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Characterizing Generic Global Rigidity

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Title: Characterizing Generic Global Rigidity


1
Characterizing Generic Global Rigidity
  • Steven J. Gortler
  • (with A. Healy and D. Thurston)

2
Global Rigidity
  • Given a graph G,
  • Given a framework p, in Rd
  • For some fixed d

R2
3
Global Rigidity
  • p is GR in Rd there is no second framework in
    Rd with the same edge lengths

R2
globally rigid
4
Global Rigidity
  • p is GR in Rd there is no second framework in
    Rd with the same edge lengths

R2
5
Global Rigidity
  • p is GR in Rd there is no second framework in
    Rd with the same edge lengths

R2
6
Global Rigidity
  • p is GR in Rd there is no second framework in
    Rd with the same edge lengths
  • Else p is GF in Rd

R2
7
Global Rigidity (notes)
  • Edge crossing is allowed
  • Euclidean rot, trans, reflect is not considered
    different

R2
8
Global Rigidity (notes)
  • Locally flexible gt globally flexible
  • Easier to characterize

R2
9
Motivation Distance Geometry
  • Input some pairwise distances
  • Output geometric framework/Eucl
  • Chemistry, sensor networks

10
Motivation Distance Geometry
  • Input some pairwise distances
  • Output geometric framework/Eucl
  • GR Well posed-ness of problem
  • GR Divide and conquer

11
Motivation Distance Geometry
  • Molecule problem
  • MDS with partial information
  • Rank constrained distance matrix completion

12
Generic
  • Given a graph G, and framework p, the GR problem
    is NP-Hard Saxe 79

13
Generic
  • Given a graph G, and framework p, the GR problem
    is NP-Hard Saxe 79
  • The reductions all involve special coincidences
    in the framework

R1
14
Generic
  • Given a graph G, and framework p, the GR problem
    is NP-Hard Saxe 79
  • The reductions all involve special coincidences
    in the framework

R1
15
Generic
  • Given a graph G, and framework p, the GR problem
    is NP-Hard Saxe 79
  • Problem seems simpler if we assume no coincidences

R1
globally rigid
16
Generic
  • Given a graph G, and framework p, the GR problem
    is NP-Hard Saxe 79
  • Problem seems simpler if we assume no coincidences

R1
globally rigid
globally flexible
17
Generic
  • Perhaps the problem is easier if we assume that
    the input framework is generic
  • Think randomly perturbed

R1
globally rigid
globally flexible
18
Generic
  • Perhaps the problem is easier if we assume that
    the input framework is generic
  • In 1D, a generic framework is GR iff the graph is
    2-connected

R1
globally rigid
globally flexible
19
Generic
  • Perhaps the problem is easier if we assume that
    the input framework is generic
  • In 1D, a generic framework is GR iff the graph is
    2-connected
  • So GGR in R1 is a property of the graph alone

R1
globally rigid
globally flexible
20
History of GGR
  • CC (Connelly condition) Sufficient for all d
    89, H95, 05
  • HC (Hendrickson condition) Necessary for all d
    88, 92

21
History of GGR
  • CC (Connelly condition) Sufficient for all d
    89, H95, 05
  • HC (Hendrickson condition) Necessary for all d
    88, 92
  • HC CC (nec suff) for d2 JJ05

22
History of GGR
  • CC (Connelly condition) Sufficient for all d
    89, H95, 05
  • HC (Hendrickson condition) Necessary for all d
    88, 92
  • HC not sufficient for d gt 3 C 91
  • HC CC (nec suff) for d2 JJ05

K5,5
23
History of GGR
  • CC (Connelly condition) Sufficient for all d
    89, H95, 05
  • HC (Hendrickson condition) Necessary for all d
    88, 92
  • HC not sufficient for d gt 3 C 91
  • HC CC (nec suff) for d2 JJ05

K5,5
24
History of GGR
  • CC (Connelly condition) Sufficient for all d
    89, H95, 05
  • HC (Hendrickson condition) Necessary for all d
    88, 92
  • HC not sufficient for d gt 3 C 91
  • HC CC (nec suff) for d2 JJ05
  • CC is necessary for all d this work

25
Main result
  • Connelly If CC is satisfied by a generic
    framework in Rd, it is globally rigid in Rd .
  • Thm If CC is not satisfied by a generic
    framework in Rd then it is globally flexible in
    Rd

26
Main result
  • Connelly If CC is satisfied by a generic
    framework in Rd, it is globally rigid in Rd .
  • Thm If CC is not satisfied by a generic
    framework in Rd then it is globally flexible in
    Rd
  • Note CC can be tested with an efficient
    randomized algorithm.

