Discovering Metrics and Scale Space

Preliminary Examination

Brittany Terese Fasy

Duke University, Computer Science Department

26 July 2010

B. Fasy (Duke CS) Prelim 26 July 2010 1 / 46

Questions

1 Tight Bound: Is the inequality

|ℓ1 − ℓ2| ≤4

π· (κ1 + κ2) · F(γ1, γ2)

a tight bound for curves in Rn for n > 3?

2 Simultaneous Scale Space: What happens if multipleagents are diffusing at the same time?

3 Understanding Vq: Does there exist a stability resultfor Vq?

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Curves and Metrics

Part 1

Curves and Metrics

My Inequality

For curves γ1 and γ2,

|ℓ1−ℓ2| ≤4

π· (κ1+κ2) ·F(γ1, γ2)

Curves and Metrics

Closed Space Curves

A curve is a continuous map γi : S1 → R

v10 v5

Curves and Metrics

Closed Space Curves

A curve is a continuous map γi : [0, 1] → Rn, such that γi (0) = γi (1).

10.2 0.4 0.6 0.80

γ(0.8)γ(0.4)

γ(0.5)γ(0.6)

γ(0.7)

γ(0.1)

γ(0) = γ(1)

γ(0.9)γ(0.3)

γ(0.2)

Curves and Metrics

Closed Space Curves

A curve is a continuous map γi : I → Rn, such that γi (0) = γi (1).

10.2 0.4 0.6 0.80

γ(0.8)γ(0.4)

γ(0.5)γ(0.6)

γ(0.7)

γ(0.1)

γ(0) = γ(1)

γ(0.9)γ(0.3)

γ(0.2)

Curves and Metrics

Inscribed Polygons

Curves and Metrics

Closed Space Curves

A curve is a continuous map γi : I → Rn, such that γi (0) = γi (1).

10.6 0.850.250

0.05 0.4 0.7 0.9

γ(0.4)

γ(0.6)

γ(0.9)

γ(0.05)

γ(0.7)

γ(0.25)

γ(0.85)

γ(0) = γ(1)

mesh(P) = max0≤i<m (ti+1 − ti )

Curves and Metrics

Arc Length

ℓi = ℓ(γi ) =

0||γ′

i (t)|| dt

ℓ(P) =∑

ℓ(ej)

Curves and Metrics

Arc Length

ℓi = ℓ(γi ) =

0||γ′

i (t)|| dt

ℓ(P) =∑

ℓ(ej)

If Pk is a sequence of polygons inscribed in a smooth closed curve γ suchthat mesh(Pk) goes to zero, then

ℓ(γ) = limk→∞

ℓ(Pk).

Curves and Metrics

Curvature

Let x ∈ I .Let rx denote the radius of the best approximating circle of γi (x).Then, the total curvature is:

κi = κ(γi ) =

01/rx dx .

Curves and Metrics

Turning Angle

v10 v5

Curves and Metrics

Curvature

κi = κ(γi ) =

01/rx dx .

κ(P) =∑

Curves and Metrics

Curvature

κi = κ(γi ) =

01/rx dx .

κ(P) =∑

If Pk is a sequence of polygons inscribed in a smooth closed curve γ suchthat mesh(Pk) goes to zero, then

κ(γ) = limk→∞

κ(Pk).

Curves and Metrics

The Frechet Distance

F(γ1, γ2) = infα : S1→S1

maxt∈S1

(γ(t) − γ(α(t)))

Curves and Metrics

Man and Dog

Curves and Metrics

F(γ1, γ2) = infα : S1→S1

maxt∈S1

(γ(t) − γ(α(t)))

Curves and Metrics

F(γ1, γ2) = infα : S1→S1

maxt∈S1

(γ(t) − γ(α(t)))

If Pk and Qk are sequences of polygons inscribed in smooth closed curvesγ1 and γ2 such that mesh(Pk) and mesh(Qk) go to zero, then

F(γ1, γ2) = limk→∞

F(Pk , Qk).

Curves and Metrics

My Inequality

For curves γ1 and γ2,

|ℓ1−ℓ2| ≤4

π· (κ1+κ2) ·F(γ1, γ2)

Curves and Metrics

Two Related Theorems

[Cha62, F50] For a closed curve contained in a disk of radius r in Rn,

ℓi ≤ r · κi .

