# Maker-Breaker domination game

We introduce the Maker-Breaker domination game, a two player game on a graph. At his turn, the first player, Dominator, select a vertex in order to dominate the graph while the other player, Staller, forbids a vertex to Dominator in order to prevent him to reach his goal. Both players play alternately without missing their turn. This game is a particular instance of the so-called Maker-Breaker games, that is studied here in a combinatorial context. In this paper, we first prove that deciding the winner of the Maker-Breaker domination game is PSPACE-complete, even for bipartite graphs and split graphs. It is then showed that the problem is polynomial for cographs and trees. In particular, we define a strategy for Dominator that is derived from a variation of the dominating set problem, called the pairing dominating set problem.

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## 1 Introduction

Since their introduction by Erdős and Selfridge in [9], positional games have been widely studied in the literature (see [11] for a recent survey book on the topic). These games are played on an hypergraph of vertex set , with a finite set of hyperedges. The set is often called the board of the game, and an element of a winning set. The game involves two players that alternately occupy a previously unoccupied vertex of . The winner is determined by a convention: in the Maker-Maker convention, the first player to occupy all the vertices of a winning set is the winner. Such games may end in a draw, as it is the case in Tic-Tac-Toe. In the Maker-Breaker convention, the objectives are opposite: one player (the Maker) aims to occupy all the vertices of a winning set, whereas Breaker wins if she occupies a vertex in every winning set. In view of the complexity of solving both kinds of games, Maker-Breaker instances are generally more considered in the literature as by definition, there is always a winner. In addition, rulesets of such games are often built from a graph. For example, one can mention the famous Shannon switching game (popularized as the game Bridg-it) [15], where, given a graph and two particular vertices and , the board corresponds to , and the winning sets are all the subsets of that form a path in . In the Hamiltonicity game [6], the winning sets are all the sets of edges containing an Hamiltonian cycle.

In view of such examples, converting a graph property into a -player game is a natural operation. Hence it is not surprising that it has also been done for dominating sets. More precisely, several games having different rulesets and known as domination game have been defined in the literature. For example, in [1, 10], a move consists in orienting an edge of a given graph and the two players try to maximize (resp. minimize) the domination number of the resulting digraph. In [5], the rules require two colors during the play. In [3], the domination game is defined in a sense where the players both select vertices and try to maximize (resp. minimize) the length of the play before building a dominating set. Since then, this version has become the standard one for the domination game, with regular progress on it [4, 8, 14, 13]. However, among the different variants of the domination game , the natural Maker-Breaker version (in the sense of Erdős and Selfridge) has never been considered in the literature. In this paper, we consider the so-called Maker Breaker Domination game, where, given a graph , the board is the set , and is the set of all the dominating sets of . In other words, the two players alternately occupy a not yet occupied vertex of . Maker wins if he manages to build a dominating set of , whereas Breaker wins if she manages to occupy a vertex and all its neighbors. In what follows and in order to be consistent with the standard domination game, Maker will be called Dominator, and Breaker will be the Staller.

When dealing with Maker-Breaker games, there are two main questions that naturally arise:

• Given a graph , which player has a winning strategy for the Maker-Breaker domination game on ?

• If Dominator has a winning strategy on , what is the minimum number of turns needed to win?

The current paper is about the first question. In the next section, we give definitions for the different cases about the winner, together with first general results. Section 3 deals with the algorithmic complexity of the problem, where the pspace-completeness is proved. In Section 4, a so-called pairing strategy is given, yielding a strategy for Dominator in graphs having certain properties. The last section is about graph operators that lead to polynomial strategies on trees and cographs.

## 2 Preliminaries

A position of the Maker-Breaker domination game is denoted by a triplet , where is a set of vertices, is a set of edges on and is a function . In other words, the function allows to describe any game position encountered during the play. If, for all in , , then is said to be a starting position. In this case, we will identify with the graph . At his turn, Dominator (respectively Staller) chooses one vertex with and changes its value to (resp. ). When there is no more Unplayed vertex, either the set of vertices forms a dominating set, and Dominator wins, or there is one vertex for which all its closed neighboorhood has value , and Staller wins. In the latter case, we say that Staller isolates . Note that whenever is a dominating set or a vertex has been isolated by Staller, the winner is already determined and cannot change, since the two conditions are complementary. Thus we will often consider that the game stops when one of the two conditions holds.

