candidacy exam talk

99
April 10, 2013 Polyhedral Computation for Characterization of Region of Entropic Vectors and Computation of Rate Regions of Coded Networks Jayant Apte ASPITRG

Upload: jayant-apte

Post on 08-Jul-2015

80 views

Category:

Technology


0 download

TRANSCRIPT

Page 1: Candidacy Exam Talk

April 10, 2013

Polyhedral Computationfor Characterization of Region of Entropic Vectors

and Computation of Rate Regions of Coded Networks

Jayant ApteASPITRG

Page 2: Candidacy Exam Talk

April 10, 2013

Introduction

Page 3: Candidacy Exam Talk

April 10, 2013

Why do we care about this object?

Kolmogorov Complexity

GroupTheory

Network Coding

Combinatorics

Probability Theory

Quantum Mechanics

Matrix Theory

Page 4: Candidacy Exam Talk

April 10, 2013

Region of entropic vectors and Network Coding

● Achievable Information Rate Region of multi-source network coding problem is the set of all possible rates at which multiple information sources can be multicast simultaneously on a network

● Most general of all network coding problems● Implicit characterization in terms of region of

entropic vectors is available

Page 5: Candidacy Exam Talk

April 10, 2013

Where does polyhedral computation come into picture?

● Finding better polyhedral inner and outer bounds on the region of entropic vectors

● Finding the the Achievable Information Rate Region of multi-source network coding problem by substituting in these better inner and outer bounds in place of exact region of entropic vectors in the implicit characterization.

● Both the problems above become problems of polyhedral computation

Page 6: Candidacy Exam Talk

April 10, 2013

Outline

● Background on Polyhedra● Representation Conversion

– Lexicographic Reverse Search

– Double Description Method

● Polyhedral Projection– Convex Hull Method(As implemented in chm0.1)

Page 7: Candidacy Exam Talk

7Jayant Apte. ASPITRGApril 10, 2013

Convex Polyhedron

Page 8: Candidacy Exam Talk

8Jayant Apte. ASPITRGApril 10, 2013

Examples of polyhedra

Bounded- Polytope Unbounded - polyhedron

Page 9: Candidacy Exam Talk

9Jayant Apte. ASPITRGApril 10, 2013

H-Representation of a Polyhedron

Page 10: Candidacy Exam Talk

10Jayant Apte. ASPITRGApril 10, 2013

V-Representation of a Polyhedron

Page 11: Candidacy Exam Talk

11Jayant Apte. ASPITRGApril 10, 2013

Representation conversion

● Given the H-representation of a polyhedron, compute V-representation: vertex enumeration

● Given the V-representation of a polyhedron, compute the H-representation: facet enumeration

Page 12: Candidacy Exam Talk

12Jayant Apte. ASPITRGApril 10, 2013

Example

(1,0,0)

(0,0,0)

(0,1,0)

(1,1,0)

(0,1,1)

(0.5,0.5,1.5)(1,1,1)

(0,0,1)

H-rep V-rep

Page 13: Candidacy Exam Talk

13Jayant Apte. ASPITRGApril 10, 2013

Polyhedral Cone

Page 14: Candidacy Exam Talk

14Jayant Apte. ASPITRGApril 10, 2013

A cone in

Page 15: Candidacy Exam Talk

15Jayant Apte. ASPITRGApril 10, 2013

Homogenization

Page 16: Candidacy Exam Talk

16Jayant Apte. ASPITRGApril 10, 2013

H-polyhedra

Page 17: Candidacy Exam Talk

17Jayant Apte. ASPITRGApril 10, 2013

Example(d=2,d+1=3)

Page 18: Candidacy Exam Talk

18Jayant Apte. ASPITRGApril 10, 2013

Example

Page 19: Candidacy Exam Talk

19Jayant Apte. ASPITRGApril 10, 2013

V-polyhedra

Page 20: Candidacy Exam Talk

20Jayant Apte. ASPITRGApril 10, 2013

Polar of a convex cone

Page 21: Candidacy Exam Talk

21Jayant Apte. ASPITRGApril 10, 2013

Polar of a convex cone

Page 22: Candidacy Exam Talk

22Jayant Apte. ASPITRGApril 10, 2013

Polar of a convex cone

H-representation V-representation

H-representationV-representation

Original space Polar/dual space

Page 23: Candidacy Exam Talk

23Jayant Apte. ASPITRGApril 10, 2013

Equivalence of vertex-enumeration and facet-enumeration

Page 24: Candidacy Exam Talk

24Jayant Apte. ASPITRGApril 10, 2013

Equivalence of vertex-enumeration and facet-enumeration

Perform Vertex Enumeration on this cone.

