Cheeger Sets for Unit Cube: A
Analytical and
Numerical Solutions for L' and £2 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
OF TEOHNOLOG
EA 02008
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Cheeger Sets for Unit Cube : Analytical and Numerical
Solutions for L' and L 2 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 numerical work on the Cheeger set for the unit cube (Q E [-0.5,0.5]3 E R3 ) 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 problem. 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 R2
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
2
3
1.1
Cheeger sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.2
Existing work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
Extension of work for R2
11
norm . . . . . . . . . . . . . . . . . . . .
13
norm . . . . . . . . . . . . . . . . . . . .
14
norm . . . . . . . . . . . . . . .
17
2.1
Simple extension for the L'
2.2
Simple extension for the
2.3
A more intuitive approach for the
£2
£2
19
Numerical Solutions
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
Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
3.4
4
7
Introduction
3.4.1
Results for
. . . . . . . . . . . . . . . . . . . . . . . . . .
26
3.4.2
Results for £2 . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
c
Analytical Solutions
4.1
31
Analytical Solution for the L£
norm
. . . . . . . . . . . . . . . . . .
31
4.1.1
General Shape . . . . . . . . . . . . . . . . . . . . . . . . . . .
31
4.1.2
Optimization
33
. . . . . . . . . . . . . . . . . . . . . . . . . . .
4
. . . . . .
34
Analytical Solution for the L2 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
Similarities between Lo and L2 Cheeger sets . . . . . . . . . . . . . .
42
4.1.3
4.2
4.3
5
Comparison of Numerical and Analytical Solutions
45
Conclusion
5.1
Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
46
List of Figures
1-1
Optimal shapes for problem (1.1) for S E R 2 using different norms
1-2
Optimal shapes for problem (1.2) for the unit square
2-1
Extending optimal shape for £'
.
8
. . . . . . . . .
9
. . . . . . . . . . . . . . . . .
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 R 2 . . . .
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
. . . . . . . . .
36
4-5
General shape for
. . . . . . . . . . . . . . . . . . . . . . . . . . .
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
L2
4-10 Two approaches for
.
norm
f2 . . . . . . . . . . .
. . . . . . .
42
. . . . . . . . . . . . . . . . . . . . . . . . .
43
4-11 The two approaches for £2 ........................
6
£2)
44
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
For R2 this means maximizing the area
fiS|
min
||S|
for
f8S1|
= L.
(1.1)
for a given perimeter 11&Sfj = L. The
optimal shape S that achieves this depends on the norm being used to measure the
perimeter.
Strang [7] uses the calculus of variations to prove a know result, that
the optimal S is a rotated ball |t(x,y)'|ID < R, in the norm dual to the norm that
defines the perimeter. Figure 1-1 shows the optimal shapes for the unconstrained
isoperimetric problem for the L1, L2 and the L
norms.
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.
This is equivalent to identifying the minimum cut 8S in a domain Q [7, 8].
(1.2)
This
important fact will be used later in Chapter 3 to convert problem (1.2) into one
7
L
4
L
4
4
27r
L
(a)
L2
(b)
£1
(c) L
Figure 1-1: Optimal shapes for problem (1.1) for S E R 2 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
D
IIDII
(1.3)
with D varying over all smooth sub-domains of Q whose boundary OD does not touch
(Q and with
II&DII
and |ID|I denoting (n - 1)- and n-dimensional Lebesgue measure
of DD and D (Kawohl [4]). Kawohl [4] proves that for Q C R 2 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 R 2 , the unique D that attains the infimum is called the Cheeger set of
Q. The Cheeger constant h(Q) provides a lower bound A, > h 2/4 for the LaplaceDirichlet 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.
8
rR
l=1
D
l=1-2r
D
l=1-2R
D
R
JD|1 = 4
|DDI = 41 ± 2,rr
(a) ,L
2
(b) L
|aDIC = 41 + 4R
(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 R 2 . As mentioned
above, Kawohl [4, 3] discusses the existence and uniqueness of the Cheeger set for
convex, non-empty Q C R 2 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 R3 )
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
DC[0,112
I|DII
IOD112
and
and
nDd
IJD~
IIDI
(1.4)
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 c 2 and the perimeters in the numerators are multiplied by c. This would give a
lower ratio in (1.4) and hence the original shape D cannot be optimal.
