1 Introduction
Let us have a graph whose vertices are occupied by guards. The graph is subject to an infinite sequence of vertex attacks. The guards may move to any neighboring vertex after each attack. After moving, a vertex attack is defended if the vertex is occupied by a guard. The task is to come up with a strategy such that the graph can be defended indefinitely.
Defending a graph from attacks using guards for an infinite number of steps was introduced by Burger et al. [2]. In this paper we study the concept of the meternal domination, which was introduced by Goddard et al. [4] (eternal domination was originally called eternal security).
The meternal guarding number is the minimum number of guards which tackle all attacks in indefinitely. Here the (slightly confusing) notion of the letter “m” emphasizes that multiple guards may move during each round. The meternal domination number is the minimum number of guards which tackle all attacks indefinitely, with the restriction that no two guards may occupy a single vertex simultaneously. We also introduce the meternal domination number with eviction , which is similar to with the additional requirement, that during each round one can decide to either attack a vertex or choose an “evicted” vertex or edge, which must be cleared of guards in the next round. There is also a variant of the problem studied by Goddard et al. [4] where only one guard may move during each round, which is not considered in our paper. We will define all concepts formally at the end of this section.
Goddard et al. [4] established exactly for paths, cycles, complete graphs and complete bipartite graphs, showing that , , and . The authors also provide several bounds for general graphs, most notably , where denotes the size of the maximum independent set in and is the size of the smallest dominating set in . Since that several results focused on finding bounds of in different conditions or graph classes.
Henning, Klostermeyer and MacGillivray [7] explored the relationship between and the minimum degree of a graph : If is a connected graph with minimum degree and has vertices, then , and this bound is tight.
Finbow, Messinger and van Bommel [3] proved the following result for grids.
For ,
Here denotes the Cartesian product of graphs and .
Van Bommel and van Bommel [12] showed for grids that
For a good survey on other related results and topics see Klostermeyer and Mynhardt [9].
Very little is known regarding the algorithmic aspects of meternal domination. The decision problem (asking if ) is obviously NPhard and belongs to EXPTIME, however, it is not known whether it lies in the class PSPACE (see [9]). On the positive side, there is a linear algorithm for computing for trees by Klostermeyer and MacGillivray [8]. Braga, de Souza and Lee [1] showed that in all properinterval graphs. Very recently Gupta et al. [5] showed that the maximum independent set in an interval graph on vertices can be solved in time , or in the case when endpoints of the intervals are already sorted. We can thus compute efficiently on properinterval graphs.
In this paper we contribute to the positive side and provide an extension of the result by Klostermeyer and MacGillivray [8]. Cactus is a graph that is connected and its every edge lies on at most one cycle. An equivalent definition is that it is connected and any two cycles have at most one vertex in common. Christmas cactus graph is a cactus in which each vertex is in at most two connected components. Christmas cactus graphs were introduced by Leighton and Moitra [10] in the context of greedy embeddings, where Christmas cactus graphs play an important role in the proof that every polyhedral graph has a greedy embedding in the Euclidean plane.
Our main result is summarized in the following theorem.
Theorem 1.1 ()
Let be a Christmas cactus graph. Then and there exists a lineartime algorithm which computes these values.
Using Theorem 1.1 we are able to devise a new bound on the meternal domination number of cactus graphs, which is stated in Theorem 3.1 in Section 3. In Section 4 we provide the lineartime algorithm for computing of Christmas cactus graphs.
Let us now introduce all concepts formally. For an undirected graph let a configuration be a multiset . We will refer to the elements of configurations as guards. Movement of a guard means changing to some element of its closed neighborhood and we denote it by . Two configurations and of are mutually traversable in if it is possible to move each guard of to obtain . A strategy in is a graph where is a set of configurations in of same size and . The order of a strategy is the number of guards in each of its configurations.
