Cheeger sets for unit cube : analytical and numerical solutions for L [infinity] and L² norms

Cheeger

Sets

for Unit Cube: A

Numerical Solutions for L' and

norms

by

Mohammad Tariq Hussain

Submitted to the School of Engineering

in partial fulfillment of the requirements for the degree of

Master of Science in Computation for Design and Optimization

at the

MASSACHUSETTS INSTITUTE OF TECHNOLOGY

February 2008

@

Massachusetts Institute of Technology 2008. All rights reserved.

Author .

School of Engineering

January 18, 2008

Certified

by...

...

Gilbert Strang

Professor of Mathematics

Thesis Supervisor

-Accepted by.

_{... .. ..}

Jaime Peraire

Professor of Aeronautics and Astronautics

Codirector, Computation for Design and Optimization Program

MAss ACHSEIN

Cheeger Sets for Unit Cube : Analytical and Numerical

Solutions for

L'

and

L

_norms

by

Mohammad Tariq Hussain

Submitted to the School of Engineering

on January 18, 2008, in partial fulfillment of the

requirements for the degree of

Master of Science in Computation for Design and Optimization

Abstract

The Cheeger constant h(Q) of a domain Q is defined as the minimum value of

IIDII/|DII

with D varying over all smooth sub-domains of Q. The D that achieves

this minimum is called the Cheeger set of Q. We present some analytical and

nu-merical work on the Cheeger set for the unit cube (Q E [-0.5,0.5]3 E R

) using

the LY and the L2 norms for measuring |IDII. We look at the equivalent max-flow

min-cut problem for continuum flows, and use it to get numerical results for the

prob-lem. We then use these results to suggest analytical solutions to the problem and

optimize these shapes using calculus and numerical methods. Finally we make some

observations about the general shapes we get, and how they can be derived using an

algorithm similar to the one for finding Cheeger sets for domains in R

Thesis Supervisor: Gilbert Strang

Title: Professor of Mathematics

Acknowledgments

I would like to thank my advisor, Professor Gilbert Strang, for suggesting the problem

and for his time and guidance. I would also like to thank my friend, Faisal Kashif, for the discussions and suggestions at different stages of the thesis. Finally I like to thank my parents for being there for me and for supporting me at all times. This thesis would not have been possible without the help of all of these people.

Contents

1 Introduction 7

1.1 Cheeger sets . . . . 8

1.2 Existing work . . . . 9

2 Extension of work for

R2

2.2 Simple extension for the £2 norm . . . . 14

2.3 A more intuitive approach for the £2 norm . . . . 17

3 Numerical Solutions 19 3.1 Max-flow min-cut problem . . . . 20

3.2 Discretization of the problem . . . . 21

3.3 Time and memory considerations . . . . 23

3.3.1 Changes . . . . 24 3.3.2 Improvements . . . . 25 3.4 R esults . . . . 26 3.4.1 Results for c . . . . 26 3.4.2 Results for £2 . . . . . . . . 27 4 Analytical Solutions 31 4.1 Analytical Solution for the L£ norm . . . . 31

4.1.1 General Shape . . . . 31

4.1.3

Comparison of Numerical and Analytical Solutions . . . .

34

4.2 Analytical Solution for the

L

_{norm . . . .}

₃₆

4.2.1

G eneral Shape . . . .

36

4.2.2

O ptim ization . . . .

40

4.2.3

Comparison of Numerical and Analytical Solutions . . . .

42

4.3 Similarities between Lo and

L

_{Cheeger sets . . . .}

₄₂

5 Conclusion 45

5.1

Future W ork . . . .

46

List of Figures

1-1 Optimal shapes for problem (1.1) for S E R2 using different norms . 8

1-2 Optimal shapes for problem (1.2) for the unit square . . . . 9

2-1 Extending optimal shape for £' norm . . . . 13

2-2 Optimal shape for general shape (2.6) . . . . 15

2-3 An unintuitive approach to extending the L2 results from R2 . . . . . 16

2-4 A more intuitive approach to extending the L2 results from R2 . . . . 18

3-1 Discretization grid with flows u, v and w . . . . 21

3-2 Numerical results for Cheeger set for L' norm . . . . 28

3-3 Convergence of Numerical results for Cheeger set for L2 norm . . . . 29

3-4 Numerical results for Cheeger set for L2 norm . . . . 30

4-1 General shape (2.6) . . . . 32

4-2 Optimal shape for general shape (4.1) . . . . 35

4-3 Comparison of numerical and analytical solutions for L . . . .. 35

4-4 Constructing general shape for f2 . . . . . . . . 36

4-5 General shape for L2 . . . . 38

4-6 Corner D etails . . . . 38

4-7 Projection of corner on xy-plane . . . . 40

4-8 Optimal shape for £2 ... 41

4-9 Comparison of numerical and analytical solutions for £2) . . . . 42

4-10 Two approaches for . . . . . 43

Chapter 1

Introduction

The unconstrained isoperimetric problem for a set S is defined as maximizing the

value of ||S1I for a given value of 110S11, or equivalently minimizing |10S11 for a given

||S||.

Unconstrained isoperimetric problem

min

for

f8S1|

= L.

(1.1)

||S|

For R2 this means maximizing the area

fiS|

optimal shape S that achieves this depends on the norm being used to measure the perimeter. Strang

[7]

< R,

What happens if we remove the "constraint" of fixed ||&S1| and instead require that S lies in a domain Q? The constrained isoperimetric problem for a domain Q is defined as minimizing the ratio of 18S1 to |ISf for all possible S C Q.

