Invariant Signatures and Histograms for Object Recognition ...olver/t_/pzhi.pdf · The Basic...
Transcript of Invariant Signatures and Histograms for Object Recognition ...olver/t_/pzhi.pdf · The Basic...
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Invariant Signatures and Histograms
for Object Recognition,
Symmetry Detection, and
Jigsaw Puzzle Assembly
Peter J. Olver
University of Minnesota
http://www.math.umn.edu/! olver
San Diego, January, 2013
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The Basic Equivalence Problem
M — smooth m-dimensional manifold.
G — transformation group acting on M
• finite-dimensional Lie group
• infinite-dimensional Lie pseudo-group
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Equivalence:Determine when two p-dimensional submanifolds
N and N " M
are congruent :
N = g ·N for g # G
Symmetry:Find all symmetries,
i.e., self-equivalences or self-congruences :
N = g ·N
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Classical Geometry
• Euclidean group: G =
!"
#
SE(m) = SO(m)! Rm
E(m) = O(m)! Rm
z $%& A · z + b A # SO(m) or O(m), b # Rm, z # R
m
' isometries: rotations, translations , (reflections)
• Equi-a!ne group: G = SA(m) = SL(m)! Rm
A # SL(m) — volume-preserving
• A!ne group: G = A(m) = GL(m)! Rm
A # GL(m)
• Projective group: G = PSL(m+ 1)acting on Rm " RPm
=' F. Klein
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Tennis, Anyone?
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Euclidean Plane Curves: G = SE(2) = SO(2)! R2
Curvature di!erential invariants:
! =uxx
(1 + u2x)
3/2,
d!
ds=
(1 + u2x)uxxx % 3uxu
2xx
(1 + u2x)
3,
d2!
ds2= · · ·
Arc length (invariant one-form):
ds =$1 + u2
x dx,d
ds=
1$1 + u2
x
d
dx
Theorem. All Euclidean di!erential invariants arefunctions of the derivatives of curvature withrespect to arc length: !, !s, !ss, · · ·
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Equi-a!ne Plane Curves: G = SA(2) = SL(2)!R2
Equi-a"ne curvature:
! =5uxxuxxxx % 3u2
xxx
9 u8/3xx
d!
ds= · · ·
Equi-a"ne arc length:
ds = 3
$uxx dx
d
ds=
13(uxx
d
dx
Theorem. All equi-a"ne di!erential invariantsare functions of the derivatives of equi-a"necurvature with respect to equi-a"ne arc length:
!, !s, !ss, · · ·
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Equivalence & Invariants
• Equivalent submanifolds N ) Nmust have the same invariants: I = I.
Constant invariants provide immediate information:
e.g. ! = 2 *' ! = 2
Non-constant invariants are not useful in isolation,because an equivalence map can drastically alter thedependence on the submanifold parameters:
e.g. ! = x3 versus ! = sinhx
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Equivalence & Invariants
• Equivalent submanifolds N ) Nmust have the same invariants: I = I.
Constant invariants provide immediate information:
e.g. ! = 2 *' ! = 2
Non-constant invariants are not useful in isolation,because an equivalence map can drastically alter thedependence on the submanifold parameters:
e.g. ! = x3 versus ! = sinhx
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Equivalence & Invariants
• Equivalent submanifolds N ) Nmust have the same invariants: I = I.
Constant invariants provide immediate information:
e.g. ! = 2 *' ! = 2
Non-constant invariants are not useful in isolation,because an equivalence map can drastically alter thedependence on the submanifold parameters:
e.g. ! = x3 versus ! = sinhx
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However, a functional dependency or syzygy amongthe invariants is intrinsic:
e.g. !s = !3 % 1 *' !s = !3 % 1
• Universal syzygies — Gauss–Codazzi
• Distinguishing syzygies.
Theorem. (Cartan)Two regular submanifolds are (locally) equivalentif and only if they have identical syzygies amongall their di!erential invariants.
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However, a functional dependency or syzygy amongthe invariants is intrinsic:
e.g. !s = !3 % 1 *' !s = !3 % 1
• Universal syzygies — Gauss–Codazzi
• Distinguishing syzygies.
Theorem. (Cartan)Two regular submanifolds are (locally) equivalentif and only if they have identical syzygies amongall their di!erential invariants.
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However, a functional dependency or syzygy amongthe invariants is intrinsic:
e.g. !s = !3 % 1 *' !s = !3 % 1
• Universal syzygies — Gauss–Codazzi
• Distinguishing syzygies.