We now define the variants of the problem which we study in our paper. For the purpose of the proof of our main result we devise a variant of the problem, where a vertex or an edge can be “evicted” during a round, that means, no guard may be left on the respective vertex or edge. We call the strategy to be defending against vertex attacks if for any the configuration and its neighbors in cover all vertices of , i.e., when a vertex is ”attacked“ one can always respond by changing to a configuration which has a guard at the vertex . Note that every configuration in a strategy which defends against vertex attacks induces a dominating set in . We call a strategy to be evicting vertices if for any and any the configuration has a neighbor in such that , i.e., when a vertex is ”to be evicted“ one can respond by changing to a configuration where no guard is present at . We call a strategy to be evicting cycle edges if for any and any edge lying in some cycle in the configuration has a neighbor in such that . That means, when an edge is ”to be evicted“ one can respond by changing to a configuration where no guards are incident to the edge.
Let the meternal guard strategy in be a strategy defending against vertex attacks in . Let the meternal guard configuration number be the minimum order among all meternal guard strategies in . Let the meternal dominating strategy in be a strategy in such that none of its configurations has duplicates and is also defending against vertex attacks. The meternal dominating set in is a configuration, which is contained in some meternal dominating strategy in . Let the meternal dominating number be the minimum order of meternal dominating strategy in . Let the meternal dominating strategy with eviction in be a strategy such that none of its configurations has duplicates, is defending vertex attacks, is evicting vertices, and is evicting cycle edges in . Let the meternal dominating with eviction number be the minimum order of meternal dominating strategy evicting vertices and edges in .
A cycle in is a leaf cycle if exactly one of its vertices has degree greater than 2. By we denote a path with edges and vertices. By we denote the subgraph of induced by the set of vertices .
2 The meternal domination of Christmas cactus graphs
In this section we prove that for Christmas cactus graphs by showing the optimal strategy. The main idea is to repeatedly use reductions of the Christmas cactus graph to produce smaller Christmas cactus graph . We prove that the optimal strategy for uses a constant number of guards more than the optimal strategy for .
This will be one part of the proof of Theorem 1.1. Before we describe the reductions in detail, we present several technical tools that are used in the proofs of validity of the reductions.
Observation 1
Every strategy used in the meternal domination with eviction can be applied in an meternal domination strategy, and every meternal domination strategy can be applied as an meternal guard configuration strategy. Every configuration in each of these strategies must induce a dominating set, therefore, they are all lower bound by the domination number . We see that the following inequality holds for every graph .
Note that we can prove bounds on all of these strategies by showing that for and its reduction it holds that and for some integer constant . If we have an exact result for the reduction gives us an exact bound on as well. This is summed up in the following lemma.
Lemma 1
Let us assume that for graphs , , and an integer constant
(1)  
(2)  
(3) 
Then .
Proof
Given the assumptions, we get in the following manner.
Recall Observation 1 where we saw that holds, giving us the desired equality. ∎
Let us have a graph with a strategy. By simulating a vertex attack, a vertex eviction, or an edge eviction on , we mean performing the attack on and retrieving the strategy’s response configuration. Simulating attacks is useful mainly in merging several strategies over subgraphs into a strategy for the whole graph.
In the following theorem we introduce a general upper bound applicable to the meternal dominating strategy with eviction.
Lemma 2
Let be a graph with an articulation such that has two connected components and such that there are exactly two vertices and in which are neighbors of . Let . If lies on a cycle in then
Proof
We will show that having two separate strategies for and we can merge them into one strategy for without using any additional guards.
We will create a strategy which keeps the invariant that in all its configurations either is occupied by a guard or the pair of vertices and are evicted. This will ensure that whenever is not occupied due to the strategy of needing a guard from to defend other vertices of , the strategy of will be in a configuration where no guard can traverse the edge.
Let the initial configuration be a combination of a configuration of which defends and a configuration of which evicts . The final strategy will consist of configurations which are unions of configurations of and which we choose in the following manner.
The vertices of were partitioned among and so a vertex attack can be distinguished by the target component. Whenever a vertex of is attacked, choose a configuration of respective component which defends . If was not attacked then simulate an attack on . If was not attacked then simulate an edge eviction on . By the configuration of the nonattacked component we ensure the invariant is true. Whenever the edge might be traversed by a guard in the ’s strategy, we use the fact that is occupied and instead of performing we move the guards and which has the same effect considering guard configuration of .