27
Main result
  • Connelly If CC is satisfied by a generic
    framework in Rd, it is globally rigid in Rd .
  • Thm If CC is not satisfied by a generic
    framework in Rd then it is globally flexible in
    Rd
  • Note CC test gives same answer for all generic
    frameworks of G in Rd

28
Main result
  • Connelly If CC is satisfied by a generic
    framework, it is globally rigid.
  • Thm If CC is not satisfied by a generic
    framework in Rd then it is globally flexible in
    Rd
  • Cor A graph is either GR for all generic fmwks
    in Rd, or is not GR for any generic fmwk in Rd.
  • So GGR in Rd is a property of the graph alone

29
On to the condition.
30
Stress Vector satisfied by a framework
  • A real number wuv on each edge euv
  • Sv wuv p(v) p(u) 0

a
b
g
c
h
f
d
R2
p(u)
e
31
Stress Vector satisfied by a framework
  • A real number wuv on each edge euv
  • Sv wuv p(v) p(u) 0

g
c
f
R2
p(u)
e
32
Stress Vector satisfied by a framework
  • A real number wuv on each edge euv
  • Sv wuv p(v) p(u) 0

b
p(u)
h
f
R2
33
Stress Vector satisfied by a framework
  • Equivalent to (symmetrically) writing each vertex
    as an affine comb of its nbrs
  • 1/ Sv wuv Sv wuv p(v) p(u)

g
c
f
R2
p(u)
e
34
Stress Vector satisfied by a framework
  • Equivalent to equilibrium point of quadratic
    spring/strut energy (no pins)
  • E(p) Su Sv wuv p(v) p(u)2

R2
35
Stress vectors easy facts
  • Stress vectors satisfied by a fixed p form a
    linear space W(p)

p
36
Stress vectors easy facts
  • Stress vectors satisfied by a fixed p form a
    linear space W(p)
  • Any affine transform T(p) will satisfy all
    stresses in W(p)

T(p)
p
37
Stress vectors easy facts
  • For some graphs, there are even more fmwks than
    the affine transforms of p,
  • that still satisfy all stresses in W(p)

p
not an affine tform of p
38
Connellys Condition
  • CC The only fmwks that satisfy all of W(p) are
    the affine transforms of p
  • This will somehow describe GGR!

39
Stresses and lengths
  • What is the relationship between stress vectors
  • Affine invariant
  • .. and lengths
  • Euclidean invariant

40
The mapping L
  • L is the mapping from d-dim fmwks to Re
  • Describing each edges squared length

L
Re
R2
41
The mapping L
  • L is the mapping from d-dim fmwks to Re
  • Describing each edges squared length

L
Re
R2
42
The set M
  • M is the image of L
  • All possible measurements

L
Re
R2
43
The set M
  • M is the image of L
  • All possible measurements
  • A semi-algebraic set
  • A smooth manifold a.e.

L
Re
R2
44
Lengths and stresses the connection
  • At a generic fmwk p,
  • span L the tangent space of M at L(p)

L
R2
Re
45
Lengths and stresses the connection
  • At a generic fmwk p,
  • span L the tangent space of M at L(p)
  • At a generic fmwk p,
  • W(p) spans the normal space of M at L(p)
  • Maxwell

L
R2
Re
46
Lengths and stresses the connection
  • So a generic fmwk that satisfies all the same
    stresses as p must have the same normal space

L
R2
Re
47
Lengths and stresses the connection
  • So a generic fmwk that satisfies all the same
    stresses as p must have the same normal space

L
R2
Re
48
Lengths and stresses the connection
  • So a generic fmwk that satisfies all the same
    stresses as p must have the same tangent space

