CALIFORNIA STATE UNIVERSITY SAN MARCOS

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THESIS SUBMITTED IN PARTIAL FULLFILLMENT OF THE REQUIREMENTS FOR THE DEGREE

MASTER OF SCIENCES

IN

MATHEMATICS

THESIS TITLE: Communication Networks

AUTHOR: Janina Silvana P. Patno

DATE OF SUCCESSFUL DEFENSE: April 19,2011

THE THESIS HAS BEEN ACCEPTED BY THE THESIS COMMITTEE IN PARTIAL FULLFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCES IN MATHEMATICS.

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Thesis Abstract

We consider the problem of sending a message in a communication network.

Our networks are modeled by graphs. In our graphs we want to send a message from a vertex s, the sender, to a vertex r, the receiver, via paths from s tor.

One difficulty we may encounter if we try sending the message along every possible s, r-path is that the receiver r will be flooded, so we focus our attention on certain finite protocols for sending the message from s to r. We study the probability that such a protocol successfully sends a message from s to r if the edges of the network live with probability p. We present some results and some open problems on this topic.

Keywords: network, communication, protocol, two-terminal graph, two-terminal reliability polynomial ii

Contents

1 Introduction 1 1.1 Introduction . 1 1.2 Basic Definitions 4

2 The Deterministic Model 11 2.1 s, r-Graphs ...... 11 2.2 Protocols for Message Transmission . 11 2.3 A Forbidden Minor Characterization 17

3 The Probabilistic Model 22 3.1 Reliability of a Protocol ...... 22 3.2 Reliability of the Complete Forwarding Protocol 26 3.3 Reliability and Deficiency of a Partial Forwarding Protocol ...... 38

4 Maximum Protocol Reliability 49 4.1 Protocols with Reliability p 49 4.2 Using Deficiencies to find p . 50

5 Comparing Graphs 69 5.1 Path and Cycle Comparison . . . . . 70 5.2 Comparing Polynomials ...... 75 5.3 Comparing the 10 Forbidden Minors 79 5.4 The Shortest s, r-Paths ...... 80

6 Kelmans' Type Construction 87 6.1 Definitions ...... 87 6.2 Operations With s, r-Graphs . 88 6.3 Multiplying Constructions . . 94 6.4 Reliable Pairs with Prescribed Crossing Profiles 96 CONTENTS iii

7 A Graph with a Piecewise Polynomial for p 100 100 701 Graph B 0 With Edge Replaced 702 Open Problems 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 108 1

Chapter 1

Introduction

1.1 Introduction

A network is an "interconnected system of things or people" [14] and networks, such as the internet, enable people to share information with one another. Com­ munication in a network has been studied by many researchers in many different forms and under different assumptions, but the goal is typically to determine the reliability of such networks.

In 1959, during the Cold War, an engineer named Paul Baran was interested in survivable communications and felt that a more reliable system of communica­ tion would help prevent unnecessary attacks during the war. Baran's first attempt with such a communication system would try to send a "Go" or "No go" message from the president to commanders via AM radio. After three years of development, he was able to create a better communication system that would send longer mes­ sages with better survivability using the idea of packet switching and distributive communication in a network [1]. CHAPTER 1. INTRODUCTION 2

In this thesis we represent networks as graphs, where the vertices are the nodes or processors and the edges are links between processors. Thus, Baran's idea for a communication network would have edges and vertices that can fail indepen­ dently without recovering, and there would be multiple distinguished vertices, the terminals, that send and receive information. The model we study can be described with graphs with no multiple edges or loops such that the vertices do not fail, but the edges can fail independently without recovering. We focus our attention on the case where we have only two specified terminals, a sender and a receiver.

Paraphrasing Douglas West (see [13], page 149), in order for a network to be able to reliably transmit information, it needs to stay connected even when some of the edges or vertices fail. So one way to study the reliability of a network is to study the number of walks with working edges between terminals. We study the reliability of a network Gin terms of the reliability polynomial, p(G,p), which is the probability that some walk remains between every two terminals if all edges can fail independently with probability p. One can study this connectivity question for any number of terminals k, where k ~ 2. So given a graph on n vertices one can look at the all terminal ( k = n) or k-terminal reliability polynomial. We will primarily study the two-terminal case.

In 1992, Brown and Colbourn [4] calculated the all terminal reliability poly­ nomial for several families of graphs, such as trees and cycles, and studied the roots and coefficients of the reliability polynomial for a connected graph. Colbourn [6] CHAPTER 1. INTRODUCTION 3 recognized that constraints on a network force restrictions on the coefficients of the reliability polynomial and that these restrictions can be more noticeable by writing the reliability polynomial in different forms.

Our study of communication in a network combines Baran's idea of packet switching in network transmission with the idea of a reliability polynomial to arrive at the concept of a protocol reliability polynomial. Like Colbourn, we calculate the reliability of specific networks and interpret our findings by writing the reliability polynomial in one or two variables.

In Chapter 2, we review the deterministic model of the network as studied by Fich, Kiindgen, Pelsmajer and Ramamurthi [8, 9]. We describe our model as a graph with a fixed sender s and a fixed receiver r, and discuss how messages are sent in our model using protocols. We also describe the main result from [8], a forbidden minor characterization. In Chapter 3, we change the model described in Chapter 2, by attaching a probability to the existence of every edge. We study how well a protocol will work in our model by determining its reliability. Chapter 4 discusses which protocol will give us the best reliability for each of the 10 forbidden minors from Chapter 2. Chapter 5 compares graphs of paths and cycles to determine which would work better as a communication network. We also give a result about polynomials that can be used to compare the reliabilities of protocols in different graphs. In Chapter 6, we use an idea from Kelmans [10] to construct a pair of graphs such that if their reliability polynomials were graphed, they would cross a CHAPTER 1. INTRODUCTION 4 specified number of times. We end the thesis with Chapter 7, in which we use our findings of paths and cycles from Chapter 5 to show that there are graphs such that the best protocol to use at low probability is different from that at high probability.

We present a number of fundamental results and some open problems on this topic.

1.2 Basic Definitions

In this section we review some basic graph theory, probability, and algebra defini­ tions and theorems we will be using in this thesis. For a more comprehensive graph theory review please refer to West [13] and Diestel [7]. A comprehensive probability review can be found in Pitman [11].

A graph G consists of a non-empty finite vertex set V(G), and an edge multiset E(G), where each edge is an unordered pair of two distinct vertices called the endpoints. If vertex vis an endpoint of edge e, then v and e are incident. When u and v are the endpoints of an edge, they are adjacent and are neighbors, and we will denote this edge by uv.

A complete graph is a graph whose vertices are pairwise adjacent.

A subgraph H of G is a graph such that the V(H) ~ V(G) and E(H) C

E(G). His said to be induced if E(H) consists of all edges in E(G) whose endpoints are both in V(H).

A walk of length k is a finite sequence v0 , v1 , v2 , ... , Vk-l, vk of vertices such that vi-l vi, for all i with 1 :<::; i :<::; k, is an edge in G. A trail is a walk such that if vi = Vj for i =/= j, then vi+l =/= Vj+l· Thus, each edge uv can be used at most twice CHAPTER 1. INTRODUCTION 5 in a trail: once as u, v and once as v, u. A walk (trail) is closed if the endpoints, v0 and vk, of the walk (trail) are the same. A path is a walk Vo, v1, v2, ... , Vk-1, vk such that v0, v1, ... , vk-l, vk are distinct vertices of G. Given two specified vertices u and v, au, v-walk (u, v-trail, u, v-path) is a walk (trail, path) vo, v1, ... , vk that starts at v0 = u and ends at vk = v. The length of a path is its number of edges, k. A walk contains a path (trail) if the sequence of vertices describing the path

(trail) is a subsequence of the sequence of vertices describing the walk, and every pair of consecutive vertices in the sequence describing path (trail) also occurs as a consecutive pair in the sequence describing the walk.

Proposition 1.2.1. Every s, r-walk contains an s, r-path.

Proof. Given an s, r-walk W0 = (s = v0 ), v1, v2, ... , Vk-1, (vk = r). If vi = Vj, fori < j, then let wl = (s = vo),vl,v2, ... ,vi-l,Vj,Vj+l, ... ,Vk-1, (vk = r). So vi, vi+l, ... , Vj-b v1 is replaced by the vertex v1 in the new walk W1. We continue in this manner until we obtain as, r-walk W1 with distinct vertices. Wz is the desired s, r-path.

D

A graph is connected if for every u, v E V(G), G contains au, v-path. The components of a graph G are its maximal connected subgraphs. A cut-edge or cut-vertex of a graph is an edge or vertex whose deletion increases the number of components. CHAPTER 1. INTRODUCTION 6

A cycle is a graph with an equal number of vertices and edges whose vertices can be placed around a circle so that two vertices are adjacent if and only if they appear consecutively along the circle.

An isomorphism from a graph G to graph H is a bijection f : V (G) --+ V (H)

such that v1v2 E E(G) if and only if f(ul)f(u2 ) E E(H). If there is an isomorphism from G to H, we say G is isomorphic to H.

A graph His a minor of a connected graph G if a copy of H can be obtained by deleting or contracting edges of G. In a graph G, the contraction of edge e with endpoints u, v is the replacement of u and v with a single vertex whose incident edges are the edges other than e that were incident to u and v.

In this thesis we will be studying a probabilistic model of a network. We will be using probabilities when exploring a network using a random graph model. A random graph, (G,p) is a graph G whose edges exist independently with probability p. This generalizes the traditional Erdos-Renyi random graph model (Kn,P) ([13],

425) by allowing any graph G as the host graph for the experiment.

Next we review some probability theory. In probability theory [11], the

outcome space 0 is the set of all possible outcomes. (In our case this will be the set of all possible combinations of edges that can be used to send the message from s to r. So 0 will always be finite in this thesis.) An event is a subset Z of the outcome space. When studying different graphs, we will want to find the probability of event Z, denoted P(Z),which is a measure of the likelihood or chance that the CHAPTER 1. INTRODUCTION 7 event occurs.

The probability function has the following Rules of Proportion and Proba-

bility:

!)Non-negative: P(Z) ;::::: 0

2)Addition: If Z1 , Z2 , Z 3 , ... , Zn is a partition of Z then

P(Z) = P(Z1 ) + P(Z2 ) + P(Z3 ) + ... + P(Zk)

3) Total One: P(O) = 1

The probability of both events Z1 and Z2 occurring simultaneously is the

probability of the intersection of events Z1 and Z2, denoted P(Z1 n Z2). The

probability of either events Z1 or Z2 or both occurring is the probability of the

union of event Z1 and event Z2, denoted P(Z1 U Z2).

Theorem 1.2.2. {Difference rule [11}) If the occurrence of Z 1 implies the occur­

rence of Z2, i.e. Z1 s:;; Z2, then P(Z1 ) :S P(Z2 ) and the difference between these

probabilities is the probability that z2 occurs and zl does not:

A basic result in probability theory is the principle of inclusion-exclusion.

Theorem 1.2.3. Suppose Z1 and Z2 are events. Then CHAPTER 1. INTRODUCTION 8

More generally, if Z1, Z2, ... , Zn are events, then

n n n n P(Uzi) = LP(Zi)- L P(Zi n z 1) + L P(Zi n Z1 n Zk)- ... i=l i=l l:Si

and will need to review some basic algebraic definitions and calculus results. To de- scribe some comparisons between polynomials the following notation is convenient:

Definition 1.2.4. For functions f, g whose domain includes [0, 1] we write f < g

(respectively f ::::; g} if for all 0 < x < 1 we have f ( x) < g( x) (respectively f ( x) ::::;

g(x)).

Definition 1.2.5. Suppose that f(x) is a polynomial written in descending pow-

ers of x. We say that f(x) has a variation in sign whenever adjacent nonzero

coefficients in f ( x) when written in descending powers are opposite in sign.

5 4 3 2 Example 1.2.6. Consider F(x) = x - x - 9x + 5x + 16x- 12. Notice that

F(x) is written in descending powers and there is a change in sign from 1 to -1

4 corresponding to the coefficients of the terms x 5 and x . We also have a change

3 2 from -9 to 5 corresponding to the coefficients of the terms x and x . Finally, there

is also a sign change from 16 to -12 corresponding to the last two terms. So we

have three variations in sign.

Definition 1.2. 7. We will be considering roots of a polynomial and their corre-

sponding multiplicity. If r is a root (or zero) of a function f, and n is the least CHAPTER 1. INTRODUCTION 9 positive integer n such that (x- r)n divides f(x), but (x- r)n+l does not, then f is said to have a root of multiplicity n at r.

Theorem 1.2.8. Descartes' Rule of Signs

Let f ( x) be a nonzero polynomial with real coefficients.

(a) The number of positive zeros off (counting multiplicity) is either equal

to the number of variations in sign of f(x) or less than this variation by

an even number.

(b) The number of negative zeros off (counting multiplicity) is either equal

to the number of variations in sign of f(-x) or less than this variation by

an even number.

5 4 3 2 Example 1.2.9. We consider F(x) = x - x - 9x + 5x + 16x - 12. When x > 0 we have three variations in sign. So F(x) has three or one positive roots.

5 4 3 2 Evaluating F at -x we have F( -x) = - x - x + 9x + 5x - 16x- 12, so

F( -x) has two variations. Thus, F( -x) has two or zero negative roots. In fact,

F(x) = (x- 1) 2 (x + 2?(x- 3) so that F(x) actually has three positive roots and

two negative roots.

The following two theorems will be needed in Chapter 5 to prove the exis­

tence of a point of intersection of two polynomials and to expand a polynomial.

Theorem 1.2.10. The Intermediate Value Theorem CHAPTER 1. INTRODUCTION 10

Iff is a continuous function on a closed interval [a, b], and cis between f(a) and f(b), then there is at least one number x E [a, b] such that f(x) =c.

Theorem 1.2.11. The Binomial Theorem

For real numbers x and y and positive integem, 11

Chapter 2

The Deterministic Model

2.1 s, r-Graphs

In this thesis we study graphs that represent a network and assume the ver­ tices (the processors) do not fail, but that the edges (the links) between processors may fail without recovering. Our graphs also have two distinguished vertices (ter­ minals), and we refer to one of these terminals as the sender, denoted by s, and the other as the receiver, r. The graphs that represent this model are called s,r-graphs

(or two-terminal graphs).

2. 2 Protocols for Message Transmission

In s, r-graphs we wish to send a message from s to r. We send the message using

packets that travel from s tor along an s, r-walk.

In public networks, intermediate processors do not store information about

the state of the communication between s and r, and our model conforms to this

restriction. Therefore, when a processor v receives a packet, it must immediately

decide to which of its neighbors w to send the packet, basing its decision only CHAPTER 2. THE DETERMINISTIC MODEL 12 on the neighbor u from which the packet arrived. This decision is encoded in an instruction, that is a list uvw such that u and w are vertices adjacent to vertex v.

Observe that the packet might be sent forward to more than one neighbor if there are multiple instructions with v as the middle vertex. A protocol for a network is a set of instructions that describes how the message is to be routed.

Example 2.2.1. Consider F0 (see Figure 2.1} with the instructions s12 and 12r.

The protocol A 1 with these two instructions informs the vertices to send the message along the walks, 1, 2, r. A1 is "good" as long as s1, 12, 2r remain alive, but if one of these edges fails, then the message will not reach r.

~----~

s r

3 4

Figure 2.1: F0 with protocol A1 = {s12, 12r}

We seek a protocol that ensures that a message sent by swill be received by r if not too many edges fail, i.e., there is still an s, r-walk with live edges that can be used to send the message. We do not know which edges fail ahead of time. We say an edge is alive or live if it did not fail and is capable of transmitting messages.

We say an edge is dead if it has failed and is not capable of transmitting messages.

Example 2.2.2. Since we would like to send a message along an s, r-walk, let us CHAPTER 2. THE DETERMINISTIC MODEL 13

consider some situations in F0 from Example 2. 2.1 in which some specific edges fail. We will give a protocol A2 for the graph that allows us to send the message in every one of these situations.

If the edges s3, 34, and 4r fail then we would like to send the message using the walks, 1, 2, r. So we would want instructions s12 and 12r in our protocol A2. If the edges s3, 14, and 2r were dead, then we would want instructions s12, 123, 234 and 34r to instruct the vertices to use the walk s, 1, 2, 3, 4, r to send the message.

Likewise, if edges s1, 23, and 4r failed then we would like instructions s34, 341,

412, 12r so that the paths, 3, 4, 1, 2, r is used to send the message. A protocol that

uses all of the above instructions is A2 = {s12, s34, 12r, 123,234,341, 34r, 412}. 2 ~ ~ --;;;;~ s r

~~ ~~

Figure 2.2: F0 with protocol A2

Suppose we were to use A 2 to send the message in F0 (see Figure 2.2}, but that none of the edges actually failed. The message is sent from s to its neighbors

1 and 3. We will just describe what happens to the message when it is sent to 1.

The instruction at vertex 1 routes the message to 2. The instructions at 2 have the message sent to r and 3. So r has received the message once. When the message is sent to 3 from 2, the instruction at 3 routes the message to 4. When the message CHAPTER 2. THE DETERMINISTIC MODEL 14 arrives at 4 from 3, the message is then sent to r and 1. Now, r has received the message twice. When the message arrives at 1 it is sent to 2 once more.

Observe that one problem with" this protocol is that if no edges fail, then the

message is sent to vertices 1, 2, 3 and 4 infinitely many times due to the closed walk

C 1 = 1, 2, 3, 4, 1 and they get flooded with many copies of the message. Note that

each time the message arrives at 2 and 4, it is sent to r. So r receives the message

infinitely many times and is flooded as well with all these packets containing the

message.

This example shows the effect of two main assumptions of our model: we

do not change our set of instruction as communication proceeds (since our commu-

nication is memoryless, i.e. we do not keep track of the status of communication)

and we do not know ahead of time which edges fails.

Definition 2.2.3. A protocol is finite if only finitely many copies of a message sent

from s reach r when no edges fail. We also say that the protocol does not flood r.

Note that we can still flood other vertices with a protocol, but as long as r is

not flooded, the protocol well be considered finite.

Thus, protocol A1 is finite and A 2 is not finite in Fa.

