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Infinitesimal rigidity of panel-hinge frameworks

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

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Infinitesimal rigidity of panel-hinge frameworks

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

  2. Theorem • A graph can be realized as an infinitesimally rigid body-hinge framework in R3 if and only if it can be realized as a panel-hinge framework in R3 • 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, • pis a mapping, (called a hinge-configuration), • e ∊ E↦ a line p(e) inR3 • 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 R3if 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 R3if 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 motionsforms 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 hingeA. • Si: a 6-vector assigned to the bodyi, (representing an infinitesimal motion). • SA: a 2-extensor (that is, a 6-vector) of the lineA. • The constraint by the hingeA: S1 – S 2= t SAfor somet ∈R • ⇔ ri・( S1 – S2 ) = 0, i=1,...,5 • {r1,r2,...,r5}: a basis of the orthogonal complement of the 1-dimensional vector space spanned bySA.

  8. Rigidity Matrix • Def.infinitesimal motionm: V→R6s.t. ri(p(e))・(m(u)- m(v)) = 0 i=1,...,5 , for each e=(u,v) ∊ E • {r1(p(e)),...,r5(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 6 columns for u 6 columns for v … m(u) 5 rows for e=uv =0 0 0 0 … m(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 kerR(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 ⇔　∃ps.t. (G,p) is infinitesimal rigid 6 columns for u 6 columns for v … m(u) 5 rows for e=uv =0 0 0 0 … m(v) …

  10. [Jackson and Jordán09] • deficiency of G, def(G):= maxP 6(|P|-1) - 5d(P ) • P :partition of V • d(P): the number edges connecting different components ofP • def(G)≧0 (since P ={V}) • There exist 6 edge-disjoint spanning trees in5G ⇔ 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-configurationp，(G,p) has k d.o.f. (i.e., rank R(G,p)=6(|V|-1)-k) • def(G)=k • The rank of matroidG6(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 R3 • 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 rigidif def(G’)=0 • a rigid subgraph is called properif1＜|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 Gvab obtained by splitting off at v is a k-graph or a (k-1)-graph. • Even though G is minimal, Gvab may not be minimal. v a b a b v a b b a

  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 Gvab is a minimal 0-graph. • (ii) If k>0, then Gvab 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 G6(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, Gvab is a minimal (k-1)-graph • if k=0, Gvab is a minimal 0-graph v a b a b

  18. Maintheorem (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 subgarphG’=(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’,p1) and (G/G’,p2) s.t. (G’,p1) is rigid and (G/G’,p2) has k-d.o.f. • Let v* be the vertex obtained by the contraction. • Idea: • We can consider the rigid framework (G’,p1) as a rigid body. Hence, replacing the panel associated with v*in (G/G’,p2) by a rigid body (G’,p1), the resulting framework has the desired property. v* G’ G G/G’

  20. Case 2: G has no proper rigid subgraph and k>0 • G has a vertex v of degree two. • For k>0, Gvab is a minimal (k-1)-dof-graph (Gvab,q) (k-1)-d.o.f. (G,p1) 0 * Gvab ((k-1)-graph) column R(G,p1) R(Gvab,q) * operations

  21. v a b R(p1(va)) v 0 0 0 va a b R(G,p1) = vb -R(p1(va)) R(p1(va)) 0 0 va R(q(ab)) R(q(ab)) -R(q(ab)) 0 vb R(p1(vb)) -R(p1(vb)) 0 0 RGvab,q[Ev,, V-{v}] 0 RG,p1[E-va-vb, V-{v}] 0 column operations v V-{v} (G,p1) va R(p1(va)) 0 v vb a b R(q(ab)) R(Gvab,q) R(p1(va)) 0 0 0 va 0 vb R(p1(vb)) R(p1(vb)) -R(p1(vb)) 0 = = RG,p1[E-va-vb,, V-{v}] 0 (Gvab,q) rank R(G,p1) ≥ rank R(p1(va)) + rank R(Gvab,q) = 5+6(|V-{v}|-1)-(k-1) = 6(|V|-1)-k

  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 • Gvabis a minimal 0-graph (G,p1) (Gvab,q) isomorphism identical (G,p2) ’ ’ (Gavc,q’) (G,p3) *** Show that at least one of (G,p1), (G,p2), and (G,p3) is rigid ***