The Maker-Breaker domination game is a finite game with perfect information and no draw. Thus, there is always a winning strategy for one of the player. There are four cases - also called outcomes - to characterize the winner of the game, according to who starts. We define , , and as the different possible outcomes for a position of the Maker-Breaker domination game.

###### Definition 1.

A position has four possible outcomes:

• if Dominator has a winning strategy as first and second player,

• if Staller has a winning strategy as first and second player,

• if the next player (i.e., the one who starts) has a winning strategy,

• otherwise (i.e., the second player wins).

Note that for proximity reasons, the notion of outcome and the last two notations are derived from combinatorial game theory

[17]. In addition, the outcome of is denoted .

The following proposition is a direct application of a general result on Maker-Breaker games stated in [2, 11]. It ensures that the outcome never occurs. For the sake of completeness, we here give a proof of this result adapted to our particular case.

###### Proposition 2 (Imagination strategy).

There is no position of the Maker-Breaker domination game such that .

Proof. Assume there is position of the Maker-Breaker domination game such that . This means in particular that Dominator wins playing second on . We next give a winning strategy for Dominator as first player, which will imply that , a contradiction.

The strategy for Dominatoras first player is the following. He first plays any vertex unplayed and then imagines he did not. He thus considers himself as the second player, seeing this vertex as an extra vertex. Whenever his winning strategy (as a second player) requires to play the extra vertex, he plays any other unplayed vertex , and considers as the new extra vertex. If Dominator was winning before all the vertices were chosen, he still wins no later than his last move in the game where he was playing second. Otherwise, when Staller chooses the last vertex of the graph, her strategy asks her to play the extra vertex since it is the only one available in the imagined game, but it means that Dominator had already won on the previous turn.

Note that this proposition is valid for any position of the game (and not only for starting positions). In other words, it ensures that a player has no interest to miss his/her turn. Figure 1 gives an example of graphs for the three remaining outcomes.

According to the three possible outcomes of a position, we now introduce an order relation on the outcomes derived from combinatorial game theory: . This allows us to state the following proposition.

###### Proposition 3.

Let be a position of the Maker-Breaker domination game and let be another position, with . Then .

Proof. A reformulation of the proposition is that if Dominator has a winning strategy on , then he also has a winning strategy on .

Assume Dominator has a winning strategy on . A winning strategy for Dominator on is to apply the same strategy as on . Indeed, for every possible sequence of moves of Staller, Dominator is able to dominate . Since every edge of is also in , Dominator is also able to dominate .

In other words, adding edges to a position can only benefit Dominator, and removing edges can only benefit Staller. Note that this property does not hold in the standard domination game.

Another result can be derived from Maker-Breaker games. The following theorem is a well known result from the early studies about positional games.

###### Theorem 4 (Erdős-Selfridge Criterion [9]).

Given a Maker-Breaker game on an hypergraph , if then Breaker wins on playing second.

In order to apply this theorem to the Maker-Breaker domination game, we need to consider a reverse version of it. Indeed, as the set corresponds to the dominating sets of , the sizes of the winning sets are not easy to control. Thus, we can also consider the Maker-Breaker domination game as the Maker-Breaker game where is the set of the closed neighborhoods of every vertex of . In that case, Dominator is the Breaker, and Staller is the Maker. Now Theorem 4 can be applied on this game:

###### Proposition 5.

Let be a starting position of the Maker-Breaker domination game and let be the minimum degree of . If then Dominator has a winning strategy for the Maker-Breaker domination game on playing second.

Proof. As stated before, the Maker-Breaker domination game on is a Maker-Breaker game played on where is the set of the closed neighborhoods of , and Staller plays the role of Maker in this game. Applying the Erdős-Selfridge Criterion, we know that if then Dominator has a winning strategy. For all in , we have , hence . Thus if then Dominator has a winning strategy.

This result can be applied to prove that some families of graphs are (e.g. -regular graphs having ). In addition, it also suggests that highly connected graphs are more advantageous for Dominator.

