Balanced Independent and Dominating Sets on Colored Interval Graphs

03/11/2020 ∙ by Sujoy Bhore, et al. ∙ University of Bonn 0

We study two new versions of independent and dominating set problems on vertex-colored interval graphs, namely f-Balanced Independent Set (f-BIS) and f-Balanced Dominating Set (f-BDS). Let G=(V,E) be a vertex-colored interval graph with a k-coloring γ V →{1,...,k} for some k ∈N. A subset of vertices S⊆ V is called f-balanced if S contains f vertices from each color class. In the f-BIS and f-BDS problems, the objective is to compute an independent set or a dominating set that is f-balanced. We show that both problems are -complete even on proper interval graphs. For the BIS problem on interval graphs, we design two algorithms, one parameterized by (f,k) and the other by the vertex cover number of G. Moreover, we present a 2-approximation algorithm for a slight variation of BIS on proper interval graphs.

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1 Introduction

A graph is an interval graph if it has an intersection model consisting of intervals on the real line. Formally, is an interval graph if there is an assignment of an interval for each such that is nonempty if and only if . A proper interval graph is an interval graph that has an intersection model in which no interval properly contains another [8]. Consider an interval graph and additionally assume that the vertices of are -colored by a mapping . We define and study color-balanced versions of two classical graph problems: maximum independent set and minimum dominating set on vertex-colored (proper) interval graphs. In what follows, we define the problems formally and discuss their underlying motivation.

-Balanced Independent Set (-BIS): Let be an interval graph with vertex coloring . Find an -balanced independent set of , i.e., an independent set that contains exactly elements from each color class.

The classic maximum independent set problem serves as a natural model for many real-life optimization problems and finds applications across fields, e.g., computer vision 

[2], information retrieval [14], and scheduling [16]. Specifically, it has been used widely in map-labeling problems [1, 18, 17, 4], where an independent set of a given set of label candidates corresponds to a conflict-free and hence legible set of labels. To display as much relevant information as possible, one usually aims at maximizing the size or, in the case of weighted label candidates, the total weight of the independent set. This approach may be appropriate if all labels represent objects of the same category. In the case of multiple categories, however, maximizing the size or total weight of the labeling does not reflect the aim of selecting a good mixture of different object types. For example, if the aim was to inform a map user about different possible activities in the user’s vicinity, labeling one cinema, one theater, and one museum may be better than labeling four cinemas. In such a setting, the -BIS problem asks for an independent set that contains vertices from each object type.

We initiate this study for interval graphs which is a primary step to understand the behavior of this problem on intersection graphs. Moreover, solving the problem for interval graphs gives rise to optimal solutions for certain labeling models, e.g., if every label candidate is a rectangle that is placed at a fixed position on the boundary of the map [9].

While there exists a simple greedy algorithm for the maximum independent set problem on interval graphs, it turns out that -BIS is much more resilient and -complete even for proper interval graphs and (Section 2.1). Then, in Section 3, we complement this complexity result with two  algorithms for interval graphs, one parameterized by and the other parameterized by the vertex cover number. We conclude with a -approximation algorithm for a slight variation of BIS on proper interval graphs.

The second problem we discuss is

-Balanced Dominating Set (-BDS): Let be an interval graphs with vertex coloring . Find an -balanced dominating set, i.e., a subset such that every vertex in is adjacent to at least one vertex in , and contains exactly elements from each color class.

The dominating set problem is another fundamental problem in theoretical computer science which also finds applications in various fields of science and engineering [5, 10]. Several variants of the dominating set problem have been considered over the years: -tuple dominating set [6], Liar’s dominating set [3], independent dominating set [11], and more. The colored variant of the dominating set problem has been considered in parameterized complexity, namely, red-blue dominating set, where the objective is to choose a dominating set from one color class that dominates the other color class [7]. Instead, our -BDS problem asks for a dominating set of a vertex-colored graph that contains vertices of each color class. Similar to the independent set problem, we primarily study this problem on vertex-colored interval graphs, which can be of independent interest. In Section 2.2, we prove that -BDS on vertex-colored proper interval graphs is -complete, even for .

2 Complexity Results

In this section we show that -BIS and -BDS are -complete even if the given graph is a proper interval graph and . Our reductions are from restricted, but still -complete versions of 3, namely 3-bounded 3 [15] and 2P2N-3 (hardness follows from the result for 2P1N- [19]). In the former 3 variant a variable is allowed to appear in at most three clauses and clauses have two or three literals, in the latter each variable appears exactly four times, twice as positive literal and twice as negative literal.

Remark 1.

