xej if v=vijand 1≤j≤k−1,
−∑ki=1−1xej if v=vik,
0 otherwise.
Note that if|e|=1, then Re=0∈R|V|. Let av∈R|V| denote the unit vector of coordinate v (av(v) =1 and av(u) =0 for u∈V−v) and let a∈R|V|the all-one vector (a(u) =1 for all u∈V ). Observe that∑v∈VRe(v) =0, that is, Reis orthogonal to a.
Claim 11 Let ⊆ . The set is independent in if and only if the vectors Re, e∈ are linearly independent.
PROOF: Suppose that the vectors Re, e∈ are linearly independent. Let /06=X ⊆V . The vectors av,v∈V−X and a are orthogonal to Re,e∈ , thus dim({Re∈R|V|: e∈ ,e⊆X})≤ |X| −1. By the independence of vectors Rewe get i (X) =|{e∈ : e⊆X}|=dim({Re∈R|V|: e∈ ,e⊆X}). Thus i (X)≤ |X| −1.
Suppose that set ⊆ is independent in the hypergraphic matroid . By Theorem 10 there exists a function f : → V2
such that f(e)⊆e for every e∈ and{f(e): e∈ }is a forest. It is easy to see that if f(e) ={u,v}holds then we can assign values to variables xej so that Re(u) =1,Re(v) =−1 and Re(w) =0 holds for all w∈V− {u,v}. With these values let M denote the| | × |V|matrix formed by the vectors Re,e∈ as rows. This matrix M is the oriented incidence matrix of the forest{f(e): e∈F}. This implies that the rows of M are linearly independent. 2
Corollary 12 The vectors Re,e∈ represent .
By using a lemma of Schwartz [14] it can be verified that there is an integer N of the order 2O(|V|)such that if we assign random integers from the interval[0,N]to the variables of the above generic representation then we obtain a matrix whose matroid will be isomorphic to with probability at least 1/2. Based on this fact, and using the matroid matching algorithm of Lov´asz [8], we obtain an efficient randomized algorithm for problem (i).
5 An algorithm for the anchor minimization problem
First we show how to solve problem (ii). Let H= (V,E)be a graph. For some X⊆V let N(X)denote the set of neighbours of X and let S(X) =X∪N(X). We say that X⊂V is tight if|N(X)|=2 and S(X)6=V . The following lemmas are easy to prove.
Lemma 13 Let H= (V,E)be 2-connected and let X,Y⊂V be distinct minimal tight sets in G. Then X∩Y=/0. Furthermore, if N(X)∩Y 6=/0 then|X|=|Y|=1.
Lemma 14 Let H= (V,E)be 2-connected and let P⊆V . Then H+K(P)is 3-connected if and only if P∩X 6=/0 for all minimal tight sets X of H.
It follows from Lemmas 13 and 14 that every inclusionwise minimal subset P for which H+K(P)is 3-connected is an optimal solution for problem (ii). This gives rise to simple polynomial algorithm for (ii).
Finally we give a sketch of the polynomial 2-approximation algorithm for the anchor minimization problem. First we check whether there exists a set P⊂V in the input graph G= (V,E)for which|P|=3 and G+K(P)is M-connected. If there is no such set, we apply the matroid matching based algorithm to find a smallest set P for which G+K(P)is M-connected.
Note that H=G+K(P)is 2-connected. Then we find a smallest set P0for which H+K(P0)is 3-connected. It is easy to see that P∪P0will be a feasible solution whose size is not more than twice the size of an optimal solution.
148 ZSOLTFEKETE, TIBORJORDAN´
References
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[2] T. EREN, D.K. GOLDENBERG, W. WHITELEY, Y.R. YANG, A.S. MORSE, B.D.O. ANDERSON, AND P.N. BEL
-HUMEUR, Rigidity, Computation, and Randomization in Network Localization, Proceedings of the International Annual Joint Conference of the IEEE Computer and Communications Societies (INFOCOM), March 2004.
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[14] J.T. SCHWARTZ, Fast probabilistic algorithms for verification of polynomial identities, J. ACM 27, 701-717, 1980.
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Servatius eds., Seattle, WA, 1995), Contemp. Math., 197, Amer. Math. Soc., Providence, RI, 1996, 171–311.
On the Rank Function of the 3-Dimensional Rigidity Matroid
BILLJACKSON∗
School of Mathematical Sciences Queen Mary, University of London Mile End Road, London E1 4NS, England
B.Jackson@qmul.ac.uk
TIBORJORDAN´ †
Department of Operations Research E¨otv¨os University
P´azm´any s´et´any 1/C, 1117 Budapest, Hungary jordan@cs.elte.hu
Abstract: A major open problem in combinatorial rigidity is to find a good characterization for independence in the 3-dimensional rigidity matroid, or more generally, to give a min-max formula for the rank function. We give a new upper bound on the rank and conjecture that our bound is tight.
