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Twenty Years of EPT Graphs: From Haifa to Rostock Martin Charles Golumbic Caesarea Rothschild Institute University of Haifa With thanks to my research.

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Presentation on theme: "Twenty Years of EPT Graphs: From Haifa to Rostock Martin Charles Golumbic Caesarea Rothschild Institute University of Haifa With thanks to my research."— Presentation transcript:

1 Twenty Years of EPT Graphs: From Haifa to Rostock Martin Charles Golumbic Caesarea Rothschild Institute University of Haifa With thanks to my research collaborators: Robert Jamison, Marina Lipshteyn, Michal Stern

2 Bell Labs in New Jersey (Spring 1981) John Klincewicz: Suppose you are routing phone calls in a tree network. Two calls interfere if they share an edge of the tree. How can you optimally schedule the calls? The Story Begins

3 Bell Labs in New Jersey (Spring 1981) John Klincewicz: Suppose you are routing phone calls in a tree network. Two calls interfere if they share an edge of the tree. How can you optimally schedule the calls? The Story Begins

4 Bell Labs in New Jersey (Spring 1981) John Klincewicz: Suppose you are routing phone calls in a tree network. Two calls interfere if they share an edge of the tree. How can you optimally schedule the calls? The Story Begins A call is a path between a pair of nodes. A typical example of a type of intersection graph. Intersection here means “share an edge”. Coloring this intersection graph is scheduling the calls. An Olive Tree Network

5 Vertex Intersection Graphs of Paths in a Tree (VPT) Edge Intersection Graphs of Paths in a Tree (EPT) Each vertex v in V(G VPT ) and V(G EPT ) corresponds to a path P v in T. (x,y)  E VPT  paths P x and P y intersect on at least one vertex in T. (x,y)  E EPT  paths P x and P y intersect on at least one edge in T. Representation P of paths in a tree T. G VPT ( P ) PcPc G EPT ( P ) a d b c

6 For VPT-representation P a and P b intersect. For EPT-representation: P a and P b do not intersect. For both VPT-representation and EPT-representation P a and P b intersect. Vertex and Edge Intersections of Paths

7 Theorem. Chordal graphs  vertex intersection graphs of subtrees of a tree. [Buneman], [Gavril], [Walter] A graph G is chordal if every cycle of size  4 has a chord, i.e., G has no induced chordless cycles C m for m  4. Chordal Graphs

8 VPT Graphs are Chordal EPT Graphs are Not Chordal A path is a subtree, therefore VPT graphs (i.e., path graphs) are chordal. However, EPT graphs may have chordless cycles of any size.

9 A First Observation (Cycles) An EPT representation of C 6 called a “6-pie”. 6 3 2 1 4 5 Chordless cycles have a unique EPT representation.

10 A First Observation (Cycles) An EPT representation of C 6 called a “6-pie”. 6 3 2 1 4 5 Theorem (Golumbic Jamison 1985): Let P be an EPT representation of G. If G contains a chordless cycle C m (m  4), then P contains an m-pie representing the cycle. Chordless cycles have a unique EPT representation.

11 Restricting the degree of the host tree Remark. If m is the maximum degree in T, then the EPT graph has no chordless (m+1) cycles (or larger).

12 Restricting the degree of the host tree Remark. If m is the maximum degree in T, then the EPT graph has no chordless (m+1) cycles (or larger). Corollary: If P is an EPT representation of G on a degree 3 tree T. Then G is chordal graph.

13 Restricting the degree of the host tree C6C6 a a a b b b c c c d d d Example. The graph C 6 requires degree 5. A 4-pie on a,b,c,d

14 Restricting the degree of the host tree C6C6 a a a b b b c c c d d d x x x y y y Example. The graph C 6 requires degree 5. Now add x and y

15 A Second Observation (Cliques) Two EPT representations of K 6 called a “claw clique” or “edge clique”. 6 3 2 1 4 5 edge All share a common edge claw All share some edge of the claw

16 A Second Observation (Cliques) Two EPT representations of K 6 called a “claw clique” or “edge clique”. 6 3 2 1 4 5 Theorem (Golumbic Jamison 1985): Let P be an EPT representation of G. If G contains a clique K m (m  3), then P contains either a claw or edge for it. Cliques have exactly two possible EPT representations.

