11email: robin.weishaupt@hhu.de, rothe@hhu.de
Stability of Special Graph Classes
Abstract
Frei et al. [6] showed that the problem to decide whether a graph is stable with respect to some graph parameter under adding or removing either edges or vertices is -complete. They studied the common graph parameters (independence number), (vertex cover number), (clique number), and (chromatic number) for certain variants of the stability problem. We follow their approach and provide a large number of polynomial-time algorithms solving these problems for special graph classes, namely for graphs without edges, complete graphs, paths, trees, forests, bipartite graphs, and co-graphs.
Keywords:
Computational Complexity Graph Theory Stability Robustness Colorability Vertex Cover Independent Set1 Introduction
Frei et al. [6] comprehensively studied the problem of how stable certain central graph parameters are when a given graph is slightly modified, i.e., under operations such as adding or deleting either edges or vertices. Given a graph parameter (like, e.g., the independence number or the chromatic number), they formally introduced the problems -Stability, -VertexStability, -Unfrozenness, and -VertexUnfrozenness and showed that they are, typically, -complete, that is, they are complete for the complexity class known as “parallel access to ,” which was introduced by Papadimitriou and Zachos [18] and intensely studied by, e.g., Wagner [21, 22], Hemaspaandra et al. [8, 10], and Rothe et al. [20]; see the survey by Hemaspaandra et al. [9].111 is contained in the second level of the polynomial hierarchy and contains the problems that can be solved in polynomial time by an algorithm that accesses its oracle in parallel (i.e., it first computes all its queries and then asks them all at once and accepts its input depending on the answer vector). Alternatively, can be viewed as the class of problems solvable in polynomial time via adaptively accessing its oracle (i.e., computing the next query depending on the answer to the previous query) logarithmically often (in the input size).
Furthermore, Frei et al. [6] proved that some more specific versions of these problems, namely --Stability and --VertexStability, are -complete for and -complete for , respectively, where was introduced by Papadimitriou and Yannakakis [17] as the class of problems that can be written as the difference of problems.
Overall, the results of Frei et al. [6] indicate that these problems are rather intractable and there exist no efficient algorithms solving them exactly. Considering the vast number of real-world applications for these problems mentioned by Frei et al. [6]—e.g., the design of infrastructure, coloring algorithms for biological networks [15, 13] or for complex information, social, and economic networks [12], etc.—these results are rather disappointing and unsatisfying.
This obstacle motivates us to study whether there are scenarios that allow for efficient solutions to these problems which in general are intractable. Our work is based on the assumption that most of the real-world applications of stability of graph parameters do not use arbitrarily complex graphs but may often be restricted to certain special graph classes. Consequently, our studies show that—despite the completeness results by Frei et al. [6]—there are tractable solutions to these problems when one limits the scope of the problem to a special graph class. We study seven different classes of special graphs: empty graphs consisting of only isolated vertices and no edges (), complete graphs that have all possible edges (), paths (), trees (), forests (), bipartite graphs (), and co-graphs (). For each such class, we study twelve different problems:
-
•
stability, vertex-stability, and unfrozenness
-
•
for the four graph parameters , , , and .
In total, we thus obtain the 84 membership results shown in Table 1, which gives the theorem, proposition, or corollary corresponding to each such result. This can be useful for real-world applications when knowledge about the stability, vertex-stability, or unfrozenness of a graph with respect to a certain graph parameter is required and graphs with such a special structure may typically occur in this application.
| Stability | E | Thm. 4.1 | Prop. 2 | Thm. 4.2 | Cor. 6 | Thm. 4.3 | Cor. 7 | Cor. 13 | |
| V | Thm. 4.1 | Cor. 4 | Thm. 4.2 | Cor. 6 | Thm. 4.3 | Cor. 7 | Thm. 4.13 | ||
| Unfrozenness | Cor. 3 | Cor. 5 | Prop. 4 | Cor. 10 | Cor. 10 | Cor. 9 | Cor. 15 | ||
| Stability | E | Thm. 4.1 | Prop. 2 | Thm. 4.2 | Cor. 6 | Thm. 4.3 | Cor. 7 | Cor. 14 | |
| V | Thm. 4.1 | Cor. 4 | Thm. 4.2 | Cor. 6 | Thm. 4.3 | Cor. 7 | Cor. 11 | ||
| Unfrozenness | Cor. 3 | Cor. 5 | Prop. 4 | Cor. 10 | Cor. 10 | Thm. 4.8 | Cor. 15 | ||
| Stability | E | Thm. 4.1 | Prop. 2 | Thm. 4.2 | Cor. 6 | Thm. 4.3 | Cor. 7 | Thm. 4.15 | |
| V | Thm. 4.1 | Cor. 4 | Thm. 4.2 | Cor. 6 | Thm. 4.3 | Cor. 7 | Cor. 12 | ||
| Unfrozenness | Cor. 3 | Cor. 5 | Prop. 3 | Prop. 5 | Thm. 4.4 | Cor. 8 | Cor. 16 | ||
| Stability | E | Thm. 4.1 | Prop. 2 | Thm. 4.2 | Cor. 6 | Thm. 4.3 | Cor. 7 | Thm. 4.14 | |
| V | Thm. 4.1 | Cor. 4 | Thm. 4.2 | Cor. 6 | Thm. 4.3 | Cor. 7 | Thm. 4.12 | ||
| Unfrozenness | Prop. 1 | Cor. 5 | Prop. 3 | Prop. 5 | Thm. 4.4 | Thm. 4.7 | Thm. 4.16 | ||
2 Preliminaries
We follow the notation of Frei et al. [6] and briefly collect the relevant notions here (referring to their paper [6] for further discussion). Let be the set of all undirected, simple graphs without loops. For , we denote by its vertex set and by its edge set; by its complementary graph with and . For , , and , let , , and , respectively, denote the graphs that result from by deleting , deleting , and adding .