[CSE07] For two closed curves in Rn,

|ℓ1 − ℓ2| ≤2 vol(Sn−1)

vol(Sn)· (κ1 + κ2 − 2π) · F(γ1, γ2).

Curves and Metrics

Questions

|ℓ1 − ℓ2| ≤4

π· (κ1 + κ2) · F(γ1, γ2)

Scale Space

Part 2

Scale Space

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Scale Space

50 100 150 200 250 300 350 400 450 500 550

H(x , 0)

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H(x , 100)

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H(x , 1000)

Scale Space

The Gaussian

−4 −2 0 2 40

G1(x , t) =1

2πt)e− x2

Gn(x , t) =1

2πt)ne−

This is the fundamental solution to the Heat Equation:

∂t(x , t)− △ u(x , t) = 0.

Scale Space

Gaussian Convolution

Blur(y , t, h0) =

x∈Rn

Gn(x − y , t)h0(x) dy

Scale Space

Homotopy

H : M × I → R is a continuous function such that

H(x , 0) = f (x)

H(x , 1) = g(x)

Scale Space

Discrete Homotopy

H : vert(K ) × {0, 1, . . . , τ} → R is a discrete function such that

H(x , 0) = f (x)

H(x , τ) = g(x)

H(·, 0)

H(·, 1)

H(·, 2)

Scale Space

Discrete Homotopy

H : K × Iτ → R is a discrete function such that

H(x , 0) = f (x)

H(x , τ) = g(x)

and the value at a general point (x , t) ∈ M × Iτ is determined by linearinterpolation.

H(·, 0)

H(·, 1)

H(·, 2)

H(x, 0.5) = 9

Scale Space

Heat Equation Homotopy

Let f (x), g(x) : R2 → R.

Let h0(x) = f (x) − g(x).Then:

H(y , t) = ht(y) = Blur(y , t, h0) =

x∈Rn

is the solution to the heat equation with initial condition H(x , 0) = h0(x).

Scale Space

Heat Equation Homotopy

Let f (x), g(x) : R2 → R.

Let h0(x) = f (x) − g(x).Then:

H(y , t) = ht(y) = Blur(y , t, h0) =

x∈Rn

is the solution to the heat equation with initial condition H(x , 0) = h0(x).

We define the Heat Equation Homotopy as:

H(x , t) := ht(x) + g(x).

Scale Space

Questions

|ℓ1 − ℓ2| ≤4

π· (κ1 + κ2) · F(γ1, γ2)

Scale Space

Color Images

Scale Space

Color Images

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t = 100

50 100 150 200 250 300 350 400 450 500 550

t = 1000

50 100 150 200 250 300 350 400 450 500 550

Scale Space

Proportion Set

A proportion set for the RGB image is the set of pixels that have the sameratios of colors. For example, the boundaries in the following imagesdepict where 4*blue = 3*green:

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t = 100

50 100 150 200 250 300 350 400 450 500 550

t = 1000

Scale Space

Questions

|ℓ1 − ℓ2| ≤4

π· (κ1 + κ2) · F(γ1, γ2)

Vineyard Distance

Part 3

Vineyard Distance

Persistence Diagrams

A set of points in R2 that

describe the changing homologyof the sublevel sets of a function

0 1 2 3 4 5 6 7 8

0 1 2 3 4 5 6 7 80

Vineyard Distance

Stacking the Persistence Diagrams

We stack the diagrams so that Dgmp(ht) is drawn at height z = t.

Vineyard Distance

Stacking the Persistence Diagrams

Then, we match the diagrams using a linear time algorithm [CSEM].

Vineyard Distance

Vineyards

The path of an off-diagonal point is called a vine. A vine isrepresented by a function s : Iτ → R

The collection of vines is called a vineyard.

Matching of Dgmp(f ) and Dgmp(g) is obtained by looking at theendpoints of the vines.

Vineyard Distance

Another Representation of a Vineyard

[Movie]

Vineyard Distance

Total Movement in a Vineyard

For a vine s, we can compute the weighted distance traveled by a point inthe persistence diagrams. Then, we sum this distance over all vines in thevineyard V .