Theorem. (Cartan)Two regular submanifolds are (locally) equivalentif and only if they have identical syzygies amongall their di!erential invariants.
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Finiteness of Generators and Syzygies
+ There are, in general, an infinite number of di!er-ential invariants and hence an infinite numberof syzygies must be compared to establishequivalence.
, But the higher order di!erential invariants arealways generated by invariant di!erentiationfrom a finite collection of basic di!erentialinvariants, and the higher order syzygies areall consequences of a finite number of loworder syzygies!
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Finiteness of Generators and Syzygies
+ There are, in general, an infinite number of di!er-ential invariants and hence an infinite numberof syzygies must be compared to establishequivalence.
, But the higher order di!erential invariants arealways generated by invariant di!erentiationfrom a finite collection of basic di!erentialinvariants, and the higher order syzygies areall consequences of a finite number of loworder syzygies!
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Example — Plane CurvesIf non-constant, both ! and !s depend on a single
parameter, and so, locally, are subject to a syzygy:
!s = H(!) (-)
But then
!ss =d
dsH(!) = H !(!)!s = H !(!)H(!)
and similarly for !sss, etc.
Consequently, all the higher order syzygies are generatedby the fundamental first order syzygy (-).
Thus, for Euclidean (or equi-a"ne or projective or . . . )plane curves we need only know a single syzygy between ! and!s in order to establish equivalence!
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Signature CurvesDefinition. The signature curve S " R2 of a plane curveC " R2 is parametrized by the two lowest order di!erentialinvariants
# : C %& S =
% &
! ,d!
ds
' (
" R2
=' Calabi, PJO, Shakiban, Tannenbaum, Haker
Theorem. Two regular curves C and C are equivalent:
C = g · C
if and only if their signature curves are identical:
S = S=' regular: (!s,!ss) .= 0.
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Symmetry and Signature
Continuous Symmetries
Theorem. The following are equivalent:
• The curve C has a 1-dimensional symmetry group H " G
• C is the orbit of a 1-dimensional subgroup H " G
• The signature S degenerates to a point: dimS = 0
• All the di!erential invariants are constant:! = c, !s = 0, . . .
=' Euclidean plane geometry: circles, lines
=' Equi-a"ne plane geometry: conic sections.
=' Projective plane geometry: W curves (Lie & Klein)
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Symmetry and Signature
Discrete Symmetries
Definition. The index of a curve C equals thenumber of points in C which map to a singlegeneric point of its signature:
"C = min)# #"1{w}
*** w # S+
Theorem. The number of discrete symmetries of Cequals its index "C.
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The Index
#
%&
N S
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The Curve x = cos t+ 15cos2 t, y = sin t+ 1
10sin2 t
-0.5 0.5 1
-0.5
0.5
1
The Original Curve
0.25 0.5 0.75 1 1.25 1.5 1.75 2
-2
-1
0
1
2
Euclidean Signature
0.5 1 1.5 2 2.5
-6
-4
-2
2
4
A"ne Signature
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The Curve x = cos t+ 15cos2 t, y = 1
2x+ sin t+ 1
10sin2 t
-0.5 0.5 1
-1
-0.5
0.5
1
The Original Curve
0.5 1 1.5 2 2.5 3 3.5 4
-7.5
-5
-2.5
0
2.5
5
7.5
Euclidean Signature
0.5 1 1.5 2 2.5
-6
-4
-2
2
4
A"ne Signature
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Canine Left Ventricle Signature
Original Canine HeartMRI Image
Boundary of Left Ventricle
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Smoothed Ventricle Signature
10 20 30 40 50 60
20
30
40
50
60
10 20 30 40 50 60
20
30
40
50
60
10 20 30 40 50 60
20
30
40
50
60
-0.15 -0.1 -0.05 0.05 0.1 0.15 0.2
-0.06
-0.04
-0.02
0.02
0.04
0.06
-0.15 -0.1 -0.05 0.05 0.1 0.15 0.2
-0.06
-0.04
-0.02
0.02
0.04
0.06
-0.15 -0.1 -0.05 0.05 0.1 0.15 0.2
-0.06
-0.04
-0.02
0.02
0.04
0.06
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Object Recognition
=' Steve Haker
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!"" #""
$#"
%""
%#"
&'()*
!"+"# " "+"# "+*!"+"*#
!"+"*
!"+""#
"
"+""#
"+"*
,-./0('12)3'142)&'()*
!"" #""
!#"
#""
##"
$""&'()5
!"+"# " "+"# "+*!"+"*#
!"+"*
!"+""#
"
"+""#
"+"*
,-./0('12)3'142)&'()5
!"+"# " "+"# "+*!"+"*#
!"+"*
!"+""#
"
"+""#
"+"*
36782/2889)"+*:%$%:
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!""#""
$""
%""
&"""
'(()*&
!"+", " "+", "+&!"+"&,
!"+"&
!"+"",
"
"+"",
"+"&
-./012345*63475*'(()*&
8"" ,""
9,"
#""
#,"
:32*&
!"+", " "+", "+&!"+"&,
!"+"&
!"+"",
"
"+"",
"+"&
-./012345*63475*:32*&
!"+", " "+", "+&!"+"&,
!"+"&
!"+"",
"
"+"",
"+"&
6;(<505<<=*"+">&!&#
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Signature Metrics
• Hausdor!
