Characterizing%20Generic%20Global%20Rigidity

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

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Motivation Distance Geometry

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

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Motivation Distance Geometry

• Molecule problem
• MDS with partial information
• Rank constrained distance matrix completion

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Generic

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

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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
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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
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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
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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
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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

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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

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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

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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.

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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

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On to the condition.
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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
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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
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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
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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
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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
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Stress vectors easy facts

• Stress vectors satisfied by a fixed p form a
  linear space W(p)

p
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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
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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
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Connellys Condition

• CC The only fmwks that satisfy all of W(p) are
  the affine transforms of p
• This will somehow describe GGR!

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Stresses and lengths

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

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The mapping L

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

L
Re
R2
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The mapping L

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

L
Re
R2
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The set M

• M is the image of L
• All possible measurements

L
Re
R2
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The set M

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

L
Re
R2
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Lengths and stresses the connection

• At a generic fmwk p,
• span L the tangent space of M at L(p)

L
R2
Re
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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
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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
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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
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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
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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
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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
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On to the proof
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Degree mod two thm
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Degree mod two thm
0
2
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Degree mod two thm
0
2
4
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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

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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

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Our plan
0
2
4
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Our plan
0
2
4
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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

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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

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Domain

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

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Domain

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

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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.

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Domain

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

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Lemma 1

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

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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

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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
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The range

• Let B(p) L(A(p))
• Achievable measurements of stress satisfiers
• Some subset of M

L
Re
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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
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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
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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
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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
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The range

• So we need to understand the shape of B(p)

L
Re
0
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Gauss fiber
Digression

• Gauss fiber points with same (not just parallel)
  tangent as chosen point

Re
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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
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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
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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
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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
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The range

• Recall the connection
• Same stresses same tangent in M
• B(p) is a gauss fiber in M

L
Re
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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
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Lemma 2

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

L
Re
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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
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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
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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
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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
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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..
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Algorithm

• Input Graph, d

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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

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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

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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?