A graph parameter is a map . We focus on the prominent graph parameters (the size of a maximum independent set), (the size of a minimum vertex cover), (the chromatic number, i.e., the minimum number of colors needed to color the vertices of a graph so that no two adjacent vertices have the same color), and (the size of a maximum clique).
For a graph parameter , an edge is said to be -stable if , i.e., remains unchanged after is deleted from . Otherwise (i.e., if is changed by deleting ), is said to be -critical. Stability and criticality are defined analogously for a vertex instead of an edge .
A graph is said to be -stable if all its edges are -stable. A graph whose vertices (instead of edges) are all -stable is said to be -vertex-stable, and -criticality and -vertex-criticality are defined analogously. Obviously, each edge and each vertex is either stable or critical, yet a graph might be neither.
Traditionally, the analogous terms for stability or vertex-stability when an edge or a vertex is added rather than deleted are unfrozenness and vertex-unfrozenness: They too indicate that a graph parameter does not change by this operation. And if, however, a graph parameter changes when an edge or vertex is added (not deleted), the notions analogous to criticality and vertex-criticality are simply termed frozenness and vertex-frozenness. Again, each edge and each vertex is either unfrozen or frozen, but a graph might be neither.
For a graph parameter , define -Stability to be the set of -stable graphs; and analogously so for the sets -VertexStability, -VertexCriticality, -Unfrozenness, -Frozenness, and -VertexUnfrozenness. These are the decision problems studied by Frei et al. [6] for general graphs in terms of their computational complexity. We will study them restricted to the graph classes mentioned in the introduction, formally defined in the subsections of Section 4.
A graph class is closed for (induced) subgraphs if for every it holds that all (induced) subgraphs of satisfy .
The notation of perfect graphs was originally introduced by Berge [2] in 1963: A graph is called perfect if for all induced subgraphs of , we have . Note that is also an induced subgraph of itself.
3 General Stability and Unfrozenness Results
In this section, we provide general results that hold for specific graph classes satisfying special requirements. These results can be used to easily determine for a given graph class whether some stability or unfrozenness results are tractable.
Theorem 3.1
Let be a graph class closed for induced subgraphs, and a tractable graph parameter for . Then - for all .
Proof. Let and compute . For all , we have , since is closed for induced subgraphs. Hence, for all , we can compute efficiently and compare it to . If there is no vertex such that the values differ, is -vertex-stable. This approach is computable in time polynomial in , so that - for all . ❑
Since every graph class that is closed for subgraphs is also closed for induced subgraphs, Corollary 1 is a simple consequence of the previous theorem.
Corollary 1
Let be a graph class closed under subgraphs and a tractable graph parameter for . Then - for all .
The first theorem made a statement related to vertex-stability about graph classes closed for induced subgraphs. Theorem 3.2 is related to edge-stability; its proof is deferred to the appendix, as is the case for most upcoming results.
Theorem 3.2
Let be a graph class closed under subgraphs and a tractable graph parameter for . Then - for all .
Some of the special graph classes we study in the next section are perfect, which is why we now provide some results for perfect graph classes.
Theorem 3.3
Let be a perfect graph. Then it holds that is -vertex-stable if and only if is -vertex-stable.
Based on this result, the next corollary follows immediately.
Corollary 2
Let be a class of perfect graphs. Then, for all graphs in , we have --VertexStability.
While the above results are related to the concepts of stability and vertex-stability, the subsequent two results address the topic of unfrozenness.
Theorem 3.4
Let be a graph class closed under complements and subgraphs. If or is tractable for , then - for all .
Note that this theorem exploits a relation between - and -Stability and -Unfrozenness. The next theorem follows by a similar approach, but exploits a relation between -Stability and - and -Unfrozenness.
Theorem 3.5
Let be a graph class closed under complements and subgraphs. If is tractable for , then - and - for all .
4 Tractability Results for Special Graph Classes
Ahead of our results for the individual graph classes, we provide two observations (proven in the appendix) which we will use multiple times in upcoming proofs.222Note that the second observation is in line with Observation 2 of Frei et al. [6].
Observation 1
--
Observation 2
Let be a graph. If an edge is -critical, then and are -critical, too.
With these two observations we can now inspect several graph classes. In the following subsections we study the problems -Stability, -VertexStability, and -Unfrozenness with , restricted to special graph classes. Frei et al. [6] showed that for we have - as well as -, where is the null graph, i.e., the graph without vertices or edges.
This is why we do not study problems related to vertex-unfrozenness, as all related questions are already answered.
4.1 Empty Graphs
Let denote the empty graph with isolated vertices and the set of all empty graphs.
Theorem 4.1
It holds that is (1) -vertex-stable for and , (2) -vertex-stable for and , (3) -vertex-stable, (4) -vertex-stable only for , and (5) -stable for .
With the previous theorem we can efficiently decide for every empty graph whether it is -stable or -vertex-stable for .
Furthermore, Theorem 4.1 also provides results for the null graph . Therefore, we exclude the null graph from all subsections hereafter in order to decrease redundancy.
Proposition 1
The only -unfrozen empty graphs are and .
The next corollary, which follows from the results in the next subsection, shows that all remaining problems related to unfrozenness are in , too.
Corollary 3
-, -, and -Unfrozenness belong to for empty graphs.
