Discrete Mathematics 329 (2014) 1–11

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Discrete Mathematics

journal homepage: www.elsevier.com/locate/disc

Edge-critical cops and robber in planar graphs✩

Shannon L. FitzpatrickUniversity of Prince Edward Island, Charlottetown, PE, C1A 4P3, Canada

a r t i c l e i n f o

Article history:Received 5 June 2012Received in revised form 3 March 2014Accepted 9 March 2014Available online 24 April 2014

Keywords:Pursuit gameCops and robberEdge-criticalPlanar graph

a b s t r a c t

The problem is to determine the number of ‘cops’ needed to capture a ‘robber’ in a gamein which the cops always know the location of the robber, and the cops and robber movealternately along edges of a reflexive graph. The cops capture the robber if one of themoccupies the same vertex as the robber at any time in the game. A cop-win graph is one inwhich a single cop has a winning strategy. A graph is cop-win edge-critical with respect toedge addition (respectively, deletion) when the original graph is not cop-win, but the ad-dition (deletion) of any edge results in a cop-win graph. In this paper, edge-critical planargraphs are characterized.

© 2014 Elsevier B.V. All rights reserved.

1. Introduction

The game of ‘Cop and Robber’ was introduced by Nowakowski and Winkler [6] and, independently, by Quilliot [5]. Thegame rules are as follows: given a connected graph, G, the cop chooses a vertex of G, then the robber chooses a vertex.Afterwards, they move alternately—each can move to an adjacent vertex or stay at his current location. The cop and robberare always aware of the other’s location. The cop wins if he ever occupies the same vertex as the robber. Graphs in whichthe cop has a winning strategy are called cop-win graphs.

Cop-win graphs have been completely characterized [5,6]. They can be recognized via a polynomial-time decompositionalgorithm that requires only knowledge of the neighbourhood of each vertex in the graph. The decomposition consists ofremoving a single vertex at each step, and the order in which the vertices are removed is referred to as a cop-win ordering.A formal description of this ordering appears at the end of this section.

In [1], Aigner and Fromme introduced a version of the game in which there is a set of cops chasing a single robber. Inthis version, the game begins with each cop choosing a vertex. Next, the robber chooses a vertex. The cops and robber thenalternate moves, but on the cops’ turn, each cop can move to an adjacent vertex or stay at his current location. The cops winif at least one cop occupies the same vertex as the robber.

The copnumber of a graph G, denoted c(G), is the least number of cops required to capture a robber on G. We say that agraph is k-cop-win if its copnumber is at most k.

For k ≥ 2, a structural characterization of k-cop-win graphs has also been found [3]. It also relies on a decompositionalgorithm with an associated vertex ordering. However, instead of an ordering on the vertices of G, as in the cop-win case,it is an ordering of the vertices in the (k + 1)-fold categorical product of Gwith itself.

In [1] it was shown that all connected planar graphs are 3-cop-win. In doing so, the authors used an argument involvingisometric paths. An isometric path or geodesic is any shortest path joining its endpoints. For any isometric path, a single copmoving on that path can, after a finite number of moves, apprehend the robber immediately should the robber move ontothe path [1]. This is because there is a graph homomorphism f from the original graph G onto the isometric path P such that

✩ Partially supported by a grant from Natural Sciences and Engineering Research Council of Canada (Grant Number 238874).E-mail addresses: sfitzpatrick@upei.ca, shannon.fitzpatrick.upei@gmail.com.

http://dx.doi.org/10.1016/j.disc.2014.03.0060012-365X/© 2014 Elsevier B.V. All rights reserved.

2 S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11

f (x) = x for every vertex x in P . Since paths are cop-win, a single copmoving on P can capture the image of the robber underf . A similar strategy will be used in Section 2 to find winning strategies in a class of 2-cop-win graphs. However, the graphhomomorphism used is not the identity when restricted the isometric path.

In [2,4], edge-criticality was examined with respect to the game of Cops and Robber. In [2], classes of graphs in whichthe copnumbers change from 1 to 2 were determined for four different operations on the edge set: addition, deletion, sub-division, or contraction of its edges. The paper then went on to characterize all regular graphs whose copnumber changesfrom 2 to 1 with the addition of any edge. In this paper, we characterize all planar graphs whose copnumber changes from2 to 1, either with the addition of any edge (Section 3), or with the deletion of any edge (Section 4).

The original motivation for examining criticality in the Cops and Robber game was to gain a greater understanding ofgraphs with copnumber 2. (Although we have the characterizations provided in [3], there is still much unknown aboutgraphs with copnumber greater than 1.) When we limit ourselves to planar graphs, each of which has copnumber at most3, a characterization of those with copnumber 2 would give us the complete picture in that class of graphs.

In what follows, we assume that all graphs are simple and finite. For a graph G, V (G) and E(G) represent the vertex andedge sets, respectively. For distinct vertices u and v, we use u ∼ v to indicate that u and v are adjacent and u � v to indicatethey are non-adjacent. For v ∈ V (G), we let NG(v) and NG[v] denote the open and closed neighbourhoods of v in G, respec-tively. When it clear that G is the graph being referred to, we will simply write N(v) and N[v]. We use d(u, v) to denote thedistance between vertices u and v, degG(v) to denote the degree of v in G, and δ(G) to denote the minimum degree of G. Todenote the addition of an edge uv to graph G, we use G+uv, and to denote the deletion of an edge uv from G, we use G−uv.For any set S ⊆ V (G), the notation G − S denotes the graph obtained by deleting the vertices of S from G.