Definition 2.2.4. An s, r-walk s = v0 , v1, ... , vk = r in G is essential for a protocol

A if vi-IViVi+I E A for all i :S k- 1, and an instruction Vi-IViVi+I in an essential

s, r-walk is called an essential instruction of A. An essential circuit wo, ... , wk is

a closed trail (w0 = wk) such that wi-l wiwi+ 1 is an essential instruction for some CHAPTER 2. THE DETERMINISTIC MODEL 15 s, r-walk for all 1 S i S k (where wk+l = WI}, and these instructions wi-I WiWi+I

are called circuit instructions.

Example 2.2.5. Consider protocol A2 = {s12, s34, 12r, 123,234,341, 34r, 412} in

Fo from Example 2.2.2. An essential walk for A 2 in F0 would be w = s, 1, 2, 3, 4, 1, 2, r

since the instructions for w are all in A2 . The instruction s12 would be an essential

instruction. With A 2 we have the essential circuit CI = 1, 2, 3, 4, 1 which has circuit

instructions 123, 234, 341, and 412, which are also essential instructions.

Theorem 2.2.6. {8] A protocol is finite if and only if it contains no essential

circuit.

Corollary 2.2. 7. Let AI, A2 be protocols with AI <;;;; A2. If A2 is finite, then AI is finite.

Proof. Every essential instruction for AI is an essential instruction for A2, so that

an essential circuit for AI would also be an essential circuit for A2. 0

Definition 2.2.8. Let I(: denote the set of all circuit instructions for an essential

circuit C. An instruction uvw contributes to an essential circuit C if it is in I(: or

if it is contained in an essential s, r-walk that contains a circuit instruction of I(:.

Let Ic denote the set of all instructions that contribute to C. So, I(:<;;;; Ic.

Example 2.2.9. Using protocol A2 = {s12, s34, 12r, 123,234,341, 34r, 412} in Fo

from Example 2.2.2, I(: {123, 234,341, 412} is the set of circuit instructions 1 =

for CI. Some instructions that contribute to CI include s 12 and 34r since the walk CHAPTER 2. THE DETERMINISTIC MODEL 16 s, 1,2,3,4,r uses the circuit instruction 123 and the two instructions s12 and 34r.

The set of all contributing instructions for C 1 is Ic1= { s12, s34, 12r, 34r} U 101 .

One of our goals is to disrupt essential circuits by removing contributing

instructions.

Definition 2.2.10. The Complete Forwarding Protocol (CFP) Aa of an s, r-graph

G consists of all instructions uvw such that there is an s, r-path in which u, v, w ap-

pear consecutively in that order. Any protocol where at least one of the instructions

from the CFP is missing will be a Partial Forwarding Protocol (PFP).

Example 2.2.11. The complete forwarding protocol of Fa is

Aa = {s12,s34, 12r,123,234,341,345,412,s32,321,214,14r,s14,143,432,32r} {see

Figure 2.3). Observe that Aa is not finite since Aa 2 A2 and A2 is not finite. In

fact Aa contains the essential circuit C1 and as well as one other essential circuit

c2 = 1, 4, 3, 2, 1.

1 ~ s r

Figure 2.3: Fa with CFP Aa

Example 2.2.12. The complete forwarding protocol of F1 shown in Figure 2.4 is

Aa = {s12, s14, s53, s54, s5r, 12r, 123, 143, 145,234, 235,341, 345, 32r, 35r, 412, CHAPTER 2. THE DETERMINISTIC MODEL 17

432,435, 453, 45r, 532,534, 541, 543}. Observe that 321 is not in the CFP even

though there is an s, r-walk with this instruction, since there is no s, r-path with

this instruction. Thus 1, 4, 3, 2, 1 is not an essential circuit for Ao.

s

2.3 A Forbidden Minor Characterization

In 2005, Fich, Kundgen, Pelsmajer, and Ramamurthi [8] studied the model of s, r­

graphs described in Section 2.2 and characterized all graphs in which the CFP is

finite by finding a list of 10 forbidden minors. (See Figure 2.5. Note that Fo and F1

are two of these minors.) They show that a graph not containing any one of these

ten graphs as a minor does not have an essential circuit and thus the CFP is finite.

Conversely, Fich, Kundgen, Pelsmajer, and Ramamurthi [8] show that a

graph containing at least one of these ten graphs as a minor would have an essential

circuit. Protocols with an essential circuit would send the message infinitely many

times to r directly or by sending it to a vertex (or vertices) that would then keep

sending it to r.

In Table 2.1 we list the CFP A0 for each of the ten forbidden minors and CHAPTER 2. THE DETERMINISTIC MODEL 18 in Table 2.2 we list the essential circuits of the CFP for each of the ten forbidden minors. CHAPTER 2. THE DETERMINISTIC MODEL 19

1. • 2 1• .2

4• •s 3. .4 Bo Fo

1. • 2 1• .2

s r s r 4.X .s 3. X.4 ~ .6 •s B1 F1 .7

11\4 s 1·x·2 r s 2 /\3si\6 r 4. .s ~8 Be .4 1 p2 s 1~t r ~ To T2 11\• 11\• r s 3\~~;·2 3·v·2 4 4 Ts Tr Figure 2.5: The 10 Forbidden Minors CHAPTER 2. THE DETERMINISTIC MODEL 20

I Forbidden Minor I Complete Forwarding Protocol Ao Fo s12,s14,s32,s34, 123, 12r, 14r, 143,214,234,321,32r,341, 34r,412,432 F1 s12,s14,s53,s54,s5r,123,12r,143,145,234,235,32r,341, 345,35r,412,432,435,453,45r,532,534,541,543 To s12,s13,s1r,s21,s23,s2r,s31,s32,s3r,123,12r,132, 13r,213,21r,231,23r,312,31r,321,32r Ts s12,s13,s1r,s31,s32,s34,s42,s43,s4r,123,124,132, 134,21r,231,234,24r,312,31r,321,324,342,34r,421, 423,431 Tr s12,s13,s1r,s42,s43,s4r, 123,124, 12r, 132, 134,21r,231, 234,24r,312,31r,321,324,32r,342,34r,421,423,42r,431 432 T2 s41,s42,s4r,s51,s53,s5r, 123,124,132, 135,14r, 15r,213, 214,215,231,235,24r,312,314,315,321,324,35r,412,413, 415,421,423,512,513,514,531,532 Bo s12,s13,s43,s45, 123,12r, 132,134,135,234,235,312,32r, 345,35r,431,432,435,453,45r,531,532 B1 s12,s13,s64,s65,s6r,123,12r,134,135,234,235,312,32r, 345,346,354,356,431,432,453,456,46r,531,532,543,546, 56r,643,645,653,654 B2 s64,s65,s6r,s71,s72,s7r, 123,127,134,135,17r,213,217, 234,235,27r,312,317,321,327,345,346,354,356,431,432, 453,456,46r,531,532,543,546,56r,643,645,653,654, 712, 713, 721, 723 Be s71,s74,s7r,s82,s83,s85,s86,s8r,123,128,132,138,145, 146,147,17r,214,217,231,238,28r,314,317,321,328,38r, 412,413,417,456,458,465,468,47r,541,547,564,568,58r, 641,647,654,658,68r, 712,713,714,741,745,746,821,823, 831,832,854,856,864,865

Table 2.1: The CFP for the Forbidden Minors CHAPTER 2. THE DETERMINISTIC MODEL 21

Forbidden Essential Essential Essential Essential Minor Circuit C1 Circuit C2 Circuit C3 Circuit C4 Fo 1,2,3,4,1 1,4,3,2, 1 F1 1,2,3,4,1 3,4,5,3 To 1,2,3,1 1,3,2,1 Ts 1,2,3,1 2,3,4,2 Tr 1,2,3,1 2,3,4,2 T2 1,2,3,1 1,3,2,1 Bo 1,2, 3,4, 5,3,1 B1 1,2, 3,4, 1, 2, 3, 5, 5,3,1 4,3,1 B2* 1, 2,3, 4, 1, 3, 5,4, 1,2,3,5, 1,3,4,5, 5,3,1 3,2,1 4,3,1 3,2,1 Be 1, 2, 3, 1, 1, 3, 2, 1, 1, 2, 3, 1, 1, 3, 2, 1, 4,5,6,4,1 4,5,6,4, 1 4,6,5,4, 1 4,6,5,4,1 * Note: B2 has two more essential circmts 1, 2, 3, 4, 5, 3, 2, 1, 3, 5, 4, 3, 1 and 1, 2, 3, 5, 4, 3, 2, 1, 3, 4, 5, 3, 1.

Table 2.2: Essential Circuits in the CFP of the Forbidden Minors 22

Chapter 3

The Probabilistic Model

3.1 Reliability of a Protocol

If the CFP A0 is finite, then the CFP should be used to route a message in G. If not, then there could be many candidates for a good protocol. To study Fo and other graphs that do not necessarily have a finite CFP, we consider the following probabilistic model. In this model we take the same descriptions and assumptions about the network as in Chapter 2, however each edge will fail (be unable to trans­ mit a message) with probability q = 1- p, where 0 < p < 1. We denote such an s, r-graph with (G,p) and refer to it as the probabilistic s, r-graph. Since edges live

(are capable of transmitting messages) with probability p, we can now evaluate the probability that a protocol has an s, r-walk with live edges by calculating what is called its reliability.

Definition 3.1.1. Given a probabilistic s, r-graph ( G, p), the reliability of a protocol

A is the probability that we can still send a message from s to r using A if the edges fail independently with probability 1 - p. We denote this protocol reliability by CHAPTER 3. THE PROBABILISTIC MODEL 23

PA(G,p), or short PA·

Observe, that the reliability of a protocol A is defined whether A is finite or

not.

Figure 3.1: Pm

Example 3.1.2. Consider a path Pm of length m, given by s = Vo, v1, ... , Vm-1,

Vm = r (see Figure 3.1}. The only protocol that can be used to send a message from

s to r is Ao = {sv 1 v2 , v1 v2v3 , ... , Vm-zVm-l r} and in order to send the message we

would need all the edges of P m to be alive. Thus

Recall that Ao is the CFP of Pm and that the reliability of a Partial Forwarding

Protocol (PFP) in this graph would be 0.

If there is more than one s, r-path, then finding a good finite protocol and

determining its reliability becomes more difficult. In the next section we will give

two general methods for computing the reliability of a protocol and use them to

compute the reliability of the CFP A0 for several graphs. Both are based on the

following result.

Proposition 3.1.3. IfT1 , ... Tk are the essentials, r-trails for a protocol A, and Zi CHAPTER 3. THE PROBABILISTIC MODEL 24 is the event that all edges in trail T; survive, then

Proof. Since by Definition 3.1.1, PA(G,p) is the probability we can still send the message from s to r, it is the probability that some s, r-walk s = vo, v1, ... , vk = r remains such that vi-lvivi+l E A for all i with 0 < i < k. However if vivi+l = VjVJ+l fori< j in such a walk, then there is also a shorter walk vo, v1, ... ,vi, VJ+l, Vj+2, ... , vk with this property. Repeating this argument we conclude that whenever an s, r­ walk remains, then an s, r-trail Ti remains. Moreover, since no trail can have length more than 2IE( G) I, the number of these trails is finite, and the equation now follows from the observation that every trail is a walk. D

Corollary 3.1.4. If A~ A', then PA::; PA'·

Proof. Suppose A ~ A'. Every essential s, r-trail for A is an essential s, r-trail

for A'. If T1, ... , TJ denote the essential s, r-trails for a protocol A, Zi denotes the event that all edges in trail Ti survive, T{, ... T~ denote the essential s, r-trails for a protocol A', and Z{ denotes the event that all edges in trail Tf survive, then

Z1 U Z2 U ... U ZJ ~ Z~ u Z~ u ... u Z£. Therefore

D

Proposition 3.1.5. For every finite protocol A there exists a finite protocol A* ~

A0 , where A0 is the CFP, such that PA ::; PA•. CHAPTER 3. THE PROBABILISTIC MODEL 25

Proof. Recall that PA(G,p) = P(Z1 U ... U Zk), where Zi denotes the event that some specific s, r-trail Ti survives. Proposition 1.2.1 implies that for each Ti there is an s, r-path Pi= v0 , ... , vk such that v1_1, v1 are consecutive on Ti in this order.

Let A* = { uvw : uvw is in some Pi} and let z; denote the event that Pi survives.

Since E(Pi) ~ E(Ti), Zi ~ z;, and, thus, Z1 u ... U Zk ~ z; U ... U Z'k so that

PA(G,p) = P(Z1 U ... U Zk)

::; P(z; U ... U Z'k)

::; PA·(G,p).

Since the CFP A0 contains all instructions corresponding to all s, r-paths in

(G, p), it contains all the instructions in A*, so A* ~ Ao.

To see that A* is finite suppose there is an essential circuit C* in A*, C* = v1, v2, ... , vk. If vi_ 1vivi+l E A* \A, then there exists an s, r-path Pi in which they appear in this order vi_ 1, vi, vi+l· However, then T; must have a subsequence si = Vi-1' Vi' w1' w2' ... ' Wm' vi' vi+ 1· Replace Vi-1 ViVi+l in C* by si. Do this for all such vi_ 1vivi+l to obtain an essential closed walk C for A. This must contain an essential circuit for A. 0

Since Proposition 3.1.5 shows that we can always find a finite protocol con­ tained in the CFP that is at least as reliable as any given finite protocol, we will from now on only need to consider PFPs that are contained in the CFP. CHAPTER 3. THE PROBABILISTIC MODEL 26 3.2 Reliability of the Complete Forwarding Pro­ tocol

The reliability of the CFP, PAo(G,p), is simply the probability that any s,r-path remains in G, since in the CFP the message is routed along every s, r-path that survives. This has been studied under the name 2-terminal reliability polynomial.

We will simply call the reliability of the CFP the reliability polynomial and denote it by p(G,p) or p for short. In Proposition 3.3.4, we will show the p to be a polynomial in p with integer coefficients. The reliability polynomial was explored by many including: Provan [2], Ball [2], Brecht and Colbourn [3], Kelmans [10], and Younes and Girgis [15]. In Chapter 6 we will discuss the constructions from

Kelmans [10] in order to prove Theorem 6.4.2.

The following result will allow us to look at s, r-paths instead of s, r-trails when calculating the protocol reliability.

Proposition 3.2.1. If there is an s, r-trail, then there is an s, r-path.

Proof. This follows from Proposition 1.2.1 since every trail is a walk and every s, r-walk contains and s, r-path. D

Proposition 3.1.3 states that the protocol reliability is the probability that one of the essential s, r-trails has live edges. However, using Proposition 3.2.1 we can prove that the protocol reliability is the probability that one of the s, r-paths has live edges. CHAPTER 3. THE PROBABILISTIC MODEL 27

Proposition 3.2.2. If H, ... , Pk are the s, r-paths in G, and Zi is the event that

Pi survives, then p(G,p) = P(Z1 U ... U Zk)·

Proof. Let T1, T2, ... , Tm_ 1, Tm be all the s, r-trails in G, where, without loss of generality, T;, = Pi for 1 :::; i :::; k. Let Zi be the event that Ti is alive. By

Proposition 3.2.1, every s, r-trail Ti contains an s, r-path Pj· So, if T;, is alive, then

Pj is alive as well, and we obtain Zi ~ Zj. So, for every i with k + 1 :S i :S m there is a j with 1 :::; j :S k such that Zi ~ Zj. Therefore, Z1 U ... U Zm = Z1 U ... U Zk.

This implies

p(G,p) = P(Z1 U ... U Zm) = P(Z1 U ... U Zk)·

0

We use the result in Proposition 3.2.2 when calculating the reliability poly­ nomial in Examples 3.2.3-3.2.6. Our examples will illustrate two ways of calculating the polynomial: Inclusion/Exclusion and Direct Count.

Example 3.2.3. (Inclusion/Exclusion)

Consider a cycle C2(m+ll consisting of two disjoint paths from s tor of length m + 1 (see Figure 3.2). Let ones, r-path be P1 given by s = Vo, v1, ... , Vm, Vm+l = r and the others, r-path be P2 given by s = u0 , u 1, ... , Um, Um+l = r. The CFP for c2(m+l) is Ao = {svlv2, vlv2v3, ... , Vm-lVmr, sulu2, ulu2u3, ... , Um-lUmr }. Using the

CFP on C2(m+l) would require at least ones, r-path with live edges in order to send the message. Let Zi be the event that Pi is alive. So we can use the principle of inclusion-exclusion to calculate the reliability polynomial. CHAPTER 3. THE PROBABILISTIC MODEL 28 Vm-I Vm

s r

Um-I U

Figure 3.2: C2(m+1)

Note that the probability that path PI can send a message is when its corre­

sponding edges are alive, so P(ZI) = pm+l. Similarly P(Z2 ) = pm+I. The probabil­

ity that both paths PI and P2 can send a message is the probability that all the edges

2 are alive, so P(ZI n Z 2 ) = p (m+l). So, by the principle of inclusion-exclusion, the

probability that PI or P2 can send a message is

We calculated p( C2(m+I), p) by considering different paths and using the

principle of inclusion-exclusion to obtain a polynomial in terms of the variable p. We

can also calculate the reliability polynomial by considering different combinations of

edges living and failing to obtain a polynomial in terms of p and q, where q = 1-p

is the probability that an edge fails. We will describe this with one simple example,

followed by examples involving two of the 10 forbidden minors.

Example 3.2.4. (Direct Count)

In this example we calculate the reliability of the Diamond graph (see Fig­

ure 3.3). We will consider different combinations of edges living and failing to CHAPTER 3. THE PROBABILISTIC MODEL 29 determine p. Our reliability polynomial will have the form

sponds to the number of ways in which we can have i edges living and 5 - i edges failing and still be able to send a message from s to r using the CFP. 1

s r

2 Figure 3.3: The Diamond Graph D

When zero or one edges are alive, we cannot send a message through the

network. Hence, a0 = 0 = a 1 . When two edges are alive, we can send the message

using paths s, 1, r or s, 2, r. Thus, a 2 = 2. When three edges are alive we can use

the paths of length two and chose any other single edge, giving us 2 · 3 = 6 ways, or

we can use the paths s, 1, 2, r or s, 2, 1, r of length three. Therefore, a3 = 6 + 2 = 8.