Page 25: Candidacy Exam Talk

25Jayant Apte. ASPITRGApril 10, 2013

Equivalence of vertex-enumeration and facet-enumeration

Then take polar again to get facets of this cone

Perform Vertex Enumeration on this cone.

Page 26: Candidacy Exam Talk

26Jayant Apte. ASPITRGApril 10, 2013

Minimality of H-representation

● If an inequality can be removed from an H-representation of a polyhedron without changing the polyhedron, then that inequality is said to be redundant.

● An H-representation is minimal if there are no redundant inequalities

Page 27: Candidacy Exam Talk

27Jayant Apte. ASPITRGApril 10, 2013

Minimality of H-representation• Magenta inequality can be removed

without changing the polyhedron• Magenta inequality is redundant

Page 28: Candidacy Exam Talk

28Jayant Apte. ASPITRGApril 10, 2013

Minimality of V-representation

● If an extreme point/extreme ray can be removed from a V-representation of a polyhedron without changing the polyhedron, then that extreme point/extreme ray is said to be redundant.

● A V-representation is minimal if there are no redundant extreme points/extreme rays

Page 29: Candidacy Exam Talk

29Jayant Apte. ASPITRGApril 10, 2013

Minimality of V-representation

Page 30: Candidacy Exam Talk

30Jayant Apte. ASPITRGApril 10, 2013

Minimality of V-representation

The red points are redundant

Page 31: Candidacy Exam Talk

31Jayant Apte. ASPITRGApril 10, 2013

Algorithm ILexicographic Reverse Search

Page 32: Candidacy Exam Talk

32Jayant Apte. ASPITRGApril 10, 2013

Lexicographic Reverse Search

● A pivoting algorithm● Based on variant of Simplex Method called

Lexicographic Simplex Method

Page 33: Candidacy Exam Talk

33Jayant Apte. ASPITRGApril 10, 2013

A linear program

(1,0,0)

(0,0,0)

(0,1,0)

(1,1,0)

(0,1,1)

(0.5,0.5,1.5)(1,1,1)

(0,0,1)

(1,0,1)

Page 34: Candidacy Exam Talk

34Jayant Apte. ASPITRGApril 10, 2013

Add slack variables

No. of variables=n=12No. of dimensions=d=3

Page 35: Candidacy Exam Talk

35Jayant Apte. ASPITRGApril 10, 2013

Co-basis(N) and Basis(B)d-subset of slack variables that are 0={ 9,10,11}: Co-basisRemaining n-d variables can be grouped together: Basis

Page 36: Candidacy Exam Talk

36Jayant Apte. ASPITRGApril 10, 2013

Co-basis(N) and Basis(B)

(0,0,1)

d-subset of slack variables that are 0={ 7,9,11}

Page 37: Candidacy Exam Talk

37Jayant Apte. ASPITRGApril 10, 2013

Degeneracy

(0,0,1)

Vertex (0,0,1) has more than one co-bases It is called a degenerate extreme point

Page 38: Candidacy Exam Talk

38Jayant Apte. ASPITRGApril 10, 2013

Lexicographic Simplex MethodOverview

● Simplex Method maximizes/minimizes a linear objective function over a polytope/polyhedron

● Uses dictionary as a primary data structure: Every basis-cobasis pair has a dictionary corresponding to it

● Choose entering basis using least subscript rule. If none is found, we've reached optimum

● Choose leaving the basis and going into co-basis using lexicographic pivot selection rule. If none is found, problem is unbounded

● Obtain the next dictionary corresponding to new basis-cobasis pair by doing the pivot operation denoted as pivot(r,s)

Page 39: Candidacy Exam Talk

39Jayant Apte. ASPITRGApril 10, 2013

Lexicographic simplex on our example

(1,0,0)

(0,0,0)

(0,1,0)

(1,1,0)

(0,1,1)

(0.5,0.5,1.5)(1,1,1)

(0,0,1)

(1,0,1)

Page 40: Candidacy Exam Talk

40Jayant Apte. ASPITRGApril 10, 2013

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

P(10,4)

P(12,8)

