Rigidity and global rigidity - WPI - Worcester Polytechnic Institute
Rigidity and global rigidity
Introduction Type of Rigidity Vertex transitive graphs Example
Bill Jackson
Brigitte Servatius
Herman Servatius
Worcester Polytechnic Institute
Example Highly connected graphs Random graphs Open problems
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Highly connected graphs
Generic rigidity in the plane is a graph theoretic property:
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Theorem 1 (Laman - 1970)[7] A graph is generically rigid in the plane if and only if it has a subset |E| of edges with |E| = 2|V (E)| − 3
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and, for every subset F ⊆ E,
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|F | ≤ 2|V (F )| − 3.
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Rigidity Different notions of rigidity.
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Rigidity Non rigid graphs have a motion.
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Non-infinitesimally rigid graphs have initial velocity candidates.
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Generic rigidity is a property of the graph, not the embedding.
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Introduction Type of Rigidity Vertex transitive graphs Example Example Highly connected graphs Random graphs
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Type of Rigidity
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We will use the term framework (in m-space) to denote a → − triple (V, E, − p ), where (V, E) is a graph and → p is an embedding (injection) of V into real m-space.
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We say that a framework is globally rigid (in m-space) if all solutions to the system of quadratic equations obtained from requiring all edge lengths to be fixed, with the coordinates of the vertices as variables, correspond to congruent frameworks; we say that a framework is rigid (in m-space) if all solutions to the corresponding system in some neighborhood of the original solution (as a point in mn-space) come from congruent frameworks. DON’T CLICK HERE!
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− We say that a given framework (V, E, → p ) is generic if all frameworks corresponding to points in a neighborhood of → → P = − p (V ) in Rnm are rigid or not rigid as is (V, E, − p ). A set of points P in m-space is said to be generic if each − − framework (V, E, → p ) with → p (V ) = P is generic.
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If G(V, E) is not rigid, we call the maximal rigid subgraphs of G the rigid components and note that rigid components partition E. Then M(G) is the direct sum over its restrictions on the rigid components. The following theorem is equivalent to Laman’s Theorem 1, it uses the rank function of M rather than independence to characterize rigidity.
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Theorem 2 [9] Let G = (V, E) be a graph. Then G is rigid if and only if for all families of induced subm graphs {Gi = (Vi, Ei)}m i=1 such that E = ∪i=1 Ei we have P m i=1 (2|Vi | − 3) ≥ 2|V | − 3.
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G(V, E) is called redundantly rigid if G(V, E − e) is rigid for all e ∈ E, i.e. the removal of a single edge e from the rigid graph G does not destroy rigidity. Redundant rigidity is a key to characterize global rigidity. Theorem 3 [5] Let G be a graph. Then G is globally rigid if and only if G is a complete graph on at most three vertices, or G is both 3-connected and redundantly rigid.
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Introduction Type of Rigidity Vertex transitive graphs Example Example Highly connected graphs Random graphs
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Vertex transitive graphs
Theorem 4 A four-regular vertex transitive graph is generically rigid in the plane if and only if it contains no subgraph isomorphic to K4, or is K5 or one of the graphs in the following figure.
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K4£K2 a)
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Vertex transitive rigid graphs containing K4.
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Let G be a vertex transitive four-regular graph on n vertices. G has 2n edges and hence must contain a cycle in the generic rigidity matroid. Let v be a vertex. If G is non-rigid, then G is not K5. The edges incident to v must be in at least two distinct rigid components, and we have several cases. If the four edges at v are all in distinct rigid components, then all rigid components would be singleton edges and the edge set of G would be independent, violating Laman’s condition 1.
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If the edges at v belong to two rigid components of equal size, then the rigid components must only have vertices of valence 2 in the component, and so must be triangles. At each vertex there are 2 triangles, and all the rigid components are triangles, by transitivity. The generic rigidity matroid is the direct sum of the rigidity matroids of the rigid components of the graph. Since a triangle is independent, the whole edge set is independent, violating Laman’s condition. If three of the edges at v belong to one rigid component, then, by transitivity, that rigid component is a 3 regular rigid graph, hence K4, K3,3, or the triangular prism. K3,3 and the triangular prism are both independent, yielding the same contradiction as before, so G contains K4.