17 No Kissing The No Kissing Lemma: If P is an EPT representation of G on a tree T, and u is any node of T. We may assume without loss of generality, and without increasing the degree of the tree, that all paths touching u continue through u. No stopping. No kissing u.

18 No Kissing The No Kissing Lemma: If P is an EPT representation of G on a tree T, and u is any node of T. We may assume without loss of generality, and without increasing the degree of the tree, that all paths touching u continue through u. No stopping. No kissing u. e c b a d e c b a d Create dummy nodes and shorten a and e

19 Degree 3 host trees If P is a deg3 EPT representation of G on a tree T, then applying the No Kissing Lemma construction to all nodes of degree 3 yields a deg3 VPT representation of G.

20 Degree 3 host trees If P is a deg3 EPT representation of G on a tree T, then applying the No Kissing Lemma construction to all nodes of degree 3 yields a deg3 VPT representation of G. i.e., deg3 EPT  deg3 VPT  deg3 EPT  chordal  EPT Now let’s prove: chordal  EPT  deg3 EPT

21 Degree 3 host trees (continued) Let P be any EPT representation of G on a tree T, and u any node of T of maximum degree d > 3. 1.Assume the no kissing lemma, and let U denote all paths passing through u. 2.Let x be a simplicial vertex of the induced subgraph G U. Thus, all paths in U share an edge with P x. 3.Perform the transformation: 4.Repeat for all nodes until max degree is 3. x y z x y z

22 Degree 3 host trees (continued) Theorem (1985): All four classes are equivalent: chordal  EPT  deg3 EPT  VPT  EPT  deg3 VPT What about degree 4?

23 Degree 3 host trees (continued) Theorem (1985): All four classes are equivalent: chordal  EPT  deg3 EPT  VPT  EPT  deg3 VPT Theorem (2005, Golumbic, Lipshteyn, Stern): weakly chordal  EPT  deg4 EPT Degree 4 host trees

24 Definition Weakly Chordal Graph No induced C m for m  5, and no induced C m for m  5. Theorem [Hayward, Hoàng, Maffray 1989] G is weakly chordal if and only if every induced subgraph of G is either a clique or has a two-pair. Weakly Chordal Graphs

25 A two-pair is a pair of vertices, such that every chordless path between them has length two edges. Remark. If {x,y} is a two-pair, then the common neighborhood of x and y is an (x,y)-minimal separator. {x,y} is a two-pair{x,y} is not a two-pair

26 Theorem [GLS 2005] A graph G has an EPT representation on a degree 4 tree if and only if G is a weakly chordal EPT graph. Degree 4 Trees Sketch of the proof. (  ) By the Pie Theorem, G has no induced C m (m  5) nor C 5 (=C 5 ). By our earlier example, C 6 requires degree 5. By a theorem of Golumbic and Jamison 1985, C m (m  7) is not an EPT graph.

27 Degree 4 Trees (continued) (  ) Let P be an EPT representation of G on tree T with maximal degree d > 4, and let u be a node of degree d. We transform P into an EPT representation P´ on T´ with fewer vertices of degree d. The full proof follows by induction. Assume the no kissing lemma at u, and let U denote all paths passing through u. The induced subgraph G U is weakly chordal, so there are two cases: G U is a clique or G U has a two-pair.