P(11,5)

V=(1 0 1)N=(4 11 8)

V=(1 1 1)N=(4 5 8)

P(r,s): pivot(r,s)

Page 41: Candidacy Exam Talk

41Jayant Apte. ASPITRGApril 10, 2013

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

P(10,4)

P(12,8)

P(11,5)

V=(1 0 1)N=(4 11 8)

V=(1 1 1)N=(4 5 8)

P(11,5)

P(10,4)

V=(0 1 0)N=(10 5 12)

V=(1 1 0)N=(4 5 12)

P(r,s): pivot(r,s)

Page 42: Candidacy Exam Talk

42Jayant Apte. ASPITRGApril 10, 2013

P(12,6)

V=(0 1 1)N=(10 5 6)

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

P(10,4)

P(12,8)

P(11,5)

V=(1 0 1)N=(4 11 8)

V=(1 1 1)N=(4 5 8)

P(11,5)

P(10,4)

V=0 1 0)N=(10 5 12)

V=(1 1 0)N=(4 5 12)

P(r,s): pivot(r,s)

Page 43: Candidacy Exam Talk

43Jayant Apte. ASPITRGApril 10, 2013

P(9,5)

V=(1 1 1)N=(6 8 5)

P(12,6)

V=(0 1 1)N=(10 5 6)

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

P(10,4)

P(12,8)

P(11,5)

V=(1 0 1)N=(4 11 8)

V=(1 1 1)N=(4 5 8)

P(11,5)

P(10,4)

V=0 1 0)N=(10 5 12)

V=(1 1 0)N=(4 5 12)

P(7,6)

V=(0.5 0.5 1.5)N=(6 8 9)

P(11,8)

V=(0.5 0.5 1.5)N=(7 8 9)

P(10,7)V=(0 0 1)N=(7 11 9)

P(12,9)

V=(0 0 1)N=(10 11 9)

P(r,s): pivot(r,s)

Page 44: Candidacy Exam Talk

44Jayant Apte. ASPITRGApril 10, 2013

P(9,5)

V=(1 1 1)N=(6 8 5)

P(12,6)

V=(0 1 1)N=(10 5 6)

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

P(10,4)

P(12,8)

P(11,5)

V=(1 0 1)N=(4 11 8)

V=(1 1 1)N=(4 5 8)

P(11,5)

P(10,4)

V=0 1 0)N=(10 5 12)

V=(1 1 0)N=(4 5 12)

P(7,6)

V=(0.5 0.5 1.5)N=(6 8 9)

P(11,8)

V=(0 0 1)N=(7 8 9)

P(10,7)V=(0 0 1)N=(7 11 9)

P(12,9)

V=(0 0 1)N=(10 11 9)

P(9,8)

V=(1 0 1)N=(8 12 9)

P(r,s): pivot(r,s)

Page 45: Candidacy Exam Talk

45Jayant Apte. ASPITRGApril 10, 2013

P(9,5)

V=(1 1 1)N=(6 8 5)

P(12,6)

V=(0 1 1)N=(10 5 6)

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

P(10,4)

P(12,8)

P(11,5)

V=(1 0 1)N=(4 11 8)

V=(1 1 1)N=(4 5 8)

P(11,5)

P(10,4)

V=0 1 0)N=(10 5 12)

V=(1 1 0)N=(4 5 12)

P(7,6)

V=(0.5 0.5 1.5)N=(6 8 9)

P(11,8)

V=(0 0 1)N=(7 8 9)

P(10,7)V=(0 0 1)N=(7 11 9)

P(12,9)

V=(0 0 1)N=(10 11 9)

P(9,8)

V=(1 0 1)N=(8 12 9)

P(11,6)

V=(0 1 1)N=(10 6 9)

P(r,s): pivot(r,s)

Page 46: Candidacy Exam Talk

46Jayant Apte. ASPITRGApril 10, 2013

P(9,5)

V=(1 1 1)N=(6 8 5)

P(12,6)

V=(0 1 1)N=(10 5 6)

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

P(10,4)

P(12,8)

P(11,5)

V=(1 0 1)N=(4 11 8)

V=(1 1 1)N=(4 5 8)

P(11,5)

P(10,4)

V=0 1 0)N=(10 5 12)

V=(1 1 0)N=(4 5 12)

P(7,6)

V=(0.5 0.5 1.5)N=(6 8 9)

P(11,8)