A graph has a cop-win ordering whenever its vertices can be ordered v1, . . . , vn so that for each index i, there is an indexj with j > i such that vi ∼ vj and NGi [vi] ⊆ NGi [vj], where Gi = G − {v1, . . . , vi−1}. In this situation, we say that vi is acorner in Gi, vi is dominated by vj in Gi, and vj dominates vi in Gi. It was proved in [5] that G is cop-win if and only if a cop-winordering of its vertices exists.

2. Preliminary results

First, we note that if the addition or deletion of an edge results in a cop-win graph, then the original graph had copnumberat most 2. This is proved in the following two lemmas. Furthermore, since the cop’s winning strategy can be described in acop-win graph, these results provide a winning strategy for two cops.

The proof of the first lemmamaps the vertices of G onto an isometric path. As discussed in the introduction, this is similarto the technique first used by Aigner and Fromme [1].

Lemma 1. If G is a connected graph and G + xy is cop-win for some pair of non-adjacent vertices x and y, then c(G) ≤ 2.

Proof. Let x and y be non-adjacent vertices in G. Let G′= G + xy. If G′ is cop-win, then a single cop C1 has a strategy to

apprehend a robber on G′ after a finite number of moves.Since G is connected, there is a shortest path P from x to y in G. Let P = v0, . . . , vk, where x = v0 and y = vk. Consider a

function f that maps the vertices of G onto P as follows

f (v) =

vd(y,v), if d(y, v) ≤ kvk, if d(y, v) > k.

This mapping accomplishes two objectives: (1) adjacent vertices in G are either mapped to the same vertex on P or toadjacent vertices on P , and (2) the endpoints of P are mapped to each other (f (x) = y and f (y) = x).

Now suppose we play the game on G, starting with C1 on x and C2 on y. Every move the robber makes on G is also a moveon G′. The strategy for C2 will be to move onto f (v) whenever C1 moves to vertex v. The strategy for C1 will be to move as ifhe were playing on G′, except when the winning strategy requires him tomove along the edge xy. If such amove is required,C1 and C2 pass on this move and switch names. Since f (x) = y and f (y) = x, this emulates a move by C1 along the edge xyin one direction, and C2 along xy in the opposite direction. Therefore, C1 can continue to play his winning strategy on G, andtwo cops are sufficient to capture the robber on G. �

Lemma 2. If G is a connected graph and G − xy is cop-win for some edge xy, then c(G) ≤ 2.

Proof. Let G∗= G−xy. If G∗ is cop-win, then a single cop C1 can apprehend the robber on G∗ after a finite number of moves.

We now play with two cops on G. Begin by placing cops C1 and C2 on x. For the remainder of the game, C2 remains on x. As aresult, the robber will be apprehended whenever he uses the edge xy. Therefore, we may assume that the robber only usesvertices on G∗. Now, C1 plays according to his winning strategy on G∗. �

3. Edge addition

A graph is cop-win edge-critical with respect to addition (CECA) when it is not itself cop-win, but the addition of any edgeresults in a cop-win graph. Among the CECA graphs, we will only consider those that have no corners. This restriction isprimarily due to the fact that the deletion of a corner from a graph has no effect on the copnumber of the graph; the new

S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11 3

graph has the same copnumber as the original. (This result can be easily proven using an argument similar to that found forcop-win graphs in [5].) Restricting our investigations to corner-free graphs is also consistent with the approach taken in [2],where it was shown that if a CECA graph has corner u, then there is a vertex v ∈ V (G) such that NG[u] = NG[v].

Let A denote the set of all CECA graphs that contain no corners. If G ∈ A, then G + uv is cop-win for any distinct non-adjacent u, v ∈ V (G). Since G has no corners, it follows that in any cop-win ordering of G + uv, the first vertex in thatordering is u, v, or a common neighbour of u and v, and the first vertex in the ordering is dominated by either u or v inG + uv. It follows that G ∼= K2 or G has diameter 2. (See [2] for further details.)

We now work towards the characterization of planar graphs in A by examining small separating cycles and using twoadditional results from [2].

Given a particular embedding of G in the plane, a separating cycle is a cycle in G that has vertices both inside and outsideof it. A separating k-cycle is a separating cycle with k vertices. We will refer to a separating 3-cycle as a separating triangle.

Lemma 3 ([2]). If G ∈ A and G � K2, then G contains an induced 4-cycle. Furthermore, if G � C4, then δ(G) ≥ 4.

Lemma 4. If G ∈ A and G is planar, then no planar embedding of G has a separating triangle.

Proof. Suppose there is some plane graphG such thatG ∈ A andG contains a separating triangle, C , with vertex set {a, b, c}.Let x be a vertex inside C , and let y be a vertex outside C .

Let G′= G + xy. By Lemma 3, δ(G) ≥ 4 and x has a neighbour w inside C . Similarly, y has a neighbour z outside C . It

follows that neither x nor y is a corner in G′. Therefore, a common neighbour of x and y is a corner in G′. Without loss ofgenerality, let a be that common neighbour. Furthermore, assume a is dominated by x in G′. It follows that x is adjacent toeach of a, b, and c , and a is not adjacent to any vertex outside of C other than y.

Now let G′′= G+wy. It follows that a, b, or c is a corner in G′′, and its dominating vertexmust be adjacent to each of a, b,

and c. However, since G is planar, w is adjacent to at most two vertices of C . Therefore, y must be the dominating vertex,and y is adjacent to each of a, b, and c.

Finally, consider G + wz. By a similar argument, it follows that z is adjacent to each of a, b, and c . This, however, yields asubgraph in G isomorphic K3,3 with bipartition {a, b, c}, {x, y, z}, contradicting that G is planar. The result follows. �

The join of graphsG andH , denotedG∨H , has vertex setV (G)∪V (H) and edge set E(G)∪E(H)∪{uv|u ∈ V (G), v ∈ V (H)}.

