An Algorithm for the Feedback Vertex Set Problem on a Normal Helly Circular-Arc Graph

The feedback vertex set (FVS) problem is to find the set of vertices of minimum cardinality whose removal renders the graph acyclic. The FVS problem has applications in several areas such as combinatorial circuit design, synchronous systems, computer systems, and very-large-scale integration (VLSI) circuits. The FVS problem is known to be NP-hard for simple graphs, but polynomial-time algorithms have been found for special classes of graphs. The intersection graph of a collection of arcs on a circle is called a circular-arc graph. A normal Helly circular-arc graph is a proper subclass of the set of circular-arc graphs. In this paper, we present an algorithm that takes ( ) O n m + time to solve the FVS problem in a normal Helly circular-arc graph with n vertices and m


Introduction
Let  be a family of nonempty sets.A simple graph G is the intersection graph of  if there exists a one-to-one correspondence between the vertices of G and the sets in  , such that two vertices in G are adja- cent if and only if their corresponding sets have a nonempty intersection.If  is a family of intervals on the real line, G is called an interval graph [1].Furthermore, a graph G is called a circular-arc graph if it is the in-tersection graph of a collection of arcs on a circle [1].Circular-arc graphs properly contain a class of interval graphs as a subclass.Circular-arc graphs have applications in areas such as genetics [2], traffic control [1], multidimensional scaling [3], compiler design [4], ring network modeling [5].In recent years, circular-arc graphs have been investigated extensively from both theoretical and algorithmic perspectives [6]- [9].Let ( ) , G V E = be a simple graph, where V is the set of vertices and E is the set of edges of G, with V n = and E m = .Suppose that V ′ is a nonempty subset of V.The subgraph of G whose vertex set is V ′ and whose edge set is the set of those edges of G that have both vertices in V ′ is called the induced subgraph on V ′ and is denoted by [ ] G V ′ [10].A cycle with no repeated vertices is a simple cycle.In this paper, the term "cycle" denotes "simple cycle".A feedback vertex set (FVS) consists of a subset F V ⊆ such that each cycle in G contains at least one vertex in F. In other words, a subset The FVS problem is to find an FVS of minimum cardinality (MFVS) in G.The FVS problem has applications in several areas such as deadlock prevention in operating systems [11], combinatorial circuit design [12], VLSI circuits [13], and information security [14].
The FVS problem is known to be NP-hard for general graphs [15] and bipartite graphs [16].In general, it is known that more efficient algorithms can be developed by restricting classes of graphs.For instance, interesting polynomial-time solutions for the FVS problem have been found for special classes of graphs, such as interval graphs [17] [18], permutation graphs [19], butterfly networks [20], hypercubes [21], star graphs [22], diamond graphs [23], and rotator graphs [24].Saha and Pal presented an algorithm that took

( )
O n m + time for the FVS problem in interval graphs using maximal clique decomposition [18].The algorithm obtains an MFVS in an interval graph by breaking all cycles for each maximal clique.Circular-arc graphs are a natural generalization of interval graphs.However, the algorithm presented by Saha and Pal [18] cannot be directly applied to circular-arc graphs because the number of maximal cliques in interval graphs is at most the number of vertices, whereas circular-arc graphs may have an exponential number of maximal cliques [25].In this paper, we propose an algorithm that takes

( )
O n m + time for the FVS problem in a normal Helly circular-arc graph.The remainder of this paper is organized as follows.We state the definitions and notations used throughout this paper in Section 2. Next, we present our algorithm for the FVS problem and analyze its complexity in Section 3. Finally, we summarize our findings in Section 4 and conclude the paper by briefly discussing the scope for future work.

Definitions and Notations
In this section, we provide the definitions and relevant notations used throughout the paper.These establish the basis of the algorithm presented in Section 3. We provide the definitions of a circular-arc model and its corresponding graph.Consider a unit circle C and a family  of n arcs 1 such that there is a one-to-one correspondence between vertex i V ∈ and i A ∈  such that an edge ( )

and only if i
A intersects with j A in the circular-arc model.Normal and Helly circular-arc models (NHCM) are precisely those without three or less arcs covering the entire circle [26].A graph that admits such a model is called a normal Helly circular-arc graph (NHCG).Examples of an NHCM and its corresponding graph are shown in Figure 1.For an NHCM consisting of n arcs, an arc ( ) Thus, ( ) 0, 0, 0, 0, 0,1, 0, 0,1,1,1, 0 ρ = , respectively.For a simple graph ( ) A chordal graph is a simple graph in which every cycle of length four or greater has a cycle chord.Interval graphs are a subclass of chordal graphs [18].Hence, an MFTS is obviously an MFVS for interval graphs.On the other hand, NHCGs are a superclass of interval graphs and not a subclass of chordal graphs.They can have some chordless cycles of length greater than three.For example, the graph 1 G shown in Figure 1 has chordless cycles 2,3,5, 6,9,10,11,12, 2 of length eight.If F is an MFTS and not an MFVS of an NHCG

