Interval-Permutation Segment Graphs
II NTERVAL - PERMUTATION SEGMENT GRAPHS
Zlatko Joveski [email protected]
Jeremy P. Spinrad [email protected]
Department of Electrical Engineering and Computer ScienceVanderbilt UniversityNashville, TN 37235 A BSTRACT
In this work we introduce the interval permutation segment (IP-SEG) model that naturally generalizesthe geometric intersection models of interval and permutation graphs. We study properties of twograph classes that arise from the IP-SEG model and present a family of forbidden subgraphs for theseclasses. In addition, we present polynomial algorithms for the clique and independent set problemson these classes, when the model is given as part of the input. K eywords Permutation graphs · Interval graphs · Geometric intersection model
Many important graph classes are defined or can be characterized by a geometric intersection model. These charac-terizations can arise from different geometric objects that are being intersected. Examples include boxicity graphs(intersection graphs of d -dimensional rectangles), circular-arc graphs (intersection graphs of arcs along a circle), andstring graphs (intersection graphs of curves in the plane) [6, 10]. Two particularly well-studied examples are the classesof interval and permutation graphs [5]. In both of their respective models, the intersecting objects are line segments inthe plane, with different restrictions imposed on their positions. In interval graphs, the line segments must lie on a singleline, while in permutation graphs, their endpoints must lie on two separate parallel lines. Because of the similarity, it isnatural to look for geometric intersection models that would generalize those of interval and permutation graphs. Oneapproach is to have geometric objects that generalize line segments. In the model of simple triangle graphs (also knownin the literature as point-interval ) [11, 3], the intersecting objects are triangles, while in the model of trapezoid graphs(also referred to as interval-interval ) [4, 3], the intersecting objects are trapezoids.Another way of generalizing the models of interval and permutation graphs is to use the same kind of intersectinggeometric objects - straight line segments, but reduce the restrictions on their possible positions. If we drop allrestrictions, then we obtain the large class of SEG graphs - the intersection graphs of straight line segments in theplane [2]. However, many of the standard optimization problems, including independent set [9], remain NP-hard onSEG graphs. Sub-classes of SEG graphs, with restrictions on the number of directions that line segments could have,have been studied, including grid intersection (or 2-DIR) graphs [7]. Such models, however, do not generalize that ofpermutation graphs since segments there can have any direction in the plane, except being parallel with the two lines.Here we introduce a new model in which the intersecting objects remain straight line segments, but they can either liealong one of two horizontal lines or go from one horizontal line to the other. This leads to two natural generalizations ofboth interval and permutation graphs, based on whether all horizontal segments lie on the same line. We formally definethe models and graph classes in Section 2. In Section 3, we show that these classes have implicit representations. Unlikesimple triangle and trapezoid graphs, the two new classes are not contained in the class of perfect graphs. However,in Section 4, we show that we are somewhat limited in how we can represent chordless cycles using the new model,which helps us identify some forbidden subgraphs for the graph classes. In Section 5, we present polynomial algorithms a r X i v : . [ c s . D M ] A ug or the clique and independent set problems, when the model is known. In Section 6, we discuss some of the openquestions on the new graph classes. Let G = ( V, E ) be a graph with a vertex set V = { v , v , . . . , v n } . An intersection model of G is a family of sets S i , i = 1 , , . . . , n, such that ( v i , v j ) ∈ E if and only if S i ∩ S j (cid:54) = ∅ . We say that G is an intersection graph of thefamily of sets S i .Many classes of graphs are defined or can be characterized as the intersection graphs of different types of families of sets.For example, line graphs are the intersection graphs of edges of graphs and circular arc graphs are the intersection graphsof arcs on a circle. In this work, we are primarily interested in interval and permutation graphs and their generalizations.Figure 1 shows the respective geometric intersection models of a graph that is both interval and permutation. Definition 2.1.
Interval graphs are the intersection graphs of intervals on the real line.
Definition 2.2.