Constrained isoperimetric problem min for S c Q. (1.2)

L L

4 4

27r 4

L

(a) £1 (b)

L2

Figure 1-1: Optimal shapes for problem (1.1) for S E R2 using different norms

which can be solved numerically.

1.1

Cheeger

sets

The Cheeger constant h(Q) of a domain Q is defined as:

h(Q) := infll&DII (1.3)

D

IIDII

with D varying over all smooth sub-domains of Q whose boundary OD does not touch

(Q and with

II&DII

of DD and D (Kawohl [4]). Kawohl [4] proves that for Q C R2 with Q convex and non-empty, there is a unique convex optimum D which can be computed by algebraic algorithms. Kawohl and Lachand-Robert present one such algorithm in [3].

For Q c R2, the unique D that attains the infimum is called the Cheeger set of

Q. The Cheeger constant h(Q) provides a lower bound A, > h2_{/4 for the} Laplace-Dirichlet operator on Q [7]. This is also the value to which the first eigenvalue of Ap(Q) of the p-Laplacian converges to as p -- 1 [4]. Ionescu and Lachand-Robert present an

interesting application of the Cheeger problem in landslides modeling. Appleton and Talbot [1] use the dual problem to the Cheeger problem to study image segmentation with medical applications.

rR

D l=1 D l=1-2r D l=1-2R

R

JD|1 = 4 |DDI = 41 ± 2,rr |aDIC = 41 + 4R

(a) ,L (b) L2 (c) LO"

Figure 1-2: Optimal shapes for problem (1.2) for the unit square

1.2

Existing work

There is extensive work on Cheeger sets and constants for Q C R2. As mentioned above, Kawohl [4, 3] discusses the existence and uniqueness of the Cheeger set for convex, non-empty Q C R2 and also provides an algebraic algorithm for calculating the Cheeger set for such planar Q.

In this thesis, we focus on the Cheeger set for a unit cube (Q = [-0.5, 0.5]3

C

using different measures of I|D 11, and so we are more interested in the different Cheeger sets (in different norms) for the unit square, rather than the Cheeger sets for general convex shapes. The Cheeger problem, for the unit square, then becomes

Minimize and

IOD11

IJD~

DC[0,112

I|DII

IIDI

Strang [8] presents the minimum cuts which solve the Cheeger problem for the unit square in different norms. The results are reproduced here in Figure 1-2.

The first thing we notice about the Cheeger sets above is that in all cases the optimal D touches the boundary &Q of the square. The proof for this is trivial and is reproduced here. Assume that a shape D that does not touch the boundary 8Q is optimal. This shape can be scaled by a factor c > 1 so that it now touches the

boundary. For this new shape, the areas in the denominators of (1.4) are multiplied

by c2

off' the corners of the square. This is a special case of the algorithm to find the Cheeger set for any planer convex domain Q presented by Kawohl [3]. The algorithm in [3] only considers measurement of ||OD1| in the L2 norm, and Strang

[7]

Lippert [14] used numerical methods to find a close approximation to the flow that fills the minimal cut and as a by-product of this exercise also calculated the Cheeger set (and constant) for the unit square for both the L' and the L2 norms. Lachand-Robert and Oudet [11] present a convex hull approach, a mixture of geometrical and numerical algorithms, to solve optimization problems on the space of convex functions and use their method to find the Cheeger set (in the L2 norm) of different shapes in

R3, including the cube.

In this thesis we work on the numerical and analytical solutions to the Cheeger problem (1.2) for the unit cube (Q = [-0.5, 0.5]3 C R3). We look only at the £" and

the L2 norms; the solution for the £l norm is trivial, consisting of the entire cube

similar to the results in Figure 1-2(a).

Chapter 1 provides a background of the work that formed the basis for the work in this thesis. Chapter 2 presents a simple extension of the results for the unit square (Figure 1-2). Chapter 3 mentions how the Cheeger problem can be converted to a continuum optimal flow problem, which can then be discretized and solved numeri-cally. Chapter 4 then uses the results from the numerical solutions from Chapter 3 to propose analytical solutions that approximate the numerical solutions and Chapter 5 provides a summary of the results and the conclusion.

Chapter 2

Extension of work for

R

2

We look at the constrained isoperimetric problem (1.2) for the unit cube with Q =

[-0.5, 0.5]3

c

_Q

is unique or convex, even with Q C Rn convex for n > 3. However Q convex implies that there exists at-least one convex optimum [2]. Lachand-Robert and Oudet [11] use a convex hull approach to calculate an approximation of a convex optimum when

Q C R' is convex, and call it the Cheeger set of Q. We will use the term Cheeger set

for Q

c

R3 in a similar context.

So what does the constrained isoperimetric problem and the Cheeger set translate to for the unit cube in R3? The Cheeger problem (1.4) now changes to

Minimize

1DI

and

aDoo(2.1)

DC[-0.5,0.5]3 ||D

II

I|

||DI in the denominator in (2.1) is the volume of the shape

D.

|I&D11

and |0DII|

in the numerator in (2.1) refer to the surface area of the shape D, measured in the appropriate norm. So the Cheeger problem (2.1) for Q = [-0.5,0.5]3 C R3 _{is simply}

finding the shape D C Q that minimizes the ratio of the surface area to the volume [11].