L
R2
Re
49
Lengths and stresses the connection
  • So a generic fmwk that satisfies all the same
    stresses as p must have the same tangent space
  • M is a cone ?M M
  • Same tangent space, not just parallel

L
R2
Re
50
Lengths and stresses the connection
  • So a generic fmwk that satisfies all the same
    stresses as p must have the same tangent space
  • This is the key connection
  • We will return to this in the proof

L
R2
Re
51
On to the proof
52
Degree mod two thm
53
Degree mod two thm
0
2
54
Degree mod two thm
0
2
4
55
Degree mod two thm
  • Typical version
  • Given smooth map from a compact manifold to a
    connected manifold of same dimension
  • Every generic point in the range has the same
    number of pre-images mod 2
  • The creases are singular
  • This number 0,1 is called the degree

56
Degree mod two thm
proper
  • General version
  • Given smooth map from a compact manifold to a
    connected manifold of same dimension
  • Every generic point in the range has the same
    number of pre-images mod 2

57
Our plan
0
2
4
58
Our plan
0
2
4
59
Our plan
  • Assume (!CC)
  • Start with the map L
  • Define a domain
  • Equate Euclidean transforms in the domain
  • Define range
  • Connected smooth manifold
  • Will need to remove some singularities while
    maintaining connectivity of range and properness
    of map
  • Show the map has degree 0
  • Each point in image has multiple pre-images
  • Framework is globally flexible

60
Domain
  • Start with stress satisfiers A(p)
  • Frameworks that satisfy all of the stresses that
    are satisfied by p
  • Affine tforms of p plus maybe more

61
Domain
  • Start with stress satisfiers A(p)
  • Mod out Euclideans A(p)/Eucl

62
Domain
  • Start with stress satisfiers A(p)
  • Mod out Euclideans A(p)/Eucl
  • Result is smooth manifold singularities

63
Domain
  • Start with stress satisfiers A(p)
  • Mod out Euclideans A(p)/Eucl
  • Result is smooth manifold singularities
  • Singularities fmwks stabilized by a n.t.
    euclidean.

64
Domain
  • Start with stress satisfiers A(p)
  • Mod out Euclideans A(p)/Eucl
  • Result is smooth manifold singularities
  • Singularities deficient affine span.

65
Lemma 1
  • Lemma If (!CC)
  • A(p) is big
  • then the singularities of A(p)/Eucl are of
    co-dimension gt 2.
  • Proof counting

66
Codimension and cutting
  • The singular co-dim will carry over to the range
  • Removal a co-dimension 2 set does not disconnect
  • Needed for degree thm

co-dim 2

67
Codimension and cutting
  • The singular co-dim will carry over to the range
  • Removal a co-dimension 2 set does not disconnect
  • Needed for degree thm

co-dim 2
co-dim 1
68
The range
  • Let B(p) L(A(p))
  • Achievable measurements of stress satisfiers
  • Some subset of M

L
Re
69
The range
  • For the degree to be well defined
  • Need to include B(p) as a full dimensional subset
    of a connected range manifold

L
Re
70
The range
  • For the degree to be well defined
  • Need to include B(p) as a full dimensional subset
    of a connected range manifold
  • To show the degree is 0
  • Sufficient for range to include pts not in B(p)

L
Re
0
71
The range
  • For the degree to be well defined
  • Need to include B(p) as a full dimensional subset
    of a connected range manifold
  • To show the degree is 0
  • Sufficient for range to include pts not in B(p)

L
Re
72
The range
  • For the degree to be well defined
  • Need to include B(p) as a full dimensional subset
    of a connected range manifold
  • To show the degree is 0
  • Sufficient for range to include pts not in B(p)

L
Re
73
The range
  • So we need to understand the shape of B(p)

L
Re
0
74
Gauss fiber
Digression
  • Gauss fiber points with same (not just parallel)
    tangent as chosen point



Re
75
Gauss fiber
Digression
  • Gauss fiber thm The Gauss fiber at a generic
    point of an irreducible algebraic variety is an
    affine space