Another thing we notice is that the optimal D is formed by "rounding-off" / "cutting-
9
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] mentions
how the optimal shape in Figure 1-2(b) can be created by taking a isoperimetrix'
(see Figure 1-1(b)), scaling it by an optimal scaling factor a < 1 and fitting its four
pieces in the four corners of the square. We leave it to the reader to see how the
same "rule" can be applied to get Figure 1-2(a) and 1-2(c) using the isoperimetrix in
Figure 1-1(a) and 1-1(c) respectively.
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 numerically. 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.
'Term used by Busemann in [9] for the optimal shape to the problem 1.1
10
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
R3 . Unlike the R 2 case, it is not known whether the optimum set D 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
DC[-0.5,0.5]3
1DI 2
||D II
and
aDoo(2.1)
||D I|
||DI in the denominator in (2.1) is the volume of the shape D. |I&D112 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 R 3 is simply
finding the shape D C Q that minimizes the ratio of the surface area to the volume
[11].
11
From calculus we know that the surface area A
Surface area in the L' norm :
of a smooth surface E with domain R is defined as :
A
=
JJ |n(u,v)|| du dv
(2.2)
R
For the surface E defined by z = f(x, y), and using x and y in place of u and v
respectively, we have :
(X, y) &(y, z) (z, x)
( , y)' a(x, y)' (x, y)
n(x, y)=
1,
Oz Oz
19 , 09
(2.3)
= (1,
z, zY)
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
iI
max{1, Izxf, zyj} dx dy
=
R
(2.4)
R
Similarly, for the |1n(x, y) 1 in the L2 norm, we get :
A
JJIn(uv)1|2 dudv
S1+z2 + zY2 dx dy
=
(2.5)
R
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 |DI11 = ||D1|2 = ||D1|00,
and so we can use standard geometry to calculate the volume in any norm.
12
a)
Z=
+
Y
X
R 1 (z
(b) Projection on
(a) General Shape
shape for L'
Figure 2-1: Extending optim
2.1
=0.5)
R1
Simple extension for the L'
xy-plane
norm
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
(2.6)
-3a E [1, 1.5]
JxJ + Jyj + Iz i<a
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 :
z
=
f(x,y) =
(0.5
.
y)
y (x,
a - x - y
13
in R 1
(x, y) in R 2
Now the (relevant) surface area of this shape, including the planes x = 0.5 and
y = 0.5, is given by :
max{1,1 z, 1z} dx dy + fmax{, IzxI, jzy
Surface Area = 3
R1
=3
R
}
dxdy
2
max{1, 0, 0} dx dy + fmax{1,1, 1} dx dy
R1
R2
ffI dxdy
= 3J 1 dx dy +
R1
(2.7)
R2
= 3(Area of R 1 )+ Area of R 2
= 3a - a 2
-
1.5
The volume of the shape can simply be calculated using geometry as
Volume = 0.5
(1.5 - a) 2 (1.5 - a))
-
(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 (the surface area) measured in the L2 norm. We do this by extending the
14
Q(a) vs a
'
5.5
5
Minimum
1.08
1.0
1.1
1.1
1.15
1-15
1.2
1.2
lOS
1.25
1.3
1.35
1.4
1.45
(a) Q(a) vs a
(b) Optimal D for a = 1
Figure 2-2: Optimal shape for general shape (2.6)
15
1.5
(a) A playing dice
(b) Trying to replicate the corner
of dice
(c) Minimal area covering the broken corner
Figure 2-3: An unintuitive approach to extending the 12 results from R2
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 2 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
however revisit this shape in Section 4.3.
'Available at http: //www. susqu. edu/brakke/evolver/evolver . html
16
2.3
A more intuitive approach for the
j2
norm
We now present a more intuitive approach to extending the L2 results for the unit
square (see Figure 1-2(b)).
Remember that the shape in Figure 1-2(b) could be
constructed by dividing a scaled version of the unit ball in the L2 norm (a circle) and
fitting the four pieces in the four corners of the square.
