Final Presentation

Edge-Disjoint Spanning Trees and InductiveConstructions

Laura Gioco

Fairfield University

August 1, 2012

Mt. Holyoke College REU 2012NSF Grant DMS - 0849637

Final Presentation

Background

A 3D body-and-bar structure is rigid ⇔ the associated graph is theunion of 6 edge-disjoint spanning trees. [Tay,1984]

Final Presentation

Background

Definitions

Definitions

A cycle is a graph with an equal number of vertices and edgeswhose vertices may be placed around a circle in such a way thattwo vertices are adjacent if and only if they appear consecutivelyalong the circle. A graph with no cycle is called acyclic.

Example

a

bb

cc

ddee

a

Final Presentation

Background

Definitions

Definitions

A graph G is connected if for all u, v ∈ V (G ) there exists a pathfrom u to v .

Example

a bb

uuv

Final Presentation

Background

Definitions

Definitions

A graph G is a tree if it is connected and acyclic.

Example

Final Presentation

Background

Definitions

Definitions

A tree T = (VT ,ET ) is a spanning tree of a graph G = (V ,E ) ifVT = V .

Example

a

b

d

c

Final Presentation

Background

Definitions

Definition

Let G = (V ,E ) be a multigraph with no loops.G is (k , k)-tight if there is a partition of the edge set such thatthere are k edge-disjoint spanning trees.

Example

a bb

c

Final Presentation

Background

Inductive Construction

Previous Work

2D Rigidity, Henneberg 0 and 1 [Henneberg 1911] [Laman1970]

new vertex

Body-and-Bar Rigidity using Henneberg operations inductivelyto form tight graphs [Tay 1991]

Generalized Counting Characterizations [Haas 2002]

Final Presentation

Background

Inductive Construction

Inductive constructions are prevalent in Rigidity Theory becausethey are often useful in proofs.

Final Presentation

Background

Inductive Construction

Henneberg 0: Inductive construction builds a rigid structure onevertex at a time.

Example

new vertexexisting G

e1, e2, ..., ek

graph G = (V ,E )new graph G ′ = (V ∪ v ,E ∪ e1, e2, ..., ek)

Final Presentation

Background

Inductive Construction

This manner of inductive construction preserves the property of kedge-disjoint spanning trees.

Final Presentation

Background

Inductive Construction

single vertex

k = 3

Final Presentation

Background

Inductive Construction

existing graph new vertex v

k = 3

Final Presentation

Background

Inductive Construction

existing graph new vertex v

k = 3

Final Presentation

Background

Inductive Construction

existing graph

new vertex v

k = 3

Final Presentation

Background

Inductive Construction

existing graph

new vertex v

k = 3

Final Presentation

Background

Inductive Construction

new vertexexisting G

e1, e2, ..., ek

graph G = (V ,E )new graph G ′ =(V ∪ v ,E ∪ e1, e2, ..., ek)

Final Presentation

Motivation

Body-and-CAD Rigidity- recent results for 2D

Combinatorial Characteristics of 2D Body-and-CAD Rigidity:associated graph G = (V ,S t D), exists S ′ ⊆ S such thatS \ S ′ is edge-disjoint union of 2 spanning trees and D ∪ S ′ isedge-disjoint union of 1 spanning tree [Sidman,Lee-St.John,LaMon 2012]

Example

We study a new class of graphs that arise from rigidity ofbody-and-CAD structures.

Final Presentation

Motivation

Body-and-CAD Rigidity- recent results for 3D

Combinatorial Characteristics of 3D Body-and-CAD Rigidity:associated graph G = (V ,S t D), exists S ′ ⊆ S such thatS \ S ′ is edge-disjoint union of 3 spanning trees and D ∪ S ′ isedge-disjoint union of 3 spanning tree [Sidman,Lee-St. John2012]

this does not include point-point coincidence constraints

Final Presentation

Motivation

Generalizing Counts

2D: 2 + 1 = 3

3D: 3 + 3 = 6

general: Let a + g = k , where a is the number of solid trees,g is the number of trees that might include dashed edges, andk is the total number of spanning trees.

Final Presentation

Motivation

Generalizing the Construction

can partition into k = a + g spanning trees

inductive constructions for (k , k)-tight graphs exist

generalize inductive construction to take into account dashedand solid edges

Final Presentation

Motivation

new vertexexisting G

e1, e2, ..., ek

graph G = (V , S t D)new graph G ′ = (V ∪ v ,S ∪ e1, ..., en t D ∪ en+1, ..., ek)

Final Presentation

Motivation

k = 3, g = 2

new vertexexisting G

e1, e2, e3

graph G = (V , S t D)new graph G ′ = (V ∪ v ,S ∪ e2 t D ∪ e1 ∪ e3)

Final Presentation

Motivation

Also k = 3, g = 2

new vertexexisting G

e1, e2, e3

graph G = (V , S t D)new graph G ′ = (V ∪ v ,S ∪ e1 ∪ e2 t D ∪ e3)

Final Presentation

Motivation

Or even k = 3, g = 2

new vertexexisting G

e1, e2, e3

graph G = (V , S t D)new graph G ′ = (V ∪ v ,S ∪ e1 ∪ e2 ∪ e3 t D)

Final Presentation

Result

This implies that this constructive step creates a new graph thatsatisfies the k = a + g spanning trees property.

Final Presentation

Future Work

There are some (k, k)-tight graphs which my constructiveoperation cannot form.

Example

Final Presentation

Future Work

Since my inductive construction does not create every (k , k)-tightgraph, I would like to modify my construction to also use the otherHenneberg move which removes edges before adding the newvertex.

new vertex

Final Presentation

Future Work

Questions?