Empowering the Voter:
A Mathematical Analysis of Borda Count
Elections with Non-Linear Preferences
A Senior Project submitted to
The Division of Science, Mathematics, and Computing
of
Bard College
by
David Polett
Annandale-on-Hudson, New York
May, 2010
Abstract
This project explores the application of non linearly-ordered sets to the field of voting
theory. In particular, a specific voting system known as Borda Count is studied. In order
to analyze the mathematical properties associated with different voting systems the alternatives on a ballot are represented as elements of an ordered set. Traditional voting theory
analysis has studied many properties of Borda Count that hold when voters’ ballots are
limited to linearly ordered sets. This project expands the Borda Count election method
to allow the use of bucket ordered and partially ordered sets as ballots instead of only
linearly ordered sets. The different mathematical properties associated with traditional
Borda Count are then analyzed in order to see if they still hold for these expanded Borda
Count voting systems.
Contents
Abstract
1
Dedication
5
Acknowledgments
6
1 Introduction
7
2 Preliminaries
12
2.1 Sets, Relations, and Orderings . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2 Voting Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.3 Linear Borda Count . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3 Bucket Borda Count
36
3.1 Bucket Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.2 Bucket Borda Score . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
4 Comparison of Bucket Borda to Linear Borda
50
4.1 Satisfied Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
4.2 Unsatisfied Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4.3 Random Thoughts on Resolvability . . . . . . . . . . . . . . . . . . . . . . . 90
5 Graded/Poset Borda Count
98
5.1 Graded Posets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
5.2 Graded/Poset Borda Score . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
6 Open Questions
110
Contents
References
3
113
List of Figures
2.1.1 Hasse Diagrams of a Set with Different Orderings . . .
2.1.2 Length and Height of a Partially Ordered Set . . . . .
2.2.1 Hasse Diagram of a Ballot . . . . . . . . . . . . . . . .
2.2.2 Same Preference Order Represented in Different Ballot
2.3.1 Allocation of Borda Scores to a Linear Ballot . . . . .
2.3.2 Linear Borda Election . . . . . . . . . . . . . . . . . .
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Formats
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3.1.1 Bucket Ordered Ballot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.1.2 Linear Extensions of a Bucket Ordered Ballot . . . . . . . . . . . . . . . . . 39
4.1.1 Monotonicity of a Bucket Borda Ballot . . . . . . . . .
4.1.2 Plurality Satisfied in Linear Borda Election . . . . . .
4.1.3 Failure of Plurality in Bucket Borda Election . . . . .
4.2.1 Insertion of a Clone in a Linear Ballot . . . . . . . . .
4.2.2 Insertion of a Clone in a Bucket Ballot . . . . . . . . .
4.2.3 Insertion of Clone Changing Outcome of Linear Borda
4.2.4 Insertion of Clone Changing Outcome of Bucket Borda
4.3.1 Inverse of a Linear Ballot . . . . . . . . . . . . . . . .
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5.1.1 Saturated and Maximal Chains in a Partially Ordered Set . . . . . . . . . . 99
5.1.2 Comparison of Graded Poset and Poset . . . . . . . . . . . . . . . . . . . . 100
5.2.1 Graded Borda Score Computations . . . . . . . . . . . . . . . . . . . . . . . 103
Dedication
To ACORN, for supporting low- and moderate-income Americans in their fight for social
justice throughout the last 40 years. In particular as it relates to this project, for greatly
expanding the American electorate by registering an incredible 1.3 million new voters
during the 2008 election season.
Acknowledgments
I would like to thank the following people for their support: John Cullinan, for spending
the year teaching me about voting theory, learning with me about posets, and making
my senior project experience more enjoyable than I could have ever imagined; Sam Hsiao,
for providing me with excellent guidance and assistance throughout my four years at
Bard; Matthew Deady, who was always willing to take a moment to discuss the game of
basketball; Ellen Lagemann, who taught me more about education in one year than I’d
learned in my entire life; Greg Spanier, for introducing me to a style of teaching math
which I can’t wait to try out for myself; my friends, for making Bard a place I’ll always
love and never forget; and Mom and Dad, for opening up so many doors of opportunity.
1
Introduction
An ordered set is a collection of elements that can be compared to one another in some
manner. Mathematicians have come up with many different types of ordered sets. The
two most commonly seen ordered sets are completely ordered sets, which we will refer to
throughout this paper as linearly ordered sets, and partially ordered sets. As there names
imply, a completely ordered set consists of elements which can all be compared to one
another while a partially ordered set contains elements which have some order but are
not necessarilly all comparable to one another. For example, the natural numbers are a
completely ordered set since they can all be compared by value: 1 < 2, 2 < 3, etc. In
contrast, we can think of the list of my favorite foods as a partially ordered set. Let’s
suppose that I like chicken better than any other meat but have no preference between
turkey and beef. Then chicken is comparable to both turkey and beef but neither of them
is comparable to the other. In a more mathematical sense common partially ordered sets,
which are also referred to as posets, include the natural numbers ordered by divisibility
and the set of subsets of a given set ordered by inclusion. As the example above shows,
ordered sets can be very useful in representing real life situations and preferences. In this
1. INTRODUCTION
8
paper, we take advantage of this aspect of ordered sets by utilizing them to model a voter’s
preferences in an election.
This senior project considers the application of partially ordered sets in the field of
voting theory. The motivation for this project initiated with my insistence that I did not
want to spend a year working on a project which I could not relate to in some way. During
my first three years at Bard I had several opportunities to speak with current seniors about
their senior projects, and they all gave me the same piece of advice – find a topic which
I could be personally invested in and excited to devote many, many hours to. Toward the
end of my junior year when I began to seriously think about what I’d like to spend the
next year studying I had two thoughts: the first was that the 300-level class I’d enjoyed the
most was Professor Hsiao’s Combinatorics class and the second was that earlier that year
Professor Cullinan taught a 100-level Voting Theory course that sounded very interesting
to me. Studying the mathematics involved in voting systems and tying in combinatorics
in some manner appealed to me as a topic which I could intuitively grasp the importance
of and hopefully become passionately committed to working on. With these two thoughts
in mind, I approached Professor Cullinan about potentially working with him on a project
that incorporated voting theory and combinatorics in some manner. Professor Cullinan
thought there was potential for a project and gave me the textbook from his voting theory
course, Math and Politics: Strategy, Voting, Power, and Proof [5], to read over the summer
in order to learn more about the field and begin formulating more specific project ideas.
While reading the voting theory textbook over the summer was invaluable for introducing me to the field of voting theory and greatly increased my knowledge of the subject,
it did not present me with any solid project ideas. I spent the first few weeks of senior
year reading a large number of economic and mathematic papers in search of a project
topic, and although I was intrigued by the topics in many of the papers I read about I
wasn’t able to find anything that seemed suitable for a project topic. Part of my dilemma
1. INTRODUCTION
9
was that I really wanted a topic which combined combinatorics and voting theory and
this was proving to be quite difficult. This all changed, however, when Professor Cullinan
introduced me to a paper written by a number of mathematicians including Bard Master
of Arts in Teaching professor Japheth Wood. This paper, titled Elections with Partially
Ordered Preferences [1], explored the application of partially ordered sets as ballots in a
number of different election methods. The motivation for this study was that traditional
ballots in which voters must compare all the candidates to one another often force voters
to overspecify their preferences in an election. Hence, it was the opinion of Wood and his
co-authors that linearly ordered ballots should be replaced by partially ordered ballots in
order to give voters the necessary freedom to express their true preferences in their ballots.
Wood’s paper studied the applicability of partially ordered sets (ballots) to a variety of
voting systems including one election method known as Borda Count. In order to count
the votes tallied in these partially ordered ballots Wood chose to look at the linear extensions of each voter’s partially ordered preference set. I found this to be an interesting
study in voting theory and appreciated that it involved some of the mathematical concepts
I’d been introduced to in combinatorics. In addition, there was a lot of room to further
investigate the ideas presented in Wood’s paper. I began studying the applicability of
non-linear ballots to the Borda Count election system and quickly fell in love with the
topic.
In this paper we will discuss how replacing the linear sets used as ballots in a Borda
Count election with non-linear sets effects the various mathematical properties associated
with Borda Count. In particular, we hope to prove that one can allow partially ordered
ballots in a Borda Count election and still retain all or nearly all of the desirable mathematical properties of Borda Count.
In Chapter 2 we begin by presenting some basic definitions and examples of ordered sets
along with the definitions of some important aspects of these sets. We follow that with
1. INTRODUCTION
10
a brief introduction to voting theory in which we list some of the key definitions we will
use to mathematically represent different aspects of an election. The chapter is concluded
with the definition and an example of the Borda Count voting system which will serve as
the basis for much of our analysis in the rest of the paper.
In Chapter 3 we begin our application of non-linear sets to Borda Count elections by
first more thoroughly defining and discussing the notion of a bucket ordered set. We then
develop a new version of Borda Count that allows for the use of bucket ordered ballots,
which we will refer to throughout the paper as Bucket Borda. Once Bucket Borda has
been defined, we will compare the way Bucket Borda counts the votes in a bucket ordered
ballot with the method of allocating points to a bucket ordered ballot using its linear
extensions presented in [?]. Bucket Borda will serve as the focus for much of our paper,
and we conclude this chapter by proving that it allocated points to the candidates in a
ballot similar to the way Borda Count allocates points in a linear election.
We begin Chapter 4 by presenting some mathematical properties of voting systems that
hold in Borda Count elections and attempt to prove that these properties also hold in
Bucket Borda elections. Through our analysis we develop three desirable mathematical
properties that hold in Linear Borda elections and which we feel all Borda Count election
methods should satisfy. Unforutnately, we then prove that it is impossible for any nonlinear Borda Count election method to satisfy these three properties simultaneously. This
is one of the most important results shown in the paper. We then present mathematical
properties which Borda Count fails to satisfy and attempt to prove that these properties
are also unsatisfied in Bucket Borda elections. The chapter is concluded with a brief look
into the resolvability of Borda Count and Bucket Borda voting systems. We were able to
prove a few interesting theorems about the resolvability of Borda Count election methods,
but there is still a lot of room here for further exploration.
1. INTRODUCTION
11
In Chapter 5 we finally are able to extend our findings to partially ordered sets. We
begin by defining a new election method, referred to as Graded Borda, which extends the
Bucket Borda election method to allow the input of graded poset ballots. This involves
determining a function to count the votes in these less restricted ballots. The rest of the
chapter is devoted to proving that Graded Borda satisfies the desirable base properties of
Borda Count and Bucket Borda discussed in Chapters 2 and 4. Throughout this chapter
we note that the Graded Borda election method can be extended to allow partially ordered
ballots without changing the way it is defined at all. This was a result which we had not
anticipated but rather that we stumbled upon when trying to prove that certain properties
of Borda Count held for Graded Borda.
The final chapter is devoted to listing some open questions and ideas for further exploration into the topic. Some of these are questions I would’ve liked to address in my
project but did not have enough time for, and others are problems which occurred to me
that I was unable to solve. It is my hope that these ideas could be useful for future Bard
students when searching for a senior project topic of their own!
2
Preliminaries
2.1 Sets, Relations, and Orderings
Before we begin our discussion of voting theory it is important to discuss some basic
properties of sets, relations and orderings. The definitions of relations, ordered sets, and
their associated properties presented in this chapter all came from [3]. In this project we
will be interested in restricting the ordering of the elements in a set by different properties.
The relations we will use to define the differently ordered sets studied throughout this
paper are listed in the following definition.
Definition 2.1.1. Let A be an ordered set of elements. For all a, b, c ∈ A the binary
relation is:
• irreflexive
if
¬(a a)
• asymmetric
if
a b, then ¬(b a)
• transitive
if
a b and b c, then a c
• transitivity of equivalence
if
a b, then either a c or c b
2. PRELIMINARIES
• complete
13
if
either a b or b a.
4
These properties can be used to define sets with different types of ordering. The three
main ordered sets we will be analyzing in this paper are listed in the definition below along
with their associated properties. We will define each of these sets more rigorously when
they are first discussed in detail in the paper. Note that in the definition below we use the
word ’strict’ before each of the sets we mention. The difference between a strictly ordered
set and a non-strictly ordered set is simply that a strictly ordered set is irreflexive while
a non-strictly ordered set is reflexive. Hence, in a non-strictly ordered set an element is
related to itself while in a strictly ordered set an element is not related to itself. Since
the sets we discuss in this paper are meant to represent voters’ preferences in an election
we do not want an element in our sets to be related to itself. For that reason we will be
working with strictly ordered sets throughout this paper. However, for ease of notation we
will drop the ’strict’ from each of these sets when discussing them throughout the paper
except in the definition that follows.
Definition 2.1.2. A binary relation on a set of alternatives A is a strict linear ordering
if it satisfies asymmetry, transitivity, and completeness; a strict bucket ordering if it
satisfies asymmetry, transitivity, and transitivity of equivalence; and a strict partial ordering
if it satisfies asymmetry and transitivity.
4
From Definition 2.1.1 it is clear that any set which is completely ordered must also satisfy
transitivity of equivalence, so it follows from Definition 2.1.2 that any linearly ordered set
is also a bucket ordered set. Additionally, since all bucket ordered sets satisfy asymmetry
and transitivity it is clear that any bucket ordered set is also a partially ordered set. By
the same argument, we see that any linearly ordered set is also a partially ordered set.
2. PRELIMINARIES
14
Thus, we can see that the set of all linearly ordered sets is a subset of the set of all bucket
ordered sets which is a subset of the set of all partially ordered sets.
Throughout the rest of this section we will present definitions for many different aspects
of the ordered sets which will be analyzed in the subsequent chapters of the paper. We
begin by defining what it means for two elements to be comparable, incomparable, and
equivalent in a given set.
Definition 2.1.3. Let P be a partially ordered set such that a, b ∈ P . If a b or b a
then a and b are said to be comparable. If ¬(a b) and ¬(b a), then a and b are
incomparable.
4
Definition 2.1.4. Let P be a partially ordered set such that a, b ∈ P . If for all c ∈ P
• a c if and only if b c,
• c a if and only if c b, and
• a and b are incomparable;
then we say that a is equivalent to b, denoted a ≡ b.
4
Of the three ordered sets we will study in this paper, linearly ordered sets are the most
restrictive, followed by bucket ordered sets, and then partially ordered sets. This is because
the elements in a linearly ordered set must be completely ranked which means each element
in the set must be comparable to every other element in the set. Bucket ordered sets allow
for a little more variation by requiring the elements in the set to only satisfy transitivity
of equivalence instead of completeness. While completeness states that any two elements
in a set must be comparable, transitivity of equivalence states that if two elements in a
set are comparable then any other element in the set must be comparable to at least one
of these alternatives. Therefore, we know that if two elements a and b are incomparable
in a bucket order B then they must be located in the same level of B. Hence, they are
2. PRELIMINARIES
15
preferred to and preferred by the same alternatives in B. It follows that if a and b are
incomparable in a bucket order B, then a ≡ b. Intuitively this means that the elements
in a bucket ordered set are partitioned into different levels of equivalent elements, and
these levels are then linearly ordered so that elements located in different levels are all
comparable to one another. Furthermore, we can view a linearly ordered set as simply a
bucket ordered set in which the sie of each equivalent level, or bucket, is one.
Partially ordered sets allow for the most variation in the ordering as there is no additional
requirement that must be met as long as the elements in the set are asymmetric and
transitive. This means the elements in a partially ordered set can be ordered in any manner
as long as transitivity and asymmetry are satisfied. Since transitivity of equivalence does
not always hold in a poset P , incomparable elements in P are not necessarily preferred to
and preferred by the same alternatives, and we can not assume that they are equivalent.
Hence, partially ordered sets are the only one of the three sets we will study that allow
truly incomparable elements. Although they may appear to be somewhat similar, the
fact that transitivity of equivalence must be satisfied in bucket ordered sets makes them
significantly easier to analyze than partially ordered sets. An example of the differences
between these three sets can be seen in Figure 2.1.1.
In order to better comprehend and view these sets we will represent them through the
use of Hasse diagrams. Intuitively, a Hasse diagram is a graphical representation of an
ordered set in which the vertices of the graph represent the elements in the set, and a line
between two vertices represents that their associated elements are related in the set. The
relation we will be working with this in this paper is and we define it below along with
a more rigorous definition of Hasse diagram.
Definition 2.1.5. Let A be an unordered set of elements and P be a partially ordered
set of A. For any a, b ∈ A, we say that a b in P if a is preferred to b.
4
2. PRELIMINARIES
16
Definition 2.1.6. The Hasse diagram of a poset P is an undirected graph representing
the poset whose vertices are the elements of some base set A and where a b if there is
a descending line connecting a to b in P .
4
From the two definitions above we see that for any two elements a and b in a partially
ordered set P , a is preferred to b if a is placed above b in the Hasse diagram of P . Hence,
we can think of the Hasse diagram of a partially ordered set as listing the elements in the
set in order from most preferred at the top to least preferred at the bottom. The base
set A mentioned in the definitions above refers to an unordered set of elements while a
partially ordered set P of A takes the elements in A and orders them in some manner. Since
linearly ordered sets and bucket ordered sets are both suborders of partially ordered sets,
the definition of Hasse diagram also applies to linearly ordered sets and bucket ordered
sets. Similar logic applies to the rest of the definitions mentioned in this section.
In Figure 2.1.1 we see examples of the Hasse diagrams of a linearly ordered set (L), a
bucket ordered set (B), and a partially ordered set (P ). Note that each of these ordered
sets consist of the same base set, A = {a, b, c, d}, but diferent orderings of the elements in
this set. In the Hasse diagram of L each alternative is connected to every other alternative,
and it follows that the elements in the set are all comparable to one another. Furthermore,
we see from the Hasse Diagram of L that a b c d. From the Hasse diagram of B
we see that b and c are equivalent since they are both preferred by a, preferred to d, and
incomparable to one another. This also follows from our statement that any two elements
in the same level of a bucket ordered set are equivalent. Thus, in B we have a b ≡ c d.
In the Hasse diagram of P we see that b and c are incomparable since there is no line
connecting b and c, and they are not equivalent since c d while b is incomparable to d.
Thus, in P we have a b and a c d.
2. PRELIMINARIES
17
Figure 2.1.1. Hasse Diagrams of a Set with Different Orderings
We now provide the definitions for different terms associated with certain elements or
suborders of an ordered set. For each of the following definitions, assume we are given a
base set of alternatives A ordered in a poset P .
Definition 2.1.7. An element a? is called the maximum of P if a? a for all a ∈ P .
An element b is called maximal if there is no a ∈ P such that a b.
4
Definition 2.1.8. An element a? is called the minimum of P if a a? for all a ∈ P . An
element b is called minimal if there is no a ∈ P such that b a.
4
It follows from the two definitions above that a poset can have many maximal or minimal
elements but only one maximum or minimum. Furthermore, the maximum of a poset is
by definition maximal as well, but a maximal element is not necessarily the maximum. If
there is more than one maximal element in a poset then by definition neither element can
be the maximum. The same holds true for minimal and minimum elements in a poset.
We will now define what is meant by the height and length of an ordered set. These
terms will be useful when determining how to allocate points to the elements in bucket
and partially ordered sets in Chapters 3 and 5 of the paper.
2. PRELIMINARIES
18
Definition 2.1.9. A chain in a partially ordered set is a subset C ⊆ P such that for all
a, b ∈ C, a and b are comparable. Hence, a chain is a linearly ordered subset of a partially
4
ordered set.