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)
+ 8(Area of spherical corner)
+ 12(Area of cylindrical edge)
(2.9)
-3r2)2) + 12( 21r(1 - 2r) - 2r(1 - 2r))
= 6 + 8(-r2
2
Similarly the volume is given by :
Volume =1 + 8( 1rr3 - r3 ) + 12(1wr2 - r 2 )(1 - 2r)
6
4
(2.10)
Now we need to minimize the ratio Q(r) of surface area (2.9) to the volume (2.10)
over 0 < r < 0.5. This was done using numerical methods and the results are shown
in Figure 2-4(b). We see that r ~~0.26 minimizes Q(r) giving a ratio of 5.396778. The
optimal shape (for r ~ 0.26) is shown in Figure 2-4(c). The unit cube (transparent)
is shown overlapping the optimal shape. Again, we will find a better ratio using a
different shape in Section 4.2.2.
17
11
(a) General Shape
Q(r) vs r
5.8
5.6
5.5
5.4
0
0.5
0.1
0.15
0.2
0.
03
0.3
0.4
0.45
i 1.5
(b) Q(r) vs r
(c) Optimal Shape for r ~ 0.26
Figure 2-4: A more intuitive approach to extending the
18
£2
results from R 2
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]. E.g. for n = 3 the flow would be f= (fi(x, y, z), f 2 (x, y, z), f 3 (x,y, z)).
The conservation of flow constraints translate to equality conditions on the divergence
of the flow:
Conservation of flow
for every point in Q.
V - f= S
(3.1)
Here S represents the sources and sinks in the domain Q.
Similarly the capacity constraints translate to
Capacity constraints
19
l|f|l < C
(3.2)
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
00 = max{ifit,
if2L, 1f3}
C
for every point in Q.
Max-flow min-cut problem
3.1
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]). The max-flow problem seeks to find
the maximum amount of flow that can be sent from s to t while maintaining flow
conservation at each node and the capacity constraints on each edge (see [13, c. 6]
for more details).
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
t,f
C
(3.3)
If we take the source S to be a net unit source, then t gives the total net flow out of
Q via
f
[14, 6]. So, in this case, (3.3) is just maximizing the total amount t of flow
while satisfying (3.1) and (3.2).
20
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
from a sequence of discrete problems. Lippert [14] approximates the
problem by a (discrete) problem in linear optimization (with quadratic constraints
for the
£2
norm). Both use the same discretization step, and we use an extension of
the discretization used in [14].
We discretize
f
= (u, v, w) into a "3D grid" over the domain Q (the cube). We
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
for (i,j,k) E [0,N] x [1,N] x [1,N], (i,jk) E [1,N] x [0,N] x [1,N] and (ij,k) c
[1, N] x [1, N] x [0, N] respectively.
21
These discrete flows are subject to the flow conservation :
Ui,j,k
-
Vi,j,k
Ui-1,j,k
AX
Vi,j-,k
f,j,k
Wi,j~k
+
Ay
+
We also need the discrete flow
-
-
s
Wi,j,k-1
Az
-
Si ,j,k
(3.4)
at any given point to satisfy the capacity con-
straints
Appleton and Talbot [1] and Lippert [14] use two different ways to handle this constraint. 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.
(UiJlki Vij~k, Wij,k)
(U-1J,k, Vij,k, WiJ,k)
As Lippert points out, for our case, we are only
[141
and so we define the magnitude of the
iI,
| 7
, lf77[II, lf7f"Ii, Iff"II, If2gI} (3.5)
interested in the maximum magnitude
flow at a point as:
= max{Iif-,
~fi,j,kII
lfII, l
Now from (3.3), (3.4) and (3.5) we can define the discretized version of the continuous
22
max-flow problem as follows :
Maximize
Subject to
t
Ui,j,k
-
Ui-ljk
+
+
V ,jk - Vi,j1,k
Wi,j,k - Wij,k-1
Ay
A~X
jk
Kui-,j,k,i,j-1,kw
V (i, j,k) E [1,
with bounds
tSio,=
0
AZtiik
i,j,k-)
I1
N] 3
0 < t
-00 < Ui~k < oo, V(iJ, k) E [0, N] x [1, N] x [1, N]
-00
<V
< oo, V(i, j, k)
-00 <Wik < o, V(i, j, k)
C [1,
N] x [0, N] x [1, N]
E [1, N] x [1, N] x [0, N]
Here i(ui,j,k, Vij,k, wi,Jk)1 in the capacity constraints is to be measured in the "appropriate" 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£
Iwi,kJ~j
< 1 and so on.
norm, II(ui,J,kvi,jk,wiJ,k)WJ1
< 1 will give lui,J,k + IVi,j,kI +
This is a simple linear programming problem with linear
constraints.
