Interval Routing Schemes for Circular-Arc Graphs
Interval routing is a space efficient method to realize a distributed routing function. In this paper we show that every circular-arc graph allows a shortest path strict 2-interval routing scheme, i.e., by introducing a global order on the vertices and assigning at most two (strict) intervals in this order to the ends of every edge allows to depict a routing function that implies exclusively shortest paths. Since circular-arc graphs do not allow shortest path 1-interval routing schemes in general, the result implies that the class of circular-arc graphs has strict compactness 2, which was a hitherto open question. Additionally, we show that the constructed 2-interval routing scheme is a 1-interval routing scheme with at most one additional interval assigned at each vertex and we outline an algorithm to calculate the routing scheme for circular-arc graphs in time, where is the number of vertices.
Keywords: interval routing, compact routing, circular-arc graphs, cyclic permutations
Routing is an essential task that a network of processors or computers must be able to perform. Interval routing is a space-efficient solution to this problem. Sets of consecutive destination addresses that use the same output port are grouped into intervals and then assigned to this port. In this way, the storage space required is greatly reduced compared to the straight forward approach, in which an output port is stored specifically for every destination address. Of course, the advantage in space efficiency depends heavily on the number of intervals assigned to the output ports, which, in turn, depends on the address and network topology. Interval routing was first introduced in [SK82, SK85]; for a discussion of interval routing we refer the reader to [vLT87, FJ88, BvLT91] and for a detailed survey to [Gav00]. When theoretical aspects of interval routing are discussed, the network is represented by a directed (symmetric) graph . As with many routing methods, the problem is that space efficiency and path optimality are conflicting goals [PU89]: as shown in [GG98], for every , there exists an -vertex graph such that for every shortest path interval routing scheme (for short IRS) for the maximal number of intervals per directed edge is only bounded by . Also, it is NP-hard to determine the most space efficient shortest path IRS for a given graph [EMZ02, Fla97]. Other worst case results can be found in [TL99]. On the other hand, there are many special graph classes [FG98] including random graphs [GP98] that are known to allow shortest path IRSs with a constantly bounded number of intervals on all directed edges. If this number is a tight bound, it is denoted the compactness of the graph. The compactness of undirected graphs is defined as the compactness of their directed symmetric version111The directed symmetric version of an undirected graph is the directed graph with and the compactness of a graph class as the smallest , such that every graph in this class has compactness at most , if such exists.
In this paper we show that the class of circular-arc graphs has (strict) compactness 2 by presenting an algorithm to construct a corresponding IRS. A different approach to realize space efficient, shortest path routing in circular-arc graphs can be found in [DL02]. In [GP08] an -bit distance labeling scheme is developed that allows for each pair of vertices in circular-arc graphs to compute their exact distance in time. In [DYL06] an -bit routing labeling scheme is developed that allows to make a routing decision in time for every vertex in an arbitrary circular-arc graph. The resulting routing path has at most two more edges than the respective shortest path. Nevertheless, our result is interesting, since interval graphs and unit circular-arc graphs are included in the class of circular-arc graphs and known to have compactness 1 [FG98, NS96], while for the compactness of circular-arc graphs hitherto only a lower bound of 2 was known (cf. the graph in Fig. 1) and the solutions from [FG98] and [NS96] cannot be extended to the class of circular-arc graphs.
While interval graphs can be represented by intervals on a line, circular-arc graphs can be represented by arcs on a circle, that is to say, every circular-arc graph is the intersection graph of a set of arcs on a circle. The first polynomial-time () algorithm to recognize circular-arc graphs and give a corresponding circular-arc model can be found in [Tuc80]. An algorithm with linear runtime for the same purpose is given in [McC03]. In circular-arc graphs maximum cliques can be computed much faster than in arbitrary graphs; an algorithm that determines a maximum clique in a circular-arc graph in time or even in time, if the circular-arc endpoints are given in sorted order, is presented in [BK97]. A discussion of circular-arc graphs can be found in [Hsu95]. Circular-arc graphs have, amongst others, applications in cyclic scheduling, compiler design [Gol04, Tuc75], and genetics [Rob76].