Definition 2.1.10. The height, H(C), of a chain C is defined by H(C) = |C| − 1 where
|C| is defined as the number of elements in the chain. The height of an element a ∈ P
is denoted Ha , and is defined to be the height of the longest C ⊆ P in which a is the
maximum. The height of a poset P is defined as the height of the maximum element or
maximal elements in P .
4
Definition 2.1.11. An antichain is a subset X of P such that for all a, b ∈ X, a and b
are incomparable.
4
Definition 2.1.12. The length, L(X), of an antichain X is defined by L(X) = |X|. The
length of an element a ∈ B is denoted La , and is defined as the number of elements located
in the same antichain as a.
4
Figure 2.1.2. Length and Height of a Partially Ordered Set
The partially ordered set P represented in Figure 2.1.2 has a maximum element a, no
minimum element, and two minimal elements f and g. Let C = {d, e, f } be a subset of
P . Then C is a chain since every element in the set is comparable to one another, and
2. PRELIMINARIES
19
C has height two since it is composed of three elements. Note that this poset has height
three since element a is the maximum of the poset and Ha = 3. The poset contains two
antichains, X1 = {b, c, d} and X2 = {f, g}, where L(X1 ) = 3 and L(X2 ) = 2. Note that P
is also a bucket order since transitivity of equivalence holds for the elements in P . Because
P is a bucket order we know that elements located in the same antichain are equivalent,
so we have b ≡ c ≡ d and f ≡ g. Finally, it should be quite clear that P is not also a linear
order since there are numerous elements in P which are not comparable to one another.
2.2 Voting Theory
Voting theory is concerned with the mathematcal analysis of different social choice procedures, which are more commonly referred to as election methods or voting systems. The
goal of all social choice procedures is to determine the will of the majority of a set of voters
when trying to choose among several candidates. When there are only two candidates in
an election this can be accomplished by simply letting each person vote for his or her
preferred candidate, and declaring the candidate who receives the most votes the winner
of the election, also referred to as the social choice. This style of voting is generally referred
to as plurality voting.
Plurality voting is one of the most well-known, widely used, and easily understood social
choice procedures in the United States and many other countries around the world. In a
plurality election each voter selects his or her single most preferred candidate, and the candidate who receives the greatest number of first-place votes is declared the winner. Clearly,
in an election consisting of only two candidates plurality does an excellent job of determining the will of the majority. However, when voters are given three or more candidates
to choose between this process becomes significantly more difficult to accomplish.
The first dilemma encountered when extending elections from two to three or more
candidates is that most traditional ballots only take into account a voter’s most preferred
2. PRELIMINARIES
20
candidate. In an election with just two candidates this is not a problem because looking at
a voter’s most preferred candidate gives us sufficient information on the voter’s preference
toward both candidates. However, in an election with three or more candidates a ballot
which only considers a voter’s most preferred candidate ignores his preferences toward
all the other candidates. The most common solution to this dilemma has been to extend
voters’ ballots from a single most preferred candidate to a ranked linear preference order
of the candidates in the election. Unfortunately, when we expand voters’ ballots in this
manner, it becomes unclear how to most effectively determine the will of the majority. As
a result of this ambiguity, many social choice procedures have been developed which when
given a set of linearly ranked ballots are said to determine the candidate most preferred
by the majority of the electorate.
Unfortunately, as was shown in a famous 1950 voting theory proof by Kenneth Arrow
there cannot possibly exist a ’fair’ social choice procedure which given an election consisting of three or more candidates is able to simultaneously satisfy three seemingly simple
and desirable properties for all voting systems to have. The three properties mentioned
in Arrow’s proof are Pareto, monotonicity, and Independence of irrelevant alternatives.
The Pareto condition states that if all of the ballots in an election are identical, then the
winner on each of these ballots should be the winner of the election. The monotonicity
condition states that if a candidate a is preferred to a candidate b in the overall election,
and a voter who had b preferred to a in his ballot switches his preferences towards a and
b (so that now a is preferred to b in his ballot), then a should still be preferred to b in the
overall election. Independence of irrelevant alternatives states that given two candidates
a and b in an election, the position of alternatives other than a and b in the ballots of
the election should be irrelevant when deciding whether a is preferred to b in the overall
election or b is preferred to a. We will discuss each of these conditions in more depth and
define them more rigorously in Chapter 4. In his impossibilty theorem, Arrow proved that
2. PRELIMINARIES
21
there can not exist a voting system which given an election consisting of at least three
candidates can simultaneously satisfy the three conditions mentioned above except for a
dictatorship (an election system in which a single voter’s ballot is always taken to represent the electorate as a whole). For a more rigorous statement and proof of this theorem,
see [5, 211-222].
In order to mathematically represent an election we must first understand and define
three important terms. The three components that come together to form an election
are a set of candidates to be voted between, a set of ballots which represent the voters’
preferences towards the candidates in the election, and a voting system which analyzes
these ballots in some pre-determined manner and outputs the winner of the election. The
rest of this section is dedicated to rigorously defining these three terms, so that we have
a good sense of how an election is represented and conducted mathematically.
We will denote the candidates in a given election by the set A whose elements will
be called alternatives most commonly denoted a1 , a2 , . . . an where n is the number of
alternatives being voted among in a given election. There will also be a set V whose
elements will be called voters denoted v1 , v2 , . . . , vm where m is the number of voters
in a given election. We will assume that in any election each voter v ∈ V arranges the
alternatives in a list, or ballot, according to his preferences such that every alternative in
A is present somewhere in his preference list. In general, a ballot is a function from the
set of all voters to any possible partial ordering of the alternatives in a given set A. Given
a specific v ∈ V , a ballot is a function from v to v’s preference ordering of the alternatives
in A. We state the definition of a ballot more rigorously below:
Definition 2.2.1. Let A be a set of n alternatives such that A = {a1 , a2 , . . . , an } and
let V be a set of m voters such that V = {v1 , v2 , . . . , vm }. We define a ballot B(v) as a
function B : V −
→ pos(A) such that the domain of our function is the set of all voters and
2. PRELIMINARIES
22
the range of our function is any possible partial ordering of A (unless specifically noted
otherwise). Thus, B(vi ) is voter vi ’s preference ordering of the alternatives in A.
4
From the definition above it is clear that a ballot is a partially ordered set of the alternatives in an election. These ballots, which we will sometimes refer to as preference lists,
can be pictured as vertical orderings with the alternatives displayed from most preferred
on top to least preferred on bottom. Recall from the previous section that the Hasse diagram of a partially ordered set P displays the elements of the set from most preferred
at the top of the graph to least preferred at the bottom. Hence, a voter’s ballot B(v) can
be represented by a Hasse diagram of the alternatives in the base set A ordered by the
relation such that for all alternatives a and b in B(v), if a is preferred to b in the ballot
then a is located above b in the Hasse diagram of B(v).
In Figure 2.2.1 we see the ballot for a voter v1 consisting of the set of alternatives
A = {a, b, c, d, e}. Since b is placed at the top of the Hasse diagram of B(v1 ) we see that it
is the most preferred alternative in the ballot. Similarly c is the least preferred alternative
in the ballot since it is placed at the bottom of the graph. Furthermore, the vertices in the
graph are all connected to one another which tells us that the alternatives in the ballot
are all comparable to one another, and it follows that B(v1 ) represents a linearly ordered
preference for the alternatives in A. From the Hasse diagram of B(v1 ) we see that v1 ’s
preference order for the alternatives in A can be expressed as b d a e c.
Considering a voter’s preferences toward a set of alternatives, it seems reasonable to
place the following restrictions on a voter’s ranked ballot. First, we will assume that a
voter who prefers alternative a to alternative b and alternative b to alternative c will
therefore prefer a to c. Hence, a voter’s ballot should satisfy transitivity. Second, we will
assume that a voter who prefers a to b will not prefer b to a. Hence, a voter’s ballot should
satisfy assymetry. At this point we feel there should be no further restrictions placed on
2. PRELIMINARIES
23
Figure 2.2.1. Hasse Diagram of a Ballot
a voter’s ballot. However, in nearly every ranked voting system utilized in today’s society
voters are forced to provide a complete preference ordering for the alternatives in the
election. These three criteria lead us to the following definition of a traditional linearly
ranked ballot.
Definition 2.2.2. Let B(v) be a traditional linear ballot in a ranked voting system.
Then for all a, b, c ∈ B(v) the following three properties must hold:
1. If a b and b c, then a c. (Transitivity)
2. If a b then ¬(b a). (Asymmetry)
3. For all a 6= b, either a b or b a . (Complete)
4
Clearly, the three restrictions placed on a traditional linear ballot in a ranked voting
system are equivalent to the mathematical properties of a linearly ordered set. Hence, we
can utilize linearly ordered sets to express a voter’s preferences in a traditional ballot.
While voters around the world are nearly universally required to vote using a ballot that
restricts their preferences to a linear ordering, we agree with the argument in [1] that these
2. PRELIMINARIES
24
ballots force voters to overspecify their preferences and often fail to accurately reflect
a voter’s true preferences in his ballot. Hence, we wish to eliminate the completeness
restriction currently placed on voters’ ballots and allow them to vote in elections utilizing
partially ordered ballots. As we’ve pointed out before, since linearly ordered sets are
a subset of partially ordered sets voters could still provide a complete ranking of the
alternatives, but they would now have much greater freedom in ranking the alternatives
in their ballots if they desired. By removing the ordering restrictions currently placed
on voters’ ballots we are improving their ability to express their true preferences for the
alternatives in an election by greatly increasing the number of possible orderings they can
utilize to rank the alternatives.
The two ways we will expand our voters ballots in this paper are through the use of
bucket ordered ballots and partially ordered ballots. The majority of this project involves
gradually removing the restrictions placed on voters’ ballots and analyzing how these
changes effect the mathematical properties of the Borda Count voting system defined in
the following section. Ideally, we would like to give voters the freedom to vote on partially
ordered ballots but these sets are significantly more difficult to analyze, so we will begin
our research in Chapter 3 by focusing on the extension from linearly ordered ballots to
bucket ordered ballots and then extend our findings to partially ordered ballots in Chapter
5.
At times throughout this paper we will want to limit the output of the ballot function
to only linear or bucket orderings of a given set of alternatives A. As we showed in Section
2.1, any linear order is also a bucket order and any bucket order is also a partial order, so
it follows that the set of all partial orders contains the set of all linear orders and the set of
all bucket orders. Hence, our general functon B : V −
→ pos(A) can be adjusted to restrict
the voters’ ballots to only linear orderings by limiting the output of B(V ) to the set of all
2. PRELIMINARIES
25
linear orderings of A. Similarly, we can also restrict voters’ ballots to bucket ordered sets
by limiting the output of B(V ) to the set of all bucket orderings of A.
We will now provide a simple example to motivate why it seems necessary to expand
the ballot types we allow voters to use in an election. Let A = {a, b, c, d} and v be a voter
whose actual preference ordering of the alternatives in A is a b, a c, and c d (so
by transitivity a d). The three ballots in Figure 2.2.2 represent voter v’s preference
ordering expressed through a linearly ordered ballot B(v1 ), a bucket ordered ballot B(v2 ),
and a partially ordered ballot B(v3 ). In the linear ballot, v is forced to select a preference
between b and c even though he views these alternatives as incomparable. The linear
orderings a c b d and a c d b could also be possible linear ballots for voter v
based on his actual preferences towards the alternatives in A. We have arbitrarily picked
one of the possible linear orderings compatible with his preferences to express his ballot. It
should be clear that in the linearly ordered ballot v is forced to overspecify his preferences
towards the alternatives in A, and thus his ballot does not accurately reflect how he truly
feels about the alternatives. In the bucket ordered ballot, v is forced to compare b and
d even though he views these alternatives as incomparable. This increases c’s weight in
the ballot and decreases the weights of b and d relative to voter v’s actual preference
ordering of the alternatives. Similar to the linear ballot, the bucket ordered ballot forces v
to compare alternatives where he actually has no preference. Only in the partially ordered
ballot is voter v’s ballot an accurate representation of his preference ordering. Since linear
ordered ballots and bucket ordered ballots would both still be allowed in a partially ordered
election we feel there is no reason to not give voters the option to use ballots which allow
for greater freedom in ranking the alternatives.
We conclude this section by mathematically defining the other two important political
terms associated with an election: a profile and a voting system. The set of all ballots in
a given election will be called a profile. This leads us to the following definition:
2. PRELIMINARIES
26
Figure 2.2.2. Same Preference Order Represented in Different Ballot Formats
Definition 2.2.3. Let A be a set of alternatives and B(vj ) be a ballot of A where j ∈
{1, 2, . . . , m} and m is the number of voters in an election. The set of all ballots in an
election is called a profile, denoted B, where B = {B(v1 ), B(v2 ), . . . , B(vm )} and ballot
B(vj ) represents the individual preference ordering of voter vj .
4
Voting theory is concerned with analyzing the mathematical properties associated with
different social choice procedures, or voting systems. These voting systems can be intuitively thought of as functions which input a profile of ballots associated with a given set
of alternatives and output the winning alternative, or social choice, of the given election.
From Alan Taylor [5] we are given the following definition of a social choice procedure:
Definition 2.2.4. A social choice procedure is a function for which a typical input
is a profile (set of preference lists) B of some set A (the set of alternatives) and the
corresponding output is either an element of A, a subset of A, or ’NW’.
4
While Taylor’s definition of a social choice procedure only returns the single alternative
in A which is the social choice according to a given profile and social choice procedure, the
definition of voting system we will use in this paper returns the complete social preference
ordering of the alternatives in A according to a given voting system. We denote the social
2. PRELIMINARIES
27
preference ordering of a given voting system as A? , where A? is a ranked ordering of the
alternatives in A based on their final preference in the overall election. If the social choice
of a voting system is a single alternative in A, we often will denote it as a? where a? ∈ A?
such that a? is the most preferred alternative in A? . It is also important to recognize that
given the same set of alternatives A and the same profile of voters B, two different voting
systems can output different social preference orderings. With these thoughts in mind, we
present the definition of a voting system, or election method, as a function from a profile
to the social preference ordering.
Definition 2.2.5. A voting system E is a function which inputs a profile B (set of
ballots) of some set A (the set of alternatives) and outputs the bucket ordered set A?
where each alternative in A appears in A? and the alternatives in A? are ordered by E.
We will call A? the social preference ordering according to E (note that we allow A? to
be a bucket ordered set in order to allow for the possibility of ties in the social preference
ordering).
4
More rigorously, we re-state the definition as follows:
Definition 2.2.6. Let A = {a1 , a2 , . . . , an } be a set of alternatives and B =
{B(v1 ), B(v2 ), . . . , B(vm )} be a profile of A. A voting system E is defined as a function
E:B−
→ A? where A? is the ranked preference ordering of the alternatives in A according
to the ballots in B and voting system E.
4
It is important to note from the two definition given above that a voting system is
a mathematical function which inputs a set of ballots and outputs the social preference
ordering of the alternatives being voted between in the election. This is the definition of a
voting system we will utilize throughout the rest of this paper to explain and analyze the
Borda Count voting system introduced in the following section.
2. PRELIMINARIES
28
2.3 Linear Borda Count
One of the more mathematically interesting and socially effective voting systems is the
ranked voting system devised by Jean-Charles de Borda in 1781 known as Borda Count.
Intuitively, Borda Count determines the social choice of an election by allocating points
to the alternatives based on how intensely they are preferred in each ballot of the election.
In a given ballot, an alternative is allocated points equal to the number of alternatives
ranked lower than it in the ballot. Hence, in a ballot with n alternatives the most preferred
alternative in the ballot receives (n − 1) points, the second most preferred alternative
receives (n − 2) points, and so on down to the least preferred alternative which receives
0 points. We can rigorously define the allocation of Borda Scores in a linear ballot in the
following manner:
Definition 2.3.1. Let A = {a1 , a2 , . . . , an } be a set of alternatives and B(v) be a linearly ordered ballot of A. Let ai be the ith most preferred alternative among the n total
alternatives in B(v). The Linear Borda Score of ai in ballot B(v) is defined as
BSai (v) = n − i.
4
Note that for ease of notation the Linear Borda Score of an alternative a in a ballot B(v)
will be written as BSa (v) rather than the more notationally complex BSa [B(v)]. Utilizing
the equation defined above, we now rigorously define a Linear Borda voting system.
Definition 2.3.2. Let A be a set of alternatives and B be a profile of A. The Linear
Borda election method, BS : B −
→ A? , is defined as the voting system which allocates
points to the alternatives in each ballot of B according to the Linear Borda Score equation
in Definition 2.3.1.
4
2. PRELIMINARIES
29
In Figure 2.3.1 we have a linearly ordered ballot B(v) of the set A = {a, b, c, d} in which
voter v’s preference ordering is b c a d. The Linear Borda Scores allocated to the
alternatives in B(v) in a Linear Borda election are as follows below:
Figure 2.3.1. Allocation of Borda Scores to a Linear Ballot
BSb (v) = n − 1 = 4 − 1 = 3
BSc (v) = n − 2 = 4 − 2 = 2
BSa (v) = n − 3 = 4 − 3 = 1
BSd (v) = n − 4 = 4 − 4 = 0
Note from this example that the most preferred alternative in the ballot is allocated a
Borda score of (n − 1) and the least preferred alternative in the ballot is allocated a Borda
score of 0. These are two properties of the Linear Borda voting system which we will want
to retain when extending Borda Count to non-linear ballots in the following chapters. The
third basic mathematical property of Linear Borda which we would like for any non-linear
2. PRELIMINARIES
30
Borda election to also satisfy involves the total amount of Borda scores allocated to the
alternatives in a single ballot and is presented in the following theorem.
Theorem 2.3.3. Let A be a set of n alternatives and B(v) be a linear ballot of A. Then
n
X
n(n − 1)
BSai (v) =
.
2
i=1
Proof. Let A = {a1 , a2 , . . . , an } be a set of alternatives and B(v) be a linear ballot of
A. Without loss of generality we can assume the alternatives in A are ordered such that
a1 a2 . . . an . By Definition 2.3.1 we know that
BSa1 (v) = n − 1
BSa2 (v) = n − 2
..
.
BSa( n−1) (v) = n − (n − 1) = 1
BSan (v) = n − n = 0.
It follows from the calculations above that
Pn
i=1 BSai (v)
= (n − 1) + (n − 2) + . . . + (n −
(n − 1)) + (n − n). Hence, we have
n
X
BSai (v) = (n − 1) + (n − 2) + . . . + (n − (n − 1)) + (n − n)
i=1
= (n + n . . . + n) − (1 + 2 + . . . + (n − 1) + n)
n(n + 1)
2
2n2 n2 + n
=
−
2
2
n2 − n
=
2
n(n − 1)
=
.
2
= n2 −
2. PRELIMINARIES
31
Thus, we see that the sum of the Linear Borda scores allocated to the alternatives in a
single linear ballot always equals
n(n − 1)
.
2
The result seen in Theorem 2.3.3 is very nice because it tells us that every ballot in
a Linear Borda election contributes the same weight toward determining the winner of
the overall election. This property along with the already mentioned facts that the most
preferred alternative in a linear ballot is allocated a Borda Score of (n − 1) and the
least preferred alternative in a linear ballot is allocated a Borda Score of 0, are the three
desirable properties of the Linear Borda voting system which we will hope to retain in
our non-linear Borda methods. These three properties will be more rigorously defined in
Section 4.1 and provide the motivation for one of our most interesting results concerning
Borda elections with non-linear preferences.