## 3 Complexity

In this section, we consider the computational complexity of deciding whether a game position of the Maker-Breaker domination game is , , or . First, remark that in the general case, deciding the outcome of a Maker-Breaker game is pspace-complete. Indeed, this game exactly corresponds to the game pos-cnf that was proved to be pspace-complete in [16].

pos-cnf is played on a formula in conjunctive normal form, with variables , where each variable is positive, that is with clauses . Two players, Prover and Disprover, alternate turns in choosing a variable that has not been chosen yet. When all variables have been chosen, variables chosen by Prover are set to true, while variables chosen by Disprover are set to false. Prover wins if is true under this valuation and Disprover wins otherwise. Without loss of generality, we can consider that each variable appears in the formula, otherwise we consider the formula . Clearly, any Maker-Breaker game is equivalent to a pos cnf game, as corresponds to the set of variables, and the winning sets correspond to the clauses. Prover has the same role as Breaker, and Maker has the role of Disprover.

The complexity of this game remains pspace-complete when reduced to instances of the Maker-Breaker domination game:

###### Theorem 6.

Deciding the outcome of a Maker-Breaker domination game position is pspace-complete on bipartite graphs.

Proof. We reduce the problem from pos-cnf. Let be a positive formula in conjunctive normal form using variables .

We build a bipartite graph from as follows. There is one vertex for each variable and two vertices for each clause:

 V={xi|1≤i≤n}∪{ckj|1≤j≤m,0≤k≤1},

and an edge between a variable vertex and a clause vertex (with ) if the variable appears in the clause . Figure 2 shows an example of such a construction, from the example where .

We now show that Prover has a winning strategy in as first player (respectively second player) if and only if Dominator has a winning strategy in as first (resp. second) player.

Assume Prover has a winning strategy in . We first consider the case where Prover is the last player to play in pos-cnf (i.e.

is odd if Prover plays first and even if Prover plays second).

Dominator builds his strategy on as follows:

• If Prover and Dominator are starting the game, Dominator chooses the vertex corresponding to the variable played by Prover in his wining strategy.

• Whenever Staller chooses a vertex , Dominator answers by choosing the vertex .

• Whenever Staller chooses a vertex , Dominator assumes Disprover chose the variable . Then he answers by choosing the vertex corresponding to the variable played by Prover in his wining strategy.

This last step is always possible since we assume that Prover is playing the last in the pos-cnf game. When all vertices are chosen, since Prover was winning in , for each vertex , there is a neighbor that was chosen by Dominator. As all variables are in a clause, and for each , Dominator chose either or , all vertices of the form are also dominated by Dominator’s choice of vertices. Hence Dominator wins the game.

If Prover is not the last player to move, Dominator follows the same strategy but when Staller is playing the last variable vertex, Dominator cannot answer a variable vertex. Then he can play any clause variable and imagines he did not, as in the Imagination strategy of Proposition 2, and goes on according to his strategy. If Staller answers the second vertex of the clause at some point, then Dominator chooses another unplayed clause vertex. At the end, we will also have, as before, one vertex of each clause plays by Dominator and the same conclusion holds.

Assume now Disprover has a winning strategy in . The strategy for Staller is exactly the same:

• Whenever Disprover’s strategy requires to choose a variable , Staller chooses the vertex .

• Whenever Dominator chooses a vertex , Staller answers by choosing the vertex .

• Whenever Dominator chooses a vertex , Staller assumes Prover chose the variable .

If the last step is not possible, this means that all the variables are chosen. Then, there exists a clause for which no variables are chosen by Prover. If and are already played, one of them has been chosen by Staller, and thus is isolated. Otherwise, Staller chooses and isolates it. In both cases, Staller wins.

###### Corollary 7.

Deciding the outcome of a Maker-Breaker domination game position is pspace-complete on chordal graphs, and also in particular on split graphs.

Proof. The proof of Theorem 6 remains valid by adding edges between the variable vertices. In particular, if they form a clique, the resulting graph is a split graph, that is special case of chordal graphs.

In view of these complexity results, the question of the threshold between pspace-completeness and polynomiality is of natural interest. The following section is a first step towards it, with a characterization of a certain structure in the graph that induces a natural winning strategy for Dominator.