Re
1d fiber
76
Gauss fiber
Digression
  • Gauss fiber thm The Gauss fiber at a generic
    point of an irreducible algebraic variety is a
    affine space

1d non-affine fiber
Re
77
Gauss fiber
Digression
  • Gauss fiber thm The Gauss fiber at a generic
    point of an irreducible algebraic variety is a
    affine space

1d non-affine fiber, exceptional
Re
78
Gauss fiber
Digression
  • Gauss fiber thm The Gauss fiber at a generic
    point of an irreducible algebraic variety is a
    affine space

0-d fiber
Re
79
The range
  • Recall the connection
  • Same stresses same tangent in M
  • B(p) is a gauss fiber in M

L
Re
80
The range
  • Recall the connection
  • Same stresses same tangent in M
  • B(p) is a gauss fiber in M
  • M is not an irreducible algebraic variety
  • But it is a full dimensional semi-algebraic
    subset of one

L
Re
81
Lemma 2
  • Lemma B(p) is a flat space
  • Full dimensional subset of an affine space

L
Re
82
Lemma 2
  • Lemma B(p) is a flat space
  • Full dimensional subset of an affine space
  • So define the range to be this whole affine space

L
Re
83
Lemma 2
  • Lemma B(p) is a flat space
  • Full dimensional subset of an affine space
  • So define the range to be this whole affine space

L
Re
84
Lemma 2
  • Lemma B(p) is a flat space
  • Full dimensional subset of an affine space
  • M is contained in first octant, the affine space
    is not
  • The degree will be zero

L
0
Re
85
Lemma 2
  • Lemma B(p) is a flat space
  • Full dimensional subset of an affine space
  • Note domain and range have same dimension

L
0
Re
86
Last step
  • Remove the images of the singularities of
    A(p)/Eucl from the range and their pre-images
  • Range remains connected if (!CC)
  • Domain and range are smooth manifolds

L
0
Re
87
Last step
  • Remove the images of the singularities of
    A(p)/Eucl from the range and their pre-images
  • Range remains connected if (!CC)
  • Domain and range are smooth manifolds

L
0
Re
88
Last step
  • Remove the images of the singularities of
    A(p)/Eucl from the range and their pre-images
  • Can now apply degree thm

4
L
2
0
Re
89
Main Theorem
4
L
2
0
Re
90
Main Theorem
4
L
2
0
Re
91
Main Theorem
  • Thm If CC is not satisfied by a generic
    framework in Rd then it is globally flexible in
    Rd

4
L
2
0
Re
92
Review
  • Defined a domain
  • A(p)/Eucl
  • Created some singularities
  • Defined same dimensional connected smooth range
  • Affine space containing B(p) (due to flatness)
  • Removed singularities
  • To get smooth domain manifold
  • Maintained range connectedness (due to high
    co-dim if !CC)
  • Now we there is a well defined degree
  • Need to know degree is 0 (due to flatness)

93
Deep Breath..
94
Algorithm
  • Input Graph, d

95
Algorithm
  • Input Graph, d
  • Pick random framework p in Rd
  • Compute stress vector space W(p)
  • Linear algebra
  • Pick a random stress vector w from W(p)
  • Compute m dimensionality of fmwks that satisfy w
  • Linear algebra
  • If m d(d1) output GR in Rd
  • If m gt d(d1) output GGR in Rd

96
Algorithm
  • With high probability, p will behave like a
    generic fmwk
  • With high probability, fmwks that satisfy w will
    satisfy all of W(p)
  • Can be done with integer linear algebra
  • The exceptions satisfy a low degree polynomial
  • No false positives
  • GGR in RP

97
One more breath
98
Bonus result
  • If a graph is generically globally flexible
  • One can continuously flex in one higher dimension
    back down to second framework

99
Future work
  • More algebraic
  • Not just the L function on graphs
  • More general field
  • More general metric signature
  • More combinatorial
  • Deterministic efficient algorithm

100
Future work
  • Given lengths, it is NP-hard to figure out the
    framework in Rd
  • Semi-definite programming will typically find
    solution in Rv
  • But sometimes it happens to give an answer in Rd
  • These are frameworks for which higher dimensions
    dont help
  • Can these cases be nicely characterized?
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