The following definition clarifies the distinction between the individual Borda Score an
alternative receives from a given ballot and the sum of the Borda Scores an alternative
receives in a given election. Each of these quantities will be mentioned numerous times
in the definitions, theorems and proofs presented throughout the rest of this paper, and
defining them now makes it much easier to discern which of the two is being discussed at
a given moment in later sections.
Definition 2.3.4. Let A be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )} be
a profile of A. Let a ∈ A and B(vj ) ∈ B. The ballot Borda Score of alternative a in
ballot B(vj ) is defined as BSa (vj ). The total Borda Score of alternative a in an election
m
X
?
BS : B −
→ A is defined as
BSa (vj ).
4
j=1
Now that we have defined a Linear Borda voting system we must also define how to
determine the winner of a given Linear Borda election. As we stated earlier, the alternative
in a Linear Borda election which receives the largest total Borda Score is declared the
winner of the election. Hence, we have the following definition.
2. PRELIMINARIES
32
Definition 2.3.5. Let A be a set of alternatives, B be a profile of A such that B =
{B(v1 ), B(v2 ), . . . , B(vm )}, and BS : B −
→ A? be a Linear Borda election. The winner of
a Linear Borda election, a? ∈ A? , is defined as the alternative whose total Borda Score is
greater than the total Borda Score of each other alternative in the election. Hence
m
X
j=1
BSa? (vj ) >
m
X
BSa (vj )
j=1
for all other a ∈ A? . We call A? the Borda Count social preference ordering and a? the
Borda Count social choice.
4
In Figure 2.3.2 we present a profile of ballots for a hypothetical Linear Borda election.
The election, BS : B −
→ A? , consists of four alternatives (A = {a, b, c, d}) and a profile of
five voters whose preference rankings of the alternatives in A are expressed through their
linearly ordered ballots (B = {B(v1 ), B(v2 ), B(v3 ), B(v4 ), B(v5 )}). From the ballots seen
in Figure 2.3.2 we compute the total Borda score for each alternative in the election:
Figure 2.3.2. Linear Borda Election
2. PRELIMINARIES
33
5
X
BSa (vj ) = 11
j=1
5
X
BSb (vj ) = 8
j=1
5
X
BSc (vj ) = 7
j=1
5
X
BSd (vj ) = 10
j=1
Hence, we see that
P5
j=1 BSa (vj )
>
P5
j=1 BSd (vj )
>
P5
j=1 BSb (vj )
>
P5
j=1 BSc (vj ).
Thus, the social preference ordering for this election is a d b c, and it follows that
a is the social choice of the Linear Borda election.
In subsequent sections of this project we will spend much time exploring the properties of
Borda Count that hold when we extend the domain of our elections to include bucket and
partially ordered ballots. We conclude this section by introducing three essential properties
which we feel are absolutely necessary for any non-linear voting system to satisfy in order
to be considered a valid extension of Linear Borda Count. These properties, which we will
refer to as the Axioms of Borda Count, are extremely simple and should not be confused
with the desirable properties of Borda Count mentioned earlier in this section.
Definition 2.3.6. Let A be a set of alternatives and B(v) be a ballot in a Borda Count
election. The Axioms of Borda Count which must be satisfied in any valid Borda Count
voting system are:
1. For all a ∈ A, BSa (v) ≥ 0.
2. If a b in B(v), then BSa (v) > BSb (v).
3. If a ≡ b in B(v), then BSa (v) = BSb (v).
2. PRELIMINARIES
34
4
It should be quite clear that these three axioms are not only desirable but necessary for
any extension of Linear Borda to be a valid Borda Count voting system. Axiom (1) states
that the minimum Borda Score an alternative can be allocated in a ballot is zero. If (1) did
not hold, then it would be possible for voter’s to severely penalize candidates by allocating
them a negative ballot Borda Score and actually lowering their total Borda Score in the
overall election. This not only seems rather unfair, but it would also make it much more
difficult to compare ballots among several voters. Axiom (2) states that if one alternative
is preferred to another in a given ballot, then the alternative that is more preferred must
be allocated a greater ballot Borda Score than the alternative that is less preferred. If (2)
did not hold, then voter’s would have no incentive to rank the alternatives in order of
preference because alternatives that were less preferred could receive larger Borda Scores
than alternatives that were more preferred. Axioms (1) and (2) can be inferred directly by
retaining desirable properties from Linear Borda. In contrast, Axiom (3) is inapplicable
when considering a linearly ordered ballot since two alternatives can never be equivalent in
a linear ordering. It is, however, extremely important to establish this axiom now because
it formalizes the notion of two alternatives being ’tied’ in a bucket ordered or partially
ordered ballot. As was mentioned in Section 2.1, if two alternatives are incomparable in
a bucket ordered ballot then they are preferred by the same set of alternatives, preferred
to the same set of alternatives, and incomparable to the same set of alternatives in the
ballot. Thus the two alternatives are equivalent in the ballot, and it seems rather obvious
that they should be allocated the same Borda Score.
The Axioms of Borda Count provide the basis for determining if a voting system which
extends Linear Borda to allow non-linear ballots is a valid Borda Count voting system. It
is important to keep these axioms in mind throughout our discussion of non-linear Borda
2. PRELIMINARIES
35
Count voting methods, and they will also be useful in proving an important result seen in
Section 4.1.
3
Bucket Borda Count
3.1 Bucket Orders
Before beginning our analysis of a Borda Count election system which allows for the use
of bucket ordered ballots, we will introduce a more rigorous definition of a bucket ordered
set along with some intuitive explanations of what is meant by a bucket ordered set.
Every ballot we discuss in this chapter will take the form of a bucket ordered set, so it is
important to grasp what exactly a bucket order is before we continue with our analysis.
Definition 3.1.1. A strict bucket ordered set, or strict bucket order (which we will refer
to as simply a bucket order), is a set B together with the order relation such that for
all a, b, c ∈ B the relation has the following three properties:
1. If a b and b c, then a c. (Transitivity)
2. If a b then ¬(b a). (Asymmetry)
3. If a b then either a c or c b. (Transitivity of Equivalence).
4
3. BUCKET BORDA COUNT
37
From this definition we see that a bucket order is similar to a linear order with the
exception that completeness has been replaced by transitivity of equivalence. In [1] a bucket
order B is described as an ordered set for which there exists a partition of B with the
property that if a and b are any two elements of B, then a and b are comparable if and only
if they belong to different blocks of the partition. Hence, a bucket order is a set of elements
partitioned into different buckets such that elements in the same bucket are incomparable
and elements in different buckets are comparable. A more intuitive description of a bucket
order is given in [2] which refers to a bucket order as a linear order with ties. From
transitivity of equivalence we know that elements located in the same partition of B are
equivalent within the bucket order, and it follows that we can think of these elements
as being tied in B. Thus, we can think of a bucket order B as a transitive, asymmetric
relation for which there are sets A0 , A1 , ...Ak−1 (the buckets) that form a partition of the
elements in the set B such that a b if and only if there are i, j ∈ N with i > j such that
a ∈ Ai and b ∈ Aj , and a ≡ c if and only if a, c ∈ Ai .
In Figure 3.1.1 we have a bucket ordered ballot B(v) consisting of the set of alternatives
A = {a, b, c, d, e, f, g}. Note that B(v) is partitioned into three distinct buckets such that
A2 = {a, b}, A1 = {c, d, e} and {f, g} = A0 . The preferences seen in B(v) can be expressed
in terms of alternatives as a ≡ b c ≡ d ≡ e f ≡ g or similarly in terms of buckets
as A2 A1 A0 . Furthermore, note that each alternative in B(v) is equivalent to every
alternative within its own bucket and comparable to every alternative not in its bucket.
In order to ease some of the computations in this section, we will often refer to the most
preferred bucket in a bucket ordered set with k buckets as Ak−1 and the least preferred
bucket in a bucket ordered set as A0 . Rather than use the term bucket, we will refer to
Ai as the ith level of a given bucket ordered ballot B(v). Note that |Ai | is equal to the
number of alternatives in level i, which is equivalent to the length of the antichain Ai .
Furthermore, we will often refer to |Ai | as Li , the length of the antichain at the ith level
3. BUCKET BORDA COUNT
38
Figure 3.1.1. Bucket Ordered Ballot
of B(v). Note that a linear order can be viewed as a bucket order in which Li = 1 for
i ∈ {0, 1, ..., k − 1}.
The motivation for this paper came from the discussion of allocating Borda Scores to
bucket ordered ballots presented in [1]. The authors of this paper developed a formula for
allocating Borda Scores to alternatives in a bucket ordered ballot by looking at the different
linear extensions of a bucket ordered ballot. Before we discuss the Linear Extension method
presented in their paper, we will define what is meant by a linear extension of a partially
ordered set.
Definition 3.1.2. Let A be a set of elements and P be a partial order of A. A linear
extension of P is a linear order LP on the elements of P such that if a b in P then
a b in LP . We denote by L(P ) the set of all linear extensions of P .
4
Intuitively, a linear extension of a partial order (and also of a bucket order) is a linear
order of the elements in the partial order such that any elements that are comparable in
the partial order must also be comparable in the linear order and their order must be
preserved. Additionally, any elements that are incomparable in the partial order must be
made comparable in the linear order but they must still be preferred by and preferred to
3. BUCKET BORDA COUNT
39
the same alternatives they were originally comarable to in the partial order. Since there
are many different ways in which the incomparable elements in a poset can be ordered,
a given partial order will have multiple linear extensions unless it is already a linearly
ordered set.
Assume we have a set of alternatives A = {a, b, c, d, e} ordered in a bucket ordered
ballot B(v) such that a b ≡ c ≡ d e. In Figure 3.1.2 we see this ballot as well as its
six possible linear extensions. Since a is preferred to every alternative in the ballot, we
know that in each linear extension a must be placed at the top of the ballot. Using similar
reasoning we know that e must be the least preferred alternative in each linear extension
of the ballot. Furthermore, alternatives b, c and d can be placed anywhere within a given
linear extension as long as they are preferred by a and preferred to e. Thus, there are six
possible linear extensions of B(v) since there are six ways to permute alternatives b, c,
and d while keeping them between a and e.
Figure 3.1.2. Linear Extensions of a Bucket Ordered Ballot
The analysis of the application of Borda Count to bucket ordered sets in [1] centered
around a formula which we will refer to as the Linear Extension Method. The Linear Extension Method allocates Borda Scores to the alternatives in a bucket order ballot by first
3. BUCKET BORDA COUNT
40
finding every possible linear extension of the bucket order. It then allocates Borda Scores
to the alternatives in each of these linear ballots according to Linear Borda Count. Next it
averages the Borda Score allocated to each alternative over all of the linear extensions of
our original ballot, and finally assigns this value as each alternative’s Borda Score in the
bucket ordered ballot. This method is effective for assigning reasonable Borda Scores to
the alternatives in a bucket ordered ballot, and one can easily see that it satisfies the Axioms of Borda Count mentioned in Chapter 2. However, this is not a very efficient method
for allocating Borda Scores to bucket ordered ballots because bucket orders in elections
with five or more alternatives can have a really large number of linear extensions, and
in such an election it could take an extremely long time to do the calculations necessary
to compute the Borda Scores for even a single ballot. In the following section we deduce
a formula for more easily and quickly computing the Borda Scores assigned through the
Linear Extension Method, but in order to show the difficulties of utilizing the Linear Extension Method we will now present a lemma concerning how many linear extensions a
given bucket order has.
Lemma 3.1.3. Let B be a bucket order consisting of n elements partitioned into k distinct
levels. Then B has |L(B)| possible Linear extensions, where |L(B)| = Lk−1 !×Lk−2 !×. . .×
L1 ! × L0 !.
Proof. Let B be a bucket order consisting of n elements partitioned into k distinct levels.
Then we know B is partitioned into levels such that Ak−1 ∪ Ak−2 ∪ . . . A1 ∪ A0 = B.
Therefore, it follows that Lk−1 + Lk−2 + . . . L1 + L0 = n where Li is the length of the
antichain at level Ai of B for i ∈ {0, 1, . . . , k − 1}. Note that in a given linear extension of
B, the Lk−1 elements in Ak−1 can be permuted in any order. Therefore, we know there are
Lk−1 ! different ways to linearly order the elements in Ak−1 . A similar argument holds for
linearly ordering the elements in levels Ak−2 to A0 . Hence, for a given level Ai there are
3. BUCKET BORDA COUNT
41
Li ! different ways to linearly order the elements in Ai . Since each of these permutations
can be combined with a given permutation of a different level to form a unique linear
extension, in order to determine the total number of possible linear extensions of B we
must simply multiply together the number of possible permutations of alternatives in each
level. Thus, we have |L(P )| = Lk−1 ! × Lk−2 ! × . . . × L1 ! × L0 !.
It should be clear from the proof above that a bucket ordered ballot with many alternatives partitioned into multiple levels can have a large enough number of linear extensions
that determining the allocation of ballot Borda Scores for the ballot will take quite a bit
of computation and time. In the following section we deduce a formula which allocates
ballot Borda Scores to the alternatives in a bucket ordered ballot similarly to the Linear
Extension Method but in a much more efficient manner.
3.2 Bucket Borda Score
In this chapter unless noted otherwise the definitions, theorems, and proofs we will consider
are all concerned with an election in which the voters’ preference orders are restricted to
bucket ordered ballots. We will introduce some notation here which we will make use of
several times throughout the chapter.
First of all, we will denote the bucket ordered ballot of a voter v as simply B(v), and
the alternative ai ∈ B(v) as an alternative in the ith level of B(v). Furthermore, we
will denote the length of level Ai in a bucket ordered ballot as simply Li . In addition,
we will let BBai (v) represent the ballot Borda Score allocated to each alternative ai in
ballot B(v). Extending this notation to an election consisting of m voters, we will let
Pm
j=1 BBa (vj ) represent the total Bucket Borda Score received by an alternative a in the
election. Similarly, we will let LEai (v) represent the ballot Borda Score allocated to an
P
alternative ai in the ith level of B(v) by the Linear Extension Method, and m
j=1 LEa (vj )
3. BUCKET BORDA COUNT
42
represent the total Borda Score allocated to alternative a by the Linear Extension Method.
In situations where the voter does not matter we will simply represent the Bucket Borda
score as BBai and the Borda Score allocated from the Linear Extension Method as LEai .
The ballot Borda Score of an alternative ai in a linearly ordered ballot can be found
by determining the number of alternatives ai is preferred to in the ballot and allocating
it one point for each of these alternatives. Since linearly ordered ballots are all organized
in the same manner and ai represents the ith most preferred alternative in a given ballot
this can be accomplished by finding the height of ai in the ballot and setting it equal to
BSai . Hence, in a linearly ordered ballot BSai = Hi = n − i.
In order to extend Borda Count to bucket ordered ballots we need to deduce a function
that accomplishes a similar task for alternatives in a bucket order. Consider an alternative
ai in a bucket ordered ballot B(v) such that ai is in the ith level of B(v). It follows that
ai ∈ Ai , and by definition we know that ai is preferred to a minimum of i alternatives
in B(v). Hence, BBai (v) ≥ Hi . We now must find a way to determine the number of
alternatives ai is preferred to that weren’t already accounted for in Hi . This means for
each Aj such that i > j we need to determine Lj − 1, and add this value to BBai (v). Thus,
so far we have BBai (v) ≥ Hi + (Li−1 − 1) + (Li−2 − 1) + . . . + (L0 − 1). Finally, we must
determine a way to calculate the value contributed to BBai (v) by each other alternative
in Ai . Since each alternative in Ai is equivalent, we know they should all receive the same
Borda Score. By allocating one half of a point to ai for each other alternative in Ai , we
ensure that one point is still allocated to the total Borda Score of B(v) for the relation
between any two alternatives in the ballot and we guarantee that the individual Borda
Score of each alternative in Ai will be equal. Hence, we want to also add
Li − 1
to BBai (v).
2
Putting all these observations together we define the Bucket Borda formula for allocating
Borde Scores to the alternatives in a bucket ordered ballot in the following manner.
3. BUCKET BORDA COUNT
43
Definition 3.2.1. Let A be a set of alternatives and B(v) be a bucket ordered ballot of
A. The Bucket Borda Score allocated to each alternative ai located in the ith level of
B(v) is defined as:
BBai (v) = Hi +
Li − 1
+ (Li−1 − 1) + (Li−2 − 1) + . . . + (L0 − 1)
2
4
Note that since each alternative located in the same level of a bucket order is equivalent,
we design our Bucket Borda formula to allocate ballot Borda Scores to an alternative ai
based on its level Ai in the ballot, and then simply apply this ballot Borda Score to every
alternative in Ai . As we saw from the description above the Bucket Borda equation should
do an excellent job of extending Linear Borda to an election with bucket ordered ballots.
We can easily check and see that given any bucket ordered ballot this formula satisfies
the three Axioms of Borda Count from Section 2.3. More importantly, given a linearly
ordered ballot B(v) and an alternative ai such that ai is the ith most preferred alternative
in B(v), we see that BSai (v) = n − i = Hi = BBai (v). Hence, given a linearly ordered
ballot our Bucket Borda formula allocates Borda Scores to the alternatives in the ballot
equivalently to Linear Borda. We now present some basic definitions pertaining to Bucket
Borda elections which are very similar to those presented for Linear Borda in the previous
chapter.
Definition 3.2.2. Let A be a set of alternatives and B be a profile of A. The Bucket
Borda election method, BB : B −
→ A? , is defined as the voting system which allocates
points to the alternatives in each ballot of B according to the Bucket Borda Score equation
in Definition 3.2.1.
4
Definition 3.2.3. Let A be a set of alternatives, B be a profile of A such that B =
{B(v1 ), B(v2 ), . . . , B(vm )}, and BB : B −
→ A? be a Bucket Borda election. Then a ∈ A is
3. BUCKET BORDA COUNT
44
the social choice of the Bucket Borda election if
m
m
X
X
BBa (vj ) >
BBb (vj )
j=1
j=1
for all b ∈ A.
4
We now present a lemma which will help us simplify and more intuitively understand
the allocation of Borda Scores by the Linear Extension Method mentioned in [1].
Lemma 3.2.4. Let A be a set of n alternatives and B(v) be a bucket ordered ballot
for voter v such that B(v) is partitioned into k distinct levels. Then the Linear Extension
Method allocates Borda Scores to each alternative ai in the ith level of B(v) in the following
manner:
LEai (v) = n − Lk−1 − Lk−2 − ... − Li+1 −
Li + 1
2
Proof. Let A be a set of n alternatives and B(v) be a bucket ordered ballot for voter
v such that B(v) is partitioned into k distinct levels. Let {A0 , A1 , . . . , Ak−1 } ⊆ B(v) be
the partitions of ballot B(v) such that Ak−1 Ak−2 . . . A1 A0 . Then we can
denote the elements in B(v) such that {a1 , a2 , . . . , aa } ∈ Ak−1 , {aa+1 , . . . , aa+b } ∈ Ak−2 ,
. . . , and {aa+b+...+1 , . . . , aa+b+···+σ } ∈ A0 where a + b + · · · + σ = n. Let i ∈ {1, . . . , n} and
αi ∈ Z where α1 , α2 , . . . , αn are the Borda Scores allocated to the a1 , a2 , . . . , aa+b+···+σ
alternatives by the Linear Extension Method such that αn < αn−1 < . . . < α1 and
αn = n − n = 0, αn−1 = n − (n − 1) = 1, and α1 = n − 1. Hence, given a linear extension
of B(v), the ith preferred alternative in the linear extension receives a ballot Borda Score
of αi = n − i.
