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Infinitesimal rigidity of panel-hinge frameworks Shin-ichi Tanigawa (joint work with Naoki Katoh) Kyoto University.

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Presentation on theme: "Infinitesimal rigidity of panel-hinge frameworks Shin-ichi Tanigawa (joint work with Naoki Katoh) Kyoto University."— Presentation transcript:

1 Infinitesimal rigidity of panel-hinge frameworks Shin-ichi Tanigawa (joint work with Naoki Katoh) Kyoto University

2 Theorem  A graph can be realized as an infinitesimally rigid body-hinge framework in R 3 if and only if it can be realized as a panel-hinge framework in R 3  Originally, posed by Tay and Whiteley in 1984, and called the Molecular conjecture.  Implying a combinatorial characterization of generic rigidity of molecular frameworks [Tay&Whiteley84, Whiteley99,04, Jackson&Jordán08]

3 3-dimensional body-hinge framework (G,p):  G = (V,E): a graph,  p is a mapping, (called a hinge-configuration), e ∊ E ↦ a line p( e ) in R 3 a vertex ⇔ a 3-d body an edge ⇔ a hinge (=a line)

4 Tay-Whiteley’s theorem (Whiteley 88, Tay 89, 91)  For a generic hinge-configuration p, (G,p) is rigid in R 3 if and only if 5G contains six edge-disjoint spanning trees. G 5G

5 Tay-Whiteley’s theorem (Whiteley 88, Tay 89, 91)  For a generic hinge-configuration p, (G,p) is rigid in R 3 if and only if 5G contains six edge-disjoint spanning trees. It cannot be applied to “special” hinge configurations

6  Infinitesimal motion of a rigid body a combination of six independent isometric motions  three translations + three rotations ⇒ The set of all infinitesimal motions forms a 6-dimensional vector space.  each rotation (transformation) can be coordinatized by a so-called 2- extensor (Plűcker coordinate) of the axis-line.

7  Hinge constraints [Crapo and Whiteley 82] Two bodies 1 and 2 are connected by a hinge A.  S i : a 6-vector assigned to the body i, (representing an infinitesimal motion).  S A : a 2-extensor (that is, a 6-vector) of the line A. The constraint by the hinge A: S 1 – S 2 = t S A for some t ∈ R ⇔ r i ・ ( S 1 – S 2 ) = 0, i=1,...,5  {r 1,r 2,...,r 5 }: a basis of the orthogonal complement of the 1-dimensional vector space spanned by S A.

8 Rigidity Matrix Def. infinitesimal motion m: V → R 6 s.t. r i (p(e)) ・ (m(u) - m(v)) = 0 i=1,...,5, for each e=(u,v) ∊ E {r 1 (p(e)),...,r 5 (p(e))} is a basis of the orthogonal complement of the vector space spanned by a 2-extensor of p(e). Rigidity matrix R(G,p): 5|E|×6|V|-matrix columns for u 5 rows for e=uv m(u)m(u) m(v)m(v) =0 … … … 6 columns for v (row space represents the orthogonal complement of a 2-extensor of p(e).)

9 Def. (G,p) is infinitesimal rigid ⇔ rank R(G,p)=6|V|-6  dim ker R(G,p) ≧ 6 the dimension of the space of all trivial motions is 6  three transformations + three rotations Def. G can be realized as an inf. rigid framework ⇔ ∃ p s.t. (G,p) is infinitesimal rigid columns for u 5 rows for e=uv m(u)m(u) m(v)m(v) =0 … … … 6 columns for v

10 [Jackson and Jordán09]  deficiency of G, def(G):= max P 6(| P |-1) - 5d( P ) P : partition of V d( P ): the number edges connecting different components of P  def(G) ≧ 0 (since P ={V})  There exist 6 edge-disjoint spanning trees in 5G ⇔ def(G)=0 (by Tutte’s tree packing theorem) Theorem(Jackson and Jordán09) The followings are equivalent for G=(V,E):  For a generic hinge-configuration p , (G,p) has k d.o.f. (i.e., rank R(G,p)=6(|V|-1)-k)  def(G)=k  The rank of matroid G 6 (5G) is equal to 6(|V|-1)-k