• Monge–Kantorovich transport
• Electrostatic/gravitational attraction
• Latent semantic analysis
• Histograms
• Gromov–Hausdor! & Gromov–Wasserstein
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Signatures
s
!
Classical Signature%&
Original curve!
!s
Di!erential invariant signature
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Signatures
s
!
Classical Signature%&
Original curve!
!s
Di!erential invariant signature
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Occlusions
s
!
Classical Signature%&
Original curve!
!s
Di!erential invariant signature
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Classical Occlusions
s
!
%&
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3D Di"erential Invariant Signatures
Euclidean space curves: C " R3
S = { (! , !s , # ) } " R3
• ! — curvature, # — torsion
Euclidean surfaces: S " R3 (generic)
S =) ,
H , K , H,1 , H,2 , K,1 , K,2
- +" R
6
or .S =) ,
H , H,1 , H,2 , H,11
- +" R
4
• H — mean curvature, K — Gauss curvature
Equi–a!ne surfaces: S " R3 (generic)
S =) ,
P , P,1 , P,2, P,11
- +" R
4
• P — Pick invariant
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Advantages of the Signature Curve
• Purely local — no ambiguities
• Symmetries and approximate symmetries
• Extends to surfaces and higher dimensional sub-manifolds
• Occlusions and reconstruction
• Partial matching and puzzles
Main disadvantage: Noise sensitivity due todependence on high order derivatives.
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Generalized Vertices
Ordinary vertex: local extremum of curvature
Generalized vertex: !s / 0
• critical point
• circular arc
• straight line segment
Mukhopadhya’s Four Vertex Theorem:
A simple closed, non-circular plane curve has n 0 4generalized vertices.
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“Counterexamples”
These degenerate curves all have the same signature:
!2 2 4 6
!8
!6
!4
!2
!2 2 4 6 8
!8
!6
!4
!2
2
!2 2 4 6 8 10
!8
!6
!4
!2
2
!2 2 4 6 8 10
!6
!4
!2
2
4
!2 2 4 6 8 10
!6
!4
!2
2
4
!2 2 4 6 8
!4
!2
2
4
!1.0 !0.5 0.5
!2
!1
1
2
$ Replace vertices with circular arcs: Musso–Nicoldi
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Bivertex Arcs
Bivertex arc: !s .= 0 everywhereexcept !s = 0 at the two endpoints
The signature S of a bivertex arc is a single arc thatstarts and ends on the !–axis.
!
!s
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Bivertex Decomposition.
v-regular curve — finitely many generalized vertices
C =m!
j=1
Bj 1n!
k=1
Vk
B1, . . . , Bm — bivertex arcs
V1, . . . , Vn — generalized vertices: n 0 4
Main Idea: Compare individual bivertex arcs, andthen determine whether the rigid equivalences are(approximately) the same.
D. Ho! & PJO, Extensions of invariant signatures for object recognition,J. Math. Imaging Vision, to appear.
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Gravitational/Electrostatic Attraction
$ Treat the two (signature) curves as masses or as oppositelycharged wires. The higher their mutual attraction, thecloser they are together.
!
!s
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Gravitational/Electrostatic Attraction
$ Treat the two (signature) curves as masses or as oppositelycharged wires. The higher their mutual attraction, thecloser they are together.
$ In practice, we are dealing with discrete data (pixels) andso treat the curves and signatures as point masses/charges.
!
!s
!
!s
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Strength of correspondence:
h(%, /%) =
!0"
0#
1
d(%, /%)! + &, d(%, /%) < 2,
0, d(%, /%) = 2.