Surface area in the L' norm : From calculus we know that the surface area A of a smooth surface

E

A

JJ

|n(u,v)|| du dv

(2.2)

For the surface E defined by z = f(x, y), and using x and y in place of u and v

respectively, we have :

&(y,

z)

a(x, y)'

n(x,

y)-

(X, y)

( , y)'

(z, x)

(x, y)

(1,

For the

I1n(x,

y)

1

in

the L' norm, we have :

Iln(x,y)lKo = max{1, Izx|, zyI}

and (2.2) simplifies to :

A

JJn(u,

v)IK

du dv

R

iI

max{1, Izxf, zyj} dx dy

Similarly, for the |1n(x, y) 1 in the L2 norm, we get :

A

J

JIn(uv)1|2

dudv

R

S1+z2

R

We will use (2.4) to calculate the surface area for the analytical solutions for the L' norm. It should also be noted that for the volume, we have |DI₁

1

and so we can use standard geometry to calculate the volume in any norm.

(2.4)

(2.5)

Oz Oz

(a) General Shape

Figure 2-1: Extending optim

R1 X Y + Z= R₁ (z =0.5) (b) Projection on

shape for L' norm

2.1

Simple extension for the

L'

norm

We first look at the Cheeger set for the unit cube in the L" norm, and try to extend the results for the square (see Figure 1-2(c)). The most obvious extension to the "cutting-off' of the corners of the square is cutting off the corner of the cube resulting in an optimal shape defined by :

D={(x, y,z)} with

{lxi

0.5,

Iyj

< 0.5,

IzI

< 0.5 and

JxJ + Jyj + Iz i <a -3a E [1, 1.5]

(2.6)

Figure 2-1(a) shows Dn [0,

0.5]1,

the part of the shape in the positive octant only.

We will use the area and volume from this shape to calculate the value of a that minimizes the ratio of surface area to volume. Due to symmetry, the total volume and total surface area of the optimal shape D will both be 8 times these values and hence the ratio remains the same. Figure 2-1(b) shows the projection of the shape onto the xy-plane. We see two distinct regions, with :

(0.5

z

f(x,y)

y

(x,

R

Now the (relevant) surface area of this shape, including the planes x = 0.5 and

y = 0.5, is given by :

Surface Area = 3

max{1,1

z,

1z}

dx dy +

IzxI,

jzy

}

dxdy

R1 R₂

= 3

max{1,

0,

0} dx

dy +

fmax{1,

1,

1} dx dy

= 3J 1 dx dy +

ffI

dxdy

)

= 3a

a

1.5

The volume of the shape can simply be calculated using geometry as

Volume

0.5

(1.5

a))

(2.7)

(2.8)

Now the ratio Q(a) of the values in (2.7) and (2.8) is to be minimized over 1 < a < 1.5. This was done using numerical methods and the results are shown in Figure 2-2(a). We see that a = 1 minimizes Q(a) giving a ratio of 4.8. The optimal shape (for a = 1) is shown in Figure 2-2(b). The unit cube (transparent) is shown overlapping the optimal shape. We should mention that we will find a better ratio using a different shape in Section 4.1.2.

Note : As a side note, we tried using the L' norm (I|n(u, v)I1i instead of |!n(u, v)II1") in (2.7) and the optimal value of a turned out to be 1.5. For the general shape (2.6), this means that we get the entire cube as the optimal D. This result is similar to the

L case for the unit square (see Figure 1-2(a)).

2.2

Simple extension for the L2 norm

Now we attempt to construct the general shape of the Cheeger set for a cube with

[aDO

Q(a) vs a

1.08 1.1 1.15 1.2 lOS

(a) Q(a) vs a

(b) Optimal D for a = 1

Figure 2-2: Optimal shape for general shape (2.6)

5.5

'

Minimum

5

(a) A playing dice (b) Trying to replicate the corner (c) Minimal area

cov-of dice ering the broken corner

Figure 2-3: An unintuitive approach to extending the 12 results from R

results for the square with perimeter measured in the L2 norm (see Figure 1-2(b)).

We first present an approach which is very close to the approach applied to the L*

norm, but which results in a very unintuitive shape for the Cheeger set. Just as in

Figure 2-1(a) the corners of the cube were "cut-off', for the

norm, we can think

of using sand-paper to "round-off" the corners of a cube.

A physical example of this can be seen in playing dice (see Figure 2-3(a)). Figure

2-3(b) shows an analytical shape that tries to replicate the corner of the dice. In that

shape, the relevant corners of the 3 squares that meet at (0.5, 0.5, 0.5) have been

"rounded-off' and have been replaced by a quarter of a circle each. Notice that the

shape in Figure 2-3(b) has a "hole" in the corner that still needs to be "covered".

One approach to covering this would be to use a surface that minimizes the area.

Brakke's Surface Evolver' is a free tool used for modelling how liquid surfaces change

shape because of various forces and constraints. It can be used to find the surface

that minimizes the total surface tension (and hence the area) of a surface between a

wire-frame. Figure 2-3(c) shows the results of evolving (100 iterations) the original

corner of the cube with the 3 quarter circles as the constraints/wire-frames.

We leave this shape here and do not pursue this approach any further. We will

2.3

A more intuitive approach for the

norm

Similarly for the unit cube, we take a sphere (of radius r < 0.5), divide it into eight parts and fit the parts into the eight corners of the cube. This also requires us to "round-off" the edges of the cube by replacing them with quarter cylinders each of radius r. Figure 2-4(a) shows the general shape, again only for the positive octant.

