mathematics & statistics | smith collegerhaas/pebbles-mathfest.pdfcreated date 7/19/2005 4:53:30...

Graphs, Trees

Pebbles, Robots

1

Outline

I. Robot Arms.

II. Rigid Graphs.

III. Characterizations of minimally rigid graphs.

IV. Trees, Arboricity and Characterizations.

V. The Pebbling Algorithm.

VI. Applications: (a) Rigid Components.

2

I. Robot arms and the Carpenter’s Rule Problem.

Can a robot arm be straightened?(bent?)

Given a planar embedding of a path can it be straightened

(without crossing)?

3

What about moving one embedding of a graph to another?

Trees?

4

Trees can Lock!

(Beidl, Demaine, Demaine, Lubiw, O’Rourke, Overmans,

Robbins, Streinu)

5

Can any planar embedding of Cn be convexified?

6

Can any planar embedding of Cn be convexified?

Yes! Connelly, Demaine, Rote; nicer algorithm by Streinu (2000).

6

II. Rigidity of Graphs

u uuLLLLLL uu u

u#

##

##

!!!!! uu

uuu

uu

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@ �

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Construct the graph G in R2 using inflexible bars for the edges

and rotatable joints for the vertices. This immersion is rigid if the

only motions of this structure are trivial (rotations and

translations).

7

A non-rigid object:

17

Examples of flexible structures

8

A non-rigid object:

18


8

A non-rigid object:

19


20


8

A non-rigid object:

20


21


8

A non-rigid object:

21


8

Intuition: How many edges are necessary to make a rigid body

with V vertices?

u-6 u-6u-6 u-6

u-6 u-6

Total degrees of freedom: 2V .

E.g. variables {xi , yi } for each vertex i.

9

Intuition: How many edges are necessary to make a rigid body

with V vertices?

u-6 u-6u-6 u-6

u- u-6�

��

Total degrees of freedom: 2V .

Adding an edge can remove a degree of freedom.

E.g. variables {xi , yi } for each vertex i.

Constraint√

(x j − xi )2 + (y j − yi )2 = c

9

Possible Positions

For each edge have an equation:√(x j − xi )2 + (y j − yi )2 = c

or (x j − xi )2+ (y j − yi )

2= c2

u uuu

��DDDDDDA

A�� u u

u

u��DDDDDD�

�AA

10

Possible Motions

Take the derivative:

ddt Position: d

dt

((x j − xi )

2+ (y j − yi )

2= c2

)

2(x j − xi )(x ′

j − x ′

i ) + 2(y j − yi )(y ′

j − y ′

i ) = 0

Solve for x ′

i and y ′

i .

Now have a system of equations with 2V variables and E

equations.

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In the end you will still be able to translate and rotate.

-

6

u uuLLLLLL

u uuLLLLLL

-

6

-

6

uu u!!!!!

Thus, if G is rigid in R2 then E ≥ 2V − 3.

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A graph is rigid in R2 if some embedding of it is.

Turns out, if some embedding is rigid then most are.

u uu uLLLLLL

HHH

H

u uu uLLLLLL

��

��

u uuu

��DDDDDDA

A��

So we can hope for COMBINATORIAL characterizations

13

Aside: When the vertices are not in general position, can get

non-rigid embedding. (rigid but not infinitesimally rigid)

We’ll be concerned with vertices in general position. (Generic

Rigidity)

14

A graph is minimally rigid in R2 if removing any edge results in a

graph that is not rigid.

u uu uLLLLLL

HHH

H

��

��

u uu uLLLLLL

��

��

HHHH

u uuu

��DDDDDDA

A��

These are not minimally rigid.

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Theorem (Laman, 1970) A graph is minimally rigid if and only if

EG = 2VG − 3 and for any subgraph H with at least 2 vertices,

EH ≤ 2VH − 3.

u uu uLLLLLL

HHHH

V = 4 then

E = 2 ∗ 4 − 3 = 5

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EH ≤ 2VH − 3.

u uu uLLLLLL

HHHH

u uu uLLLLLL

HHHH

u�

��

�

V = 4 then V = 5 then

E = 2 ∗ 4 − 3 = 5 E = 2 ∗ 5 − 3 = 7

16



EH ≤ 2VH − 3.

u uu uLLLLLL

HHH

H

u uu uLLLLLL

HHHH

u�

��

�

u uu uLLLLLL

HHHH

u,

,,

,,

,

V = 4 then V = 5 then

E = 2 ∗ 4 − 3 = 5 E = 2 ∗ 5 − 3 = 7

16

Rigidity in 3-d.

Now every vertex has 3 degrees of freedom.

17

Rigidity in 3-d.


There are 6 independent translations and rotations in R3

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Rigidity in 3-d.



Theorem: If a graph is minimally rigid in R3 then EG = 3VG − 6

and for any subgraph H with at least 2 vertices, EH ≤ 3VH − 6.

