Candidacy Exam Talk

April 10, 2013

Polyhedral Computationfor Characterization of Region of Entropic Vectors

and Computation of Rate Regions of Coded Networks

Jayant ApteASPITRG

April 10, 2013

Introduction

April 10, 2013

Why do we care about this object?

Kolmogorov Complexity

GroupTheory

Network Coding

Combinatorics

Probability Theory

Quantum Mechanics

Matrix Theory

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

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

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)

7Jayant Apte. ASPITRGApril 10, 2013

Convex Polyhedron


Examples of polyhedra

Bounded- Polytope Unbounded - polyhedron


H-Representation of a Polyhedron


V-Representation of a Polyhedron


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


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


Polyhedral Cone


A cone in


Homogenization


H-polyhedra


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


Example


V-polyhedra


Polar of a convex cone





H-representation V-representation

H-representationV-representation

Original space Polar/dual space


Equivalence of vertex-enumeration and facet-enumeration



Perform Vertex Enumeration on this cone.



Then take polar again to get facets of this cone

Perform Vertex Enumeration on this cone.


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


Minimality of H-representation• Magenta inequality can be removed

without changing the polyhedron• Magenta inequality is redundant


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





The red points are redundant


Algorithm ILexicographic Reverse Search


Lexicographic Reverse Search

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

Lexicographic Simplex Method


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)


Add slack variables

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


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


Co-basis(N) and Basis(B)

(0,0,1)

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


Degeneracy

(0,0,1)

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


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)


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)


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)


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)


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)


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)


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)


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)


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)


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


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


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)


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)


Problems with pivoting methods

● Degeneracy● Duplicate output of extreme points


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


Algorithm IIDouble Description Method


Definitions


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.


How it works?


Example


Example


Example

Consider a DD pair:

Insert new constraint:


Example


Example


Example


Example


Compute new rays(DD Lemma)


New DD pair


New cone


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


Algorithm IIIConvex Hull Method


Polyhedral Projection


Example


CHM intuition (12,6,6)

(12,6)


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


Finding the first d+1 points of projection

initialhull( )














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


CHM

?

?

?


CHMNot a facet of


CHM


CHM


Updating the current hull to include new extreme

point of projectionupdatehull( )


CHM

Existing hull

New Vertex


CHM

Existing hull

New Vertex


Updating hull via iteration of DD Method

Homogenization Polar

DD Iteration

Polar Again

ReverseHomogenization

Old Hull

New Hull


CHM


CHM


CHM


Runtime Comparison


Demonstration


Questions


Vertices of


Vertices of


Vertices of


Vertices of


Vertices of

Technology

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