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 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-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 R3

a vertex 　⇔　 a 3-d body an edge 　⇔　 a hinge (=a line)

Tay-Whiteley’s theorem (Whiteley 88, Tay 89, 91) For a generic hinge-configuration p, (G,p) is rigid in R3

if and only if 5G contains six edge-disjoint spanning trees.

G 5G

Tay-Whiteley’s theorem (Whiteley 88, Tay 89, 91) For a generic hinge-configuration p, (G,p) is rigid in R3

if and only if 5G contains six edge-disjoint spanning trees.

It cannot be applied to “special” hinge configurations

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.

Hinge constraints [Crapo and Whiteley 82] Two bodies 1 and 2 are connected by a hinge A.

Si : a 6-vector assigned to the body i, (representing an infinitesimal motion).

SA: a 2-extensor (that is, a 6-vector) of the line A.

The constraint by the hinge A: S1 – S 2= t SA for some t ∈ R

⇔ ri ・ ( S1 – S2 ) = 0, i=1,...,5 {r1,r2,...,r5}: a basis of the orthogonal complement of the 1-dimensional vector

space spanned by SA.

　

Rigidity Matrix Def. infinitesimal motion m: V → R6 s.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

))(( eR p 000

6 columns for u

5 rowsfor e=uv

m(u)

m(v)

=0

……

…

))(( eR p

6 columns for v

))((

))((

:))((

5

1

er

er

eR

p

p

p (row space represents the orthogonal complement of a 2-extensor of p(e).)

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

))(( eR p 000

6 columns for u

5 rowsfor e=uv

m(u)

m(v)

=0

……

…

))(( eR p

6 columns for v

))((

))((

:))((

5

1

er

er

eR

p

p

p

[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 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 G6(5G) is equal to 6(|V|-1)-k

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...

　　 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.

　　

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.

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 Gv

ab 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 a b

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 Gv

ab is a minimal 0-graph.

(ii) If k>0, then Gvab is a minimal (k-1)-graph.

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

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, Gv

ab is a minimal (k-1)-graph

if k=0, Gvab is a minimal 0-graph

v

a b a b

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

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’,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.

G G/G’

v*

G’

Case 2: G has no proper rigid subgraph and k>0

G has a vertex v of degree two. For k>0, Gv

ab is a minimal (k-1)-dof-graph

(G,p1)(Gv

ab,q)(k-1)-d.o.f.

column

operationsR(G,p1) R(Gv

ab,q)

* 0

*Gv

ab

((k-1)-graph)

R(G,p1) =vavb

v a b

R(p1(va)) -R(p1(va)) 0

0

0

0

0R(p1(vb)) -R(p1(vb))

RG, p1

[E-va-vb, V-{v}]

column operations

vavb

v a b

R(p1(va)) 0

0

0

0R(p1(vb)) -R(p1(vb))

RG, p1

[E-va-vb,, V-{v}]

0R(p1(vb))

vavb

a b

R(p1(va)) 0

0

0

0R(q(ab)) -R(q(ab))

RGvab, q[E

v,, V-{v}]

0R(q(ab))

v

=

va

vb

V-{v}

R(p1(va))

0

R(q(ab))

0

v

R(Gvab,q)=

rank R(G,p1) ≥ rank R(p1(va)) + rank R(Gv

ab,q)

= 5+6(|V-{v}|-1)-(k-1) = 6(|V|-1)-k

(G,p1)

(Gvab,q)

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 Gv

ab is a minimal 0-graph

(Gvab,q)

(G,p2)

(G,p3)

(G,p1)

isomorphism

*** Show that at least one of (G,p1), (G,p2), and (G,p3) is rigid ***

(Gavc,q’)

’ ’

identical

R(G,p1)=

va

vb

v V-{v}

R(p1(va))

0

R(p1(vb))

0

R(Gvab,q)

rank R(G,p1) ≥ rank R(p1(va)) + rank R(Gvab,q)

= 5+6(|V-{v}|-1)=6(|V|-1)-1

We need to show rank R(G,p1)≥ 6(|V|-1)

(G,p1)

Claim: there exists a redundant row in R(Gvab,q) among those

associated with ab. (Sketch)

From a combinatorial argument, there exists a redundant edge among 5ab in the combinatorial matroid G6(5Gv

ab)

This redundancy also happen in the rigidity matrix by induction

At most 4 edges are used among 5abv

a b ba

R(G,p1)=R(q(ab))

va

vb

V-{v}

R(p1(va))

0

0

v

R(Gvab,q)

row operations a linear comb. of rows of R(q(ab)), denoted r1.

Note: r1 is nonzero and is determined by R(Gv

ab,q)

va

(vb)1*

v V-{v}

R(p1(va))

0

0

R(Gvab,q)

- (the row (ab)1)*

r1

redundant row, say (ab)1

If is non-singular, we are done!!

Analysis of R(G,p1)

q(ab) in (Gvab, q) = p1(vb) in (G,p1)

R(G,p2)=

vb

va

V-{v}

R(p2(vb))

0

R(q(ab))

0

v

R(Gvab,q)

each row of ab in R(Gvab, q)

　⇔ each row of va in R(G,p2)

row operationsa linear comb. of rows of

R(q(ab)), which is equal to r1.

vb

(va)1*

v V-{v}

R(p2(vb))

0

0

R(Gvab,q)

- (the row (ab)1)*

r1

the redundant row (ab)1

If is non-singular, we are done!!

Analysis of R(G,p2)

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)

Analyzing R(G,p3)

vb

ac

va

V-{v,a,b,c}

p3(ac)

0

q(ac)

0

a v

0

c b

0

0

0

0

***

-q(ac) 0

q(ab) -q(ab))

-p3(ac)

vb

ac

va

V-{a}

R(p3(ac))

0

R(q(ac))

a

R(Gvab,q)

0

0

add columns of a to those of c

the redundant row (ab)1

ac

va

V-{a}

R(p3(ac))

0

R(q(ac))

a

R(Gvab,q)

R(G,p2)=

q(ab) in R(Gvab, q) = p3(vb) in R(G,p2)

q(ac) in R(Gvab, q) = p3(va) in R(G,p2)

row operations a linear comb. of rows of R(q(ac)), denoted by r3.

ac

(vb)1*

v V-{v}

R(p3(ac))

0

0

R(Gvab,q)

- (the row (ab)1)*

r3

the redundant row (ab)1

Analyzing R(G,p3)

vb

0

0

Claim: r1 + r3=0

(intuition): r1 can be considered as a force applied to the panel Π(a) in (Gv

ab, q) through the hinge q(ab)

r3 can be considered as a force applied to the panel Π(a) in (Gv

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(Gvab,q) =

ab

ac

a b c

R(q(ab)) -R(q(ab)) 0

0

0

0

0

R(q(ac)) -R(q(ac))

***

If at least one of is non-singular, we are done.

v V-{v}

R(p3(ac))

0

0

R(Gvab,q)

- (the row (ab)1)*

r3=-r1

Summary of matrix-transformations

v V-{v}

R(p1(va))

0

0

R(Gvab,q)

- (the row (ab)1)*

r1

v V-{v}

R(p2(vb))

0

0

R(Gvab,q)

- (the row (ab)1)*

r1

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(p3(ac))

r3=-r1

Last stepR(p1(va))

r1

R(p2(vb))

r1

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 graph and 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!!