The eviction of vertices and edges present in and is solved in the same way as vertex attacks. The only attack which remains to be solved is an edge eviction of or . Both of these are defended by simulating an eviction of in and in . The two strategies will ensure there are no guards on either , , nor and the invariant is still true. ∎
Let for denote contraction of the edge in . We show that contracting an edge will not break an meternal guard configuration strategy.
Lemma 3
Let be a graph and be its edge. Then for a graph ( after contraction of )
Moreover, there is a strategy which differs only in vertices incident to .
Proof
Let and let be the vertex after the contraction of . Let the meternal guard configuration for be the meternal guard configuration of where in every configuration each and is substituted with . Any movement of guards in the original strategy is still possible in the new one. The only change is that instead of moving among and the guard will stay on . Hence the traversable configurations will stay traversable. We devised a strategy for using the same number of guards as was used in the original strategy for . ∎
We may now proceed with the reductions.
Lemma 4
Let be a cycle on vertices. Then .
Proof
It is easy to see, that the domination number of a cycle on vertices is hence it suffices to show that the meternal domination with eviction number is and by Observation 1 we get the desired equality.
First, note that, in every dominating configuration, two guards can be at most three edges apart. Any vertex attack can be defended by moving all guards in the configuration along the cycle in one direction or the other.
Evicting a vertex can be done by evicting an incident edge since every vertex is on the cycle. To evict an edge we look at guards on the four closest vertices (in order), see Figure 2. If one incident pair of these vertices is not guarded we can rotate the configuration to move the guard gap over to . If that is not the case perform if is guarded and if is guarded. Note that the new configuration has guards on and because if there was no guard on and then we would rotate the configuration (and similarly for and ). We did not move any guard which dominates vertices from the rest of the graph, and the four closest vertices are dominated by and , hence the final configuration is dominating. ∎
Reduction 1
Let be a Christmas cactus graph and be a leaf vertex which is connected to a vertex of degree . Remove and from .
Lemma 5
Proof
Let us take a graph with a vertex of degree which is connected to a leaf vertex and an articulation of . Let be where vertices and are removed.
Assume we have an optimal meternal domination strategy for . Extend it by adding one guard on who can both defend and evict its vertices. This guard does not interfere with the rest of the strategy in any way. Therefore .
Assume we have an optimal meternal guard configuration for . For to be dominated there needs to be a guard on either or . Remove vertices and along with the guard which is always present there from to obtain and move any excess guards to . All remaining guards can substitute any movement among and by staying on . Therefore .
Having the two inequalities we use the Lemma 1 to obtain the desired equality. ∎
Reduction 2
Let be a Christmas cactus graph and be a leaf vertex which is connected to a vertex of degree . Let the vertex has neighbors such that and are not connected. Remove and from and connect by an edge.
Lemma 6
Proof
Let us take a graph with a leaf vertex which is connected to by an articulation . Denote the vertices incident to on the Christmas cactus graph’s cycle and .
Upper bound: We show an upper bound using the meternal domination with edge eviction. We split into two components: the leaf vertex and a graph . We also see that of a single vertex is 1.
Use Lemma 2 on decomposition of into and to get .
Lower bound: Now we show that .
Assume that we have an optimal strategy for meternal guard configuration on . Note that we can obtain by contracting and while adapting the strategy as shown in Lemma 3. We obtain a strategy for which uses exactly guards. Additionally, we see that if is to be dominated in there needs to be at least one guard in . This means at least one guard is present at in all configurations of strategy on .
The only case when the guard on needed to move from to is when he was immediately substituted, i.e., and moves were performed. The contraction transforms this move to and , however, in the guard from can move directly to not using the guard on . Since we have a stationary guard at which is not critical for defending any vertices of we can remove him from all the configurations of the strategy for . Therefore . ∎
Reduction 3
Let be a Christmas cactus graph and be a leaf cycle on vertices where . Let be the only articulation on this cycle. Remove and create a new vertex and the edge in .