Permutation graphs are the intersection graphs of line segments whose endpoints lie on two parallellines L and L , so that for each segment s , its endpoints s and s lie on L and L , respectively. Figure 1: A graph that is both permutation and interval, along with the respective geometric representations.Note that we can obtain an equivalent definition of interval graphs by dropping the requirement that the line in questionis real and by substituting intervals with line segments . This allows for multiple natural ways of simultaneouslygeneralizing the geometric intersection models of interval and permutation graphs. These include point-interval (or simple triangle ) and interval-interval (or trapezoid ) graphs introduced by Corneil and Kamula [3], in which thegeometric objects corresponding to vertices are formed by combining multiple line segments. In this work, we study adifferent generalization of permutation and interval graphs in which the geometric objects corresponding to verticesremain individual line segments.
Definition 2.3.
Let L and L and be two parallel lines. We say that a line segment s is an interval segment if both itsendpoints s and s lie on the same line L i . We say that s is a permutation segment if s lies on L and s lies on L .Let I be a set of interval segments and P a set of permutation segments, and let G be the intersection graph of the set ofsegments I ∪ P . We call I ∪ P an interval-permutation-segment ( IP-SEG ) model of G . If all interval segments in I lieon the same parallel line L i , we call I ∪ P an IP-SEG* model of G . Definition 2.4.
A graph G is an IP-SEG graph if it has an IP-SEG model. G is an IP-SEG* graph if it has an IP-SEG*model.Note that under the above definitions, an IP-SEG* model is also an IP-SEG model, which indicates that IP-SEG* graphsare contained in IP-SEG graphs. We will later show that this containment is proper.We know that other generalizations of permutation and interval graphs like simple triangle or trapezoid graphs remainperfect. One important characteristic that distinguishes IP-SEG and IP-SEG* graphs is that they may contain odd-holes,meaning they are not contained in the class of perfect graphs. IP-SEG* and IP-SEG models of a cycle on five verticesare shown in Figure 2. It is easy to see how these models can be extended to models of larger cycles.2
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Figure 2: An IP-SEG* model of a C is shown on the left. An alternative IP-SEG model is given on the right. Note thatthe numbers shown next to line segments indicate the corresponding vertex labels from a natural labelling of C . Each line segment s in the plane is uniquely determined by its endpoints s and s , each of which can be specifiedby their x and y coordinates. We will use the notation s = ( x ( s ) , y ( s ) , x ( s ) , y ( s )) . Thus, when we say we aregiven the model or representation of an IP-SEG or an IP-SEG* graph G , we mean that for each vertex v of G we aregiven a four-tuple of numbers ( x ( v ) , y ( v ) , x ( v ) , y ( v )) . In this form, this is not a representation that characterizesIP-SEG or IP-SEG* graphs. In fact it is the characterizing representation of the class of SEG graphs. However, weknow that for IP-SEG and IP-SEG* graphs, the y coordinate identifies the horizontal line that an endpoint belongsto, i.e. y ( v ) , y ( v ) ∈ { , } . y ( v ) = y ( v ) = y ( v ) indicates that we are dealing with an interval segment and y ( v ) (cid:54) = y ( v ) means we have a permutation segment. In addition and without loss of generality, we may assumethat for each permutation segment p the first endpoint is positioned on L i.e. p = ( x ( p ) , , x ( p ) , . Further, wemay assume that for each interval segment, its left endpoint comes first, i.e. for i = ( x ( i ) , y ( i ) , x ( i ) , y ( i )) , we have x ( i ) ≤ x ( i ) . We also show that we can limit the range of values that x ( v ) and x ( v ) can take.Recall that in the model of interval graphs, whether two intervals share a non-empty intersection depends entirely onthe relative ordering of their endpoints along the horizontal line. Similarly, whether two permutation segments of apermutation graph representation intersect depends entirely on the relative orderings of their endpoints along L and L . A similar observation can be made for IP-SEG graphs. In particular, for two segments s and q , the following holds:i) If s = ( x ( s ) , y ( s ) , x ( s ) , y ( s )) and p = ( x ( p ) , y ( p ) , x ( p ) , y ( p )) are two interval segments, they intersect ifand only if • y ( s ) = y ( p ) ( s and p lie on the same horizontal line), and • x ( p ) ≤ x ( s ) ≤ x ( p ) or x ( s ) ≤ x ( p ) ≤ x ( s ) ( s and p overlap).ii) If s = ( x ( s ) , , x ( s ) , and p = ( x ( p ) , , x ( p ) , are two permutation segments, they intersect if and only if • x ( s ) ≤ x ( p ) and x ( p ) ≤ x ( s ) , or • x ( p ) ≤ x ( s ) and x ( s ) ≤ x ( p ) .iii) If s = ( x ( s ) , y ( s ) , x ( s ) , y ( s )) is an interval segment and p = ( x ( p ) , , x ( p ) , is a permutation segment,they intersect if and only if s contains one of the endpoints of p , that is • y ( s ) = 1 and x ( s ) ≤ x ( p ) ≤ x ( s ) , or • y ( s ) = 2 and x ( s ) ≤ x ( p ) ≤ x ( s ) .Thus, all we need to know to determine adjacency between two segments s and p from an IP-SEG or an IP-SEG* modelis to know what types of segments p and s are, on which line L i each of their endpoints is located on, and what are therelative orderings of their endpoints on L and/or L . The last part implies that the number of different values the x endpoint coordinates could take is bounded by the maximum number of endpoints positioned on a single line and, byextension, the total number of endpoints. In other words, we may assume that in an IP-SEG or an IP-SEG* model of agraph G , we always have x ( v ) , x ( v ) ∈ { , , . . . , n } .With the above limitations on the range of x and y values, it follows that we need only O ( logn ) bits of informationper vertex to properly represent an IP-SEG or an IP-SEG* graph, meaning these two classes, similar to interval andpermutation graphs, have an implicit representation [8, 10]. Therefore, the number of IP-SEG and IP-SEG* graphs on n vertices is bounded by O ( nlogn ) Note that in the above discussion we have allowed for the possibility of segments sharing one or both endpoints. Weknow that permutation graphs are usually defined so that no two permutation segments share an endpoint. While this isnot imposed as a requirement in the definition of interval graphs, the consecutive cliques arrangement [1] of an intervalgraph implies that every interval graph has a representation in which no two intervals share an endpoint. This is also thecase for IP-SEG and IP-SEG* graphs. 3ndeed, suppose that we have an IP-SEG model of a graph G in which line segments share endpoints and let e be one ofthose endpoints. Without loss of generality, we may assume e lies on L . Let P be the set of permutation segmentshaving e as an endpoint. Similarly, let I L and I R be the sets of interval segments having e as a left and right endpoint,respectively. We can modify the IP-SEG model of G so that e is no longer a shared endpoint, in the following way. Usethe open interval i e around e that does not contain any other endpoints. In it, arrange the L endpoints of permutationsegments in P in a reverse order from the one their corresponding L endpoints have. In case some of the permutationsegments also share an endpoint on L , resolve the tie arbitrarily. Then, we can extend the interval segments in I L to the left, so that they all include new endpoints of permutation segments in P , end at different endpoints, but theirendpoints still remain within i e . We can achieve the same thing with interval segments in I R by extending them to theright. With this transformation, we ensure that permutations that shared e as an endpoint now either intersect between L and L , or still have a shared endpoint on L . In addition, all interval segments in I L and I R do contain the newendpoints of permutation segments P and all interval segments in I L and I R still contain the point e , meaning theyhave non-empty pairwise intersections. Finally, no new pairwise intersections are created as segments in P ∪ I L ∪ I R are the only ones that have an endpoint in the open interval i e .Thus, the above transformation leads to a new IP-SEG model of G that has one fewer shared endpoint. Doing thisrepeatedly, we can obtain an IP-SEG model in which no two segments share an endpoint. The following lemmasummarizes the results in this section. Lemma 3.1.
Every IP-SEG graph has an IP-SEG model such that for every vertex v , x ( v ) , x ( v ) ∈ { , , . . . , n } and no two segments of the model share an endpoint. In an IP-SEG model, two interval segments that do not lie on the same parallel line cannot intersect. Let G be a graphwith an IP-SEG model consisting of interval segments only. If G is connected, then all of the interval segments in itsIP-SEG model must lie on the same line. By extension, if G has more than one connected component, the intervalsegments of a single component C must lie on the same line. It is easy to see that the interval segments of a component C on L can be translated to L , while ensuring that they do not intersect with other interval segments already on L .This means that a graph G has an IP-SEG model consisting of interval segments only, if and only if it has an IP-SEG*model of only interval segments. In other words, G has an IP-SEG model consisting of only interval segments if andonly if G is an interval graph. We also have the more straightforward analogous observation for permutation segments:a graph G has an IP-SEG model consisting of only permutation segments if and only if G is a permutation graph.Interval graphs are chordal and therefore an IP-SEG or an IP-SEG* model of a chordless cycle of length greater thanthree cannot consist exclusively of interval segments. Similarly, permutation graphs cannot contain an induced cycle onmore than four vertices. Therefore, an IP-SEG or an IP-SEG* model of a chordless cycle of length greater than fourcannot consist exclusively of permutation segments. A cycle on four vertices, however, may be represented using onlypermutation segments, as shown in Figure 3.