Since the L2 norm is dual to itself, so for the Cheeger set in the
(Ui~J,k, Vi,,kiWiJ,k)2 < I will become u 2,
,
k
+
W2
£2
norm,
< 1 and so on. This is
a quadratically constrained problem.
3.3
Time and memory considerations
Since the problem is in 3D, so it is obvious that any discretization will result in
a large number of variables.
For the discrete problem mentioned above, we have
23
approximately 3N 3 variables (of the form
9N
3
Uij,k, Vi,j,k
constraints, for the problem in the L2 norm.
and wij,k) and approximately
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
number of variables by noting that Ui,jkk:Ui,k,j. This is because u denotes the
24
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
Thus replacing Ui,j,k by
Ui,k,j
Ujjk
=
VJ,i,k
W,k,i
W
= Uikj).
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 "Formulation2" 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.
corresponding times for the
£2
The
norm for N = 20 were a little more than 21 minutes
and less than 25 seconds; a decrease in time by a factor of 50! This can have a
25
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,
the flow fi,,k is equal to the capacity Ci,j,k
j,
k) where the magnitude of
1. So we look at only those points
with |ffij,kII > 1 - 108. This gives the approximate Cheeger set Dapp
26
using (3.5).
= {(i, j, k)
||ika||
The "outer surface" defined by this set of points is shown in
> 1 - 10-8}.
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£
3.4.2
norm.
Results for L2
Similarly for the
C2
>
>|fjkII
1 - 10-.
norm, we look at only those points with
gives the approximate Cheeger set Dapprox
{(i,j, k) :IIfi,j,kII
>
1
-
10
3
This
}. The
"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 L 2 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, k) : I IfiJ,k I
1>-- 10-} contains almost
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
the unit square. We discuss the shapes from these results in much more detail in the
next Chapter.
27
Surface for numerical solution (N = 40)
0.4
0.450.4-0.350.30.250.20.15
0.1
0.05
0
00
.
-
.0
.1
0.21
2
.1 2 02
..
0.1135
0.
4
.-
(a) Surface
0.5
Level curves for numerical solution (N = 40)
0.5
0.45 -
0.45
0.4 -
0.4
0.35 -
0.35
0.3 -
O3
S0.25
0.25
-
0.2
0.2-
0.15 -
0.15
0.1 -
0.1
0.05 -
0.05
0x
(b) Contours
Figure 3-2: Numerical results for Cheeger set for L'
28
norm
Convergence of
t
53 -
S25-
51
N
Figure 3-3: Convergence of Numerical results for Cheeger set for L2 norm
29
Surface of numerical solution (N = 50)
0.5E0.450--
0.4,.
0.35 .
0.3,
S0.25
0.2
0.15
0.1
0.05
0
00
0.1
0.1
0.2
0.3
0.2
x
0.3
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
-
S
0
0.1
0.2
0.3
0.4
0.05
.
(b) Contours
Figure 3-4: Numerical results for Cheeger set for
30
f2
norm
Chapter 4
Analytical Solutions
We now use the results from Chapter 3 to improve on the analytical solutions, discussed 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
4.1.1
Analytical Solution for the L' norm
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|
D={(x, y, z)} with
0.5, IyI
lxI+Iyl ! b, jyj+ z
IxI+|yI +z
and
0.5, Izl < 0.5
<a
b, zl +x
b
bE [0.5, 1]
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
with the part of cube it will replace. It should be noted that for b = 1, this shape
31
+
z
+z-b
x~y+z=a
x
(a)0.5
(b) General Shape in relation to
"corner" of cube
(a) General Shape
0.