Next observe that if only one edge is dead we will always be able to send the

message, so a4 = 5. Lastly, there is only one way to have all edges alive, giving us

Example 3.2.5. (Direct Count) Let us consider the reliability polynomial of the

forbidden minor T0 (see Figure 3.4). We use the same method as in Example 3.2.4. CHAPTER 3. THE PROBABILISTIC MODEL 30 ai corresponds to the number of ways in which we can have i edges living and 9 - i edges failing and still be able to send a message from s tor under the CFP. 1

Figure 3.4: To

Observe that a0 = 0 since if all the edges fail we cannot send a message.

The coefficient a 1 is 0 as well since there is no s, r-path of length one and, hence,

we cannot send the message if only one edge is alive. To calculate the coefficient to

the term p2q7 we just count the number of paths of length two. Observe that there

are three paths of length two, and they are P1 = s, 1, r; P2 = s, 2, r; and P3 = s, 3, r.

So a2 = 3.

When three edges are alive, to send the message we can use a path with three

live edges or use a path of length two with live edges together with any other edge.

We noted in the last paragraph that there are three paths of length two and 9-2

=7 edges to pick as the third edge. Also, observe that we have six paths of length

three: P4 = s,1,2,r; P 5 = s,1,3,r; P6 =s,3,2,r; P7 = s,2,1,r; Ps = s,3,1,r;

and P9 = s, 2, 3, r. Thus, the coefficient a3 is 3 · 7 + 6 = 27.

For s, r-transmissions to fail, we must have at least one edge fail from each

of the paths s, 1, r; s, 2, r; and s, 3, r. If s1, s2, s3 all fail (or 1r, 2r, 3r all fail}

then s, r-transmission fails, but otherwise we need at least five edges to fail: If s1 CHAPTER 3. THE PROBABILISTIC MODEL 31

and 2r are alive, then for s, r-transmission to fail, s2, 1r, 12, at least one of s3 or

3r, and at least one of 13 or 23 must fail; in fact, if s3 survives (respectively 3r ),

then 32 (respectively 13) must fail.

When only two edges fail, s, r-transmission is always possible and the coef-

ficient a7 is the number of ways to choose two edges from the nine to fail. That is,

a7 = ( ~ ) = 36. Likewise, when one edge fails, a message can still be sent from s

to r, and, hence, as is just the number of ways to choose one out of the nine edges

to be the one that is dead. So as = 9. For the coefficient a9 we must have all the

edges alive, and since there is only one way of having this occur we see that ag = 1.

For the case when five edges are alive and 4 have failed, we first consider

all the different possible ways of having five edges alive, which is ( ~ ) = 126.

From our total possible combinations we subtract 6 · 2 = 12, since the only way

to stop s, r-transmission would be if all three edges adjacent to s orr were dead,

and then we would only need one out of the other six edges to be dead. Hence,

a5 = 126 - 12 = 114.

When calculating a6 , we observe that there are only two combination of edges

in which s, r-transmission is not possible if three edges have failed. Communication

between s and r will not occur when all three edges adjacent to r or those adjacent

to s fail. Thus, we calculate the possible ways of picking six edges out of the nine

and then subtract two. Therefore, a6 = ( ~ ) - 2 = 84- 2 = 82.

For the term p4 q5 we count the total number of ways to choose five dead CHAPTER 3. THE PROBABILISTIC MODEL 32 edges and then subtract the number of combinations of five dead edges that do not

allows, r-transmission. There are ( ~ ) = 126 ways of having five dead edges. If

all three edges adjacent to s (similarly r) fail, then s, r-transmission is not possible.

For the fourth and fifth failing edge not adjacent to s ( similarly r) in this case,

we have ( ~ ) = 15 choices, giving us a total of 2 · 15 = 30 in this case. So, we

will have to subtract 30 from 126. Observe that if at most one edge incident with

s and r fails, then s, r-transmission remains possible. Suppose now, without loss of

generality, that only two edges adjacent to s have failed. Say edges s 1 and s2 fail,

but s3 is alive. Then 3r must fail to stop the message from reaching r. Moreover,

we may assume that 1r is alive, so that 31 must be dead, and it then follows that

23 must be the last dead edge since there are two possible paths from 2 to r. (See

Figure 3. 5.) There are ( ~ ) = 3 ways to pick two dead edges incident with s, and

4 5 the same applies for r. So, the coefficient for the term p q is 126- 30- 6 = 90. 1

Figure 3.5: T0 with five edges failing.

Therefore, our reliability polynomial for T0 is

Observe that we can re-write p(T0 , p) in terms of p instead of in terms of CHAPTER 3. THE PROBABILISTIC MODEL 33 p and q by replacing all the q 's with 1 - p, and then expand using the Binomial

Theorem. This way we obtain

If pis in terms of p, as in Example 3.2.3, then for a graph G, we can write pin

terms of p's and q's by multiplying term pi by (p+q)IE(G)I-i, fori= 1, 2, ... , IE( G) I-

1. We can do this since q = 1-p and hence (p+q)m-i = (p+ 1-p)m-i = 1. Observe

that qiE(G)I = 1. qiE(G)I = paqiE(G)I and piE(G)I = piE(G)I . 1 = piE(Gllqa. Thus,

= 2pm+l. (p + q)m+l _ p2(m+l)qa

~ 2pm+l (~ ( m 71 ) piqm+H) _ p'(m+l)q'

= f 2 ( m 71 ) pm+Hiqm+l-i + p2(m+l)qa. t=a

Example 3.2.6. We use the principle of inclusion-exclusion to find the reliability

polynomial of Fa (see Figure 3.6). Since we will have to consider paths in Fa, we

list all s, r-paths in Table 3.1. We let Zi be the event that the path Pi survzves.

3 5 From the Table 3.1 we obtain 2::::::~= 1 P(Zi) = 4p + 4p . 1 2

3

Figure 3.6: Fa CHAPTER 3. THE PROBABILISTIC MODEL 34

Path Length Probability of Path of Path, P(Zi) P1 = s, 1, 2, r 3 p3 P2 = s,3,4,r 3 p3 ?3 = s,3, 2,r 3 p3 P4 = s, 1,4,r 3 p3 Ps = s,3,4, 1,2,r 5 p5 P6 = s,1,2,3,4,r 5 p5 P7 = s,3,2, 1,4,r 5 p5 Ps = s,1,4,3,2,r 5 p5

Table 3.1: All s, r-Paths of F0

Note that F0 has eights, r-paths, and in order to use the inclusion-exclusion principle, we will also have to evaluate the event that more than one path can be used to send the message. Specifically, we will have to evaluate when 2, 3, 4, 5, 6,

7, and 8 paths of F0 can be used to send the message. We will write some of the probabilities of these events in Tables 3. 2 - 3.4.

Observe that each edge of F0 is used in exactly four of the eight s, r-paths, and, hence, not using an edge e would only allow us to use at most 8 - 4 = 4 paths.

For example, the edge s 1 is in paths P1, P4, P6, P8 and the edge 23 is in path

P3 , P6, P7,P8 . When considering the event that four paths have live edges, eight

7 combinations of the paths have probability p . They are listed in the Table 3.4.

8 The other combinations of four paths have probability p . For example, the event

8 Z1 n Z4 n Z6 n Z7 requires all edges to be alive and hence has probability p . Since -- ~--·~~------~---

CHAPTER 3. THE PROBABILISTIC MODEL 35

Zi1 n Zi2 Number P(Zi1 n Zi2) zh n Zi2 Number P(Zh n Zi2) of Edges of Edges Z1 n Z2 6 p6 z3nz5 6 p6 Z1 n z3 5 p5 z3 n z6 7 p7 Z1 n z4 5 p5 Z3 n z7 6 p6 Z1 n z5 6 p6 Z3nZs 6 p6 Z1 n z6 6 p6 z4 n z5 7 p7 Z1 n Z1 7 p7 z4 n z6 6 p6 Z1 n Zs 6 p6 z4nz1 6 p6 Z2 n z3 5 p5 Z4 n Zs 6 p6 Z2 n z4 5 p5 z5nz6 8 p8 Z2 n z5 6 p6 z5 n z7 7 p7 Z2 n z6 6 p6 Z5 n Zs 7 p7 Z2 nz1 6 p6 z6nz1 7 p7 Z2 n Zs 7 p7 z6 nZs 7 p7 z3 n z4 6 p6 Z1nZs 8 p8

Table 3.2: Probability of Two Specified Paths in F0 Surviving there are ( ! ) = 70 total possible combinations of four paths, there are 70-8 = 62

8 combinations of paths with probability p . Therefore 2: P(Zh n Zi 2 n Zi3 n Zi4 ) =

The probability of the event that any five or more paths have live edges will be p8 since in order to use five or more paths we need all the edges to be alive. To see this, suppose we did not need all eight edges to be alive, then there is a combination of five or more paths that does not use at least one edge. However, we noted that each edge is found in four paths. Hence we will need all edges available when considering five or more paths. Observe that there are ( ~ ) = 56 ways of choosing five out of

8 the eight paths and each combination has probability p , so 2: P(Zi1 n Zi 2 n Zi3 n CHAPTER 3. THE PROBABILISTIC MODEL 36

Z1 n Z2 n Z3 p7 Z2 n Z4 n z6 p7 Z1 n Z2 n Z4 p7 Z2 n z4 n Z1 PI Z1 n Z2 n z5 p7 Z2 n Z4 n Zs p7 Z1 n Z2 n z6 PI Z2 n Z5 n z6 p8 Z1 n Z2 n Z1 p8 Z2 n z5 n Z1 PI Z1 n Z2 n Zs p8 Z2 n Z5 n Zs p8 Z1 n z3 n Z4 PI Z2 n z6 n Z1 p7 Z1 n z3 n z5 p7 Z2 n z6 n Zs pl5 Z1 n z3 n z6 p7 Z2 n Z1 n Zs p8 Z1 n z3 n Z1 p7 Z3 n Z4 n z5 p8 Z1 n Z3 n Zs p7 z3 n z4 n z6 p8 Z1 n z4 n z5 p7 z3 n z4 n Z1 PI Z1 n Z4 n z6 p7 Z3 n Z4 n Zs p7 Z1 n z4 n Z1 p7 z3 n z5 n z6 p8 Z1 n Z4 n Zs p7 z3 n z5 n Z1 p7 Z1 n z5 n z6 p8 Z3 n Z5 n Zs PI Z1 n Z5 n Z1 p8 z3 n z6 n Z1 p8 Z1 n Z5 n Zs p{ Z3 n z6 n Zs p8 Z1 n z6 n Z1 p8 Z3 n Z1 n Zs p8 Z1 n z6 n Zs p7 z4 n z5 n z6 p8 Z1 n Z1 n Zs p8 z4 n z5 n Z1 p8 Z2 n z3 n z4 p7 Z4 n Z5 n Zs p8 Z2 n z3 n z5 p7 Z4 n z6 n Z1 p7 Z2 n z3 n z6 p{ Z4 n z6 n Zs p7 Z2 n Z3 n Z1 p7 Z4 n Z1 n Zs p8 Z2 n Z3 n Zs p7 z5 n z6 n Z1 p8 Z2 n z4 n Z5 p7 Z5 n z6 n Zs p8 z6 n Z1 n Zs p~'> Z5 n Z1 n Zs p8

Table 3.3: Probability of Three Paths in F0 CHAPTER 3. THE PROBABILISTIC MODEL 37

Z1 n Z2 n z3 n z6 p7 Z1 n Z2 n z4 n z5 p7 Z1 n z3 n z4 n Z1 p7 Z1 n Z3 n Z5 n Zs p7 Z1 n z4 n z6 n Zs p7 Z2 n Z3 n Z4 n Zs p7 Z2 n Z3 n z5 n Z1 p7 Z2 n z4 n z6 n Z1 p7

Table 3.4: Some Probabilities of Four Paths in F0

8 ) . Zi 4 n Zi5 = 56p There are ( ~ ) = 28 ways of choosing six out of the eight paths

8 and each combination has probability p , so 2:: P( Zi 1 n Zi 2 n Zi3 n Zi 4 n Zi5 n Zi6 ) =

28p8 . There are ( ~ ) = 8 ways of choosing seven out of the eight paths and each

8 8 combination has probability p , so 2:: P( Zi 1 n Zi 2 n Zi3 n Zi4 n Zi5 n Zi7 n Zi7 ) = 8p .

Note that there is only one combination of having all eight paths, and, therefore,

8 the sum of the probabilities of eight paths is just p .

Thus, by the inclusion-exclusion principle

With a similar effort the reliability polynomials for the other minors can also be computed and we simply state the first few terms for each them in Table 3.5. CHAPTER 3. THE PROBABILISTIC MODEL 38

I Minor Graph I Reliability Polynomial p I 2 3 4 To 3p + 6p - 13p ... Ts(Tr) 2p2 + 2p3 + 5p4 + ... T2 2p2 + Op3 + 1p4 + ... Be 2p2 - 1p4 + 8p5 + ... B2 2p2- 1p4 + Ops + ... F1 1p2 + 1p3 + 2p4 + ... B1 1p2 + 1p3 + 1p4 + ... Fo 4p3 - 14p6 + 16p7 + ... Bo 2p3 + 4p4 - 2p5 + ...

Table 3.5: Table of Reliabilities

3.3 Reliability and Deficiency of a Partial Forwarding Protocol

Recall that the reliability polynomial is the probability we can still send a message from s to r using the CFP (i.e, the protocol that has all instructions corresponding to an s, r-path). Observe that by definition, a PFP is missing at least one instruc- tion from the CFP, and all instructions in the CFP are essential. So, for any PFP A of a (G, p) the reliability polynomial would be greater than the protocol reliability of the PFP A.

We consider PFP's since in the 10 forbidden minors the CFP is not finite.

As noted earlier, graphs in which the CFP is infinite are characterized by the existence of essential circuits, and the 10 forbidden minors are the smallest graphs that contain such circuits. In this section we consider optimal finite protocols for these 10 forbidden minor graphs and we discuss ways to describe those protocols.

Example 3.3.1. In Example 3.2.5 we calculated the reliability polynomial of To. CHAPTER 3. THE PROBABILISTIC MODEL 39

Table 2.2 in Section 2.3 describes the two essential circuits in To, C1 = 1, 2, 3, 1 and

C2 = 1, 3, 2, 1. We will now study which instruction we have to remove to interrupt

C1 and C2 to arrive at a finite PFP A.

When we consider the PFP where one of the instructions s1r, s2r, or s3r

is removed from the CFP, our protocol reliability will be lower than the reliability

polynomial since the term of the coefficient p2 q7 will be 2 instead of 3 and none of

the essential circuits will be affected. Thus, it is not useful to remove any of the

three instructions s1r, s2r, or s3r, so all three instructions are in A.

The only instructions that contribute to C1 are the following: s12, s23,

s31, 123, 231,312, 12r, 23r, 31r. Instructions that contribute to C2 are: s13, s21,

s32, 132, 213, 321, 32r, 21r, 13r.

Observe that since we have no instructions of the form uvu in A, every

essential s, r-trail for A in T0 is an s, r-path, or is of the form s, i, j, k, i, r where

{ i, j, k} = {1, 2, 3}. However, if an s, r-trail of the latter form survives, then so does

the s, r-path s, i, r which is essential since we already observed that sir E A. Thus,

it suffices to consider which s, r-paths are essential for A. We list all s, r-paths,

some of which were already described in Example 3. 2. 5, in Table 3. 6.

We first focus on C 1 : Observe that when the instruction s12 is removed

from the CFP, the paths P4 and P11 cannot be used to send a message. When the

instruction s23 is removed from the CFP, the paths P7 and P 13 cannot be used to

send a message. The removal of the instruction s31 from the CFP will not permit CHAPTER 3. THE PROBABILISTIC MODEL 40

Path I Length I Path I Length I of Path of Path P1 = s, 1, r 2

P2=s,2,r 2 P9 = s,3,2,r 3 P3=s,3,r 2 P10 = s,3,2, 1,r 4 P4 = s, 1, 2, r 3 P11 = s, 1,2,3,r 4 P5=s,1,3,r 3 P12 = s,1,3,2,r 4 P6 = s,2, 1, r 3 P13 = s,2,3,1,r 4 P7 = s, 2,3,r 3 P14 = s,2,1,3,r 4

P8 = s, 3, 1, r 3 P15 = s,3,1,2,r 4

Table 3.6: All s, r-Paths of T0

messages to be sent using paths P8 and P 15 . If the instruction 12r is removed, then the paths P4 and P15 are affected. The removal of instructions 23r affects P7 and

Pu. The removal of instruction 31r does not allow the use of Ps and P13· If the

instruction 123 is not in a PFP, the path P 11 cannot be used to send a message.

Thus, it is better to remove 123 than s12 and 23r. When removing the instruction

231 from the C F P, the path H 3 cannot be used to send a message (so this is better than s23 and 31r ). Not having the instruction 312 in a PFP will not allow the use of the path P15 (which is better than s31 and 12r ). Moreover, removing any one of 123,231,312 results in a protocol in which C1 is not an essential circuit. Thus, removing one of the latter three instructions is the best option to stop C1 and to permit the use of as many paths as possible.

Now we consider instructions that would stop the essential circuit C2 . Ob- serve that when the instruction s13 is removed, the paths P5 and H 2 cannot be used to send a message. When the instruction s21 is removed, the paths P6 and P14 CHAPTER 3. THE PROBABILISTIC MODEL 41

cannot be used to send a message. When the instruction s32 is removed, the paths

P9 and P10 cannot be used to send a message. If the instruction 13r is removed,

then the paths P 5 and P14 are affected. The removal of instructions 32r affects Pg

and g 2 • The removal of instruction 21r does not allow the use of P6 and P10. If the

instruction 132 is removed, the path P 12 cannot be used to send a message. When

removing the instruction 321, the path P10 cannot be used to send a message. The

removal of instruction 213 will not allow the use of the path g 4 . Thus, eliminating

the instructions 132, 321, or 213 from the CFP to form a PFP would stop C2 and

eliminate the use of only one path, rather than two. Therefore, removing one of

these three instructions is the better option to stop C2 , as that would permit the use

of as many paths as possible.