This time finding the surface area and the volume is a much easier task, and we use simple geometry to find the two in terms of the radius r. The surface area of the complete shape is given by :

Surface Area = 6(Area of top surface)

+

(2.9)

+ 12(Area of cylindrical edge)

= 6 + 8(-r2

12( 1r(1

2r)

2r(1

2r))

Similarly the volume is given by :

Volume =1 + 8(

1rr3

6 4

Now we need to minimize the ratio

Q(r)

Q(r)

11

(a) General Shape

Q(r) vs r

0.5 0.1 0.15 0.2 0. 0 3 1.5

(b) Q(r) vs r

(c) Optimal Shape for r ~ 0.26

Figure 2-4: A more intuitive approach to extending the £2 results from R2 0.3 0.4 0.45 i

Chapter 3

Numerical Solutions

Continuum flow problems are the continuous analog of the discrete network flow problems. In the discrete case, an undirected network consists of a set N of nodes and a set E of edges. Each edge has a positive capacity c,, > 0 and a feasible flow

through the edge cannot exceed the capacity of the edge. In addition to this we have conservation of flow at each node, with "flow-in=flow-out". We generally also have one or more "source" and one or more "sink" nodes. A flow through the network is defined as a set of flows on the edges. A feasible flow will fulfill the capacity constraints on edges as well as the flow conservation for each node including the source and the sink nodes.

In the continuum case, instead of being defined as a set of nodes and edges, the "network" is defined as a closed set Q C R'. The flows are given by vector fields in Q [12, 14, 6,

8].

n

f=

f

The conservation of flow constraints translate to equality conditions on the divergence of the flow:

Conservation of flow V - f= S (3.1)

for every point in Q. Here S represents the sources and sinks in the domain Q. Similarly the capacity constraints translate to

for every point in Q. Here C represents the capacity over the domain Q, and 11ffl represents the magnitude of the flow in the appropriate norm. E.g. for n = 3, if the magnitude of the flow is measured in the L& norm, (3.2) changes to

Illo1

=

max{ifit,

1f3}

for every point in Q.

3.1

Max-flow min-cut problem

For the general (discrete) network defined at the start of this chapter, let s and t be the source and sink nodes respectively (it is easy to show that a network with more than one source or sink nodes can be converted to a network with exactly one source node and exactly one sink node, e.g. see

[15]).

Suppose we divide the nodes in the network into two disjoint sets S and T with

S E S, t E T and S U T = N. The capacity of this cut is defined as the sum

of the capacities of the edges which cross it. The min-cut problem seeks to find, among all possible s-t cuts, the cut with the minimum value. Ford and Fulkerson [10] demonstrated that the maximal s-t flow in a network equals the minimal s-t cut in the network.

Strang [6, 8] and Iri [12] describe the maximal flow and the minimal cut problems for continuous flows. For the continuous case the maximum flow problem is

max t: V - f= tS and 1ifi C (3.3)

t,f

If we take the source S to be a net unit source, then t gives the total net flow out of Q via

f

w - -

1-V

U

-Figure 3-1: Discretization grid with flows u, v and w

Iri [12] showed that, under very general continuity assumptions, the maximal flow is strictly equal to the minimal surface. For such a flow and surface, the flow saturates the surface uniformly. Strang [7, 8] shows how, for a net unit source S, the minimal cut is the same as the Cheeger set for Q and Grieser [5] mentions that the Cheeger constant h(Q) (see (1.3)) is equal to the optimal t that satisfies (3.3).

3.2

Discretization of the problem

Appleton and Talbot [1] have proposed an algorithm for computing the maximum flow vector

f

the discretization used in [14].

We discretize

f

divide the cube into N equal parts along the 3 axis. This gives (N + 1)3 distinct nodes. The flow is now considered along the edge between two consecutive nodes (see Figure 3-1). This leads to the primary variables Ui,j,k, Vi,j,k and Wi,j,k defined

These discrete flows are subject to the flow conservation :

Ui,j,k - Ui-1,j,k Vi,j,k - Vi,j-,k Wi,j~k - Wi,j,k-1

s

AX + Ay + Az - Si ,j,k

(3.4)

We also need the discrete flow f,j,k at any given point to satisfy the capacity

con-straints

Appleton and Talbot [1] and Lippert [14] use two different ways to handle this con-straint. The magnitude of the discrete flow at any given point can be obtained by considering the following approximations to the flow near a vertex.

f,7PP=

fJ7

to participate in the algorithm. As Lippert points out, for our case, we are only

interested in the maximum magnitude

[141

~fi,j,kII

Iff"II, If2gI}

max-flow problem as follows :

Maximize t

Subject to Ui,j,k - Ui-ljk + V ,jk - Vi,j1,k + Wi,j,k - Wij,k-1 tSio,= 0

A~X Ay AZtiik

j k Kui-,j,k,i,j-1,kw i,j,k-) I 1

V

(i, j,k) E [1,

with bounds 0 < t

-00 < Ui~k < oo, V(iJ, k) E

[0, N] x [1, N] x [1, N]

C

[1,

[1, N]

Here i(ui,j,k, Vij,k, wi,Jk)1 in the capacity constraints is to be measured in the

"ap-propriate" norm. Since finding the Cheeger set of a domain Q is identical to finding the minimum cut of Q, and the maximum flow problem is the dual of the minimum cut problem, so the "appropriate" norm for the flow is actually the dual of the norm used to find the Cheeger set (see [6]).

This means that for the Cheeger set in the L' norm, we will need to use the dual norm, the L1 norm, in the capacity constraints in the discretized problem. So, for the Cheeger set in the L£ norm, II(ui,J,kvi,jk,wiJ,k)WJ1

<

Iwi,kJ~j

constraints.