Let ai be an alternative in level Ai of B(v). Let Li+1 = λ and |Ai | = Li = φ. Note from
above that that Lk−1 = a, Lk−2 = b, . . ., and L0 = σ. Now, let Lk−1 + Lk−2 + . . . + Li+1 =
a + b + . . . + λ = ρ. Then {aρ+1 , aρ+2 , . . . , aρ+φ } ∈ Ai .
Recall from Lemma 3.1.3 that there are Li ! ways to linearly order the φ alternatives in
Ai . Furthermore, each ai ∈ Ai is allocated each possible Borda Score for alternatives in
3. BUCKET BORDA COUNT
45
this level (Li − 1)! times. Hence, the ballot Borda Score allocated to each alternative ai
based on calculations from the linear extensions of B(v) can be expressed by the following
equation:
LEai (v) =
(Li − 1)!αρ+1 + (Li − 1)!αρ+2 + . . . + (Li − 1)!αρ+φ
Li !
Thus, we have
(Li − 1)!αρ+1 + (Li − 1)!αρ+2 + . . . + (Li − 1)!αρ+φ
Li !
(Li − 1)!(αρ+1 + αρ+2 + . . . + αρ+φ )
=
Li !
αρ+1 + αρ+2 + . . . + αρ+φ
=
Li
[n − (ρ + 1)] + [n − (ρ + 2)] + . . . + [n − (ρ + φ)]
=
Li
φn − φρ − (1 + 2 + . . . + φ)
=
Li
φ(φ + 1)
φn − φρ −
2
=
Li
Li (Li + 1)
Li n − Li ρ −
2
=
Li
Li+1
Li (n − ρ −
)
2
=
Li
Li+1
=n−ρ−
2
Li + 1
= n − Lk−1 − Lk−2 − . . . − Li+1 −
.
2
LEai (v) =
We conclude that the theorem holds and
LEai (v) = n − Lk−1 − Lk−2 − . . . − Li+1 −
Li + 1
2
is a valid representation for how the Linear Extension Method allocates Borda Scores to
each of the alternatives in the ith level of a bucket ordered ballot.
Comparing the Bucket Borda formula described earlier in this section to the Linear
Extension formula shown above, we see that while the Bucket Borda formula calculates
3. BUCKET BORDA COUNT
46
an alternative’s Borda Score by summing together the number of elements the alternative
is preferred to in a ballot the formula derived from the Linear Extension Method determines an alternative’s Borda Score by considering the total number of alternatives in the
ballot and then subtracting the number of alternatives which are preferred to the given
alternative. In the following theorem, we show that these two methods of allocating Borda
Scores to the alternatives in a bucket ordered ballot are equivalent.
Theorem 3.2.5. Let A be a set of n alternatives and B(v) be a bucket ordered ballot for
voter v partitioned into k distinct levels. Given any alternative ai ∈ A, LEai (v) = BBai (v).
Proof. Let A be a set of n alternatives and B(v) be a bucket ordered ballot for voter v
partitioned into k distinct levels. Note that the sum of the length of each anti-chain in
B(v) must equal the total number of alternatives in A. Hence L0 + L1 + . . . + Li−1 + Li +
Li+1 + . . . + Lk−2 + Lk−1 = n. By Lemma 3.2.4, we know that
LEai (v) = n − Lk−1 − Lk−2 − . . . − Li+1 −
Li + 1
.
2
We can rewrite this expression in the following manner:
Li + 1
2
Li + 1
= n − (Li+1 + Li+2 + . . . + Lk−1 ) −
2
LEai (v) = n − Lk−1 − Lk−2 − . . . − Li+1 −
= n − (n − Li − Li−1 − Li−2 − . . . − L0 ) −
= Li + Li−1 + Li−2 + . . . + L0 −
Li + 1
2
Li + 1
2
2Li Li + 1
−
+ Li−1 + Li−2 + . . . + L0
2
2
Li − 1
=
+ Li−1 + Li−2 . . . + L0
2
=
Note that in any bucket ordered ballot B(v), Hi = i for all ai in B(v). From Definition
3.2.1, we have
BBai (v) = Hi +
Li − 1
+ (Li−1 − 1) + (Li−2 − 1) + . . . + (L0 − 1).
2
3. BUCKET BORDA COUNT
47
It follows from this equation that:
Li − 1
+ (Li−1 − 1) + (Li−2 − 1) + . . . + (L0 − 1)
2
Li − 1
= Hi +
+ Li−1 + Li−2 + . . . + L0 − i
2
Li − 1
= Hi − i +
+ Li−1 + Li−2 + . . . + L0
2
Li − 1
=
+ Li−1 + Li−2 + . . . + L0
2
BBai (v) = Hi +
Thus, we conclude that LEai (v) = BBai (v) =
Li − 1
+ Li−1 + Li−2 + . . . + L0 for any
2
voter v ∈ V .
From Theorem 3.2.5, we see that the Bucket Borda function defined earlier in this section
always allocates Borda Scores to the alternatives in a bucket ordered ballot equivalently
to the Linear Extension Method presented in [1]. Furthermore, we see that simplest way
to calculate the Bucket Borda Score of an alternative ai in a bucket ordered ballot is to
allocate
1
of a point to ai for each alternative it is equivalent to (number of alternatives
2
located in Li minus 1 divided by 2) and 1 point to ai for each alternative it is preferred to
(sum of the total number of alternatives located in levels preferred by Ai ). In order for the
Bucket Borda voting system to be consistent with Linear Borda, given any bucket ordered
ballot B(v) containing n alternatives, the sum of the Bucket Borda Score allocated to each
alternative in the ballot should equal
n(n − 1)
. Note that since BBai (v) only represents
2
the ballot Borda Score allocated to each alternative on level Ai of a given ballot, we
must multiply this quantity by the length of each level Li in order to determine the total
amount of ballot Borda Scores allocated to a given bucket ordered ballot. Hence, we have
the following theorem:
3. BUCKET BORDA COUNT
48
Theorem 3.2.6. Let A be a set of n alternatives and B(v) be a bucket ordered ballot
k−1
X
n(n − 1)
partitioned into k distinct levels. Then
.
BBai Li =
2
i=0
Proof. Let A be a set of n alternatives and B(v) be a bucket ordered ballot partitioned
into k distinct levels. Note once again that L0 + L1 + . . . + Lk−1 = n. By Theorem
3.2.5 we know that BBai (v) = LEai (v) for all ai ∈ A and any v ∈ V . It follows that
Pk−1
Pk−1
i=0 LEai (v)Li . Furthermore, from Lemma 3.2.4 we have
i=0 BBai (v)Li =
LEai (v) = n−Lk−1 −Lk−2 −. . .−Li+1 −
k−1
X
BBai Li =
i=0
k−1
X
Li + 1
. Hence, we have the following computations:
2
LEai Li
i=0
= (n −
Lk−1 + 1
Lk−2 + 1
L0 + 1
)Lk−1 + (n − Lk−1 −
)(Lk−2 ) + . . . + (n − Lk−1 − Lk−2 − . . . −
)(L0 )
2
2
2
= nLk−1 −
Lk−2 2 + Lk−2
L0 2 + L0
Lk−1 2 + Lk−1
+ nLk−2 − Lk−1 Lk−2 −
+ . . . + nL0 − Lk−1 L0 − Lk−2 L0 − . . . −
2
2
2
= n(Lk−1 + Lk−2 + . . . + L0 ) −
= n2 −
Lk−1 2 + Lk−1
Lk−2 2 + Lk−2
L0 2 + L0
− Lk−1 Lk−2 −
− . . . − Lk−1 L0 − Lk−2 L0 − . . . −
2
2
2
Lk−1 2 + Lk−1
Lk−2 2 + Lk−2
L0 2 + L0
− Lk−1 Lk−2 −
− . . . − Lk−1 L0 − Lk−2 L0 − . . . −
2
2
2
=
2n2 − Lk−1 2 − Lk−1 − 2Lk−1 Lk−2 − Lk−2 2 − Lk−2 − . . . − 2Lk−1 L0 − 2Lk−2 L0 − . . . − L0 2 − L0
2
=
2n2 − Lk−1 2 − 2Lk−1 Lk−2 − Lk−2 2 − . . . − 2Lk−1 L0 − 2Lk−2 L0 − . . . − L0 2 − Lk−1 − Lk−2 − . . . − L0
2
=
2n2 − Lk−1 2 − 2Lk−1 Lk−2 − Lk−2 2 − . . . − 2Lk−1 L0 − 2Lk−2 L0 − . . . − L0 2 − (Lk−1 + Lk−2 + . . . + L0 )
2
=
2n2 − Lk−1 2 − Lk−2 2 − . . . − L0 2 − 2Lk−1 Lk−2 − . . . − 2Lk−1 L0 − 2Lk−2 L0 − . . . − 2L1 2L0 − n
2
=
2n2 − (Lk−1 2 + Lk−2 2 + . . . + L0 2 + 2Lk−1 Lk−2 + . . . + 2Lk−1 L0 + 2Lk−2 L0 + . . . + 2L1 2L0 ) − n
2
=
2n2 − (Lk−1 + Lk−2 + ... + L0 )2 − n
2
=
2n2 − (n)2 − n
2
=
n2 − n
n(n − 1)
=
.
2
2
Thus, we conclude that
k−1
X
i=0
BBai (v)Li =
n(n − 1)
for any voter v ∈ V .
2
We conclude this section by noting once again how well the Bucket Borda voting system
developed in this chapter has expanded the concept of Linear Borda. While Linear Borda
3. BUCKET BORDA COUNT
49
allocates one point to an alternative for each alternative it is preferred to in a ballot,
Bucket Borda allocates one point to an alternative for each alternative it is preferred to
and an additional half of a point for each alternative it is equivalent to in a ballot. In the
next chapter we present many mathematical properties associated with voting systems
and try to prove that Linear Borda and Bucket Borda respond in the same manner to
these properties.
4
Comparison of Bucket Borda to Linear Borda
4.1 Satisfied Properties
It is well known that Linear Borda Count satisfies many properties devised by voting
theorists which are desirable for any voting system to have. In this section it is our goal to
state many of the properties which hold for Linear Borda, and then either prove or disprove
that they hold for the Bucket Borda Count voting system devised in the previous chapter.
The majority of the general voting properties discussed throughout this chapter come
from [7] in which mathematician Douglas Woodall defined many mathematical properties
that a preferential election rule should satisfy and then analyzed them in reference to his
preferred voting system, Single Transferrable Vote. However, we will begin our analysis in
this section and the following one by looking at the three conditions used by Arrow in his
impossibility theorem referred to in Section 2.2.
The first property we will look at is the Pareto criterion. It is perhaps the simplest
and most obvious voting system property we will present in this paper and is satisfied by
virtually every possible election method. Intuitively, Pareto states that if every voter in an
election prefers an alternative a to an alternative b, then a should be preferred to b in the
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
51
social preference list. We define the Pareto criterion rigorously below and then provide a
simple proof that it is satisfied by Bucket Borda. It is well-known that the Pareto criterion
is satisfied by Linear Borda, and since the proof of this is practically identical to the proof
for Bucket Borda we choose not to provide it here.
Definition 4.1.1. Let A be a set of alternatives such that a, b ∈ A, and let B be a profile
of ballots such that for every ballot in the profile a b. Then an election system, E : B
−
→ A? , satisfies the Pareto criterion if it is always the case that a b in A? .
4
Theorem 4.1.2. Bucket Borda satisfies the Pareto criterion.
Proof. Let A be a set of alternatives such that a, b ∈ A, and let B = {B(v1 ), B(v2 ), . . . , B(vm )}
be a profile of m ballots. Let j ∈ {1, . . . , m}. Assume for all B(vj ), voter vj prefers a to
b in his ballot. Then given a Bucket Borda election, BB : B −
→ A? , we know that for
Pm
P
each B(vj ) ∈ B, BBa (vj ) > BBb (vj ), and it follows that m
j=1 BBb (vj ).
j=1 BBa (vj ) >
Thus, we conclude that a b in A? and the Pareto criterion holds.
The second voting system property we will discuss is the monotonicity criterion. We feel
that Woodall’s statement of monotonicity is a slightly stronger version of monotonicity
than Taylor’s, which we briefly mentoned in the discussion of Arrow’s impossibility theorem in Section 2.2. For this reason we will utilize Woodall’s definition while analyzing
monotonicity in this section. Intuitively, Woodall’s monotonic property states that if the
winner of an election has its preference improved on one of the ballots in the profile of
the election then that alternative should remain the winner of the election. This property
is desirable because if a voter increases his preference toward an alternative on his ballot,
then that should never hurt the alternative in the overall election. The formal definition
of a monotonic voting system is presented below.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
52
Definition 4.1.3. Let A be a set of alternatives, B be a profile of ballots, and E : B
−
→ A? be an election system such that a? ∈ A? is the social choice of the election. Then
E satisfies the monotonicity criterion if when one of the voters in B adjusts his ballot
by increasing a? ’s preference (that is, exchanging a? ’s position with that of the alternative
immediately above a? ), a? always remains the social choice of the election.
4
It is fairly obvious that Linear Borda is a monotonic voting system. Since improving
the position of an alternative in a linear ballot always increases the ballot Borda Score of
that alternative by one, it follows that the total Borda Score of that alternative will also
increase by one. Hence, it seems clear that an alternative which is already the social choice
of a Borda Count election will remain the social choice when its total Borda Score increases
by one and nothing else in the profile changes. We will now show that this property also
holds for Bucket Borda elections.
Theorem 4.1.4. Bucket Borda Count is a montonic voting system.
Proof. Let A be a set of alternatives and B be a profile of bucket ordered ballots consisting
of m voters such that B = {B(v1 ), B(v2 ), . . . , B(vm )}. Let a? ∈ A such that a? is the
P
winner of a Bucket Borda election. Since a? is the social choice we know m
j=1 BBa? (vj ) >
Pm
j=1 BBa (vj ) for all other a ∈ A.
Assume voter vi where i ∈ {1, . . . , m} adjusts his ballot from B(vi ) to B 0 (vi ) by improving his preference toward a? . In a bucket ordered ballot this corresponds to moving a? from Ai to Ai+1 where Ai , Ai+1 ⊆ B(v) and Ai+1 Ai . Hence, from the definition of Bucket Borda and the Axioms of Borda Count we have that BBa? (vi0 ) >
BBa? (vi ). Let B 0 = {B 0 (v1 ), B 0 (v2 ), . . . , B 0 (vm )} where B 0 (vj ) = B(vj ) for all j ∈
Pm
Pm
0
{1, . . . , m} except j = i. Then
j=1 BBa? (vj ) >
j=1 BBa? (vj ) since for all j 6= i,
BBa? (vj0 ) = BBa? (vj ) and when j = i, BBa? (vj0 ) > BBa? (vj ). Therefore, we have
Pm
Pm
Pm
Pm
0
0
j=1 BBa (vj ) ≥
j=1 BBa (vj ) for all other a ∈ A.
j=1 BBa? (vj ) >
j=1 BBa? (vj ) >
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
Hence,
Pm
0
j=1 BBa? (vj )
>
Pm
0
j=1 BBa (vj )
53
and it follows that a? remains the social choice
in B 0 . Thus, we conclude that Bucket Borda is a monotonic voting system.
We now provide an example to make the idea of a monotonic Bucket Borda election
clear. Let A = {a, b, c, d} be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )} be
a profile of m ballots in a Bucket Borda election. Assume b is the social choice of the
election. Then we know that the total Borda Score of b is greater than that of the other
three alternatives. Assume voter vi where i ∈ {1, . . . , m} monotonically adjusts his ballot
from B(vi ) to B 0 (vi ) as shown in Figure 4.1.1.
Figure 4.1.1. Monotonicity of a Bucket Borda Ballot
In Figure 4.1.1 voter vi improves his preference toward b by moving it from level A1 in
B(vi ) to level A2 in B 0 (vi ). Note that BBb (vi ) = 1.5 and BBb (vi0 ) = 2.5. Since BBb (vi0 ) >
BBb (vi ) and b’s ballot Borda Score for each other ballot in the profile has remained
constant, it is clear that b must remain the social choice of the Bucket Borda election.
At this point, we should recognize that Bucket Borda satisfies both the Pareto and
monotonicity criterions mentioned in the discussion of Arrow’s impossibility theorem in
Section 2.2. Hence, by Arrow’s impossibility theorem it must be the case that Bucket
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
54
Borda fails to satisfy Independence of irrelevant alternatives. The proof of this fact will
be presented in the following section.
The next property satisfied by Linear Borda we will look at is the consistency criterion.
The consistency criterion, which Woodall refers to as convexity, states that if the ballots in
an election are arbitrarily divided into two distinct profiles and separate elections utilizing
each of these profiles result in the same alternative being elected as the social choice,
then an election under the profile as a whole will also select that alternative as the social
choice. Intuitively, this criterion states that splitting the electorate into disjoint profiles
which each select the same winner should result in an overall election which also selects
this alternative as the winner. We rigorously define consistency below.
Definition 4.1.5. Let A be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )}
be a profile of m ballots. Let k ∈ N such that 1 ≤ k < m. Partition the orignal profile into two distinct profiles, so there now exist B1 = {B(v1 ), . . . , B(vk )} and
B2 = {B(vk+1 ), . . . , B(vm )}. If the partitioned elections E : B1 −
→ A? and E : B2 −
→ A?
both select a ∈ A as the social choice, then E satisfies the consistency criterion if the
complete election E : B −
→ A? always selects a as the social choice.
4
This property is satisfied by any ranked voting system which determines the social choice
by allocating points to the alternatives in each ballot. It follows that both Linear Borda
and Bucket Borda are consistent voting systems, and we provide the proof for Bucket
Borda below.
Theorem 4.1.6. Bucket Borda satisfies the consistency criterion.
Proof. Let A be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )} be a profile of
m bucket ordered ballots. Let k ∈ N such that 1 ≤ k < m. Assume we split the original
profile B in two, so we now have B1 = {B(v1 ), . . . , B(vk )} and B2 = {B(vk+1 ), . . . , B(vm )}
for some k. Let BB : B1 −
→ A? and BB : B2 −
→ A? be two separate Bucket Borda
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
55
P
elections which each select a ∈ A as the social choice. Then we know kj=1 BBa (vj ) >
Pk
Pm
Pm
j=1 BBb (vj ) and
j=k+1 BBa (vj ) >
j=k+1 BBb (vj ) for all other b ∈ A. It follows
from these two inequalities that:
k
X
BBa (vj ) +
j=1
m
X
j=k+1
m
X
j=1
BBa (vj ) >
BBa (vj ) >
k
X
j=1
m
X
BBb (vj ) +
m
X
BBb (vj )
j=k+1
BBb (vj ).
j=1
Thus we see that a must be the social choice of the election utilizing the complete profile,
and we conclude that Bucket Borda satisfies the consistency criterion.
Another important property satisfied by Linear Borda Count is the participation criterion. Intuitively, a voting system satisfies the participation criterion if adding a ballot
to a profile in which alternative a is preferred to alternative b never changes the winner
of the election from a to b. This criterion is important for a voting system to satisfy because inserting a new ballot into an election in which a is preferred to b should never hurt
a’s chance of winning the election relative to b. We rigorously define participation in the
following manner.