V=(0 0 1)N=(7 8 9)

P(10,7)V=(0 0 1)N=(7 11 9)

P(12,9)

V=(0 0 1)N=(10 11 9)

P(9,8)

V=(1 0 1)N=(8 12 9)

P(11,6)

V=(0 1 1)N=(10 6 9)

P(r,s): pivot(r,s)

Page 47: Candidacy Exam Talk

47Jayant Apte. ASPITRGApril 10, 2013

P(9,5)

V=(1 1 1)N=(6 8 5)

P(12,6)

V=(0 1 1)N=(10 5 6)

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

P(10,4)

P(12,8)

P(11,5)

V=(1 0 1)N=(4 11 8)

V=(1 1 1)N=(4 5 8)

P(11,5)

P(10,4)

V=0 1 0)N=(10 5 12)

V=(1 1 0)N=(4 5 12)

P(7,6)

V=(0.5 0.5 1.5)N=(6 8 9)

P(11,8)

V=(0 0 1)N=(7 8 9)

P(10,7)V=(0 0 1)N=(7 11 9)

P(12,9)

V=(0 0 1)N=(10 11 9)

P(9,8)

V=(1 0 1)N=(8 12 9)

P(11,6)

V=(0 1 1)N=(10 6 9)

P(r,s): pivot(r,s)

●Tree formed by tracing all possible pathsof simplex method

●Reverse the direction of edges to get the reverse search tree

Page 48: Candidacy Exam Talk

48Jayant Apte. ASPITRGApril 10, 2013

ЯEVERSE Search

1. Start with dictionary corresponding to optimum vertex

2. Let current basis be B

3. For a certain and any is there a valid simplex pivot from dictionary corresponding to to the current dictionary?

4. Denoted as reverse(s), for and returns if answer is yes else returns 0

5. If do pivot(r,s), go down the reverse search tree by recursively performing 2-5

6. If reverse(s) returns 0 for all go back 1 level up the tree using ordinary simplex pivot

Page 49: Candidacy Exam Talk

49Jayant Apte. ASPITRGApril 10, 2013

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

R(10)=4P(4,10)

R(11)=5p(5,11)

R(12)=9P(9,12)

R(12)=8P(8,12)

R(11)=5P(5,11)

R(10)=4P(4,10)

R(12)=6P(6,12)

R(11)=6P(6,11)

R(10)=7P(7,10)

R(9)=8P(8,9)

R(11)=8P(8,11)

R(7)=6P(6,7)

R(9)=5P(5,9)

V=(1 0 1)N=(4 11 8)

V=0 1 0)N=(10 5 12)

V=(0 0 1)N=(10 11 9)

V=(1 1 0)N=(4 5 12)

R(5)=0

R(4)=0

R(11)=0

R(8)=0

R(9)=0

V=(0 1 1)N=(10 5 6)

V=(0.5 0.5 1.5)N=(7 8 9)

V=(1 0 1)N=(8 12 9)

V=(1 1 1)N=(6 8 5)

V=(1 1 1)N=(5 6 9)

V=(0 1 1)N=(10 6 9)

V=(0 0 1)N=(7 11 9)

R(9)=0

R(5)=0 R(6)=0 R(9)=0

R(7)=0

R(8)=0 R(12)=0

R(9)=0

V=(0.5 0.5 1.5)N=(6 8 9)

R(6)=0 R(8)=0 R(5)=0

R(6)=0R(8)=0 R(8)=0 R(12)=0 R(9)=0

R(4)=0 R(11)=0

R(4)=R(5)=R(12)

R(10)=R(5)=R(6)

R(s): reverse(s)P(r,s): pivot(r,s)

Page 50: Candidacy Exam Talk

50Jayant Apte. ASPITRGApril 10, 2013

V=(0 0 0)N=( 10 11 12)

V=(1 0 0)N=(4 11 12)

R(10)=4P(4,10)

R(11)=5p(5,11)

R(12)=9P(9,12)

R(12)=8P(8,12)

R(11)=5P(5,11)

R(10)=4P(4,10)

R(12)=6P(6,12)

R(11)=6P(6,11)

R(10)=7P(7,10)

R(9)=8P(8,9)

R(11)=8P(8,11)

R(7)=6P(6,7)

R(9)=5P(5,9)

V=(1 0 1)N=(4 11 8)

V=0 1 0)N=(10 5 12)