Algorithm and Its Correctness
In this section, we present an algorithm for solving the FVS problem for an NHCG.We will concisely describe the outline of our algorithm.First, we decompose a given NHCG into maximal cliques.An FTS is obtained by removing 2 j N − vertices from each maximal clique j MC .An MFTS is constructed by minimizing the number of removed vertices.At this point, if the constructed MFTS includes no periphery, it is an MFVS.Otherwise, we can obtain an MFVS by including a vertex for breaking the periphery in the MFTS.Let G shown in Figure 1 as an example to illustrate Algorithm 1 step by step (the updated part is underlined).
Step 2 of Algorithm 1 repeats the processes described above until 1 U becomes an empty set.The method described above thus constructs an MFTS F of the NHCG G.
Here, we explain how ( ) i ρ is used to find an MFTS in Step 2 of Algorithm 1.In the example shown in MC N ≥ and a periphery have a vertex v in common.Clearly, a periphery is broken by removing a vertex v ("5" in Figure 2(a)).Such vertex v containing j MC and a periphery in common must be included in some ( ) where . By executing the method described above, Step 2 outputs an MFTS F such that an NHCG [ ] G V F − has neither a triangle nor a periphery, if possible.
Thus far, we have presented an example where an MFTS of an NHCG G is also its MFVS.However, there exist cases where an MFTS of G obtained by executing Step 2 of Algorithm 1 is not an MFVS of G.We describe the procedure to construct an MFTS of NHCG 2 G shown in [ ] 0,0,0,0,0,0,0,0,0,0,0,0 [ ]

Concluding Remarks
In this paper, we proposed an algorithm that takes

( )
O n m + time to find an MFVS on an NHCG with n vertices and m edges.Our algorithm employs other algorithm to find maximal cliques [27] according to a method that can be understood intuitively.The complexity of our algorithm depends on the number of maximal cliques in an NHCG.Reducing the complexity of the algorithm and extending the results to other graphs are issues to be explored in future research.

Figure 1 ,Figure 1 . 1 M
Figure 1.Normal Helly circular-arc model 1 M and its corresponding graph 1 G .(a) A normal Helly circular-arc model σ is the total number of triangles including vertex i in G.For the sake of convenience, in the example shown in Figure1, we denote the σ and ρ value sequences of G by = be an NHCG corresponding to a model M. Algorithm 1 receives as an input the endpoints , i i a b of each i A and back-arc set BA, and outputs an MFVS F of G.We use the graph 1 all maximal cliques can be generated in( )O n m + time for an NHCG G with n vertices and m edges[27].Moreover, we compute ( ) , 1 j r ≤ ≤ , has no triangle.In the example 1 G shown in Figure1, we first select vertices "1" and "4" and add them to F because ( ) ( ) the maximum values of all σ values.In the next step, for 7 MC , we select vertex "9" and add it to F because ( ) ( )

2 jN − vertices in descending order of σ for each 1 ) 1 MC and add them to F for 1 1 G MC . Because all triangles in 1 MC
is an FTS of G.It is obvious that the cardinality of F can be reduced by including vertices that appear in many triangles in G. ( ) i σ is, by definition, the total number of triangles including vertex i in G.An MFTS can be obtained by selecting e., vertex k is contained in the largest number of triangles, and 2 U is the set of maximal cliques containing k.Therefore, we break all triangles containing k in 2 j MC U ∈ by priority to reduce the cardinality of F. Here, we assume that 2 U consists of m maximal cliques, i.e. of Step 2, we select all vertices except two minima with σ values in and F is clearly an MFTS of [ ] are broken by removing vertices in F, ( ) similar argument, following the execution of the m-th step, no triangle, and F is an MFTS of [ ] the case of the first iteration, we select all vertices except two minima with σ values in j MC and add them to F for each 2 j

Figure 2 5 , 6 in
containing vertices u ("2", "3", "4" in Figure2(a)) except v in j MC .This is because it is clear from the corresponding model that if there exists a vertex x adjacent to u ( ) v ≠ in j MC , G must have a triangle uvx .Next, we consider the case of a maximal clique have two vertices , v w in common.It is obvious that a periphery is broken by removing either v or w ("2", "5" in Figure 2(b)).In this case, for v and w, there exist maximal cliques Figure 2(b)) containing v and w, respectively.Moreover, we have no maximal clique vertex except v and w ("3", "4" in Figure in j MC .This is because it is clear from the corresponding model that if there exists a vertex x adjacent to ( ) , u v w ≠ in k MC , G must have a triangle uvx or uwx .Therefore, in Step 2

Figure 3 Figure 2 .Figure 3 .
Figure 2. A maximal clique and a periphery sharing vertices.(a) MC shares a vertex with a periphery; (b) MC shares two vertices with a periphery.
Let G be an NHCG.Following the execution of Step 2 of Algorithm 1, F is an MFTS of G. Proof: Each triangle contained in G is a subset of any maximal clique no vertex is added to F. This implies that the elimination of vertices in F obtained in the previous step breaks all triangles of 2MC F− and add them to F. It is obvious that the cardinality of F can be reduced by adding vertices that appear in several triangles in F. Following this step, [ ]