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On the other hand, if G is a generically rigid four regular, vertex transitive graph, then the generic rigidity matroid can have at most three independent cycles. The number of K4’s at each vertex is the same. If there are 4, then the graph is K5. The case of 3 K4’s is vacuous. If there are two K4’s, then they share an edge (a, b) not incident to v. The union of the two K4’s contains three vertices of valence 4, {v, a, b} as well as two vertices of valence 3, {x, y}. By vertex transitivity, x must be incident to two K4’s, however the vertices v, a and b cannot be adjacent to the fourth neighbor of x since they are already of degree 4. If there is only one K4 at each vertex, then we know that there can be at most three independent cycles in the generic rigidity matroid, so there can be either exactly two or three K4’s, both of which are shown in Figure 1.
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Introduction Type of Rigidity Vertex transitive graphs Example Example Highly connected graphs Random graphs Open problems
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Lemma 1 The rigid components of a vertex transitive nonrigid graph which are not single edges belong to the same isomorphism class.
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Proof: Any graph automorphism maps rigid components to rigid components. If G is non-rigid and vertex transitive, then every vertex v must be the vertex of attachment of two or more rigid components. There can not be two nontrivial components joined at v, because any basis for a nontrivial component has minimum vertex degree at least two. The union over these bases is an independent set in G and can therefore not have minimum degree 4 at each vertex. If there is exactly one nontrivial component joined to single edge components at each v, then, for the same reason as before, the number of single edge components is exactly 1 at each vertex. If the edge set of G is independent, the valence of each vertex is at most 3.
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Theorem 5 Let G be a vertex transitive non-rigid graph. Then G is k-regular with k ≤ 6, and contracting the nontrivial rigid components of G produces a vertex transitive graph of regularity at most 5.
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Proof: By the lemma, there is only one kind of non-trivial rigid component, say on s vertices. Such a component is attached to its complement in G by s independent edges. Contracting the non-trivial rigid components will, by our transitivity assumptions, produce an s-regular graph on v vertices, where v is the number of non-trivial rigid components in G. The contracted graph has sv/2 edges. The rank of G is v(2s−3)+sv/2, which must be smaller than 2sv −3. This yields the inequality 6 < v(6 − s), thus each rigid component has at most 5 vertices and the regularity of G is at most 6.
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Theorem 6 Let G be a connected k-regular vertex transitive graph on n vertices. Then G is not rigid if and only if either: (a) k = 2 and n ≥ 4. (b) k = 3 and n ≥ 8. (c) k = 4 and G has a factor consisting of s disjoint copies of K4 where s ≥ 4 (d) k = 5 and G has a factor consisting of t disjoint copies of K5 where t ≥ 8.
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A
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Two embeddings which are rigid, but neither infinitesimally rigid nor globally rigid. A
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Two embeddings which are rigid and infinitesimally rigid but not globally rigid. •First •Prev •Next •Last •Go Back •Full Screen •Close •Quit
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Two embeddings which are rigid, but not globally rigid.
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We observe that for a rigid G which is not redundantly rigid, M(G) is not connected. It is in fact the direct sum over the maximal redundantly rigid subgraphs (or singleton edges). The arguments in the preceding proofs are unaltered if we replace rigid components by redundantly rigid subgraphs and we obtain
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Theorem 7 A vertex transitive rigid graph is also globally rigid unless it has a factor consisting of 3 copies of K4 or 6 copies of K5.
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Proof: For rigid but not globally rigid graphs, equality holds in the last inequality of the proof of Theorem 5, and the two solutions yield the two exceptions stated.
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4.
Example
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5.
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Let G be a graph and k be a positive integer. The graph G is k-connected if for all pairs of subgraphs G1, G2 of G such that G = G1 ∪ G2, |V (G1) − V (G2)| ≥ 1 and |V (G2) − V (G1)| ≥ 1, we have |V (G1) ∩ V (G2)| ≥ k. Lov´asz and Yemini [9] showed that every 6-connected graph G is rigid. Planar graphs are at most 5-connected and one might wonder if the connectivity requirement can be lowered in order to imply rigidity of a planar graph. However, the following figure shows a 5connected planar nonrigid graph. The rigidity properties are easily checked using Theorem 2.