Since the removal of one of the three circuit instructions from each essen­

tial circuit would allow us to use more paths than removing other instructions, we

determine that the 3 · 3 = 9 PFP's that arise in this way yield the highest protocol

reliability. We will now calculate the protocol reliabilities of the PFP's when two

circuit instructions are removed from the CFP, one corresponding to C1 and the

other to C2 . Note that the removal of an instruction has an effect on the coefficients

of the terms of the reliability polynomial. So, we consider what we would have to

subtract from the reliability polynomial to find the protocol reliability of the PFP.

Note that each of the instructions 123, 231, 312, 132, 321, and 213 all

correspond to an s, r-path of length four. Hence, the first term in the reliability CHAPTER 3. THE PROBABILISTIC MODEL 42 polynomial that would be affected by the removal would be the term p4 q5 and since

4 5 we would be removing two of these instructions, we would have to subtract 2p q from the reliability polynomial. Observe that the terms piq9-i for i = 5, 6, 7, 8, 9 would not be affected by the removal of any of these instructions. This is because if we consider the s, r-paths of length four with one of these instructions, then when considering piq9-i fori= 5, 6, 7, 8, 9, we would have to have at least another live edge in addition to the four that must be alive for the path. The addition of this edge allows the use of an s, r-path of length two or three since the instructions corresponding to paths of lengths two or three were not affected. 1 1 1 / r s / r

1

r

Figure 3.7: Combinations of P10 and an extra live edge in To

For example, let us look at path P10 = s, 3, 2, 1, r (see Figure 3. 1). If we

add the edge s1, then we can use path P 1 = s1r. If we add edge s2, then we can

use path P2 = s21r. The addition of edge 13 would allow the use of Ps = s, 3, 1, r.

Adding edge 3r permits the use of path P3 = s, 3, r, and finally, the addition of edge CHAPTER 3. THE PROBABILISTIC MODEL 43

2r would let us use the path P9 = s, 3, 2, r.

Therefore, the protocol reliability of the PFP A obtained from the CFP Ao by removing one instruction from C1 and one from C2 would be

PA(To,P) = p(To,P)- 2p4q5

= 3p2q7 + 27p3q6 + 90p4q5 + 114p5q4 + 82p6q3 + 36p7 q2 + 9p8q

+ lpg- 2p4q5

= 3p2q7 + 27p3q6 + 88p4q5 + 114p5q4 + 82p6q3 + 36p7q2 + 9p8q + p9.

We also showed that this is the highest protocol reliability PA for any finite protocol

4 5 A for T0 . The calculated term 2p q is the deficiency polynomial with respect to the two instructions removed from the CFP to stop the essential circuits of To.

Definition 3.3.2. Given an s, r-graph (G,p), the deficiency of a protocol A, de­ noted dA(G,p), is the probability that some s, r-trail remains in G, but that none of these trails are essential for A. That is, dA(G,p) = p(G,p)- PA(G,p). Thus, the protocol reliability of A would be the deficiency subtracted from the reliability polynomial p.

Example 3.3.3. Consider the graph F0 . Recall F0 has two essential circuits C1 and

C2 (See Figure 2.2}. In Table 3. 7 we list all the instructions of the CFP of the graph

F0 . Observe that the instructions in first four rows correspond to instructions that contribute to C1 being an essential circuit, whereas the instructions in the second four rows contribute to C2 . Note that instructions in the first column and first CHAPTER 3. THE PROBABILISTIC MODEL 44

Instructions Contained Instructions Contained in Path(s) in Path(s) s34 P2,Ps 412 Ps s12 Pl,P6 123 p6 12r Pl,Ps 341 Ps 34r P2,P6 234 p6 s32 P3,P7 321 p7 s14 P4,Ps 432 Ps 14r P4,P7 214 p7 32r P3,Ps 143 Ps

Table 3.7: CFP Instructions of F0

four rows correspond to two paths in the CFP, and the instructions in the first column and last four rows correspond to two other paths. However, instructions in the sets 1(; (the set of circuit instructions for and 1 = {123, 234,341, 412} C1)

) 1(;2 = {143, 432,321, 214} (the set of circuit instructions for C2 only correspond to one path, and moreover removing any instruction from I(;i stops Ci from being an essential circuit. In order to stop the essential circuits in F0 we would like to remove instructions from the CFP that allow us to use the largest number of paths

• possible. So, preferably, we would like to remove instructions from 1(;1 U 1(;2

Note that the instructions in 1(; correspond to the instructions in the essen- 1

. tial circuit C1 and affect path P5 or P6 The instructions in 1(;2 correspond to the

instructions in the essential circuit C2 and affect path P7 or P8 . Also, note that 1(;1 and 1(; are disjoint. So, we can just consider removing instructions from these two 2 sets to stop the essential circuits, and only two paths would be affected.

Observe that P5 , P6, P7, and P8 are paths of length five, and when one CHAPTER 3. THE PROBABILISTIC MODEL 45 instruction c1 in I01 and c2 in 102 is removed from the CFP, two of these paths are

5 3 affected. So the removal of c1 and c2 will reduce the coefficient of the term p q in p by 2. Therefore, if A0 is the CFP of F0 , c1 E 101 , and c2 E 102 , then the deficiency

5 3 dAo\{q,c2}(Fo,p) is 2p q . So, we obtain

PAo\{CI,c2}(Fo,p) = p(Fo,P)- dA\{cl,c2}(Fo,P)

= 4p3q5 + 20p4q4 + 38p5q3 + 26p6q2 + 8p7q + p8.

In an s, r-graph (G,p) with an essential circuit, the deficiency of a finite protocol must always be nonzero. So, as observed in Examples 3.3.1 and 3.3.3, and as a result from Corollary 3.1.4, PA < p for every PFP A. We also observed in the examples that the protocol reliability and the deficiency polynomials are polynomials in p.

Proposition 3.3.4. Given (G,p), the reliability polynomial p(G,p), protocol reli­ ability PA(G,p), and deficiency dA(G,p) for some protocol A, are polynomials in p.

Proof. Given (G,p), to prove that p(G,p), PA(G,p), and dA(G,p) for some protocol

A, are polynomials in p, we must show that p(G,p), PA(G,p), and dA(G,p) can be written in terms of one variable p as follows: there exist constants m, ao, a1, ... ,am

such that PA(G,p) = ampm + ... + a1p + ao.

Since dA(G,p) = p(G,p)- PA(G,p), if p(G,p) and PA(G,p) are polynomial

in p, then dA(G,p) is also a polynomial in p. Furthermore, p(G,p) = PA0 (G,p), so CHAPTER 3. THE PROBABILISTIC MODEL 46 that it suffices to show PA(G,p) is a polynomial for every protocol A. Observe that

PA(G,p) determines different possible combinations of dead and live edges such that protocol A has surviving essential trails. When counting different possible combi- nations of live and dead edges we obtain terms in PA (G, p) of the form akpkqiE(G)I-k, where q = 1 - p and 0 ::; k ::; IE (G) 1. In PA (G, p), the coefficient ak is the number of combinations of edges we can have in ( G, p) such that k edges are alive, IE (G) I - k edges are dead, and the message can still be sent from s to r using the protocol A.

Since q = 1- p, our term akpkqiE(G)I-k = akpk(1 - p)IE(G)I-k is a term written in only one variable. So, using the Binomial Theorem,

Therefore p(G,p) and PA(G,p) are polynomials in p. 0

Lemma 3.3.5. Deficiency Lemma

Given an s,r-graph (G,p), let A0 be the CFP ofG and It:;;;; Ao. If P1, ... ,Pk

are the s, r-paths that use at least one instruction from I, and Zi is the event that CHAPTER 3. THE PROBABILISTIC MODEL 47 path Pi is alive, but every s, r-trail not using an instruction from I is dead, then

Proof. Let T1, T2, ... , Tm be the s, r-trails in G using at least one instruction from

I, where, without loss of generality, Ti =Pi for 1 :::; i :::; k. Let Zi be the event that

Ti is alive, but every s, r-trail other than T1, T2, ... , Tm is dead.

First, we show that dAo\I(G,p) = P(Z1 U ... U Zm)· Let Z denote the event that some s, r-trail survives, and let Z' denote the event that some s, r-trail

that is essential for the protocol A 0 \ I survives. Note Z' is contained in Z and

Z- Z' = Z1 U ... U Zm, since Z- Z' is the event that only those s, r-trails that use an instruction in I survive.

Note p(G,p) = P(Z) and PAo\I(G,p) = P(Z'). So, by the Difference Rule,

dAo\I(G,p) = p(G,p)- PAo\I(G,p)

= P(Z)- P(Z') = P(Z- Z')

= P(Z1 U ... U Zm)·

Now, recall by Proposition 3.2.1, each trail Ti with i > k contains an s, r­

path P. If P = Pj for some 1 :::; j :::; k, then Zi ~ Zj. Otherwise, Pis an s, r-trail other than T1, T2 , ••• Tm, and thus Zi = 0.

Altogether we conclude that Z1 u Z 2 u ... u Zm = Z1 u ... U Zk and the result

follows. 0

Theorem 3.3.6. Monotonicity of the Deficiency CHAPTER 3. THE PROBABILISTIC MODEL 48

Suppose that (G,p) is an s, r-graph and A1 , A2 , A0 are protocols such that

Ao is the CFP of G. If A1 ~ A2 ~ Ao then dA1 2: dA2.

Proof. By Theorem 1.2.2, since A1 ~ A2 , PA1(G,p):::; PA2(G,p). Thus, by Defini­ tion 3.3.2 of Deficiency,

dA2(G,p) = p(G,p)- PA2

:S: p(G,p)- PA1

= dAJG,p).

0 49

Chapter 4

Maximum Protocol Reliability

4.1 Protocols with Reliability p

In this section we would like to find finite protocols in as, r-graph G that gives us the maximum protocol reliability, p.

Definition 4.1.1. For every p we define

p(G,p) = max{pA(G,p): A is a finite protocol}.

In Examples 3.1.2 and 3.2.3 we noticed that the CFP of a path and that of a cycle is finite and so the reliability polynomial p is p. In general, in graphs in which the CFP is finite, i.e. those without a minor from the forbidden minor characterization, p = p. However, if the CFP is not finite, p(G,p) < p(G,p) for all p with 0 < p < 1. Thus, it is interesting to determine p for the 10 forbidden minors in Section 2.3.

In Example 3.3.1 we describe finite PFP's A in T0 that have a circuit in­ struction removed from each corresponding essential circuit and give us a protocol CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 50 reliability of

We also showed that PA is at least as large as the protocol reliability for any other finite protocol, and thus p(T0 ,p) = PA(T0 ,p). Hence, each of these finite PFP's in To, and similarly in F0 in Example 3.3.3, simultaneously gives us the maximum protocol reliability for all values of p. In each of the other 8 forbidden minors we can also find PFP's that give us p simultaneously for all values of p. However, in general we may have different protocols A for different values of p that achieve p.

4.2 Using Deficiencies to find p

We can use deficiencies to find the PFP A in a graph that achieves p for any given graph. The reason we would like to use deficiencies instead of calculating the reliability of a protocol is because deficiencies are a lot easier to calculate, and there is less room for error. When calculating the deficiency of a PFP without a certain circuit instruction, we noticed that a live path with the circuit instruction usually cannot have other live edges not on that path. Then other paths that do not use the circuit instruction would be alive as well. So, we are usually just calculating the probability of having one or two paths alive instead of a combination of all the possible paths being alive.

Example 4.2.1. Let us consider the graph B 0 and calculate the deficiency poly­ nomial with respect to a PFP A. We remove instructions that contribute to es- CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 51

sential circuits from the CFP A 0 to find the protocol A that gives us p(Bo,p). As noted in Table 2.2, in Section 2.3, B 0 has one essential circuit C = 1, 2, 3, 4, 5, 3, 1 with circuit instructions I(; = {123, 234, 345,453, 531, 312}. The contributing in­ structions are Ic = I(; U { s12, s13, s43, s45, 12r, 134, 235, 32r, 35r, 45r, 431, 532}

= A 0 - {132, 135,432, 435}. Observe that 132 is in an s, r-walk s, 1, 3, 2, 1, 3, 4, 5, r which contains the circuit instruction 345. However, 312 ¢ A 0 . Likewise, the instruction 135 is in the s, r-walk s, 1, 3, 5, 4, 3, 1, 2, r which contains the circuit in­ struction 312, but 354 ¢ A 0 . All other s, r-walks with 132 or 135 and a circuit instruction require an instruction not in A 0 . Thus, 132, 135 ¢ Ic. We want to remove instructions in Ic to find p.

Suppose we wish to remove the instruction c1 . By the Deficiency

Lemma (3.3. 5), to calculate dAo\{q} (B0 , p) it is enough to consider the paths affected by the removal of c1 . If there is more than one path affected, we will use the principle of inclusion-exclusion to calculate the deficiency. Let us examine the different possible PFP's and calculate their corresponding deficiencies.

If the instruction 123 is removed from the CFP A 0 two paths will be affected,

P 1 = s, 1, 2, 3, 4, 5, r and P2 = s, 1, 2, 3, 5, r, since it is found in both paths, whereas no others, r-path uses 123. Observe that H has length six and Pz has length five.

If the edges of P1 were alive along with other edges in B 0 , then there would be other paths in B 0 with live edges, but the only other path that can be alive with P1 is

Pz. The probability of the event of only having path P1 and P2 with live edges is CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 52

7 3 p q . So, having H with live edges, but no others, r-trail except possibly P2 , would

6 4 7 3 contribute p q + p q to the deficiency. Observe that P2 can only be alive with edge

s4, 34, or 45 without creating other paths, and we cannot use more than one of

these three edges. Thus, having g alive, but no other s, r-trail, would contribute

5 5 4 6 4 7 3 dAo\{123} (Bo,P ) =p q + P q +p q ·

A similar argument can be made for the instructions 345, 312 and 453.

Circuit Path Containing Path Containing Deficiency Instruction Instruction Instruction Polynomial 123 s,1,2,3,4,5,r s, 1,2,3,5,r p5q5 + 4p6q4 + p7 q3 234 s,1,2,3,4,5,r p6q4 345 s,1,2,3,4,5,r s, 1,3,4,5,r p5q5 + 4p6q4 + p7 q3 453 s,4,5,3,1,2,r s,4,5,3,2,r p5q5 + 4p6q4 + p7q3 531 s,4,5,3,1,2,r p6q4 312 s,4,5,3,1,2,r s,4,3,1,2,r p5q5 + 4p6q4 + p7 q3

Table 4.1: Circuit Instructions for B 0

In Table 4.1, we list the circuit instructions, paths affected by the removal of

the corresponding circuit instruction, and the deficiency with respect to the protocol

that does not include the circuit instruction, but contains all other instructions

from the CFP. From the table we observe that the instructions 123, 345 , 312 and

453 have the same deficiency. Also, the instructions 234 and 531 have the same

deficiency. Observe that the removal of the instruction 234 would only affect a

single path of length six, namely P 1. Thus, dAo\{243} is the probability that P1 is the

6 4 6 4 only live trail, and so it equals p q . Similarly dAo\{531} = p q . CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 53

Contributing s, r-Path s, r-Path (and Edge) Instruction Containing Instruction Containing Instruction s12 s,1,2,r s1,2,3,4,5,r s13 s,1,3,2,r s, 1, 3, 4, 5, r and 12 s43 s,4,3,2,r s,4,3,1,2,r and 45 s45 s,4,5,r s,4,5,3,1,2,r 12r s, 1,2,r s,4,5, 1,2,r 134 s, 1,3,4,5,r s, 1, 3, 4, 5, r and 12 235 s, 1,2,3,5,r s,1,2,3,5,r and 45 32r s,4,3,2,r s, 4, 5, 3, 2, r and 12 35r s,1,3,5,r s, 1, 2, 3, 5, r and 45 431 s,4,3,1,2,r s, 4, 3, 1, 2, r and 45 45r s,4,5,r s,1,2,3,4,5,r 532 s,4,5,3,2,r s, 4, 5, 3, 2, r and 12

Table 4.2: Contributing Non-Circuit Instructions for Bo

To find p(B0 ,p) we want to find instructions whose removal from the CFP would stop the essential circuit and one with the least deficiency. Let us consider the remaining instructions that contribute to C. In Table 4.2 we list the contributing instructions that are not circuit instructions in column one. The paths listed in the

without corresponding instruction. The path and edge combinations listed in the third column contribute the term p6q4 to the deficiency of the PFP. So for any c1 E

6 4 Ic-10, dAo-{cl} > p q . Thus, by Theorem 3.3.6 (Monotonicity of the Deficiency), CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 54

Ao \ {234} and Ao \ {531} are the only protocols A with PA(Bo,P) = p(Bo,p) and

p(Bo,p) = p(Bo,P)- dA(Bo,P)

= (2p3q7 + 18p4q6 + 40p5q5 + 210p6q4

+ 100p7 q3 + 43psq2 + 10pgq + p1o) _ p6q4.

= 2p3q7 + 18p4q6 + 40p5q5 + 209p6q4

7 3 8 2 9 10 + 100p q + 43p q + 10p q + p .

Similar considerations show that each of the other members of the list of

10 forbidden minors from Section 2.3 also has a specific PFP that achieves p for all values of p simultaneously. We now look at more examples of p for the other forbidden minors.

Example 4.2.2. T8 (rr is similar.)

Observe that Ts has two essential circuits C1 = 1, 2, 3, 1 and C2 = 2, 3, 4, 2.

Contributing instructions are Ic = {s12, s13, s31, s34, s42, s43, 123, 124, 134,231,

234,312,321,342,421,423,431, 21r, 24r, 31r, 34r }. Observe that we cannot remove only one instruction from the CFP A 0 to obtain a finite PFP. We must remove at least two instructions, ideally we want one instruction that will stop cl and one that will stop c2.

Consider A 0 \ {123, 234}. Note that the removal of 123 stops the essential circuit C1 , and 234 stops C2 . The only path with instruction 123 or 234 is the path g = s, 1, 2, 3, 4, r. In fact, P1 contains both instructions. Since P 1 is of length 5 CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 55

and no other edge can be alive with P1 without creating another s, r-path other than

When we consider removing instructions 231 and 423 from A 0 we notice that the PFP would be finite. The path P2 = s, 4, 2, 3, 1, r uses instructions 231 and 423, there is no other path with either one of these instructions, and we cannot add any

5 5 edges without creating another path. So, dAo\{231,423}(Ts) = p q as well.