This paper is organized as follows. In Section 2 we define cyclic permutations, circular-arc graphs, and interval routing schemes. In Section 3 we prove the main result of this paper, namely that every circular-arc graph allows a shortest path strict 2-interval routing scheme. To this end, we choose an arbitrary circular-arc graph and fix a cyclic permutation on , in Section 3.1. In Section 3.2 we fix an arbitrary vertex and partition into the three sets , , and ; the vertices in these sets appear consecutively in and are referred to as the vertices to the right of , the vertices face-to-face with , and the vertices to the left of , respectively. In Section 3.3 we show that the vertices in the three sets , , and can be assigned to the edges incident to , such that the conditions of a shortest path strict 2-interval routing scheme are fulfilled. In Section 4 we show that the upper bound of the number of intervals in a shortest path strict 2-interval routing scheme constructed in this way is close to the number in an 1-IRS. In Section 5 we outline how to implement the implied IRS in time.
2.1 Cyclic Permutations
Cyclic permutations are relevant for the two main subjects of this paper, namely circular-arc graphs and non-linear interval routing schemes. Just like a permutation, a cyclic permutation implies an order on a set with the only difference being that there is no distinct first, second, , or -th element. We also define cyclic permutations on multisets, where we assume that multiple appearances of an element are distinguishable, i.e., a multiset on elements is identified with .
Definition 2.1 (Cyclic Permutation)
A cyclic permutation on a (multi-)set with elements is a function such that .222If the condition holds for one element in , it holds for all elements in . We call the successor of (in ) and we write .
Every cyclic permutation is a bijective mapping. Let be a cyclic permutation on a (multi-)set . One could imagine as an arrangement of on a clock face, for an -hour clock. To go through beginning at means to consider the elements in as they appear in , beginning at . Next we define subsets of that appear consecutively in .
Definition 2.2 (Ring-Interval, Ring-Sequence)
For a cyclic permutation on a (multi-)set and , we recursively define the ring-interval from to (in ) as
By we indicate that the left endpoint is excluded, by that the right endpoint is excluded, and by that both endpoints are excluded. Therefore we have
The ring-sequence from to (in ), denoted by , is the order in which elements of appear when going through beginning at .
Let be a cyclic permutation on a (multi-)set and . If and , we have .
2.2 Circular-Arc Graphs
We assume that the reader is familiar with basic graph theoretical definitions. The intersection graph of a family of sets is the graph where the vertices are the sets, and the edges are the pairs of sets that intersect. Every graph is the intersection graph of some family of sets. A graph is an interval graph if it is the intersection graph of a finite set of intervals (line segments) on a line and a unit interval graph if these intervals have unit length. A graph is a circular-arc graph if it is the intersection graph of a finite set of arcs on a circle; the latter we call an arc model of . Since an interval graph is a special case of a circular-arc graph, namely a circular-arc graph that can be represented with a set of arcs that do not cover the entire circle, we define the set of strict circular-arc graphs as the set of circular-arc graphs that are not interval graphs. Every strict circular-arc graph is connected. A unit circular-arc graph is a circular-arc graph that has an arc model in which the arcs have unit length. For a survey on circular-arc graphs see [LS09] and for the definition of further special graph classes see [BLS99]. Fig. 1 illustrates a strict circular-arc graph.
A difference between interval graphs and circular-arc graphs, that is worth mentioning, is that the maximal cliques of interval graphs can be associated to points of the “interval model” and therefore an interval graph can have no more maximal cliques than vertices. In contrast, circular-arc graphs may contain maximal cliques that do not correspond to points of some arc model [LS09]. In fact, just as in arbitrary graphs, the number of maximal cliques in circular-arc graphs can grow exponentially in the size of the graph [Tuc80]. For the special case that there is an arc model where no three arcs cover the whole circle the number of maximal cliques is bounded by the number of vertices of the graph [CFZ08]. Also maximal cliques may occur several times within some arc model, which is addressed by multisets in this paper.
The following definition and the subsequent corollary formalize important graph-theoretic properties of circular-arc graphs.
