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Pebble games for rigidity Overview. The game of pebbling was first suggested by Lagarias and Saks, as a tool for solving a particular problem in number.

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Presentation on theme: "Pebble games for rigidity Overview. The game of pebbling was first suggested by Lagarias and Saks, as a tool for solving a particular problem in number."— Presentation transcript:

1 Pebble games for rigidity Overview

2 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. ) Introduction

3 Rigidity in 2D – Bar and Joint Introduced by Jacobs & Hendrickson 1997, based on Laman conditionJacobs & Hendrickson 1997 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.

4 What does Laman mean? For the whole graph |e|=2|v|-3 For subgraphs |e|≤2|v|-3 |e|+3=2|v| Number of edges (constrains) Trivial DOFs of a body in 2D Number of vertices x 2 DOFs of a point in 2D The edges (constrains) are well distributed

5 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

6 Example Let’s take the 2D undirected graph and run pebble game in 2D Corresponds to 2D bar & joint floating framework

7 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.

8 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 Begin with a single vertex (clearly independent) Quadruple the edge

9 Direct an edge move 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). 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.

10 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.

11 Result The algorithm ends when all the edges have been processed. Result : 3 pebbles (only trivial DOFs)

12 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.

13 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)

14 Partition to Assur graphs A directed cut defines a partition to Assur graphs.

15 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)

16 Rigidity in 2D – Body & Bar Body & Bar – subgroup of Bar & Joint Graph representation: ◦ Every rigid body = vertex ◦ The rest are bars

17 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

18 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)

19 Example Bb 2D 1 2 3 4 Grounded mechanism with 1 DOF Partition to Assur groups : 4 can be analyzed as stand alone (dyad) Grounded body (vertex)

20 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

21 Examples Minimally rigid Flexible with 1 DOF

22 Thank You !


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