28 Degree 4 Trees (continued) Case 1. G U is a clique. If it were a claw clique, then u would have degree 3, and we are done. Otherwise, G U is an edge clique, and all paths in U share an edge, say (v 1,u). Perform the transformation: v1v1 v1v1 v2v2 v2v2

29 Degree 4 Trees (continued) Case 2. G U has a two pair {x,y}. The common neighborhood S of {x,y} is a minimal separator and splits G U into (at least) 2 connected components: G X containing x and G Y containing y. The star edges centered at u, are now painted. The two contained in P x are red; those in P y blue. Propagate the coloring to other star edges via the paths P z (z  U \ S): if P z has one red edge, then paint its other star edge red; if P z has one blue edge, then paint its other star edge blue. No star edges gets two colors!

30 Degree 4 Trees (continued) Subcase 2a. G S is a clique. Perform the transformation: S

31 Degree 4 Trees (continued) Subcase 2b. G S is not a clique. There exists a path P v (v  U \ S) that contains only one of the edges (v 1,u),(v 2,u), (v 3,u),(v 4,u), say (v 1,u). Let  be the non- empty collection of such paths, which thus form an edge clique containing (v 1,u). Color the star edges as follows: S

32 Subcase 2b, continued. 1. (v 1,u) is not colored. 2. (v i,u), i  5 is colored pink if it is contained in a path in . 3. (v i,u) is colored if is contained in a path that already has a pink edge. Lemma: The edges of P y are not colored. Example. [v 1,u,v 5 ]  P 1 [v 1,u,v 7 ]  P 2 [v 5,u,v 6 ]  P 3  = {P 1,P 2 } Q.E.D.

33 From Haifa to Rostock (Spring 1985?) The Story Continues

34 Algorithmic Aspects of EPT Graphs The recognition and coloring problems of an EPT graph are NP-complete (Golumbic and Jamison, 1985). There is a 3/2-approximation algorithm for coloring EPT graphs. (Tarjan, 1985) On deg3EPT or deg4EPT graphs, coloring is polynomial since, respectively, they are chordal (GJ 1985) or weakly chordal (GLS 2005). Max-clique and Max-stable set are polynomial (GJ 1985).

35 [h,s,t] Graphs and Representations A collection of subtrees of a tree T satisfying: h:T has maximum vertex degree h s:Each subtree has maximum vertex degree s t:An edge (x,y) in G if T x and T y share t vertices

36 Interval graphs  [2,2,1] EPT  [ , 2, 2] Chordal graphs  [ , ,1]  [3,3,1]  [3,3,2] (MS,JM) [h,s,t] Graphs and Representations A collection of subtrees of a tree T satisfying: h:T has maximum vertex degree h s:Each subtree has maximum vertex degree s t:An edge (x,y) in G if T x and T y share t vertices

37 Interval graphs  [2,2,1] EPT  [ , 2, 2] Chordal graphs  [ , ,1]  [3,3,1]  [3,3,2] (MS,JM) Any graph  [ , ,2] [ ,2,2]  chordal  [3,2,2]  [3,2,1] (GJ) [ ,2,2]  weakly chordal  [4,2,2] (GLS) [h,s,t] Graphs and Representations A collection of subtrees of a tree T satisfying: h:T has maximum vertex degree h s:Each subtree has maximum vertex degree s t:An edge (x,y) in G if T x and T y share t vertices

38 Two Directions to Generalize EPT 1.Keep t = 2 (edge intersection) Increase s (generalizing to subtrees) but Bound max degree h. 2.Increase t (constant tolerence) Keep s = 2 (paths)

39 A New Characterization Theorem Golumbic, Lipshteyn, Stern [WG2006]: The class [4,4,2]-graphs is equivalent to weakly chordal  (K 2,3, P 6, 4P 2, P 2  P 4, H 1, H 2, H 3 )-free. K 1 and K 2 are cliques of size at most 2.