Definition 2.4 (Clique-Cycle, Left Clique , and Right Clique )
Let be a circular-arc graph, be an arc model of , and be the circle of . To each point on corresponds a clique that contains the vertices whose corresponding arcs contain . Let multiset contain all cliques that correspond to points on and define as the multiset of cliques we obtain, when we remove all cliques from that are not maximal with respect to inclusion among the cliques in and all but one cliques that are equal. Pair , where is the cyclic permutation on that is implied by ordering the elements in as their corresponding points appear (clockwise) on , is called a clique-cycle for . For every vertex , there exist two cliques such that a clique contains if and only if . We call the left clique of , denoted by , and the right clique of , denoted by .
When circular-arc graphs are discussed in this paper, the argumentation is based on one of their clique-cycles deduced from some arc model.
A frequently used notation in our paper is that of a dominating vertex, which is a vertex adjacent to all other vertices of the graph. The set of all dominating vertices of graph is denoted by . Since a dominating vertex is adjacent to all the vertices, all the maximal cliques contain it, wherefore ’s left and right clique form a cyclic interval that covers the entire clique cycle.
Let be a clique-cycle for a circular-arc graph and a dominating vertex. The left clique and the right clique of are not unique, but can be chosen arbitrarily with being the only constraint.
Consider the arc model of a circular-arc graph and two intersecting arcs . The arcs and can intersect in the following two ways: 1. The intersection of and constitutes another arc. • This is the case, (a) if is included in or vice versa, (b) if and are congruent, i.e., cover the exact same part of the cycle, or (c) if exactly one endpoint of lies in and vice versa. 2. The intersection of and constitutes two arcs (the corresponding vertices will be defined as counter vertices in the subsequent definition). This is the case, if and jointly cover the circle and meet at both ends.
Definition 2.6 (Counter Vertex )
Let be a clique-cycle for a circular-arc graph . We call a counter vertex of , if and and (or, equivalently, ). We call a pair of counter vertices and denote the set of all counter vertices of by .333Note that, .
The next corollary follows directly from the arc model of a circular-arc graph.
Let be a circular-arc graph and be a vertex with . Then every vertex in is adjacent to or to every vertex in , in other words, and any of its counter vertices constitute a dominating set.
Definition 2.8 (Reaching to the Left/Right)
Let be a clique-cycle for a circular-arc graph , two adjacent vertices with and a clique with . We say, reaches at least as far to the right as if , and reaches further to the right than if . Analogously, we define reaches at least as far to the left as by and reaches further to the left than by .
Although the arcs corresponding to counter vertices cover very different areas of the circle, it is impossible to say which of the two arcs “reaches further to the left or right”, when the point of view is the middle of the circle.
2.3 Interval Routing Schemes
We assume that the reader is familiar with basic concepts of interval routing and refer to [Gav00] for an exhaustive introduction and survey. If an IRS assigns at most intervals to each (directed) edge and only implies shortest paths, we denote it by -IRS. If this -IRS is furthermore strict, which means that for every vertex no interval assigned to the outgoing edges of contains ’s number, it is denoted by -SIRS. Definition 2.9 is needed for Definition 2.10, which is an alternative definition for shortest path strict interval routing schemes that is equivalent to definitions in literature but better suited for our purposes. To apply Definition 2.10 to an undirected graph (as for example a circular-arc graph), the graph is converted to its directed, symmetric version. While directed edges are often called arcs, we choose the former term to avoid confusion with the arcs of a graph’s arc model.
Definition 2.9 (First Vertex )
Let be a directed graph and . Vertex is a first vertex from to , if there exists a shortest directed path , i.e., there is a shortest directed path from to , that first traverses . The set of first vertices from to is denoted by
Definition 2.10 (Shortest Path Strict Interval Routing Scheme)
Let be a directed graph. A shortest path strict interval routing scheme for is a pair , where is a cyclic permutation on , called the vertex order, and , called the directed-edge-labeling, maps every directed edge to a set of ring-intervals in such that for every vertex ,
maps the outgoing directed edges of to a set of ring-intervals in , such that the intervals assigned to different directed edges never intersect,
for every vertex one of these these ring-intervals contains (vertex must not appear in one of these intervals), and
if is contained in a ring-interval in , then is a first vertex from to .
