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Final Presentation
Edge-Disjoint Spanning Trees and InductiveConstructions
Laura Gioco
Fairfield University
August 1, 2012
Mt. Holyoke College REU 2012NSF Grant DMS - 0849637
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Background
A 3D body-and-bar structure is rigid ⇔ the associated graph is theunion of 6 edge-disjoint spanning trees. [Tay,1984]
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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
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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
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Background
Definitions
Definitions
A graph G is a tree if it is connected and acyclic.
Example
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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
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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
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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]
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Background
Inductive Construction
Inductive constructions are prevalent in Rigidity Theory becausethey are often useful in proofs.
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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)
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Background
Inductive Construction
This manner of inductive construction preserves the property of kedge-disjoint spanning trees.
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Background
Inductive Construction
single vertex
k = 3
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Background
Inductive Construction
existing graph new vertex v
k = 3
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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)
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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.
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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
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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.
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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
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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)
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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)
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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)
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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)
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Result
This implies that this constructive step creates a new graph thatsatisfies the k = a + g spanning trees property.
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Future Work
There are some (k, k)-tight graphs which my constructiveoperation cannot form.
Example
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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
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Future Work
Questions?
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