xy=b
= b)
R 4 (y
R 3 (x + y + z =
a)
+x=a-05
R 1 (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)
32
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 :
z = f(x, y) =
0.5
(x,y) in R1
b- x
(x,y) in R 2
a- x- y
(x,y) in R 3
b- y
(x,y) in R 4
Now the relevant area of this surface, including the planes x = 0.5 and y = 0.5, is
given by :
f max{1,IzxI1,zy I}dx dy + 3 ff max{1,Jz
+ ff max{1, Iz.i, IzI} dx dy
Surface Area = 3
Jzj} dx dy
R2
R3
J
= 3ff max{1, 0,0} dx dy + 3
7max1, 0, 1} dx dy +
=
3f
1dx dy+3 ff1 dxdy +
R1
=
R2
II max{1, 1,
1} dx dy
R3
ff1 dxdy
R3
3(Area of R1 ) + 3(Area of R 2 ) + Area of R3
=3ab - a 2-
1.5b 2
(4.2)
33
The volume of the shape can be calculated as :
Volume = fz
dx dy + 2f z dx dy + Jfz dx dy
R1
R2
R3
= ff0.5 dx dy + 2ff b- x
dx dy +
R1
=
R2
((b - 0.5)2
-
a- x
-
y dxdy
43
R3
1(2b - a - 0.5)2)) + ((b - b2 )(a - b))
+ ( a3 + ab - 1a2
a(b2)
-
-a)
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'
bounded by :
4.381 < II
|ID|I
34
4.4495
norm is
Q(a, b)
for (a, b)
5.8
0.95-
0.9-
-
5.6
0.85-
-
5.4
0.8-
-
5.2
-1 0.75-
-
5
-
4.8
0.65-
-
4.6
0.6.
-
4.4
0.7
0.55
-
4.2
-
0.7
0
0.8
09
1
1.1
1.2
1.3
1.4
1.5
(a) Q(a, b) against a and b
(b) Optimal
(0.83,0.78)
D for (a, b)
Figure 4-2: Optimal shape for general shape (4.1)
Surfacfo
numrerical
solut on (N
=10)
02402
x
x
(b) Analytical Solution
(a) Numerical Solution
Figure 4-3: Comparison of numerical and analytical solutions for LG
35
(b) Step 2
(d) Step 4
(c) Step 3
Figure 4-4: Constructing general shape for
4.2
£2
Analytical Solution for the L 2 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
fact to construct part of the general shape shown in Figure 4-4(a). As part of the
36
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 r 1 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 r 2 and length (0.5 - r1
-
r 2 ).
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/ 8
th of a sphere formed the
corner of the general shape in Figure 2-4(a). The quarter circle with radius r 2 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, - r 2}. This is the line parallel to the z-axis which passes through
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 r 1 = 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
II&D112/ 1D||.
We ignore this issue for the calculation of the parameters for the optimal
shape and suggest an improvement later in Section 4.3.
37
(a) General Shape
(b) Problem with the corner
Figure 4-5: General shape for L2
(b) Part of "corner"
(a) Dividing the "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 cylindrical 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.
38
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 - r 1 , z = y - r 1 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 1/6 th 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 (transparent) regions is given by z2
= x -
r1. We now have all the pieces to calculate the area
and the volume of the general shape in terms of r 1 and r 2 .
39
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 J2
dx dy+ J
R1
1
8
dx dy+ J
R2
2
dx dy)
(4.4)
R3
where 17rr2 gives the area of the circular portion at the top, and
+
zi
ax
(y
Similarly the volume of the corner piece is given by
Volume of Corner = 6 (8rrr 2 +
zi dx dy + JJz1 - z 2 dx dy +
R1
R2
zi
-
z 2 dx dy)
R3
(4.5)
We do not attempt to simplify the expressions any further, and use numerical integration 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, r2 ), 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.
40
Q(ri,r 2 ) for (ri, r )
2
,7
0.3
6
5.5
0.0
0.05
0.1
0,15
(a)
0.25
Q(ri, r 2 ) against
0.4
045
,.I
4.5
r1 and r 2
(b) Optimal D for (ri,r 2 )
(0.29,0.21)
Figure 4-8: Optimal shape for L2
Now
Q(ri, r2 )
has to be minimized over valid values of r, and r 2 . This was
done numerically and the results are shown in Figure 4-8(a). We see that (ri, r 2 ) ~
(0.29, 0.21) minimizes Q(ri, r2 ) giving an optimal value of Q(ri, r2 ) = 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 r1 and r 2 is shown in Figure 4-8(b).
41
05
Y
4
(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
£2
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
£2
norm is bounded
by:
5.319 < IDH2 < 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).