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A planar 5-connected non-rigid graph.
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However,Lov´asz and Yemini [9] note that their proof technique will show that G−{e1, e2, e3} is rigid for all e1, e2, e3 ∈ E, and hence that G is redundantly rigid. This result was combined with Theorem 3 in [5] to deduce Theorem 8 Every 6-connected graph is globally rigid.
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We say G is essentially 6-connected if: (a) G is 4-connected, (b) for all pairs of subgraphs G1, G2 of G such that G = G1 ∪G2, |V (G1)−V (G2)| ≥ 3 and |V (G2)−V (G1)| ≥ 3, we have |V (G1) ∩ V (G2)| ≥ 5, and (c) for all pairs of subgraphs G1, G2 of G such that G = G1 ∪G2, |V (G1)−V (G2)| ≥ 4 and |V (G2)−V (G1)| ≥ 4, we have |V (G1) ∩ V (G2)| ≥ 6. Theorem 9 Every essentially 6-connected graph is redundantly rigid.
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As an example we note that the complete bipartite graph K4,m, m ≥ 4, satisfies the hypotheses of Theorem 9, and hence is globally rigid, but does not satisfy the hypotheses of Theorem 8.
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An even weaker connectivity condition is sufficient to imply that 4-regular graphs are globally rigid. A graph G = (V, E) is said to be cyclically k-edge-connected if for all X ⊆ V such that G[X] and G[V − X] both contain cycles, we have at least k edges from X to V − X.
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Theorem 10 Let G = (V, E) be a cyclically 5-edgeconnected 4-regular graph. Then G is globally rigid.
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Proof: Suppose that G is not globally rigid. The facts that G is cyclically 5-edge-connected and 4-regular imply that G is 3-connected. Hence, by [5], G is not redundantly rigid. Let G1 = (V1, E1) be a largest redundantly rigid component of G. Since |E| = 2n, G contains an M -circuit and hence |V1| ≥ 4 and |E1| ≥ 2|V1| − 2. Since G is 4-regular this implies that dG(V1) ≤ 4 and |E(G − V1)| ≥ 2|V − V1| − 2. Thus G1 and G − G1 both contain cycles. Now the fact that dG(V1) ≤ 4 contradicts the hypothesis that G is cyclically 5-connected.
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Let Gn,d denote the probability space of all d-regular graphs on n vertices with the uniform probability distribution. A sequence of graph properties An holds asymptotically almost surely, or a.a.s. for short, in Gn,d if limn→∞ PrGn,d (An) = 1. Graphs in Gn,d are known to be a.a.s. highly connected. It was shown by Bollob´as [1] and Wormald [12] that if G ∈ Gn,d for any fixed d ≥ 3, then G is a.a.s. d-connected. This result was extended to all 3 ≤ d ≤ n − 4 by Cooper et al. [3] and Krivelevich et al. [6]. Stronger results hold if we discount ‘trivial’ cutsets. In [13], Wormald shows that if G ∈ Gn,d for any fixed d ≥ 3, then G is a.a.s. cyclically (3d − 6)-edge-connected. Together with Theorem 10, this immediately gives:
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Theorem 11 If G ∈ Gn,4 then G is a.a.s. globally rigid.
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In fact this result holds for all d ≥ 4. Theorem 12 If G ∈ Gn,d and d ≥ 4 then G is a.a.s. globally rigid.
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Proof: If d ≥ 6 then G is a.a.s. 6-connected by [3, 6] and the result follows from Theorem 8. If d = 5 then the result follows from Theorem 11 by contiguity, see [13].
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Let G(n, p) denote the probability space of all graphs on n vertices in which each edge is chosen independently with probability p. In the following we will assume that all logarithms are natural. We will need the following results on G(n, p).