11 Panel-hinge frameworks 3-dimentional panel-hinge framework: (G,p)  G=(V,E) : a graph  p : hinge-configuration satisfying “hinge-coplanarity condition” (i.e., all hinges incident to a body lie on a common hyperplane.) vertex ⇔ panel = a hyperplane in R 3 edge ⇔ hinge Rem. “hinge-coplanarity” is a special geometric relation among hinges; Tay-Whiteley’s characterization may be false...

12 Theorem  G can be realized as an infinitesimally rigid panel-hinge framework if and only if 5G contains six-edge-disjoint spanning trees  Def. G is a k-graph if def(G)=k ; it is a minimal k-graph if def(G)=k and removal of any edge results in a non k-graph. Theorem: If G is a minimal k-graph, then it can be realized as a panel-hinge framework with k degree of freedom. (Outline)  combinatorial part : propose an inductive construction of minimal k- graphs w.r.t. # vertices.  algebraic part : provide an explicit construction of a k-dof panel- hinge framework (G,p) following the inductive construction given in combinatorial part.

13 Rigid subgraphs (combinatorial part) a subgraph G’=(V’,E’) is called rigid if def(G’)=0 a rigid subgraph is called proper if 1 < |V’| < |V|. Lemma : For a minimal k-graph G, the graph obtained by contracting a proper rigid subgraph is a minimal k-graph.

14 Splitting-off operation A splitting-off at a vertex of degree two: Lemma : Let G be a k-graph with a vertex v of degree 2. Then the graph G v ab obtained by splitting off at v is a k-graph or a (k-1)- graph. Even though G is minimal, G v ab may not be minimal. v ab ab v a b a b

15 Splitting-off operation Lemma : Let G = (V,E) be a minimal k-dof-graph which has no proper rigid subgraph.  (i) If k=0, then G v ab is a minimal 0-graph.  (ii) If k>0, then G v ab is a minimal (k-1)-graph.

16 Small degree vertices Lemma : Let G = (V,E) be a minimal k-graph which has no proper rigid subgraph. Then, G has two consecutive vertices of degree two. (sketch of the existence of a degree two vertex)  If no proper rigid subgraph exists, there is a basis of G 6 (5G) that contains 5(E-e) for any edge e  ⇒ 5|E-e| ≦ 6(|V|-1) ⇒ 5|E| ≦ 6|V|  ⇒ (average degree) = 2|E|/|V| ≦ 2.4  ⇒ there exists a vertex of degree two

17 Summary of combinatorial part For a minimal k-graph G =( V, E )  if G contains a proper rigid subgraph G ’ G/G ’ is a minimal k-graph  otherwise, G contains a vertex v of degree two if k>0, G v ab is a minimal (k-1)-graph if k=0, G v ab is a minimal 0-graph v ab ab

18 Main theorem (algebraic part) Theorem : Let G=(V,E) be a minimal k-graph. Then G=(V,E) can be realized as a panel-hinge framework (G,p) with rank R(G,p)=6(|V|-1)-k. By induction on |V|  |V|=2  |V| > 2 Case 1 : G has a proper rigid subgraph G’ Case 2 : G has no proper rigid subgraph with k>0 Case 3: G has no proper rigid subgraph with k=0

19 Case 1: G has a proper rigid subgarph G’=(V’,E’) G/G’ is a minimal k-graph (by Lemma) (Also, G’ is minimal 0-graph.) By induction, we have panel-hinge realizations (G’,p 1 ) and (G/G’,p 2 ) s.t. (G’,p 1 ) is rigid and (G/G’,p 2 ) has k-d.o.f. Let v* be the vertex obtained by the contraction. Idea:  We can consider the rigid framework (G’,p 1 ) as a rigid body. Hence, replacing the panel associated with v* in (G/G’,p 2 ) by a rigid body (G’,p 1 ), the resulting framework has the desired property. GG/G’ v* G’