Lemma 7
Proof
First, we will show the bounds for the strategies and use Lemma 1 to get the desired result for . After the proof we will briefly discuss that exactly the same proof technique is applicable to .
Let be present in which is to be substituted with for . Let the vertices of be denoted for the articulation and for the leaf.
Upper bound: We will show that the meternal dominating set with edge eviction strategy for can be extended to adding guards.
Let be vertices in . Let us create a strategy for defending by merging the strategy defending with the strategy defending from Lemma 4. We will keep an invariant that at least one of the following pairs is occupied: , , or , see Figure 5.
Let us divide the attacks on into ones which target the vertices and edges of and the rest which target vertices and edges present in .
If the cycle configuration changed while defending an attack, we will simulate the vertex attack on if is occupied, or the eviction of if either or is occupied. On the other hand if configuration of changed while defending an attack on we simulate the vertex attack on if is occupied or on if is not occupied. Note that the invariant is always met so when the strategies are merged (note that is represented by either or ) we can remove the guard from strategy of which is paired up with a guard from . Since the strategy of Lemma 4 does not use two guards on incident vertices we cannot have a pair with and present at the same time. Therefore, the merged strategies do not have multiple guards at the same vertex.
The total number of guards is .
Lower bound: We will show that the meternal guard configuration strategy on can be reduced to removing guards.
Assume we have an optimal strategy for the meternal guard configuration on . The leaf must be dominated by at least guards on distinct positions. Let us label vertices next to on the as (in order), see Figure 6. Let be a path and let be a path on vertices. has to be dominated by at least guards on in each configuration of the strategy. Let us alter all the configurations by contracting all the edges of obtaining a vertex . All the configurations in the strategy of contain guards on because was occupied by guards in order to dominate . These guards are always present in and translate to guards in . Even if one of these guards leaves, another one must enter. This translates to swapping guards between and in which is an excess movement and we substitute it with stationary guards. Since these stationary guards are not necessary for dominating we remove them from resulting in the final strategy for .
Having a proof of the exact bound for in hand, let us discuss that the same argument works for . The upper bound can change in a way that it is possible for the strategy to occupy and at the same time, and still cannot occupy at the same time as or . This does not break the invariant because if and are occupied both the attacks are simulated on in . The lower bound changes in a way that is a path on vertices which still has to be dominated by at least on . ∎
Reduction 4
Let be a Christmas cactus graph and be a leaf cycle on vertices. Let be the only articulation on this cycle. Remove from .
Lemma 8
Proof
We will show the bounds on the strategies and use the Lemma 1 to get the desired result.
Let be the Christmas cactus graph and be its leaf cycle on vertices. Let be the vertex of of degree bigger than . Let be where is removed.
Upper bound: We extend an optimal meternal domination with edge eviction strategy on to using guards.
Note that is an articulation and that has two connected components. Let and be neighbors of on the cycle . Let . Let us define and . Observe that as defined in Lemma 2 is a cycle on vertices. We apply Lemma 2 on decomposition and of to get
Lower bound: We will show that . Let be a path on vertices. Consider an optimal meternal guard strategy . Let us alter this strategy by contracting all edges of resulting in a vertex of , altering the strategy due to Lemma 3. In all configurations of the new strategy the vertex is occupied by at least guards because must have contained at least guards to be dominated. If we assume that the guards move only if necessary to allow change between configurations then we have stationary guards on who never move away. However they do not defend any attacks because they only defended so we can remove them from every configuration of the strategy for obtaining a strategy which uses at most guards. ∎
Reduction 5
Let be a Christmas cactus graph and be a cycle on three vertices , let be leafs in , such that connects to , to , and is connected to the rest of the graph (no other edges are incident to ). We call the subgraph a bull graph. Remove from .
Lemma 9
Let be a Christmas cactus graph. Let be a bull graph connected to the rest of via a vertex of degree . Let be after application of Reduction 5 on . Then is a Christmas cactus graph and .