In Table 4.3, we list all the contributing instructions and paths that require the listed instruction. Note that there is more than one path listed for all the in-

Contributing s, r-Path s, r-Path Instruction s12 s, 1,2,4,r s, 1,2,3,4,r s13 s, 1,3,4,r s, 1,3,2,4,r s31 s,3, 1,r s,3,1,2,4,r s34 s,3,4,r s,3,4,2, 1,r s42 s,4,2,1,r s,4,2,3,1,r s43 s,4,3,1,r s,4,3,2,1,r 123 s, 1,2,3,4,r 124 s, 1,2,4,r s,3,4,2,4,r 134 s,1,3,4,r 231 s,4,2,3, 1,r 234 s,1,2,3,4,r 312 s,3,1,2,4,r 321 s,3,2,1,r s,4,3,2,1,r 342 s,3,4,2, 1,r 421 s,4,2,1,r s,3,4,2, 1,r 423 s,4,2,3, 1,r 431 s,4,3,1,r 21r s,3,2, 1,r s,4,3,2, 1,r 24r s,3,2,4,r s, 1,3,2,4,r 31r s,3, 1,r s,4,2,3, 1,r 34r s,3,4,r s,1,2,3,4,r

Table 4.3: Contributing Instructions for Ts CHAPTER .4. MAXIMUM PROTOCOL RELIABILITY 56 structions except for 123, 134, 231, 234, 312, 342, 423 or 431. Also, one of the paths listed for each instruction, except for 134 and 431, has length 5. Notice that

P 1 has the instructions s12, 123, 234, and 34r, but s12 and 34r are in other paths besides P1 . P2 has the instructions s42, 423, 231, and 31r, but s42 and 31r are in other paths besides P2 .

Also, removing just one of the instructions 124, 134, 312, or 342 from the

CFP A 0 is not enough to stop both essential circuits, so any finite protocol missing either of these instructions is also missing another instruction from Ao. If we remove both 134 and 431, the essential circuits remain, and we would still have to remove another instructions from the CFP to find a finite PFP without 124 and

134.

Thus, the removal of any other combinations of contributing instructions from A 0 , besides 123 and 234 or 231 and 423, to find a finite PFP would give a defi­

5 5 ciency that will be more thanp q . Thus, PAo\{123,234}(Ts) = p(Ts) = PAo\{123,234}(Ts)·

Example 4.2.3. T2

The forbidden minor T2 has two essential circuits C1 = 1, 2, 3, 1 and C2 =

1, 3, 2, 1. The contributing instructions are listed in the first and fourth column of Table 4.4. In columns 2, 3, and 5 we list paths that contain the corresponding instruction.

Note that the instructions in the fourth column are only found in one s, r­ path. The instructions 123, 132, 213, 214, 231, 312, 315, 321, 412, and 513 are CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 57 in one path of length 6, and so would contribute the term p6q5 to the deficiency of the PFP without exactly one of these instructions. Also, the addition of any edge with one of these paths of length six, would create another live path that does not use one of the instructions 123, 132, 213, 214, 231, 312, 315, 321, 412, and 513.

So, the deficiency of the PFP without exactly one of the instructions 123, 132, 213,

6 5 214, 231, 312, 315, 321, 412, and 513 would just be p q .

Contributing s, r-Path s, r-Path Contributing s, r-Path Instruction Instruction s41 s,4,1,5,r s,4, 1,2,3,5,r 123 s,4,1,2,3,5,r s42 s,4,2,3,5,r s,4,2, 1,3,5,r 132 s,5,1,3,2,4,r s51 s,5,1,4,r s,5,1,3,2,4,r 213 s,4,2,1,3,5,r s53 s,5,3,2,4,r s,5,3,2,1,4,r 214 s,5,3,2,1,4,r 124 s,5,1,2,4,r s,5,3, 1,2,4,r 215 s,4,2,1,5,r 135 s,4,1,3,5,r s,4,2, 1,3,5,r 231 s,4,2,3, 1,5,r 14r s,5,1,4,r s,5,3,2,1,4,r 312 s,5,3,1,2,4,r 15r s,4,1,5,r s,4,2,3,1,5,r 314 s,5,3, 1,4,r 24r s,5,1,2,4,r s,5, 1,3,2,4,r 315 s,4,2,3,1,5,r 235 s,4,2,3,5,r s,4, 1,2,3,5,r 321 s,5,3,2, 1,4,r 324 s,5,3,2,4,r s,5, 1,3,2,4,r 412 s,4, 1,2,3,5,r 35r s,4,2,3,5,r s,4,2, 1,3,5,r 413 s,4,1,3,5,r 421 s,4,2,1,5,r s,4,2,1,3,5,r 415 s,4,1,5,r 423 s,4,2,3,5,r s,4,2,3,1,5,r 512 s,5,1,2,4,r 531 s,5,3,1,4,r s,5,3, 1,2,4,r 513 s,5,1,3,2,4,r 532 s,5,3,2,4,r s,5,3,2,1,4,r 514 s,5,1,4,r

Table 4.4: Contributing Instructions for T2

Observe that the removal of only one of the instructions 214, 315, 412, and

513 from the CFP does not stop any essential circuit. However, the removal of any

one instruction in 101 = {123, 231, 312} from the CFP would not allow the use of the

essential circuit C1 and the removal of any one instructions in 102 = {132, 321, 213} CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 58

would stop the use of the essential circuit C2. Thus, for any c1 E Ic1 and c2 E Ic2 ,

A= A 0 - { c1, c2} is finite. Since c1 and c2 are in different paths and we cannot have both paths alive without creating another path that does not use the instructions c1 or c2,

Note that the path P 1 = s, 5, 3, 2, 1, 4, r has the instructions 214 and 321 and when 321 is removed, the instruction 214 cannot be used. So, dAo-{321}(T2) = dAo-{321,214}(T2)· The path P2 = s, 5, 1, 3, 2, 4r has the instructions 513 and 132 and when 132 is removed, the instruction 513 cannot be used. So, dAo-{132}(T2) = dAo-{132,513}(T2)· The instructions 315 and 231 are in the same path, and when 231 is removed we cannot use the instruction 315. Thus, dAo-{231} (T2) = dAo-{231,315} (T2).

We have a similar result with the instructions 513 and 132 since they lie on the same path, so dAo-{132}(T2) = dAo-{132, 51 3}(T2)· Therefore, the PFP's A1 = Ao­

{321,214,c2}, A2 = A 0 - {132,513,c2}, A 3 = A 0 - {c1,213,315}, and A4

Ao - { c1, 132, 513} are finite, and their corresponding deficiency is 2p6q5.

If all the instructions 215, 314, 413, 415, 512, and 514 in column 4 are

removed from A 0 , the essential circuits remain. Thus, the removal of any combi­ nation of the instructions 215, 314, 413, 415, 512, and 514 from the CFP is not a finite protocol, and we would have to remove another contributing instruction. So the deficiency of a finite PFP without one of these instructions would have a p4 q7 or p5q6 plus the deficiency of the other instruction that is not 215, 314, 413, 415, CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 59

512, or 514.

The contributing instructions in column one are in at least two paths. Note that one of the paths listed is of length 6, and, so, the contributing instructions in column 1 would contribute more than p6 q5 to the deficiency of the PFP without those instructions.

6 5 Thus, 2p q is the minimal deficiency for T2 and is obtained with protocol

A, A1 , A 2 , A3 , or A4.

Example 4.2.4. F1

Notice that F 1 has two essential circuits C1 = 1, 2, 3, 4, 1 and Cz = 3, 4, 5, 3.

In Table 4. 5 we list the contributing instructions in the first column. In the second column we list one s, r-path that uses the corresponding contributing instructions.

In the third column we list a path of length 6 or a path of length 5 and an edge that show the existence of the term p6 q4 in the deficiency of the PFP without the listed instruction. Note that there is exactly one path of length 6, and no other s, r-path, that uses the instructions 234, 345, 341, 453, or 534. The addition of any other edge with the listed path for the instructions 234, 345, 341, 453, and 534 would create another path that does not use the corresponding instruction. Therefore the

6 4 deficiency of A 0 - {c} forcE {234,345,341,453,534} is p q . Also, the instruc­ tions 234, 345, 341, 453, and 534 are circuit instructions and the removal of the instructions would stop the corresponding essential circuit. So, if we remove one instruction in I = {234, 341} and one in 1'(; = {345, 453, 534} from the CFP Ao, 01 2 CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 60

Contributing s, r-Path s, r-Path (and Edge) Instruction s12 s, 1,2,r s,1,2,3,4,5,r s14 s,1,4,5,r s, 1, 4, 3, 5, r and 12 s53 s,5,3,2,r s,5,3,4,1,2,r s54 s,5,4,1,2,r s,5,4,3,2,r and 12 123 s, 1,2,3,5,r s, 1,2,3,4,5,r 143 s,1,4,3,2,r s, 1,4,3,5,r and 12 145 s,1,4,5,r s,1,4,5,3,2,r 12r s, 1,2,r s,5,3,4,1,2,r 235 s,1,2,3,5,r s,1,2,3,5,r and 34 32r s,5,3,2,r s,5,4,3,2,r and 12 35r s,1,2,3,5,r s, 1,4,3,5,r and 12 412 s,5,4,1,2,r s,5,3,4, 1,2,r 432 s,5,4,3,2,r s, 5, 4, 3, 2, r and 12 435 s,1,4,3,5,r s, 1, 4, 3, 5, r and 12 45r s,1,4,5,r s,1,2,3,4,5,r 532 s,5,3,2,r s,1,4,5,3,2,r 541 s,5,4, 1,2,r s,5,4,1,2,r and 34 543 s,5,4,3,2,r s, 5, 4, 3, 2, r and 14 234 s, 1,2,3,4,5,r 345 s, 1,2,3,4,5,r 341 s,5,3,4,1,2,r 453 s, 1,4,5,3,2,r 534 s,5,3,4,1,2,r

Table 4.5: Contributing Instructions for F1

the PFP obtained would be finite.

Observe that 234 and 345 are in the same path, and so are 341 and 534.

So, if we let Z234 be the event that only s, r-paths with instruction 234 remain and let Z345 be the event that only s, r-paths with instruction 345 remain, then CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 61

Note that if we remove any instruction from the CFP that has two paths

6 4 listed in Table 4. 5, then we will get a deficiency greater than p q , since the second path alone contributes that term. Consider the deficiency of the PFP without the instructions 453 and 234 (or 341}. We cannot have the s, r-path with 453 and a s, r-path with instruction 234 (or 341} alive since then we would have a live s, r-path that does not require one of the two instructions. So, the deficiency of

6 4 dAo-{453,234} = 2p q = dAo-{453,341}·

6 4 Therefore, the minimal deficiency for F1 is p q , and this deficiency is only achieved by removing {341, 534} or {234, 341} from the CFP.

Example 4.2.5. B 2

In this example we will just show the computation of the minimal deficiency, which will give us the finite PFP with maximum reliability. Observe the Bz has six essential circuits C1 = 1, 2, 3, 4, 5, 3, 1, C2 = 1, 3, 5, 4, 3, 2, 1, C3 = 1, 2, 3, 5, 4, 3, 1,

C4 = 1,3,4,5,3,2,1, C5 = 1,2,3,4,5,3,2,1,3,5,4,3,1, andC6 = l,2,3,5,4,3,2,1,

3, 4, 5, 3, 1 and all of these have to be stopped. Notice that C 5 and C6 share some common instructions with C1 , C2 , C3 and C4 . In Table 4.6 we list the circuit in­ structions since the removal of a combination of these instructions from Ao will give us the PFP with minimal deficiency. Observe that the removal of a circuit instruction will affect 2 or 4 of the essential circuit. So, the finite PFP would have

2 or more instructions removed from the CFP. Note that the removal of any one in­ structions listed in Table 4. 6 will affect 4 paths. However, some of the instructions CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 62

Circuit c1 c2 c3 c4 c5 c6 Number of Paths Instruction with Instruction 123 J J J J 4 234 J J 4 345 J J J J 4 453 J J J J 4 531 J J 4 312 J J J J 4 135 J J 4 354 J J J J 4 543 J J J J 4 432 J J 4 321 J J J J 4 213 J J J J 4 235 J J 4 431 J J 4 134 J J 4 532 J J 4

Table 4.6: Circuit Instructions for B 2

share a common path. None of the instructions that give us the PFP with the max- imum protocol reliability have a common s, r-path since the instructions removed from CFPA0 are found in two essential circuits that do not share any common instructions.

The finite PFP's that do give us the maximum protocol reliability are A1, which has all instructions in the CFP except for c1 E {123, 312} and c2 E {321, 213}, and A2 , which does not contain the instructions c3 E {345, 453} and C4 E {354, 543} and has all other instructions in A 0 . Both A1 and A2 are finite. We will only con- sider one possible such protocol, since the other possibilities for A1 and A2 are quite a similar. CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 63

Consider A 1 = A0 -{123, 321 }. We will use the inclusion-exclusion principle to calculate the deficiency of A 1 . In order to calculate dAo-{321}(B2) and so we want to calculate the different possible ways of sending a message from s to r requiring the use of a path with instruction 321. In order to use 321, we must have live edges

5 4 s6, 12, 23, 17, 7r and dead edges 6r, 13, 27, and s7, giving us the factor p q .

This leaves us the choice of the edges 34, 35, 45, 46, 56, which induces a subgraph isomorphic to the Diamond graph in Example 3.2.4 that has a protocol reliability of

2 3 3 2 4 5 5 4 D = p(D) = 2p q + 8p q + 5p q + p . Thus, dAo-{321}(B2) = p q D.

When calculating dAa-{123} ( B 2), in order to use 123, we must have live edges

5 4 s7, 17, 12, 23, 6r and dead edges s6, 27, 13, 7r, giving us the factor p q . This leaves a choice of the edges 34, 35, 45, 46, 56, which, as above, has a protocol

2 3 3 2 4 5 5 4 reliability of D = p(D) = 2p q + 8p q + 5p q + p . Thus, dAo-{321}(B2) = p q D.

Note that if we wish to use instructions 123 and 321 to send a message from s tor we would require a trail from 6 to 3 and edges 16, 26, 13, and 23 along with s6 or s7 and 6r or 7r to be alive. Without loss of generality, suppose s7 and 6r are alive, and say the path 6, 4, 3 is alive as well. Observe that we would have a live s, r-path s, 6, 4, 3, 1, 7, r that does not use the instruction 123 or 321. Therefore, we cannot have an s, r-trail with instructions in A 0 that must use the instructions 123 and 321 without creating a path that does not use the two i.nstructions 123 and 321.

Therefore, by the inclusion-exclusion principle, CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 64

5 4 The deficiency 2p q D can be shown to be minimal. So A0 - {123, 321} is a fi­ nite protocol that gives us p. The deficiency for the other possible A1 's and A2 's, described earlier are also minimal and give us p.

Example 4.2.6. Be

In the graph Be, a PFP that yields minimal deficiency is A= Ao- {c1,c2},

forcl E {123,231} and c2 E {132,321}. Note that the removal of123 or231 from

A 0 stops two essential circuits C1 = 1, 2, 3, 1, 4, 5, 6, 4, 1 and C3 = 1, 2, 3, 1, 4, 6, 5, 4, 1.

The removal of 132 or 321 stops the essential circuits C2 = 1, 3, 2, 1, 4, 5, 6, 4, 1 and

C4 = 1, 3, 2, 1, 4, 6, 5, 4, 1. Thus, A is finite. We will only consider one possible protocol for A, since the other possibilities for A are treated similarly.

Consider A= A0 - {231, 132}. To calculate dA(Be) we require the use of an s, r-trail with instruction 231. In order to use 231, we must have live edges s8, 28,

23, 13, 7r and dead edges 8r, 38, 12 and s7, and a live path from 1 to 7. So, for

5 4 the required live and dead edges we have the factor p q , and this will be multiplied by the reliability we obtain for the path from 1 to 7.

If in the diamond portion of the graph (the subgraph induced by vertices 4,

5, 6, and 8} there is a live path from 8 to 4 (which happens with probability D} then

2 the only appropriate choice is 17 alive and 14,47 dead, yielding a factor of Dpq .

If the diamond does not have a live path then we can have one, two or three of the edges 14, 17, and 47 alive. We can have just 17 alive; 17 with 14 or 47 alive; live path I, 4, 7; or all three edges alive. So, if the diamond has no live path the factor CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 65 is (1 - D)(pq2 + 3p2q + p3).

Therefore

5 4 2 2 2 3 dAo-{231}(Be) = p q (pq D + (1- D)(pq + 3p q + p ))

= p5q4(pq2 _ (1 _ D)pq2 + (1 _ D)(pq2 + 3p2q + p3))

= p5q4(pq2 + (1 _ D)(3p2q + p3))

= p6q6 + p5q4(1 _ D)(3p2q + p3)

= p6q6 + p7q4(1 _ D)(3q + p).

6 6 7 4 Similarly, dAa-{132}(Be) = p q + p q (1- D)(3q + p), except that this time s8 and

7r are dead and s7 and 8r are alive.

Note that if we wish to be able to use instructions 231 and 132 to send a

message forms tor, then by what we said above, s7, s8, 7r, 8r would all have to be

alive. However, this would mean that we have the s, r-path s, 7, r, which does not use

either of these instructions. Therefore, dA(Be) = dAo-{231}(Be) + dAo-{132}(Be) =

2p6q6 + 2p7q4 (1- D)(3q + p), which can be verified to be the minimal deficiency

for Be.

Example 4.2. 7. B 1

In the graph B 1, there are several combinations of circuit instructions whose

removal from A0 will give us a finite PFP with minimal deficiency. Let c1 E

{234,345}, c2 E {431,543}, c3 E {453,531}, and c4 E {235,354}. Note that

the removal of c1 or c3 will stop the essential circuit C1 , and the removal of the CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 66

instructions c2 or c4 will stop C2 . Some of the combinations of instructions whose

removal from A 0 will give us a PFP with minimal deficiency include: c1 and c2;

345 and 354; c3 and c4 ; and 453 and 543.