Since the L2 norm is dual to itself, so for the Cheeger set in the £2 norm,

(Ui~J,k, Vi,,kiWiJ,k)2 < I will become u 2, , k + W2 < 1 and so on. This is

a quadratically constrained problem.

approximately 3N3 _{variables (of the form Uij,k, Vi,j,k and wij,k) and approximately}

9N 3 constraints, for the problem in the L2 norm. However for the L' norm, the number of variables and constraints is larger because of the type of constraints that the Cplex/MOSEK input file formats allows. This means that even for a small N like 100, we have about 3 million variables and about 9 million constraints! It is certainly not feasible to try solving such huge problems even on commercial software

like Cplex.

3.3.1

Changes

Below we mention how we use the symmetry of the problem to make some simple modifications to the problem, making it smaller for any given N (unfortunately, the size of the problem still grows as O(N)). These changes apply to the problem in either norm.

1. The problem is symmetric about the x-, y- and the z-axis and so the first and

most obvious modification is to look at only the problem in the positive octant (like we have done in the numerical case). This does require us to define the boundary conditions, and due to the symmetry of the problem, we can see that along the z-axis, the flow in the z direction has to be zero (and similarly for the other two axes). This immediately reduces the value of N, required for any given resolution of the solution, to half, thus reducing the number of variables and constraints by a factor of 8.

2. The problem is also symmetric in the x, y and z directions, and so we can

replace Vi,j,k by Uji,k and Wi,j,k by Ukj,,. This reduces the number of variables

by a factor of 3. We call this formulation "Formulation 1" in the comparisons

below. We did not try to solve any formulations of the problem without the first two changes.

3. Now, we have only the variables in u. For these we can further decrease the

flow in the x direction and so u is symmetric in the y and z directions (also, using the relations in the previous point, we see that Ujjk = VJ,i,k W,k,i W = Uikj).

Thus replacing Ui,j,k by Ui,k,j Vj > k, we can further reduce the number of

variables by a factor of about 2.

4. The final step involves removing all the redundant constraints produced because of the replacements of variables above. This step does increase the time of the generation of the Cplex input file, but that is countered by the fact that this allows us to load and solve larger problems (meaning larger N) in Cplex. We refer to this formulation as "Formulation 2" below.

3.3.2

Improvements

The changes mentioned in the previous section lead to improvements in the following areas of the problem solution.

As mentioned above, these changes directly reduce the number of variables by a factor of 6, and the number of constraints by a factor of about 5.5. E.g. for the L2 problem with N = 50, Formulation 1 results in a total of more than 1 million

constraints, which are reduced to less than 0.2 million for Formulation 2. This directly effects the size of the input file that is generated by a similar factor. E.g The size of the generated input file for the L2 problem for N = 50, dropped from around 85MB to only 15MB when switching from Formulation 1 to Formulation 2.

The difference in file size then has a much larger effect on the time it takes for Cplex to load/read the problem. E.g. for the L2 problem with N = 50, the time it took Cplex to read the problem from the input file jumped from close to 5 minutes for Formulation 1, to under 10 seconds for Formulation 2.

Finally, there was also a gain in the time Cplex took to actually solve the problems.

E.g for the £L problem for N = _{30 the solution time dropped from just under 5}

minutes to 1 minute when switching from Formulation 1 to Formulation 2. The corresponding times for the norm for N =

considerable effect for larger N, keeping in mind that for N = 50 , Cplex took slightly less than 45 minutes to solve the L2 problem. This was for Formulation 2; for the same problem in Formulation 1, Cplex gave an out-of-memory error even before starting any iterations for the calculation of the solution.

Note : All stats mentioned above for the problem in the Ll norm are for the barrier

method in Cplex. The simplex method was extremely inefficient compared to the barrier method. E.g. for N = 20 (using Formulation 2), the Cplex simplex method took more than 12 minutes to find the optimal solution. The barrier method, however took less than 6 seconds to solve the same problem.

3.4

Results

The linear and quadratically constrained problems defined above were solved using MOSEK and Cplex. Due to the large number of variables and constraints (even after the improvements mentioned above), we use relatively small values of N and so the results do not have a very high resolution.

Any algorithm that solves a maximal flow problem for a discrete network, also gives the minimum cut since the "max flow saturates the minimum cut". So once we have a maximum flow, then the minimum cut is simply the set of edges that have flow equal to their capacity. The same applies to the continuous maximum flow problem and we use this to find the Cheeger set (=minimum cut) from the results of the optimization problem defined in Section 3.2.

3.4.1

Results for

L'

Solving the discretized version of the continuous maximum flow problem gives values for flow variables Ujk, Vi,j,k and Wi,j,k. Based on these values we find

fk

Ideally speaking, we are interested only in points (i,

j,

the flow

fi,,k

Ci,j,k

{(i,

j,

||ika||

Figure 3-2. The value of the objective function t, i.e. the maximum flow, is 4.38114966 (for N = 40).

We should note that any feasible flow gives a lower bound on the value of the Cheeger constant. This is because this is the maximum flow for (the discretized version of) the continuous max-flow problem, which is the dual problem to finding the minimum cut. Therefore this is actually just a lower bound on the value for the Cheeger constant for the unit cube in the L£ norm.

3.4.2

Results for L2

Similarly for the C2 norm, we look at only those points with > >|fjkII 1 - 10-. This

gives the approximate Cheeger set Dapprox {(i,j, k) :I

Ifi,j,kII

}.

"outer surface" defined by this set of points is shown in Figure 3-4. The value of the objective function t, i.e. the maximum flow, is 5.3187700759 (for N = 50). Again,

this is actually just a lower bound on the value of the Cheeger constant for the unit cube in the L2 _norm.