Proof
Let be a Christmas cactus graph and be a leaf bull graph connected to the rest of by its vertex . Let and .
Upper bound: Assuming an optimal meternal dominating strategy on we can extend it to by adding two guards who only guard one edge each to the bull graph. One guard will be on or and one on either or . Attacks on will be resolved by the strategy for and attacks on vertices of will be resolved by the new guards.
Suppose that has a neighbor not in . Then the edge evictions of the new cycle are resolved by simulating a vertex eviction on in and moving the new guards into and .
If there is no neighbor of outside of , then is exactly the bull graph and is solved as a trivial case.
Therefore .
Lower bound: Let us contract all edges of the subgraph of to obtain and alter the configurations in accordance to Lemma 3. We note that the vertex inherited all the guards of the original which must contain at least guards in all configurations. Two guards from will be stationary at and are not necessary for defending . This shows that .
∎
Let the pan graph be a with one leaf attached.
Reduction 6
Let be a Christmas cactus graph and be a cycle on three vertices , let be a leaf in , such that connects to and is connected to the rest of the graph (no other edges are incident to ). The is a pan graph. Remove from .
Lemma 10
Let be a Christmas cactus graph. Let be a pan graph connected to rest of the graph via a vertex of degree . Let be after application of Reduction 6 on . Then is a Christmas cactus graph and .
Proof
Let be a Christmas cactus graph and be a leaf pan connected to the rest of the graph by its degree vertex . Let and .
Upper bound: Let us have an optimal meternal dominating strategy with edge eviction for . We will extend this strategy to by adding a guard to defend the edge . These strategies cover all vertex attacks, however, since by adding vertices and we created a cycle , we have to show how to perform edge evictions on the cycle edges. If any edge eviction occurs, move the new guard to . If the edge eviction targets or then simulate eviction on the vertex or in , respectively. If should be evicted, then simulate eviction on the vertex in , which would force a guard to stand at in the next configuration. We move him to instead. This can be done because is a neighbor to and its only neighbor where the guard could have come from. Even though the guard is at in the next move we can think of him as if he was at because he has the same set of possible moves.
Lower bound: Let us contract the edges of in resulting in a single vertex in , see Figure 10. Since can be dominated only by a guard at or , then contains a stationary guard which is not crucial for its defense. We remove this guard from all configurations of the strategy to obtain strategy which uses at most one less guard than the optimal guard configuration of . Therefore .
∎
Using the reductions we are ready to prove the part of Theorem 1.1 stating that for all Christmas cactus graphs.
A block or a connected component of graph is a maximal connected subgraph of .
Lemma 11
In a nonelementary christmas cactus graph with no leaf cycles, no leaf vertices connected to a vertex of degree , and no leaf vertices connected to a block of size bigger than , there is at least one leaf bull or one leaf pan graph.
Proof
Let us call the blocks of size triangles. Removing edges of all the triangle subgraphs would split the christmas cactus into connected components of blocks. Let us choose a triangle and traverse the graph in the following way. If the current triangle is a bull or a pan we end the traversal and have a positive result. Otherwise, choose the component we have not visited yet and find a different triangle graph incident to it. Such triangle must exist otherwise it would be a leaf component. Mark this component as visited and repeat the process. See Figure 11. ∎
Theorem 2.1
Let be a Christmas cactus graph. Then .
Proof
A Christmas cactus graph always contains either a leaf or a leaf cycle. This will be shown by contradiction. If all vertices have degree at least two and each cycle has at least two neighboring blocks then the chain of blocks would either never end or it must close itself, creating another big cycle, contradicting that the graph is a cactus.
We will use reductions until we obtain an elementary graph for which the optimal strategy is known. The graph is called elementary if it is a cycle, single edge, a path on three vertices, a bull, or a pan. The proper reduction is chosen repeatedly in the following manner, which is also depicted in Figure 13.

If is elementary we return the optimal strategy.

Otherwise there is a leaf vertex in and its neighbor is an articulation.