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Lemma 2 Let G ∈ G(n, p), where p = (log n + k log log n + w(n))/n, k ≥ 2 is an integer and limn→∞ w(n) = ∞. For each fixed integer t, let St be the set of vertices of G of degree at most t. Then, a.a.s. (a) Sk−1 is empty, (b) no two vertices of St are joined by a path of length at most two in G, (c) G − St−1 is non-empty and t-connected. Proof: Facts (a) and (b) are well known, see for example [2]. Fact (c) follows from (a), (b) and [10, Theorem 4]
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Theorem 13 Let G ∈ G(n, p), where p = (log n + k log log n + w(n))/n, and limn→∞ w(n) = ∞. (a) If k = 2 then G is a.a.s. rigid. (b) If k = 3 then G is a.a.s. globally rigid.
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Proof: (a) We adopt the notation of Lemma 2. It follows from Lemma 2 that a.a.s. S1 = ∅ and G − S5 is a.a.s. 6connected. Hence G−S5 is a.a.s. (globally) rigid by Theorem 8. Since adding a new vertex joined by at least two new edges to a rigid graph preserves rigidity, it follows that G is a.a.s. rigid. (b) This follows in similar way to (a), using the facts that S2 = ∅ and that adding a new vertex joined by at least three new edges to a globally rigid graph preserves global rigidity.
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The bounds on p given in Theorem 13 are best possible since if G ∈ G(n, p) and p = (log n + k log log n + c)/n for any constant c, then G does not a.a.s. have minimum degree at least k, see [2].
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Let Geom(n, r) denote the probability space of all graphs on n vertices in which the vertices are distributed uniformly at random in the unit square and each pair of vertices of distance at most r are joined by an edge. Suppose G ∈ Geom(n, r). Li, Wan and Wang [8] have shown that if nπr2 = log n + (2k − 3) log log n + w(n) for k ≥ 2 a fixed integer and limn→∞ w(n) = ∞, then G is a.a.s. k-connected. As noted by Eren et al. [4], this result can be combined with Theorem 8 to deduce that if nπr2 = log n+9 log log n+w(n) then G is a.a.s. globally rigid. On the other hand, it is also shown in [8] that if nπr2 = log n+(k −1) log log n+c for any constant c, then G is not a.a.s. k-connected. It is still conceivable, however, that if nπr2 = log n+log log n+w(n) then G is a.a.s. rigid, and that if nπr2 = log n+2 log log n+w(n) then G is a.a.s. globally rigid.
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References ´s, Random graphs, in Surveys in Combinatorics 1999 (N.V. Temperley, ed.), [1] B. Bolloba LMS Lecture Note Series 52, Cambridge University Press, Cambridge, UK, 1981, 80–102. Introduction
´s, Random graphs Academic Press, New York, 1985. [2] B. Bolloba
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[3] C. Cooper, A. Frieze and B. Reed, Random regular graphs of non-constant degree, SIAM J. Algebraic Discrete Methods 3 (1982), no. 1, 91–98. [4] T. Eren, D.K. Goldenberg, W. Whiteley and Y.R. Yang, Rigidity, computation, and randomization in network localization, preprint.
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´n, Connected rigidity matroids and unique realizations of [5] B. Jackson and T. Jorda graphs, J. Combinatorial Th., Series B, to appear.
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[6] M. Krivelevitch, B. Sudakov, V.H. Vu and N.C. Wormald, Random regular graphs of high degree, SIAM J. Algebraic Discrete Methods 3 (1982), no. 1, 91–98.
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[7] G.Laman On Graphs and rigidity of plane skeletal structures, J. Engrg. Math. 4 (1970), 331–340. [8] X-Y. Li, P-J. Wan and Y. Wang, Fault tolerance deployment and topology control in wireless networks, in Proceedings of the ACM Symposium on Mobile Ad Hoc Networking and Computing (Mobihoc) (2003), Annapolis, MD, June 2003. ´sz and Y. Yemini, On generic rigidity in the plane, SIAM J. Algebraic Discrete [9] L. Lova Methods 3 (1982), no. 1, 91–98. [10] T. Luczak, Size and connectivity of the k-core of a random graph, Discrete Math. 91 (1991), 61–68. [11] J.Graver, B. Servatius and H.Servatius Combinatorial rigidity, American Mathematical Society (1993).