## 4 Pairing strategy

A natural winning strategy for Breaker in a Maker-Breaker game is the so-called pairing strategy as defined in [11]. This strategy can be applied when a subset of the board can be partitioned into pairs such that each winning set contains one of the pairs. In that case, a strategy for Breaker as a second player consists in occupying the other element of the pair that has been just occupied by Maker. By doing so, Breaker will occupy at least one element in each winning set and thus win the game. In the context of the Maker-Breaker domination game, such a subset correspond to a special dominating set that we introduce below.

###### Definition 8.

Given a graph , a subset of pair of vertices of is a pairing dominating set if all the vertices are distinct and if the intersection of the closed neighborhoods of each pair covers the vertices of the graph:

 V=k⋃i=1N[ui]∩N[vi].

Figure 3 shows an example of a pairing dominating set. Clearly, if one chooses a vertex in for each pair of a pairing dominating set, the resulting set is a dominating set of .

From this definition, we will say that a vertex is pairing dominated if there exists a pair from a pairing dominated set such that . In addition, all the pairs satisfying are useless in the construction of a pairing dominating set. Note that a pair of a pairing dominating set is not nessarily an edge of the graph.

The pairing strategy applied to the Maker-Breaker domination game can be translated into a strategy on a pairing dominating set:

###### Proposition 9.

If a graph admits a pairing dominating set, then .

Proof. If admits a pairing dominating set, then Dominator applies the following strategy as a second player: each time Staller occupies a vertex of a pair for some , Dominator answers by occupying the other vertex of the same pair if it is not yet occupied. Otherwise, Dominator plays randomly. By definition of a pairing dominating set, it ensures that the vertices chosen by Dominator form a dominating set of . Hence Dominatorhas a strategy as second player, thus also as first player using Proposition 2.

This result induces the following corollary that ensures a winning strategy for Dominator as a first player.

###### Corollary 10.

Given a graph , if there exists a vertex of such that admits a pairing dominating set, then .

Proof. If such a vertex exists, then Dominator starts and occupy it. He then applies his pairing strategy on as a second player to dominate the rest of the graph.

From this property, a natural question that arises is the detection of graphs having a pairing dominating set. An example of such graphs is when the vertices of the graph can be partitioned into cliques of size at least . In that case, a trivial pairing dominating set consists in choosing any two vertices in each clique. Note that the question of the existence of such a partition is often referred to as the packing by cliques problem (with cliques of size at least ). It was proved to be polynomial by Hell and Kirkpatrick in [12]. A particular case of this decomposition is when the graph admits a perfect matching. As an example, Proposition 9 ensures that paths or cycles of even size are as they have a perfect matching.

###### Remark 11.

The condition of Proposition 9 is not necessary. Indeed, the graphs of Figure 4 are examples with outcome and it can be shown that they do not admit a pairing dominating set. Yet, we will see in Section 5 two families of graphs (cographs and trees) for which there is an equivalence between the existence of a winning strategy for Dominator and the existence of a pairing dominating set.

We conclude this section with a study of the complexity of the pairing dominating set problem.

###### Theorem 12.

Given a graph , it is np-complete to decide whether admits a pairing dominating set.

Proof. Let be a graph. By definition, the problem is clearly in np. It remains to prove the np-hardness of the problem by reducing it from 3-sat. Let be an instance of 3-sat over the variables . Without loss of generality, one can assume that all the variables appear in both their positive and negative version in , but not in the same clause. From , we build the following graph as illustrated by Figure  5.

• Each clause , , is associated to a vertex .

• Each variable , is associated to a gadget over seven vertices such that and are two triangles, and is adjacent to both and . The pairs and will be denoted and respectively.

• For each variable and clause , we add the two edges and (resp. and ) if appears in clause in its positive (resp. negative) form.

We first claim that any assignment of the variables that makes satisfiable induces a pairing dominating set in . Let be such an assignment. We build the following set of pairs of vertices: for each variable , we add the pairs to if is true in , and the pairs otherwise. It now suffices to check that is a pairing dominating set. First of all, one can easily remark that all the vertices of the gadgets (i.e., vertices different from the clauses ) are pairing dominated by . In addition, as each clause is satisfied by , each vertex is adjacent to at least one pair or of . Hence any choice of vertex in such a pair allows to dominate .