V=(0 0 1)N=(10 11 9)

V=(1 1 0)N=(4 5 12)

R(5)=0

R(4)=0

R(11)=0

R(8)=0

R(9)=0

V=(0 1 1)N=(10 5 6)

V=(0.5 0.5 1.5)N=(7 8 9)

V=(1 0 1)N=(8 12 9)

V=(1 1 1)N=(6 8 5)

V=(1 1 1)N=(5 6 9)

V=(0 1 1)N=(10 6 9)

V=(0 0 1)N=(7 11 9)

R(9)=0

R(5)=0 R(6)=0 R(9)=0

R(7)=0

R(8)=0 R(12)=0

R(9)=0

V=(0.5 0.5 1.5)N=(6 8 9)

R(6)=0 R(8)=0 R(5)=0

R(6)=0R(8)=0 R(8)=0 R(12)=0 R(9)=0

R(4)=0 R(11)=0

R(4)=R(5)=R(12)

R(10)=R(5)=R(6)

R(s): reverse(s)P(r,s): pivot(r,s)

Page 51: Candidacy Exam Talk

51Jayant Apte. ASPITRGApril 10, 2013

Problems with pivoting methods

● Degeneracy● Duplicate output of extreme points

Page 52: Candidacy Exam Talk

52Jayant Apte. ASPITRGApril 10, 2013

How Lexicographic Simplex deals with them

● Degeneracy– Lexicographic Simplex Method visits only a subset of

bases called Lex-positive Bases

● Duplicate output extreme points– Out of the lex-positive basis we can identify a unique basis

called Lex-min Basis corresponding to each extreme point

– Output extreme point only if current basis is lex-min

● These features make Lexicographic simplex best choice for reverse search

Page 53: Candidacy Exam Talk

53Jayant Apte. ASPITRGApril 10, 2013

Algorithm IIDouble Description Method

Page 54: Candidacy Exam Talk

54Jayant Apte. ASPITRGApril 10, 2013

Definitions

Page 55: Candidacy Exam Talk

55Jayant Apte. ASPITRGApril 10, 2013

Double Description Method:The High Level Idea

● An Incremental Algorithm

● Starts with certain subset of rows of H-representation of a cone to form initial H-representation

● Adds rest of the inequalities one by one constructing the corresponding V-representation every iteration

● Thus, constructing the V-representation incrementally.

Page 56: Candidacy Exam Talk

56Jayant Apte. ASPITRGApril 10, 2013

How it works?

Page 57: Candidacy Exam Talk

57Jayant Apte. ASPITRGApril 10, 2013

Example

Page 58: Candidacy Exam Talk

58Jayant Apte. ASPITRGApril 10, 2013

Example

Page 59: Candidacy Exam Talk

59Jayant Apte. ASPITRGApril 10, 2013

Example

Consider a DD pair:

Insert new constraint:

Page 60: Candidacy Exam Talk

60Jayant Apte. ASPITRGApril 10, 2013

Example

Page 61: Candidacy Exam Talk

61Jayant Apte. ASPITRGApril 10, 2013

Example

Page 62: Candidacy Exam Talk

62Jayant Apte. ASPITRGApril 10, 2013

Example

Page 63: Candidacy Exam Talk

63Jayant Apte. ASPITRGApril 10, 2013

Example

Page 64: Candidacy Exam Talk

64Jayant Apte. ASPITRGApril 10, 2013

Compute new rays(DD Lemma)

Page 65: Candidacy Exam Talk

65Jayant Apte. ASPITRGApril 10, 2013

New DD pair

Page 66: Candidacy Exam Talk

66Jayant Apte. ASPITRGApril 10, 2013

New cone

Page 67: Candidacy Exam Talk

67Jayant Apte. ASPITRGApril 10, 2013

Minimality of representation

● New ray AD generated above is redundant● What to do?

– Generate new rays for only those positive-negative ray pairs that are adjacent

– Can check adjacency using either

combinatorial adjacency oracle or algebraic adjacency oracle

● Prevents combinatorial explosion of number of extreme rays

Page 68: Candidacy Exam Talk

68Jayant Apte. ASPITRGApril 10, 2013

Algorithm IIIConvex Hull Method

Page 69: Candidacy Exam Talk

69Jayant Apte. ASPITRGApril 10, 2013

Polyhedral Projection

Page 70: Candidacy Exam Talk

70Jayant Apte. ASPITRGApril 10, 2013

Example

Page 71: Candidacy Exam Talk

71Jayant Apte. ASPITRGApril 10, 2013

CHM intuition (12,6,6)

(12,6)

Page 72: Candidacy Exam Talk

72Jayant Apte. ASPITRGApril 10, 2013

How it works...