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Figure 3: An IP-SEG* model of a C consisting of permutation segments onlyWe have already seen that a C is an IP-SEG and an IP-SEG* graph. It is easy to see how the models in Figure 2 can beextended by adding more interval segments to represent larger chordless cycles. Our goal is to the show that the ways inwhich such a cycle can be represented is in a certain sense limited and we will use that to identify examples of graphsthat do not belong to these two classes.Let u and v be vertices of an IP-SEG graph G corresponding to interval segments i u and i v , respectively, in a givenIP-SEG representation of G . Further, suppose that i v is fully contained in i u . Then, then u and v are neighbors andall other neighbors of v must also be neighbors of u in G . In other words, N [ v ] ⊆ N [ u ] must hold for the closedneighborhoods of v and u .Consider a path P n = ( v , v , . . . , v n ) . Clearly, N [ v ] ⊆ N [ v ] , N [ v n ] ⊆ N [ v n − ] , and the v and v n are the onlyvertices in P n with the property of having a closed neighborhood that is contained within a closed neighborhood ofanother vertex. Now consider an IP-SEG representation of P n using only interval segments. From the above discussion4t follows that an interval segment could be contained within another only if it corresponds to v or v n . We say that anIP-SEG representation of P n using only interval segments is an active interval representation if no interval segmentis contained within another. Assuming that the interval segment corresponding to v is the one with the leftmost leftendpoint, it is easy to see that in an active interval representation of P n , the left-to-right ordering of all left endpointswould coincide with the ordering of the vertices in P n . The same is true for the left-to-right ordering of all rightendpoints of interval segments. An example is shown in Figure 4. Figure 4: An active (left) and an inactive (right) interval representation of a P .Suppose we are given an IP-SEG representation IP( C n ) of a chordless cycle C n = ( v , v , . . . , v n ) , with n ≥ . Let s ( v i ) denote the line segment in IP( C n ) corresponding to vertex v i . Further, suppose that not all s ( v i ) ’s are permutationsegments, i.e. that IP( C n ) consists of both interval and permutation segments. Let S i,j = ( s ( v i ) , s ( v i +1 ) , . . . , s ( v j )) be a maximal sequence of consecutive interval segments in IP( C n ), i.e. such that s ( v i − ) and s ( v j +1 ) are permutationsegments and for each k ∈ { i, i + 1 , . . . , j } , s ( v k ) is an interval segment (indices are modulo n ). We call S i,j an interval arc of IP( C n ). Analogously, we can define the notion of a permutation arc of IP( C n ). Using the above notation,any IP-SEG representation IP( C n ) of a chordless cycle C n can be described as consisting of interval and permutationarcs. Clearly, the number of interval arcs must equal that of permutation arcs. We also have the following result. Lemma 4.1.
Let C n be a chordless cycle, with n ≥ . C n has an IP-SEG representation with t interval arcs of lengths { l , l , . . . , l t } if and only if C n − n (cid:48) , where n (cid:48) = (cid:80) ti =1 ( l i − , has an IP-SEG representation with t interval arcs eachof length .Proof. Let S i,i + k be an interval arc of a given IP-SEG representation of a chordless cycle C n . Note that S i,i + k ,being formed by k + 1 segments, must be an active interval representation of the path P k +1 . Suppose we take oneinterval segment s j in S i,i + k and replace it with two partially overlapping interval segments s j (cid:48) and s j (cid:48)(cid:48) such that s j = s j (cid:48) ∪ s j (cid:48)(cid:48) and s j (cid:48) ∩ s j (cid:48)(cid:48) does not overlap any of the other segments in S i,i + k . With this we are transformingan IP-SEG representation of C n into an IP-SEG representation of C n +1 . Similarly, assuming k ≥ , we can taketwo consecutive interval segments s j and s j +1 in S i,i + k and replace them with one new segment s j ∗ such that s j ∗ = s j ∪ s j +1 . This allows us to transform an IP-SEG representation of C n with an interval arc of length k + 1 intoan IP-SEG representation of C n − with an interval arc of length k .We are interested in determining how many interval arcs an IP-SEG representation of a chordless cycle can have. Recallthat in Figure 2 we saw one IP-SEG* representation of a C consisting of one interval arc and one permutation arc aswell as an IP-SEG representation consisting of two interval arcs and two permutation arcs. We will show that these arein fact the only two possible numbers of interval arcs we could have. Lemma 4.2.