-completeness of -balanced independent (dominating) set implies the -completeness of -balanced independent (dominating) set for . Let be the interval graph in an -balanced independent (dominating) set instance. We construct an interval graph consisting of independent copies of . Clearly has -balanced independent (dominating) set if and only if has an -balanced independent (dominating) set.

2.1 -Balanced Independent Set

We first describe the reduction. Let be a 3-bounded 3 formula with variables and clause set . From we construct a proper interval graph and a vertex coloring of as follows. We choose the set of colors to contain exactly colors, one for each clause in and we number these colors from to . We add a vertex for each occurrence of a variable in a clause in . Furthermore, we insert an edge whenever was inserted because of a positive occurrence of and was inserted because of a negative occurrence of . Finally, we color each vertex with color . See Figure 1 for an illustration. It is clear that the construction is in polynomial time. The graph created from is a proper interval graph as it consists only of disjoint paths of length at most three and can clearly be constructed in polynomial time and space.

Figure 1: The graph resulting from the reduction for -balanced independent set in Theorem 1 depicted as interval representation with the vertex colors being the colors of the intervals.
Theorem 1.

The -balanced independent set problem on a graph with vertex coloring is -complete, even if is a proper interval graph and .

Proof.

The problem is clearly in since for a given solution it can be checked in linear time if it is an independent set and contains vertices of each color.

We already described the reduction. It remains to argue the correctness. Assume was constructed as above from a 3-bounded 3 formula and let be a solution to the -balanced independent set problem on . We construct a variable assignment for as follows. By definition we find for each color precisely one vertex . If was inserted for a positive occurrence of we set to true and else to false. Since is an independent set in this assignment is well defined. Now assume it was not satisfying, then there exists a clause for which none of its literals evaluates to true. Hence, none of the at most three vertices corresponding to the literals in is in . Recall that there is a one to one correspondence between clauses and colors in the instance of -balanced independent set we created. Yet, does not contain a vertex of that color, a contradiction.

For the opposite direction assume we are given a satisfying assignment of the 3-bounded 3 formula with clauses . Furthermore let be the graph with vertex coloring constructed from as described above. We find a -balanced independent set of from the given assignment as follows. For each clause we choose one of its literals that evaluates to true and add the corresponding vertex to the set of vertices . Since there is a one-to-one correspondence between the colors and the clauses and the assignment is satisfying, clearly contains one vertex per color. It remains to show that contains only independent vertices. Assume for contradiction that there are two vertices and . Then, by construction of , we know that and further that correspond to one positive and one negative occurrence of in . By the construction of this implies a contradiction to the assignment being satisfying. ∎

2.2 -Balanced Dominating Set

We reduce from 2P2N-3 where each variable appears exactly twice positive and twice negative. Let be a 2P2N-3 formula with variables and clause set . For variable in we denote with the four clauses appears in, where are clauses with positive occurrences of and are clauses containing negative occurrences of .

Figure 2: Illustrations of three variable gadgets and a clause gadget from Theorem 3 as interval representations. The vertex colors are given as the colors of the intervals.

We construct a graph from as follows. For each variable we introduce six vertices and and for each clause with occurrences of variables , , and we add up to three vertices for each (In case a clause has less than three literals we add only one or two vertices). If the connection to the variable is clear, we also write , , , and for the vertices introduced for this variable’s occurrences in the clauses and , respectively. Furthermore, we add for each variable the edges , , , and , as well as for each clause all possible edges between the three vertices introduced for . For each variable we introduce five colors, namely , , , , and . We set . Finally, we set . Equivalently for , , and . See Figure 2 for an example.

In total we create many vertices and many edges, thus the reduction is in polynomial time. All variable and clause gadgets are independent components and only consist of paths of length three and triangles, hence is a proper interval graph. Furthermore, can clearly be constructed in polynomial time and space.

To establish the correctness of our reduction for 1-BDS we first introduce a canonical type of solutions for the graphs produced by our reduction. We call canonical, if for each variable we either find or . If for a variable and a -balanced dominating set we find one of the two above sets in , we say is in canonical form in . The next lemma shows that if has a -balanced dominating set we can turn it into a canonical one.

Lemma 2.

Let be a graph generated from a 2P2N-3SAT formula with clause set as above and a -balanced dominating set, then can be transformed into a canonical -balanced dominating set in time.

Proof.