4.2 Complete Graphs
Since we studied empty graphs, it was an immediate consequence that we also study complete graphs. A complete graph with vertices is defined as and we denote the set of all complete graphs by . We know that -VertexStability holds. Together with the first statement of Proposition 6 from [6] we obtain that -VertexCriticality, so every complete graph is -vertex-critical. Furthermore, we know that -VertexStability. Applying the second statement of Proposition 6 from [6], we obtain that -VertexStability as well as -VertexCriticality. Consequently, every complete graph which does not possess exactly one vertex is -vertex-critical and -vertex-stable.
Observation 3
It holds that -VertexCriticality.
So far we know for all parameters whether complete graphs are vertex-stable or not, summarized in the subsequent corollary.
Corollary 4
For every , the problem -VertexStability belongs to for complete graphs.
Next, let us take a look at the edge-related stability problems.
Observation 4
For , we have , , , and .
Since and , these cases were already covered in the previous subsection, so the next result is stated only for .
Proposition 2
Let with . Then is -critical for .
We call a complete graph complete because all possible edges (ignoring loops) are present in the graph. Hence, for every , we have , so the next corollary is immediately clear.333This corollary is in line with [6, Proposition 5(2)], as .
Corollary 5
For all , every is -unfrozen.
4.3 Paths
Denote by the path with vertices and by the set of all paths. Again, all proofs are deferred to the appendix.
Observation 5
For , we have , , and and , , and . Additionally, for , we have and .
Having made this straightforward observation, we can formulate the following stability results for paths. Thereby, is ignored, as argued earlier.
Theorem 4.2
Let . is -stable and -vertex-stable but not -vertex-stable for . is neither -stable nor -vertex-stable for , but it is -vertex-stable. is -stable but not -vertex-stable. For , is - and -stable as well as - and -vertex-stable; it is not -vertex-stable; and it is neither -stable nor -stable but -vertex-stable if is even, and it is -stable and -stable but not -vertex-stable if is odd.
This theorem yields for all paths and that -Stability and -VertexStability are in . We exclude onwards.
Observation 6
For , is -unfrozen and is -frozen.
It remains to study the unfrozenness of paths with vertices.
Proposition 3
For , is neither -unfrozen nor -unfrozen.
Proposition 4
For , is neither - nor -unfrozen if is odd, and is - and -unfrozen if is even.
Again, -Unfrozenness is in for all paths and .
4.4 Trees and Forests
We say is a tree (i.e., ) if has no isolated vertices and no cycles of length greater than or equal to . Furthermore, is a forest (i.e., ) if there exist trees such that . For every tree , it holds that (see, e.g., Bollobás [3]). So, we have if , and if .
Also, there exists a tractable algorithm to determine for trees (for example, as , we can simply use the algorithm for bipartite graphs from 8). Thus we can compute for trees using Gallai’s theorem [7] (stated as Theorem 0.H.1 in the appendix), and all four graph parameters , , , and are tractable for trees.
Now, let with and , , be a forest. It is easy to check that , , , and . Furthermore, it is known that the class of forests is closed under subgraphs and induced subgraphs. From these observations we have the following results (with proofs in the appendix).
Theorem 4.3
Let be a graph parameter. Then the problems -Stability and -VertexStability are in for all forests.
With the next corollary follows immediately.
Corollary 6
For all and , the problems -Stability and -VertexStability belong to .
We now focus on the unfrozenness problems. All trees and forests with fewer than three vertices (, , and ) were already covered in previous sections. It remains to study trees and forests with at least three vertices.
Proposition 5
Every tree with is neither - nor -unfrozen.
Based on this result we can deduce whether forests are - or -unfrozen. As forests without edges are empty graphs, we study forests with at least one edge.
Theorem 4.4
If contains but no as induced subgraphs, is - and -unfrozen. If contains as an induced subgraph, is not - nor -unfrozen.
- and -Unfrozenness are covered in Corollary 10 of the next subsection.
4.5 Bipartite Graphs
is a bipartite graph if and . Denote the set of all bipartite graphs by . We begin with two simple observations. Again, most proofs are deferred to the appendix.
Observation 7
Let be a bipartite graph. Then if , and if .
Consequently, we can efficiently calculate and for all bipartite graphs. Next, we describe a tractable method to calculate and for bipartite graphs.
Observation 8
We can calculate and efficiently for .
Hence, we can efficiently calculate for every and . Furthermore, as the class of bipartite graphs is closed under subgraphs and induced subgraphs, the following corollary follows from Theorem 3.1.
Corollary 7
For every , the problems -Stability and -VertexStability are in for all bipartite graphs.
Next, we discuss approaches for how to decide whether a bipartite graph is stable. If a bipartite graph has no edges, we have . For bipartite graphs with one edge, we have the following simple result.
Proposition 6
Let be a bipartite graph with . Then is neither -stable nor -vertex-stable for .
Next, we provide results for bipartite graphs with more than one edge.
Lemma 1
Every bipartite graph with is -stable.
With Lemma 1 we can characterize -vertex-stability.
Theorem 4.5
Let be a bipartite graph with . is -vertex-stable if and only if for all it holds that .
The proof of the following lemma is similar to that of Lemma 1.
Lemma 2
Every bipartite graph with is -stable.
With Lemma 2 we also can characterize -vertex-stability.
Theorem 4.6
Let be a bipartite graph with . is -vertex-stable if and only if for all it holds that .
Besides these (vertex-)stability characterizations for bipartite graphs, we now address unfrozenness for them.