Lemma 5 ([2]). G ∨ H ∈ A if and only if G ∈ A and H ∈ A.

Lemma 6. Let G ∈ A, and assume G is planar. If G contains an induced separating 4-cycle, then G ∼= C4 ∨ K2.

Proof. Assume that there is a plane graph G such that G ∈ A, and G contains an induced separating 4-cycle C with verticesa, b, c , and d, in that order. Let O be the set of vertices outside of C , and let I be the set of vertices inside of C . Without lossof generality, assume 1 ≤ |I| ≤ |O|. Note that, since G has diameter 2, every pair of non-adjacent vertices has a commonneighbour. For any pair x, y such that x ∈ I and y ∈ O, that common neighbour must be in C . We now prove two claims,which will be used later in the proof.

Claim 1. There cannot be two vertices x ∈ I and y ∈ O such that NG(x) ∩ C = NG(y) ∩ C = C unless G ∼= C4 ∨ K2. To provethis, we consider the graph induced by {x, y} ∪ C. If NG(x) ∩ C = NG(y) ∩ C = C, then this subgraph is isomorphic to C4 ∨ K2and every region is bounded by a 3-cycle. It follows that G ∼= C4 ∨K2, since the addition of any new vertex to this subgraph wouldcreate a separating triangle, which is a contradiction.

Claim 2. If u and v are consecutive vertices in C, then |NG(u)∩NG(v)∩O| ≤ 1. Otherwise, there are two vertices y, z ∈ NG(u)∩NG(v) ∩ O, and either {u, v, y} is a separating triangle with z inside, or {u, v, z} is a separating triangle with y inside. In eithercase, we have a contradiction.

In addition to these claims, we will also be using the results from Lemmas 3 and 4 frequently in this proof, specificallythat δ(G) ≥ 4 and G has no separating triangle. Also recall that for any vertices u and v in V (G) such that u � v and u = v,G + uv has a corner dominated by either u or v, and that corner is either u, v, or a common neighbour of u and v. We nowproceed by considering three cases based on the size of I and O.Case 1. |I| ≥ 2 and |O| ≥ 2.

Let {x, w} ⊆ I and {y, z} ⊆ O. Suppose some vertex in G is adjacent to every vertex of C . Without loss of generality, letx be that vertex. It follows that w lies inside one of the 3-cycles formed by x and two consecutive vertices on C . Since y liesoutside this cycle, we have a separating triangle, a contradiction. Therefore, no vertex in G is adjacent to every vertex on C .Since δ(G) ≥ 4, x has a neighbour in I, and y has a neighbour in O.

Now let G′= G + xy. In G′, x cannot dominate y since y has a neighbour in O. Similarly, y cannot dominate x, and any

corner in G′ is a common neighbour of x and y. Without loss of generality, assume that a is that corner, and a is dominatedby x in G′. It follows that {a, b, d} ⊆ N(x) and y ∼ a. Furthermore, x � c and NG(a) ∩ O = {y}. Hence, a � z.

Now let G′′= G + xz. As before, we can conclude that a common neighbour of x and z is a corner in G′′. Since x � c , it

follows that either b or d is the corner, and the corner is dominated by x or z. This is a contradiction since the corner will beadjacent to both a and c , but x � c and z � a.

4 S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11

Case 2. |I| = 1 and |O| ≥ 3.If some vertex y ∈ O has three neighbours in C , say a, b, and c , thenwe have a 4-cycle C ′ induced on {a, y, c, d}. Note that

x and b are inside C ′. Since G has no separating triangle, it follows that there are two vertices, say z, u ∈ O that are outsideC ′. This implies that C ′ has a separating 4-cycle with at least two vertices inside that cycle and at least two vertices outsidethat cycle. Hence, we have Case 1. Therefore, we may assume that no vertex in O has more than two neighbours in C . Sinceδ(G) ≥ 4, this also implies that every vertex in O has at least two neighbours in O.Case 2a. |O| ≥ 4.

Let I = {x} and {y, z, u, v} ⊆ O. Since δ(G) ≥ 4,NG(x) = {a, b, c, d}. Let G′= G + xy. Since y is adjacent to at most two

vertices in C , x cannot be dominated by y in G′. Furthermore, y has a neighbour in O, so y cannot be dominated by x in G′.Hence, a common neighbour of x and y in C is a corner in G′. Without loss of generality, assume a is a corner in G′. It followsthat amust be dominated by x, since |NG(y) ∩ C | ≤ 2. Therefore, NG(a) ∩ O = {y}.

By using a similar argument on each of G + xz,G + xu, and G + xv, we can conclude that each vertex of C is adjacentto exactly one vertex in O, and that these neighbours in O are distinct. Since G has diameter 2 and is planar, every vertexin O shares a common neighbour with x that is in C . Hence, |O| = 4. Since δ(G) ≥ 4, and each vertex in O has exactlyone neighbour in C , the set O induces K4. This, together with the disjoint paths from x to each of y, z, u, and v, completes aK5-subdivision in G, contradicting that G is planar.Case 2b. |O| = 3.

Let I = {x}. Since δ(G) ≥ 4, each vertex in O has exactly two neighbours in C , and O induces K3. Let G′= G + xy for any

y ∈ O. The corner in G′ must be a vertex in C . Suppose a is a corner in G′. Since y has only two neighbours in C , it follows thata is dominated by x, and NG(a) ∩ O = {y}. Without loss of generality, NG(b) ∩ O = {z} and NG(c) ∩ O = {u}. However, thisimplies that y, z, and u all have d as their second neighbour in C . Hence, {y, z, u, d} induces K4. Together with the internallydisjoint paths from x to these vertices, we see that G contains a K5-subdivision, which is a contradiction.Case 3. |I| = 1 and |O| = 2.