Let be a vertex order. We say that given suffices a shortest path -SIRS, if there exists a directed edge-labeling for that maps every outgoing directed edge of to at most ring-intervals in , such that satisfies the three constraints above.
A directed graph supports a shortest path -SIRS, if a vertex order exists, such that every vertex given suffices a shortest path -SIRS.
3 Main result
The compactness of the class of circular-arc graphs has not been determined until now, which is somehow surprising, since the class of circular-arc graphs is closely related to the class of interval graphs and unit circular-arc graphs and both classes are known to have strict compactness [NS96, FG98]. However, for the compactness of the class of circular-arc graphs only a lower bound of 2 was known. In particular, circular-arc graphs exist that do not allow for optimal 1 interval routing schemes, as, for example, a wheel graph, i.e., a cycle together with a dominating vertex, with six outer vertices [FG98]. Also the circular-arc graph shown in Fig. 1 does not allow optimal 1 interval routing schemes. In this section we prove the main result of this paper, which is given by the following theorem and shows that the lower bound of 2 for the compactness of circular-arc graphs is indeed sharp.
Theorem 3.1 (Main Theorem)
The class of circular-arc graphs has strict compactness 2.
Since every non-strict circular-arc graph is an interval graph and therefore has strict compactness 1 [NS96], we only need to consider strict circular-arc graphs in the proof. For the rest of this section let be a clique-cycle for an arbitrary strict circular-arc graph and be the directed symmetric version of .
Outline of the proof
To show that there exists a shortest path -SIRS for and thus Theorem 3.1 holds, we proceed as follows. In Section 3.1 we construct a cyclic permutation on . In Section 3.2 we choose an arbitrary vertex and partition the vertices in in three disjoint ring-intervals in in dependency on . In order to show that given suffices a shortest path 2-SIRS, we show how to define an directed edge-labeling for the outgoing directed edges of that satisfies the corresponding constraints given in Definition 2.10. We consider each of the three ring-intervals in Section 3.3.1 and 3.3.2 and show how the vertices in the respective ring-interval can be sufficiently mapped to by . Since we have chosen arbitrarily, it follows that every vertex in suffices a shortest path -SIRS given and therefore, by Corollary 2.11, supports a shortest path -SIRS. Since is an arbitrary strict circular-arc graph, this proves Theorem 3.1.
The following notations are straight forward but may be formalized for convenience. Let be a directed edge with , where is a set of ring-intervals in , and be a ring-interval in . Assigning to directed edge means to define . Of course, when we start constructing , every directed edge is mapped to the empty set. Let and be two ring-intervals that can be joined to one ring-interval by Construction 2.3. If and are assigned to the same directed edge , we can redefine
and therefore save one ring-interval on directed edge . This redefinition is called compressing the (two) ring-intervals on directed edge (to one ring-interval). Let be a set of vertices. To distribute (over the outgoing directed edges of ) means to partition into ring-intervals and assign these to outgoing directed edges of (sufficiently). For convenience, we sometimes refer to ring-intervals as intervals.
3.1 Definition of the Vertex Order
In this section we consider a circular-arc graph together with a clique-cycle and show how to construct a cyclic permutation on that serves as the given vertex order. The ordering is obtained by sorting the arcs using their left cliques as primary sort key and right cliques as secondary sort key, making the first arc the successor of the last. A pseudocode for this purpose is presented in Listing 1 and its idea is explained next.
By Corollary 2.5 the left and right clique of a dominating vertex are not unique. Since it simplifies the proof, if all dominating vertices have the same left and thus also the same right clique, these cliques are unified in Line 1. For a fixed in Line 1, the generated cyclic permutation is fully deterministic, except for the ordering of true twins.444Two vertices in a graph are called true twins if they are adjacent to the same set of vertices and to each other. The dummy vertex introduced in Line 4 is needed to close the cyclic order once all vertices are integrated in . The loop in Line 6 runs once through all cliques in (beginning at , that is arbitrarily chosen in Line 2) in the order defined by (Line 13). For every visited clique the loop in Line 8 integrates every vertex, whose left clique is , in . By Line 9, vertices with the same left clique are ordered with respect to their right clique. An example for the defined vertex ordering can be found in Fig. 1.
The following observations are crucial for the rest of this paper.