Theorem 4.7
Let be a bipartite graph. is -unfrozen if and only if possesses no as an induced subgraph.
Proof. We prove both directions separately. First, assume is -unfrozen but contains as an induced subgraph. Write and for the corresponding vertices and edges. Then . However, adding to we obtain , as forms a -clique in , a contradiction to the assumption that is -unfrozen. Next, assume that possesses no as an induced subgraph but is not -unfrozen. Hence, there must exist such that . Denote the two disjoint vertex sets of by . Obviously, and cannot be true, since then would hold. Therefore, without loss of generality, we assume . Adding to must create a cycle of odd length in , as cycles of even length as well as paths can be colored with two colors. Consequently, possesses with , , as a subgraph. This implies that must possess as an induced subgraph, again a contradiction. ❑
Slightly modifying (the direction from right to left in) the previous proof yields Corollary 8. This time, adding to must create a -clique in .
Corollary 8
is -unfrozen if and only if possesses no as an induced subgraph.
Both results show that - and -Unfrozenness belong to for bipartite graphs. For the last results of this section we require the following lemma.
Lemma 3
Let be a bipartite graph and . If , then there exists some vertex cover with and .
Theorem 4.8
For every , the problem -Unfrozenness belongs to .
The previous proof allows for every nonedge of a bipartite graph to decide if it is -unfrozen such that - follows for . Gallai’s theorem [7] immediately yields - for bipartite graphs.
Corollary 9
-Unfrozenness and - for all .
Since , the next corollary follows as well.
Corollary 10
The problems - and -Unfrozenness as well as the problem -Frozenness belong to for all trees and forests.
4.6 Co-Graphs
First of all, we recursively define co-graphs, following a slightly adjusted definition by Corneil et al. [4].
Definition 1 (co-graph)
The graph is a co-graph. If and are co-graphs, then and are co-graphs, too.
We denote the set of all co-graphs by and use the operators and as is common (see, e.g., [6]). We will use the following result by Corneil et al. [4].
Theorem 4.9
Co-graphs are (i) closed under complements and (ii) closed under induced subgraphs, but (iii) not closed under subgraphs in general. Furthermore, is a co-graph if and only if possesses no as an induced subgraph.
Property (iii) is not proven in their work [4]. However, is an easy example since is a co-graph (see Example 1 below), and removing one edge yields . Since every co-graph can be constructed by and , we can identify a co-graph by its co-expression.
Example 1 (co-expression)
The co-expression describes the graph .
Obviously, we can build a binary tree for every co-graph via its co-expression. The tree’s leaves correspond to the graph’s vertices and the inner nodes of the tree correspond to the expression’s operations. For example, the tree in Figure 1 corresponds to the co-expression from Example 1 and, thus, describes a . We call such a tree a co-tree. To formulate our results regarding stability and unfrozenness of co-graphs, we need the following result of Corneil et al. [5].
Theorem 4.10
For every graph , we can decide in time whether is a co-graph and, if so, provide a corresponding co-tree.
Combining the previous results with the next one by Corneil et al. [4], we can efficiently determine a co-graph’s chromatic number.
Theorem 4.11
Let be a co-graph and the corresponding co-tree. For a node from , denote by the graph induced by the subtree of with root . To every leave of we add as a label . For every inner node of we add, depending on the inner node’s type, the following label: (1) If is a -node with children and , , and (2) if is a -node with children and , . If is the root node of , then it holds that .
A result similar to the previous one for was given by Corneil et al. [4].
Remark 1
We label all leaves of with . Each inner node of with children and is labeled by if contains the -operation, and by if contains the -operation. For the root of , it then holds that .
By the previous remark we can efficiently calculate for co-graphs. Based on these results, we can state the following theorems whose proofs again are deferred to the appendix.
Theorem 4.12
For every , the problem -VertexStability is in .
With a similar proof as for the previous theorem, we obtain the next result.
Theorem 4.13
For every , the problem -VertexStability is in .
We can use the same proof as for Theorem 4.13 to obtain the next corollary. However, this time we additionally use Gallai’s theorem [7] to calculate out of for and all induced subgraphs with one vertex removed.
Corollary 11
For every co-graph, the problem -VertexStability is in .
Although Frei et al. [6, Proposition 5(5)] have already shown that the problem -VertexStability is in for co-graphs, the next corollary provides an alternative because - is true and co-graphs are closed under complements.
Corollary 12
For all the problem -VertexStability is in .
Next, let us study the edge-related stability problems for co-graphs. To obtain our results, we need the following two auxiliary propositions.
Proposition 7
Let with and let be -critical for . There exist two co-graphs such that or . Assuming, without loss of generality, that , is -critical for .
Proposition 8
Let and . If and are -critical for , then is -critical for as well.
Having these results, we are now able to provide our stability-related results.
Theorem 4.14
For all co-graphs, the problem -Stability is in .
Next, we want to study the problem of -Stability for co-graphs. To do so, we need the following lemmas with their proofs deferred to the appendix.
Lemma 4
Let be a co-graph. We can compute all cliques of size for in time polynomial in .
Lemma 5
Let be a graph and and . Then it holds that and are in .
Having these results, we can show the next theorem.
Theorem 4.15
The problem -Stability is in for co-graphs.
As we now know that we can efficiently determine whether a given co-graph is -stable, we can exploit the fact that co-graphs are closed under complements to obtain the following corollary (whose proof is deferred to the appendix).
Corollary 13
The problem -Stability is in for co-graphs.
Corollary 14
The problem -Stability is in for co-graphs.
