Pebble Game for rigidity presentation
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Pebble games for rigidity
Overview
Introduction
The game of pebbling was first suggested
by Lagarias and Saks, as a tool for solving
a particular problem in number theory.
The algorithm can be used for analyzing
the rigidity and flexibility of a variety of
structures (mechanisms, molecules exc. )
Rigidity in 2D – Bar and Joint
Introduced by Jacobs & Hendrickson 1997
, based on Laman condition
Laman graph of n vertices (1970)
The whole graph has exactly 2n −3 edges
For all k, every k-vertex subgraph has at most 2k −3
edges
The graph describes a minimally rigid systems of rods
(edges) and joints (vertices) in the plane.
What does Laman mean?
For the whole graph
|e|+3=2|v|
|e|=2|v|-3
Number of edges
(constrains)
For subgraphs
|e|≤2|v|-3
Trivial DOFs of a
body in 2D
Number of vertices
x 2 DOFs of a point
in 2D
The edges (constrains) are
well distributed
Laman condition
Taking Laman condition literally leads to
exponential time complexity
◦ Every sub-graph must be tested for the number of
vertices vs. edges
Alternate Laman condition:
◦ The edges in G are independent in 2D if and only if
for each edge (a, b), the graph formed by quadrupling
(a, b) has no induced subgraph of k nodes and >2k
edges
=> Building the graph from sub-graph and testing
each new edge for independence leads to
polynomial time complexity
Example
Let’s take the 2D undirected graph and run pebble game in 2D
Corresponds to 2D bar & joint floating framework
Example
Let’s take a floating bar and joint framework in 2D
Each vertex is given two pebbles, corresponding to the two
degrees of freedom of a point in 2D
A vertex can use its pebbles to cover any two edges which are incident to that
vertex.
Starting point is arbitrary.
Building directed graph
We will build directed edges by extending
independent sub-group
◦ Quadrupling each new edge to test
independence with the sub-group we already
have
Quadruple
the edge
Begin with a single vertex (clearly independent)
Direct an edge move
• General direct an edge move
•If i and j are vertices with two pebbles on each
(corresponding to independent constraint),
add the edge ij (oriented toward j) and pick up one of the pebbles from
i.
In case 2 pebbles cannot be found at the ends of that constraint after an
exhaustive search, the constraint is redundant (and therefore can to be
directed).
Pebble slide move
Pebble slide move
If ij is an edge in the pebble game's graph and there is a
pebble on j, reverse ij and move the pebble from j to i.
Result
The algorithm ends when all the edges have been
processed.
Result : 3 pebbles (only trivial DOFs)
Redundant edge
What happens to over constraints ?
example with SS :
•Any edge that can’t be directed corresponds to redundant constraint.
This happens when the length between it’s end point is already set by
other constrains.
•Inside the rigid region any edge can be redundant.
DOF number
The resulting number of pebbles after all
edges were processed, gives the number of
DOFs of the corresponding mechanism.
◦ For floating mechanism there are 3 trivial DOFs
=> 3 remaining pebbles is an indicator for rigidity
◦ For grounded mechanism 0 pebbles indicate
rigidity (edges linked to gnd vertices are directed
in the end, with any free pebble)
Actually this is the number of non preset
vertices (that have pebbles = DOFs)
Partition to Assur graphs
A directed cut defines a partition to Assur
graphs.
Algorithm summary
Find a subset of independent edges.
Use an incremental algorithm: pick an edge, test if it’s independent of the
current subset.
Alternate Laman theorem: The edges in G are independent in 2D if and
only if for each edge (a, b), the graph formed by quadrupling (a, b) has no
induced subgraph of k nodes and >2k edges
Grow a maximal set of independent edges one at a time.
New edge is added if it is independent of existing set.
◦
Quadruple the new edge and test the Laman condition.
If Laman failed, then output “not rigid”.
If 2n-3 independent edges are found, then graph is rigid (n = number of
vertices).
Testing an edge for independence takes O(n) time: we do 3 depth-first
search in a graph with O(n) edges.
At most m edges will be tested. The total running time is O(nm)
Rigidity in 2D – Body & Bar
Body & Bar – subgroup of Bar & Joint
Graph representation:
◦ Every rigid body = vertex
◦ The rest are bars
Laman for Body & Bar in 2D
Laman graph for this case:
◦ The whole graph has exactly 3n −3 edges
◦ For all k, every k-vertex subgraph has at most
3k −3 edges
|e|+3=3|v|
Number of edges
(constrains)
Trivial DOFs of a
body in 2D
Number of vertices
x 3 DOFs of a body
in 2D
Pebble game for Body & Bar 2D
Every vertex is given 3 pebbles
The condition to direct an edge is to have
≥ 4 pebbles on the incident vertices
The rest is the same (example follows)
Example Bb 2D
4
1
Grounded
body (vertex)
2
3
• Grounded mechanism with 1 DOF
• Partition to Assur groups : 4 can be analyzed as stand alone (dyad)
Rigidity in 3D – Body and bar
Introduced by Tay and Whiteley
◦ Vertices -> rigid bodies
◦ Edges -> bars
◦ Hinge = 5 bars
Laman graph for this case:
◦ The whole graph has exactly 6n −6 edges
◦ For all k, every k-vertex subgraph has at most
6k −6 edges
Need at least 7 pebbles to direct an edge
Examples
Minimally rigid
Flexible with 1 DOF
Thank You !