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[12] N.C. Wormald, The asymptotic connectivity of labelled regular graphs, J. Combinatorial Th., Series B, (1981), 156–167. [13] N.C. Wormald, Models of random regular graphs, in Surveys in Combinatorics 1999 (J.D. Lamb and D.A. Preece, eds.), (1999), 239–298.
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Generic rigidity in the plane is a graph theoretic property:
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Theorem 1 (Laman - 1970)[7] A graph is generically rigid in the plane if and only if it has a subset |E| of edges with |E| = 2|V (E)| − 3
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|F | ≤ 2|V (F )| − 3.
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Rigidity Different notions of rigidity.
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Rigidity Non rigid graphs have a motion.
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Non-infinitesimally rigid graphs have initial velocity candidates.
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Generic rigidity is a property of the graph, not the embedding.
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We will use the term framework (in m-space) to denote a → − triple (V, E, − p ), where (V, E) is a graph and → p is an embedding (injection) of V into real m-space.
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We say that a framework is globally rigid (in m-space) if all solutions to the system of quadratic equations obtained from requiring all edge lengths to be fixed, with the coordinates of the vertices as variables, correspond to congruent frameworks; we say that a framework is rigid (in m-space) if all solutions to the corresponding system in some neighborhood of the original solution (as a point in mn-space) come from congruent frameworks. DON’T CLICK HERE!
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− We say that a given framework (V, E, → p ) is generic if all frameworks corresponding to points in a neighborhood of → → P = − p (V ) in Rnm are rigid or not rigid as is (V, E, − p ). A set of points P in m-space is said to be generic if each − − framework (V, E, → p ) with → p (V ) = P is generic.
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If G(V, E) is not rigid, we call the maximal rigid subgraphs of G the rigid components and note that rigid components partition E. Then M(G) is the direct sum over its restrictions on the rigid components. The following theorem is equivalent to Laman’s Theorem 1, it uses the rank function of M rather than independence to characterize rigidity.
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Theorem 2 [9] Let G = (V, E) be a graph. Then G is rigid if and only if for all families of induced subm graphs {Gi = (Vi, Ei)}m i=1 such that E = ∪i=1 Ei we have P m i=1 (2|Vi | − 3) ≥ 2|V | − 3.
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G(V, E) is called redundantly rigid if G(V, E − e) is rigid for all e ∈ E, i.e. the removal of a single edge e from the rigid graph G does not destroy rigidity. Redundant rigidity is a key to characterize global rigidity. Theorem 3 [5] Let G be a graph. Then G is globally rigid if and only if G is a complete graph on at most three vertices, or G is both 3-connected and redundantly rigid.
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Theorem 4 A four-regular vertex transitive graph is generically rigid in the plane if and only if it contains no subgraph isomorphic to K4, or is K5 or one of the graphs in the following figure.
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K4£K2 a)
b)
Vertex transitive rigid graphs containing K4.
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Let G be a vertex transitive four-regular graph on n vertices. G has 2n edges and hence must contain a cycle in the generic rigidity matroid. Let v be a vertex. If G is non-rigid, then G is not K5. The edges incident to v must be in at least two distinct rigid components, and we have several cases. If the four edges at v are all in distinct rigid components, then all rigid components would be singleton edges and the edge set of G would be independent, violating Laman’s condition 1.
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If the edges at v belong to two rigid components of equal size, then the rigid components must only have vertices of valence 2 in the component, and so must be triangles. At each vertex there are 2 triangles, and all the rigid components are triangles, by transitivity. The generic rigidity matroid is the direct sum of the rigidity matroids of the rigid components of the graph. Since a triangle is independent, the whole edge set is independent, violating Laman’s condition. If three of the edges at v belong to one rigid component, then, by transitivity, that rigid component is a 3 regular rigid graph, hence K4, K3,3, or the triangular prism. K3,3 and the triangular prism are both independent, yielding the same contradiction as before, so G contains K4.