The vertices are ordered primarily by their left clique, that is to say, vertices having the same left clique appear consecutively in .
Vertices with the same left clique are ordered in ascending order by their right clique.
For two vertices with , we know that and are adjacent and either
and reaches at least as far to the right as , or
, in other words, reaches one clique further to the left than .555Because the relation “reaching to the right” is not defined, if and are counter vertices, we intuitively extend Definition 2.8 to this case by fixing .
The next definition is based on Assertion 1 of the preceding remark, that is to say, on the fact that, since vertices having the same left clique appear consecutively in , for a given clique , two vertices exist, such that the vertices in ring-interval are exactly those with left clique .
Definition 3.3 (Head Vertex and Tail Vertex )
Let be a clique-cycle for some circular-arc graph , given by Listing 1, , and the unique vertices such that ( and are not necessarily distinct). We call the head vertex of and the tail vertex of and denote them by and , respectively.
Let be a clique-cycle for a circular-arc graph , given by Listing 1, and . The successor (in ) of the tail vertex of is the head vertex of the successor (in ) of , i.e., .
3.2 Partition of the Vertex Order
The cyclic permutation on allows to state further definitions. We now fix arbitrarily for the rest of the proof and show that given suffices a shortest path -SIRS. If is a dominating vertex, we can define for every vertex and have thereby shown that given suffices a shortest path -SIRS. Therefore, we now assume that is not a dominating vertex. Fig. (a)a illustrates the subsequent Definition.
Definition 3.5 (Left Vertex )
Let be a clique-cycle for some circular-arc graph , given by Listing 1, and not a dominating vertex. • Let be the union of the following two vertex sets. (i) The set of vertices that are adjacent to and reach further to the left than but are neither counter vertices of nor dominating vertices. (ii) The set of vertices in .666We have and . If , we define the left vertex of , denoted by , as the unique vertex in such that
Let be a clique-cycle for a circular-arc graph , given by Listing 1, not a dominating vertex, the left vertex of , and be a vertex that is adjacent to and reaches further to the left than but is neither a dominating vertex nor a counter vertex of . Then reaches at least as far to the left as .
Proof Assume reaches further to the left than . Then by Definition 2.8 we know that , which implies that
Since is primary ordered by the left cliques of the vertices, it follows
In Definition 3.8 we partition the vertices in into three ring-intervals in . The following argumentation gives the idea behind this partitioning and also proves Theorem 3.9. Consider we go through beginning at . Since the vertices in are primary ordered by their left cliques, the left clique of the vertices first considered is (or , if is the tail vertex of clique ). The next vertices traversed have left clique , followed by vertices with left clique , and so on. We eventually reach the set of vertices with left clique . The last vertex we come across in this set is and fixed in Definition 3.7. The vertices we came across so far (excluding , including ) will be defined as the “vertices to the right of ” in Definition 3.8 and are adjacent to , as was contained in their left clique.
Definition 3.7 (Middle Vertex )
Let be a clique-cycle for a circular-arc graph , given by Listing 1, and not a dominating vertex. We define the middle vertex of , denoted by , as the tail vertex of the right clique of .
When we continue visiting the vertices after in the same manner, the next vertex we come across that is not a dominating vertex but adjacent to has to be , as we could find a contradiction in the same manner as in the proof of Theorem 3.6, if we had . These vertices after and before are defined as the “vertices face-to-face with ” in Definition 3.8. When we continue to go through , we eventually reach again. These vertices from to (excluding ) are defined as the “vertices to the left of ” in Definition 3.8, which is illustrated in Fig. (b)b.
Definition 3.8 (, , )
Let be a clique-cycle for a circular-arc graph , given by Listing 1, not a dominating vertex, the left vertex of , if exists, and the middle vertex of . We define
the vertices to the right of as , and
the vertices face-to-face with
as , if does not exist, or else
as , and
the vertices to the left of
as , if exists, or else
as .777 We have from Definition 3.5.
Let be a clique-cycle for a circular-arc graph , given by Listing 1, and but not a dominating vertex.
Every vertex to the right of is adjacent to .
Every vertex face-to-face with that is not a dominating vertex is not adjacent to .