If the vertex is connected to the rest of the graph by only one edge then use Reduction 1,

Vertex is connected to two vertices and which are different from .

If there is no edge between and then they must be connected by a path in , otherwise, would be in more than two blocks. Use Reduction 2.

If there is an edge between and then it cannot be on any other cycle than . Note that vertices form a triangle which is be connected to at most other blocks.
Using the reductions we eventually end up in a situation where the Christmas cactus graph is an elementary graph. The optimal strategy for cycle was shown in Lemma 4, all the configurations of optimal strategies of all the remaining graphs are depicted in Figure 12.
In each of these elementary graphs, allowing eviction attacks does not increase the necessary number of guards. Also, allowing more guards at one vertex does not add any advantage and does not decrease the necessary number of guards. Therefore, for all of these cases it holds that . ∎
3 Upper bound on the meternal domination number of cactus graphs
Definition 1
Let us have a cactus graph . Let us color vertices of in the following way. Let a vertex be colored red if it is contained in more than two connected components of , otherwise it is colored black. Let denote the number of red vertices, and denote the number of red connected components (e.g. and in Figure 14).
Let be a graph created from by contracting each red connected components into a red vertex.
Let be a set of maximal connected components of black vertices in . Let . Note that contains only red vertices. Let the Christmas cactus decomposition be a disjoint union of graphs induced by for all . See Figure 14.
Theorem 3.1
The meternal domination number of a cactus graph is bounded by
where are the components of the Christmas cactus decomposition, is the number of red vertices in and is the number of connected components of red vertices.
Proof
Let be a graph where all connected components of red vertices are contracted creating one red group vertex for each component as shown in Figure 14. Let be the Christmas cactus decomposition of .
First, find an optimal strategy for each Christmas cactus graph in separately by the process presented in Section 2. We will show how to merge these disjoint strategies into one strategy for the whole graph and subsequently generalize it for .
Assume that all red vertices of are always occupied, hence we have to show how to defend black vertices. Let us reverse the process of Christmas cactus decomposition and merge disjoint Christmas cactus graphs by the red vertices to obtain . When a black vertex is attacked we simulate an attack on the respective Christmas cactus graph to get a configuration which defends the vertex as shown in Fig. 15. If any of the red vertices of the Christmas cactus graph are not occupied then we simulate an attack on the red vertex in all other Christmas cactus graphs which contain it.
The process ensures that in each configuration all but one Christmas cactus graph incident to each red vertex has a guard on it. A red vertex incident to black components in is always occupied by exactly guards. So we can remove guards and it remains always occupied by exactly one guard. This strategy for uses guards.
Now we get from by expanding the red vertices back into the original red connected components. Add guards such that all red vertices are occupied. The strategy will be altered slightly. When we defend by moving a guard from red vertex then another guard from a different Christmas cactus component is forced to move to by a simulated attack. However in the left vertex and the attacked vertex might not coincide so we pick a path from to in the red component and move all the guards along the path.
The change in number of guards can be imagined as removing guards on red vertices of and adding guards on all red vertices of . We devised a strategy for which uses guards. ∎
4 Lineartime algorithm
We present a description of a lineartime algorithm, which computes in Christmas cactus graphs in linear time. The algorithm applies previously presented reductions on the blockcut tree of the input graph.
Definition 2 (F. Harary [6])
Let the blockcut tree of a graph be a graph , where is the set of articulations in and is the set of biconnected components in . A vertex is connected by an edge to a biconnected component if and only if in .
The high level description of the algorithm is as follows.

Construct the Christmas cactus decomposition of and iterate the following for each component .

Construct the blockcut tree of the Christmas cactus .

Repeatedly apply the reductions on the leaf components of .

If a reduction can by applied, appropriately modify in constant time, so that it represents with the chosen reduction applied. At the same time, increase the resulting appropriately.

If is empty return the resulting and end the process.


Use Theorem 3.1 to get the upper bound.
This result is summed up in the following theorem. We also present the detailed pseudocode of the lineartime algorithm for finding the meternal domination number for Christmas cactus graphs.