We believe that this lower bound is far from tight. The optimal t will converge to the Cheeger constant as we increase N, but N = 50 is too small a value to give a very

good approximation to the Cheeger constant. This can also be seen from the fact that the approximate Cheeger set Dapprox =

{(i, j,

all of the points "inside" the surface shown in Figure 3-4(a) and so the minimum cut (consisting of all saturated edges) is not a good approximation to the surface &D of the optimal shape. However we look at only the "outer surface" made by the points to approximate the Cheeger set. Also Figure 3-3 shows how the optimal value of t, the maximum flow, changes with N. We can clearly see that at N = 50, the lower

bound defined by the maximal flow is still rising.

The first thing that is to be noticed about the two shapes is that none of them is identical to the shapes proposed in Chapter 2, which simply extended the results for

Surface for numerical solution (N = 40) 0.4 0.45- 0.4-- 0.35- 0.3- 0.25- 0.2-0.15 0.1 0.05 0 0 0 . -.0 0.21.1 2 .1 2 02 .. 0.1135 0. .-4 (a) Surface

Level curves for numerical solution (N = 40)

x

(b) Contours

Figure 3-2: Numerical results for Cheeger set for L' norm

0.5 0.45 -0.4 -0.35 -0.3 -S0.25 - 0.2-0.15 0.1 0.05 - 0-0.5 0.45 0.4 0.35 O 3 0.25 0.2 0.15 0.1 0.05

Convergence of t

53

-

S25-51

N

Surface of numerical solution (N = 50) 0.5 E- 0.450--0.4,. 0.35 . 0.3, S0.25 0.2 0.15 0.1 0.05 0 00 0.1 0.1 0.2 0.2 0.3 0.3 x 0.4 0.4 0. 0.5 (a) Surface

Fig -Level r curves for numerical solution (N =50)

0.45- 0.45 0.4-- 0.4 0.35- 0.35 0.3- 0.3 A 0.25 -0.25 0.2- 0.2 0.15- 0.15 0.1 - 0.1 0.05 - 0.05 S 0 0.1 0.2 0.3 0.4 . (b) Contours

Chapter 4

Analytical Solutions

We now use the results from Chapter 3 to improve on the analytical solutions,

dis-cussed in Chapter 2, for the Cheeger problem (2.1) for the unit cube. We first present

the general shape and then the optimized shape for both the L' and the L2 norms.

As in Chapter 2, we only look at D n [0, 0.5]3, the part of the optimal shape in the

positive octant.

4.1

Analytical Solution for the L' norm

4.1.1

General Shape

We start with the shape for the LO norm as it is much simpler than the shape for the

£2

norm. It is obvious from the numerical results (see Figure 3-2), that the Cheeger

set for the unit cube in the L* norm is defined by:

x|

0.5, IyI

0.5,

0.5

and

D={(x, y, z)} with

lxI+Iyl

!

b,

jyj+ z

b,

zl +x

b

bE [0.5, 1]

IxI

+|yI +z

<

a

3a E [b,2b - 0.5]

(4.1)

Figure 4-1(a) shows the general shape, and Figure 4-1(b) shows the general shape

+ z

+z-b

x~y+z=a

(a)0.5 x

(a) General Shape

0.

(b) General Shape in relation to

"corner" of cube R4 (y = b) xy=b R3 (x + y + z = a) +x=a-05 R1 (z = 0.5) R2 ( + z =b) x (a- b b- 0.5 0.5 (c) Projection on xy-plane

Figure 4-1: General shape (2.6)

simplifies to the general shape discussed in Section 2.1 (see Figure 2-1(a)).

4.1.2

Optimization

Figure 4-1(c) shows the projection of the shape on the xy-plane. We see the following

distinct regions :

f(x, y)

0.5

b - x

(x,y)

in R

(x,y) in

(x,

y) in

Now the relevant area of this surface, including the planes x

=

0.5 and y

=

0.5, is

given by :

Surface Area = 3

f

IzxI1,zy I}

3

ff

max{1,

Jz

R2

+

ff

max{1,

Iz.i,

IzI}

dx

dy

R3

max{1,

0,

0}

dx dy + 3

J

7max1, 0, 1} dx dy +

II

_{max{1, 1,}

_{dx dy}

=

3f

1 dx dy+3

ff1

dxdy +

ff1

dxdy

R1 R2 R3

=

3(Area of R

_{) + 3(Area of R}

_{) + Area of R}

3

=3ab

-

(4.2)

The volume of the shape can be calculated as :

Volume

=

fz

dx dy + 2f z dx dy + Jfz dx dy

ff0.5

_{+ 2ff}

_b

_x

a

x

y dxdy

((b

1(2b

a

+ ((b

b

_b))

+

(

a3 + ab

1a2

a(b2)

Some of the above expressions were calculated using Mathematica; we don't attempt

to simplify the expression for the volume any further since we will use numerical

methods to find the optimal value of Q(a, b), the ratio of surface area to volume. The

results are shown in Figure 4-2(a). The white region in the figure consists of invalid

pairs of (a, b) (see (4.1)).

We see that (a, b) ~ (0.83,0.78) minimizes Q(a, b) giving an optimal value of

P 4.4495 which is an improvement over the optimal value of 4.8 for the general shape

in Figure 2-1(a). The optimal shape corresponding to these values of a and b is shown

in Figure 4-2(b).