An IP-SEG* model of a chordless cycle must consist of exactly one interval arc. I I B ( P ) p , Figure 5: An IP-SEG* representation of a chordless cycle cannot consist of two interval arcs because the broken line B ( P ) induced by the first permutation arc P will have to intersect a segment of the other permutation arc P . Proof.
Suppose there exists a chordless cycle C n with an IP-SEG* model that has two interval arcs. Then, by Lemma4.1, there exists a chordless cycle C m , m ≤ n , with an IP-SEG* model that has two one-segment interval arcs I and I . Without loss of generality, we may assume that in this model I and I both lie on L and I is to the left of I .Let P = ( p , , p , , . . . , p ,q ) and P = ( p , , p , , . . . , p ,r ) be the two permutation arcs so that p , and p , are the5ermutation segments that intersect I and the intersection point p , ∩ I is to the right of p , ∩ I . Consider the points z = p , ∩ I , z q +1 = p q, ∩ I , and z k = p ,k ∩ p ,k +1 , ≤ k ≤ q . The straight-line segments connecting z i with z i +1 , ≤ i ≤ q , form a broken line B ( P ) connecting z with z q +1 . B ( P ) is a continuous curve connecting z and z q +1 that is bounded between L and L . As such, one of its constituting straight-line segments must intersect p , (seeFigure 5). Since each of these segments must be contained within some permutation segment of P , that would implythat there are permutation segments p ,x and p ,y that intersect. However, these two permutation segments correspondto two non-consecutive vertices of a chordless cycle that cannot be adjacent, which is a contradiction. Therefore, a C n cannot have an IP-SEG* model with two interval arcs. Furthermore, if a C n could have an IP-SEG* model with k ≥ interval arcs, we could turn this into an IP-SEG* model with k − interval arcs by collapsing one of the interval arcsinto a single point and effectively merging two permutation arcs into one. Therefore, a C n cannot have an IP-SEG*model with more than one interval arc. Lemma 4.3.
An IP-SEG model of a chordless cycle must consist either of exactly one interval arc or of exactly twointerval arcs positioned on different horizontal lines. I I I p , B ( P ) I I I B ( P ) B ( P ) Figure 6: An IP-SEG representation of a chordless cycle cannot consist of two interval arcs on L and one on L because: (1) the broken line B ( P ) induced by the permutation arc P will have to intersect a segment of the permutationarc P (model on the left) or (2) a segment of the permutation arc P would have to intersect either B ( P ) , B ( P ) , or I (model on the right). Proof.