3.3 Definition of the Directed Edge-labeling
Next we show that given suffices a shortest path -SIRS by constructing a mapping from the set of outgoing directed edges of to at most two ring-intervals in , according to Definition 2.10. We investigate the vertices to the left of and the vertices to the right of in Section 3.3.1 and the vertices face-to-face with in Section 3.3.2. For every vertex in the respective interval, we determine a first vertex from to such that and every vertex that is hitherto assigned to directed edge can be embraced by at most two ring-intervals in . This is a simple task for the vertices in and and even for the vertices in the approach is straight forward, if there is a dominating vertex or a counter vertex of . In fact, it only gets tricky, if there are no dominating vertices and no counter vertices.
3.3.1 Vertices to the Left and Right
By the first assertion of Theorem 3.9 every vertex to the right of is adjacent to , wherefore these vertices can be distributed by assigning to directed edge , for every vertex . Assume that we have . In general, not every vertex in is adjacent to . Although is adjacent to all vertices in , it is generally not possible to assign to directed edge , since may contain vertices that are adjacent to (other than ), and therefore must be assigned to “their own directed edge”. Let be a vertex that is adjacent to and assume we start at to go through until we come across the next vertex that is adjacent to or itself. Denote this vertex . We can show that is adjacent to every in , wherefore we can assign to directed edge . Therefore the vertices to the left of can be distributed by assigning every vertex to directed edge , if is adjacent to , and else to directed edge , where is the first vertex that precedes in and is adjacent to . Theorem 3.10 proves the outlined idea formally. When distributing the vertices in sets and as just outlined, every outgoing directed edge of (except for edges incident to dominating vertices face-to-face with , which are not yet labeled) gets one ring-interval assigned.
Let be a clique-cycle for a circular-arc graph , given by Listing 1, but not a dominating vertex, and the left vertex of (then the vertices in are the vertices to the left of ). Let be the sequence of vertices that we obtain, when we order the vertices in that are adjacent to as they appear in ring-sequence and append . For and , vertex is a first vertex from to .
Proof For some , , let . Clearly the theorem holds if . If , does not appear in and therefore is not adjacent to . Since the vertices in are primary ordered by their left clique and appears before in , it follows that the left clique of is in . Since is adjacent to , is contained in every clique in . This implies that is contained in the left clique of and thus and are adjacent. Therefore is a shortest path between and and is a first vertex from to .
3.3.2 Vertices Face-to-Face
This section discusses the distribution of vertices face-to-face with . It is probably the common case, that neither counter nor dominating vertices exist, since only then path lengths are unbounded. Thus, this case is discussed in the next paragraph and the subsequent paragraph discusses the remaining cases.
188.8.131.52 Neither Dominating nor Counter Vertices exist
In this case, always reaches further to the left than and there is at least one vertex adjacent to that reaches further to the right. Let be a vertex that is not adjacent to , be the set of vertices adjacent to that reach farthest to the left and be the set of vertices adjacent to that reach farthest to the right ( and might intersect). As evident from the arc model of , every vertex in or every vertex in is a first vertex from to (cf. Corollary 3.13). Since we have , we now fix a vertex adjacent to that reaches farthest to the right.
Definition 3.11 (Right Vertex )
Let be a clique-cycle for a strict circular-arc graph without dominating vertices and without counter vertices, given by Listing 1, , the left vertex of , the middle vertex of , and be the set of vertices that are adjacent to and reach farthest to the right. We define the right vertex of , denoted by , as , if , or as , if , or else as an arbitrary vertex in .
In order to show that suffices a shortest path 2-SIRS given , we could choose arbitrarily in even if or . However, we will show that given suffices a shortest path -SIRS, if or . The next definition redefines the notation of the left and right vertex as a function, in order to allow recursive usage to easily determine a first vertex from to every vertex. The subsequent corollary is clear when illustrated.
Definition 3.12 (, and , )
Let be a clique-cycle for a strict circular-arc graph without dominating vertices and without counter vertices, cyclic permutation given by Listing 1, and . We define as the left vertex of and as the right vertex of .888It follows that is the left vertex of the left vertex of and is the right vertex of the right vertex of , etc. For , we define as the smallest such that is adjacent to and as the smallest such that is adjacent to .