We only look at one example of a finite PFP, A= A 0 - {234,431}, with minimal deficiency (so, c1 = 234 and c2 = 431}. Note that we cannot use in­ structions 234 and 431 simultaneously without creating a s, r-path that does not use either one of these instructions. So dA(B1) = dAo-{234}(Bl) + dAo-{431}(Bl)·

We look at the deficiency of A0 - {234}. Observe that in order to use the in­ struction 234, we must have live edges s1, 12, 23, 34, 6r and dead edges s6, 13, 2r, giving us the factor p5q3. Then we have different ways of choosing from the re­

maining edges. We can have one live edge 46 with dead edges 35, 45, 56. We can

have two live edges 45 and 56 with dead edges 35 and 46. Other possible live edges

are 46 and one of the edges 35, 45 or 56 and the other edges dead. The only

possible combination of three live edges are 45, 46, and 56 with dead edge 35. So,

dAo-{234}(B1) = p5q3(pq3 + 4p2q2 + p3q) = p6q4(q2 + 4pq + p2).

When calculating dAo-{431 }, we must have live edges s6, 12, 13, 34, and 2r

and dead edges sl, 23, and 6r, giving us the factor p5q3. We can have the one edge

46 alive and dead edges 35, 45, 56. Two edges could be alive 45,56 or the edge 46 and

one of the edges 35, 45, 56. We can have three edges alive 45, 46, and 56 and dead

edge 35. Therefore, dAo-{431}(Bl) = p5q3(pq3 + 4p2q2 + p3q) = p6q4(q2 + 4pq + p2).

4 Thus, dA = 2p6q (q2 + 4pq + p2). CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 67

Example Minor Instruction to Instruction to Deficiency Graph Remove from C1 Remove from C2 3.3.3 Fo 123,234,341,412 143,432,321,214 2p5q3 4.2.4 F1 234 345 p6q4 (or 341) (or 534) 3.3.1 To 123,231,312 132,321,213 2p4q5 4.2.2 Ts 123 234 p5q5 (or 231) (or 423) 4.2.2 Tr 123 234 p5q5 (or 231) (or 423) 4.2.3 T2 123,231,312 132,321,213 2p6qb 4.2.1 Bo 234,531 poq4 4.2.7 B1 234,345 543,431 2p8q4 + 8p7 q5 (or 345) (or 354) +2p6q6 (or 453, 531) (or 235, 354) (or 453) (or 543) 4.2.5 B2 123,312 213,321 2Dp5q4 (or 345, 453) (or 354, 543) 4.2.6 Be 123,231 132,321 2p6q6 + 2p7 q4 ·(1- D)(3q + p)

Table 4.7: Minimum Deficiency for Finite PFP

In Table 4.7 we list the ten forbidden minors with circuit instruction(s) whose removal yields p together with their corresponding deficiency, where D = p( D, p) from Example 3.2.4.

From Table 4. 7 we notice that we can find a protocol for each graph minor by removing an instruction listed in the first column corresponding to C1 and one

listed in the second column corresponding to C2 for each of the graphs. In Bo we notice we only have to remove one instruction since there is only one essential

circuit. We also notice that in F0 , To and T2 it does not matter which one of the circuit instructions is removed from each essential circuit to find a PFP with p. CHAPTER 4. MAXIMUM PROTOCOL RELIABILITY 68

However, in the other forbidden minors we cannot simply remove any of the circuit instructions because we need to remove one of the listed circuit instructions from each essential circuit to find p. 69

Chapter 5

Comparing Graphs

In Chapter 3 we observed that in the 10 forbidden minors pis a polynomial. Since, according to Chapter 4, the deficiencies of each of the 10 forbidden minors are polynomials, p is also a polynomial. Recall that in order to find p, we had to find the protocol that would give us the maximum protocol reliability for a given p and that this protocol always existed and worked for all values of p. However, in some

graphs perhaps we would want to use a protocol A1 for values of p between 0 and c,

and protocol A2 may be better to use when pis between c and 1, for some 0 < c < 1.

So, then there would not be one protocol A that would give us the maximum and p would be a piecewise function. To explore the idea of p being a piecewise function, we examine graphs that yield a better reliability polynomial p for some values of p when the graphs are compared to one another. This comparison can then inform us how best to form a network if we know in advance whether the value for p is close to 0 or close to 1.

Definition 5.0.8. Let c be a real number. Two polynomials f, g have a crossing CHAPTER 5. COMPARING GRAPHS 70 at c if f(c) = g(c). The order of the crossing is the multiplicity of x = c as a zero

of f(x) - g(x). Two s, r-graphs G, H have a p-crossing at c if p(G, c) = p(H, c)

and have a p-crossing at c if p(G, c)= p(H, c), where 0 < c < 1.

5.1 Path and Cycle Comparison

Let us compares, r-paths and s, r-cycles to see if one of these graphs yields a better

reliability polynomial for certain values of p.

Example 5.1.1. From Example 3.1.2 and 3.2.3, we know that the reliability poly-

2 nomials are p(Pm,P) = pm and p(C2(m+1),p) = 2p(m+l) -p (m+l). Since the CFP for

a path and cycle is finite, then p = p. Observe that when m = 1 we compare C2(2)

2 4 and P1 , where p(C2(2),p) = 2p -p and p(P1 ,p) = p. These reliability polynomials

intersect when p = c ~ 0.618 (see Figure 5.1).

Protocol Reliability OJl

p

Figure 5.1: P 1 and C4 p-Crossing CHAPTER 5. COMPARING GRAPHS 71

Therefore, for p between 0 and 0.618 it is best to use Pt as the network and for p between 0.618 and 1 it is best to use C 2(2) as the network (see Figure 5.1).

We find a similar result when we look at the case form= 2. When m = 2, P2 and

C 2(3l have a p-crossing when p is approximately 0.544. These discoveries lead us to the following theorem:

Theorem 5.1.2. Path and Cycle Reliability Polynomial

There is a rm such that

if 0 < p < rm: p(Pm,P) > p(C2(m+1),p). (5.1)

if rm < p < 1: p(C2(m+l),P) > p(Pm,p). (5.2)

Moreover, rm is decreasing in m and rm is a p-crossing of multiplicity 1.

The following definition will be important in the proof of the theorem as

well as in Chapter 6.

Definition 5.1.3. Let G, H be two s, r-graphs. The p-difference, respectively p­

difference, of G and H is

b.p(G,H,p) = p(G,p)- p(H,p)

b.p(G,H,p) = p(G,p)- p(H,p).

Proof. (Theorem 5.1.2) Consider a path Pm of length m and a cycle C2(m+l) of

length 2(m + 1), form> 0. We wish to show that C2(m+l) and Pm has a p-crossing

point between 0 and 1, indicating that at some values of p, the maximum protocol CHAPTER 5. COMPARING GRAPHS 72 reliability of Pm may be better than the maximum protocol reliability of C2(m+l) and vise versa.

2 Therefore, a p-crossing would be a root of the polynomial 9m (p) = pm+ -

2p + 1. (Observe that p = 0 is a root of the polynomial, but we are looking at values between 0 and 1.) We consider all positive values of p to see if there is

a p-crossing. Note that 9m(P) has two variations between the coefficients of the

terms pm+2 and p, and the coefficients of p and 1. By Descartes' Rule of Signs, we

conclude that 9m(P) has zero or two positive roots for all positive p. Observe that

9m(1) = 1m+2 - 2 · 1 + 1 = 0. Thus, p = 1 is a root of 9m(P) and also a root of the

p-difference. This implies that the polynomial must have exactly one other positive

root.

Consider the p-difference of Pm and C2(m+ll when p = ~· Note that

1)2m+2 =- ( -2 < 0 . CHAPTER 5. COMPARING GRAPHS 73

Now, we consider the p-difference of Pm and C2(m+l) when p = ~- Note that

3) (3)2m+2 (3)m+1 (3)m !:::..P ( C2(m+l),Pm,4 =- 4 +2 4 - 4

= _ ( ~) 2m+2 + 2 ( ~) ( ~) m _ ( ~) m

= _ ( ~) 2m+2 + ( ~) . ( ~) m _ ( ~) m

= _ (~)2m+2 + (~)m (~ _ 1)

= _ (~) 2m+2 + (~) m (~)

~ mm (- mm+2 + ~) .

Proof of Claim: First, observe that ~m > 0 for all positive p. So, if we show

that ( -(~)m+ 2 + ~) > 0, then, since (~m) > 0, (~m)( -(~)m+ 2 + ~) > 0. Observe

that ( -(~)m+ 2 + ~) > 0 if and only if (~)m+ 2 < ~-

We let P(n) be the statement (~)n+ 2 < ~- We will use mathematical indue-

tion on n to prove that P(n) holds true for all positive n. 2 As a base case, evaluate the inequality P( n) when n = 1. Then ~ 1+ = ~~ <

32 1 64 2.

For the induction step, suppose the statement P(k) holds true for some value

of k, 1 S k. That is, assume ~k+ 2 < ~ for some k, 1 S k. We wish to show that

the case for n = k + 1 holds true as well.

By the induction hypothesis we obtain ~(k+1)+ 2 = (~)(~k+ 2 ) < (~)(~). How-

3)(1) 3 1 Th · 3(k+1)+2 1 S b h t· 1 . d t. ever, (4 2 = 8 < 2. at 1s, 4 < 2 . o, y mat ema 1ca m uc wn, CHAPTER 5. COMPARING GRAPHS 74

( -(~)k+2 + ~) > 0.

Therefore, since ( -(~)n+ 2 + ~) > 0 and (r) > 0, this implies

as desired. (QED for our claim.)

Since .6.p(C2(m+l), Pm, ~) < 0 and .6.p(C2(m+l), Pm, ~) > 0, by the Intermedi­

ate Value Theorem, there exists a 'Ym between~ and~ such that .6.p( C2(m+1)' Pm, '"tm)

= -('"Ym) 2m+2+2('"Ym)m+l_ bm)m = 0. That is, 1m is a root of -p2m+2+2pm+l_Pm

for p. Observe that 'Ym is the only root between 0 and 1 and '"fm has multiplicity 1.

Hence, for probability p, where 0 < p < 1, Pm and C2(m+l) have only one p-crossing

point strictly between 0 and 1, namely 'Ym·

As .6.p(C2(m+l), Pm, ~) < 0 this implies that for values of p between 0 and

'"fm, Pm has a better reliability polynomial and hence

Seeing that .6.p(C2(m+l), Pm, ~) > 0, for p between '"Ym and 1, C2(m+l) has a better

reliability polynomial and, thus,

Finally, to see that 'Ym > '"Ym+l for all m 2: 1 observe that

So, by the Intermediate Value Theorem, 'Ym+l is in fact in the interval (~, '"fm)·

0 CHAPTER 5. COMPARING GRAPHS 75

5.2 Comparing Polynomials

In Section 5.1 we looked at two specific graphs and their corresponding reliability polynomials. However, what if we were just given two polynomials in one variable

(x) to compare? Can we determine which one would yield a higher value for positive x close to zero and use this information to compare the reliability polynomials of the 10 forbidden graphs or any other graphs for that matter?

Definition 5.2.1. Let A and B be functions whose domain contains [0, 1],

A 0 such that for all X with 0 A(x).

A :::;0 B if there exists an E > 0 such that for all X with 0

A =o B if there exists an E > 0 such that for all X with 0

A <1 B if there exists an r:; > 0 such that for all X with 1- E A(x).

A :::;1 B if there exists an r:; > 0 such that for all X with 1- E

A =1 B if there exists an r:; > 0 such that for all X with 1- E

k m Theorem 5.2.2. Given two polynomials A(x) = Laixi and B(x) = Lbixi, i=O i=O where m, k > 0. Suppose there is a j such that 0 = b0 - ao = b1 - a1 = ...

= bj-1- aj-1· If 0 < b1 - a1, then A(x)

k m Proof. Given A(x) = L aixi and B(x) = L bixi, for 0 < x < 1. Suppose there is i=O i=O a value j such that 0 = b0 - a0 = b1 - a 1 = ... = b1_1- a1_1 and 0 < b1- aj. We wish to show 0 < B(x)- A(x) for positive x near zero. It suffices to consider B(x):dA(x). CHAPTER 5. COMPARING GRAPHS 76

k Define f(x) = L (bi - ai)xi-(j+l) and let M = 1 + max{\f(x)\ : 0 ::; x :'S 1 }. i=j+l Consider f = min { bj2~j, 1} and observe that for all x with 0 < x < f we get

m k :i (B(x)- A(x)) = L bixi-j- L aixi-j i=O i=O k = (bj- aj) + L (bi- ai)xi-j i=j+l

>- (b J -a)- J x · M

> (b1 - a1)- f · M b -a > (b -a)- M 1 1 - 1 1 2M b·- a = 1 1 > 0, as desired. 2

D

m m Lemma 5.2.3. If f(p) = L aipiqm-i then f(p) = L bkpk where q = 1- p and i=O k=O

Proof. In f(p) we replace q by 1- p and use the Binomial Theorem to expand the CHAPTER 5. COMPARING GRAPHS 77 polynomial. Therefore

m f(p) = L aipi(1- p)m-i i=O

~ ~"

~ ~ (~ ( m j i) (-l)'a,j)H)

~ ~ ( ~ ( '; ~: ) ( -1 )'-' "

= f (t (: =~) (-1)k-iai) pk. k=O t=O

0

m m Corollary 5.2.4. Let A(p) = L aipiqm-i and B(p) = L bipiqm-i, for 0 < p < 1 i=O i=O and where q = 1 - p.

a)If there is a j such that ai = bi for all 0 ::::; i < j, but a1 < b1 then A

b)If there is a l such that ai = bi for alll < i :S m, but at < bt then A <1 B.

Proof. a) By Lemma 5.2.3 CHAPTER 5. COMPARING GRAPHS 78

Suppose there is a j such that ai = bi for all 0 :::;; i < j, but aj < bj. So, for all 0 :::;; k < j, we have that the coefficient of pk satisfies

For k = j the coefficient of pk satisfies the following inequality:

For b) we replace p by 1 - q and use the Binomial Theorem to expand the polynomial.

m A( I - q) = l: ai(l - q)i(q)m-i i=O

= f (k-~=i ( k-~+i) (-l)k-m+iail) (letk=m+j-i) t=O k-m+t=O

= ~ ( ~ ( i . ) ( -l)k-(m-i)aiqk) ~ ~ k- (m-2) i=O k=m-i

= ~ ( ~ ( i . ) ( -l)k-(m-i)aiqk) ~ ~ k- (m-2) k=O i=m-k = ~ ( ~ ( i . ) ( -l)k-(m-i)ai) qk ~ ~ k- (m-2) k=O i=m-k CHAPTER 5. COMPARING GRAPHS 79

Similarly, B(l- q) ~ t, (1:, (k _ (~ _ i) ) ( -l)Hm-ilb,) q' So, for all 0 :::; k < m - l, we have that the coefficient of qk satisfies

~ ( i . ) ( _ 1)k-(m-i)ai = ~ ( i . ) ( _ 1)k-(m-i)bi . . ~ k-(m-z) .~ k-(m-z) •=m-k •=m-k

For k = m - l the coefficient of qk satisfies the following inequality:

f ( i ~ l) (-1)i-lai = f ( i ~ l) (-1)i-lbi •=l •=l 1 1 + ( l ~ l ) ( -1) - (at- bt)

< f ( i ~ l) (-1)i-lbi. •=l Thus, if we view A(1 - q) and B(1 - q) as polynomials in q, then Theorem 5.2.2 implies that B(1- q)

D

5.3 Comparing the 10 Forbidden Minors

Using Theorem 5.2.2, we can now compare the reliability polynomials p and the maximum protocol reliability p of the 10 forbidden minors. In Table 5.1 we list the values for p for the minor graphs in ascending order for p close to 0. So, p(To, p) is the largest reliability polynomial and p( B 0 , p) is the smallest reliability polynomial when compared to the p's of the other forbidden minors.

If we wish to compare the p of the ten forbidden minors, we must subtract the deficiencies from Table 4.7 from the reliabilities in Table 3.5. Observe that the terms explicitly stated in Table 5.1 are the same as in Table 3.5 since the the first CHAPTER 5. COMPARING GRAPHS 80

Ranking Minor Maximum Finite Graph Protocol Reliability 1 To 3p2 + 6p3 + ... 2 Ts(Tr) 2p2 + 2p3 + 5p4 + ... 3 T2 2p2 + Op3 + 1p4 + ... 4 Be 2p2 + Op3- 1p4 + 8p5 + ... 5 B2 2p2 + Op3- 1p4 + Op5 + ... 6 F1 1p2 + 1p3 + 2p4 + ... 7 B1 1p2 + 1p3 + 1p4 + ... 8 Fa Op2 + 4p3 + ... 9 Bo Op2 + 2p3 + ...

Table 5.1: Table of Maximum Reliabilities terms of the p's of the ten forbidden minors are unchanged after subtracting their deficiencies. So, for p near zero p(T0 , p) is the largest maximum protocol reliability and p( B0 , p) is the smallest maximum protocol reliability when compared to the p's of the other forbidden minors.

5.4 The Shortest s, r-Paths

In order to use Theorem 5.2.2 to compare the reliability of the 10 forbidden minors near 0, we had to write the polynomials in ascending order of the exponents of p. When a polynomial is written in ascending order, the first nonzero term in the polynomial is the dominant term.

n Definition 5.4.1. Given a polynomial L akpk qn-k, suppose the polynomial is writ­ k=O ten in ascending powers of p and aipiqn-i is the first nonzero term. Then aipiqn-i is the dominant term and ai is the dominant coefficient. n Given a polynomial L akpk, suppose the polynomial is written in ascending k=O CHAPTER 5. COMPARING GRAPHS 81 powers of p and aipi is the first nonzero term. Then aipi is the dominant term and ai is the dominant coefficient.

Lemma 5.4.2. Given (G,p) and a protocol A. Suppose ajpiqn-j is the dominant term of PA(G,p), then when PA(G,p) is written in one variable p, ajpi will be the dominant term.

m Proof. Suppose PA(G,p) = L aipiqm-i, and the dominant term is ajpiqm-j. There- i=O fore, ai = 0 for 0 ::; i < j, and the smallest power of p is j. If PA is written only in terms of p, we will show that pi will still be the smallest power of p in PA and that ajpi is the dominant term.