Separation:
d(%, /%) =
!0"
0#
3% % /% 3D % 3% % /% 3
, 3% % /% 3 < D,
2, 3% % /% 3 0 D,
Scale of comparison:
D(C, /C) = (D"(C,/C),D"s
(C, /C) ),
D"(C,/C) = max
%
maxz#C
(!|z)%minz#C
(!|z), maxz#/C
(!|z)%minz#/C
(!|z)(
,
D"s
(C, /C) = max
%
maxz#C
(!s|z)%minz#C
(!s|z), maxz#/C
(!s|z)%minz#/C
(!s|z)(
.
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Piece Locking
$ $ Minimize force and torque based on gravitationalattraction of the two matching edges.
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The Ba#er Jigsaw Puzzle
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The Ba#er Solved
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The Rain Forest Giant Floor Puzzle
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The Rain Forest Puzzle Solved
=' D. Ho! & PJO, Automatic solution of jigsaw puzzles,preprint, 2012.
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The Rain Forest Puzzle Solved
=' D. Ho! & PJO, Automatic solution of jigsaw puzzles,preprint, 2012.
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The Distance Histogram
Definition. The distance histogram of a finite set ofpoints P = {z1, . . . , zn} " V is the function
'P (r) = #)(i, j)
*** 1 4 i < j 4 n, d(zi, zj) = r+.
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The Distance Set
The support of the histogram function,
supp 'P = $P " R+
is the distance set of P .
Erdos’ distinct distances conjecture (1946):
If P " Rm, then #$P 0 cm,# (#P )2/m"#
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The Distance Set
The support of the histogram function,
supp 'P = $P " R+
is the distance set of P .
Erdos’ distinct distances conjecture (1946):
If P " Rm, then #$P 0 cm,# (#P )2/m"#
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Characterization of Point Sets
Note: If /P = g · P is obtained from P " Rm by arigid motion g # E(n), then they have the samedistance histogram: 'P = '/P .
Question: Can one uniquely characterize, up to rigidmotion, a set of points P{z1, . . . , zn} " Rm
by its distance histogram?
=' Tinkertoy problem.
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Characterization of Point Sets
Note: If /P = g · P is obtained from P " Rm by arigid motion g # E(n), then they have the samedistance histogram: 'P = '/P .
Question: Can one uniquely characterize, up to rigidmotion, a set of points P{z1, . . . , zn} " Rm
by its distance histogram?
=' Tinkertoy problem.
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Yes:
' = 1, 1, 1, 1,(2,
(2.
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No:
Kite Trapezoid
' =(2,
(2, 2,
(10,
(10, 4.
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No:
P = {0, 1, 4, 10, 12, 17}
Q = {0, 1, 8, 11, 13, 17}" R
' = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 16, 17
=' G. Bloom, J. Comb. Theory, Ser. A 22 (1977) 378–379
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Characterizing Point Setsby their Distance Histograms
Theorem. Suppose n 4 3 or n 0 m+ 2.Then there is a Zariski dense open subset in the space of npoint configurations in Rm that are uniquely characterized,up to rigid motion, by their distance histograms.
=' M. Boutin & G. Kemper, Adv. Appl. Math. 32 (2004) 709–735
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Limiting Curve Histogram
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Limiting Curve Histogram
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Limiting Curve Histogram
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Sample Point HistogramsCumulative distance histogram: n = #P :
%P (r) =1
n+
2
n2
1
s$r
'P (s) =1
n2#)(i, j)
*** d(zi, zj) 4 r+,
Note:'P (r) =
12n2[%P (r)% %P (r % () ] ( 5 1.
Local cumulative distance histogram:
)P (r, z) =1
n#)j*** d(z, zj) 4 r
+=
1
n#(P 6 Br(z))
%P (r) =1
n
1
z #P
)P (r, z) =1
n2
1
z #P
#(P 6 Br(z)).
Ball of radius r centered at z:
Br(z) = { v # V | d(v, z) 4 r }
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Limiting Curve Histogram Functions
Length of a curve
l(C) =2
Cds < 2
Local curve distance histogram function
hC(r, z) =l(C 6 Br(z))
l(C)
=' The fraction of the curve contained in the ball of radius rcentered at z.
Global curve distance histogram function:
HC(r) =1
l(C)
2
ChC(r, z(s)) ds.