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On the other hand, if G is a generically rigid four regular, vertex transitive graph, then the generic rigidity matroid can have at most three independent cycles. The number of K4’s at each vertex is the same. If there are 4, then the graph is K5. The case of 3 K4’s is vacuous. If there are two K4’s, then they share an edge (a, b) not incident to v. The union of the two K4’s contains three vertices of valence 4, {v, a, b} as well as two vertices of valence 3, {x, y}. By vertex transitivity, x must be incident to two K4’s, however the vertices v, a and b cannot be adjacent to the fourth neighbor of x since they are already of degree 4. If there is only one K4 at each vertex, then we know that there can be at most three independent cycles in the generic rigidity matroid, so there can be either exactly two or three K4’s, both of which are shown in Figure 1.
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Lemma 1 The rigid components of a vertex transitive nonrigid graph which are not single edges belong to the same isomorphism class.
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Proof: Any graph automorphism maps rigid components to rigid components. If G is non-rigid and vertex transitive, then every vertex v must be the vertex of attachment of two or more rigid components. There can not be two nontrivial components joined at v, because any basis for a nontrivial component has minimum vertex degree at least two. The union over these bases is an independent set in G and can therefore not have minimum degree 4 at each vertex. If there is exactly one nontrivial component joined to single edge components at each v, then, for the same reason as before, the number of single edge components is exactly 1 at each vertex. If the edge set of G is independent, the valence of each vertex is at most 3.
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Theorem 5 Let G be a vertex transitive non-rigid graph. Then G is k-regular with k ≤ 6, and contracting the nontrivial rigid components of G produces a vertex transitive graph of regularity at most 5.
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Proof: By the lemma, there is only one kind of non-trivial rigid component, say on s vertices. Such a component is attached to its complement in G by s independent edges. Contracting the non-trivial rigid components will, by our transitivity assumptions, produce an s-regular graph on v vertices, where v is the number of non-trivial rigid components in G. The contracted graph has sv/2 edges. The rank of G is v(2s−3)+sv/2, which must be smaller than 2sv −3. This yields the inequality 6 < v(6 − s), thus each rigid component has at most 5 vertices and the regularity of G is at most 6.
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Theorem 6 Let G be a connected k-regular vertex transitive graph on n vertices. Then G is not rigid if and only if either: (a) k = 2 and n ≥ 4. (b) k = 3 and n ≥ 8. (c) k = 4 and G has a factor consisting of s disjoint copies of K4 where s ≥ 4 (d) k = 5 and G has a factor consisting of t disjoint copies of K5 where t ≥ 8.
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Two embeddings which are rigid, but neither infinitesimally rigid nor globally rigid. A
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Two embeddings which are rigid, but not globally rigid.
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We observe that for a rigid G which is not redundantly rigid, M(G) is not connected. It is in fact the direct sum over the maximal redundantly rigid subgraphs (or singleton edges). The arguments in the preceding proofs are unaltered if we replace rigid components by redundantly rigid subgraphs and we obtain
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Theorem 7 A vertex transitive rigid graph is also globally rigid unless it has a factor consisting of 3 copies of K4 or 6 copies of K5.
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Proof: For rigid but not globally rigid graphs, equality holds in the last inequality of the proof of Theorem 5, and the two solutions yield the two exceptions stated.
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Let G be a graph and k be a positive integer. The graph G is k-connected if for all pairs of subgraphs G1, G2 of G such that G = G1 ∪ G2, |V (G1) − V (G2)| ≥ 1 and |V (G2) − V (G1)| ≥ 1, we have |V (G1) ∩ V (G2)| ≥ k. Lov´asz and Yemini [9] showed that every 6-connected graph G is rigid. Planar graphs are at most 5-connected and one might wonder if the connectivity requirement can be lowered in order to imply rigidity of a planar graph. However, the following figure shows a 5connected planar nonrigid graph. The rigidity properties are easily checked using Theorem 2.
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A planar 5-connected non-rigid graph.
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However,Lov´asz and Yemini [9] note that their proof technique will show that G−{e1, e2, e3} is rigid for all e1, e2, e3 ∈ E, and hence that G is redundantly rigid. This result was combined with Theorem 3 in [5] to deduce Theorem 8 Every 6-connected graph is globally rigid.