At this point, we finish the study of stability problems for co-graphs, as all open questions are answered, and turn to the problems related to unfrozenness. The next two corollaries exploit the fact that co-graphs are closed under complements and follow a similar argumentation.
Corollary 15
The problems -Unfrozenness and -Unfrozenness are in for co-graphs.
Corollary 16
The problem -Unfrozenness is in for co-graphs.
Finally, we answer the last remaining open question related to unfrozenness and co-graphs.
Theorem 4.16
The problem -Unfrozenness is in for co-graphs.
5 Conclusion
We have provided 84 tractability results regarding the stability, vertex-stability, and unfrozenness problems when restricted to special graph classes. In particular, we studied these three problems for seven important graph classes and four central graph parameters. Doing so, our work provides some baseline for further, more expanding work along this line of research. For future work, we propose to study further special graph classes that are not covered here. Besides the study of stability for other graph classes, one can also study the concept of cost of stability:444Bachrach et al. [1] study a related notion of “cost of stability” for cooperative games. Given a graph, the question is how costly it is to stabilize it. In other words, what is the smallest number of vertices or edges to be added to or removed from the graph such that the resulting graph is stable or unfrozen with respect to some graph parameter. Relatedly, it would make sense to combine these two approaches and study the cost of stability for special graph classes.
Acknowledgments
This work was supported in part by Deutsche Forschungsgemeinschaft under grants RO 1202/14-2 and RO 1202/21-1.
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Appendix 0.A Deferred Proofs from Section 3: General Stability and Unfrozenness Results
Proof of Theorem 3.2. Let be a graph and a graph parameter that can be computed efficiently for graphs in . Now, compute . Since is closed under subgraphs, for every every edge it holds that is in . Hence, we can compute efficiently. Simply check for every edge whether holds. If no such edge exists, we know that is -stable. Consequently, we can solve -Stability for all graphs in in . ❑
Proof of Theorem 3.3. Let be a perfect graph. We have as well as for every vertex it holds that as is an induced subgraph of . Consequently, we have
This completes the proof. ❑
Proof of Theorem 3.4. Without loss of generality, assume that we can efficiently compute for all . Since is closed under subgraphs, applying Theorem 3.2 yields that -Stability belongs to for all graphs in . Let be a graph. As is closed under complements, also belongs to . Consequently, we can efficiently compute . From Proposition 5 1. in [6] we know that -- holds. Hence, if is -stable, it immediately follows that is -unfrozen. Therefore, for graphs in we can decide efficiently whether they are -unfrozen, such that -Unfrozenness belongs to for all graphs in . ❑
Proof of Theorem 3.5. The argumentation is similar to the argumentation of Theorem 3.4. We know that is closed under subgraphs and we can compute for all efficiently. Consequently, by Theorem 3.2 it follows that -Stability belongs to for all graphs in . From Proposition 5 2. in [6] we know that
holds. Consequently, let be a graph. As is closed under complements, is in , too. Now, efficiently decide whether is -stable. If that is the case, it immediately follows that is - and -unfrozen. Hence, - and -Unfrozenness belong to for all graphs in . ❑
Proof of Observation 1. Let -VertexStability be a graph. Consequently, is -vertex-stable, i.e., every is -stable. Together with Observation 3 in [6] it follows immediately that every edge is -stable, since all its incident vertices are -stable. If all edges of are -stable, it follows that is -stable and thus -Stability holds. ❑
Appendix 0.B Deferred Proofs from Section 4.1: Empty Graphs
Proof of Theorem 4.1. We prove every item on its own.
-
1.
We have and for . Furthermore, for it holds that , such that holds. Consequently, for or is -vertex-stable, but not for since the only vertex is -critical.
-
2.
It holds that and for . Hence, and for are -vertex-stable but isn’t, since its only vertex is -critical.
-
3.
We have . Therefore, for every we have , so that all vertices are -stable and, thus, is -vertex-stable.
-
4.
It holds that . Hence, for every we have , so is not -vertex-stable for .
-
5.
Since for has no edges, all edges of are -stable and, thus, is -stable.
This completes the proof. ❑
Proof of Proposition 1. Since , it obviously follows that both are -unfrozen. For all other with , we just add one arbitrary nonedge to so that holds. Consequently, these graphs cannot be -unfrozen. ❑
Proof of Corollary 3. It is trivial to see that for all . Consequently, by Proposition 5 2. from [6] it follows that --Unfrozenness belong to for empty graphs, since -Stability is in for all complete graphs, c.f. Proposition 2. With a similar argumentation it follows from Proposition 5 1. that -Unfrozenness is in for all empty graphs, since -Stability is in for all complete graphs. ❑
Appendix 0.C Deferred Proofs from Section 4.2: Complete Graphs
Proof of Observation 3. Let be a complete graph. Obviously, for every vertex we have as well as . Consequently, for every we have
such that is -vertex-critical. ❑
Proof of Proposition 2. For every we have , such that is -critical. Furthermore, we have , and , so that is -, - and -critical. The previous argumentation holds for all edges in , and therefore, is -critical. ❑
Appendix 0.D Deferred Proofs from Section 4.3: Paths
Proof of Observation 5. Obviously, and are true. Furthermore, for we simply color the vertices of alternatingly in two colors. Additionally, one can immediately see , as well as for . For the last statements, we argue as follows. The graph has vertices and edges. The first and the last vertex of the graph can each cover at most one edge. All vertices in between can cover at most two edges. Consequently, we need vertices for a vertex cover of . Applying Gallai’s theorem [7] yields from the previous result, as must hold. ❑
Proof of Theorem 4.2.
