Classifying Trees with EdgeDeleted - PowerPoint PPT Presentation

1 / 15
About This Presentation
Title:

Classifying Trees with EdgeDeleted

Description:

... deleted eccentricity g(v) of v is defined to be the maximum eccentricity of v ... G with minimum edge-deleted eccentricity are called edge-deleted central ... – PowerPoint PPT presentation

Number of Views:46
Avg rating:3.0/5.0
Slides: 16
Provided by: cent175
Category:

less

Transcript and Presenter's Notes

Title: Classifying Trees with EdgeDeleted


1
  • Classifying Trees with Edge-Deleted
  • Central Appendage Number 2
  • Shubhangi Stalder
  • University of Wisconsin Waukesha
  • Linda Eroh, John Koker,
  • Hosien S. Moghadam, Steven J. Winters
  • University of Wisconsin Oshkosh

2
Basic Definitions
  • A graph G is 2-edge-connected if the removal
    of any edge of G never results in a
    disconnected graph.

3
  • For a vertex v in a 2-edge-connected graph G,
    the edge-deleted eccentricity g(v) of v is
    defined to be the maximum eccentricity of v in
    G e over all edges e of G.
  • Example - g(w)9
  • and g(x)g(y)10

4
  • The vertices of G with minimum edge-deleted
    eccentricity are called edge-deleted central
    vertices while the vertices of maximum
    edge-deleted eccentricity are called edge-deleted
    peripheral vertices.
  • The subgraph induced by the edge-deleted central
    vertices of G is called the edge-deleted center
    EDC(G) of G and the subgraph induced by the
    edge-deleted peripheral vertices EDP(G) is
    called the edge-deleted periphery.

5
  • The central appendage number of a graph G is
    the minimum difference V(H) V(G) over all
    graphs H with C(H) ?? G.
  • The edge-deleted central appendage number A(G)
    of a graph G is the minimum difference V(H)
    V(G) over all graphs H with EDC(H) ?? G.
  • A branch of a tree T is a component of
    T V(C(T)).

6
Background for Results
  • Previously know results for central appendage
    number
  • For every graph G, there exists a connected
    graph H such that C(H) ? G.
  • 2 ? central appendage number of connected graph ?
    4.
  • There are no trees with central appendage number
    3.
  • If in a tree all end vertices are equidistant
    from the center, then the central appendage is 2,
    and otherwise 4.

7
  • Previously know results for A(G) (Koker,
    Moghadam, Stalder, and Winters 2002)
  • For every graph G, there exists a 2-
    edge-connected graph H such that EDC(H) ? G.
  • 2 ? A(T) ? 3.
  • If T is a tree with radius n that satisfies all
    the properties below, then A(T) 2.
  • C(T) lt wgt.
  • deg(w) ? 4.
  • For z ? V(T), if 1 ? dT(z, w) lt n 1, then
    degT z 2, and if dT(w, z) n 1, then degT z
    ? 2.

8
Basic Assumptions
  • When T is a tree with A(T) 2, let H be a
    graph with V(H) V(T) U x, y and EDC(H)
    T.
  • Since g(x) g(y) (call this number k), there
    exists an e ? E(H) such that dH-e(x, y) k.
  • Let D be the set of peripheral vertices of T.

9
Result 1
  • If T is a tree with C(T) ltw, w?gt, then
    A(T) ? 2.
  • Proof Let if possible A(T) 2. Now prove some
    general results for all trees with A(T) 2
  • g(u) k-1 for all u in T.
  • e cannot be an edge of T.
  • For peripheral vertices u,v of T, with ux, vy ?
    E(He)
  • Shortest u-v path lies entirely in T, and
    dHe(u, v) k 2 diam T, eH-e(v) eH-e(u)
    k-1
  • u and v cannot be end vertices of the same
    branch of T.

10
  • D must contain at least 2 vertices u,v ? V(He)
    with ux, vy ? E(He).
  • All end vertices are equidistant from the center
    (k 2 rad(T) 1 when C(T)ltw,w? gt k2 rad(T)
    2 when C(T)ltwgt)
  • e xy.
  • Since C(T) ltw, w?gt, without loss of
    generality, these results give us that all end
    vertices of w must be adjacent to x and those of
    w? to y. Therefore, dHww?(w, w?) gt k 1 which is
    a contradiction to the fact g(w)k-1, proving
    Result 1 A(T) ? 2.

11
Result 2
  • Let T be a tree with C(T) ltwgt. Then A(T)
    2 if and only if the following are satisfied
  • All end vertices are equidistant from the
    center.
  • deg w ? 4.
  • For z ? V(T), if 1 ? dT(z, w) lt rad T 1,
    then degT z 2, and if dT(z,w) rad T 1,
    then degTz ? 2.

12
  • Proof ?(? Koker, Moghadam, Stalder, and Winters,
    2002)
  • ?
  • The general results necessary for Result 1 give
    us that all end vertices are equidistant from the
    center, and that e xy.
  • Show deg w ? 4. (Let if possible deg w lt 4.)
  • Show for z ? V(T), if 1 ? dT(z, w) lt n 1,
    then degT z 2, and if dT(w, z) n 1, then
    degT z ? 2.

13
(No Transcript)
14
  • Lemma 1. gH(u) k 1 for all u ? V(T).
  • Lemma 2. dHe(x, y) k, then e ? E(T).

15
  • Lemma 3. If u, v ? V(T) such that ux and vy
    are edges in He, then
  • a shortest u-v path in He lies entirely in T
  • dHe(u, v) k 1 or k 2
  • eHe(u) eHe (v) k 1.
Write a Comment
User Comments (0)
About PowerShow.com