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Convergence of Histograms
Theorem. Let C be a regular plane curve. Then, for bothuniformly spaced and randomly chosen sample pointsP " C, the cumulative local and global histograms convergeto their continuous counterparts:
)P (r, z) %& hC(r, z), %P (r) %& HC(r),
as the number of sample points goes to infinity.
D. Brinkman & PJO, Invariant histograms,Amer. Math. Monthly 118 (2011) 2–24.
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Square Curve Histogramwith Bounds
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Kite and Trapezoid Curve Histograms
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Histogram–Based Shape Recognition500 sample points
Shape (a) (b) (c) (d) (e) (f)
(a) triangle 2.3 20.4 66.9 81.0 28.5 76.8
(b) square 28.2 .5 81.2 73.6 34.8 72.1
(c) circle 66.9 79.6 .5 137.0 89.2 138.0
(d) 27 3 rectangle 85.8 75.9 141.0 2.2 53.4 9.9
(e) 17 3 rectangle 31.8 36.7 83.7 55.7 4.0 46.5
(f) star 81.0 74.3 139.0 9.3 60.5 .9
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Curve HistogramConjecture
Two su"ciently regular plane curves C and /C have
identical global distance histogram functions, so
HC(r) = H/C(r) for all r 0 0, if and only if they are
rigidly equivalent: C 8 /C.
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Possible Proof Strategies
• Show that any polygon obtained from (densely) discretizing acurve does not lie in the Boutin–Kemper exceptional set.
• Polygons with obtuse angles: taking r small, one can recover(i) the set of angles and (ii) the shortest side length fromHC(r). Further increasing r leads to further geometricinformation about the polygon . . .
• Expand HC(r) in a Taylor series at r = 0 and show that thecorresponding integral invariants characterize the curve.
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Taylor Expansions
Local distance histogram function:
LhC(r, z) = 2r + 112!2 r3 +
,140!!ss +
145!2s +
3320
!4-r5 + · · · .
Global distance histogram function:
HC(r) =2r
L+
r3
12L2
3
C!2 ds+
r5
40L2
3
C
,38!4 % 1
9!2s
-ds+ · · · .
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Space CurvesSaddle curve:
z(t) = (cos t, sin t, cos 2 t), 0 4 t 4 2*.
Convergence of global curve distance histogram function:
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SurfacesLocal and global surface distance histogram functions:
hS(r, z) =area (S 6 Br(z))
area (S), HS(r) =
1
area (S)
2 2
ShS(r, z) dS.
Convergence for sphere:
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Area HistogramsRewrite global curve distance histogram function:
HC(r) =1
L
3
ChC(r, z(s)) ds =
1
L2
3
C
3
C+r(d(z(s), z(s
!)) ds ds!
where +r(t) =
%1, t 4 r,
0, t > r,
Global curve area histogram function:
AC(r) =1
L3
3
C
3
C
3
C+r(area (z(s), z(s
!), z(s!!)) ds d s! ds!!,
d s — equi-a"ne arc length element L =2
Cds
Discrete cumulative area histogram
AP (r) =1
n(n% 1)(n% 2)
1
z %=z! %=z!!#P
+r(area (z, z!, z!!)),
Boutin & Kemper : The area histogram uniquely determinesgeneric point sets P " R2 up to equi-a"ne motion.
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Area Histogram for Circle
$ $ Joint invariant histograms — convergence???
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Triangle Distance Histograms
Z = (. . . zi . . .) " M —sample points on a subset M " Rn (curve, surface, etc.)
Ti,j,k — triangle with vertices zi, zj, zk.
Side lengths:
%(Ti,j,k) = ( d(zi, zj), d(zi, zk), d(zj, zk) )
Discrete triangle histogram:
S = %(T ) " K
Triangle inequality cone:
K = { (x, y, z) | x, y, z 0 0, x+ y 0 z, x+ z 0 y, y + z 0 x } " R3.
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Triangle HistogramDistributions
Circle Triangle Square
Convergence to measures . . .
=' Madeleine Kotzagiannidis
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Practical Object Recognition
• Scale-invariant feature transform (SIFT) (Lowe)
• Shape contexts (Belongie–Malik–Puzicha)
• Integral invariants (Krim, Kogan, Yezzi, Pottman, . . . )
• Shape distributions (Osada–Funkhouser–Chazelle–Dobkin)Surfaces: distances, angles, areas, volumes, etc.
• Gromov–Hausdor! and Gromov-Wasserstein distances (Memoli)=' lower bounds & stability