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We say G is essentially 6-connected if: (a) G is 4-connected, (b) for all pairs of subgraphs G1, G2 of G such that G = G1 ∪G2, |V (G1)−V (G2)| ≥ 3 and |V (G2)−V (G1)| ≥ 3, we have |V (G1) ∩ V (G2)| ≥ 5, and (c) for all pairs of subgraphs G1, G2 of G such that G = G1 ∪G2, |V (G1)−V (G2)| ≥ 4 and |V (G2)−V (G1)| ≥ 4, we have |V (G1) ∩ V (G2)| ≥ 6. Theorem 9 Every essentially 6-connected graph is redundantly rigid.
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As an example we note that the complete bipartite graph K4,m, m ≥ 4, satisfies the hypotheses of Theorem 9, and hence is globally rigid, but does not satisfy the hypotheses of Theorem 8.
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An even weaker connectivity condition is sufficient to imply that 4-regular graphs are globally rigid. A graph G = (V, E) is said to be cyclically k-edge-connected if for all X ⊆ V such that G[X] and G[V − X] both contain cycles, we have at least k edges from X to V − X.
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Theorem 10 Let G = (V, E) be a cyclically 5-edgeconnected 4-regular graph. Then G is globally rigid.
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Proof: Suppose that G is not globally rigid. The facts that G is cyclically 5-edge-connected and 4-regular imply that G is 3-connected. Hence, by [5], G is not redundantly rigid. Let G1 = (V1, E1) be a largest redundantly rigid component of G. Since |E| = 2n, G contains an M -circuit and hence |V1| ≥ 4 and |E1| ≥ 2|V1| − 2. Since G is 4-regular this implies that dG(V1) ≤ 4 and |E(G − V1)| ≥ 2|V − V1| − 2. Thus G1 and G − G1 both contain cycles. Now the fact that dG(V1) ≤ 4 contradicts the hypothesis that G is cyclically 5-connected.
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Let Gn,d denote the probability space of all d-regular graphs on n vertices with the uniform probability distribution. A sequence of graph properties An holds asymptotically almost surely, or a.a.s. for short, in Gn,d if limn→∞ PrGn,d (An) = 1. Graphs in Gn,d are known to be a.a.s. highly connected. It was shown by Bollob´as [1] and Wormald [12] that if G ∈ Gn,d for any fixed d ≥ 3, then G is a.a.s. d-connected. This result was extended to all 3 ≤ d ≤ n − 4 by Cooper et al. [3] and Krivelevich et al. [6]. Stronger results hold if we discount ‘trivial’ cutsets. In [13], Wormald shows that if G ∈ Gn,d for any fixed d ≥ 3, then G is a.a.s. cyclically (3d − 6)-edge-connected. Together with Theorem 10, this immediately gives:
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Theorem 11 If G ∈ Gn,4 then G is a.a.s. globally rigid.
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In fact this result holds for all d ≥ 4. Theorem 12 If G ∈ Gn,d and d ≥ 4 then G is a.a.s. globally rigid.
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Proof: If d ≥ 6 then G is a.a.s. 6-connected by [3, 6] and the result follows from Theorem 8. If d = 5 then the result follows from Theorem 11 by contiguity, see [13].
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Let G(n, p) denote the probability space of all graphs on n vertices in which each edge is chosen independently with probability p. In the following we will assume that all logarithms are natural. We will need the following results on G(n, p).
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Lemma 2 Let G ∈ G(n, p), where p = (log n + k log log n + w(n))/n, k ≥ 2 is an integer and limn→∞ w(n) = ∞. For each fixed integer t, let St be the set of vertices of G of degree at most t. Then, a.a.s. (a) Sk−1 is empty, (b) no two vertices of St are joined by a path of length at most two in G, (c) G − St−1 is non-empty and t-connected. Proof: Facts (a) and (b) are well known, see for example [2]. Fact (c) follows from (a), (b) and [10, Theorem 4]
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Theorem 13 Let G ∈ G(n, p), where p = (log n + k log log n + w(n))/n, and limn→∞ w(n) = ∞. (a) If k = 2 then G is a.a.s. rigid. (b) If k = 3 then G is a.a.s. globally rigid.
