Candidacy Exam Talk

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

April 10, 2013
Why do we care about this object?
Kolmogorov
Complexity
Group
Theory
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 I
Lexicographic 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=12
No. of dimensions=d=3

Co-basis(N) and Basis(B)
d-subset of slack variables that are 0={ 9,10,11}: Co-basis
Remaining 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 Method
Overview
● 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)
●Tree formed by tracing all possible paths
of 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)=4
P(4,10)
R(11)=5
p(5,11)
R(12)=9
P(9,12)
R(12)=8
P(8,12)
R(11)=5
P(5,11)
R(10)=4
P(4,10)
R(12)=6
P(6,12)
R(11)=6
P(6,11)
R(10)=7
P(7,10)
R(9)=8
P(8,9)
R(11)=8
P(8,11)
R(7)=6
P(6,7)
R(9)=5
P(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)=0
R(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 II
Double 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
Consider a DD pair:
Insert new constraint:

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 III
Convex 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( )

Finding the first d+1 points of
projection

projection

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
?
?
?

CHM
Not a facet of

CHM

Updating the current hull
to include new extreme
point of projection
updatehull( )

CHM
Existing hull
New Vertex

Updating hull via iteration of DD
Method
Homogenization Polar
DD Iteration
Polar Again
Reverse
Homogenization
Old Hull
New Hull

CHM

Runtime Comparison

Demonstration

Questions

Vertices of

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