Let be a clique-cycle for a strict circular-arc graph without dominating vertices and without counter vertices, cyclic permutation given by Listing 1, and . The following three assertions hold for every vertex that is not adjacent to .
If , then is a first vertex from to .
If , then is a first vertex from to .
If , then and are first vertices from to .
Since no vertex in is adjacent to , the corollary implies that or is a first vertex from to every vertex in . Since sequence “runs” through the clique-cycle as implied by and sequence “runs” through the clique-cycle as implied by , that is to say, in the other direction, there exists an such that and “meet”, or more formally, are adjacent or equal. This number is fixed in the following definition.
Definition 3.14 (Apex Number)
Let be a clique-cycle for a strict circular-arc graph without dominating vertices and without counter vertices, given by Listing 1, and . If we have or we set , otherwise we define as the smallest number greater than 1 such that and are adjacent or equal. We call integer the apex number of .
The separate treatment for apex number in Definition 3.14 ensures that the arcs of and intersect/meet adverse to on the circle and therefore the arcs of , , and to cover the entire circle. In particular, if and intersect in ’s arc they cannot cover the entire cycle, since they are not counter vertices, but the apex number would be defined as 1, if only the second part of the case distinction were in place.
By the definition, for apex number vertices and are adjacent if and only if and are adjacent. Furthermore, for vertices and are not adjacent.
The following Theorem 3.15 shows that there exists a vertex with , where is the appex number of , such that is a first vertex from to every vertex in and vertex is a first vertex from to every vertex in . Since Theorem 3.15 implies that the vertices in can be distributed by assigning to directed edge and to directed edge , the main theorem (Theorem 3.1) follows for the case in which neither dominating vertices nor counter vertices exist.
Let be a clique-cycle for a strict circular-arc graph without dominating vertices and without counter vertices, given by Listing 1, , the left vertex of , the middle vertex of , and the right vertex of . There exists a vertex such that, for every vertex , we have , and for every vertex , we have .
Proof Let the apex number of .
If the theorem holds for , since every vertex in is adjacent to .
The following argumentation is partially illustrated in Fig. 3.
If and , let
and . We have , because is adjacent to a vertex in . When reaches further to the right than , might appear in a clique in and, therefore, be adjacent to , which implies . However, in this case, we also have . Therefore, we could only have , when reaches so far to the right, such that it is adjacent to . However, in this case would reach further to the left than , because is adjacent to and is “at most” adjacent to . Because is the left vertex of this contradicts Theorem 3.6. Thus, Corollary 3.13 implies
Now we go through beginning at ,999This equality holds by Corollary 3.4. in other words, we begin at the first vertex after the “last” vertex in , until we eventually come across a vertex that either is or adjacent to .101010This vertex may be . Also there may be vertices in that are adjacent to , in other words, we may have “passed by” a vertex that is adjacent to already. We define
i.e. is the set of vertices after and before in . Note that, by the choice of , no vertex in is adjacent to .
Assume there is a vertex that is not adjacent to . When we go through beginning at we eventually come across (since and are adjacent, we have ). Before we come across , we come across , since otherwise and would be adjacent. We do not come across until we came across , since otherwise and would be adjacent. It follows that and are not adjacent and, because is adjacent to and , it follows that reaches further to the right than . Because and are adjacent, must appear in . It cannot appear in , because then is adjacent to , which contradicts Definition 3.11, since reaches further to the right than and is the right vertex of . Therefore, we have . Because we came across before , we have and, therefore,
Since the vertices in are primary ordered by their left clique, Inclusion (3) implies
Remember that we began at to go through until we found a vertex that is adjacent to . Inclusion (4) implies that we must have come across before . But then we have . Since
contradicts the initial assumption, every vertex in is adjacent to .
Since by the choice of no vertex in is adjacent to , we have
which, by Corollary 3.13, implies
We define . Since the ring-intervals and appear consecutively in , by Construction 2.3, we have
Since the vertices in are primary ordered by their left cliques, no vertex
is adjacent to and therefore we have . We define
i.e. is the set of vertices after and before in . Note that
since otherwise would reach further to the left than