Now consider a pairing dominating set of . We first show that for each gadget associated a variable , up to symmetry, there are only four cases to pairing dominate the vertices , and , depicted by Figure 6. Indeed, since each vertex has degree , there are three cases for it to be pairing dominated by : either the pair , or , or must belong to .

(i) The pair belongs to . Then, by considering the vertex , of degree , the pair must belong to . Concerning the vertex , it is necessarily dominated by vertices from the triangle , leading to the three cases , and of Figure 6.

(ii) The pair belongs to . By symmetry of the gadget, this case is similar to the previous one and we get the symmetric pairs from Figures , and .

(iii) The pair belongs to (Figure 6 (d)). Then both vertices and must belong to in the pairs and .

In order to find an assignment for , we now show that can be transformed into a pairing dominating set where each pair is as in Figure 6 (a) (or its symmetrical, according to case ). Consider first that for the gadget associated to some variable , the pairs of are those depicted by Figure 6 . As the vertex has no other neighbor than and , replacing a pair by the pair in remains a valid pairing dominating set since both and are adjacent to . This operation is clearly possible if is not in . In the case where is already in , say in a pair , remark that removing this pair from does not break the pairing dominating property of if is added. Indeed, since, by definition of , and have the same neighborhood (except , that is already in a pair), we have that . Since and play a symmetrical role, we can use the same argument to replace the pairs of Figure 6 by those of in . The last case is when the pairs of are those of Figure 6 for the variable . Since and (as and cannot be in the same clause), we can replace the pairs of Figure 6 by those of Figure 6 without breaking the pairing dominating property of . In case was already in , say in a pair , once again this pair can be removed from as is either empty or at most a subset of , which is already pairing dominated by the pairs of Figure 6 .

Hence we have transformed such that all the vertices different from the are pairing dominated by the pairs of vertices of Figure 6 . In addition, if admits other pairs than those depicted by Figure 6 , then these pairs are necessarily of the form , , or . The last two types of pairs can be removed from as and are already pairing dominated. Concerning the pairs , they can also be removed from as the sets belong to the gadgets (and are different from the clause vertices), and are thus already pairing dominated.

We now build the following assignment of the variables of : for all , the variable is set to true if and only if the pair belongs to . As each vertex is pairing dominated in by at least a pair or for some , it means that each corresponding clause has at least a variable equal to true, which concludes the proof.

## 5 Graph operations

In the first part of this section, we study the outcome of operations of graphs for which the outcome is already known. This will lead to polynomial time algorithms to solve the Maker-Breaker domination game on cographs and forests, as these families can be built from joins, unions and by adjoining pendant edges.

### 5.1 Union and join

Let and be disjoint graphs. The union of and is the graph with vertex set and edge set . The join of and is the graph with vertex set and edge set .

###### Theorem 13.

Let and be two starting positions of the Maker-Breaker domination game.

• If or then .

• If then .

• If then .

• Otherwise, .

This result is summarized in Table 1. Note that the outcome is absorbing for the union, while the outcome is neutral.

Proof. Assume Staller has a winning strategy on or . Then she has a winning strategy on . Indeed, without loss of generality assume that she has a winning strategy on . Her strategy on is to play only on following her winning strategy. If at some point Dominator is playing on , this can be considered as a passing move in and by Proposition 2 this does not compromise Staller’s strategy. At some point she will isolate a vertex in and thus in .

Thus if or has outcome , then whatever Dominator plays as a first move, Staller still has a winning strategy on this graph. If both positions have outcome then after Dominator’s first move, Staller can play on the other component and also wins. This proves the first two points.

If both positions have outcome , then Dominator has a winning strategy on both graphs playing second. He can answer to every move of Staller on the component she plays with his winning strategy on this component. If one of the graph is full, Dominatorcan play any vertex on the other graph and imagines he did not, as in the imagination strategy of Proposition 2 At the end, Dominator dominates both components and so has outcome .