Definition 4.1.7. Let A be a set of alternatives such that a, b ∈ A and B =
{B(v1 ), B(v2 ), . . . , B(vm )} be a profile of m ballots. Asssume a is the social choice of
a given election method, E : B −
→ A? . The voting system E satisfies the participation
criterion if the addition of a ballot to the profile in which a b does not change the
winner of the election from a to b.
4
Similar to consistency, participation is satisfied by any ranked voting system which
determines the social choice by allocating points to the alternatives in each ballot. We
provide the proof for Bucket Borda below, and the extension to Linear Borda is trivial.
Theorem 4.1.8. Bucket Borda satisfies the participation criterion.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
56
Proof. Let A be a set of alternatives such that a, b ∈ A and B = {B(v1 ), B(v2 ), . . . , B(vm )}
be a profile of m ballots. Assume a is the social choice of the Bucket Borda election,
Pm
P
BB : B −
→ A? . Then we know m
j=1 BBb (vj ). Suppose a new ballot,
j=1 BBa (vj ) >
B(vm+1 ), is added to the original profile in which a b. Denote the new profile as B 0 where
B 0 = {B(v1 ), B(v2 ), . . . , B(vm+1 )}. Since a b in B(vm+1 ), it follows from the Axioms of
Borda Count that BBa (vm+1 ) > BBb (vm+1 ). Therefore, we have
m
m
X
X
[BBa (vj )] + BBa (vm+1 ) >
[BBb (vj )] + BBb (vm+1 )
j=1
j=1
m+1
X
BBa (vj ) >
j=1
m+1
X
BBb (vj ).
j=1
Hence, we see that a is preferred to b in the Bucket Borda election utilizing the profile
with the new ballot. Since a remains preferred to b, we see it is impossible for b to be
the winner of the new election. Thus, the insertion of B(vm+1 ) into the profile has not
improved b’s chance of winning the election relative to a, and we conclude that Bucket
Borda satisfies the participation criterion.
Every property we have listed so far has held in both Linear Borda and Bucket Borda
elections. However, as we will see with the next property this is not always the case. The
plurality criterion intuitively states that if the number of ballots ranking a as their first
overall preference is greater than the number of ballots on which alternative b is shown any
preference at all, then a should always have a greater probability than b of being elected
(or we can say that a should always be preferred to b in the overall election).
Definition 4.1.9. Let A be a set of alternatives such that a, b ∈ A and B be a profile.
Assume that the number of ballots in B which rank a as the single most preferred alternative is greater than the number of ballots in B on which alternative b is shown any
preference over another alternative. A voting system E : B −
→ A? under these conditions
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
57
satisfies the plurality criterion if a is always preferred to b in A? (in other words, a is
always preferred to b in the overall election).
4
In order to better understand what is meant by the plurality criterion we provide
an example of a Linear Borda election which satisfies plurality in Figure 4.1.2. Assume we have a Linear Borda election, BS : B −
→ A? , such that A = {a, b, c},
B = {B(v1 ), B(v2 ), B(v3 ), B(v4 ), B(v5 ), B(v6 )} and the ballots in B are given by the Hasse
diagrams seen in Figure 4.1.2.
Figure 4.1.2. Plurality Satisfied in Linear Borda Election
Note that a is the most preferred alternative on four of the ballots in the profile
{B(v1 ), B(v2 ), B(v3 ), B(v4 )} while b is shown any preference at all on only three of the
ballots in the profile {B(v4 ), B(v5 ), B(v6 )}. Hence, in order to satisfy the plurality criterion
P
P
a must finish ranked ahead of b in A? . We have 6i=1 BSa (vi ) = 9 and 6i=1 BSb (vi ) = 4.
It follows from these computations that a b in A? which means that a is preferred to
b in the overall election. Thus, we see that the Linear Borda election presented in this
example satisfies the plurality criterion.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
58
It is fairly obvious from the example above that Linear Borda satisfies the plurality
criterion since an alternative receiving more first-place votes than another alternative
receives any votes at all must be allocated a higher total Borda Score in a Linear Borda
election. For completeness sake, and in order to better see why this property holds for
Linear Borda and fails Bucket Borda, we provide the proof for Linear Borda below.
Theorem 4.1.10. Linear Borda satisfies the plurality criterion.
Proof. Let A be a set of n alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )} be a profile
of m linearly ordered ballots. Let a, b ∈ A. Assume the number of ballots in B which have
a as the most preferred alternative is greater than the number of ballots in B on which
alternative b is shown any preference. Let k be the number of ballots in B which have a as
the most preferred alternative. Then we know that for some j such that k > j, b is shown
any preference at all on j ballots in B.
From the definition of Linear Borda Count we know that a receives a Borda Score of
k(n−1) from the k ballots in which it is the most preferred alternative. The minimum total
Borda Score a can receive occurs if it is the least preferred alternative in the (m−k) ballots
P
in which it is not most preferred. Therefore, min[ m
i=1 BSa (vi )] = k(n − 1). Furthermore,
we know b must receive a Borda Score of zero on the (m−j) ballots in which it isn’t shown
any preference over another alternative. Hence, the maximum Borda Score b can receive
from this election occurs if it is the most preferred alternative on the j ballots on which
P
it is shown any preference. Therefore, max[ m
i=1 BSb (vi )] = j(n − 1). Note that k > j, so
P
Pm
we have that min[ m
i=1 BSa (vi )] = k(n − 1) > j(n − 1) = max[ i=1 BSb (vi )]. Since the
minimum Borda Score alloted to a is greater than the maximum Borda Score alloted to b
we see that a must receive a higher total Borda Score than b for any election in which this
assumption holds. Hence, a is preferred to b in any Linear Borda election for which these
assumptions hold, and we conclude that the plurality criterion holds for Linear Borda.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
59
This criterion seems reasonable because if an alternative a is the first choice on more
ballots than an alternative b is shown any preference at all, then we should be able to
safely assume that our voting system will rank a above b. Woodall himself has called
the plurality criterion ”‘a rather weak property that must hold in any real election.”’
However, as we will see below, there is an interesting property of Bucket Borda which
makes it incompatible with the plurality criterion.
Theorem 4.1.11. Bucket Borda does not satisfy the plurality criterion.
Proof. In order to prove Bucket Borda does not satisfy the plurality criterion it will
suffice to show there exists a set of alternatives and profile of ballots which when applied
to a Bucket Borda election contradict the plurality criterion.
Let A = {a, b, c} be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(v19 )} be a
profile of 19 bucket ordered ballots such that ten of the voters vote one way and nine vote
another way as represented by the two ballots shown in Figure 4.1.3. Note from Figure
Figure 4.1.3. Failure of Plurality in Bucket Borda Election
4.1.3 that alternative a is the single most preferred candidate on 10 of the 19 ballots while
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
60
alternative b is shown any preference at all on only 9 of the 19 ballots (since it is not
preferred over any alternative in {B(v1 ), . . . , B(v10 )}). Hence, it is clear that our profile
satisfies the assumptions necessitated by the plurality criterion. However, using Bucket
Borda to allocate points among the alternatives in the profile we have that
19
X
BBa (vj ) = 10(2) + 9(0) = 20.
j=1
19
X
BBb (vj ) = 10(1/2) + 9(2) = 23.
j=1
19
X
BBc (vj ) = 10(1/2) + 9(1) = 14.
j=1
Since
19
X
j=1
BBb (vj ) >
19
X
BBa (vj ) >
j=1
19
X
BBc (vj ), it follows that b is the winner of the
j=1
election despite the fact that alternative a was the most preferred candidate on more
ballots than b was shown any preference at all. Thus, we conclude that Bucket Borda does
not satisfy the plurality criterion.
The failure of Bucket Borda to satisfy the plurality criterion is due to the fact that in
ballots with more than one alternative in the bottom level of the ballot, Bucket Borda
allocates a small amount of points to each of the alternatives in this level. This is because in
Bucket Borda ballots an alternative does not need to be preferred over another alternative
in order to receive a ballot Borda Score greater than zero; it can receive points just for
being incomparable to other alternatives in the ballot. This means that an alternative in
the bottom level which is shown no preference over any other alternative can still receive
a Borda Score greater than zero since it is also allocated one-half of a point for each other
alternative in the bottom level that it is equivalent to (as we saw for ballots 1-10 in the
previous proof). In contrast, alternatives in a Linear Borda ballot only receive points for
being preferred to other alternatives in the ballot. Hence, Linear Borda always allocates a
score of zero to an alternative that is shown no preference on a ballot. This key difference
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
61
is what allows the plurality criterion to hold true for linearly ordered ballots but not for
bucket ordered ballots.
The failure of Bucket Borda to satisfy the plurality criterion made me wonder if there
was something flawed with the way our Bucket Borda formula allocated Borda Scores to
the alternatives in a bucket ordered ballot. This motivated me to want a different Bucket
Borda formula which, more similarly to Linear Borda, would allocate a Borda Score of
zero to alternatives in a bucket ordered ballot which are shown no preference in the ballot.
In addition to this property, my work with Borda Count led me to believe there are two
other fundamental properties of Linear Borda which any Bucket Borda Count function
should strive to satisfy as well. All three of these are mathematical properties of Linear
Borda which seem desirable to retain when extending Borda Count to non-linear ballots.
We mentioned each of these properties in our initial discussion of Linear Borda in Section
2.3 and define them rigorously below:
Definition 4.1.12. Let A be a set of n alternatives, B be a profile of bucket ordered
ballots, and BB : B −
→ A? be an election method which allocates points to the alternatives
in the ballots of B based on the Borda Count. The three desirable properties of Linear
Borda for any bucket ordered Borda voting system BB to satisfy are:
1. For all B(v), B(u) ∈ B,
n
X
i=1
BBai (v) =
n
X
BBai (u).
i=1
2. If there exists an a? ∈ B(v) such that ¬(a? a) for all a ∈ B(v), then BBa? (v) = 0.
3. If there exists an a? ∈ B(v) such that a? a for each alternative a ∈ B(v) and there
exists b? ∈ B(u) such that b? b for each alternative b ∈ B(u), then BBa? (v) =
BBb? (u).
4
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
62
The first property states that every ballot in a bucket ordered Borda Count election
should contribute the same weight to the overall Borda Score of the election. Hence, the
sum of the Borda Scores allocated to the alternatives in each ballot of a given election
should be the same. This seems reasonable because every ballot in a given Bucket Borda
election should carry the same weight in determining the social choice. It would be undesirable for voters to have the ability to increase (or decrease) the weight of their ballot in
an election depending on how they rank the alternatives.
The second property states that if there is an alternative in a bucket ordered ballot
which is not preferred to any other alternative in its ballot, then that alternative should
be allocated a ballot Borda Score of zero. This seems reasonable because if a voter does
not show any preference for an alternative in his ballot then the total Borda Score for
that alternative should not benefit from the inclusion of the ballot in the election. The
motivation for this property comes largely from the failure of our original Bucket Borda
formula to satisfy the plurality criterion.
The third property states that the ballot Borda Score allocated to an alternative which
is preferred to every other alternative in its ballot should be the same for every ballot in
a given election. This property places a maximum value on the amount of points a voter
can allocate to his single most preferred alternative regardless of how he ranks the other
alternatives in his ballot. Furthermore, it states that this maximum should be constant
for every ballot in an election. This property is desirable because we need to establish
a maximum value allocated to the most preferred alternative in a ballot or voters could
manipulate their ballots in order to increase the weight of the ballot Borda Score allocated
to their most preferred alternative.
After much computation attempting to devise a formula which could simultaneously
satisfy these three properties, I came to the rather disappointing conclusion that it was
impossible to devise a system of allocating Borda Scores to a profile of bucket ordered
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
63
ballots such that all three of these properties were fulfilled. This observation gave rise to
the following impossibility theorem.
Theorem 4.1.13. Let A be a set of n alternatives, B be a profile of bucket ordered ballots,
and BB : B −
→ A? be a social choice procedure which allocates points to the alternatives
in the ballots of B based on the Borda Count. There does not exist a voting system BB : B
−
→ A? that satisfies the three desirable properties of Linear Borda stated in Definition
4.1.12.
Proof. We will prove this result by showing that when any two of these properties are
assumed to hold for a given set of alternatives and profile of ballots it becomes impossible
for the third property to hold. Note that we will refer many times throughout this proof
to the Axioms of Borda Count first stated in Definition 2.3.6. Thus, we have three cases:
Case 1: Assume (1) and (2) hold. Then we know for all possible B, every ballot in B
is allocated the same weight and any alternative which receives no preference in a given
ballot is allocated a score of zero.
Let B = {B(v1 ), B(v2 )} where B(v1 ) and B(v2 ) are represented by the following Hasse
diagrams:
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
64
By property (2) alternatives in a ballot which are shown no preference must receive a
score of zero, and it follows that BBa4 (v1 ) = BBa2 (v2 ) = BBa3 (v2 ) = BBa4 (v2 ) = 0. By
property (1) the sum of the Borda Scores from each ballot must be equal, so we have
4
X
i=1
BBai (v1 ) =
4
X
BBai (v2 )
i=1
BBa1 (v1 ) + BBa2 (v1 ) + BBa3 (v1 ) + BBa4 (v1 ) = BBa1 (v2 ) + BBa2 (v2 ) + BBa3 (v2 ) + BBa4 (v2 )
BBa1 (v1 ) + BBa2 (v1 ) + BBa3 (v1 ) = BBa1 (v2 )
BBa1 (v1 ) = BBa1 (v2 ) − BBa2 (v1 ) − BBa3 (v1 )
By Axioms (1) and (2) of Borda Count we know that BBa2 (v1 ), BBa3 (v1 ) > 0, and it
follows that BBa1 (v1 ) < BBa1 (v2 ). Hence BBa1 (v1 ) 6= BBa1 (v2 ), and we conclude there
exists a profile in which the Bucket Borda function does not allocate an equal Borda Score
to the single most preferred alternative in each ballot. Thus, property (3) does not hold.
Case 2: Assume (1) and (3) hold. Then we know for all possible B, every ballot in B is
allocated the same weight and any ballot with a single most preferred alternative allocates
an equal score to that alternative.
Let B = {B(v1 ), B(v2 )} where B(v1 ) and B(v2 ) are represented by the following Hasse
diagrams:
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
65
Then by property (3) we know that BBa1 (v1 ) = BBa1 (v2 ) and by property (1), we have
P4
P4
i=1 BBai (v1 ) =
i=1 BBai (v2 ). Therefore,
BBa2 (v1 ) + BBa3 (v1 ) + BBa4 (v1 ) = BBa2 (v2 ) + BBa3 (v2 ) + BBa4 (v2 ).
Note that by Axioms (1) and (2) of Borda Count Ballots we know that BBa2 (v1 ) >
BBa3 (v1 ) > BBa4 (v1 ) ≥ 0. Hence BBa2 (v1 ) + BBa3 (v1 ) + BBa4 (v1 ) = BBa2 (v2 ) +
BBa3 (v2 ) + BBa4 (v2 ) > 0. Furthermore, by Axioms (2) and (3) of Borda Count Ballots
we know that alternatives cannot be allocated a negative Borda Score and alternatives
located on the same level of a bucket ordered ballot must receive the same Borda Score. It
follows that BBa2 (v2 ) = BBa3 (v2 ) = BBa4 (v2 ) > 0. We conclude there exists a profile in
which the Bucket Borda function allocates Borda Scores greater than zero to alternatives
that are shown no preference. Thus, property (2) does not hold.
Case 3: Assume (2) and (3) hold. Then we know for all possible B, every alternative
which receives no preference in a given ballot is allocated a score of zero and each ballot
with a single most preferred alternative allocates an equal score to that alternative.
Let B = {B(v1 ), B(v2 )} where B(v1 ) and B(v2 ) are represented by the following Hasse
diagrams:
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
66
Then by property (2), we have BBa4 (v1 ) = BBa2 (v2 ) = BBa3 (v2 ) = BBa4 (v2 ) = 0
and by property (3), we know that BBa1 (v1 ) = BBa1 (v2 ). Accounting for the alternaP
tives which are allocated a score of zero, we have that 4i=1 BBai (v2 ) = BBa1 (v2 ) and
P4
i=1 BBai (v1 ) = BBa1 (v1 ) + BBa2 (v1 ) + BBa3 (v1 ).
By Axioms (1) and (2) of Borda Count Ballots we know that BBa2 (v1 ) > BBa3 (v1 ) > 0,
and we already stated that BBa1 (v1 ) = BBa1 (v2 ). It follows that
4
X
BBai (v2 ) = BBa1 (v2 )
i=1
< BBa1 (v2 ) + BBa2 (v1 ) + BBa3 (v1 )
= BBa1 (v1 ) + BBa2 (v1 ) + BBa3 (v1 ) + BBa4 (v1 ).
Hence,
P4
i=1 BBai (v2 )
<
P4
i=1 BBai (v1 ),
and we conclude there exists a profile in which
the sum of the Borda Scores allocated to each ballot in the profile is not equal. Thus,
property (1) does not hold.
Combining these three cases we conclude it is imposssible for a Borda Count voting
system applied to a profile of bucket ordered ballots to satisfy the three desirable properties
of Linear Borda stated in Definition 4.1.12.
The impossibility theorem presented above proves that there is no Bucket Borda formula
which perfectly mimics the three most desirable mathematical properties of Linear Borda.
Given this limitation, we feel that the Bucket Borda voting system already devised in this
paper is the most effective method for allocating Borda Scores to the alternatives in a
bucket ordered ballot. One could very well argue that it is quite sensible to allocate a nonzero score to the alternatives on the bottom level of a bucket ordered ballot. Since none of
these alternatives are the single least preferred alternative in the ballot, it seems that they
deserve to be rewarded by the ballot in some manner. However, this is a question of more
concern for social scientists than mathematicians and we will address it no further here.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
67
The important aspect of our impossibility theorem is that when extending our analysis
from linearly ordered sets to bucket ordered sets some desirable properties are inevitably
lost due to the somewhat more complicated structure of our sets. We will now shift our
focus and discuss some unsatisfied properties of Linear Borda in the next section.
4.2 Unsatisfied Properties
As was proven by Arrow in his famous impossibility theorem of the 1950s, no voting system in an election with three or more alternatives can simultaneously satisfy the Pareto,
monotonicity, and independence of irrelevant alternatives criterions. As we have already
shown that Bucket Borda satisfies the Pareto and monotonicity criterions, we will begin
this section by proving that Bucket Borda fails to satisfy Independence of irrelevant alternatives. We will then introduce some other mathematical properties which do not hold in
Linear Borda elections, and we will see how these properties are handled in Bucket Borda
elections.
I feel that Taylor’s definition of independence of irrelevant alternatives (IIA) is easier to
grasp and more interesting mathematically than Woodall’s so we will be working with it
in this section. In section 2.2 we noted that IIA generally states that given two candidates
a and b in an election, the position of alternatives other than a and b in the ballots of
the election should have no impact when determining whether a is preferred to b or b is
preferred to a in the social preference list. Another way to phrase this property is that if a
is the winner of an election, and one or more voters change the preferences in their ballots
but none change their mind about whether a is preferred to b or b to a, then b should
not be the winner of the new election [5]. Intuitively, this means that when voters change
their preferences toward alternatives other than a and b in their ballots, this should be
irrelevant to the question of social preference between a and b. We rigorously state the
definition of independence of irrelevant alternatives below and then provide the proof that
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
68
it fails to hold for both Linear and Bucket Borda. Note that in order to make or definition
of IIA applicable to bucket ordered ballots, we include the case where a is equvialent to b
in a given ballot.
