Common vertex of longest cycles in circular arc graphs

Transcription

1 McGill University Vanderbilt University, June 2nd, 2012 Joint work with Guantao Chen

2 McGill University Vanderbilt University, June 2nd, 2012 Joint work with Guantao Chen

3 Introduction Theorem (Gallai 1955) All longest paths in a connected graph have a common vertex. Example (Zimfirescu 1978) Counter-Example: The graph obtained from Splitting a vertex of peterson graph into three vertices. Chordal graphs: a graph without an induced cycles of length more than three. Conjecture (Lehel) All longest paths in a connected chordal graph have a common vertex. Conjecture Any three longest cycles in a connected graph have a common vertex.

4 Previous Results Outerplanar: all vertices belong to the unbounded face of a planar embedding. Theorem (Axenovich) Any three longest paths in a connected outerplanar graph have a common vertex. Intersection graph: G forms from a collection of sets C : G = (V ; E) where V = C and AB E for two elements A, B C if and only if A B. Chordal graph: the intersection graph of subtrees of a hosting tree. Interval graph: the intersection graph of intervals on the real line. Circular arc graph: the intersection graph of arcs on a circle. Theorem (Balister-Győri-Lehel-Schelp 2004) All longest paths in a connected circular arc graph have a common vertex. Theorem (Balister-Győri-Lehel-Schelp 2004) All longest cycles in a connected interval graph have a common vertex.

5 Main Results Theorem (Chen-W.) All longest cycles in a connected chordal graph have a common vertex. Spider graph: A tree with at most one vertex having degree more than two. Theorem (Chen-W.) All longest paths in the intersection graph of a collection of subtrees of a spider graph have a common vertex.

6 Nondecreasing path Let G be an interval graph, For X V (G), let [L(X ), R(X )] be the interval form of X. For A, B V (G), we say A < B if L(A) < L(B) and R(A) < R(B). A path A 1A 2... A k in an interval graph is nondecreasing if A j A i whenever i < j. Lemma Any path in an interval graph can be reordered into a nondecreasing path. Given a path P with length k in an interval graph, we have the following algorithm to rearrange it into a nondecreasing path I 1I 2... I k. Algorithm: Initial: 1. S 0 = V (P); 2. I 1 is the vertex in S 0 with minimum right endpoint; Repeat: 3. S j = S j 1 I j ; 4. I j+1 is the vertex in S j intersecting I j with minimum right endpoint.

7 Circular arc graph Theorem (Chen-W.) All longest cycles in a connected circular arc graph have a common vertex. Sketch of the proof: Let A be the collection of arcs. Let M be the arcs in A which are not included in any other arc in A. Observation Arcs in M can be labeled as M 1, M 2,..., M t in the clockwise order in the circle. Observation Given any longest cycle C = A 1,..., A m and B A. 1. If (A i A i+1 ) B, then B V (C). 2. If A i B for some i, then B V (C). Claim For any longest cycle C, elements in M appear in C are consecutive in the list in M, i.e. M V (C)= {M a, M a+1,..., M b 1, M b }.

8 Suppose the conjecture is not true. Let C 1 be a longest cycle with fewest elements in M. Assume C 1 M = {A,..., B} (list in clockwise order). Let C 2 be a longest cycle that does not contain A. Assume C 2 M = {C,..., D} (list in clockwise order). Observation A C B D A in clockwise order. Claim There is a nondecreasing path from A to D with length C 1. Rearrange C 1 to paths: Nondecreasing: J 1(= A), J 2,..., J b (= B),..., J m. (1) Nonincreasing: I 1(= B), I 2,..., I a(= A),..., I m. (2) Rearrange C 2 to paths: Nonincreasing: J 1(= D), J 2,..., J c(= C),..., J m. (3) Nondecreasing: I 1(= C), I 2,..., I d(= D),..., I m. (4) Using (1) and (4), we get a nondecreasing path: A, I a 1,..., I 2, B, J k,..., J 2, D with length m + a d. Using (2) and (3), we get a nonincreasing path D, I d 1,..., I 2, C, J j, dots, J 2, A with m + d a.

9 Spider graph Theorem (Chen-W.) All longest paths in the intersection graph of a collection of subtrees of a spider graph have a common vertex. Skectch of the proof: The center of the spider graph is the only vertex that have degree at least three. A branch of the spider graph is a path from the center to a leaf. Let H be the set of subtrees containing the center. Suppose the theorem is not true for some graph. Claim All the longest paths have a vertex which contains the center as a vertex in the corresponding subtree.

10 Spider graph, cont... The hub subtree of a branch is the element in H which has the longest subpath in this branch. A segment of a path is a maximal subpath which does not have elements in H. Observation Any segment of a path appear in a single branch. Claim For any longest path, if the hub subtree for a branch does not appear in the branch, then there is no segment appear in this branch. Consider S which is the longest segment among all the segments in all the longest paths. Let T be the corresponding hub tree for this segment. Claim T appears in all the longest paths. Proof. Suppose T does not appear in some longest path P, then neither does any element in S. Let S be the last segment in this longest path. Replace S by T plus S, we get a new path P, which is longer. Contradiction!

11 Thank you for your attention!