Finally, assume without loss of generality that and . If Staller plays first, as in the first case, by applying her winning strategy as the first player in she will be able to isolate a vertex and to win. On the other hand, if Dominator plays first, he can play his winning move on and then answers to Staller on the component she has played on with his winning strategy. So the first player has a winning strategy and the outcome is .

###### Theorem 14.

Let and be two starting positions of the Maker-Breaker domination game.

• If and (or and )
then .

• Otherwise, .

Proof. (i) Assume that and . If Dominator starts, he will win by playing on the unique vertex of and dominates the join, so he has a winning strategy as a first player. However, since , if Staller starts, she can play on the only vertex of and then apply her winning strategy as second player on . So she wins on as first player as well as Dominatorand .

(ii) Since we are not in the first case, there are two possibilities : Either both and have at least two vertices or, without loss of generality, and .

Assume first that both and have more than two vertices. Let , be two vertices of and , two vertices of . Since every vertex of is a neighbor of every vertex of and conversely, forms a pairing dominating set for and the outcome is according to Proposition 9.

Assume now that and . Note that Dominator has a winning strategy on as first player. Assume that Staller is the first player. If on her first move she does not play on the vertex of , then Dominator wins immediately by playing on it. If she does play on it, then Dominator will apply his winning strategy as first player on . This will allow him to dominate and, since each vertex of dominates , all the vertices of will be dominated. Dominator has a winning strategy as second player, hence .

The combination of these two results gives a complexity result on the class of cographs. Recall that cographs (or -free graphs) can be inductively built from a single vertex by taking the union of two cographs or the join of two cographs. In addition, from a given cograph, recovering this construction from unions and joins can be found with a linear time algorithm [7]. Since we know the outcome of Maker-Breaker domination game for and for the union and the join operators, we can deduce the following corollary.

###### Corollary 15.

Deciding the outcome of the Maker-Breaker domination game on cographs can be done in polynomial time.

As stated in Remark 11, for some families of graphs the outcome of a starting position is if and only if it admits a pairing dominating set. We show that the family of cographs satisfies this property.

###### Theorem 16.

A cograph has outcome if and only if it admits a pairing dominating set.

Proof. By Proposition 9 that if a graph admits a pairing dominating set, then it has outcome . It remains to prove that all cographs with outcome admits a pairing dominating set.

The proof is done by induction on the number of vertices of .

First note that the result is true when . The only such cographs are , and , and among them the only graph with outcome is . admits a perfect matching and thus a pairing dominating set.

Assume now that every cograph of outcome with a number of vertices less or equal to admits a pairing dominating set. Let be a cograph of outcome with vertices. By definition of a cograph, is either the union or the join of two smaller cographs.

If is the union of two cographs and , they necessarily have outcome by Theorem 13. By induction hypothesis, they both admit a pairing dominating set, which union is a pairing dominating set for .

Assume now that is the join of two cographs and .

If both and have more than two vertices, then if , are any two vertices of and , are any two vertices of , forms a pairing dominating set for .

Assume now that and let be its unique vertex. Then has either outcome or by Theorem 14. If has outcome then by induction hypothesis, it admits a pairing dominating set. Every vertex of this pairing dominating set is a neighbor of and it remains also a pairing dominating set for .

Assume now that . is either the union of two cographs or the join of two cographs.

If is the join of two cographs, by Theorem 14, it must be the join of a graph with vertex and of a graph with outcome . Notice that and are both universal vertices so is a pairing dominating set for . If is the union of and then, without loss of generality, by Theorem 13 and . By induction hypothesis, admits a pairing dominating set . Note also that by Theorem 14, has outcome , so by induction hypothesis it admits a pairing dominating set . Since pairing dominates and pairing dominates , forms a pairing dominating set for .

### 5.2 Glue operator and trees

We now study the operator consisting of gluing two graphs on a vertex. This operator will be useful in the study of trees. A more formal definition is the following:

###### Definition 17.

Let and be graphs and let and be two vertices. The glued graph of and at and is the graph with vertex set (where is a new vertex) and for which is an edge if and only if is an edge of or or and is and edge of or is an edge of .

If the vertex is clear from the context or does not matter, the glue will be denoted by . Similarly if the vertex is not useful in the notation, we might also remove it. Figure 7 gives a representation of the glued of two graphs.