Let be not in canonical form in . Since is a -balanced dominating set we know that either or of is in . Without loss of generality assume that . Consequently, we find that and . Now, let be the set that we obtain by deleting the elements of from and adding all missing elements of to . Clearly is in canonical form in . We need to show that is still a -balanced dominating set. It is straight forward to verify that every color appears exactly once in if was -balanced. Now assume there was a vertex that is not dominated by any vertex in . Yet, we at most deleted and in but since both and all their neighbors are dominated. As our operations only affected vertices introduced for and occurrences of we can simply iterate this process for each variable until every variable is in canonical form. ∎

Theorem 3.

The -balanced dominating set problem on a graph with vertex coloring is -complete, even if is a proper interval graph and .

Proof.

The problem is clearly in  as we can verify if a given set of vertices is a -balanced dominating set in by checking if it is a dominating set and if it contains vertices of each color in linear time.

Let be constructed from a 2P2N-3SAT formula with clause set as above and let be a -balanced dominating set of . By Lemma 2 we can assume is canonical. We construct an assignment of the variables in by setting to true if its and to false otherwise. Assume this assignment was not satisfying, i.e., there exists a clause such that none of the literals in evaluates to true. For each positive literal of we then get that the corresponding variable was set to false. Hence, for and consequently . Equivalently for each negative literal we find and . As a result we find that none of the vertices introduced for literals in is in and especially that none of them is dominated as they are each others only neighbors. Yet, is a -balanced dominating set by assumption, a contradiction.

In the other direction, assume we are given a satisfying assignment of a 2P2N-3 formula with clause set . Furthermore, let be the graph constructed from as above. We form a canonical -balanced dominating set of in the following way. For every variable that is set to true in the assignment we add to and for every variable that is set to false we add . This clearly is a -balanced set and it is canonical. It remains to argue that it dominates . For the vertices introduced for variables this is clear, since we pick either or , as well as or for every variable . Now, assume there was a clause and none of the vertices introduced for literals in was in . Then, by construction of , we find that for any positive (negative) occurrence of a variable in the variable was set to false (true). A contradiction to the assignment being satisfying. ∎

3 Algorithmic Results for the Balanced Independent Set Problem

In this section, we first take a parameterized perspective on -BIS and provide two  algorithms222 is the class of parameterized problems that can be solved in time for input size , parameter , and some computable function . with different parameters. Then we give a -approximation algorithm for the related problem of maximizing the number of different colors in the independent set.

3.1 An FPT Algorithm Parameterized by

Assume we are given an instance of -BIS with being an interval graph with vertex coloring . We can construct an interval representation , , from in linear time [12]. Then our algorithm works as follows. To start with, we sort the right end-points of the intervals in in ascending order. Next, we define for all intervals with the index as the index of the interval whose right endpoint is rightmost before ’s left endpoint. For each color , let denote the

-dimensional unit vector of the form

, where the element at the -th position is and the rest are . For a subset we define a cardinality vector as the -dimensional vector , where each element represents the number of intervals of color in . We say is valid if all and the set is independent.

The key observation here is that there are at most many different valid cardinality vectors as there are only colors and we are interested in at most intervals per color. In the following let , , be the union of all valid cardinality vectors of the first intervals in . Let in the beginning. To compute an -balanced independent set the algorithm simply iterates over all right endpoints of the intervals in and in the -th step computes as . Finally, we check the cardinality vectors in and return true in case there is one with entries being all and false otherwise. An -balanced independent set can be easily retrieved by backtracking the decisions we made to compute the cardinality vector.

Theorem 4.

Let be an interval graph with a vertex coloring . We can compute an -balanced independent set of or determine that no such set exists in time, where is the inverse Ackermann function.

Proof.

Let be an interval representation of on which we execute our algorithm. For the set just contains the valid cardinality vector with all zeros which is clearly correct. Let be the set of valid cardinality vectors computed after step . Now, in step we calculate the set as the union of and the potential new solutions based on independent sets of intervals containing . Assume is an independent set of intervals such that its cardinality vector is valid, but there is no valid cardinality vector such that is larger or equal in every component than . Since contained all valid cardinality vectors for the intervals in we know that is such that . Yet, the set contained all valid cardinality vectors for the set of intervals . Since has overlaps with all intervals in and hence cannot be in any independent set with any such interval we can conclude that . Though, we also find , a contradiction.

Next we consider the running time. The key observation is that there are at most different valid cardinality vectors. Checking the validity can be done in for each new vector as only one entry changes. Furthermore, using a standard disjoint-set data structure, we can guarantee that we do not store any duplicates and update the set of valid cardinality vectors in time where is the inverse Ackermann function. Finally, we have to sort the intervals. Using standard sorting algorithms this works in . Altogether, this results in a running time of . ∎

3.2 An FPT Algorithm Parameterized by the Vertex Cover Number

Here we will give an alternative  algorithm for -BIS, this time parameterized by the vertex cover number of , i.e., the size of a minimum vertex cover of .