Keywords: rigid graphs, rigidity matroid, rank function
1 Introduction
A framework(G,p)in d-space is a graph G= (V,E)and a map p : V →Rd. The rigidity matrix of the framework is the matrix R(G,p)of size|E|×d|V|, where, for each edge vivj∈E, in the row corresponding to vivj, the entries in the d columns corresponding to vertices i and j contain the d coordinates of(p(vi)−p(vj))and (p(vj)−p(vi)), respectively, and the remaining entries are zeros. See [14] for more details. The rigidity matrix of(G,p)defines the rigidity matroid of(G,p)on the ground set E by linear independence of rows of the rigidity matrix. A framework(G,p)is generic if the set of coordinates of the points p(v), v∈V , is algebraically independent over the rationals. Any two generic frameworks(G,p)and(G,p0)have the same rigidity matroid. We call this the d-dimensional rigidity matroid d(G) = (E,rd)of the graph G. We denote the rank of d(G)by rd(G).
Lemma 1 [14, Lemma 11.1.3] Let(G,p)be a framework inRd. Then rank R(G,p)≤S(n,d), where n=|V(G)|and S(n,d) =
nd− d+12
if n≥d+2
n 2
if n≤d+1.
We say that a graph G= (V,E)is rigid in Rdif rd(G) =S(n,d). (This definition is motivated by the fact that if G is rigid and(G,p)is a generic framework on G, then every continuous deformation of(G,p)which preserves the edge lengths
||p(u)−p(v)||for all uv∈E, must preserve the distances||p(w)−p(x)||for all w,x∈V , see [14].) We say that G is M-independent inRd if E is independent in d(G). For X⊆V , let EG(X)denote the set, and iG(X)the number, of edges in G[X], that is, in the subgraph induced by X in G. We use E(X)or i(X)when the graph G is clear from the context. A cover of G is a collection of pairwise incomparable subsets of V , each of size at least two, such that∪X∈ E(X) =E.
Lemma 1 implies the following necessary condition for G to be M-independent.
Lemma 2 If G= (V,E)is M-independent inRdthen i(X)≤S(|X|,d)for all X⊆V . It also gives the following upper bound on the rank function.
Lemma 3 If G= (V,E)is a graph then
rd(G)≤min
∑
X∈
S(|X|,d) where the minimum is taken over all covers
of G.
The converse of Lemma 2 also holds for d=1,2. The case d=1 follows from the fact that the 1-dimensional rigidity matroid of G is the same as the cycle matroid of G, see [3, Theorem 2.1.1]. The case d=2 is a result of Laman [8].
Similarly, the inequality given in Lemma 3 holds with equality when d=1,2. The case d=2 is a result of Lov´asz and Yemini [9]. Neither of these statements hold for d≥3. Indeed, it remains an open problem to find good characterizations for independence or, more generally, the rank function in the d-dimensional rigidity matroid of a graph when d≥3.
∗Supported by the Royal Society/Hungarian Academy of Sciences Exchange Programme and the Finite Structures project of the R´enyi Institute of Mathematics, Budapest within the 6th Framework Programme of the EU.
†Supported by the MTA-ELTE Egerv´ary Research Group on Combinatorial Optimization and the Hungarian Scientific Research Fund grant no. T037547, T049671.
150 BILLJACKSON, TIBORJORDAN´
In 1983, Dress, Drieding and Haegi [2, equation (39)], [13, Conjecture 3] conjectured that 2-thin covers could be used to determine the rank function of (G): if G= (V,E)is a graph and E0⊆E then the rank r(E0)is equal to
min{val( )}, (1)
where the minimum is taken over all 2-thin covers of G[E0]. This conjecture, which would have provided a good charac-terization for the rank function of (G), was recently disproved in [6].
At a conference on rigidity held in Montreal in 1987, Dress conjectured that equality is obtained in (1) for the special 2-thin cover defined as follows. For u,v∈V , the edge uv is an implied edge of G if uv6∈E and r(E+uv) =r(E). The closure
b
G of G is the graph obtained by adding all the implied edges to G. A rigid cluster of G is a set of vertices which induce a maximal complete subgraph ofG. It is not difficult to see that any two rigid clusters of G intersect in at most two vertices.b Thus the set of rigid clusters of G is a 2-thin cover of G.
Conjecture 4 (see [1],[3, Conjecture 5.6.1], and [11, Conjecture 2.3]) Let G= (V,E)be a graph and be the set of rigid clusters of G. Then
r(E) =val( ). (2)
This conjecture is still open. Note however, that even if Conjecture 4 was shown to be true, it would not provide a good characterization for the rank function.
It is conceivable that Conjecture 4 is true because of the special intersection properties of rigid clusters. If so, then it may be possible to resurrect the first conjecture of Dress et al. by only considering 2-thin covers whose intersection properties reflect those of rigid clusters. Note that for graphs of bounded maximum degree the rank function has been determined in [7].
We say that a 2-thin cover
of a graph G= (V,E) is independent if the subgraphs of (V,H( ))induced by the sets Xi∈ are M-independent. The cover is closed if(V,H( ))is a subgraph of G. The following lemma shows that independent 2-thin covers of G can be used to give an upper bound on r(G)(c.f. [6, Lemma 3.4]). (The hypothesis that the cover is closed is not crucial since an independent 2-thin cover of G is an independent closed 2-thin cover of a supergraph of G.)
Definition 6 Aniterated2-thin coverof Gof depthm is a rooted tree of depth m whose nodes are induced subgraphs of G and is such that
(i) the root of is G,
(ii) each leaf of is at distance m from the root,
(iii) for each node W of which is not a leaf, the vertex sets of the children of W is an independent closed 2-thin cover W of W .