By Lemma 5.2.3, PA = """~m(k( """~ mm _- k~.) (-1) k-i ai ) p.k Observe that since ai = 0 for 0 ::; i < j, the jth term would be

Therefore, pi is still the smallest power of p and ajpi is the leading term. D

We will show that the exponent of the dominant term will tell us the length of the shortest s, r-path in its corresponding graph, and the coefficient is the number of these shortest paths. This is also true when examining the polynomial when CHAPTER 5. COMPARING GRAPHS 82 written in only p or p, q form. If it is written in p, q form, then we would write the polynomial in ascending orders of p.

Thus, another way of comparing two graphs is by looking at the number and length of the shortest s, r-paths.

Theorem 5.4.3. Given an s,r-graph (G,p) with m edges and protocol A, suppose cp1qm-l is the dominant term of the protocol reliability PA(G,p). The coefficient c is the number of essentials, r-trails of length l and there is no essentials, r-trail of length less than l.

Proof. Given an s, r-graph (G,p) with m edges, we calculate the protocol reliability

1 of (G, p) with protocol A and write it in ascending order. Suppose that cp qm-l is the dominant term and therefore is the first nonzero term in PA when written in ascending order. This term corresponds to l edges being alive and m - l edges being dead. If there is no essential s, r-trail with live edges when only l edges are alive, then we cannot send the message and the coefficient to that term must be

1 zero. However, this is a contradiction to our assumption that cp qm-l is the first nonzero term. So, there must be an essential s, r-trail with live edges when ( G, p) has llive edges. By Proposition 3.2.1, suppose that the essentials, r-trail with live edges has length k.

If k > l then the essential s, r-trail with k edges cannot possibly have live edges when only l edges are alive.

If k < l then when k edges are alive there is at least one possible way of CHAPTER 5. COMPARING GRAPHS 83 sending the message, using this essential trail of length k. This implies that the term pkqm-k exists in PA(G,p). However, since k < l, the term pkqm-k would come before cp1qm-l when PA(G,p) is in ascending order in terms of the exponents of p.

This contradicts the assumption that cp1qm-l is the dominant term.

So, k = l. By the same argument, any other live essential s, r-trail when l edges are alive must have length l. Thus, the dominant coefficient of the dominant term counts the number of essentials, r-trails of length l and these trails would be the shortest essential trails in (G, p) with protocol A.

Similarly, when PA ( G, p) is written only in terms of p, the dominant term tells us the length of the shortest essential s, r-trail and its coefficient counts the number of essential s, r-trails of that length. D

Corollary 5.4.4. Given (G,p) with m edges, suppose cp1qm-l is the dominant term of the reliability polynomial p( G, p). The coefficient c is the number of s, r-paths of length l and there is no s, r -path of length less than l.

Proof. Given (G,p) and the CFP A0 , recall that p(G,p) = PA0 (G,p). Letting

A= A 0 , by Theorem 5.4.3, the coefficient cis the number of s, r-trails of length l, and there is no s, r-trail of length less than l. By Proposition 3.2.1, since an s, r­ trail contains an s, r-path, if the length of the shortest s, r-trail is l, the s, r-path contained in the trail has length less than or equal to l. However, every path is a trail, and, so, the shortest s, r-paths must have length l and there must be c of them. D CHAPTER 5. COMPARING GRAPHS 84

We will show a similar result when we consider p for p is close to zero.

However, we first prove a lemma that will help us to show the result for p.

Lemma 5.4.5. Suppose G is an s, r-graph such that the length of a shortest path is l. Let A*= {uvw: u,v,w appear in some s,r-path of length lin that order}.

Then A* ~ A 0 and A* is finite.

Proof. Given a graph G and protocol A*, since the instructions uvw in A* appear in s, r-paths, then A* ~ A 0 . Suppose A* is not finite. So, there is an essential circuit

C. Note that since A* ~ A 0 , the instructions in A* are not of the form uvu, and, so,

C must have length at least 3. Let C = x0 , x1, x2, ... , xk_ 1 , Xk, x 0 for k 2:: 3. Let Pi be an s, r-path containing the instruction Xi-1, xi, Xi+l, and say that Xi is at distance k in ?;, from s. It follows that the distance d( s, xi) from s to xi in G is also k, since a shorter s, xi-path in G could be combined with the xi, r-path contained in Pi to get an s, r-walk (and thus an s, r-path) of length less than l in G, a contradiction.

Similarly, since the distance in Pi from s to xi+ 1 is k + 1, that is d( s, xi+ 1) = k + 1, we can deduce d(s, xi) < d(s, xi+l)· Considering Pi fori= 0, 1, 2, 3, ... , k, it follows that d(s, xo) < d(s, x1) < d(s, x2) < d(s, x 3 ) < ... < d(s, xk) < d(s, x 0 ), a contradiction.

So, A* does not contain an essential circuit, and, therefore, A* is finite.

D

With A* defined as in Lemma 5.4.5 we can now prove:

Theorem 5.4.6. Suppose (G,p) is an s, r-graph with m edges, whose shortest s, r­ path has l edges and there are c such paths. If A is a protocol such that p(G,p) =o CHAPTER 5. COMPARING GRAPHS 85

PA(G,p), then A*~ A, and the dominant term of PA is cp1qm-l.

Proof By Lemma 5.4.5, A* is finite and thus PA• ::; p =o PA· Let k be the length of a shortest s, r-trail that is essential for A, and suppose there are d such trails.

By Theorem 5.4.3, the dominant term of PA(G,p) is dpkqm-k, and the dominant term for A* is cp1qm-l. Since l is the length of the shortest path in G, l ::; k.

If l < k, then using Corollary 5.2.4, we have PA d, then by Corollary 5.2.4, we have PA

By Theorem 5.4.3 the coefficient cis the number of essentials, r-paths of length l, and there is no s, r-path of length less than l. So, A* ~ A and there are c paths of length l with corresponding instructions in A.

0

Theorem 5.4. 7. Suppose ( Gi, p) is as, r-graph with mi edges, i = 1, 2. If a shortest s,r-path in G2 has length less than a shortest s,r-path in G1 , then p(G1,p)

Proof Suppose (Gi,p) is a s,r-graph with mi edges, i = ·1,2. Define ci as the

CHAPTER 5. COMPARING GRAPHS 86

number of shortest s, r-paths in G i. Let protocol A 1 and Az be such that /3( G 1, p) =o

PA 1 (G1,P) and p(Gz,p) =o PA2 (Gz,p). Suppose a shortest path in G1 has length h and a shortest path in G2 has length l 2 • Then by Theorem 5.4.6, the dominant

1 term in PA 1 (G 1 ,p) is of the form c1phqm -h, c1 > 0, and the dominant term in

PA 2 (Gz,p) is of the form c2plzqmz-lz, c2 > 0. By Lemma 5.4.2, the dominant term would just be c1ph and c2plz. If the shortest path in G2 has length less than that

12 of G1, then we must have lz < h and the coefficient of p in PA 1 is zero, whereas

in PA 2 is Cz > 0. So, by Theorem 5.2.2, p(Gz,p)

If shortest paths in G 1 and G2 are of the same length, say l, but Gz has more of these short paths, then c2 > c1 . Thus, by Theorem 5.2.2, P(G 2 ,p)

D 87

Chapter 6

Kelmans' Type Construction

Sections 6.1, 6.2, and 6.3 introduce the basic definitions and results of the p-crossing properties of s, r-graphs developed and studied by Kelmans [10]. We use these results in the construction of graphs with prescribed ,8-crossings in Section 6.4.

6.1 Definitions

Definition 6.1.1. A pair H = (H1 , H2 ) is an ordered pair of probabilistic s, r­

graphs (H1,p), (H2,p). A pair is called reliable if p(H1,p) = p(H1,p) and p(H2,p) =

p(H2,p).

Definition 6.1.2. The p-difference of the pair H = (H1, H2) is

The ,8-difference of the pair H = (H1, H2) is

Note that the p-difference and the p-difference are functions of p. CHAPTER 6. KELMANS' TYPE CONSTRUCTION 88

Definition 6.1.3. Let ¢ be a polynomial and { x 1 < x 2 < ... < xk} be the set of all

roots of¢ in (0, 1) and let mi be the multiplicity of Xi, 1 ::::; i ::::; k. The profile of¢

is Prj[¢]= (m1, ... ,mk)·

We call a root r E (0, 1) of b.p(H, p) a p-crossing point of H, and Pr f[b.p(H, p)]

the p-crossing profile of H. Since p(H1,p0 ) = p(H2,Po) precisely when Po is a p­

crossing point of H, the p-crossing profile of H gives us the number of intersections

of p(H1,p) and p(H2,p) with multiplicities.

We call a root r' E (0, 1) of b.p(H, p) a p-crossing point of H, and Pr f[b.p(H,p)]

the p-crossing profile of H. Since p(H1,p) = p(H2,p) precisely when p is a i)­

crossing point of H, the p-crossing profile of H gives us the number of intersections

of p(H1,p) and p(H2,p) with multiplicities.

In this chapter we will be using probabilistic s, r-graphs (G, p) and from now

on will not be writing p. So, an s, r-graph G will be a probabilistic s, r-graph (G,p)

and p(G) = p(G,p).

6.2 Operations With s, r-Graphs

Givens, r-graphs G1, G2 with terminal vertices s1 and r1, and s2 and r2 correspond­

ingly, we can obtain an s, r-graph H by series operation, written H = G1 o G2, by

settings= s1, r = r2 , and identifying r1, s2 with a new vertex x.

Example 6.2.1. T0 o F0

To describe T0 o F0 , we relabel the vertices ofT0 with s = s', 1 = 1', 2 = 2', CHAPTER 6. KELMANS' TYPE CONSTRUCTION 89

3 = 3', r = r' and we relabel the vertices of F0 with s = s", 1 = 1", 2 = 2", 3 = 3",

4 = 4", and r = r". The vertices r' of T0 and s" of F0 are identified as a single

vertex x in T0 o F0 . The vertices ofT0 o F0 are (V(T0 ) U V(F0 ) U {x})- {r', s"}. The

neighbors of x in T0 o F0 are the neighbors of r' in T0 together with the neighbors

of s" in Fa, i.e., Nr00p0 (x) = Nr0 (r') U Np0 (s"). Otherwise, adjacencies in Too Fa

between vertices other than x are exactly as in T0 or in F0 (see Figure 6.1). 1' 1" 2"

Figure 6.1: T0 o Fa

We obtain H by parallel operation, written G 1IIG2 , by identifying s = s1 = s2

Example 6.2.2. To liFo

Relabel the vertices in T0 and F0 as we did in Example 6.2.1. In To liFo we

identify s' and s" as a single vertex s, and we identify r' and r" as a single vertex

r. So, we obtain NroiiFo(s) = Nr0 (s') U Np0 (s") and NroiiFo(r) = Nr0 (r') U Np0 (r").

Other adjacencies are as in T0 and F0 (see Figure 6.2).

Given protocols A1 for G 1 and A2 for G2. Let A1 o A 2 be the protocol for

G1 oG2 obtained from A1 UA2 by replacing s 1, s 2, r 1, r 2 by s, x, x, r respectively and

adding all instruction of the form uxv where u is a neighbor of r1 in G1 and v is a

neighbor of s2 in G2 . CHAPTER 6. KELMANS' TYPE CONSfRUCTION 90

Figure 6.2: ToiiFo

Let A1IIA2 be the protocol for G1IIG2 obtained from A1 U A2 by replacing

Proposition 6.2.3.

PA1oA2(G1 ° G2) = PA1(G1) · PA2(G2) (6.1)

p(G1 o G2) = p(G1) · p(G2) (6.2)

p(G1 o G2) = p(G1) · p(G2) (6.3)

PA1IIA2(G1IIG2) = PA1(G1) + PA2(G2)- PA1(G1) · PA2(G2) (6.4)

p(G1IIG2) = p(G1) + p(G2)- p(G1) · p(G2) (6.5)

p(G1IIG2) = p(G1) + p(G2)- p(G1) · p(G2) (6.6)

Proof. Let V(G1) = { s1, v1, v2, ... , Vk-1, r1} and V(G2) = { s2, U1, u2, ... , Un-1, r2}.

(6.1) To calculate PA1oA2(G1 o G2) observe that every s, r-path must go via

x, so we first calculate the probability we can send a message from s to x and then CHAPTER 6. KELMANS' TYPE CONSTRUCTION 91

from x to r with the protocol A1 o A 2 . Note that (G1 o G 2 ) uses only protocol

A1 from s to x since this is G1 with vertices s1 and r 1 relabeled. ( G1 o Gz) uses

only protocol A 2 from x to r since this is G 2 with vertices s 2 and rz relabeled.

So, if we let Zi be the event that an essential s, r-path for Ai in Gi remains,

since the protocol from s to x is independent of the protocol used from x to r,

PAloA2(Gl 0 Gz) = P(Zl n Zz) = P(ZI)P(Zz) = PAl(GI). PA2(Gz). Observe that

the protocol A 1 o A2 includes instructions of the form uxv with u in G 1 and v

in Gz, so if there is a live essential s, r-path P1 = s, v1, ... , vk, x in A1 and a live

essentials, r-path P2 = x, u 1 , ... , u1, r in A 2 , then there is a live essential s, r-path

(6.2) If we let A1 be the CFP of G1 and A 2 be the CFP of Gz, then A1 o Az

is the CFP of G 1 o G 2 and the result follows from (6.1).

(6.3) We let A1 be the protocol that yields the maximum protocol reliability

in G1 and A2 be the protocol that gives the maximum protocol reliability in Gz.

Then we have

p(G1 o Gz) 2: PA oA (GI o Gz) PA (GI) · PA (Gz) p(G1) · p(Gz). 1 2 = 1 2 =

To show p(G1 o G2 ) :::; p(G1) · p(G2 ), suppose A is the maximum protocol

of G 1 o G 2 . Then A would have to consist of a protocol A1 that routes a message

from s to x then a protocol A2 to route from x tor. Thus, A~ A 1 o Az and, using

(6.1), CHAPTER 6. KELMANS' TYPE CONSTRUCTION 92

p(GI) · p(G2) ~ PA 1 (GI) · PA 2 (G2) = p(G1 o G2).

So, p(G1 o G2) = p(G1) · p(G2).

(6.4) Let Zi be the event that an essential s, r-path for Ai in Gi remains.

Observe that part of the graph ( G1ll G2) is isomorphic to G1 with s1 mapped to

s, r1 mapped to r, and all other vertices mapped to itself. Also note that the

part of the graph isomorphic to G1 uses protocol A1 to route the message. So, the

protocol reliability for this part of the graph (G1IIG2) is PA 1 (GI) = P(ZI)· Also,

part of the graph (G1IIG2) is isomorphic to G2 with s2 mapped to s, r2 mapped

to r, and all other vertices mapped to itself. The part of the graph isomorphic to

G2 would use protocol A2 to route the message and, thus, has protocol reliability

PA2 (G2) = P(Z2). Observe that P(Z1) and P(Z2) are independent of each other,

so that P(Z1 n Z2) = P(Z1)P(Z2).

To find PA1 IIA2 (GIIIG2), we calculate the probability we can send a message

in (G1IIG2) with protocol AI!IA2 using the part of the graph isomorphic to G1

or G2, hence PA1 IIA2 (GIIIG2) = P(Z1 U Z2). By Theorem 1.2.3 (the principle of

inclusion-exclusion), P(Z1 U Z2) = P(Z1) + P(Z2) - P(Z1)P(Z2). This implies

(6.5) If we let A1 be the CFP of G1 and A2 be the CFP of G2, then A1IIA2

is the CFP of G1IIG2 and the result follows from (6.4). CHAPTER 6. KELMANS' TYPE CONSTRUCTION 93

(6.6) We let A1 be the protocol that yields the maximum protocol reliability in G 1 and A2 be the protocol that gives the maximum protocol reliability in G2.

Then we have

p(G1IIG2) ~ PA 1 (Gl) + PA2(G2)- PA 1 (G1) · PA2(G2)

= p(GI) + p(G2)- p(G1) · p(G2).

To show p(G1jjG2):::; p(G1) + p(G2)- p(G1) · p(G2), suppose A is the maxi­

mum protocol of G1jjG2. Then A would have to consist of a protocol A1 that routes

a message from s to r along G1 in G1jjG2 and then a protocol A 2 to route from s

tor along G2 in G1IIG2. So A= A1IIA2, and by (6.4)

PAl(Gl) + PA2(G2)- PAl(Gl)PA2(G2)

= 1- 1 +PAl (Gl) + PA2(G2)- PAl (Gl)PA2(G2)

= 1-(I- PAl(GI)- PA2(G2) + PAl(Gl)PA2(G2))

= 1- (1- PA 1 (Gl))(I- PA 2 (G2)

:::; 1- (I- P(G1))(I- p(G2)

= p(G1) + p(G2)- p(G1)p(G2).

So, p(G1IIG2) = p(G1) + p(G2)- p(G1) · p(G2). D CHAPTER 6. KELMANS' TYPE CONSTRUCTION 94

6.3 Multiplying Constructions

Definition 6.3.1. Let G = (G1 , G2 ) and F = (F1 , F2 ) be ordered pairs of s, r- graphs. The combination R(G, F) of G and F is the pair of s, r-graphs R(G, F) =

(R1(G, F), R 2 (G, F)), where

Figure 6.3: Combination of G = (G1, G2) and F = (F1, F2)

Theorem 6.3.2. Let G = (G1 , G2 ) and F = (F1 , F2 ) be two ordered pairs of s, r- graphs. Then

b.p(R(G, F))= b.p(G) · b.p(F) (and we say that R is p- multiplying.), and (6.7) b.p(R(G, F))= b.p(G) · b.p(F) (and we say that R is p- multiplying). (6.8)

If G and F are reliable pairs, then so is R(G, F).

