Lemma 1 (Nearness Lemma). the cop that was at must follow along the red path to guard (i.e. prevent the robber from moving safely to ), after which the robber would be steps away from and the cops would all still be at least steps away from , since every cop in and was initially at least steps away from , so by the Lemma 1 (Nearness Lemma). the robber can move safely to .

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But if the cop in is at , then the robber can instead use the green path as shown on the right to move safely to one of the green-circled key vertices ( is the one on the bottom).

More precisely, when the robber takes the first steps along the green path, the cop that was at must follow along the red path to guard by the Lemma 1 (Nearness Lemma)., after which the robber would be steps away from and the cops would all still be at least steps away from , so by the Lemma 1 (Nearness Lemma). the robber can move safely to .

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3.1.2 Case 2

If the cop in is not at the X-marked vertex in the diagram on the right, then (exactly like in case 1a) the robber can use the green path to move safely to one of the green-circled key vertices ( is the one on the bottom) or back to .

More precisely, after the robber takes the first step along the green path, if the cop at immediately starts moving along the red path, then the robber can move back to . Otherwise, the robber can continue along the green path, and by the Lemma 1 (Nearness Lemma). the cop that was at must follow along the red path to guard , after which the robber would be steps away from and the cops would all still be at least steps away from , since every cop in and was initially at least steps away from , so by the Lemma 1 (Nearness Lemma). the robber can move safely to .

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But if the cop in is at , then the robber can instead use the green path as shown on the right to move safely to the green-circled vertex , after which either the robber can move safely to , or the cop that was initially at must next move to the red-circled vertex, and the other cops are still in (and go to 3.1.4 Case 4).

More precisely, when the robber moves along the green path, by the Lemma 1 (Nearness Lemma). the cop that was at must follow along the red path in order to guard , during which no other cop can move.

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3.1.3 Case 3

If the cop in is not at the X-marked vertex in the diagram on the right, then the robber can use the green path to move safely to the green-circled vertex or back to . And in the former case, either the robber can move safely to , or the cop that was initially at must next move to the red-circled vertex, and the other cops are at most step outside (and go to 3.1.4 Case 4).

More precisely, after the robber takes the first step along the green path, if the cop at immediately starts moving along the red path, then the robber can move back to . Otherwise, the robber can continue along the green path, and by the Lemma 1 (Nearness Lemma). the cop that was at must follow along the red path in order to guard , during which no other cop can move.

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But if the cop in is at , then (essentially like in case 1b) the robber can instead use the green path as shown on the right to move safely to one of the green-circled key vertices ( is the one on the left).

More precisely, when the robber takes the first steps along the green path, the cop that was at must follow along the red path to guard by the Lemma 1 (Nearness Lemma)., after which the robber would be steps away from and the cops would all still be at least steps away from , so by the Lemma 1 (Nearness Lemma). the robber can move safely to .

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3.1.4 Case 4

The two unfinished cases above (i.e. 2b and 3a) can be handled in the same way. The robber is now at vertex as shown on the right (solid green circle) and is next to move. One cop is at a nearby vertex (solid red circle), and the other two cops are each at a vertex in one of the two red-dotted regions. From here, the robber can move safely to one of the green-circled key vertices (named in clockwise order around the ‘hexagon’ from the top-left).

To establish this, first observe that the robber can move safely along the green path to the thin-green-circled vertex. After that, the cop that was in must within the next move get to within steps from (i.e. reach or pass a thin-red-circled vertex) in order to guard it by the Lemma 1 (Nearness Lemma)..

So if the robber cannot reach safely, the cop that was at can move at most step so far, and hence the robber can continue moving safely along the green path to the dotted-green-circled vertex. At this point, the robber is only steps away from , so the cop from must within the next move get to within steps from (i.e. reach a dotted-red-circled vertex) in order to guard it by the Lemma 1 (Nearness Lemma)., and hence must have taken at least steps. But if the cop from does move in this manner, then no other cop can have moved so far, and hence the robber can safely move along the rest of the green path to by the Lemma 1 (Nearness Lemma)..

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3.2 No cop in some region

We finally deal with the situation where there is no cop in for some . By symmetry, and since the cop at must be within steps from in order to guard by the Lemma 1 (Nearness Lemma)., we only need to consider cases:

  1. has no cop, and is in and just outside .