42
(b) Analytical shape from Section
4.1
(a) Naive approach from Section
2.1
(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
1
in Figure 4-10(a) is acting as the "inner Cheeger set" 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
acting as the inner Cheeger set for the optimal shape in Figure 4-11(b).
'Term used by Kawohl in [3]
43
doldloffisob--
--
- L - I, -
. -
-
Iz
(p
(b) Analytical shape from Section 4.2
(a) Naive approach from Section
2.2
Figure 4-11: The two approaches for L2
44
-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 R2 , 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 problems in linear and non-linear optimization. These problems 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 constant 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
presented earlier. These optimized shapes then provided a better upper bound on
the value of the Cheeger constant for the unit cube in the two norms respectively.
45
We also mentioned how the shape used for the L2 norm was not optimal and made
a suggestion as to how it could be improved, without pursuing this improvement in
this work.
We concluded by informally presenting an algorithm (without any attempt at a
proof) that could give us the general shape of the Cheeger set of the unit cube for both
norms. We tried to relate the steps of the algorithm with the steps of the algorithm
for finding the Cheeger set for a general convex domain in R2 (see Kawohl [3]).
5.1
Future Work
We can see three possible tasks for future work.
1. Better lower bounds : Due to the huge number of variables and constraints in
the discretized (linear and non-linear optimization) problems, we were unable
to get a very high resolution solution, even after using some simple "tricks" to
reduce the size of the problem. We might be able to get a higher resolution
solution and hence a better lower bound by using existing polynomial time
algorithms for maximal flows in discrete networks (see [13, c. 7]). We might
also be able to use an approach similar to Appleton and Talbot
[1].
2. Better upper bounds for L2 norm : We mentioned how our "optimal" analytical
shape for the
£2
norm can be improved if we cover the non-convex part. One
way for this could be to use a "minimal surface" or convex hull as the covering
surface. It would be nice to have an analytical expression for such a surface,
and to see how that affects the optimal parameters and the Cheeger constant.
3. Verification of the algorithm :
In the end, we made a few observations about
how the general shapes we got could be derived using an "algorithm" similar
to the one presented by Kawohl [3]. It might be an interesting exercise to see if
this algorithm (or a variation) can be applied to more general shapes in R3 . One
could use the algorithms and results from Lachand-Robert and Oudet [11] to
get shapes that can be used for comparisons of the results from this algorithm.
46
Bibliography
[1] B. Appleton & H. Talbot. Globally minimal surfaces by continuous maximal
flows. IEEE Transactions on Pattern Analysis and Machine Intelligence, 2004.
[2] B. Kawohl. On a family of torsional creep problems. J. Reine Angew. Math.,
1990.
[3] B. Kawohl & T. Lachand-Robert. Characterization of Cheeger sets for convex
subsets of the plane. Pacific J. Math., to appear.
[4] B. Kawohl & V. Fridman. Isoperimetric estimates for the first eigenvalue of the
p-Laplace operator and the Cheeger constant. Comment. Math. Univ. Carolinae,
2003.
[5] D. Grieser. The first eigenvalue of the Laplacian, isoperimetric constants, and
the max flow min cut theorem. Archiv der Mahematik, 87:75-85, 2006.
[6] G. Strang. Maximal flow through a domain. Math. Program., pages 26:123-143,
1983.
[7] G. Strang. Maximum area with Minkowski measures of perimeter. Proc. Royal
Soc. Edinburgh, 2007.
[8] G. Strang. Maximum flows and minimum cuts in the plane. Journal of Global
Optimization, 2007.
[9] H. Busemann. The isoperimetric problem in the Minkowski plane. American J.
Math, 1949.
[10] L. Ford & D. Fulkerson. Flows in networks. Princeton University Press, 1962.
[11] T. Lachand-Robert and E. Oudet. Minimizing within convex bodies using a
convex hull method. SIAM J. on Optimization, 2005.
[12] M. Iri. Theory of flows in continua as approximation to flows in networks. Survey
of Mathematical Programming, 2:263-278, 1979.
[13] R. Ahuja, T. Magnanti & J. Orlin.
applications. Prentice Hall, 1997.
47
Network flows: Theory, algorithms and
[14] R. Lippert. Discrete approximations to continuum optimal flow problems. Studies
in Applied Mathematics, 2006.
[15] R. Sedgewick. Algorithms in C. Princeton University Press, 1962.
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