-
1.
has no edges, so is -stable. Furthermore, since and holds, is -vertex-stable, too. Obviously, is not -vertex-stable, as . The same argumentation holds with respect to and .
-
2.
We have as well as , so is not -stable. Furthermore, we have and , so is neither -vertex-stable. However, , so is -vertex-stable.
-
3.
We have and . For all we have , and , as , i.e., is -stable. Since we have and , it follows that is not -vertex-stable.
-
4.
Let . For all we have , as . Consequently, is - and -vertex-stable. The same argument holds for all , so is - and -stable, too. From 5 we know that holds. Consequently, we have
so is not -vertex-stable.
If for a , we have as well as
so is not -stable. Furthermore, it holds that and
so is neither -stable. However, let . Consequently, we have
with , assuming w.l.o.g. . Then it follows that
so is -vertex-stable. Next, assume . We have and for every it holds that
for with . Without loss of generality we can assume that and for . Then it holds that
Therefore, and, thus, is -stable. Additionally, it holds that and for every we have
with , assuming without loss of generality that and for . Then it follows that
Therefore, is -stable. Finally, it holds that
so is not -vertex-stable. ❑
Proof of Observation 6. The first observation related to immediately follows from the fact that holds, because then every nonedge of is -unfrozen and hence is -unfrozen, too. For the second observation one must note that holds, i.e., there is only one nonedge. When we add this nonedge to we obtain . As , , and hold, it follows that is -frozen, such that is -frozen, too. ❑
Proof of Proposition 3. From 5 we know . When we add the nonedge to , we obtain as well as , since and then build a -clique. Thus is - and -frozen such that can neither be - nor -unfrozen. ❑
Proof of Proposition 4. For this result we require two facts. For denote by a circle with vertices. Then we have as well as . Furthermore, a path with vertices has two vertex covers of size , namely and . We prove both statements separately. First, assume for . In this case is a nonedge for . Adding to results in . With the previous remark we know that
such that is -frozen and, thus, cannot be -unfrozen. A similar argument with respect to ,
yields that is -frozen and consequently, cannot be -unfrozen.
Now, assume that for holds, i.e., is even. Denote by a nonedge of , i.e., for . As previously mentioned, there are two optimal vertex covers for , , containing all evenly numbered vertices, and , containing all oddly numbered vertices. Selecting , such that holds, results in an optimal vertex cover for , also covering the newly introduced edge . Consequently, , such that is -unfrozen. Since this argument holds for an arbitrary nonedge , it follows that all nonedges are -unfrozen and therefore, is -unfrozen, too. If is -unfrozen, for every nonedge we have
such that is -unfrozen, too. ❑
Appendix 0.E Deferred Proofs from Section 4.4: Trees and Forests
Proof of Theorem 4.3. Let and be a forest. As previously stated, we can efficiently calculate . Now, for every and it holds that , such that we can also efficiently compute and . Consequently, we can decide in time polynomial in whether is -vertex-stable or -stable and therefore, both problems, -Stability and -VertexStability belong to . ❑
Proof of Proposition 5. With and it follows that must contain as an induced subgraph. Denote the corresponding vertices by , , and . Then is true. Furthermore, as does not contain any cycle, must be a nonedge of . Adding to creates the -clique in , such that we obtain
Hence, is -frozen and thus cannot be -unfrozen. A similar argument yields that cannot be -unfrozen. ❑
Proof of Theorem 4.4. We prove both statements separately. If contains but no as an induced subgraph, we have . Let be a nonedge of . Both vertices satisfy one of two cases: Either the vertex is isolated or part of some in . In both cases, adding to does not create a -clique, such that still holds and is - and -unfrozen. Since that holds for all nonedges of , it follows that is - and -unfrozen. Contrarily, if contains as an induced subgraph, we can follow the same arguments as in the proof of Proposition 5 to see that is neither - nor -unfrozen. ❑
Appendix 0.F Deferred Proofs from Section 4.5: Bipartite Graphs
Proof of Observation 7.
-
1.
If , it obviously holds that , as we can color all vertices with the same color, and the largest clique has size , such that is true, too.
-
2.
