Post on 22-Dec-2015
Global Rigidity
• p is GR in Rd : there is no second framework in Rd with the same edge lengths
R2 globally rigid
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
Global Rigidity (notes)
• Edge crossing is allowed
• Euclidean: rot, trans, reflect is not considered different
R2
Motivation: Distance Geometry
• Input: some pairwise distances
• Output: geometric framework/Eucl– Chemistry, sensor networks
Motivation: Distance Geometry
• Input: some pairwise distances
• Output: geometric framework/Eucl– GR: Well posed-ness of problem– GR: Divide and conquer
Motivation: Distance Geometry
• Molecule problem
• MDS with partial information
• Rank constrained distance matrix completion
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
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
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
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
Generic
• Perhaps the problem is easier if we assume that the input framework is “generic”– Think: randomly perturbed
R1 globally rigid globally flexible
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
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
History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]
History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]
HC = CC (nec & suff) for d=2 [JJ’05]
History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]
HC not sufficient for d >= 3 [C ’91]
HC = CC (nec & suff) for d=2 [JJ’05]
K5,5
History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]
HC not sufficient for d >= 3 [C ’91]
HC = CC (nec & suff) for d=2 [JJ’05]
K5,5
History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]
HC not sufficient for d >= 3 [C ’91]
HC = CC (nec & suff) for d=2 [JJ’05]
CC is necessary for all d [this work]
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
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.
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
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
• A real number wuv on each edge euv
Σv wuv [p(v) – p(u)] = 0
Stress Vector satisfied by a framework
R2
f
a
c
b
e
dh
g
p(u)
u
• A real number wuv on each edge euv
Σv wuv [p(v) – p(u)] = 0
Stress Vector satisfied by a framework
R2p(u)e
f
g
c
u
• A real number wuv on each edge euv
Σv wuv [p(v) – p(u)] = 0
Stress Vector satisfied by a framework
u
R2
f
b
h
p(u)
Stress Vector satisfied by a framework
• Equivalent to (symmetrically) writing each vertex as an affine comb of its nbrs
[1/ Σv wuv] Σv wuv p(v) = p(u)
R2
u
fc
e
g
p(u)
Stress Vector satisfied by a framework
Equivalent to equilibrium point of quadratic spring/strut energy (no pins)
E(p)= Σu Σv wuv |p(v) – p(u)|2
R2
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)
p T(p)
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
Connelly’s Condition
• CC := The only fmwks that satisfy all of W(p) are the affine transforms of p
• This will somehow describe GGR!
Stresses and lengths
• What is the relationship between stress vectors– Affine invariant
• ….. and lengths– Euclidean invariant
The mapping “L”
• “L” is the mapping from d-dim fmwks to Re
• Describing each edge’s squared length
Re
R2
L
The mapping “L”
• “L” is the mapping from d-dim fmwks to Re
• Describing each edge’s squared length
Re
R2
L
The set “M”
• “M” is the image of “L” – All possible measurements– A semi-algebraic set
• A smooth manifold a.e.
Re
R2
L
Re
Lengths and stresses: the connection
• At a generic fmwk p,
span L* = the tangent space of M at L(p)
R2
L
Re
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]
R2
L
Re
Lengths and stresses: the connection
• So a generic fmwk that satisfies all the same stresses as p must have the same normal space
R2
L
Re
Lengths and stresses: the connection
• So a generic fmwk that satisfies all the same stresses as p must have the same normal space
R2
L
Re
Lengths and stresses: the connection
• So a generic fmwk that satisfies all the same stresses as p must have the same tangent space
R2
L
Re
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: λM = M– Same tangent space, not just parallel
R2
L
Re
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
R2
L
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
Degree mod two thm
• 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
proper
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
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
Domain
• Start with stress satisfiers: A(p)
• Mod out Euclideans: A(p)/Eucl
• Result is smooth manifold + singularities
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.
Domain
• Start with stress satisfiers: A(p)
• Mod out Euclideans: A(p)/Eucl
• Result is smooth manifold + singularities– Singularities: deficient affine span.
Lemma 1
• Lemma: If (!CC)– A(p) is “big”
• then the singularities of A(p)/Eucl are of co-dimension >= 2.– Proof: counting
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
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
The range
• Let B(p) := L(A(p))– Achievable measurements of stress satisfiers– Some subset of M
Re
L
The range
• For the degree to be well defined– Need to include B(p) as a full dimensional
subset of a connected range manifold
Re
L
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)
Re
L
0
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)
Re
L
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)
Re
L
Gauss fiber
Re
Digression
…
…
• Gauss fiber: points with same (not just parallel) tangent as chosen point
Gauss fiber
• Gauss fiber thm: The Gauss fiber at a generic point of an irreducible algebraic variety is an affine space
Re1d fiber
Digression
…
…
Gauss fiber
• Gauss fiber thm: The Gauss fiber at a generic point of an irreducible algebraic variety is a affine space
Re
1d non-affine fiber
Digression
Gauss fiber
• Gauss fiber thm: The Gauss fiber at a generic point of an irreducible algebraic variety is a affine space
Re
1d non-affine fiber, exceptional
Digression
Gauss fiber
• Gauss fiber thm: The Gauss fiber at a generic point of an irreducible algebraic variety is a affine space
Re
0-d fiber
Digression
The range
• Recall “the connection” – Same stresses = same tangent in M
• B(p) is a gauss fiber in M
L
Re
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
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
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
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
Re
0
Lemma 2
• Lemma: B(p) is a flat space– Full dimensional subset of an affine space
• Note: domain and range have same dimension
L
Re
0
Last step
L
• 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
0
Re
Last step
L
• 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
0
Re
Last step
L
• Remove the images of the singularities of A(p)/Eucl from the range and their pre-images
• Can now apply degree thm
0
Re
24
Main Theorem
L
0
24
•[Thm] If CC is not satisfied by a generic framework in Rd then it is globally flexible in Rd
Re
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)
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(d+1) output “GR in Rd”
• If m > d(d+1) output “GGR in Rd”
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
Bonus result
• If a graph is generically globally flexible
• One can continuously flex in one higher dimension back down to second framework
Future work
• More algebraic: – Not just the L function on graphs– More general field– More general metric signature
• More combinatorial:– Deterministic efficient algorithm