Since δ(G) ≥ 4, each vertex in O is adjacent to exactly three vertices in C , and the two vertices of O are adjacent. ByClaim 2, we can assume that NG(y) ∩ C = {a, b, c} and NG(z) ∩ C = {c, d, a}, where y, z ∈ O. It follows that G ∼= K2 ∨ C5.By Lemma 5, G is not in A since C5 is not in A. This is a contradiction.Case 4. |I| = 1 and |O| = 1.

Since δ(G) ≥ 4, it follows that G ∼= K2 ∨ C4, and our proof is complete. �

We are ready to prove the following theorem, which characterizes all planar graphs in A.

Theorem 7. If G ∈ A and G is planar, then G ∈ {C4, K2, K2 ∨ C4}.

Proof. Let G ∈ A, where G is planar. By Lemma 4, G contains no separating triangle. By Lemma 6, if G contains an inducedseparating 4-cycle, then G ∼= C4 ∨ K2. We will therefore assume that G contains no separating 4-cycle.

Assume G is a plane graph such that G ∈ {K2, C4}. By Lemma 3, G contains an induced 4-cycle C , and hasminimumdegreeat least 4. Let C have vertex set {a, b, c, d}. Without loss of generality, we may assume there are no vertices inside C , sinceG has no induced separating 4-cycle.

Let G′= G + ac . Recall that G has diameter 2. We also note that G′ has diameter 2. Otherwise, G′ would have diameter

1, which means G is a complete graph minus an edge. This is a contradiction since the only graph of that type that is notcop-win is K 2.

Now, since G′ is cop-win, there is a cop-win ordering. Let x be the first vertex in such an ordering. It follows that x isdominated by either a or c . (Otherwise, xwould have been a corner in G.) Without loss of generality, assume x is dominatedby a in G′. This is means x = c , or x is a common neighbour of a and c. We consider three cases: (1) x = c , (2) x = b or x = d,and (3) x ∼ a and x ∼ c , but x is neither b nor d.Case 1. x = c .

It follows that NG′ [c] ⊆ NG′ [a] and, therefore, NG(c) ⊆ NG(a). Since degG(c) ≥ 4, there are two vertices, v and w, thatare common neighbours of c and a, but are not on C . Since there are no vertices inside C , we may assume that b, v, w, and dappear in clockwise order around c. Since v and d are on different sides of the cycle with vertex set {a, b, c, w}, v and d arenot adjacent. Therefore, the vertex set {a, v, c, d} induce a 4-cycle in G, with vertices both inside and outside this cycle. Thiscontradicts the fact that G has no induced separating 4-cycle.Case 2. x = b or x = d.

Without loss of generality, assume x = b. It follows that NG′ [b] ⊆ NG′ [a]. Since degG(b) ≥ 4, there are two vertices, v andw, that are common neighbours of b and a, but not on C . Since there are no vertices inside C , we may assume that a, v, w,and c appear in clockwise order around b. Since the vertex set {a, b, w} induces a triangle in G, with v on one side and c onthe other, we have a separating triangle, which is a contradiction.Case 3. x is a common neighbour of c and a, but x = b and x = d.

Recall that NG′ [x] ⊆ NG′ [a]. As before, we have degG(b) ≥ 4. So, there is at least one vertex, w, such that w ∼ b, butw ∈ {a, c, x}. We note that b ∼ x, otherwise the vertex set {a, x, c, b} induces a separating 4-cycle, with w on one side andd on the other, which is a contradiction. However, this means either {a, x, b} or {b, x, c} induces a separating triangle, withw and d on different sides (see Figs. 1–3).

Hence G ∈ {C4, K2}. �

S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11 5

Fig. 1. Case 1.

Fig. 2. Case 2.

Fig. 3. Case 3.

Recall thatA represents the set of all CECA graphs that contain no corners. Furthermore, for any corner u in a CECA graph,there is another vertex, v, in that graph such that N[u] = N[v].

For a given graph G and a vertex u in V (G), let Gu be the graph such that V (Gu) = V (G) ∪ {v} for some v ∈ V (G), andE(Gu) = E(G) ∪ {vw|w ∈ N[u]}. We will say that Gu is obtained from G by duplicating u. Therefore, for any CECA graph,G, and any vertex u in G, the graph obtained by duplicating u is also CECA. By Theorem 7, if G ∈ A and G is planar, thenG ∈ {C4, K2, K2 ∨ C4}. Therefore, all planar CECA graphs can be obtained by repeatedly duplicating vertices of graphs in theset {C4, K2, K2 ∨ C4}, provided planarity is maintained.

Duplicating vertices in K2 results in the disjoint union of two complete graphs. So, to maintain planarity, we will berestricted to complete graphs on at most four vertices. Duplication of any vertex of K2 ∨ C4 results in a nonplanar graph.There are two planar graphs that result from duplication of vertices of C4. These are the centre two graphs in Fig. 4. Any newgraphs that result from further duplications are nonplanar. We, therefore, have the following characterization of all planarCECA graphs.