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Rigidity in 3-d.



Theorem: If a graph is minimally rigid in R3 then EG = 3VG − 6

and for any subgraph H with at least 2 vertices, EH ≤ 3VH − 6.

Converse is false...

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III. Characterizations of minimally rigid graphs.



EH ≤ 2VH − 3.

uu

uuu

uu

��

��

@ �

@

Algorithm? Not very efficient...

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EH ≤ 2VH − 3.

Trees

Trees are minimally connected graphs.

Theorem G is a tree if and only if EG = VG − 1 and for any

subgraph H with at least 2 vertices, EH ≤ VH − 1.

2V − 3 = 2(V − 1) − 1

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Characterizations of minimally rigid graphs.

Theorem (Lovasz - Yemini, 1982) A graph is minimally rigid if

and only if adding any edge (including doubling an existing edge)

results in a graph that is two edge-disjoint spanning trees.

uu

uuu

uu

��

��

@ �

@ uu

uuu

uu

��

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@ �

@��

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Algorithm? Edmonds Matroid Partitioning Algorithm

-polynomial in V, E. Run it C(n,2) times!

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Theorem (Crapo, 1992) A graph is minimally rigid if and only if

its edges are the disjoint union of 3 trees such that each vertex is

incident with exactly 2 trees and in any subgraph the trees have

different spans.

uu

uuu

uu

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@ �

@ uu

uuu

uu

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This characterization can be tested in one application of a

modified Matroid partitioning algorithm.

25


Theorem (Henneberg, 1860): A graph is minimally rigid if and

only if it can be constructed using a Henneberg 2-sequence.

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Begin with the graph consisting of 2 vertices and a single edge.

1. Add a new vertex of degree 2.

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2. Delete an edge {v1, v2}, and add a new vertex v adjacent to

each of {v1, v2} adjacent to any additional vertex.

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The characterizations of minimally rigid graphs in the plane.

Theorem The following are equivalent.

1. EG = 2VG − 3 and every subgraph, H, with VH ≥ 2,

EH ≤ 2VH − 3 = 2(VH − 1) − 1

2. Adding any one edge to G results in a graph that can be

decomposed into 2 spanning trees.

3. G can be decomposed into 3 trees such that each vertex is

incident with exactly 2 trees.

4. G can be obtained from a single edge by Henneberg

construction.

5. G is minimally rigid in the plane.

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IV. Trees, Arboricity and Characterizations

• Trees are minimally connected graphs.

• Trees have n − 1 edges.

Given a (multi-)graph G, can the edges of G be partitioned into k

spanning trees? For what k?

Forests

• Forests are acyclic graphs with at most n − 1 edges.

Given a graph G, for what k can the edges of G be partitioned

into k forests?

Equivalently, if EG = k(VG − 1) − l does there exist l edges that

can be added to G to form a graph whose edges can be

partitioned into k spanning trees?

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Nash-Williams defined:

The arboricity of the graph G is the minimum number of forests

into which the edges of G can be partitioned.

Theorem(Nash-Williams, Tutte ’64): A graph G, has arboricity

≤ k if and only if for all subgraphs H ⊂ G

EH ≤ k(VH − 1).

The Matroid Partitioning Algorithm (Edmonds) produces the k

forests or shows that k is not sufficient, in polynomial time.

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Graphs for which adding some l edges results in k spanning trees.

Theorem The following are equivalent for a graph G, and

integers k > 0 and l ≥ 0.

1. EG = k(VG − 1) − l , and for subgraphs H ⊂ G with at least

2 vertices EH ≤ k(VH − 1).

2. There exist some l edges which when added to G results in a

graph that can be decomposed into k spanning trees.

3. G can be decomposed into k + l trees such that each vertex is

incident with exactly k trees.

4. G can be constructed using Henneberg type construction.

32

Graphs for which adding any l edges results in k spanning trees.

Theorem The following are equivalent for a graph G, and

integers k > 0 and 0 ≤ l < k.

1. EG = k(VG − 1) − l , and for subgraphs H ⊂ G with at least

2 vertices EH ≤ k(VH − 1) − l .

2. Adding any l edges to G results in k spanning trees.

3. G can be decomposed into k + l trees such that each vertex is

incident with exactly k trees, such that any subgraph H

contains at least k + l subtrees as well.

4. G can be constructed using Henneberg type construction.

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Nice theorems, not so nice algorithms.

Best algorithm for determining if a graph is the union of k-trees

and some related problems.

Gabow and Westermann ’88 O((km)3/2)

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V. The Pebbling Algorithm.

First developed for basic rigidity problem by Hedrickson and

Jacobs ’97.

New version by Lee and Streinu. O((kn)2)

Determines if a given graph is the disjoint union of k trees.