  2. has no cop, and is in and just outside .

  3. has no cop, and is in and just outside .

3.2.1 Case 0

Before analyzing those cases, we shall show how to handle a common subcase. Here we assume case , but it is essentially the same in the other cases.

If no cop is exactly steps from , then the robber can oscillate between the green-circled vertices (see right) after moving to the nearest one, as long as the cop that was at also oscillates between the red-circled vertices. If the cops deviate from this, the robber can thereafter move safely to either or by the Lemma 1 (Nearness Lemma)..

Henceforth in all the subsequent cases we can assume that some cop is exactly steps from .

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3.2.2 Case 1

By the Lemma 1 (Nearness Lemma). there must be a cop in , since the cop at cannot guard . Moreover, there must be a cop in the yellow-dotted region (see right), which encloses vertices outside that are within steps from the key vertex at the end of the green path, otherwise the robber can use the green path to move safely to either or by the Lemma 1 (Nearness Lemma)., since when the robber takes the first steps along the green path, the cop that was at must follow along the red path to guard , after which the robber is only steps away from .

Henceforth for the rest of this case we can assume that there is exactly one cop in each of .

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Since the cop in is exactly steps from , and in particular not at the X-marked vertex (see right), the robber can use the green path to move safely to one of the key vertices or back to , exactly like in Section 3.1.1 (though the third cop is in rather than ).

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3.2.3 Case 2

By the Lemma 1 (Nearness Lemma). there must be a cop in , since the cop at cannot guard . Moreover, there must be a cop in the yellow-dotted region (see right), which encloses vertices outside that are within steps from the key vertex at the end of the green path, otherwise the robber can use the green path to move safely to either or by the Lemma 1 (Nearness Lemma)., since when the robber takes the first steps along the green path, the cop that was at must follow along the red path to guard , after which the robber is only steps away from .

Henceforth for the rest of this case we can assume that there is a cop in each of .

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If the cop exactly steps from is in , then there is exactly one cop in each of , and in particular there is no cop at the X-marked vertex (see right), and so the robber can use the green path to move safely to one of the green-circled key vertices ( is the one on the left) or back to , exactly like in Section 3.1.2 (though the third cop is in rather than ).

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But if the cop exactly steps from is in , then there are already cops in , and so the third cop must be in . Thus the robber can use the green path (see right) to move safely to the green-circled vertex , and the cop at must follow along the red path to the red-circled vertex in order to guard by the Lemma 1 (Nearness Lemma)., during which the other cops cannot move and hence remain within .

Thus after moving along the green path to , if the robber cannot reach safely in the next move, then on that move it must be that one cop is at and the other two cops are in , and this situation is covered by Section 3.1.4.

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3.2.4 Case 3

By the Lemma 1 (Nearness Lemma). there must be a cop in , since the cop at cannot guard . Moreover, there must be a cop in the yellow-dotted region (partly shown on the right) that encloses vertices outside that are within steps from the key vertex at the end of the green path, otherwise the robber can use the green path to move safely to either or by the Lemma 1 (Nearness Lemma)., since when the robber takes the first steps along the green path, the cop that was at must follow along the red path to guard , after which the robber is only steps away from .

Since are disjoint, we can for the rest of this case assume that there is exactly one cop in each of .

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Since the cop in is exactly steps from , and in particular is at least steps away from the green-circled key vertex (see right), the robber can use the green path to move safely to either or back to .

More precisely, after the robber takes step along the green path, if the cop that was at moves away from then the robber can move safely back to , otherwise the robber can continue moving safely along the green path since it is already only steps away from .

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4 7 Cops Catch the Robber

Even though it seems like the robber can barely manage to escape from cops using the strategy given in the previous section, it is partly because in that strategy the robber waits until a cop is right next to it, and there does not seem to be a concise strategy for cops to catch the robber, if there is one at all.