Theorem 8. The graph G is a planar CECA graph if and only G is isomorphic to one of the following:1. Km,n for any m and n such that 1 ≤ m, n ≤ 42. one of the four graphs in Fig. 4

4. Edge deletion

In this section, we consider criticality with respect to edge deletion in planar graphs; that is, the deletion of any edgechanging a planar graph with copnumber 2 into a graph with copnumber 1. Let B be the set of graphs that are not cop-win,but the deletion of any edge results in a cop-win graph. We refer to these as CECD graphs. The main result is given in Theo-rem 18, which shows that the set of CECD graphs is exactly the set {Ck, Ck ∨ K2 | k ≥ 4} where Ck is a cycle on k vertices. Inworking towards this result, we find specific properties of the corners that result from edge deletion in a planar CECD graph(Lemmas 12 and 14), and verify that every CECD graph has the wheel on 5 vertices is an induced subgraph (Lemma 15).

6 S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11

Fig. 4. Planar CECA graphs.

Fig. 5. W4 .

Recall that all graphs in B have copnumber 2. As before, for any G ∈ B, G has no corners. However, unlike in A, this isnot a restriction imposed to simplify the problem.

Proposition 9. If G ∈ B , then G contains no corner.

Proof. Suppose G ∈ B and G has a corner, x, which is dominated by y. If x has degree 1, then G− xy consists of the subgraphG − {x} and the isolated vertex x. Since c(G − {x}) = c(G), it follows that c(G − xy) = 3, which is a contradiction. Hence,x has degree at least 2. Let z ∈ NG(x) where z = y. It follows that G − xz is cop-win and, since x is also a corner in G − xz,there is a cop-win ordering for G − xz that starts with x. However, this ordering is also a cop-win ordering in G, which is acontradiction. �

If G ∈ B, then G′= G − uv is cop-win for any u, v ∈ V (G), u ∼ v, u = v. Since G has no corners and only the

neighbourhoods of u and v are affected by the edge deletion, it follows that G′ has only u or v (possibly both) as a corner.The wheel on n vertices consists of a cycle on n − 1 vertices and a single vertex, called the hub, that is adjacent to every

vertex in the cycle. It is denoted Wn−1. We are interested in the wheel on five vertices, denoted W4 and shown in Fig. 5. InW4, the vertex of degree 4 is the hub.

We will show in Lemma 15 that every planar graph G in B is either a cycle or contains W4 as an induced subgraph. Thekey to locating an induced W4 in G is to first find a vertex of degree 4 that becomes a corner when one of its incident edgesis removed.

We nowpresent a series of three lemmas that, togetherwith their corollaries, will establish the existence of such a vertexof degree 4 in G.

Lemma 10. Let G ∈ B and v ∈ V (G). If degG(v) = 2 or degG(v) = 3, then v does not lie on a 3-cycle in G.

Proof. Assume that v lies on a 3-cycle in G. Since v is not a corner, degG(v) > 2.Assume deg(v) = 3 and NG(v) = {x, y, z}. If v lies on a 3-cycle, then two of x, y and z are adjacent. Assume x ∼ y. Since v

is not a corner in G, it follows that x � z and y � z. Now let G′= G− xy. Without loss of generality, assume x is a corner in G′

dominated byw. Note thatw = v, otherwise xwould be a corner inG dominated by v. Since x ∼ v, it follows thatw ∼ v inG′

(and also in G). However, this impliesw = y orw = z, which is a contradiction since neither y nor z is adjacent to x in G′. �

Corollary 11. If G ∈ B and degG(v) = 3, then for any u ∈ N(v), v is not a corner in G − uv.

Proof. This follows from Lemma 10 since any vertex w that dominates v would form a 3-cycle together with v and a vertexfrom N(v) \ {u}. �

Lemma 12. Let G ∈ B and assume G is planar. If x is a corner in G − xy, then degG(x) = 2 or degG(x) = 4.

Proof. Assume xy is some edge in G, x is a corner in G− xy, and degG(x) ≥ 5. Since x is not a corner in G, there is a vertex inz ∈ N(x) − {y} such that z dominates x in G′, but z � y.

Let a, b, and c be three distinct vertices in NG(x) − {y, z}. It follows that z is adjacent to a, b and c. Now let G′′= G − xz.

Either x or z is a corner in G′′. If x is a corner, then there is a vertex w such that w dominates x in G′, but w � z. Hence,

S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11 7

Fig. 6. Case 1.

w ∈ {a, b, c}, but w is adjacent to each of a, b, and c. (It is possible that w = y.) However, this means that the subgraphinduced on {x, z, w, a, b, c} contains a K3,3. This contradicts the planarity of G. Hence, z is a corner in G′, and there is a vertexw such that w dominates z in G′ and w � x. This again implies that each of a, b, and c is adjacent to w and we again haveK3,3 as a subgraph, which is a contradiction. Therefore, degG(x) ≤ 4. By Corollary 11, degG(x) = 3, and since x is not a cornerin G, degG(x) = 1. The result follows. �

Corollary 13. Let G ∈ B and assume G is planar. For any edge e in G, at least one endpoint of e has degree 2 or 4.

Proof. This follows since for any edge xy in G, either x or y is a corner in G − xy. �

Lemma 14. Let G ∈ B and assume G is planar. If G is not a cycle, there exists an edge e such that at least one endpoint of e hasdegree 4, and that vertex is a corner in G − e.

Proof. Assume xy is an edge of Gwhere G ∈ B, but G is not a cycle. It follows that either x or y is a corner in G− xy. Assumex is the corner. By Lemma 12, degG(x) = 2 or degG(x) = 4. If x has degree 4, then the proof is complete. Hence, we mayassume that x has degree 2. Let NG(x) = {y, z} where y = z.

Form a maximal path in G starting with the edge xy so that all interior vertices are degree 2 in G. Let P be that path, butlabelled so that it ends with the edge yx. That is, P = v1, . . . , vℓ where vℓ = x. Let Q be a similar maximal path starting withthe edge xz, and let Q = vℓvℓ+1 . . . vk. Note that, since G is not a cycle, neither v1 nor vk has degree 2. Furthermore, since Gis not cop-win, neither has degree 1. Also note that we may have v1 = vk.