If , denote . Consequently, we can color all vertices in in one color and all vertices in in a second color, since there are no edges among the vertices of nor . Furthermore, there can not exist a clique of size three or larger in , as bipartite graphs do not possess cycles of odd length as induced subgraphs, but every clique of size three or larger possesses a cycle of length three as an induced subgraph. ❑
Proof of Observation 8. Using the Hopcroft-Karp algorithm [11] we can efficiently compute a maximum matching for . Applying König’s theorem [14], we know that holds. Hence, can be computed efficiently for every . With Gallai’s theorem we obtain from and, thus, can compute efficiently for , too. ❑
Proof of Proposition 6. Denote ’s only edge by for . Then is neither -stable nor -vertex-stable, as as well as and for hold. Consequently, is neither -stable, following from [6, Proposition 5], nor -vertex-stable because of and . Furthermore, we have as well as , such that is neither -stable nor -vertex-stable. Lastly, but , such that is not -stable nor -vertex-stable, too. ❑
Proof of Lemma 1. Let be an arbitrary edge of . Since , it holds that and, thus, is -stable. ❑
Proof of Theorem 4.5. Assume to be -vertex-stable. Furthermore, as we assume that , it holds that . Then there cannot exist a vertex with , as such a vertex would be -critical, since because of . For the opposite direction, assume that for all vertices we have . Hence, no matter what vertex we remove from , it always holds that , so and, thus, is -vertex-stable. ❑
Proof of Lemma 2. For all we have as holds, such that is -stable. ❑
Proof of Theorem 4.6. Assume that is -vertex-stable. Consequently, for all it holds that . If there is one with , we have as , a contradiction to ’s -vertex-stability. Contrarily, assume that for all it holds that . Then, for all , it follows that . Consequently, and, hence, is -vertex-stable. ❑
Proof of Lemma 3. Denote by a minimum vertex cover with and write . Then and holds. As is a minimum vertex cover for , all edges in are covered by . All remaining edges in are covered by , so that covers all edges in and is a vertex cover of . ❑
Proof of Theorem 4.8. Let be a bipartite graph with and . Then holds and, according to 8, we can calculate efficiently. For any nonedge either (1.) or (2.) , , must hold. We study both cases separately: (1) If , then is a bipartite graph, such that we can efficiently calculate and compare it with to determine whether is -unfrozen or -frozen. (2) Without loss of generality assume . Then two cases are possible: (a) can be rearranged, such that it is bipartite. This is possible if and only if , which can be checked efficiently. In this case we can compute efficiently to determine whether is -unfrozen or -frozen. (b) is no bipartite graph since it contains a cycle of odd length as subgraph. In this case we check with Lemma 3 for , and afterwards for , whether there exists some minimum vertex cover for with or respectively. If one of these two checks is positive, we know that holds and hence, is -unfrozen. Otherwise, must hold, such that is -frozen. Doing so, we can check every nonedge efficiently for -unfrozenness, such that -Unfrozenness is in for all graphs in . ❑
Appendix 0.G Deferred Proofs from Section 4.6: Co-Graphs
Proof of Theorem 4.13. Let be a co-graph. According to Theorem 4.10 we can calculate the graph’s co-tree efficiently. Now, calculate according to Theorem 4.11. Since co-graphs are closed under induced subgraphs, for every it holds that is a co-graph, too. Thus we can calculate efficiently. If there is a vertex such that holds, we immediately know that is not -vertex-stable. Otherwise, if for all it holds that , it directly follows that is -vertex-stable. Consequently, we can decide for every co-graph whether it is -vertex-stable or not in and, therefore, it follows that -VertexStability is in for co-graphs. ❑
Proof of Theorem 4.13. Let be a co-graph. Calculate according to Remark 1. Now, for every we calculate as previously described. If there exists at least one vertex such that , it follows immediately that is not -vertex-stable. Otherwise, is -vertex-stable. Hence, this results in -VertexStability belonging to for co-graphs. ❑
Proof of Corollary 12. Let be a co-graph. Then its complement is a co-graph, too. Hence, we can exploit the fact that holds and reuse the same idea as in Theorem 4.13 to decide whether is -vertex-stable. ❑
Proof of Proposition 7. We prove both cases separately.
-
1.
If , it holds that . Furthermore, if is -critical for , then it holds that . As we assume , the removal of from only affects , i.e., . Therefore, it follows that must hold, as otherwise the removal of would not affect . Consequently, is true and is -critical for .
-
2.
If holds, we have
such that is -critical for . ❑
Proof of Proposition 8. Let be a co-graph and an edge with two -critical vertices . First, we study the case that as well as and holds. Afterwards, we explain how to generalize the proof.
From the previous Proposition 7 we know that must be -critical for and -critical for . According to Observation 4 from [6] there exists an optimal coloring for , such that for all it holds that . In other words, there is a coloring for , such that is the only vertex in of its color. A similar, optimal coloring must exist for with respect to . For the combined graph with removed, i.e., , according to Observation 1 from [6], it must hold that . Consequently, we can reuse and from and , assuming distinct colors sets for and , to obtain a legal coloring of with colors. However, we can color in the same color , as is colored, and thus obtain a legal coloring for with colors. This is possible because
-
1.
is the only vertex in colored in by definition of ,
-
2.
no vertex is colored with , as holds, and
-
3.
is the only vertex in with this color, by definition of , and there is no edge between and .
Consequently, after removing from , we can color with one color less than before, such that holds and is -critical.
Initially, we assumed that with and holds. If , there cannot exist any edge between vertices from and . Hence, the only cases left are or with both vertices in or . Without loss of generality, let us assume that both vertices are in . Following Proposition 7, we know that both vertices are -critical for , as they are -critical for . When we can show that is -critical for , it immediately follows that is also -critical for . That is because if and is -critical for , we have , such that
holds. If , there is one more argument to add. We know that and are -critical for and . Consequently, must hold, as otherwise, or cannot be -critical for , since is true. But then, it is enough to show that is -critical for , since reducing by one via removing also causes a reduction of by one and hence, is -critical for , too.