Nevertheless, it is not too hard to give a winning strategy for cops, establishing that , which we shall do in this section. The intuitive idea behind this strategy is to use some cops to ‘guard’ some vertices so as to restrict the robber to certain possible regions. At the start we move the cops into an initial ‘guarding’ configuration, and thereafter in each phase we keep the robber ‘confined’ to a region using some cops while moving the other cops to new ‘guarding’ positions to ‘divide’ that region, so that the robber would now be ‘confined’ to a smaller region.

4.1 Hexagon Guarding

We begin with a lemma concerning how one cop can be used to guard a hexagon (shown below in blue), namely to guard the three ‘sides’ of a hexagonal face of the truncated icosahedron that are adjacent to the neighbouring pentagonal faces, in the sense of preventing the robber from ‘crossing over’. The rough idea is that the cop will try to stay in the central vertices of the hexagon, namely at one of the three vertices of the triangle in the centre of the hexagon, and move towards one ‘side’ only when the robber gets close to that side.

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For a cop to guard the blue hexagon, it must stay within the red-outlined region, and its position must be as follows.

If the robber is within a proximal region of the hexagon, shown as the green-dotted region (the other two proximal regions are positioned symmetrically around the hexagon), it is labelled according to the number on its vertex as shown, and the cop must be either on the shown red path and labelled according to the number on its vertex, or at the central vertex beside the red path (next to both the -labelled and -labelled vertices) and labelled or , whichever is furthest from the robber’s label. Define the deviation of such a cop to be the (absolute) difference between the cop’s label and the robber’s label. (The labelling is different for a cop guarding another hexagon.) If the robber is not within any proximal region or its corresponding red path, it is labelled , and the cop’s deviation is defined to be where is its distance from the nearest central vertex. If the robber is on the red path, the cop’s deviation is defined to be where is its distance from the robber. Intuitively, the cop’s deviation captures roughly how far it is from being able to guard its hexagon, where any deviation of or less is optimal.

We say that the cop guards this hexagon iff the cop is positioned as stated above and its deviation is at most . Clearly, if such a cop can move on every turn, then it can preserve this invariant and hence prevent the robber from crossing the red path (i.e. moving to any of its vertices) without getting caught. In general, we want to maintain this even with multiple cops each guarding some hexagon. To do so, we say that a set of cops strongly guard their hexagons iff the cops in guard distinct non-adjacent hexagons and furthermore at most one cop in has deviation more than , and we shall prove the following crucial lemma.

Lemma 2 (Hexagon Guard Lemma).

Take any game state where it is the robber’s turn and some set of cops strongly guard their hexagons. Then from that point onwards, the cops can either win or indefinitely preserve the invariant that the cops in strongly guard their hexagons after the cops’ turn. If additionally no three cops in guard hexagons that are all adjacent to the same hexagon, then in the latter case the cops can move in such a way that infinitely often on their turn no cop in moves.

Proof ().

We can assume that the robber does not move next to a cop, otherwise the cops can immediately win. Thus the robber cannot move onto the red path for any cop in . Observe that when the robber moves, the deviation of each cop in is still well-defined, and there are at most two relevant cops in , where a cop in is relevant iff the robber moved within, into or out of a proximal region of the hexagon guarded by that cop. So when the robber moves, only relevant cops in can have their deviations changed, and by at most , whereas irrelevant cops in have their deviation remaining at . Also, if the new deviation for a relevant cop in is more than , then either the robber is now in a proximal region for that cop, in which case it is possible for the cop to move (in one step) to a vertex on the red path to adjust its deviation from to , or the robber has just left a proximal region, in which case it is possible for the cop to move to a central vertex to adjust its deviation in the same way. There are two main cases:

  1. There is at most one cop in whose deviation changed. If now , then the invariant still holds. But if now , then that cop can move to adjust its deviation to , hence preserving the invariant since implies .

  2. There are exactly two relevant cops in with deviations where initially , and their deviations changed by at most each. By the strong guarding invariant, initially and , so now and . If now , the invariant already holds. But if now , then the cop with deviation can move to adjust its deviation to , hence preserving the invariant since implies .