Now consider the graph G − v1v2. It has a cop-win ordering that begins with the sequence v2, v3, . . . , vk−1. The nextvertex in this ordering must be either v1 or vk. Otherwise, that vertex would have been a corner in G. In other words, eitherv1 or vk is a corner in G − {v2, v3, . . . , vk−1}.Case 1: Assume v1 = vk. Since y = z, and, by Lemma 10, y and z are not adjacent, we have a cycle on {v1, v2, . . . , vk−1}wherek ≥ 5. Also note that, since degG(v1) ≥ 3, v1 is a cut vertex in Gwith the path v2v3 . . . vk−1 being one of the components ofG − {v1}. (See Fig. 6.)

Letw be the vertex that dominates v1 in G−{v2, v3, . . . , vk−1}. Note thatw appears in a different component of G−{v1}

than v2, v3, . . . , vk−1.NowconsiderG−v1w. Since k ≥ 5, v2 and vk−1 are not adjacent. Therefore, v1 is not a corner inG−v1w. So,w is a corner in

G−v1w, and in the cop-win ordering ofG−v1w, v1 must appear before any of v2, v3, . . . , vk−1. This is impossible, however,since no neighbour of v1 is adjacent to both of v2 and vk−1. Therefore, G − v1w is not cop-win, which is a contradiction.Case 2: Assume v1 = vk and v1 is a corner in G−{v2, v3, . . . , vk−1}. Since NG(v1)∩{v2, v3, . . . , vk−1} = {v2}, it follows thatv1 is also a corner in G − v1v2. Since degG(v1) = 2, by Lemma 12, degG(v1) = 4, and the proof is complete.Case 3: Assume v1 = vk and vk is a corner in G−{v2, v3, . . . , vk−1}. As in the previous case, it follows that vk is also a cornerin G − vk−1vk. Since degG(vk) = 2, then by Lemma 12, degG(v1) = 4, and the proof is complete. �

Now that we can locate the required vertex of degree 4, we can use it to locate an inducedW4 in any planar CECD graphthat is not a cycle.

Lemma 15. Assume G ∈ B and G is planar. If G is not a cycle, then G contains an induced W4 such that the hub of W4 hasdegree 4 in G.

Proof. Let xy be an edge such that degG(x) = 4 and x is a corner in G − xy. We know such an edge exists by Lemma 14. LetNG(x) = {a, b, c, y}. Without loss of generality, assume that x is dominated by b in G − xy. It follows that b ∼ a and b ∼ c.

8 S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11

Note that b � y, otherwise xwould also be a corner in G. We now let G′= G− ab. We now consider the two possible cases:

a is a corner in G′ or b is a corner in G′.Case 1: Assume a is a corner in G′. Therefore, by Lemma 12, a has degree 2 or 4 in G. Since a lies on the 3-cycle with vertexset {a, b, x} in G, by Lemma 10, degG(a) = 4. Furthermore, in G′, a is dominated by a vertex adjacent to both a and x, but notadjacent to b. It follows that y must dominate a in G′, y ∼ a and y � b.Case 1a: Assume x is a corner in G − xa. It follows that xmust be dominated by c , since all other neighbours of x are adjacentto a. Therefore, c ∼ y. Since c ∼ b, c ∼ y and c � a, we conclude that {a, b, c, y} induces a 4-cycle. Since each of thesevertices is adjacent to x, we have an inducedW4 in G in which x is the hub and degG(x) = 4.Case 1b: Assume a is a corner in G−xa dominated by vertexw. Therefore, in G, w is adjacent to a, b, and y, but not x. Therefore,w, b and y are distinct vertices, {w, b, x, y} induces a 4-cycle, and all of these vertices are adjacent to a. Hence, we have aninducedW4 in G in which a is the hub. Since degG(a) ≥ 4 and a is a corner in G∗, by Lemma 12, degG(a) = 4.Case 2: Assume b is a corner in G′, but a is not. Recall that, in G, b is adjacent to a, c , and x, but b � y. It follows from Lemma 12that degG(b) = 4, so b has exactly one neighbour that is not in NG[x]. Call that neighbour w. Now b must be dominated inG′ by a vertex that is adjacent or equal to each of x, b and c , but not a. It follows that the required vertex is adjacent to bothx and b in G′. The only vertex that fits that criteria is c . Therefore, b is dominated by c in G′. Hence, c is adjacent to w. Nowconsider G − bc .Case 2a: Assume c is a corner in G − bc. It follows that c must be dominated by a vertex adjacent to x, but not b. Therefore, ydominates c in G− bc , and y is adjacent to c and w. It follows that {x, y, w, b} induces a 4-cycle, and all of these vertices areadjacent to c . Hence, we have an inducedW4, and by Lemma 12, degG(c) = 4.Case 2b: Assume b is a corner in G − bc. It follows that b must be dominated by a vertex adjacent to x, but not adjacent to c.Therefore, b is dominated by a in G−bc. Hence, a is not adjacent to c , but a is adjacent tow. So, {a, x, c, w} induces a 4-cyclewith every vertex in this set adjacent to b. Hence, we have an inducedW4. Again, by Lemma 12, degG(b) = 4.

In all cases we have shown that the graph G contains an inducedW4 in which the hub has degree 4 in G. �

Let H = P4 ∨ K2 where P4 is a path on 4 vertices. Call the two interior vertices of the original P4 the interior vertices of H .(See Fig. 7.)

Lemma 16. Let G ∈ B and assume G is planar. If G is not a cycle, then H is a subgraph of G where the interior vertices of H havedegree 4 in G.