4.1.3

Comparison of Numerical and Analytical Solutions

Figure 4-3 shows the numerical and the analytical Cheeger sets (shown only for the

positive octant) for the L' norm. As we mentioned before, the maximal flow of

- 4.381 found in Section 3.4.1 gives a (not so tight) lower bound on the value of the

Cheeger constant, and the value calculated in Section 4.1.2 gives an upper bound.

So we can say that the solution to the Cheeger problem (2.1) for the L' norm is

bounded by :

4.381

<

II

4.4495

Q(a, b) for (a, b) 0.95- 5.8 0.9- - 5.6 0.85- - 5.4 0.8- - 5.2 -1 0.75- - 5 0.7 - - 4.8 0.65- - 4.6 0.6. - 4.4 0.55 - 4.2 0 0.7 0.8 09 1 1.1 1.2 1.3 1.4 1.5

(a) Q(a, b) against a and b

(b) Optimal D for (a, b) (0.83,0.78)

Figure 4-2: Optimal shape for general shape (4.1)

Surfacfo numrerical solut on (N =10)

024-02

x x

(b) Step 2

(c) Step 3 (d) Step 4

Figure 4-4: Constructing general shape for £2

4.2

Analytical Solution for the

L

_norm

Now we come to the analytical solution for the £2 norm. The analytical expression for the general shape is not obvious from the numerical results (see Figure 3-4). So we first present the rationale behind the suggested general shape and then present the results of the optimization.

4.2.1

General Shape

Constructing the General Shape

The first thing that we notice about the optimal numerical solution (see Figure 3-4) is the 3 "circular" regions on the x = 0.5, y = 0.5 and z = 0.5 planes. We use this

first step, we replace the three squares that intersect at the point (0.5, 0.5, 0.5), by a smaller "squares", each with a quarter circle of radius r1 in one corner (see top

surface in Figure 4-4(a)). The 3 (visible) edges of the original cube are replaced by (quarter) cylinders of radius r2 and length (0.5 - r1 - r2). This gives us the shape in Figure 4-4(a).

Now all we need to do is to find a surface that covers the "hole" we see in Figure 4-4(a), and whose normal is parallel to the normals of all surfaces it meets (so the resulting figure is smooth). We focus on this "hole" in Figure 4-4(b) and this will be referred to as the "corner" of the optimal shape just as 1/8th of a sphere formed the

corner of the general shape in Figure 2-4(a). The quarter circle with radius r₂ marks the area where the quarter cylinders (that replace the edges) meet the "corner".

Next, we take this quarter circle (on the surface parallel to the xz-plane), where the cylinder meets the "corner", and rotate it about the line defined by {(x, y, z) :

x = y = 0.5 - r, - r2

}.

the center of the quarter circle (of radius ri) on the top surface of the cube. Figure 4-4(c) shows the resulting figure.

Finally we repeat the last step symmetrically for the other two sides, to get the shape shown in Figure 4-4(d). This step does result in some overlap of the 3 surfaces created by the rotations, and this will have to be taken into account when calculating the surface area and the volume of the resulting figure.

We have now finished the construction of the corner. This gives us the completed shape shown in Figure 4-5(a). It should be noted that for r1 = 0, this shape simplifies

to the general shape discussed in Section 2.3 (see Figure 2-4(a)).

At this point we should also point out that this shape is not a perfect choice for the Cheeger set, as the view of the "corner" in Figure 4-5(b) shows. We see that the resulting figure is not convex and that by covering the "dent" in the shape, we can decrease the surface area and increase the volume thus getting a smaller value for

(a) General Shape (b) Problem with the corner

Figure 4-5: General shape for L2

(a) Dividing the "corner" (b) Part of "corner"

Figure 4-6: Corner Details

Analytical Expression for General Shape

We can think of the general shape as consisting of three planar surfaces, three

cylin-drical edges and a corner. Finding the contributions that the first two make to the

surface areas and the volumes is simple enough; geometry will give us the answers

easily. The corner is not that simple because of the overlapping mentioned above. So

before we begin the calculations of the parameters of the optimal shape, we would

like to present a method of breaking up the corner into 6 identical pieces, with each

piece being defined by surfaces with analytical expressions. We can then use calculus

to find the surface area and the volume of one piece and multiply these by 6 to get

the surface area and the volume for the corner.

Note : While describing the analytical shape for the corner, and also while cal-culating the surface area and volume of the corner, we will use the shape in Figure 4-4(d) as a reference, ignoring the fact that it is actually not positioned at the origin in the general shape (see Figure 4-4(a)). This means that we will use the corner opposite the curved surface (clearly visible in Figure 4-4(b)) as the origin, and all lengths/equations will be with reference to that. It might be a little difficult for the reader to remember this as he reads the rest of this chapter, but it simplifies the expressions a lot and we sacrifice readability over simplicity.

Figure 4-6(a) shows how the corner piece can be broken into 3 identical parts, such that in each part we have only one rotated surface to deal with. This is done by

cutting the corner with parts of the planes z = x - r1, z = y - r1 and x = y. Figure

4-6(b) then shows the details of the "top" piece (of the three parts in Figure 4-6(a)). The reader should note that this is exactly the surface we got in Figure 4-4(c) by the first of 3 rotations. The transparent part represents the portion of this piece that is covered by the other two rotated surfaces and so does not participate in the surface area or the volume.

Finally, we divide this top piece into two equal parts by the plane x = y. This gives us exactly 1/6th of the shape we refer to as the "corner" of the optimal shape.