Given Lemma 4.2, we only need to consider proper IP-SEG models having at least one interval arc on eachhorizontal line. Suppose there exists a chordless cycle C n with an IP-SEG model that has two interval arcs I and I on L and one interval arc I on L . As in the proof of Lemma 4.2, we may assume that each of these areone-segment interval arcs. Let P = ( p , , p , , . . . , p ,q ) be the arc between I and I , P = ( p , , p , , . . . , p ,r ) thearc between I and I and P = ( p , , p , , . . . , p ,s ) the arc between I and I . Further, assume that p , and p , arethe permutation segments of P and P , respectively, intersecting with I . If the intersection point p , ∩ I is to theleft of p , ∩ I , then the broken line B ( P ) induced by the intersection points of permutation segments of P (seeproof of Lemma 4.2) must intersect p , . This would imply an intersection between two permutation segments thatcorrespond to non-consecutive vertices in C n , a contradiction. Suppose the intersection point p , ∩ I is to the left of p , ∩ I and consider the endpoint of p , that lies on L . It cannot lie within I as p , and I do not correspond toconsecutive vertices in the chordless cycle. But if it were to lie to the left or right of I , then B ( P ) or B ( P ) wouldneed to intersect with p , , both of which lead to a contradiction. Therefore, a C n cannot have an IP-SEG model withtwo interval arcs on one horizontal line and one interval arc on the other. Recalling the observation about collapsing arcs,this also implies that an IP-SEG model of a chordless cycle C n cannot have more than one interval arc per horizontalline.We can use Lemma 4.3 to identify examples of graphs that do not belong to the classes of IP-SEG* and IP-SEG graphs.Consider the graph G on 21 vertices formed by a cycle C = ( v , v , . . . , v ) and vertices w i and z i , ≤ i ≤ , suchthat w i is pendant to v i and z i is pendant to w i . If G has an IP-SEG* model, it must consist on one interval arc I andone permutation arc P . Let P = ( p , . . . , p k ) , with consecutive segments corresponding to consecutive vertices in C and p intersecting I to the left of p k . It is easy to see that ≤ k . We also have that k ≤ , since we cannot representcycles of length more than four with permutation segments only. Let O and O be the two permutations of (1 , . . . , k ) that define the intersections of permutation segments of P . must appear before k in both O and O . Further, since p and p k are the only permutation segments intersecting I , and k must appear consecutively in O . If k ≥ , then p cannot intersect neither p nor p k . However, the only way to achieve this is if both O and O contain the sub-ordering (1 , , k ) , which contradicts and k being consecutive in O . Therefore, k ≤ .From the above, it follows that the IP-SEG* representation of C must have an interval arc composed of at least threesegments. Let v t be a vertex corresponding to an interior interval segment s t of I . Note that since s t is between theendpoints of the broken line B ( P ) induced by the permutation arc, the segment corresponding to v t ’s neighbor w t cannot be a permutation segment. This, combined with the fact that s t ’s ends are overlapped by the two neighboringinterval segments, w t must correspond to an interval segment that is fully contained in s t . However, w t has a neighbor6igure 7: The graph G (left) is not an IP-SEG* graph, but is an IP-SEG graph. One possible IP-SEG model of G isshown on the right. z t that it does not share with v t , a contradiction. Therefore, G is not an IP-SEG* graph. Note that this would be truefor any graph G n constructed in an analogous way from a cycle C n with n ≥ . G is also a minimal non-member ofIP-SEG* under vertex removal. G and G , can be represented using an IP-SEG model, which demonstrates that the class IP-SEG* is properly containedin IP-SEG. However, this is not true for graphs G n with n ≥ . One can show that in an IP-SEG model of a chordlesscycle that consists of two interval and two permutation arcs, the permutation arcs cannot consist of more than twosegments. Thus, at least one interval arc of an IP-SEG representation of a C n with n ≥ would need to be of length atleast three. This would lead us to the same contradiction when trying to assign segments to w t and z t for which v t corresponds to an interior interval segment. The respective geometric intersection models of interval and permutation graphs immediately imply very naturalpolynomial algorithms for solving several important optimization problems, including clique, independent set, andgraph coloring. For example, given the model of an interval graph, we can find a maximum independent set byusing a greedy algorithm which at each step selects the interval with the leftmost right endpoint, while removing thatinterval and all other that intersect it from consideration. Given the model of a permutation graph, we can easily find amaximum clique by recovering the defining permutation of the graph from the model and finding the longest decreasingsubsequence in it. These algorithms are not of great practical importance, as there exist linear-time algorithms for therespective problems on larger graph classes, that do not require a model as part of the input. Nevertheless, they point usto an initial direction in the study of such problems on the new classes of IP-SEG and IP-SEG* graphs. In particular,we look at the clique and independent set problems on the classes, when the IP-SEG model is given.