(or related conditions)

35

Pebbling Algorithm (Streinu & Lee):

Basic algorithm to determine if a given graph is the disjoint

union of k trees.

• Place k pebbles on each vertex.

• Accept an edge if there are at least k + 1 pebbles total at its

ends.

• Delete one pebble and orient edge away from that vertex.

• Pebbles can be moved against directed edges, and then edges

reversed.

• If can add all edges and end with k pebbles then YES k-trees.

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k = 3 example


37

k = 3 example

• Accept an edge if at least k + 1 pebbles at its ends.


• Pebbles can move against directed edges, then edges reversed.

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k = 3 example

• Accept an edge if at least k + 1 pebbles at its ends.


• Pebbles can move against directed edges, then edges reversed.

• If can add all edges and end with k pebbles then YES k-trees.

37

Add some s for k-trees Algorithm.

The (k, k + 1) pebble game.


• Accept an edge if there are at least k + 1 pebbles total at its

ends.

• If can add all edges and end with k + s then YES.

Determines whether this is a graph for which adding some s

edges gives a graph that is the disjoint union of k spanning trees.

38

Rigidity Algorithm.

The (2, 4) pebble game.

• Place 2 pebbles on each vertex.

• Accept an edge if there are at least 4 pebbles total at its ends.

• If can add all edges and end with 3 pebbles then YES

minimally rigid.

Determines whether this is a graph for which

adding any edge results in a graph that is the disjoint union of 2

spanning trees.

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General Algorithm.

The (k, k + 1 + r ) pebble game.


• Accept an edge if there are at least k + 1 + r pebbles total at

its ends.

• If can add all edges and end with k + r then YES.


adding any r edges results in a graph that is the disjoint union of

k spanning trees.

40

General Algorithm.

The (k, k + 1 + r ) pebble game.


• Accept an edge if there are at least k + 1 + r pebbles total at

its ends.

• If can add all edges and end with k + r + s then YES.


adding some s edges gives a graph for which

adding any r edges results in a graph that is the disjoint union of

k spanning trees.

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VI. Applications: (a)Rigid Components.

If a graph does not represent a rigid body, what motions does it

allow? The pebble game can find rigid components.

41

In a rigid component for each pair of vertices can have 2 pebbles

on one and 1 on the other and no more.

These are in the same rigid component.

42



These are NOT in the same rigid component.

(The algorithm updates the components as edges are added so is

efficient.)

43



The rigid components.

44




45




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VI(b) Protein Folding (in 3-d)

Bonds between pairs of atoms often provide a stronger constraint

(lengths, angles).

Join together bodies not just points in space.

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Body and bar frameworks

Theorem (Tay) A body and bar framework in d space is

generically minimally rigid if the corresponding graph

decomposes into d(d-1)/2 edge disjoint spanning trees.

Theorem(Tay) A body and hinge framework is generically

minimally rigid if the corresponding graph satisfies

E = k(V − 1) − 1 and EH ≤ k(VH − 1) − 1 . (For

k = d(d + 1)/2 − 1)

-The pebble game tests for these.

48

Basic Rigidity:

Graver, Jack; Servatius, Brigitte; Servatius, Herman Combinatorial rigidity.

Graduate Studies in Mathematics, 2. AMS 1993.

Robot Arms and Carpenters Rule:

Ileana Streinu, A Combinatorial Approach to Planar Non-Colliding Robot Arm

Motion Planning in Proc. 41st ACM (FOCS) 2000, pp. 443-453.

Arboricity and Trees

R Haas, Characterizations of arboricity of graphs. Ars Combinatorica, 63:129 -

137, 2002.

Pebble Games

Audrey Lee, Ileana Streinu, Pebble Game Algorithms or Body-and-Bar Rigidity

preprint Smith College.

Application to Protein Flexibility

Donald J. Jacobs, A.J. Rader, M.F. Thorpe, and Leslie A. Kuhn. Protein flexibilty

predictions using graph theory. Proteins 44, 150-165, 2001.

49

Convexification Problem:

Add extra edges so that the object has restricted motion.

Use pseudotriangles.

50

A pseudotriangle has 3 convex corners:

51

A pseudotriangulation is an embedding of a planar graph in

which all interior faces are pseudotriangle. The exterior face is

convex.

52

Theorem: Minimal pseudotriangulations are minimally rigid.

Pseudotriangulations provide stucture to convexify embedded

cycles.

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Algorithm for Convexifying:

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Algorithm for Convexifying:

Pseudotriangulate the polygon, including hull edges.

55

Remove one hull edge. Now have one-degree of freedom and its

an expansive motion.

Open the object.

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Repeat.

http://cs.smith.edu/ streinu/Research/Motion/motion.html

http://cs.smith.edu/ streinu/Research/Motion/Animation/mech1a.html

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mathematics & statistics | smith collegerhaas/pebbles-mathfest.pdfcreated date 7/19/2005 4:53:30...

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