Definition 4.2.1. Let A be a set of alternatives such that a, b ∈ A, let B be a profile of
A, and let a be the social choice of a given election E : B −
→ A? . Let X = {B(vj )|a b
in B(vj )}, Y = {B(vj )|b a in B(vj )}, and Z = {B(vj )|a ≡ b in B(vj )} where j ∈
{1, . . . , m}. Then E satisfies the independence of irrelevant alternatives criterion if
adjusting the individual preferences of the ballots in B while keeping a b in X, b a in
Y , and a ≡ b in Z never results in b being included as the social choice.
4
Theorem 4.2.2. Bucket Borda does not satisfy independence of irrelevant alternatives.
Proof. In order to prove Bucket Borda does not satisfy independence of irrelevant alternatives (IIA) it will suffice to show there exists a set of alternatives and profile of ballots
which when applied to Bucket Borda contradict the IIA criterion. Furthermore, we will
construct our counter-example utilizing only linearly ordered ballots in order to show that
it simultaneously does not hold for Linear Borda Count.
Let A = {a, b, c, d} be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(v6 )} be a
profile in a Bucket Borda election such that the voter’s preferences are expressed in the
ballots shown below:
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
69
Note that the total Bucket Borda Score for alternative a is 14 while the total Bucket Borda
Score for alternative b is 12. Hence, a is the social choice of this Bucket Borda election.
Now, assume that voter’s v1 , v3 , and v4 adjust their ballots so the new profile for the
election looks as follows:
Note that the preferences toward alternatives b and c have been swapped in B(v1 ), the
preferences between b and d have been swapped in B(v3 ), and the preferences between a
and c have been swapped in B(v4 ). Note further that the preferences between a and b have
not been swapped in any of the ballots. Hence, a is still preferred to b in the first three
ballots of the profile, and b is still preferred to a in the second three ballots. However, b
now receives a total Borda Score of 14 while a only receives a total Borda Score of 12.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
70
Therefore, the social choice of the Bucket Borda (or Linear Borda) election has changed
from a to b although none of the voters changed their mind about whether a is preferred to
b or b is preferred to a. Thus, we conclude that Bucket Borda fails to satisfy independence
of irrelevant alternatives.
The second unsatisfied property of Linear Borda elections we will discuss is the later-noharm criterion. The later-no-harm criterion is very similar to independence of irrelevant
alternatives, and it inuitively states that increasing the preference of a less preferred alternative in a ballot should not cause a more preferred alternative in the ballot to lose the
election.
Definition 4.2.3. Let A be a set of alternatives such that a, b ∈ A, B be a profile of A,
and a be the social choice of a given election E : B −
→ A? . Let B(vi ) ∈ B be a ballot in
the profile such that a b in B(vi ). A voting system under these conditions satisfies the
later-no-harm criterion if increasing the preference of b in B(vi ) but keeping a b in
B(vi ) never results in a being replaced as the social choice.
4
This criteria cannot hold in any ranked election system which allocates points to the
alternatives since improving a losing alternatives position in a ballot will improve its score
from the ballot and based on the composition of the rest of the profile could surely give
it enough points to win the election. It is unclear whether this is even a very desirable
property for voting systems to satisfy since it seems perfectly reasonable that if a voter
improves his preference toward an alternative in a ballot then that should benefit the
alternative in the overall election. We will assume the property is unsatisfied by Linear
Borda elections and provide a proof below that it also does not hold in Bucket Borda
elections.
Theorem 4.2.4. Bucket Borda fails to satisfy the later-no-harm criterion.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
71
Proof. In order to prove Bucket Borda does not satisfy the later-no-harm criterion it will
suffice to show there exists a set of alternatives and profile of ballots which when applied
to Bucket Borda contradict the later-no-harm criterion.
Let A = {a1 , a2 , . . . , a6 } be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(v5 )} be
a profile in a Bucket Borda election such that the voter’s preferences are expressed in the
ballots shown below:
Note that a1 is the social choice over a6 since
P5
j=1 BBa1 (vj )
= 21 >
P5
j=1 BBa6 (vj )
= 20.
Assume voter v1 adjusts his ballot by increasing his preference toward a6 such that the new
preference order for B(v1 ) is a1 a6 a2 a3 a4 a5 . Note that the preference of a6
has greatly improved in B(v1 ) but a1 is still preferred to a6 in the ballot. From Definition
4.2.3, we see that we have satisfied the conditions necessary for the later-no-harm criterion
to apply.
However, note that a6 is now preferred to four alternatives in B(v1 ) while before v1
adjusted his ballot it was preferred to none. Hence, we must add four points to the ballot
Borda Score of a6 from B(v1 ), and therefore must also add four points to the total Borda
Score of a6 . Furthermore, note that the position of a1 in B(v1 ) has not changed so the
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
72
P
total Borda Score of a1 remains constant. Thus we have 5j=1 BBa6 (vj ) = 20 + 4 = 24 >
P5
j=1 BBa1 (vj ) = 21, and it is clear that the adjustment of B(v1 ) has changed the social
choice of the Bucket Borda election from a1 to a6 . From this result we conclude that
Bucket Borda does not satisfy the later-no-harm criterion.
The next voting criterion we will discuss is the indepedence of clones criterion. The
independence of clones criterion, formulated by Nicholas Tideman in [6], states that the
insertion of an alternative identical to one already present in an election, a clone, will not
cause the winner of the election to change. Before we begin our analysis of the clone independence of Linear Borda and Bucket Borda elections, we must first discuss what exactly
is meant by the insertion of a clone into a linear election and a non-linear election. The
clone of an alternative a in a linearly ordered ballot is well-defined and can be intuitively
thought of as an alternative that is:
1. Preferred to all alternatives preferred by a and
2. Preferred by a and all alternatives preferred to a.
We rigorously define the clone of an alternative in a linear ballot in the following manner:
Definition 4.2.5. Let A be a set of alternatives and B(v) be a linearly ordered ballot of
A. Let a0 ∈ A. We define the clone a00 of a0 as the alternative in B(v) such that a0 a00
and the following are true for all other a ∈ A:
• if a a0 then a a00
• if a0 a then a00 a.
Note that the insertion of a00 into the original ballot B(v) will be denoted by the cloned
ballot B 0 (v).
4
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
73
In Figure 4.2.1 we have a ballot B(v) consisting of the set A = {a, b, c, d} such that
a b c d and a cloned ballot B 0 (v) in which the clone of alternative b, b0 , has
been inserted into the original ballot. By definition of a linear clone, in B 0 (v) we have
a b b0 c d. Note that b0 is preferred to every alternative b is preferred to and
preferred by b and each alternative preferred to b.
Figure 4.2.1. Insertion of a Clone in a Linear Ballot
Intuitively, a clone in a bucket ordered ballot should be quite similar to a clone in a
linearly ordered ballot. However, since bucket ordered ballots allow for alternatives to be
equivalent to one another there is no longer any reason for an alternative to be preferred
to its clone as we saw in linear ballots. Hence, a clone in a bucket order ballot should be
defined very similarly to a clone in a linear ballot with the exception that an alternative
and its clone should be equivalent in a bucket ordered ballot. Extending the concept of
clones to bucket ordered ballots, we have the following definition:
Definition 4.2.6. Let A be a set of alternatives and B(v) be a bucket ordered ballot of
A. Let a0 ∈ A. We define the clone a00 of a0 as the alternative in B(v) such that a0 ≡ a00
and the following are true for all other a ∈ A:
• if a a0 then a a00
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
74
• if a0 a then a00 a.
4
Note that these definitions of a clone are nearly identical except in linear ballots a
clone is preferred by the alternative it’s cloning and in bucket ordered ballots a clone is
equally preferred to the alternative it’s cloning. In some ways this difference makes bucket
ordered ballots a more natural place for clones to be inserted since it makes more sense
for an alternative to be equivalent to its clone rather than preferred to its clone.
In Figure 4.2.2 we have the same set of alternatives from Figure 4.2.1 now ordered in a
bucket ordered ballot B(v). Once again the clone of b, b0 , has been inserted into the cloned
ballot B 0 (v). Note that b b0 in Figure 4.2.1 but b ≡ b in 4.2.2. This is the key difference
between the role of a clone in a linearly ordered election and a bucket ordered election.
The importance of this difference will be seen in the theorems and proofs presented later
in this section.
Figure 4.2.2. Insertion of a Clone in a Bucket Ballot
Independence of clones is a desirable property for a voting system to have because
if the insertion of a cloned alternative can greatly influenced the outcome of an election
then political factions could strategically place similar alternatives onto a ballot in order to
help ensure their political goals. Election methods that fail indendence of clones can either
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
75
be clone negative (the insertion of an identical alternative will decrease an alternative’s
chance of winning the election) or clone positive (the insertion of an identical alternative
will increase a candidate’s chance of winning the election).
It is a rather well known result that Linear Borda Count is clone positive, and we provide
our proof of this result below.
Theorem 4.2.7. Linear Borda is a clone positive voting system.
Proof. Let A = {a1 , a2 , . . . , an } be a set of n alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )}
be a profile of m linearly ordered ballots of A. Let a? ∈ A be the social choice of a Linear
Pm
P
Borda election. Then we know m
j=1 BBa (vj ) for all other a ∈ A.
j=1 BBa? (vj ) >
Let the clone a0i of ai be inserted into the election for some ai ∈ A such that ai 6=
a? . From the definition of a linear clone we know ai a0i in each ballot of the profile.
Therefore, the insertion of a0i will increase the Borda Score of ai by one for each ballot
in the profile. Hence, the insertion of the clone will increase the total Borda Score of ai
by m since there are m ballots in the profile and ai ’s ballot Borda Score increases by one
in each ballot. Thus, the total Borda Score of ai in the cloned election can be given by
Pm
Pm
0
j=1 [BBai (vj )] + m.
j=1 BBai (vj ) =
There are now two cases we must consider for each B(v) in the election:
• Assume a? ai in B(v). Then a? a0i in B 0 (v) and the insertion of a0i increases the
ballot Borda Score of a? by one in B 0 (v).
• Assume ai a? in B(v). Then a0i a? in B 0 (v) and the insertion of a0i has no effect
on the ballot Borda Score of a? in B 0 (v).
Let k be the number of ballots in the election such that a? ai . Note that k ≤ m since
m equals the total number of ballots in the election. It follows from above that
m
X
j=1
BBa? (vj0 ) =
m
X
j=1
[BBa? (vj )] + k ≤
m
X
[BBa? (vj )] + m.
j=1
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
76
In the previous paragraph we saw that the insertion of a0i increases the total Borda Score
of ai by m in the cloned election. Now, we see that this same insertion increases the total
Borda Score of a? by some number k where k ≤ m. Hence, the insertion of the clone
increases the total Borda score of ai by as much or more than it increases the total Borda
Score of a? . Thus, it is clear that the insertion of a0i increases ai ’s chances of winning the
election and we conclude that Linear Borda is a clone positive voting system.
The following corollary uses what we learned from Theorem 4.2.7 to prove an even
more powerful result concerning the insertion of a clone into a Linear Borda election. In
particular, this corollary tells us exactly when the insertion of a clone in a Linear Borda
election can be influential enough to change the social choice to the cloned alternative.
Corollary 4.2.8. Let A = {a1 , a2 , . . . , an } be a set of n alternatives and B =
{B(v1 ), B(v2 ), . . . , B(vm )} be a profile of m linearly ordered ballots of A. Let a? ∈ A
be the winner of a Linear Borda election. Then for any ai ∈ A such that ai 6= a? , the insertion of a0i into the election will change the social choice to ai if and only if the number
of ballots on which ai a? is greater than the difference of the total Borda Scores between
a? and ai .
Proof. Let A = {a1 , a2 , . . . , an } be a set of n alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )}
be a profile of m linearly ordered ballots of A. Let a? ∈ A be the winner of a Linear Borda
P
Pm
count election. Then we know m
j=1 BBa? (vj ) >
j=1 BBa (vj ) for all other a ∈ A.
Let ai ∈ A such that ai 6= a? . Then we can partition B into the following two sets C and
D. Let C = {B(v)|a? ai in B(v)} and D = {B(v)|ai a? in B(v)}. Clearly, C ∪ D = B
since either a? ai or ai a? for all B(v) ∈ B. Note that |C| = number of ballots in the
profile such that a? ai and |D| = number of ballots in the profile such that ai a? . The
general form of the Hasse diagrams for ballots in C and D are shown below
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
77
Let the clone a0i of ai be inserted into the election. By definition we know that for each
B(v) ∈ C, a? ai a0i , and it follows that for all B(v) ∈ C:
BBai (v 0 ) = BBai (v) + 1 and
BBa? (v 0 ) = BBa? (v) + 1.
Similarly, for each B(v) ∈ D, ai a0i a? , and it follows that for all B(v) ∈ D:
BBai (v 0 ) = BBai (v) + 1 and
BBa? (v 0 ) = BBa? (v).
Hence, we have
m
X
BBai (vj0 ) =
j=1
m
X
BBai (vj ) + |C| + |D|
j=1
and
m
X
j=1
BBa? (vj0 ) =
m
X
BBa? (vj ) + |C|.
j=1
With this information clearly established we will now split our proof into two cases and
solve for each.
⇐ Assume that the number of ballots in which ai a? is greater than the difference
Pm
of the total Borda Scores between a? and ai . Thus, we have |D| >
j=1 BBa? (vj ) −
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
Pm
j=1 BBai (vj ).
m
X
78
It follows from the equations deduced in the previous pragraph that:
BBai (vj0 ) =
j=1
>
m
X
j=1
m
X
BBai (vj ) + |C| + |D|
BBai (vj ) + |C| +
j=1
m
X
BB (vj ) −
a?
m
X
j=1
= |C| +
= |C| +
m
X
j=1
m
X
BBai (vj )
j=1
BBa? (vj )
BBa? (vj0 ) − |C|
j=1
=
m
X
BBa? (vj0 ).
j=1
Therefore, we see that
m
X
BBai (vj0 ) >
j=1
m
X
BBa? (vj0 ) in the cloned election, and it follows
j=1
that the insertion of a0i has changed the social choice from a? to ai .
⇒ Assume that the insertion of a0i into the election changes the social choice to ai . Then
Pm
P
0
0
we know m
j=1 BBa? (vj ). Thus, we have
j=1 BBai (vj ) >
m
X
BBai (vj0 )
=
j=1
>
=
m
X
j=1
m
X
j=1
m
X
BBai (vj ) + |C| + |D|
BBa? (vj0 )
BBa? (vj ) + |C|.
j=1
Hence,
m
X
BBai (vj ) + |C| + |D| >
j=1
m
X
BBa? (vj ) + |C|,
j=1
|D| >
m
X
j=1
BBa? (vj ) −
m
X
BBai (vj ).
j=1
Therefore, we see that if the insertion of a0i into the election changes the social choice to
m
m
X
X
ai , then |D| >
BBa? (vj ) −
BBai (vj ).
j=1
j=1
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
79
Combining both of these cases we conclude that the insertion of a0i into a Linear Borda
election changes the social choice to ai if and only if the number of ballots on which ai a?
is greater than the difference of the total Borda Scores between a? and ai in the original
election.
Hence, we have shown that Linear Borda is a clone positive voting system. Furthermore,
the insertion of a clone into a Linear Borda election only changes the outcome if the number
of ballots in which the cloned alternative is preferred to the social choice is greater than
the difference in their original Borda Scores. We now present an example of an election in
which Corollary 4.2.8 holds.
Let A = {a, b, c} be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(v5 )} be a profile of ballots in a Linear Borda election such that the voters’ preferences are expressed
P5
through the Hasse diagrams shown in Figure 4.2.3. Note that
j=1 BBb (vj ) = 7 and
Figure 4.2.3. Insertion of Clone Changing Outcome of Linear Borda
P5
j=1 BBa (vj )
= 6. Hence, b is the social choice of the original election. Note further that
the number of ballots in which a b is greater than the difference in total Borda score
between b and a. Inserting a0 into the election gives us the following cloned ballots:
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
In the cloned election we now have that
P5
0
j=1 BBb (vj )
= 9 and
80
P5
0
j=1 BBa (vj )
= 11.
Thus, the outcome of the election has changed and the insertion of a0 has led to a being
the social choice of the cloned election. It is important to reemphasize here that the result
of the cloned election would not have changed if the number of ballots in which a b had
been less than the difference in total Borda Scores between a and b in the original election.
Assuming that Bucket Borda and Linear Borda would respond in the same manner to
the insertion of a clone, one would think that Bucket Borda is also clone positive. However,
reanalyzing the previous example using Bucket Borda instead of Linear Borda gives us a
somewhat surprising result.
Let A = {a, b, c} be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(v5 )} be a profile of ballots in a Bucket Borda election such that the voter’s preferences are expressed
through the ballots shown in Figure 4.2.3. In the previous example, we inserted the clone
of a into the election and the social choice of the election changed from b to a. However,
utilizing bucket ordered ballots instead of linearly ordered ballots we obtain the following
ballots in the cloned election.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
81
P5
0
In the cloned Bucket Borda election we now have that
j=1 BBb (vj ) = 9 and
P5
0
0
j=1 BBa (vj ) = 8.5. Thus, b remains the social choice despite the insertion of a into
the election. We conclude from this example that Bucket Borda and Linear Borda respond differently to the insertion of clones, and it initially appears that Bucket Borda
may be clone independent.
After computing many other examples, I began to convince myself that Bucket Borda
was clone independent and set out to prove this fact. However, while working through the
proof I realized in my computations that there were certain bucket ordered profiles for
which the insertion of a clone could change the outcome of a Bucket Borda election. While
it is much less likely for the insertion of a clone to change the outcome of a Bucket Borda
election than a Linear Borda election (as we saw in the previous example), it is possible
for the insertion a clone to change the outcome of a Bucket Borda election. An example
of such a profile is given in the following proof.
Theorem 4.2.9. Bucket Borda is not a clone independent voting system.
Proof. In order to prove that Bucket Borda is not clone independent, it will suffice to
provide one example in which the profile of a Bucket Borda election is clone positive.
Let A = {a1 , a2 , . . . , a6 } be a set of alternatives and B = {B(v1 ), B(v2 ), . . . , B(v5 )}
be a profile in a Bucket Borda election such that the voter’s preferences are expressed
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
82
in the ballots shown in Figure 4.2.4. Note that a1 is the social choice over a6 since
Figure 4.2.4. Insertion of Clone Changing Outcome of Bucket Borda
P5
j=1 BBa1 (vj )
= 21 >
P5
j=1 BBa6 (vj )
= 20. Now we will adjust the profile by inserting
the clone of a6 , a06 , into the election. Thus, we have the following cloned ballots:
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
It follows from the ballots above that
P5
j=1 BBa1 (vj )
= 22 <
83
P5
j=1 BBa6 (vj )
= 22.5.
Thus, the introduction of a06 changed the social choice from a1 in the original election
to a6 in the cloned election. By definition, therefore, we have shown that this Bucket
Borda election must be clone positive. Thus, we conclude that Bucket Borda is not clone
independent.