We now list the remaining cases for two relevant cops in the following tables. By symmetry we can assume that one of the relevant cops in guards the blue hexagon, and the robber is in the upper half of the green-dotted region. The first table is for when the other relevant cop in guards the purple-dotted hexagon (partly shown), and the second and third tables are for when the other relevant cop in guards the yellow-dotted hexagon. Each case is given on a separate row, characterized mainly by the change in the labels for the robber given in the first column (with respect to those two cops). For each pair of robber label changes, there are only a few cases in which we need to move a cop to preserve the invariant, and in each case we can indeed move just one of those two cops to do so, resulting in the cop label changes given in the second column.

Robber labels Cop labels
Robber labels Cop labels
Robber labels Cop labels

Finally, under the additional assumption that no three cops in guard hexagons that are all adjacent to the same hexagon, we shall prove that player (Cops) can move the cops in as stipulated above to preserve the invariant, such that after finitely many turns the game will reach a state where it is ’s turn and does not need to move any cop in to preserve the invariant (i.e. the invariant is already satisfied). To do so, we shall consider any robber strategy where always has to move some cop in to preserve the invariant, and show that it is impossible.

Each robber move must change the cop deviations, so we can assume that one of the relevant cops in guards the blue hexagon, and it is not hard to verify that:

  1. The robber cannot move to an adjacent vertex with the same robber labels (including moving to or from a central vertex or along one of the broken edges in the below-left diagram).

  2. The robber cannot move to a vertex with the same robber labels as one move ago (including moving backwards along the same edge that it used in the previous turn), otherwise the invariant would still be satisfied without any cop moving. This entails checking each of the above cases one by one:

    1. First main case: Only one cop in has deviation changed after the (previous) robber move. With respect to that cop, let be the labels for the cop and robber respectively after that move, and be the robber label before that move. It must be that to make that cop move, hence by symmetry we can assume , and the new cop label is . Trivially (see the diagram in Section 4.1), yielding . And since that cop was guarding its hexagon before the robber move. Thus and hence , so if the robber returns to the previous labels, then no cop needs to move.

    2. Second main case: Exactly two cops in have deviations changed after the (previous) robber move, each by at most , where initially and so . It must be that after that move to make the corresponding cop move. With respect to that cop, let be the labels for the cop and robber respectively after that robber move. Then , so by symmetry we can assume , and the new cop label is . It cannot be that , otherwise , which forces and implies that are distinct non-integers, which is impossible (see the diagram in Section 4.1). Thus and hence , so the robber must not return to the previous labels, otherwise we once again have and and hence no cop needs to move.

    3. Remaining cases: We can easily check that, in each row of the above tables, the new cop labels satisfy the desired invariant with the old robber labels.

From these we can infer that the robber also cannot move to the vertices erased in the below-right diagram. For convenience, we also mark three of the -labels as to distinguish those vertices for easy reference later.

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It remains to analyze all ways the robber can enter the proximal region for

(classified by the robber label sequence):

  1. The robber moves . For the invariant to be violated after that, there must be a cop in guarding the yellow hexagon, and the robber label for changes , and the cop labels for must be initially . But the cops can move , after which the robber cannot move (vertically) otherwise the cops do not need to move. So the robber must continue (horizontally), and the cops continue . After that, the robber is forced to continue , and the cops continue and then do not need to move (on the next turn).

  2. The robber moves . For the invariant to be violated after that, there must be a cop in guarding the purple hexagon, and the robber label for changes , and the cop labels for must be initially . But the cops can move , after which the robber cannot move , and so must continue , and the cops continue , and then do not need to move.

  3. The robber moves . Then for the invariant to be violated after that move, there must be a cop in guarding either the purple hexagon or the yellow hexagon.

    1. If is guarding the purple hexagon, then the robber label for changes , and the cop labels for must be initially or . moves , after which the robber cannot move , and so must continue . It must be that the cop labels are now to force them to continue . But after that the robber must continue or , and then the cops do not need to move.

    2. If is guarding the yellow hexagon, then the robber label for changes , and the robber must move , and the cop labels for must be initially or . moves , after which the cop labels must be to force them to continue . After that, the robber cannot move , and so must continue , and the cops continue . Again, the robber cannot move , but must continue , and the cops continue . Again, the robber cannot move , nor otherwise the cops continue and then do not need to move. Hence the robber must continue on the path (labels for ). On the next move after that, if the robber does not enter a new proximal region, it must move to force to move , but on the subsequent turn no cop needs to move. Therefore there must be another cop in guarding the hexagon just below the bottommost hexagon in the diagram, and the robber must enter its proximal region.