20 Case 2: G has no proper rigid subgraph and k>0 G has a vertex v of degree two. For k>0, G v ab is a minimal (k-1)-dof-graph (G,p 1 ) (G v ab,q) (k-1)-d.o.f. column operations R(G,p 1 ) R(G v ab,q) * 0 * G v ab ((k-1)-graph)

21 R(G,p 1 ) = va vb v ab R(p 1 (va)) -R(p 1 (va)) R(p 1 (vb))-R(p 1 (vb)) R G, p 1 [E-va-vb, V-{v}] column operations va vb v ab R(p 1 (va)) R(p 1 (vb))-R(p 1 (vb)) R G, p 1 [E-va-vb,, V-{v}] 0 R(p 1 (vb)) va vb ab R(p 1 (va)) R(q(ab))-R(q(ab)) R G v ab, q [E v,, V-{v}] 0 R(q(ab)) v = va vb V-{v} R(p 1 (va)) 0 R(q(ab)) 0 v R(G v ab,q) = rank R(G,p 1 ) ≥ rank R(p 1 (va)) + rank R(G v ab,q) = 5+6(|V-{v}|-1)-(k-1) = 6(|V|-1)-k (G,p 1 ) (G v ab,q)

22 Case 3: G has no proper rigid subgraph (k=0) G has two vertices v and a of degree two which are adjacent to each other G v ab is a minimal 0-graph (G v ab,q) (G,p 2 ) (G,p 3 ) (G,p 1 ) isomorphism *** Show that at least one of (G,p 1 ), (G,p 2 ), and (G,p 3 ) is rigid *** (G a vc,q’) ’ ’ identical

23 R(G,p 1 )= va vb v V-{v} R(p 1 (va)) 0 R(p 1 (vb)) 0 R(G v ab,q) rank R(G,p 1 ) ≥ rank R(p 1 (va)) + rank R(G v ab,q) = 5+6(|V-{v}|-1)=6(|V|-1)-1 We need to show rank R(G,p 1 )≥ 6(|V|-1) (G,p 1 )

24 Claim: there exists a redundant row in R(G v ab,q) among those associated with ab. (Sketch)  From a combinatorial argument, there exists a redundant edge among 5 ab in the combinatorial matroid G 6 (5G v ab )  This redundancy also happen in the rigidity matrix by induction At most 4 edges are used among 5ab v abba

25 R(G,p 1 )= R(q(ab)) va vb V-{v} R(p 1 (va)) 0 0 v R(G v ab,q) row operations a linear comb. of rows of R(q(ab)), denoted r 1. Note: r 1 is nonzero and is determined by R(G v ab,q) va (vb) 1* v V-{v} R(p 1 (va)) 0 0 R(G v ab,q) - (the row (ab) 1 ) * r1r1 redundant row, say (ab) 1 If is non- singular, we are done!! Analysis of R(G,p 1 ) q(ab) in (G v ab, q) = p 1 (vb) in (G,p 1 )

26 R(G,p 2 )= vb va V-{v} R(p 2 (vb)) 0 R(q(ab)) 0 v R(G v ab,q) each row of ab in R(G v ab, q) ⇔ each row of va in R(G,p 2 ) row operations a linear comb. of rows of R(q(ab )), which is equal to r 1. vb (va) 1* v V-{v} R(p 2 (vb)) 0 0 R(G v ab,q) - (the row (ab) 1 ) * r1r1 the redundant row (ab) 1 If is non- singular, we are done!! Analysis of R(G,p 2 )