We will now introduce two similar properties neither of which holds in Borda Count
elections. However, using these properties to restrict the possible profiles of our election
will allow us to determine exactly when a Bucket Borda election is clone independent. In
the proof above note that although a1 was the Bucket Borda social choice of the original
election, it was preferred by a6 on four of the five ballots in the election. This phenomenon
refers to a somewhat unsettling property of Borda Count (and Bucket Borda) elections
which is that they fail to satisfy the majority criterion. The majority criterion states that
if an alternative is preferred to every other alternative on more than half of the ballots in
an election, then that alternative should win the election. More formally, it can be defined
as follows:
Definition 4.2.10. Let A be a set of alternatives and B be a profile. Assume there is a
single alternative a ∈ A such that a is the most preferred alternative on more than half of
the ballots in B. A voting system E : B −
→ A? satisfies the majority criterion if a must
be the winner of the election.
4
The failure of Borda Count to satisfy the majority criterion can be clearly seen from the
profile shown in Figure 4.2.4 in which a6 is the most preferred alternative on a majority of
the ballots but still fails to win the election. The majority criterion appears to be a very
important property for voting systems to satisfy, and one of the main arguments used
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
84
to question the validity of Borda Count is its failure to satisfy this property. However,
this is not necessarily a fault at all since Borda Count voting systems by design place
less weight on voters’ most preferred alternatives and attempt to consider the preference
rankings across their entire ballots when determining the social choice. Furthermore, in
order for an alternative to be a majority winner and still lose a Borda Count election it
must be viewed very unfavorably by the voters in the election that do not rank it first
in their ballots (we see this happening in both Figure 4.2.3 and Figure 4.2.4). In these
situations there is often another alternative in the election that is viewed more favorably
overall by the voting population and that alternative becomes the Borda Count social
choice. It is important to note that in most instances an alternative that is most preferred
on a majority of the ballots in an election will be the Borda Count social choice but in
situations such as the two examples mentioned above this is not always the case for good
reason.
A voting criterion similar to the majority criterion but somewhat stronger was made
popular by the eighteenth-century mathematician Marie Jean Antoine Nicolas Caritat [7].
Also referred to as the Marquis de Condorcet, Caritat developed a voting method known
as the Condorcet method based on his own Condorcet criterion. Intuitively, the Condorcet
criterion states that the alternative in an election which wins in a head-to-head matchup
with each other alternative should be the winner of the election. A Condorcet voting system
simply always elects this alternative as the winner of the election. This is a strong method
to elect the social choice but unfortunately a Condorcet winner does not always exist, and
if no alternative satisfies the Condorcet criterion then the Condorcet method fails to elect
a winner. We rigorously define the Condorcet criterion in the following manner:
Definition 4.2.11. Let A be a set of alternatives and B be a profile of ballots. Assume
there exists an alternative a ∈ A such that a b on more ballots in B than b a for each
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
85
other alternative b ∈ A. Then the voting system E : B −
→ A? satisfies the Condorcet
criterion if a must be the social choice of E.
4
From the profile in Figure 4.2.4 we see that a6 wins in a head to head matchup with
each other alternative in the profile, but we already know that a1 is the social choice of the
Linear or Bucket Borda election associated with this profile. Hence, it is clear that Linear
and Bucket Borda both fail to satisfy the Condorcet criterion. Stated below is a relatively
simple theorem relating the majority criterion and the Condorcet criterion. Intuitively, it
states that the majority criterion implies the Condorcet criterion.
Theorem 4.2.12. Let A be a set of alternatives and B be a profile of ballots. If the Bucket
Borda election BB : B −
→ A? satsfies the majority criterion, then BB also satisfies the
Condorcet criterion.
Proof. Let A = {a1 , a2 , . . . , an } be a set of n alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )}
be a profile of m bucket ordered ballots of A. Let BB : B −
→ A? satisfy the majority criterion. Then there is some a? ∈ A such that a? is the social choice. Without loss of generality,
let a? = a1 . Then by the majority criterion a1 ai for all i ∈ {2, 3, . . . , n} in more than
m
m
of the ballots in B. Clearly, this implies a1 ai on more than
of the ballots in
2
2
the profile for each i ∈ {2, 3, . . . , n}, and it follows that a1 ai on more ballots than
ai a1 for each i ∈ {2, 3, . . . , n}. Therefore, we conclude that BB : B −
→ A? satisfies the
Condorcet criterion.
By restricting the profiles of our voters’ to those that satisfy the Condorcet criterion, we
are able to prove an interesting theorem relating Bucket Borda and clone independence.
Theorem 4.2.13. If the ballots in a Bucket Borda election satisfy the Condorcet criterion,
then the election is clone independent.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
86
Proof. Let A = {a1 , a2 , . . . , an } be a set of n alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )}
be a profile of m bucket ordered ballots of A. Let a? ∈ A be the winner of a Bucket Borda
m
m
X
X
election. Then
BBa? (vj ) >
BBa (vj ) for all other a ∈ A.
j=1
j=1
Let the clone a0i of ai be inserted into the election for some ai ∈ A such that ai 6= a? .
By definition we know ai ≡ a0i in each ballot of the profile. Recall that the Bucket Borda
score of an alternative can be given by
BBai (vj ) = Hi +
Li−2 − 1
Li−i − 1
Li − 1 Li−1 − 1
+
(Li−1 ) +
(Li−2 ) + ... +
(Li−i )
2
Li−1
Li−2
Li−i
Note that the insertion of a0i will increase Li by one and not effect the size of any other
antichain in the ballot. Hence, L0i = Li +1 and for every other partition k in a given ballot,
L0k = Lk . Furthermore, the insertion of a0i will not effect the height of ai in any ballot, so
Hi = Hi0 for each ballot in the election. Therefore, the insertion of a0i will increase the ballot
1
1
for each ballot in the election. Hence, BBai (vj0 ) = BBai (vj ) +
2
2
m
m
X
X
m
for each j ∈ {1, . . . , m}. It follows that
BBai (vj0 ) =
BBai (vj ) + .
2
Borda Score of ai by
j=1
j=1
Note further that for each B(v) ∈ B, either a? ai , ai a? , or a? ≡ ai . Thus, we
can partition the m ballots in the election into three disjoint sets X, Y , and Z. Let
X = {B(v)|a? ai in B(v)}, Y = {B(v)|ai a? in B(v)}, and Z = {B(v)|a? ≡ ai in
B(v)}.
We then have the following three possible scenarios concerning the total Bucket Borda
Score of a? in the cloned election. If a? ai in B(v), then B(v) ∈ X and inserting a0i
will increase the ballot Borda Score of a? by 1 in B 0 (v). If ai a? , then B(v) ∈ Y and
inserting a0i will not effect the ballot Borda Score of a? in B 0 (v). If a? ≡ ai , then B(v) ∈ Z
and inserting a0i will increase the ballot Borda Score of a? by
m
X
j=1
BBa? (vj0 ) =
m
X
j=1
BBa? (vj ) + |X| +
1
in B 0 (v). Thus, we have
2
|Z|
.
2
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
87
Since the set of all ballots was partitioned into X, Y, and Z, we know that m = |X| +
|Y | + |Z|. It follows that
m
X
BBai (vj0 )
=
j=1
m
X
j=1
m
m X
|X| + |Y | + |Z|
=
.
BBai (vj ) +
BBai (vj ) +
2
2
j=1
Furthermore, since a? is the social choice of the original election, we know
m
X
BBa? (vj ) >
j=1
m
X
BBai (vj ).
j=1
Substituting the two equations shown above into this inequality gives us the following
m
X
m
BBa? (vj0 )
j=1
m
X
|X| + |Y | + |Z|
|Z| X
>
BBai (vj0 ) −
− |X| −
2
2
BBa? (vj0 ) −
j=1
m
X
BBa? (vj0 ) −
j=1
m
X
|X|
>
2
BBai (vj0 ) >
j=1
j=1
m
X
BBai (vj0 ) −
j=1
|Y |
2
|X| |Y |
−
.
2
2
Because the election satisfies the Condorcet criterion, we know that the number of ballots
in the original election in which a? ai must be greater than the number of ballots in
which ai a? . Hence |X| > |Y |, and it follows that
m
X
j=1
m
X
BBa? (vj0 )
−
BBa? (vj0 ) −
j=1
m
X
j=1
BBa? (vj0 ) >
m
X
j=1
m
X
j=1
m
X
|X| |Y |
−
> 0. Therefore, we have
2
2
BBai (vj0 ) >
|X| |Y |
−
>0
2
2
BBai (vj0 ) > 0
BBai (vj0 ).
j=1
Thus, a? remains the social choice of the cloned election despite the insertion of a0i into
the election. We conclude that given a Bucket Borda election which satisfies the Condorcet
criterion, the election will always be clone independent.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
88
From the proof above one might now assume that if a Bucket Borda election does not
satisfy Condorcet, then the election is clone positive. However, recall from the profile
in 4.2.3 that there are some Bucket Borda elections which don’t satisfy the Condorcet
criterion yet are still clone independent. Hence, the profile in a Bucket Borda election
must satisfy a stronger criteria in order to ensure that the election will be clone positive.
This property is shown in the following theorem.
Theorem 4.2.14. Let A = {a1 , a2 , . . . , an } be a set of n alternatives, B = {B(v1 ), B(v2 ), . . . , B(vm )}
be a profile of m ballots of A, and BB : B −
→ A? be a Bucket Borda election on B. Let
a? ∈ A be the social choice of the Bucket Borda election. Then for any ai ∈ A such that
ai 6= a? , the insertion of a0i into the election changes the social choice to ai if and only if
the difference between the number of ballots in which ai a? and the number of ballots in
which a? ai is greater than twice the difference between the total Borda Scores of a? and
ai .
Proof. Let A = {a1 , a2 , . . . , an } be a set of n alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )}
be a profile of m bucket ordered ballots of A. Let a? ∈ A be the winner of a Bucket Borda
m
m
X
X
Count election. Then
BBa? (vj ) >
BBa (vj ) for all other a ∈ A.
j=1
j=1
Let the clone a0i of ai be inserted into the election for some ai ∈ A such that ai 6=
a? . Partition B into three sets X, Y and Z such that X = {B(v)|a? ai in B(v)},
Y = {B(v)|ai a? in B(v)}, and Z = {B(v)|a? ≡ ai in B(v)}. Note that X, Y , and Z
constitute the entire set of ballots in the profile, and it follows that m = |X| + |Y | + |Z|.
From the proof of Theorem 4.2.13 we know that
m
X
BBa? (vj0 ) =
j=1
m
X
BBa? (vj ) + |X| +
j=1
|Z|
2
and
m
X
j=1
BBai (vj0 ) =
m
X
j=1
m
BBai (vj ) +
m X
|X| + |Y | + |Z|
=
BBai (vj ) +
.
2
2
j=1
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
89
Since a? is the social choice of the original election, we also know that
m
X
BBa? (vj ) >
j=1
m
X
BBai (vj ).
j=1
⇐ Assume the difference between the number of ballots in which ai a? and the
number of ballots in which a? ai is greater than twice the difference between the total
Pm
P
|Y | |X|
−
> m
Borda Scores of a? and ai . Then
j=1 BBai (vj ). Hence,
j=1 BBa? (vj ) −
2
2
we have
Pm
|Y | Pm
|X|
> j=1 BBa? (vj ) +
.
j=1 BBai (vj ) +
2
2
From the equations listed above it follows that
m
X
m
BBai (vj0 ) −
j=1
m
X
m |Y | X
|Z| |X|
+
>
+
BBa? (vj0 ) − |X| −
2
2
2
2
j=1
BBai (vj0 ) −
j=1
|X| |Z|
−
>
2
2
m
X
j=1
BBai (vj0 ) >
m
X
j=1
m
X
BBa? (vj0 ) −
|X| |Z|
−
2
2
BBa? (vj0 ).
j=1
Thus, the insertion of the clone into the election has increased the total Borda Score of ai
over a? , and it follows that ai is the social choice of the cloned election.
⇒ Assume the insertion of a0i into the election changes the social from a? to ai . Then
P
Pm
0
0
we know m
j=1 BBa? (vj ). It follows that
j=1 BBai (vj ) >
m
X
j=1
m
X
j=1
BBai (vj ) +
m
X
m
|Z|
BBai (vj ) + + >
BBa? (vj0 ) + |X| +
2
2
|X| |Y | |Z|
+
+
>
2
2
2
|Y | |X|
−
>
2
2
j=1
m
X
BBa? (vj ) + |X| +
j=1
m
X
BBa? (vj ) −
j=1
m
X
|Z|
2
BBai (vj ).
j=1
From the inequality above we know that the difference between the number of ballots
in which ai a? and the number of ballots in which a? ai is greater than twice the
difference between the total Borda Scores of a? and ai .
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
90
Combining our two cases, we conclude that the insertion of a0i into the election changes
the social choice to ai if and only if the difference between the number of ballots in which
ai a? and the number of ballots in which a? ai is greater than twice the difference
between the total Borda Scores of a? and ai .
4.3 Random Thoughts on Resolvability
Here we include some initial analysis on the resolvability of Linear and Bucket Borda
voting systems. I did not have time to finish the observations began in this section so it
consists of some preliminary results but is also in need of further exploration. Intuitively,
a voting system satisfies resolvability if given enough voters participating in an election
it is very unlikely for the election to finish in a complete tie. More formally, according
to Woodall a voting system is resolvable if given a set of alternatives, a profile, and an
election system, the proportion of profiles resulting in a complete tie approaches zero
as the number of voters in the election approaches infinity. Before we rigorously define
resolvability we will discuss what it means for a ballot and a profile to produce a complete
tie in an election. We begin by defining what it means for a Borda Count election to result
in a complete tie and then introduce the concept of the inverse of a ballot to show how
such a result can be obtained in a profile consisting of only two ballots.
Definition 4.3.1. Let A be a set of n alternatives, B be a profile of m voters, and BS : B
−
→ A? be a Borda Count election. The election results in a complete tie (which will also
be referred to at times as simply a tie) if for all ai in A,
m
X
BSai (vj ) = t,
j=1
where t is a constant representing the total Borda Score received by each alternative in
the election.
4
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
91
Definition 4.3.2. Let A be a set of n alternatives and B(v) a ballot of A. We define the
inverse of a ballot, B −1 (v), as the ballot which when entered into a profile with only
B(v) results in a tie among every alternative in A? .
Hence, for each ai ∈ A,
BBai (v) + BBai (v −1 ) = t
where t is a constant representing the total Borda Score allocated to each alternative from
4
the two ballots.
Lemma 4.3.3. The inverse of a linear ballot is unique.
Proof. Let A = {a1 , a2 , . . . , an } be a set of alternatives and B(v) be a linear ballot of
A. Without loss of generality we can assume the alternatives in A are ordered such that
P
n(n − 1)
a1 a2 . . . an . By Theorem 2.3.3 for any voter v, ni=1 BSai (v) =
. It
2
P
P
follows that ni=1 BSai (v) + ni=1 BSai (v −1 ) = n(n − 1). Hence, n(n − 1) is the sum of
the total Borda Scores allocated to the n alternatives in A by B(v) and B(v −1 ). Similarly,
we know from Definition 4.2.1 that each of the n alternatives in A must receive a Borda
Score of t from B(v) and B(v −1 ). Therefore, we have tn = n(n − 1), and it follows that
t = n − 1. Furthermore, from Definition 4.2.1 we know the alternatives of A are ordered
in B(v −1 ) such that for all ai ∈ A:
BSai (v) + BSai (v −1 ) = t.
Combining these two observations, we have
(n − 1) + BSa1 (v −1 ) = n − 1
(n − 2) + BSa2 (v −1 ) = n − 1
..
.
(n − (n + 1)) + BSa2 (v −1 ) = n − 1
(n − n) + BSan (v −1 ) = n − 1.
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
92
Solving each of these equations for BSai (v −1 ) we see that
BSa1 (v −1 ) = 0
BSa2 (v −1 ) = 1
..
.
BSan−1 (v −1 ) = n − 2
BSan (v −1 ) = n − 1.
Since B −1 (v) must allocate ballot Borda Scores to the alternatives in A as described
above, it follows that B −1 (v) must be the linear ordering such that an an−1 . . . a2 a1 . The Hasse diagrams of these two ballots are shown in Figure 4.3.1. Thus, we
conclude that given a linear ballot B(v) there exists a unique inverse B −1 (v).
Figure 4.3.1. Inverse of a Linear Ballot
We now introduce the idea of the inverse of a profile in an election. Intuitively, the
inverse of a given profile is a set of ballots which when put into a Borda Count election
with the original profile results in a complete tie. It is not clear to me whether a profile and
its inverse should necessarily consist of the same number of ballots, but in the definition
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
93
below I defined it as such. This could be an interesting idea to explore further in another
paper.
Definition 4.3.4. Let A be a set of n alternatives and B = {B(v1 ), B(v2 ), . . . , B(vm )} be
a set of m ballots. We define the inverse of a profile, B −1 as the set of ballots such that
|B| = |B −1 | and BS : B ∪ B −1 −
→ A? results in a tie among every alternative in A? .
4
While the inverse of a ballot is always unique, the same argument cannot be made about
the inverse of a profile. I am sure there exist some profiles that do indeed have unique
inverses, but the following theorem shows that this is not always the case.
Theorem 4.3.5. The inverse of a profile in a Borda Count election containing more than
one ballot is not necessarily unique.
Proof. In order to prove that a profile in a Borda Count election does not always have a
unique inverse it will suffice to show there exists a set of alternatives and profile of ballots
which have more than one inverse.
Let A be a set of alternatives such that A = {a, b, c} and B be a profile of voters such
that B = {B(v1 ), B(v2 ), B(v3 )}. Let the ballots in B be represented by the following Hasse
diagrams:
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
Then
P3
j=1 BSa (vj )
= 5,
P3
j=1 BSb (vj )
= 2 and
P3
j=1 BSc (vj )
94
= 2. From the definition
of the inverse of a linear ballot, we know that B −1 (v1 ), B −1 (v2 ), and B −1 (v3 ) can be
represented by the following Hasse diagrams:
Hence, we can define B −1 as the set B −1 = {B −1 (v1 ), B −1 (v2 ), B −1 (v3 )}. Note that BB :
B ∪ B −1 −
→ A? allocates Borda Scores to the alternatives in A? such that
3
X
BSa (vj ) +
j=1
3
X
3
X
j=1
BSb (vj ) +
3
X
j=1
j=1
3
X
3
X
j=1
BSa (vj−1 ) = 6
BSc (vj ) +
BSb (vj−1 ) = 6
BSc (vj−1 ) = 6.
j=1
Now, let B 0 = {B 0 (v1 ), B 0 (v2 ), B 0 (v3 )} be represented by the following Hasse diagrams:
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
95
Then BB : B ∪ B 0 −
→ A? allocates Borda Scores to the alternatives in A? such that
3
X
BSa (vj ) +
j=1
3
X
j=1
BSa (vj−1 ) = 6
j=1
BSb (vj ) +
j=1
3
X
3
X
3
X
BSb (vj−1 ) = 6
j=1
BSc (vj ) +
3
X
BSc (vj−1 ) = 6.
j=1
Therefore, we see that B 0 allocates Borda Scores to the alternatives in A equivalently to
B −1 , and thus is also a valid inverse of the profile B. We conclude that B −1 is not unique
for every B.