  4. The robber moves . There must be a cop in guarding either the purple hexagon or the yellow hexagon, otherwise the cop label for must be initially , and moves , after which the robber must continue or so no cop needs to move.

    1. If is guarding the purple hexagon, then by symmetry the situation is exactly as in item 3a.

    2. If is guarding the yellow hexagon, then the robber label for remains at , so the cop label for must be initially and moves . At this point, the cop labels for cannot be , otherwise the robber cannot move or , and so must continue , but the cops continue and then do not need to move. Therefore the cop labels must be , so the robber cannot move or , and must continue , and the cops continue . After that, as in the later half of item 3b, the robber must continue on the path of robber labels for , and then enter a proximal region of a hexagon that is guarded by a cop in and just below the bottommost hexagon in the diagram.

Therefore the robber must indefinitely repeat item 3b or item 4b. But this is impossible, because it requires three cops in guarding hexagons that are all adjacent to the same hexagon (the bottommost one in the diagram).       

Remark 0 ().

Incidentally, if three cops in guard hexagons that are all adjacent to the same hexagon, then even if the robber is ‘confined’ inside the region around the central hexagon, the robber can indefinitely repeat the path to force those three cops to keep moving, which implies that the cops cannot catch the robber unless they break out of this guarding pattern! This is one reason it seems difficult to ascertain whether or not .

The next lemma captures how we can expand strong guarding of some hexagons to an extra hexagon (using an extra cop), while still strongly guarding the original hexagons. Consequently, once the cops have confined the robber to a region by strongly guarding some hexagons, then the cops can keep the robber confined to that region while moving an extra cop to strongly guard yet another hexagon, to confine the robber even further.

Lemma 3 (Guard Expansion Lemma).

Take any set of distinct non-adjacent hexagons, no three of which are adjacent to the same hexagon. And take any game state, where some set of cops strongly guard all the hexagons in except some hexagon , and there is another cop not in . Then the cops can move in such a way that the cops in still always strongly guard their hexagons (after their turn) and yet eventually the cops in strongly guard all the hexagons in .

Proof ().

We start by gradually moving to a central vertex of while maintaining the invariant that the cops in strongly guard their hexagons (after their turn), by the Lemma 2 (Hexagon Guard Lemma).. After that, if the robber is not in a proximal region of , then the cops in strongly guard all the hexagons in and we are done. But if the robber is in a proximal region of , then there is a vertex on the corresponding red path (see the diagram in Section 4.1) such that would strongly guard all the hexagons in if is at . Place a guide at . From then on, after each robber’s turn, we perform the following steps:

  1. If does not strongly guard their hexagons (treating as an actual cop), move one cop/guide in so that they (again) strongly guard their hexagons, by the Lemma 2 (Hexagon Guard Lemma).. Otherwise move nothing.

  2. If in step 1 we moved or nothing at all, then move towards (if it is not already at the same vertex).

Note that this yields valid moves because on each cops’ turn we move only one cop. Also, the distance from to (measured after the cops’ turn) never increases (since is moved whenever is moved), and it decreases repeatedly until it is zero because infinitely often no cop/guide is moved in step 1, by the Lemma 2 (Hexagon Guard Lemma). again. Thus eventually is at the same vertex as and hence the cops in strongly guard all the hexagons in .       

4.2 The Winning Strategy

Now we are ready to present the winning strategy for cops. Identify of the hexagons underlying whose centres form a cube, and divide them into red and yellow hexagons, where the yellow hexagons are opposite the centre of the cube (as in the diagram on the right). Note that no three of the red hexagons are adjacent to the same hexagon.

Let be a set of of the cops. At the start, place each cop in at a central vertex of a different red hexagon, and place the th cop anywhere. Once the opponent has placed the robber, let be the set of cops in with deviation at most , and note that the cops in strongly guard their hexagons. The idea is roughly to expand strong guarding from those hexagons to all red hexagons to confine the robber to ‘half of the cube’, and then keep the robber there while expanding also to the yellow hexagon in that ‘half’ to further confine the robber to a ‘square of the cube’. After that, we use the corresponding cops to continue strongly guarding that ‘square’ while moving other cops to divide the confinement region in half, and then gradually reduce it further.