Theorem 4.1
Let be a cactus on vertices and edges. Then there exists an algorithm which computes an upper bound on in time . Moreover, this algorithm computes the of Christmas cactus graphs exactly.
Proof
First step of the algorithm is to create the Christmas cactus decomposition of the input graph . For each of these Christmas cactus graphs we run Algorithm 1 and then output the answer devised by Theorem 3.1.
The construction of a Christmas cactus decomposition of is done by constructing the blockcut tree of and coloring red all the vertices which are present in at least blocks, and coloring black all other vertices. We use the DFS algorithm to find all the connected components of red vertices and contract each of these components into a single red vertex. Next, we use the DFS algorithm to retrieve all connected components of black vertices along with their incident red vertices. Note that every edge of is contained at most once in the Christmas cactus decomposition, hence the total number of vertices and edges in the decomposition is bounded by .
We run the Algorithm 1 for each component separately. Now we will show its correctness. The algorithm performs reductions in the while loop at line 10. In each iteration it processes one leaf block vertex in .
First, we argue that the algorithm correctly counts the number of guards on all elementary graphs. In case consists of a single block, it is detected on line 12. The block is removed and the number of guards increases by on line 58, which is consistent with the result for elementary cycle and edge. In the other case the consists of several blocks. If represents a path on three vertices, one guard will added by line 27 and one by line 58 and both blocks are removed. If represents a pan it will count guards by first reducing one block by line 31 or 27, and then reducing a single block of size at most on line 58. If represents a bull the algorithm will find a leaf block of size reducing the bull to a path on vertices on line 31, counting correctly guards.
Now we show that if the algorithm performs an operation on a leaf block it resolves in the correct number of guards at the end. Let be a leaf block vertex processed in the loop. Note that each block of a Christmas cactus graph is either a cycle or an edge.
Consider the case where is a leaf cycle. If the cycle has , then the block is removed entirely adding guards on line 40, exactly as in Reduction 4. Otherwise the cycle is contracted to a block of size and guards are added on line 44, as in Reduction 3.
Consider the case where the leaf block has , representing a leaf vertex. Let be the block that shares an articulation with . If block has then we remove both of these blocks and add a guard on line 27, as in Reduction 1. If block has then is a leaf vertex connected to a cycle. On line 31 is removed and is decreased by one. This is consistent with Reductions 2, 5, and 6. Note that reducing a leaf incident to a component on three vertices first, yields the same result as reducing the graph first and waiting for the leaf be reduced in either Reduction 2, 5, or 6.
The algorithm performs reductions which were proved to be correct. This concludes the proof of correctness.
Let be the number of vertices of the Christmas cactus graph, and be the number of its edges. Now we show that Algorithm 1 runs in time . Using Tarjan’s algorithm [11], we can find the blocks of a graph in linear time. By the straightforward augmentation of the algorithm we obtain the blockcut tree where every block contains additional information with the number of vertices it contains. Note that the number of vertices and edges of is bounded by . Therefore and .
Now consider the while loop at line 10. We claim that every vertex in is processed at most twice in the loop and every iteration takes constant time. Let be the currently processed vertex. During one iteration of the main cycle either a block of size at most is deleted or a block of size at least is either shrunk to size or deleted. Therefore, the algorithm for Christmas cactus graphs runs in time .
For the cactus graph we need to create the Christmas cactus decomposition which uses Tarjan’s algorithm [11] for creating the blockcut tree, and the DFS, which both runs in . As stated, the running time of the algorithm is linear in the size of the Christmas cactus. Therefore, the total running time is bound by sum of their sizes . ∎
5 Future work
The computational complexity of the decision variant of the meternal domination problem is still mostly unknown as mentioned in the introduction.
The natural extension of the algorithm from cactus graphs is to the more general case of graphs with treewidth . It is also an interesting question if we can design an algorithm, whose running time is parameterized by the treewidth of the input graph.
Acknowledgments. We would like to thank Martin Balko and an anonymous referee for their valuable comments and insights.
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