Lemma 5.

Let be a graph. Consider a vertex cover in and its complement . Then any maximal independent set of can be constructed from by adding the subset of and removing its neighborhood in , namely

Proof.

For a fixed but arbitrary maximal independent set , in the following, we denote the set ( as .

We first prove the independence of . Note that by the definition of a vertex cover is an independent set. Furthermore, the set , as a subset of the independent set , is also independent. Then, in the union of these two independent sets, any adjacent pair of vertices must contain one vertex in and one in . Hence, after removing all the neighboring vertices of , the set is independent.

Next we prove that . Assume there exists one vertex in but not in . Since it must also be in the set . With the assumption that , we get that must be in . Consequently, is in the independent set and is at the same time a neighbor of vertices in , a contradiction.

Finally we prove . We showed above that is an independent set and also . Since is a maximal independent set by assumption we get . ∎

Lemma 6.

Let be a graph with vertex cover number . There are maximal independent sets of .

Proof.

Consider a minimum vertex cover in and its complement . Note that since is a (minimum) vertex cover, is a (maximum) independent set. Furthermore, any maximal independent set of can be constructed from by adding and removing its neighborhood in , namely by Lemma 5. Thus there are maximal independent sets of . ∎

Theorem 7.

Let be an interval graph with a vertex coloring . We can compute an -balanced independent set of or determine that no such set exists in time.

Proof.

According to Lemma 6, there are maximal independent sets of . The basic idea is to enumerate all the maximal independent sets and compute their maximum balanced subsets. Enumerating all maximal independent sets of an interval graph takes time per output [13]. Given an arbitrary independent set of we can compute an -balanced independent subset in time or conclude that no such subset exists. Therefore, the running time of the algorithm is . ∎

3.3 A 2-Approximation for 1-Max-Colored Independent Sets

Here we study a variation of BIS, which asks for a maximally colorful independent set.

-Max-Colored Independent Set (-MCIS): Let be a proper interval graph with vertex coloring . Find a -max-colored independent set of , i.e., an independent set , whose vertices contain a maximum number of colors.

We note that the -completeness of 1-BIS implies that -MCIS is an -hard optimization problem as well. In the following, we will show a simple sweep algorithm for -MCIS with approximation ratio .

Our algorithm selects one interval for each color greedily. We maintain an array of size  to store the selected intervals. After sorting the right end-points in ascending order, we scan the intervals from left to right. For each interval in this order, we check if the color of is still missing in our solution (by checking if is not yet occupied). If so, we store in and remove all remaining intervals overlapping . Otherwise, if is not empty, we remove and continue scanning the intervals. This is repeated until all intervals are processed. Using that is a proper interval graph and a charging argument on the colors in an optimal solution that are missing in the greedy solution, we obtain our approximation result.

Figure 3: Comparison of a solution of the algorithm and an optimal solution . Subset contains two colors (red and blue) missing from , but the endpoint of each interval in is contained in a different interval from .
Theorem 8.

Let be a proper interval graph with a vertex coloring . In time, we can compute an independent set with at least colors, where is the number of colors in a 1-max-colored independent set.

Proof.

It is clear that the greedy algorithm described above finds an independent set. By maintaining the array , checking if a color is already selected takes constant time. Therefore the entire algorithm runs in time.

To prove the approximation factor , we compare the solution of our greedy sweep algorithm with a fixed 1-max-colored independent set (see Figure 3). Let be the subset of consisting of intervals in missing colors of . Consider an interval . In our solution , there must be one interval containing the left endpoint of . Otherwise, since the color of is not contained in , the greedy algorithm would scan and select it in the solution . Since is an independent set, for each in , we can find exactly one interval containing its left endpoint. And since is an independent set and is a proper interval graph, each in contains at most one left end-point of intervals in . By the injectivity of the “left endpoint contained in” relation, we can conclude that the cardinality of the set is greater than or equal to the cardinality of . Now note that , hence has size at least . ∎

4 Conclusions

In this paper, we have studied the -Balanced Independent and Dominating set problem for proper interval graphs. We proved that these problems are -complete and obtained algorithmic results for the -Balanced Independent Set problem. An interesting direction is to obtain algorithmic results for -Balanced Independent Set problem for other geometric intersection graphs, e.g., unit square intersection graphs, unit disk graphs etc. Another interesting problem is to design approximation or parameterized algorithm for the -Balanced Dominating Set problem for proper interval graphs.

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