Suppose that C is a clique in an IP-SEG graph G with a given IP-SEG model. Since interval segments lying on differentparallel lines cannot intersect, all interval segments in C , if any, must lie on a single line. Without loss of generality, wemay assume that is L . Denote by C int the set of interval segments in C and by C prm the set of permutation segmentsin C .Consider the case when C int (cid:54) = ∅ and let s C be the intersection of interval segments in C int . As such, s C is fullycontained in each interval segment in C int and it must contain the top endpoint of each permutation segment in C prm .In addition, the endpoints of s C must be endpoints of one or more interval segments in C int . s C Figure 8: An IP-SEG representation of a clique on 5 vertices7f we knew what s C was, it would not be difficult to recover a clique of maximum size. First, set C int to be the setof interval segments fully containing s C . Then, identify all permutation segments that have an endpoint in s C andfind a clique of maximum size C prm on the permutation graph they induce. Note that, since there may be multiplepossible options for C prm , there may be multiple cliques of maximum size that correspond to s C . Nevertheless, theabove procedure will recover a maximum clique, when we know s C .This leads to a simple polynomial algorithm for the clique problem, when the IP-SEG model is known. First, identifythe maximum clique on the graph G p induced by the permutation segments of G . Then, go over all pairs of endpoints s , s of interval segments along L and identify the maximum clique formed by interval segments containing theinterval s C = [ s , s ] and permutation segments with an endpoint in s C . Repeat the same for pairs of endpoints along L . The clique of largest size found in this procedure is a maximum clique of G . MaxClique (given the model) Find the maximum clique C p of the graph induced by all permutation segments of G Set c max = | C | and C max = C p For i in { , } For each pair of endpoints s , s on L i Identify the interval segments on L i containing the interval [ s , s ] and store them in C int Find a maximum clique formed by permutation segments with an endpoint in [ s , s ] and store it in C prm Combine C int and C prm into C If | C | > c max then set c max = | C | and C max = C Return C max Depending on what subroutine we apply to find a maximum clique C prm on the permutation graphs induced bypermutation segments, the overall running time of the algorithm would be O ( n ( n + m )) or O ( n logn ) . We say that a line segment p (interval or permutation) in an IP-SEG model is to the right of another line segment q , andwrite q < p , if for each parallel line L i , either at least one of p and q does not have an endpoint on L i or each endpointof p on L i is to the right of each endpoint of q on L i . We say that a line segment r is between two segments p and q if p < r and r < q . Clearly, two line segments p and q do not intersect if and only if q < p or p < q .Suppose G is an IP-SEG graph with a given model IP ( G ) . As the model is fixed, we will interchangeably use thenotions of vertices in G and segments in IP ( G ) , as well as the notions of induced subgraphs of G and sets of linesegments in IP ( G ) , depending on the context. Let I be a maximum independent set in G and I prm and I int be thesets of permutation and interval segments, respectively, that form I . From the above observation, it follows that thepermutation segments in I prm must form a sequence { p i } such that each p i +1 is to the right of p i . p n p n − I ( p n − ) I ( p n − , p n ) I ( p n − , p n ) Figure 9: An IP-SEG representation of an independent set with at least two permutation segmentsLet p n − and p n be the next to last and last permutation segments in the sequence { p i } , respectively. Let G ( p n − ) bethe set of all segments, interval and permutation, that p n − is to the right of and let I ( p n − ) = I ∩ G ( p n − ) . It is easyto see that if I = I ( p n ) is of maximum size in G , then I ( p n − ) must be an of maximum size in the subgraph G ( p n − ) of G .Denote by G i ( p n − , p n ) the set of all interval segments along L i that are between p n − and p n and let I i ( p n − , p n ) = I ∩ G i ( p n − , p n ) . It is again easy to see that if I = I ( p n ) is of maximum size in G , then I i ( p n − , p n ) must be ofmaximum size in the subgraph G i ( p n − , p n ) of G . 