  23. V-{v} v va R(p1(va)) 0 vb R(G,p1)= R(p1(vb)) R(Gvab,q) 0 (G,p1) R(Gvab,q) rank R(G,p1) ≥ rank R(p1(va)) + rank = 5+6(|V-{v}|-1)=6(|V|-1)-1 We need to show rank R(G,p1)≥6(|V|-1)

  24. Claim: there exists a redundant row in R(Gvab,q)among those associated withab. • (Sketch) • From a combinatorial argument, there exists a redundant edge among 5ab in the combinatorial matroid G6(5Gvab) • This redundancy also happen in the rigidity matrix by induction At most 4 edges are used among 5ab v a b a b

  25. Analysis of R(G,p1) v V-{v} va R(p1(va)) 0 q(ab) in (Gvab, q) = p1(vb) in (G,p1) (vb)1* 0 R(Gvab,q) - (the row (ab)1) * v V-{v} redundant row, say (ab)1 va R(p1(va)) 0 R(G,p1)= vb R(q(ab)) R(Gvab,q) 0 row operations a linear comb. of rows of R(q(ab)), denoted r1. r1 If is non-singular, we are done!! Note: r1 is nonzero and is determined by R(Gvab,q)

  26. Analysis of R(G,p2) v V-{v} vb R(p2(vb)) 0 each row of ab in R(Gvab, q) 　⇔ each row of va in R(G,p2) (va)1* 0 R(Gvab,q) - (the row (ab)1) * v V-{v} the redundant row(ab)1 R(G,p2)= vb R(p2(vb)) 0 va R(q(ab)) R(Gvab,q) 0 row operations a linear comb. of rows of R(q(ab)), which is equal to r1. r1 If is non-singular, we are done!!

  27. Analyzing R(G,p3) q(ab) in R(Gvab, q) = p3(vb) in R(G,p2) q(ac) in R(Gvab, q) = p3(va) in R(G,p2) c b v V-{v,a,b,c} a a V-{a} 0 -p3(ac) ac ac R(p3(ac)) p3(ac) 0 0 q(ab) -q(ab)) 0 vb R(G,p3)= 0 0 -q(ac) 0 0 R(Gvab,q) va va R(q(ac)) q(ac) *** 0 add columns ofa to those of c 0 the redundant row(ab)1 vb 0

  28. Analyzing R(G,p3) v V-{v} ac R(p3(ac)) 0 (vb)1* q(ab) in R(Gvab, q) = p3(vb) in R(G,p2) q(ac) in R(Gvab, q) = p3(va) in R(G,p2) 0 R(Gvab,q) - (the row (ab)1) * a V-{a} the redundant row(ab)1 0 R(p3(ac)) ac 0 R(G,p2)= vb R(Gvab,q) va R(q(ac)) 0 a linear comb. of rows of R(q(ac)), denoted by r3. row operations r3

  29. a b c • Claim: r1 + r3=0 • (intuition): • r1 can be considered as a force applied to the panel Π(a) in (Gvab, q) through the hinge q(ab) • r3 can be considered as a force applied to the panel Π(a) in (Gvab,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(q(ab)) R(q(ab)) 0 0 ab ac R(q(ac)) -R(q(ac)) 0 0 0 R(Gvab,q) = ***

  30. v v v Summary of matrix-transformations V-{v} V-{v} V-{v} R(p2(vb)) R(p1(va)) R(p3(ac)) 0 0 0 • If at least one of is non-singular, we are done. 0 0 0 R(Gvab,q) - (the row (ab)1) R(Gvab,q) - (the row (ab)1) R(Gvab,q) - (the row (ab)1) * * * r1 r1 r3=-r1

  31. Last step • Suppose is singular. Then, r1 is orthogonal to a 2-exntesor of p1(va). • Suppose is singular for any choice of p1(va) on Π(a). Then, r1 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 p1(va), p2(vb), p3(ac). Then, r1 is orthogonal to 2-extensors of all lines on Π(a), Π(b), or Π(c). ⇒ These 2-extensors span 6-dimentional space. • ⇒r1=0, a contradiction. R(p1(va)) R(p3(ac)) R(p2(vb)) r1 r3=-r1 r1

  32. Unsolved Problems • Corollary: Let G2 be the square of a graph G. Then, the rank of G2 in generic 3-rigidity matroid is 3|V|-6-def(G) (by Jackson&Jordán 07, 08) • Conjecture (Jacobs, Jackson&Jordán07) : Let G be a graphand let u,v∈V. Then, r(G2+uv) = r(G2) if and only if u and v belong to the same rigid component of G2 (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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