40 Forbidden Subgraphs

41 Definition of k-EPT Graphs The k-Edge Intersection Graphs of Paths in a Tree (x,y)  E  paths P x and P y intersect on at least k edges in T. Def. G is a k-EPT graph if G has a k-EPT representation. k-EPT representation tree T of G G = (V,E) for k = 4 edges k-EPT  [ , 2, k+1] (i.e., share k+1 vertices)

42 Examples of Intersections For VPT representation: paths a and b intersect. For k-EPT representation, k>0: paths a and b do not intersect. For VPT and 1-EPT representation: paths a and b intersect. For k-EPT representation, k>1: paths a and b do not intersect. For VPT and k-EPT representation, k 4: paths a and b intersect. For k-EPT representation, k>4: paths a and b do not intersect.

43 Properties of k-EPT 1-EPT k-EPT, for any fixed k > 1. - Divide each edge into k edges, by adding k-1 dummy vertices. When restricted to degree 3 trees, the containment is also strict. 1-EPT k-EPT, for any fixed k > 1.

44 New Properties of k-EPT VPT graphs are incomparable with k-EPT graphs, for any fixed k  1. When restricted to degree 3 trees, VPT  k-EPT, for any fixed k  2. Chordless cycles are degree 3 k-EPT for k  2.

45 Recognition of k-EPT Important Properties Any maximal clique of a k-EPT graph is either a k-edge clique or a k-claw clique. A k-EPT graph G has at most maximal cliques. k-edge clique k-claw clique

46 Recognition of k-EPT Branch Graphs Definition: Let C be a subset of vertices of G. The branch graph B(G/C): G B(G/C) Theorem: Let C be a maximal clique of a k-EPT graph G. Then the branch graph B(G/C) can be 3-colored.

47 Recognition of k-EPT NP-Completeness Theorem: It is an NP-complete problem to decide whether a VPT graph is a k-EPT graph. Proof: An arbitrary undirected graph H is 3-colorable iff a certain graph G=(V,E) is a k-EPT graph. H T P ij path in T (i,j) E(H) Q i edge in T i V(H)

48 {P ij } corresponds to a maximal clique C of the VPT graph G. B(G/C) is isomorphic to H. If G is VPT and k-EPT  H is 3-colorable If H is 3-colorable and G is VPT  G is VPT and 1-EPT  G is k-EPT. Therefore, G is VPT and k-EPT  G is VPT and H is 3-colorable T HP ij path in T (i,j) E(H) Q i edge in T i V(H)

49 Recognition of k-EPT Corollaries Corollary: Recognizing whether an arbitrary graph is a k-EPT graph is an NP-complete problem. Corollary: Let G be a VPT graph. Then G is a 1-EPT graph iff G is a k-EPT graph, (hence: iff G is chordal).

50 Coloring of k-EPT Theorem: The problem of finding a minimum coloring of a k-EPT graph is NP-complete. Same proof as Golumbic & Jamison (1985) for the case k = 1.

51 Forbidden Subgraph Theorem: The following graph is not a k-EPT graph, for any fixed k > 1.

52 Open Problem The relationships between k-EPT graphs and (k+1)-EPT graphs, for any fixed k. Is k-EPT  (k+1)-EPT? We have graphs that are not 1-EPT but are a (k+1)-EPT graph, for any fixed k.

53 C 4 is not VPT but is k-EPT k  1

54 D is k-EPT k  2 is not 1-EPT Graph D

55 Orthodox Representations P A representation for G is orthodox if For each path, its endpoints are leaves (leaf generated), and Two paths P i, P j share a leaf if and only if vertices i and j are adjacent in G.

56 Subtrees of a Tree (i)  (ii)  (iii) McMorris & Scheinerman 1991  (iv)  (v) Jamison & Mulder 2000 ( , ,1 )

57 Orthodox Representations orth( ,2,1)  orth(3,2,1)  orth(3,2,2) Theorem 6.8

58 The Complete Heirarchy

59 6.4 6.7 6.8

60 The Complete Heirarchy

61

62 More on Algorithmic Graph Theory

63 Further Research Characterize families of [h,s,t] graphs for various values of h, s and t. Find intersection models as (h,s,t)-representations for known families of graphs. weakly chordal  [?, ?, ?]

64 Thank You!


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