Proof. Assume G is not a cycle. By Lemma 15, G contains an induced W4, the hub of which has degree 4. Let x be the hubof an induced W4 in G, and let NG(x) = {a, b, c, d}. Assume the vertices of NG(x) appear in this order on the 4-cycle theyinduce. Now consider G − ab. Without loss of generality, assume that a is a corner in G − ab. By Lemma 12, degG(a) = 4.Furthermore, a is dominated by d in G − ab. It follows that a and d have a common neighbour w such that w ∈ N[x].

Next consider G − bc. Either b or c is a corner in this graph.Case 1: Assume b is a corner in G − bc. It follows that degG(b) = 4, and b is dominated by a in G − bc. Therefore, b has aneighbour that is also adjacent to a, but not in N[x]. Since degG(a) = 4, this vertex must be w. We now have the requiredsubgraph, H , as seen in Fig. 8. Furthermore, degG(a) = degG(x) = 4.Case 2: Assume c is a corner in G − bc. It follows that degG(c) = 4, and c is dominated by d in G − bc. If c is adjacent to w,then we have the subgraph H . (Interchange the (ordered) pair a, c with the (ordered) pair d, b in Case 1.) Otherwise, c hasa neighbour v such that v ∈ N[x] and v = w. Since v is adjacent to d, we know that the degree of d is at least 5. Hence, byLemma 12, c is a corner in G − cd, and c is dominated by b. Therefore, v is also adjacent to b, and we have the subgraph H ,as seen in Fig. 9. Furthermore, degG(c) = degG(x) = 4. �

Let Dk = Ck−2 ∨ K2 where k ≥ 6, as seen in Fig. 10. (Note that D6 is the tetrahedron.) We now show that these planargraphs are contained in B.

Lemma 17. For any k ≥ 6, Dk is in B .

Proof. Consider the graph Dk for any k ≥ 6. Note that Dk is not cop-win since it contains no corner. Let x and y be two non-adjacent vertices of maximum degree. (If k > 6, then this is a unique (unordered) pair of vertices. If k = 6, there are threesuch pairs, but by symmetry, all choices are analogous.) Then G− {x, y} is a cycle on k− 2 vertices. Let {v1, v2, . . . , vk−2} bethe vertex set of that cycle, with vertices appearing in that order. We now have two general cases to consider with respectto edge deletion: (1) deleting an edge from {x, y} to {v1, v2, . . . , vk−2} or (2) deleting an edge of the type vivi+1 on the cycle.Case 1: Consider G− xv1. This graph is cop-win with cop-win ordering v1, v2, . . . , vk−3, x, vk−2, y. Note that in this orderingv1 is dominated by y, and vi is dominated by vi+1 for all i = 2, 3, . . . , k − 3.Case 2: Consider G − v1v2. This graph is also cop-win with cop-win ordering v2, v3, . . . , vk, x, v1, y. Note that vi is againdominated by vi+1 for all i = 2, 3, . . . , k − 3, and vk−2 is dominated by v1. �

It turns out that the set of graphs Dk, k ≥ 6 together with the cycles of length at least 4 constitute all of the graphs in B.This characterization of B is now given.

S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11 9

Fig. 7. H .

Fig. 8. Case 1.

Fig. 9. Case 2.

Fig. 10. Dk .

Theorem 18. A graph G is a planar CECD graph if and only if G ∼= Ck for some k ≥ 4 or G ∼= Dk for some k ≥ 6.

Proof. (⇐) It is straightforward to see that all cycles on at least four vertices are both planar and inB. By Lemma 17,Dk ∈ Bfor all k ≥ 6, and the planarity of Dk is obvious.

(⇒) Assume G is both planar and in the set B. If G is a cycle, then G ∼= Ck for some k ≥ 4, otherwise G is cop-win.Assume G is not a cycle. By Lemma 16, G has H as a subgraph. Let u and v be the two interior vertices of subgraph H , and

let the outside cycle have vertex set {a, b, c, d}, as in Fig. 11. It also follows from Lemma 16 that we may assume degG(u) =

degG(v) = 4. We know that degG(b) ≥ 4 and degG(d) ≥ 4, since b and d are not corners in G. We will now associate H withthis particular subgraph.

Assume G has exactly six vertices. It follows that V (G) = V (H) and E(H) ⊂ E(G) (since H is cop-win). When examiningcandidates for edges in E(G)−E(H), we can eliminate any edges incidentwith u or v since each has degree 4 inG.We can alsoeliminate ac , since this would result in both u and v being corners in G. This leaves only bd. Therefore, E(G) = E(H) ∪ {bd}and G ∼= D6.

We now proceed by induction on the number of vertices of G. We assume that any planar graph on k ≥ 6 vertices in Bthat is not a cycle is isomorphic to Dk.

10 S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11

Assume G is a planar graph in B that is not a cycle and has k + 1 vertices, k ≥ 6. It follows that G has a subgraph iso-morphic to H with vertex set {u, v, a, b, c, d}, as before, with degG(u) = degG(v) = 4. We claim that vertices b and d arenon-adjacent in G.

To prove this claim, suppose b and d are adjacent in G, and let K denote the subgraph induced on V (H). Note that K ∼= D6,since any additional edges would violate either G having no corners or both u and v having degree 4 in G.

Since G is connected, there is an edge e such that e ∈ E(G) − E(K). It follows that G′= G − e is a cop-win graph, and has

a cop-win ordering v1, . . . , vn. Suppose vi is the first vertex in the ordering such that vi ∈ V (K). Therefore, vi is a corner inGi = G − {v1, . . . , vi−1}. Since NGi(u) = NK (u) = NG(u), it follows that vi = u. Otherwise u would be a corner in G. By asimilar argument, vi = v.