Now the equation of the curved surface (not including the circular area on the top) is given by :

z 2 + (V/X2 + y2 - rl)2=r2

The equation of the plane dividing the valid (non-transparent) and invalid (transpar-ent) regions is given by z2 = x - r1. We now have all the pieces to calculate the area

R₁ R₂ R₃

Figure 4-7: Projection of corner on xy-plane

4.2.2

Optimization

Figure 4.2.2 shows the projection of the corner on the xy-plane. The surface area of

the corner (Figure 4-4(d)) is then given by :

Surface Area of Corner = 6

(1irr2+J

R1 R2 R3

1

where 17rr2 gives the area of the circular portion at the top, and

8

+

zi

ax

(y

Similarly the volume of the corner piece is given by

Volume of Corner =

6

+

+

R1 R2 R3

(4.5)

We do not attempt to simplify the expressions any further, and use numerical

integra-tion to find the values. Based on these expressions for surface area and the volume of

the corner piece, we can find the expression for Q(ri, r

), the ratio of the total surface

area to total volume. The contribution to the surface area and the volume due to

the cylindrical edges and the planar surfaces are simple to calculate and we do not

present them here.

Q(ri, r2) for (ri, r2)

0.0

0.05 0.1 0,15 0.4 045

5.5

4.5

(a)

Q(ri,

(b) Optimal D for (ri, r₂)

(0.29,0.21)

Figure 4-8: Optimal shape for L2

Now

Q(ri,

r

) has to be minimized over valid values of r, and r

.

This was

done numerically and the results are shown in Figure 4-8(a). We see that (ri, r

) ~

(0.29, 0.21) minimizes Q(ri, r

) giving an optimal value of Q(ri, r

) = 5.3815 which is

slightly better than the optimal value for general shape in Figure 2-4(a). The optimal

shape corresponding to these values of r

and

is shown in Figure 4-8(b).

,7

0.3

6

,.I

05

4 Y

(a) Numerical Solution (b) Numerical Solution

Figure 4-9: Comparison of numerical and analytical solutions for L2)

4.2.3

Comparison of Numerical and Analytical Solutions

Figure 4-9 shows the numerical and the analytical Cheeger sets (shown only for the

positive octant) for the

norm. As we mentioned before, the maximal flow of

~ 5.319 found in Section 3.4.2 gives a (not so tight) lower bound on the value of the

Cheeger constant, and the value calculated in Section 4.2.2 gives an upper bound. So

we can say that the solution the Cheeger problem (2.1) for the

norm is bounded

by:

5.319

<

IDH

< 5.382

4.3

Similarities between

L'

and

L2

Cheeger sets

In the end we mention some observations about the shape of the analytical solutions.

Figure 4-10(a) shows the general shape based on the simple extension of the results

for the unit square. We got this shape by "cutting-off' the corner of the cube just like

the Cheeger set for the square (in the L norm) has the corners of the square cut-off.

Another way of looking at the same shape is to think of each of the 6 squares (that

form the cube) being replaced by (the general shape of) its Cheeger set (remember

that the figure only shows the first octant). This step results in "holes" where the

corners of the cube used to be. We cover these holes with a "minimal surface" to get

the shape in Figure 4-10(a).

(a) Naive approach from Section 2.1

(b) Analytical shape from Section

4.1

(c) Unit ball for L£ in R3

Figure 4-10: Two approaches for L'

Now if we "surround" this shape (similar to Kawohl's algorithm [3] for convex shapes in R2) with (a scaled) unit ball in the dual norm (shown in Figure 4-10(c)), we get the general shape in Figure 4-10(b). So (a scaled version of) the simple shape in Figure 4-10(a) is acting as the "inner Cheeger set"1 for the Cheeger set in Figure

4-10(b).

Applying the same approach to the L2 norm, the first step, of replacing each side of the cube with the (general shape of) its Cheeger set, gives us the shape in Figure 4-11(a), the shape we saw in Section 2.2. If we ignore the covering-of-the-hole step, and "surround" this shape by the unit ball in the dual norm (which is a sphere), we get exactly the shape in Figure 4-11(b), the shape we suggested and optimized in Section 4.2 above. If however we first cover the hole with a "minimal surface" (maybe as in Figure 2-3(c)), we will get a shape similar to the Figure 4-11(b), but without the non-convex portion (see Figure 4-5(b)). Again, the simple shape in Figure 4-11(a) is

(a) Naive approach from Section 2.2

(p

(b) Analytical shape from Section 4.2

Figure 4-11: The two approaches for L2

doldloffisob-- -- - L - I , - - . - -Z dw

Chapter 5

Conclusion

In this thesis we presented work on the Cheeger sets for the unit cube using two

different measures of surface area. We first looked at the existing work for Cheeger sets

(and Cheeger constants) for domains in R

, specially the square. We then presented

an approach which extended the results for the Cheeger sets for the square to the

Cheeger sets for a cube. These general shapes were then optimized using calculus

and numerical methods. The optimized shapes gave us an upper bound on the value

of the Cheeger constant.

After this, we looked at the continuous maximum flow and minimum cut problems,

and how these relate to Cheeger sets and Cheeger constants for general domains. This

continuous network optimization problem was then discretized and converted to

prob-lems in linear and non-linear optimization. These probprob-lems were then solved using

commercial mathematical programming optimizers and the results were presented.

These results not only gave us some lower bounds on the value of the Cheeger

con-stant for the unit cube, but also gave an approximation to the general shape of the

Cheeger set for each of the two norms.

These results were then used to present a different set of analytical shapes for

the two norms. These shapes were again optimized using calculus and numerical

methods, and the results were found to be better than the ones for the simpler shapes