27 R(G,p 3 )= q(ab) in R(G v ab, q) = p 3 (vb) in R(G,p 2 ) q(ac) in R(G v ab, q) = p 3 (va) in R(G,p 2 ) Analyzing R(G,p 3 ) vb ac va V-{v,a,b,c} p 3 (ac) 0 q(ac) 0 av 0 cb *** -q(ac) 0 q(ab)-q(ab)) -p 3 (ac) vb ac va V-{a} R(p 3 (ac)) 0 R(q(ac)) a R(G v ab,q) 0 0 add columns of a to those of c the redundant row (ab) 1

28 ac va V-{a} R(p 3 (ac)) 0 R(q(ac)) a R(G v ab,q) R(G,p 2 )= q(ab) in R(G v ab, q) = p 3 (vb) in R(G,p 2 ) q(ac) in R(G v ab, q) = p 3 (va) in R(G,p 2 ) row operations a linear comb. of rows of R(q(ac)), denoted by r 3. ac (vb) 1* v V-{v} R(p 3 (ac)) 0 0 R(G v ab,q) - (the row (ab) 1 ) * r3r3 the redundant row (ab) 1 Analyzing R(G,p 3 ) vb 0 0

29 Claim : r 1 + r 3 =0 (intuition):  r 1 can be considered as a force applied to the panel Π(a) in (G v ab, q) through the hinge q(ab)  r 3 can be considered as a force applied to the panel Π(a) in (G v ab,q) through the hinge q(ac)  Since the panel Π(a) is incident to only q(ab) and q(ac), these two forces must be in sign-inverse. R(G v ab,q) = ab ac a bc R(q(ab)) -R(q(ab)) R(q(ac))-R(q(ac)) ***

30 R(G,p 3 )= row operations ac (vb) 1* v V-{v} R(p 3 (ac)) 0 0 R(G v ab,q) - (the row (ab) 1 ) * r 3 =-r 1 the redundant row (ab) 1 Analyzing R(G,p 3 ) vb ac va V-{a} R(p 3 (ac)) 0 R(q(ac)) a R(G v ab,q) 0 0

31 If at least one of is non-singular, we are done. v V-{v} R(p 3 (ac)) 0 0 R(G v ab,q) - (the row (ab) 1 ) * r 3 =-r 1 Summary of matrix-transformations v V-{v} R(p 1 (va)) 0 0 R(G v ab,q) - (the row (ab) 1 ) * r1r1 v V-{v} R(p 2 (vb)) 0 0 R(G v ab,q) - (the row (ab) 1 ) * r1r1

32 Suppose is singular. Then, r 1 is orthogonal to a 2-exntesor of p 1 (va). Suppose is singular for any choice of p 1 (va) on Π(a). Then, r 1 is orthogonal to every 2-extensors of a line on Π(a). (Similar for the other two matrices.) Suppose all of are singular for every choice of p 1 (va), p 2 (vb), p 3 (ac). Then, r 1 is orthogonal to 2-extensors of all lines on Π(a), Π(b), or Π(c). ⇒ These 2-extensors span 6-dimentional space. ⇒ r 1 =0, a contradiction. R(p 3 (ac)) r 3 =-r 1 Last step R(p 1 (va)) r1r1 R(p 2 (vb)) r1r1

33 Unsolved Problems Corollary: Let G 2 be the square of a graph G. Then, the rank of G 2 in generic 3-rigidity matroid is 3|V|-6-def(G) (by Jackson&Jordán 07, 08) Conjecture (Jacobs, Jackson&Jordán07) : Let G be a graph and let u,v ∈ V. Then, r(G 2 +uv) = r(G 2 ) if and only if u and v belong to the same rigid component of G 2 (where r is the rank function of 3-rigidity matroid).  A rigid component is an inclusionwise-maximal rigid subgraph. Conjecture : Let (G, p) be a panel-hinge framework. Suppose two panels Π(u) and Π(v) are relatively flexible. If connecting between these panels by a bar lying on the intersection of them, then the degree of freedom always decreases. Problem: Efficient computation of the decomposition into redundantly rigid components Open Problem: Provide a simpler proof!!


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