At some point, we must arrive in the case that one vertex is in and the other vertex is in and holds, since the -operation is the only possibility to add edges between vertices in co-graphs. ❑
Proof of Theorem 4.14. Let be a co-graph. We can compute efficiently and, according to Observation 1 in [6], for every edge and every vertex it holds that Thus, for every edge , we proceed as follows to efficiently determine whether is -critical or -stable for : Denote for . Then it follows that and are induced subgraphs of and is a subgraph of . According to the earlier referenced Observation 1, it must hold that
Hence, if or , which we can compute efficiently, it immediately follows that . In other words, if or is -stable, we know that must be -stable, too.555This is in line with Observation 3 from [6]. If and are -critical, it follows by Proposition 8 that is -critical. Since we can determine for every node in efficiently, whether it is -stable, we can also efficiently determine for every edge in whether it is -stable. Consequently, we can decide in polynomial time whether is -stable. Thus - for co-graphs follows. ❑
Proof of Lemma 4. For a co-graph , let us denote by the set of all cliques of of size . Then the result easily follows by the recursive nature of co-graphs. To begin, if , obviously it holds that . If , we have
If , we have
Since we can compute a co-graph’s co-tree as well as the size of its biggest clique(s) efficiently, it follows that the previously described algorithm to compute can be executed in polynomial time, too. ❑
Proof of Lemma 5. First of all, it is obvious that by removing a vertex or an edge from we cannot increase . Hence, holds. Furthermore, when we remove from , either is part of all cliques in of size , so that holds or is not part of all of them, so that holds. Generally speaking, by removing a vertex from , we can either reduce a clique’s size by one or leave it as it is. Now, let be an edge of . Either is between two vertices of a clique in of size or not. If that is the case, we reduce the clique’s seize by one or leave it unchanged. Hence, by removing an edge from , we either reduce by one or do not alter it at all. Therefore, for all and it holds that . ❑
Proof of Theorem 4.15. Let be a co-graph. By Theorem 4.11 we can compute efficiently for and all induced subgraphs. In order to decide whether is -stable, we proceed as follows for every edge :
Case 1: for two co-graphs , and either or holds, since there are no edges between and . Assume without loss of generality that . As , we efficiently check whether holds. In this case, we know that cannot be critical to , because even if would be -critical to , using Lemma 5, we still have . Otherwise, if holds, we study whether is -critical for by recursively selecting the appropriate case, this time with as . This is sufficient because if is -critical for , it is also -critical for .
Case 2: and or . In this case, it is sufficient to check whether is -critical for the partial graph, i.e., or , containing and . That is because holds and so, if is -critical for one of the two partial graphs, is also critical for . Once again, we check this by recursively applying the appropriate case for the corresponding partial graph.
Case 3: and are in different partial graphs. Assume that and holds. Now, in order for to be -critical, there must exist only one clique with as well as and only one clique with and . We can check both conditions efficiently using Lemma 4. If this is the case, then all other cliques in are strictly smaller than and all other cliques in are strictly smaller than . Hence, the only clique of size in is , containing and . Removing from causes to be reduced by one since there is only one clique of size in , and afterwards, it is missing the edge in . Therefore, only in this case is -critical.
The number of recursive calls is limited by , since every inner node of ’s co-expression combines at least two nodes. Every case can be computed efficiently, such that we can determine for a co-graph in time in for some whether is -stable. Consequently, -Stability is in for all co-graphs. ❑
Proof of Corollary 13. Let be a co-graph. Then is a co-graph, such that holds. If is -stable, which we can determine efficiently by Theorem 4.15, it follows immediately that is -stable. Hence, we can efficiently determine whether a co-graph is -stable, such that -Stability is in for co-graphs. ❑
Proof of Corollary 15. Let be a co-graph. Co-graphs are closed under complements, so is a co-graph as well and we can compute from efficiently. According to Theorem 4.15, we can check in time polynomial in whether is -stable. Using Proposition 5 2. from [6], we immediately know that is - and -unfrozen if is -stable. Hence, both problems, - and -Unfrozenness are in for all co-graphs. ❑
Proof of Corollary 16. From Corollary 13 we know that we can efficiently decide for a co-graph whether it is -stable. Hence, let be a co-graph. Consequently, is a co-graph, too, and we can calculate for in time polynomial in whether it is -stable. Applying Proposition 5 1. from [6], we know that if is -stable, it follows that is -unfrozen. Therefore, we can efficiently calculate whether is -unfrozen. ❑
Proof of Theorem 4.16. Let be a co-graph and a nonedge of . Since has at least two vertices, and , either or for two co-graphs holds. We handle both cases separately:
-
1.
If is true, then must belong either to or to , since , such that . Without loss of generality assume that holds. If is -unfrozen for , i.e., , then is -unfrozen for , since follows. Contrarily, if is -frozen for , i.e., , then is -frozen for as well, as holds. Hence, it is enough to determine whether is -unfrozen or -frozen for and we can follow the argumentation of this proof recursively for .
-
2.
If is true, can belong to or . We split this into two sub-cases:
-
(a)
If or , we proceed as follows. Without loss of generality assume . Since holds, an increase of affects only if . Otherwise, is -unfrozen for (but not necessarily for ). If , then it holds that if is -unfrozen for , it follows that is -unfrozen for , since is true. Similarly, if is -frozen for , it follows that is -frozen for , since . Consequently, it is enough to determine whether is -unfrozen or -frozen for and we can follow the argumentation of this proof recursively for .
-
(b)
If , then and follows. Now, if is true, it follows that is -frozen for , since . Contrarily, if or , it follows that is -unfrozen for since and share no edge but . Because of that we can arrange the colors for and in such a way that both vertices incident to have different colors, resulting in .
-
(a)
Following these cases, we can efficiently determine for every nonedge whether it is -frozen or -unfrozen for , resulting in - for all co-graphs. ❑
Appendix 0.H Gallai’s Theorem
For the sake of self-containment, we here state Gallai’s theorem [7], which is used to obtain several of our results, and we also provide its proof.
Theorem 0.H.1 (Gallai’s theorem)
For every graph , it holds that
Proof. Let be a vertex cover for of size and assume that there are two vertices which are adjacent, i.e., . This contradicts the fact that is a vertex cover for , as would not be covered by . Consequently, must be an independent set for and we obtain
| (1) |
Let be an independent set for of size . For every edge it must hold that either or is not in , as this would contradict the fact that is an independent set. Hence, must be a vertex cover for and we obtain
| (2) |
Equation 1 yields and Equation 2 yields , such that we obtain
This completes the proof. ❑