● If projection dimension=d, first find d+1 extreme points of projection and their convex hull using procedure called initialhull()

● Initialhull() gives us first approximation of projection ● Every iteration find one new extreme point of projection

and compute convex hull corresponding to pre-existing extreme points and the new extreme point

● We stop when all the facets of current approximation are facets of

Page 73: Candidacy Exam Talk

73Jayant Apte. ASPITRGApril 10, 2013

Finding the first d+1 points of projection

initialhull( )

Page 74: Candidacy Exam Talk

74Jayant Apte. ASPITRGApril 10, 2013

Finding the first d+1 points of projection

Page 75: Candidacy Exam Talk

75Jayant Apte. ASPITRGApril 10, 2013

Finding the first d+1 points of projection

Page 76: Candidacy Exam Talk

76Jayant Apte. ASPITRGApril 10, 2013

Finding the first d+1 points of projection

Page 77: Candidacy Exam Talk

77Jayant Apte. ASPITRGApril 10, 2013

Finding the first d+1 points of projection

Page 78: Candidacy Exam Talk

78Jayant Apte. ASPITRGApril 10, 2013

Finding the first d+1 points of projection

Page 79: Candidacy Exam Talk

79Jayant Apte. ASPITRGApril 10, 2013

Finding the first d+1 points of projection

Page 80: Candidacy Exam Talk

80Jayant Apte. ASPITRGApril 10, 2013

Fact

● The cost functions for finding the extreme points of projection can be obtained from facets of that are not the facets of

● Checking whether a facet of is a facet of can be accomplished by simply running a linear program over

Page 81: Candidacy Exam Talk

81Jayant Apte. ASPITRGApril 10, 2013

CHM

?

?

?

Page 82: Candidacy Exam Talk

82Jayant Apte. ASPITRGApril 10, 2013

CHMNot a facet of

Page 83: Candidacy Exam Talk

83Jayant Apte. ASPITRGApril 10, 2013

CHM

Page 84: Candidacy Exam Talk

84Jayant Apte. ASPITRGApril 10, 2013

CHM

Page 85: Candidacy Exam Talk

85Jayant Apte. ASPITRGApril 10, 2013

Updating the current hull to include new extreme

point of projectionupdatehull( )

Page 86: Candidacy Exam Talk

86Jayant Apte. ASPITRGApril 10, 2013

CHM

Existing hull

New Vertex

Page 87: Candidacy Exam Talk

87Jayant Apte. ASPITRGApril 10, 2013

CHM

Existing hull

New Vertex

Page 88: Candidacy Exam Talk

88Jayant Apte. ASPITRGApril 10, 2013

Updating hull via iteration of DD Method

Homogenization Polar

DD Iteration

Polar Again

ReverseHomogenization

Old Hull

New Hull

Page 89: Candidacy Exam Talk

89Jayant Apte. ASPITRGApril 10, 2013

CHM

Page 90: Candidacy Exam Talk

90Jayant Apte. ASPITRGApril 10, 2013

CHM

Page 91: Candidacy Exam Talk

91Jayant Apte. ASPITRGApril 10, 2013

CHM

Page 92: Candidacy Exam Talk

92Jayant Apte. ASPITRGApril 10, 2013

Runtime Comparison

Page 93: Candidacy Exam Talk

93Jayant Apte. ASPITRGApril 10, 2013

Demonstration

Page 94: Candidacy Exam Talk

94Jayant Apte. ASPITRGApril 10, 2013

Questions

Page 95: Candidacy Exam Talk

95Jayant Apte. ASPITRGApril 10, 2013

Vertices of

Page 96: Candidacy Exam Talk

96Jayant Apte. ASPITRGApril 10, 2013

Vertices of

Page 97: Candidacy Exam Talk

97Jayant Apte. ASPITRGApril 10, 2013

Vertices of

Page 98: Candidacy Exam Talk

98Jayant Apte. ASPITRGApril 10, 2013

Vertices of

Page 99: Candidacy Exam Talk

99Jayant Apte. ASPITRGApril 10, 2013

Vertices of