Pebbles, Trees and Rigid Graphs. May, 2005

Transcription

1 Pebbles, Trees and Rigid Graphs May,

2 Rigidity of Graphs Construct the graph G in 2 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). 1

3 Intuition: How many edges are necessary to make a rigid body with V vertices? Total degrees of freedom: 2V. 2

4 Intuition: How many edges are necessary to make a rigid body with V vertices? Total degrees of freedom: 2V. Adding an edge can remove a degree of freedom. 2

5 In the end you will still be able to translate and rotate. Thus, if G is rigid in 2 then E 2V 3. 3

6 A graph is rigid in 2 if some embedding of it is. Turns out, if some embedding is rigid then most are. 4

7 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) 5

8 A graph is minimally rigid (2-isostatic) in 2 if removing any edge results in a graph that is not rigid. These are not minimally rigid. 6

9 Theorem (Laman, 1970) A graph is minimally rigid if and only if E G = 2V G 3 and for any subgraph H with at least 2 vertices, E H 2V H 3. V = 4 then E = = 5 7

10 Theorem (Laman, 1970) A graph is minimally rigid if and only if E G = 2V G 3 and for any subgraph H with at least 2 vertices, E H 2V H 3. V = 4 then V = 5 then E = = 5 E = = 7 7

11 Theorem (Laman, 1970) A graph is minimally rigid if and only if E G = 2V G 3 and for any subgraph H with at least 2 vertices, E H 2V H 3. V = 4 then V = 5 then E = = 5 E = = 7 7

12 Rigidity in 3-d. Now every vertex has 3 degrees of freedom. 8

13 Rigidity in 3-d. Now every vertex has 3 degrees of freedom. There are 6 independent translations and rotations in 3 9

14 Rigidity in 3-d. Now every vertex has 3 degrees of freedom. There are 6 independent translations and rotations in 3 Theorem: If a graph is minimally rigid in 3 then E G = 3V G 6 and for any subgraph H with at least 2 vertices, E H 3V H 6. 10

15 Rigidity in 3-d. Now every vertex has 3 degrees of freedom. There are 6 independent translations and rotations in 3 Theorem: If a graph is minimally rigid in 3 then E G = 3V G 6 and for any subgraph H with at least 2 vertices, E H 3V H 6. Converse is false... 11

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17 Current Applications of Rigidity. robotics (Streinu, Connelly, Demaine, Rote) geometrical properties of molecular, conformations including protein folding (Whitelely, Thorpe) molecular modelling, (Bezdek, Streinu, and others) 13

18 Characterizations of minimally rigid graphs. Theorem (Laman, 1970) A graph is minimally rigid if and only if E G = 2V G 3 and for any subgraph H with at least 2 vertices, E H 2V H 3. Algorithm? Not very efficient... 14

19 Theorem (Laman, 1970) A graph is minimally rigid if and only if E G = 2V G 3 and for any subgraph H with at least 2 vertices, E H 2V H 3. Trees Trees are minimally connected graphs. Theorem G is a tree if and only if E G = V G 1 and for any subgraph H with at least 2 vertices, E H V H 1. 2V 3 = 2(V 1) 1 15

20 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. Algorithm? Edmonds Matroid Partitioning Algorithm -polynomial in V, E. Run it C(n,2) times! 16

21 Characterizations of minimally rigid graphs. 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. This characterization can be tested in one application of a modified Matroid partitioning algorithm. 17

22 Characterizations of minimally rigid graphs. Theorem (Henneberg, 1860): A graph is minimally rigid if and only if it can be constructed using a Henneberg 2-sequence. 18

23 Begin with the graph consisting of 2 vertices and a single edge. 1. Add a new vertex of degree 2. 19

24 2. Delete an edge {v 1, v 2 }, and add a new vertex v adjacent to each of {v 1, v 2 } adjacent to any additional vertex. 20

25 The characterizations of 2-isostatic (minimally rigid) graphs. Theorem The following are equivalent. 1. E G = 2V G 3 and every subgraph, H, with V H 2, E H 2V H 3 = 2(V H 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 2-isostatic. 21

26 Trees 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 E G = k(v G 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? 22

27 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 E H k(v H 1). The Matroid Partitioning Algorithm (Edmonds) produces the k forests or shows that k is not sufficient, in polynomial time. 23

28 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 E G = k(v G 1) l, and for subgraphs H G with at least 2 vertices E H k(v H 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. 24

29 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. E G = k(v G 1) l, and for subgraphs H G with at least 2 vertices E H k(v H 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. 25

30 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 O((km) 3/2 ) Pebbling Algorithm: First developed for basic rigidity problem by Hedrickson and Jackson. New, general version by Lee and Streinu. O((kn) 2 ) 26

31 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. 27

32 k = 3 example Place k pebbles on each vertex. 28

33 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

34 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

35 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

36 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

37 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

38 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

39 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

40 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

41 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

42 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

43 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

44 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

45 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

46 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

47 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. 28

48 k = 3 example Accept an edge if at least k + 1 pebbles at its ends. Delete one pebble and orient edge away from that vertex. Pebbles can move against directed edges, then edges reversed. If can add all edges and end with k pebbles then YES k-trees. 28

49 Add some s for k-trees Algorithm. The (k, k + 1) pebble game. Place k pebbles on each vertex. 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. 29

50 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. 30

51 General Algorithm. The (k, k r) pebble game. Place k pebbles on each vertex. Accept an edge if there are at least k r pebbles total at its ends. If can add all edges and end with k + r then YES. Determines whether this is a graph for which adding any r edges results in a graph that is the disjoint union of k spanning trees. 31

52 General Algorithm. The (k, k r) pebble game. Place k pebbles on each vertex. Accept an edge if there are at least k r pebbles total at its ends. If can add all edges and end with k + r + s then YES. Determines whether this is a graph for which 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. 31