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We shall now go into the details of how to move the cops.

4.2.1 Movement Phase 1

Move the cops in to eventually strongly guard the red hexagons, by applying the Lemma 3 (Guard Expansion Lemma). (creftype 3) to expand strong guarding from the hexagons guarded by the cops in to all the red hexagons, one hexagon at a time. After this, the robber will be confined to one of the two possible ‘halves of the cube’ on either ‘side’ of the ‘ring’ of red hexagons, where one ‘side’ is represented on the right by the coloured hexagons. (Of course, the robber is confined to only one ‘side’ of each red hexagon.)

Note that, during this phase, we do not care if the robber ‘escapes’ past any of the vertices that the cops are eventually supposed to guard. All that matters is that after finitely many moves, these cops strongly guard their red hexagons and hence the robber will be confined to one ‘side’ of that ‘ring’ of hexagons.

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4.2.2 Movement Phase 2

Keep the robber confined to its current ‘half of the cube’, while moving the remaining th cop to ‘expand’ strong guarding to the yellow hexagon in that ‘half’, again by the Lemma 3 (Guard Expansion Lemma). (creftype 3). After this, the robber will be confined to one of three possible ‘squares of the cube’, represented on the right by the coloured hexagons (the other cases are symmetric).

Now keep the robber confined to its current ‘square’, using the cops that had been strongly guarding the red/yellow hexagons, while moving of the other cops to the key vertices shared by blue hexagons (as indicated by the pink circles), by the Lemma 2 (Hexagon Guard Lemma). (creftype 2). After this, the robber will be confined to ‘half of that square’ or between the pink-circled key vertices.

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4.2.3 Movement Phase 3

More precisely, we can assume that the robber never moves next to a cop, so the robber is now confined to one of the green-dotted regions in the diagram on the right, with red cops strongly guarding the red hexagons (one on each red path) and pink cops shown as solid pink circles with on the left. If the robber is in-between the pink cops, we can trivially move a third cop to catch the robber. Otherwise we maintain strong guarding of the red hexagons, while moving a th cop to the pink-circled vertex, by the Lemma 2 (Hexagon Guard Lemma). (creftype 2).

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Maintain strong guarding of the red hexagons, until we do not have to move any red cop, by the Lemma 2 (Hexagon Guard Lemma). (creftype 2).

By left-right symmetry we can assume that the robber is at this point in the green-dotted region as shown on the right. Move the pink cop one step along the pink path, and continue gradually moving it along the path while maintaining guarding of the left red hexagon. Observe that after that first step along that path, there is no need to guard the right red hexagon anymore, and that on each subsequent step, the robber is confined to a smaller region.

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Finally, as shown on the right, gradually move the pink cop along the given path (labelled “1”), followed by the pink cop along the given path (labelled “2”), all the while maintaining guarding of the left red hexagon, and the robber will be caught.

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5 1-Cop-Move Number 4 on a Small Graph

Although it seems difficult to find a strategy for fewer cops to win on , we can easily show that there is some connected (planar) subgraph of such that . This follows readily from the following lemma.

Lemma 4 (Vertex Guard Lemma).

Take any graph and a vertex in . Then .

Proof ().

Let . Then we can use cop to guard by staying there without moving, forcing the robber to never move to , and hence we can use other cops to catch the robber on the graph .       

Theorem 5 ().

There is a connected planar graph with at most vertices such that .

Proof ().

Let and let be the number of vertices in . For each let where is a vertex in that is not a cut vertex (i.e.  is still connected). Clearly , so there is some minimum such that , If , then . Otherwise if , then and hence by the Lemma 4 (Vertex Guard Lemma).. In either case, and is a connected (planar) subgraph of .       

6 Open Questions

It is not clear what the true value of is, and it would be very interesting if it was more than , because then the robber’s winning strategy against cops would have to be very different from the one given in this paper against cops. One also hopes that we will eventually find an explicit simpler and smaller finite connected planar graph with -cop-move number exactly , and get a better understanding of whether finite connected planar graphs have bounded -cop-move number or not.

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