Proof. (6.7) We give Kelmans' [10] proof (with few modifications) that R is p- CHAPTER 6. KELMANS' TYPE CONSTRUCTION 95 multiplying. Using Proposition 6.2.3 (6.2) and (6.5) we obtain:

p(RI(G, F))= p((Gl 0 FI)II(G2 0 F2))

= p(G1 o F1) + p(G2 o F2)- p(G1 o FI)p(G2 o F2)

= p(G1)p(F1) + p(G2)p(F2)- p(GI)p(F1)p(G2)p(F2),

p(R2(G, F))= p((G1 o F2)II(G2 o F1))

= p(G1 o F2) + p(G2 o FI)- p(G1 o F2)p(G2 o F1)

= p(G1)p(F2) + p(G2)p(F1)- p(G1)p(F2)p(G2)p(F1),

and thus

!J.p(R(G, F))= !J.p(R1(G, F), R2(G, F))

= p(R1(G, F))- p(R2(G, F))

= p(G1)p(F1) + p(G2)p(F2)- p(G1)p(FI)p(G2)p(F2)

- [p(G1)p(F2) + p(G2)p(F1)- p(G1)p(F2)p(G2)p(F1)]

= p(GI)p(F1) + p(G2)p(F2)- p(GI)p(F2)- p(G2)p(F1)

= [p(G1)p(F1)- p(G1)p(F2)] + [p(G2)p(F2)- p(G2)p(F1)]

= p(GI)[p(FI)- p(F2)]- p(G2)[p(F1)- p(F2)]

= (p(GI)- p(G2))(p(F1)- p(F2))

= !J.p(GI, G2). !J.p(FI, F2)

= !J.p(G) · !J.p(F).

The proof of (6.8) is similar. CHAPTER 6. KELMANS' TYPE CONSTRUCTION 96

If G and F are reliable pairs then by Definition 6.1.1

p(F2) = p(F2).

So, by Proposition 6.2.3 (6.2), (6.3), (6.5) and (6.6) we have

Similarly, p(R2(G,F)) = p(R1(G,F)). Therefore, R(G,F) = (R1(G,F),R2(G,F))

is a reliable pair. D

6.4 Reliable Pairs with Prescribed Crossing Pro­ files

We observed in Section 5.1 that the reliability polynomial of a path Pm and the

reliability polynomial of a cycle C2(m+l), for a fixed m, intersect at a single value

/m, and for distinct values of m the 'Ym are distinct. Thus, the pair (Pm, C2(m+l))

has one p-crossing. Kelmans was able to prove that we can build pairs of s, r-graphs

such that the profile of their p-difference is any given sequence of positive integers CHAPTER 6. KELMANS' TYPE CONSTRUCTION 97

Theorem 6.4.1. For every finite sequence M of positive integers, there exists a pair (G, H) of s, r-graphs such that the p-crossing profile of (G, H) isM.

We will give a slight modification of Kelmans' [10] proof by instead producing a reliable pair.

Theorem 6.4.2. For every sequence m 1, ... , mk of positive integers, there is a re­

liable pair (G, H) with p-crossings /'l, ... , /'k of multiplicity m1, ... , mk respectively,

and no other p-crossings.

We first prove a proposition and use it as a base case in the proof of Theo­

rem 6.4.2.

Proposition 6.4.3. For every m there is a reliable pair ( Gm, Hm) with a p-crossing

of multiplicity m, but no other p-crossing.

Proof. We use mathematical induction on m. For our base case consider the pair

(Pn, Cz(nH)) (for any n > 0). By Theorem 5.1.2, this pair is reliable and has a

p-crossing of multiplicity 1.

Suppose that m is some integer, m 2: 1, and the statement for m holds

true. Then there is a reliable pair of s, r-graphs (Gm, Hm) whose only p-crossing

has multiplicity m. We wish to show that there is a reliable pair (Gm+l, HmH)

with a single p-crossing of multiplicity m + 1. CHAPTER 6. KELMANS' TYPE CONSTRUCTION 98

The combination R((Pn, C2(n+l)), (Gm, Hm)) gives us a new pair (R1, R2) such that

R1 = R1((Pn, C2(n+1)), (Gm, Hm))

= (Pn o Gm)II(C2(n+l) oHm) and

R2 = R2((Pn, C2(n+l)), (Gm, Hm))

= (Pn o Hm)II(Gm o C2(n+1l)

By Theorem 6.3.2, (R1 , R2 ) is a reliable pair with

!:l.p(R1, R2) = !:l.p[R((Pn, C2(n+l)), (Gm, Hm))]

= !:l.p(Pn, C2(n+1)) · !:l.p( Gm, Hm)·

So, the pair (R1 , R2 ) has only one p-crossing with multiplicity m + 1. We set

(Gm+l, Hm+l) = (R1, R2) D

Proof. (Theorem 6.4.2)

We use mathematical induction on the number of p-crossings. Suppose that

n 1 < n 2 < ... < nk are any integers, ni > 0, i = 1, ... , k. For our base case

we consider k=l. By Proposition 6.4.3 there is a reliable pair (Gi, Hi), using the

combination R(C2(n;+l)> PnJ, with only one p-crossing of multiplicity mi.

Suppose that k is some integer, k > 1, and that the statement for k - 1

holds true. Then there is a reliable pair (G', H') with p-crossings 11, ... , {k-1 of

multiplicity m1, ... , mk_1 respectively. We will show that there is a pair (G, H) with CHAPTER 6. KELMANS' TYPE CONSTRUCTION 99

p-crossings 11, ... , rk of multiplicity m 1 , ... , mk respectively. By Proposition 6.4.3 there is a reliable pair (Gk, Hk) with one p-crossing at rk of multiplicity mk. Let

(G, H) = R((G', H'), (Gk, Hk)).

The combination R((G', H'), (Gk, Hk)) gives us a new reliable pair (R1, Rz) such that

R1 = (G' o Gk)II(H' o Hk),

R2 = (G' o Hk)li(Gk o H').

By Theorem 6.3.2,

6.p(R1, Rz) = 6.p[R((G', H'), (Gk, Hk))]

= 6.p(G', H') · 6.p(Gk, Hk)·

So, the pair (R1 , R2 ) hasp-crossings 1 1 , ... , rk of multiplicity m 1 , ... , mk respectively,

and no other p--crossings. D 100

Chapter 7

A Graph with a Piecewise Polynomial for p

7.1 Graph B0 With Edge Replaced

In this chapter we explore graphs where the protocol that gives us the maximum

reliability when p is close to 0 is different from the protocols that give us the

maximum reliability when p is close to 1.

Example 7.1.1. Consider the graph B 0 and two new vertices 6 and 7. Replace

the edge 13 in B 0 with the edges 16, 17, 36, and 37 (see Figure 7.1}. Call this new

graph B~. 1

s r

4 5

Figure 7.1: B~

In this chapter A 0 will always refer to the CFP of B~. The addition of these CHAPTER 7. A GRAPH WITH A PIECEWISE POLYNOMIAL FOR p 101 edges creates two essential circuits in the CFP of B0. The two essential circuits are

C1 = 1,2,3,4,5,3,6,1 andC2 = 1,2,3,4,5,3,7,1. In Table 7.1 we list the circuit instructions, and we observe that we can remove one or two instructions from the

CFP to find a finite protocol that eliminates the essential circuits.

Observe that 617, 637, 716, and 736 are not essential instructions since they do correspond to an s, r-path, and, hence, they are not in the CFP A 0 . There- fore, these instructions do not create more essential circuits, and C1 and Cz are the only essential circuits. Note that the instructions s12, s16, s17, s43, s45, 12r,

32r, 35r, 45r contribute to C1 and C2 . However if only one of these instructions is removed, C1 and C2 are still a problems, and the protocol will keep flooding r. So, these instructions are not listed in the Table 7.1, but will be addressed.

Essential Circuit Essential Circuit Circuit Instructions Circuit Instruction c1,C2 123 c1 536 c1,c2 234 c1 361 c1,C2 345 c1 612 c1,C2 453 Cz 537 Cz 371 c2 712 Table 7.1: B0 Circuit Instructions

Proposition 7.1.2. Let A1 = A 0 - {234} be a protocol for B0. A1 is finite and

Proof. Since the circuit instruction 234 is in lc1 and in Ic2 , its removal from Ao would stop both essential circuits in B0. Thus, A1 is finite. CHAPTER 7. A GRAPH WITH A PIECEWISE POLYNOMIAL FOR p 102

When calculating dA 1 notice that certain edges must be alive, and other edges must be dead in order to have an s, r-path using the instruction 234, and no s, r-trail without the instruction 234. The edges that must be alive include edges s1, 12, 23, 34, 45, 5r, and the edges that must be dead are edges s4, 35, 2r.

6 3 This requirement of live and dead edges yields the factor p q . From the remaining

edges 16, 17, 36,37 we can have different combinations of live and dead edges that

will not create a live trail that does not use the instruction 234. If all four edges

4 16, 17, 36, 37 are dead we have the term q . If only 1 of the 4 edges is alive, then

3 the 4 combinations will give us the term 4pq . If 2 of the 4 edges are alive, only

four of the possible six combinations yield terms: 16 and 17, 16 and 37, 17 and 36,

2 2 36 and 37. This last possibility yields the term 4p q . Therefore,

0

Proposition 7.1.3. Let A2 = A 0 - {536, 537} be a protocol for B0. A2 is finite

and

Proof. Since the circuit instruction 536 is in ! 01 and 537 E Ic2 , their removal from

A 0 would stop both essential circuits in B0. Thus, A2 is finite.

When calculating dA 2 notice that certain edges must be alive and others

must be dead in order to require a trail to use instructions 536 or 537. The edges CHAPTER 7. A GRAPH WITH A PIECEWISE POLYNOMIAL FOR p 103 that must be alive are: s4, 45, 53, 12, 2r and either pair 16,36 or 17, 37. The edges that must be dead are: s1, 23, 34, 5r. These requirements of live and dead edges,

7 6 in addition to having either 16, 36 or 17, 37 dead, yields our first term 2p q in dA 2 •

When calculating the coefficient for the term p8 q5 we observe that we can have either pair 16, 36 or 17, 37 alive, giving us two choices. If 16, 36 were alive, then we have two choices to pick from for the last remaining edge to be alive, 17 or 37.

Similarly, if 17,37 were alive, we have two choices. Thus, the coefficient to p8 q5 is

4. The term p9 q4 correspond to the edges s4, 45, 12, 2r, 16, 36, 17,37 being alive and

all others being dead, and, so, the coefficient would be 1. Observe that if we have

more than 9 edges alive in B 0 we would create a live path that does not use the

instructions 536 or 537. Therefore,

D

Observe that by Corollary 5.2.4, dA 2

more is true, and we show that p =o A2 and p =1 A1.

Example 7.1.4. LetA3 = A0-{s16} be a protocol forB~. We will showdA2

· , and dA 1 <1 dA 3 To calculate the first term of dA 3 when pis written in ascending

order, we consider the shortest path with the instruction s16. Note that one of the

shortest paths is P 1 = s, 1, 6, 3, 5, r. (There is also the paths, 1, 6, 3, 2, r of the same

5 8 length as P1 .} Thus, P1 contributes the term p q . Observe that in order for a finite CHAPTER 7. A GRAPH WITH A PIECEWISE POLYNOMIAL FOR p 104

PFP to be better than the others when p is close to zero, we need the p in the first term of the deficiency (when written in ascending powers of p) to be the greatest exponent possible, hence subtracting less from the first terms of p when calculating the protocol reliability of a finite PFP. So we would need the exponent of p in the

first term to be higher than the other PFP's. This is not the case for dA 3 since the

7 6 • first term for dA 2 is 2p q . Thus, by Corollary 5.2.4, dA 2

Let us consider the last term of dA 3 when the deficiency is written in as­

cending powers of p. We would look at the smallest number of removable edges to

calculate this term. A set of edges is removable for some given instruction, here

s16, if the fact that these edges are dead requires that any s, r-trail must use this

instruction. A set of removable edges for s16 is s4, 12, 37. This would contribute

10 3 • the term p q to dA 3 Observe that in order for a finite PFP to be better than other

finite PFP when p is close to 1, the last term must have q with the largest possible

10 exponent. Since dA 3 has a term of the form cp q3, where c is a constant, and the

8 5 • last term of dA 1 is 4p q , then by Corollary 5.2.4, dA 1 <1 dA 3

In Table 7.2 we list all instructions from A0 , the shortest path with this

instruction and a set of removable edges for the instruction. Using Table 7.2 we

will prove the following theorem:

Theorem 7.1.5. If A is a finite protocol for B0 then ------

CHAPTER 7. A GRAPH WITH A PIECEWISE POLYNOMIAL FOR p 105

I Instruction I Short Path I Removable Edges I s12 s,1,2,r s4,23,5r s16 s, 1,6,3,5,r s4, 12,17 s17 s, 1, 7,3,5,r s4, 16,12 s43 s,4,3,2,r s1,45 s45 s,4,5,r s1,34 123 s,1,2,3,5,r 16, 17,s4,2r 12r s,1,2,r s4,23,5r 163 s, 1,6,3,5,r s4,37, 12 173 s,1, 7,3,5,r s4,36, 12 234 s, 1,2,3,4,5,r s4, 16, 17,35,2r 235 s, 1,2,3,5,r s4,45, 16, 17,2r 32r s,4,3,2,r 12,5r 345 s, 1,6,3,4,5,r s4,35,2r 35r s,4,3,5,r 45,2r 361 s,4,3,6,1,2,r s1,17,23,5r 371 s,4,3,7,1,2,r s1, 16,23,5r 432 s,4,3,2,r s1, 12,45,5r 435 s,4,3,5,r s1,12,23,45 436 s,4,3,6,1,2,r s1,23,35,37,45 437 s,4,3, 7, 1,2,r s1,23,35,36,45 453 s,4,5,3,2,r s1,34,5r 45r s,4,5,r 35,2r 532 s,4,5,3,2,r s1, 12,34,5r 536 s,4,5,3,6,1,2,r s1,17,23,34,5r 537 s,4,5,3, 7, 1,2,r s1,16,23,34,5r 612 s,4,3,6,1,2,r s1, 17,23,5r 632 s, 1,6,3,2,r s4, 12, 17,45,5r 634 s,1,6,3,4,5,r s4,12,23,37,35 635 s,1,6,3,5,r s4,34,37,23,2r 712 s,4,3, 7,1,2,r s1,16,23,5r 732 s,1,7,3,2,r 12,34,35,36,45 734 s,1, 7,3,4,5,r s4,36,35,23,2r 735 s,1,7,3,5,r s4,34,36,23,2r Table 7.2: B0 Instructions CHAPTER 7. A GRAPH WITH A PIECEWISE POLYNOMIAL FOR p 106

Proof. Let A be a finite protocol forB~. Using Table 7.2 we can determine the first and last terms for the deficiencies of the PFP's not containing said instruction(s).

Observe that the longest path listed as one of the shortest paths for an instruction has length 7 and corresponds to paths that contain the instruction 536 or 537. All

3 10 4 9 other short paths listed have length 3, 4, 5, or 6, contributing a term p q , p q , p5q8 , or p6q7 . As stated in Example 7.1.4, in order for a finite PFP to be better than the others when p is close to zero, we need to subtract less from the first terms of the reliability polynomial and would, therefore, need the p in the first term of

the deficiency (when written in ascending powers of p) to be the term with the

greatest exponent possible for p. However, A0 - {536} and Ao - {537} are not

finite. Proposition 7.1.3, however, showed A 2 = A0 - {536, 537} to be finite and

4 7 6 8 5 9 . dA 2 = 2p q + 4p q + p q Since for any finite protocol A, we need one or more

instructions in 10 removed from A0 , the p's in the first term of dA 2 have higher

degree than any dA. Thus, there is some c E A 0 - A with c ~ 536,537, such that

dA 2 :SodA.

Now we consider the last terms of a possible finite PFP. As stated in Exam­

ple 7.1.4, in order for a finite PFP to be better than another finite PFP when pis

close to 1, the last term must have q with the largest possible exponent. Observe

that the largest number of removable edges listed is five and the set of instructions

that have five removable edges isS= {234, 235,436,437,536,537,632,634,635,732,

734, 735}. All other instructions have two, three, or four removable edges, and hence CHAPTER 7. A GRAPH WITH A PIECEWISE POLYNOMIAL FOR p 107

11 2 10 3 9 4 would have terms p q , p q , or p q . Note also that A0 - c is infinite for c E S except if c = 234. In fact, the only subsets S' C S such that A 0 - S' is finite must contain 234 or both 536 and 537, since these are the only circuit instructions in S, and S does not contain s12 and 12r. Thus, by monotonicity of discrepancy,

A1 = A 0 - {234}, and A2 = A 0 - {536, 537} are the finite protocols with the lowest

8 5 coefficient for p q .

6 7 7 6 8 5 By Proposition 7.1.2 and 7.1.3, dA1 = p q + 4p q + 4p q and dAz =

2p7 q6 + 4p8 q5 + p9q4 . Hence dAz would be subtracting less from the last terms

of the reliability polynomial than any other protocol A. Thus, by Corollary 5.2.4,

PA1 :S1 PA for all finite protocols A. 0

From Theorem 7.1.5 we can make some conclusions about the maximum

protocol reliability of B0.

Corollary 7.1.6. Let A1 = A 0 - {234} and A2 = A0 - {536, 537} be protocols for

B0. For any finite protocol A:

Moreover, PA1

Proof. By Theorem 7.1.5, for any finite protocol A, dAz :So dA and dA 1 :S1 dA. This

implies that PA = p- dA :Sop- dAz =PAz and PA = p- dA :S1 P- dA1 = PA1· We CHAPTER 7. A GRAPH WITH A PIECEWISE POLYNOMIAL FOR p 108

conclude that there exists an E1 > 0 and E2 > 0 such that

p(B~) = PA 2 , for 0 < p < E1,

p(B~) = PA 1 , for 1 - E2 < p < 1.

. By Proposition 7.1.2 and 7.1.3,

PA = p _ p7 q6 + psq5 + p9q4 PAl = p- p6q7 + 4p7 q6 + 4p8q5 2 2 4

Thus pis not a polynomial in [0, 1]. D

7.2 Open Problems

Some questions we can ask regarding a piecewise polynomial for p are the following:

Is it possible to characterize all graphs for which p is not a polynomial?

For every m, can we construct graphs for which p is a piecewise function

consisting of m polynomials? 109

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