0ωs(t) ·

∂s(t)

∂tdt

Vq(H) =

Vineyard Distance

Total Movement in a Vineyard

For a vine s, we can compute the weighted distance traveled by a point inthe persistence diagrams. Then, we sum this distance over all vines in thevineyard V .

Ds =∑

i∈{1,2,...,τ}

ωs(i) · ||s(ti ) − s(ti−1)||∞

Vq(H) =

Vineyard Distance

Questions

|ℓ1 − ℓ2| ≤4

π· (κ1 + κ2) · F(γ1, γ2)

Vineyard Distance

Related Metrics

Let A = Dgm(f ) and B = Dgm(g).We find a bijection between A and B by minimizing some quantity, suchas:

Bottleneck Distance

Wasserstein Distance

Vineyard Distance

Bottleneck Matching

The bottleneck cost of a matching is the maximum L∞ distance betweenmatched points:

W∞(P) = max(a,b)∈P

||a − b||∞.

We seek to minimize the bottleneck distance over all perfect matchings:

W∞(A, B) = minP

{W∞(P)}.

Vineyard Distance

Wasserstein Matching

The Wasserstein cost is measures the cumulative distance as follows:

Wq(P) =

(a,b)∈P

||a − b||q∞

We seek to minimize the Wasserstein distance over all perfect matchings:

Wq(A, B) = minP

{Wq(P)}.

Vineyard Distance

Related Stability Results

We say that the matching of persistence diagrams is stable if the cost ofthe matching is bounded by some reasonable function of ||f − g ||∞.

[CSEH] The Bottleneck Distance is stable for monotone Functionsf , g : M → R.

W∞(A, B) ≤ ||f − g ||∞[CSEHM10] The Wasserstein Distance is stable for tame LipschitzFunctions with bounded degree k total persistence.

Wq(A, B) ≤ C 1/q||f − g ||1−k/q∞

Vineyard Distance

Related Stability Results

We say that the matching of persistence diagrams is stable if the cost ofthe matching is bounded by some reasonable function of ||f − g ||∞.

[CSEH] The Bottleneck Distance is stable for monotone Functionsf , g : M → R.

W∞(A, B) ≤ ||f − g ||∞[CSEHM10] The Wasserstein Distance is stable for tame LipschitzFunctions with bounded degree k total persistence.

Wq(A, B) ≤ C 1/q||f − g ||1−k/q∞

Is the Vineyard Metric stable too?

Vq(f , g) ≤ ???

Vineyard Distance

Preliminary Findings

Let A = Dgm(f ) and B = Dgm(g).

W1(A, B) ≤ V1(f , g)

W∞(A, B) ≤ V∞(f , g)

Vineyard Distance

Questions

|ℓ1 − ℓ2| ≤4

π· (κ1 + κ2) · F(γ1, γ2)

Questions

|ℓ1 − ℓ2| ≤4

π· (κ1 + κ2) · F(γ1, γ2)

50 100 150 200 250 300 350 400 450 500 550

Thank You

My adviser, Herbert Edelsbrunner

My committee: Hubert Brey, John Harer, and Carlo Tomasi

Those who read my prelim document and provided comments,including Michael Kerber and Amit Patel.

Michelle Phillips (for making the graphics of dog and person)

Everyone here!

Questions?

References

G. Donald Chakerian, An Inequality for Closed Space Curves, Pacific J. Math. 12 (1962),no. 1, 53–57.

David Cohen-Steiner and Herbert Edelsbrunner, Inequalities for the Curvature of Curves

and Surfaces, Found. Comput. Math. 7 (2007), no. 4, 391–404.

David Cohen-Steiner, Herbert Edelsbrunner, and John Harer, Stability of persistence

diagrams, 2005, pp. 263–271.

David Cohen-Steiner, Herbert Edelsbrunner, John Harer, and Yuriy Mileyko, Lipschitz

Functions have Lp-Stable Persistence, Found. Comput. Math. 10 (2010), no. 2, 127–139.

David Cohen-Steiner, Herbert Edelsbrunner, and Dmitriy Morozov, Vines and Vineyards by

Updating Persistence in Linear Time, 2006, pp. 119–126.

Istvan Fary, Sur Certaines Inegalites Geometriques, Acta Sci. Math. (Szeged) 12 (1950),no. aa, 117–124.

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