Let x ∈ {a, b, c, d}. Since x is a corner in Gi, but not K (since K is not cop-win), x is dominated by some vertex y ∈

V (G)−V (K). This, however, is impossible since NGi(x) contains at least one of u and v, and each of u and v has no neighbouroutside of K . We therefore conclude that b and d are non-adjacent in G. It follows that G has H as an induced subgraph.

Next, we claim that either a or c must have degree at least 5. To prove this, Assume degG(a) = degG(c) = 4. It followsthat G− uv is cop-win, and there is a cop-win ordering that begins with u, followed by v. Let G′ be the graph resulting fromremoving vertices u and v from G − uv. It follows that G′ contains a 4-cycle on {a, b, c, d} where degG′(a) = degG′(c) = 2.However, we see that in any cop-win ordering of G′, both a and c must be removed before b or d can be a corner, but neithera nor c can be a corner until either b or d is removed. This contradicts the fact that G′ is cop-win. We may therefore assume,without loss of generality, that degG(a) ≥ 5.

We know that degG(b) ≥ 4. Since a ∼ b in G, by Lemma 12, degG(b) = 4, and b is a corner in G − ab. Since b must bedominated in G − ab by a vertex adjacent to v, it follows that b is dominated by c. Let w ∈ N(b) − V (H). It follows thatc ∼ w and degG(c) ≥ 5. By a similar argument, b is a corner in G − bc dominated by a, and a ∼ w. Thus, we have the graphseen in Fig. 12 as a subgraph of G.

Consider the vertex u. Assumewe remove u and then add the edge dv to obtain the newgraphG′. (G′= (G−{u})+dv.)We

claim thatG′ is also inB. We do so by showing thatG′ is not cop-win, and the removal of any edge results in a cop-win graph.Assume G′ is cop-win, and there is a cop-win ordering of the vertices v1, . . . , vn. Let vi be the first vertex in S = {v, b, a,

w, c, d} to appear in this ordering. Recall that degG(v) = degG(b) = 4, b � d, and a � c . Given these conditions, it isimpossible to find vi ∈ S such that vi is a corner in Gi. Therefore, G′ is not cop-win.

We now verify that G′− e is cop-win for any edge e in G′.

Case 1: Assume e is some edge that is in G, but not in the induced subgraph H . It follows that e is also an edge in G′. The graphG−e is cop-win and has a cop-win ordering v1, v2, . . . , vn. Suppose vi is the first vertex of V (H) to appear in this ordering. Itis straightforward to see that vi = b or vi = d. Without loss of generality assume that vi = b. In other words, b is a corner inGi. It then follows that v is a corner in Gi −{b}, and u is a corner in Gi −{b, u}. We can therefore assume that vi = b, vi+1 = vand vi+2 = u. It follows that v1, v2, . . . , vi, vi+1, vi+3, . . . , vn is a cop-win ordering of G′

− e. Hence, G′− e is cop-win.

Case 2: Assume e = av or e = cv. We know that G − ua is cop-win, and u is a corner in that graph. Removing u from G − uaresults in the graph G− {u}, which has v as a corner. It follows that G− {u, v} is cop-win. Now consider G′

− av. Vertex v isa corner in this graph. With its removal we have (G′

− av) − {v} = G− {u, v}, which is cop-win. Hence, G′− av is cop-win.

Similarly G′− cv is cop-win.

Case 3: Assume e = bv or e = dv. The graph G − du is cop-win with u as a corner. Removing u from G − du results in thecop-win graph G − {u} in which v is a corner. It follows that G − {u, v} is a cop-win graph. The graph G′

− dv has v as acorner, and the removal of v from G′

− dv results in the graph G′− {v} = G − {u, v}, which is cop-win. Hence, G′

− dv iscop-win. Similarly, G′

− bv is cop-win.Case 4: Assume e = ab, e = ad, e = cb or e = cd. The graph G − ab is cop-win, and b is a corner. Removal of b from G − abresults in G − {b} in which v is a corner. In G − {b, v}, u is a corner. Therefore, G − {b, u, v} is cop-win. Now, in G′

− ab, bis a corner. Its removal gives G′

− {b} in which v is a corner. Since G′− {b, v} = G − {b, u, v}, it follows that G′

− {b, v} iscop-win, and G′

− ab is cop-win. Similarly, G′− cb,G′

− cd, and G′− ad are all cop-win.

It follows that G′ is also in B. Obviously, G′ is planar, and G′ is not a cycle. Therefore, G′ ∼= Dk. It follows from the con-struction that G ∼= Dk+1. Hence, every planar graph in B that is not a cycle is isomorphic to Dk for some k ≥ 6. The resultfollows. �

5. Further research

Some possibilities for future investigations are summarized in the following problems.

Problem 1. When discussing planar graphs, edge contraction and edge subdivision are of particular interest since planarityis preserved. What can we say about planar graphs whose copnumber changes from 2 to 1 under one of these operations?

Problem 2. We have determined when copnumber changes from 2 to 1 under the operations of edge addition and deletionin planar graphs. What can be said about planar graphs whose copnumber changes from 3 to 2 under these operations?

Problem 3. The original motivation for looking at cop-win critical graphs was to aid in finding a characterization of 2-cop-win graphs. Given any 2-cop-win graph, is that graph a subgraph of some CECA (CECD) graph?

Problem 4. Can this lead to a characterization of planar graphs with copnumber 2?

S.L. Fitzpatrick / Discrete Mathematics 329 (2014) 1–11 11

Fig. 11. H .

Fig. 12. Resulting subgraph.

References

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