8hile finding I ( p n − ) amounts to solving the original problem of finding I = I ( p n ) , given that the subgraph G i ( p n − , p n ) is an interval graph, we can find I i ( p n − , p n ) by applying an existing algorithm for the independent seton interval graphs.This leads to a simple dynamic programming algorithm for the independent set problem in which for each permutationsegment p we keep track of the largest independent set I ( p ) formed by p along with segments that p is to the right of.Begin by obtaining a left-to-right topological ordering T of the set of all permutation segments using the "to the right"relation. Then, process segments in T in a left-to-right order. If a segment p is not to the right of any of the alreadyprocessed segments from T , then I ( p ) is simply the union of p and the largest independent sets on L and L formedby interval segments that p is to the right of. Otherwise, we also need to consider the sets I ( p (cid:48) ) ∪ I ( p (cid:48) , p ) ∪ I ( p (cid:48) , p ) for each permutation segment p (cid:48) that p is to the right of. We also need to account for the possibility that the largestindependent set of G consists only of interval segments in IP ( G ) . For this, we simply need to combine the largestindependent set of the interval subgraph of G induced by interval segments lying on L with the correspondingindependent set on L . MaxIS (given the model) Find the largest independent sets I and I of interval segments on L and L , respectively Set I max = I ∪ I and i max = | I max | Find a topological ordering T of the permutation segments of G For p in T Set I ( p ) = I ( p ) ∪ I ( p ) , where I i ( p ) is a largest ind. set of interval segments s on L i such that s < p For p (cid:48) in T such that p (cid:48) < p I ∗ ( p ) = I ( p (cid:48) ) ∪ I ( p (cid:48) , p ) ∪ I ( p (cid:48) , p ) If | I ( p ) | < | I ∗ ( p ) | then set I ( p ) = I ∗ ( p ) If | I ( p ) | > i max then set I max = I ( p ) and i max = | I ( p ) | Return I max Since we are given the model and thus we have the interval segments in sorted order, we can find each of the independentsets of subgraphs in the algorithm in O ( n ) time. This leads to an overall running time of O ( n ) . In this work we introduced two graph classes, IP-SEG* and IP-SEG, generalizing the classes of permutation andinterval graphs, based on their geometric models. We showed that these graph classes have an implicit representation.In addition, we saw that unlike earlier generalizations such as simple triangle and trapezoid graphs, these classes are notcontained in the class of perfect graphs. Nonetheless, we are somewhat limited in how we can represent a chordlesscycle using an IP-SEG model, which leads to some forbidden subgraphs for the two classes. We have also discussedalgorithms for the clique and independent set problems on the classes, when an IP-SEG model is given.The recognition problem is a natural question that remains open. Finding alternative characterizations would be of valuein tackling the recognition problem. In particular, given that interval and permutation graphs have nice vertex orderingcharacterizations and a similar result has been recently obtained for one of their generalizations - simple triangle graphs- by Takaoka [11], it would be worth exploring if such a characterization can be found for IP-SEG* or IP-SEG graphs.Another avenue would be to identify other forbidden subgraphs for the two classes.Future work should also be done on studying other optimization problems on the class. A good candidate wouldbe coloring, given the simple algorithms for this problem on interval and permutation graphs arising naturally fromtheir geometric intersection models. Finally, it would be interesting to know if we can design robust algorithms foroptimization problems such as clique and independent set, for when the IP-SEG model is not given as part of the input.
References [1] B
OOTH , K. S.,
AND L UEKER , G. S. Testing for the consecutive ones property, interval graphs, and graphplanarity using PQ-tree algorithms.
Journal of Computer and System Sciences 13 (1976), 335–379.[2] B
RANDSTÄDT , A., L E , V., AND S PINRAD , J.
Graph Classes: A Survey . SIAM, Philadelphia, 1999.93] C
ORNEIL , D.,
AND K AMULA , P. Extensions of permutation and interval graphs.
Congressus Numerantium 58 (1987), 267–275.[4] D
AGAN , I., G
OLUMBIC , M.,
AND P INTER , R. Trapezoid graphs and their coloring.
Discrete Applied Mathematics21 (1988), 35–46.[5] F
ISHBURN , P. C.
Interval orders and interval graphs: A study of partially ordered sets . Wiley, New York, 1985.[6] G
OLUMBIC , M. C.
Algorithmic Graph Theory and Perfect Graphs . Academic Press, New York, 1980.[7] H
ARTMAN , I. B.-A., N
EWMAN , I.,
AND Z IV , R. On grid intersection graphs. Discrete Mathematics 87 , 1(1991), 41–52.[8] K
ANNAN , S., N
AOR , M.,
AND R UDICH , S. Implicit representation of graphs.
SIAM Journal on DiscreteMathematics 5 , 4 (1992), 596–603.[9] K
RATOCHVIL , J.,
AND N ESETRIL , J. Independent set and clique problems in intersection defined graphs.
Commentationes Mathematicae Universitatis Carolinae 31 , 1 (2016), 85–93.[10] S
PINRAD , J.
Efficient Graph Representations . American Mathematical Society, Providence, Rhode Island, 2003.[11] T
AKAOKA , A. A vertex ordering characterization of simple-triangle graphs.