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Proof: (a) We adopt the notation of Lemma 2. It follows from Lemma 2 that a.a.s. S1 = ∅ and G − S5 is a.a.s. 6connected. Hence G−S5 is a.a.s. (globally) rigid by Theorem 8. Since adding a new vertex joined by at least two new edges to a rigid graph preserves rigidity, it follows that G is a.a.s. rigid. (b) This follows in similar way to (a), using the facts that S2 = ∅ and that adding a new vertex joined by at least three new edges to a globally rigid graph preserves global rigidity.
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The bounds on p given in Theorem 13 are best possible since if G ∈ G(n, p) and p = (log n + k log log n + c)/n for any constant c, then G does not a.a.s. have minimum degree at least k, see [2].
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Let Geom(n, r) denote the probability space of all graphs on n vertices in which the vertices are distributed uniformly at random in the unit square and each pair of vertices of distance at most r are joined by an edge. Suppose G ∈ Geom(n, r). Li, Wan and Wang [8] have shown that if nπr2 = log n + (2k − 3) log log n + w(n) for k ≥ 2 a fixed integer and limn→∞ w(n) = ∞, then G is a.a.s. k-connected. As noted by Eren et al. [4], this result can be combined with Theorem 8 to deduce that if nπr2 = log n+9 log log n+w(n) then G is a.a.s. globally rigid. On the other hand, it is also shown in [8] that if nπr2 = log n+(k −1) log log n+c for any constant c, then G is not a.a.s. k-connected. It is still conceivable, however, that if nπr2 = log n+log log n+w(n) then G is a.a.s. rigid, and that if nπr2 = log n+2 log log n+w(n) then G is a.a.s. globally rigid.
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References ´s, Random graphs, in Surveys in Combinatorics 1999 (N.V. Temperley, ed.), [1] B. Bolloba LMS Lecture Note Series 52, Cambridge University Press, Cambridge, UK, 1981, 80–102. Introduction
´s, Random graphs Academic Press, New York, 1985. [2] B. Bolloba
Type of Rigidity Vertex transitive graphs Example Example Highly connected graphs
[3] C. Cooper, A. Frieze and B. Reed, Random regular graphs of non-constant degree, SIAM J. Algebraic Discrete Methods 3 (1982), no. 1, 91–98. [4] T. Eren, D.K. Goldenberg, W. Whiteley and Y.R. Yang, Rigidity, computation, and randomization in network localization, preprint.
Random graphs
´n, Connected rigidity matroids and unique realizations of [5] B. Jackson and T. Jorda graphs, J. Combinatorial Th., Series B, to appear.
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[6] M. Krivelevitch, B. Sudakov, V.H. Vu and N.C. Wormald, Random regular graphs of high degree, SIAM J. Algebraic Discrete Methods 3 (1982), no. 1, 91–98.
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[7] G.Laman On Graphs and rigidity of plane skeletal structures, J. Engrg. Math. 4 (1970), 331–340. [8] X-Y. Li, P-J. Wan and Y. Wang, Fault tolerance deployment and topology control in wireless networks, in Proceedings of the ACM Symposium on Mobile Ad Hoc Networking and Computing (Mobihoc) (2003), Annapolis, MD, June 2003. ´sz and Y. Yemini, On generic rigidity in the plane, SIAM J. Algebraic Discrete [9] L. Lova Methods 3 (1982), no. 1, 91–98. [10] T. Luczak, Size and connectivity of the k-core of a random graph, Discrete Math. 91 (1991), 61–68. [11] J.Graver, B. Servatius and H.Servatius Combinatorial rigidity, American Mathematical Society (1993).
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[12] N.C. Wormald, The asymptotic connectivity of labelled regular graphs, J. Combinatorial Th., Series B, (1981), 156–167. [13] N.C. Wormald, Models of random regular graphs, in Surveys in Combinatorics 1999 (J.D. Lamb and D.A. Preece, eds.), (1999), 239–298.
Introduction Type of Rigidity Vertex transitive graphs Example Example Highly connected graphs Random graphs Open problems
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