At this point there is a lot of further investigation that could be done to see if there
are certain kinds of profiles whose inverses are unique. Additionally, it is quite clear that
the inverse of a bucket ordered ballot in a Borda Count election is unique but I am not
sure if this also holds for a partially ordered ballot. However, I was not able to address
these questions in my project and they may prove to be useful topics for the launching
of another senior project or other research paper. Furthermore, while it seems intuitively
obvious that Linear and Bucket Borda are resolvable voting systems, I did not rigorously
prove either of these results. For the benefit of the reader, the definition of resolvability
is presented below along with a result concerning the resolvability of certain profiles in a
Borda Count election.
Definition 4.3.6. Let A be a set of alternatives, B be a profile, and E : B −
→ A? be a
voting system. Then E satsifies the resolvability criterion if inserting additional ballots
into B increases the likelihood that E : B −
→ A? does not result in a complete tie.
4
We conclude this section with an interesting result concerning a certain form of profile
which can never produce a Borda Count election resulting in a tie. This result could
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
96
potentially be used as part of a proof that Linear Borda is a resolvable voting system and
that was the motivation for providing it here.
Theorem 4.3.7. Let A be a set of n alternatives and B be a profile of m voters. If m is
odd and n is even then there are no possible ordering of the alternatives in B such that
the outcome of the election BS : B −
→ A? results in a complete tie. Hence, the insertion
of B into BS : B −
→ A? must result in a resolvable election (one that does not result in a
complete tie).
Proof. Let A be a set of n alternatives and B be a profile consisting of m voters such
that m is odd and n is even. The total amount of Borda Scores allocated among the
n alternatives in any Borda Count election can be determined by multiplying the total
Borda Score allocated to each ballot times the number of voters. By Theorem 2.3.3 we
know the total Borda Score allocated to the alternatives in any single ballot is
n(n − 1)
.
2
Thus, we can express the total amount of Borda Scores allocated in a given Borda Count
election as:
n(n − 1)
· m.
2
Note that in order to attain a complete tie we must be able to equally distribute the total
amount of Borda Scores allocated in the election among the n alternatives in the election.
Thus, we need the expression shown above to be divisible by n. Rearranging the terms in
this expression we see that the total amount of Borda Scores allocated in the election is
equivalent to
n·
m(n − 1)
2
In order for this expression to be divisible by n we simply need
m(n − 1)
to be a whole
2
number. Since n−1 and m are both odd, we know that m(n−1) is odd. Hence, m(n−1) = d
for some d = 2k + 1 where k ∈ N ∪ {0}. Therefore, we can express the total amount of
4. COMPARISON OF BUCKET BORDA TO LINEAR BORDA
97
Borda Scores as
n·
d
2k + 1
=n·
.
2
2
2k + 1
∈
/ N , and we can see from this that there does not exist an odd whole
2
d
m(n − 1)
number d such that is a whole number. Since d = m(n − 1) it follows that
is
2
2
Clearly,
not a whole number. Therefore, we have that
n·
cannot possibly be divisible by n since
is odd and n is even then
m(n − 1)
2
m(n − 1)
is never a whole number. Thus, if m
2
n(n − 1)
· m can never be divisible by n, and we conclude it
2
is impossible to obtain a complete tie in a Borda Count election consisting of an even
number of alternatives and an odd number of voters.
5
Graded/Poset Borda Count
5.1 Graded Posets
Recall from Chapter 1 that a partially ordered set is a set of elements, P , togther with an
ordered relation, , such that for all a, b, c ∈ P transitivity and asymmetry hold. Posets
are similar to linearly ordered and bucket ordered sets with the exception that they place
no restrictions on how elements must be ordered within the set. In this chapter we will
initially extend our voting analysis to a subset of posets referred to as graded posets, and
then we will apply our findings for graded posets to any form of poset. The set of all graded
posets is smaller than the set of posets but larger than the set of bucket ordered sets, so it
is a good place for us to continue our analysis of non-linear Borda Count election methods.
Before presenting the definition of a graded poset, we provide the definitions for two kinds
of chains seen in partially ordered sets which will be useful in forming an intuitive idea of
what a graded poset is.
Definition 5.1.1. Let P be a partially ordered set and C be a chain in P . Recall that a
chain, C, is a subset of P such that for all a, b ∈ C, a and b are comparable. Hence, for
5. GRADED/POSET BORDA COUNT
99
all a, b ∈ C, either a b or b a. A chain is called a saturated chain if there does not
exist c ∈ P − C such that a c b and C ∪ {c} is a chain.
4
Definition 5.1.2. A chain C in a poset P is a maximal chain if C is saturated, and
for its minimal element a? ∈ C there does not exist a ∈ P such that a? a, and for its
maximal element a? ∈ C there does not exist a ∈ P such that a a? .
4
Let A = {a, b, c, d, e, f } be a set of alternatives and P be a partially ordered set of A
expressed through the Hasse diagram in Figure 5.1.1. Note that C = {a, b, f } is a chain
since any two elements of C are comparable. However, C is not a saturated chain since
b d f and C 0 = C ∪ {d} = {a, b, d, f } is still a chain. Note that C 0 is a maximal
chain since it is saturated and since there are no alternatives in P that are preferred to
its maximal element a or less preferred than its minimal element f .
Figure 5.1.1. Saturated and Maximal Chains in a Partially Ordered Set
Intuitively, a graded poset is simply a poset in which every maximal chain has the same
height. This means that the poset shown in Figure 5.1.1 is also a graded poset since each
of it’s maximal chains ({a, b, d, e}, {a, b, d, f }, {a, c, d, e}, {a, c, d, f }) have height three. It
follows from this notion of graded posets that all bucket ordered sets are also graded posets
since every maximal chain in a bucket ordered set must have the same height. However,
5. GRADED/POSET BORDA COUNT
100
there are many ordered sets which are graded posets that are not bucket ordered sets. We
will now provide a rigorous definition of a graded poset followed by an example.
Definition 5.1.3. Let P be a partially ordered set ordered by the relation . Then for
all a, b, c ∈ P the relation has the following two properties:
1. If a b and b c, then a c. (Transitivity)
2. If a b then ¬(b a). (Asymmetry)
If every maximal chain in P has the same height, then we say that P is a graded poset
G.
4
Let A = {a, b, c, d, e, f } be a set of alternatives. Let G be a graded poset of A and P be
a partially ordered set of A expressed by the Hasse diagrams seen in Figure 5.1.2. Since
Figure 5.1.2. Comparison of Graded Poset and Poset
every maximal chain in G has height two, we know it must be a graded poset. Note that
P is not a graded poset, because C1 = {a, b} and C2 = {a, d, e} are both maximal chains
in P but their heights are not equal. Furthermore, note that G is not a bucket ordered set
because every alternative in the second level of the Hasse diagram is not preferred to every
5. GRADED/POSET BORDA COUNT
101
alternative in the third level. In particular we see that there is no relation between b and
f or d and e in G. Furthermore, these elements are not equivalent since b is clearly more
preferred overall than f and d is more preferred overall than e. Hence, for the first time we
are introduced to the idea of two alternatives in a ballot being completely incomparable.
In linear ordered sets this was never an issue because every alternative is comparable to
each other alternative in the set, and in bucket ordered sets we didn’t have to deal with
incomparable alternatives because every alternative is either comparable or equivalent to
each other alternative in the set. In order to develop a formula for allocating Borda Scores
to the alternatives in a graded poset ballot (which will be referred to as simply graded
ballots for the rest of the chapter) we must deduce a way to deal with alternatives in a
ballot which are incomparable. In the following section we will focus on addressing this
problem.
5.2 Graded/Poset Borda Score
Our goal in this section is to deduce a formula for allocating Borda scores to alternatives in
a graded ballot. We would like the formula we develop to be compatible with the fomulas
we’ve already discussed for Linear Borda and Bucket Borda voting systems. This means
we want our Graded Borda function to allocate scores to linear and bucket ordered ballots
in the same manner that Linear and Bucket Borda do. Since linear and bucket ordered
sets are both subsets of graded posets, it is sensible that the formula we develop should
allocate accurate scores to ballots of this manner. Hence, we want our formula to take the
ideas we used to develop Bucket Borda from Linear Borda and extend them to graded
ballots.
In linear ballots, we have seen that Borda Count simply gives an alternative one point
for each alternative it is preferred to. In bucket ordered ballots, we have seen that Bucket
Borda gives an alternative one point for each alternative it is preferred to and half of a
5. GRADED/POSET BORDA COUNT
102
point for each alternative it is equivalent to. Extending these ideas to graded ballots, it
seems natural that each alternative in a graded ballot should be given one point for each
alternative it is preferred to, and half of a point for each alternative it is either equivalent
or incomparable to.
Definition 5.2.1. Let A be a set of alternatives and B(v) be a graded poset ballot of A.
The Graded Borda Score allocated to each alternative ai in B(v) is defined as:
GBai (v) = (# of alternatives preferred by ai ) +
1
(# of alternatives incomparable to ai )
2
4
Note here that the definition we presented for allocating Graded Borda scores to the
alternatives in a given ballot makes no mention of the ’graded’ aspect of the ballot. The
same is true for the rest of the theorems and proofs mentioned in this section. Hence,
while we set out to provide a Borda method first for graded posets and then finally for
any kind of poset, the graded formula we will work with throughout this chapter actually
applies equally well to posets of any form. Thus, every result that follows in this section
referring to Graded Borda can also be extended to Poset Borda without having to change
any of the calculations.
Now that we have defined a formula for allocating Graded Borda scores to the alternatives in a ballot, we can introduce the definition of a Graded Borda voting system.
Definition 5.2.2. Let A be a set of alternatives and B be a profile of A. The Graded
Borda election method, GB : B −
→ A? , is defined as the voting system which allocates
points to the alternatives in each ballot of B according to the Graded Borda Score equation
in Definition 5.2.1.
4
The social choice of a Graded Borda election is determined in the same manner as the
winner of a Linear or Bucket Borda election, so we find it unnecessary to redefine that
5. GRADED/POSET BORDA COUNT
103
term once again. Just recall that the winner of any Borda Count election is the alternative
which receives the greatest total Borda score among all the alternatives in the election.
In Figure 5.2.1 we have the Hasse diagrams of three seperate ballots. Note that B(v1 )
is a linear ballot, B(v2 ) is a bucket ballot and B(v3 ) is a graded ballot. Since linear and
bucket ordered sets are both subsets of graded posets, we know that we can include these
ballot types in our Graded Borda elections. In B(v1 ) Linear Borda would allocate one
point to each alternative for every alternative it is preferred to in the ballot. From our
definition of Graded Borda Score, it is clear that in a Graded Borda election the exact
same score would be allocated to each of these alternatives since they would each receive a
Borda score equal to the number of alternatives they are preferred to in the ballot (these
alternatives wouldn’t receive any points for being incomparable to other alternatives in
the ballot since in a liner ballot the alternatives are all comparable to one another).
Figure 5.2.1. Graded Borda Score Computations
In B(v2 ) we see that Bucket Borda would allocate ballot Borda scores to the alternatives in the following manner: BBa (v2 ) = 5, BBb (v2 ) = BBc (v2 ) = BBd (v2 ) = 3,
and BBe (v2 ) = BBf (v3 ) = .5. Since alternative a is preferred to every alternative
in the ballot and incomparable to none, its Graded Borda score can be expressed as
5. GRADED/POSET BORDA COUNT
104
1
GBa (v2 ) = 5 + (0) = 5. Since alternatives b, c, and d are each preferred to two alterna2
tives and incomparable to two alternatives, their Graded Borda scores can be expressed
1
as GBb (v2 ) = GBc (v2 ) = GBd (v2 ) = 2 + (2) = 2 + 1 + 3. Finally, alternatives e and f
2
are each preferred to zero alternatives and incomparable to one, so their Graded Borda
1
scores can be expressed as GBe (v2 ) = GBf (v2 ) = 0 + (1) = .5. Hence, we see that for
2
bucket ordered ballots Bucket Borda and Graded Borda allocate the same Borda scores
to the alternatives in the ballot.
In B(v3 ) we have the same graded ballot we saw earlier in Figure 5.1.2. In bucket
ballots, alternatives on the same level always receive the same Borda score because by
definition they are equivalent in the ballot. However, as we see in B(v3 ), this is not the
case in graded posets, and we want our Graded Borda formula to be able to allocate
different Borda scores to alternatives on the same level if they are differently preferred in
the ballot. For example alternatives b, c, and d are all located at height one in the ballot
but they are not all equally preferred. Alternatives b and d are preferred to one alternative
in the ballot and incomparable to three while alternative c is preferred to two alternatives
and incomparable to two. Hence, we want our Graded Borda election to allocate a higher
Borda score to alternative c than alternatives b and d since it has an overall greater level of
preference in the ballot. Computing the Graded Borda scores of these alternatives we see
1
1
that GBb (v3 ) = GBd (v3 ) = 1 + (3) = 2.5 while GBc (v3 ) = 2 + (2) = 3. Furthermore,
2
2
compare alternatives e and f in B(v2 ) to alternatives e and f in B(v3 ). Note that these
alternatives are slightly more preferred in B(v3 ) than B(v2 ) since in both ballots they
are preferred to no other alternatives but in B(v3 ) they are each incomparable to two
alternatives while in B(v2 ) they are incomparable to only one alternative. Calculating
1
their Graded Borda scores from B(v3 ) we see that GBe (v3 ) = GBf (v3 ) = 0 + (2) = 1.
2
Hence GBe (v3 ) = GBf (v3 ) = 1 > GBe (v2 ) = GBf (v2 ) = .5 which makes intuitive sense
5. GRADED/POSET BORDA COUNT
105
when we compare the two ballots. From computing the Graded Borda scores for these
three ballots we see that Graded Borda accurately allocates Borda scores to linear and
bucket ballots while also providing more flexibility in order to allow our elections to consist
of graded ballots as well.
Now that we understand how Graded Borda allocates points to the alternatives in a
ballot, we will prove some important general results about Graded Borda elections.
Theorem 5.2.3. Graded Borda satisfies the three Axioms of Borda Count from Definition
2.3.6.
Proof. Let A be a set of alternatives and B(v) be a graded ballot of the alternatives in
A. Let a, b ∈ A such that a b. Assume b is preferred to m alternatives in B(v). Since
a b we know that a is preferred to b and every alternative that b is preferred to. Hence,
a is preferred to a minimum of m + 1 alternatives. Now, assume b is incomparable to k
alternatives in B(v). Since a b any alternatives that are incomparable to b are either
preferred by a or incomparable to a. Then a is either incomparable or preferred to k other
k
points from these k alternatives
2
k
k
in the ballot. By definition of Graded Borda, we have Gb (v) = m + < (m + 1) + =
2
2
alternatives in B(v). Hence, a is allocated a minimum of
min(Ga (v)), and it follows that Ga (v) > Gb (v). Thus, the Graded Borda Score allocated
to alternative a is always greater than the Graded Borda Score allocated to alternative b,
and we see that Axiom 1 holds for ballots in a Graded Borda election.
From the definition of Graded Borda, it is clear that every alternative in the ballot is
either allocated a score of 0 or some value greater than zero. Thus, we have that Axiom 2
holds for ballots in a Graded Borda election.
Let a, b ∈ A such that a ≡ b. Let m be the number of alternatives in B(v) that a is
preferred to and k be the number of alternatives in B(v) that a is incomparable to. Since
a and b are equivalent, we know b is also preferred to m alternatives and incomparable
5. GRADED/POSET BORDA COUNT
to k alternatives. Hence, Ga (v) = m +
106
k
= Gb (v). Thus, we see that Axiom 3 holds for
2
ballots in a Graded Borda election. Therefore we conclude that Graded Borda satisfies the
three Axioms of Borda Count and is a valid Borda Count voting system.
In addition to satisfying the three Axioms of Borda Count, we would also like for
the Graded Borda voting system to satsify as many of the desirable properties of Borda
Count defined in Section 4.1 as possible. In particular, we would like for Graded Borda to
satisfy the two desirable properties of Borda Count which were also satisfied by Bucket
Borda. This means we want to show that every ballot in a Graded Borda election with n
alternatives allocates the same number of ballot Borda Scores among the n alternatives in
the ballot, and we want the maximum ballot Borda Score allocated to a single alternative
in a ballot of a Graded Borda election with n alternatives to be the same for any possible
ballot containing n alternatives. We will begin with the proof of the latter since it is
somewhat easier to comprehend and helps us to gain a better understanding of how Graded
Borda elections work before we begin the more difficult proof.
Theorem 5.2.4. The ballot Borda Score allocated to the single most preferred alternative
in any ballot of a Graded Borda election consisting of n alternatives is the same.
Proof. Let A be a set of n alternatives and B be a profile of graded ballots in a Graded
Borda election. Let B(v) be a ballot in the profile, and a? ∈ B(v) such that a? is the single
most preferred alternative in the ballot. Then a? a for all other a ∈ B(v). Since there
are n alternatives in the ballot, it follows that a? is preferred to (n−1) alternatives. By the
definition of Graded Borda, a? is allocated one point for each of the (n − 1) alternatives
it is preferred to in B(v) regardless of how the alternatives are ordered below a? . Thus,
Ga? (v) = n − 1. Since the alternatives preferred by a? can be arranged in any manner
5. GRADED/POSET BORDA COUNT
107
without effecting Ga? (v), we conclude that (n−1) is the Borda Score allocated to the single
most preferred alternative in any ballot of a Graded Borda election with n alternatives.
Note here that (n−1) is also the ballot Borda Score allocated to the single most preferred
alternative in any ballot of a Linear or Bucket Borda election. This is significant because
it provides further evidence that our Graded Borda function is perfectly compatible with
linear and bucket ballots while also allowing us to extend the domain of our elections to
include graded ballots.
We have already seen that the ballots in Linear and Bucket Borda elections consisting
of n alternatives allocate
n(n − 1)
points among the alternatives in the ballot. In the
2
following proof, we show that that any ballot in a Graded Borda election consisting of n
alternatives also allocates
n(n − 1)
points among the alternatives in the ballot.
2
Theorem 5.2.5. Let A = {a1 , a2 , . . . , an } be a set of n alternatives and B(v) be a partially
n
X
n(n − 1)
.
ordered ballot. Then
GBai (v) =
2
i=1
Proof. Let B(v) be a graded ballot consisting on n alternatives. We will utilize proof
by induction to prove this result. Let n = 2. Then the possible graded ballots can be
expressed by the following two Hasse diagrams:
Clearly, for each of these ballots
P2
i=1 GB(ai )(v)
=1=
2(2 − 1)
.
2
5. GRADED/POSET BORDA COUNT
108
Now, assume our theorem holds for graded ballots consisting of n alternatives. Then
P
n(n − 1)
given any graded ballot, B(v), we know that ni=1 GBai (v) =
. Insert an alter2
native x randomly into B(v) and denote this new ballot as B(v 0 ). For each ai ∈ B(v 0 ), x is
either preferred to ai , preferred by ai , or incomparable to ai . Note that if x is preferred to
ai then inserting x into the ballot contributes one point to GBx (v 0 ), and if x is preferred by
ai then inserting x into the ballot increases GBai (v 0 ) by one. Hence, if x is comparable to
ai then inserting x into B(v) increases the total Borda Score allocated to the alternatives
in B(v 0 ) by one. If x is incomparable to ai then inserting x into the ballot contributes
of a point to GBx (v 0 ) and also increases GBai (v 0 ) by
1
2
1
. Hence, if x is incomparable to
2
ai then inserting x into B(v) increases the total Borda Score allocated to the alternatives
in B(v 0 ) by one. Let c be the number of alternatives in B(v 0 ) which are comparable to x
and let i be the number of alternatives in B(v 0 ) which are incomparable to x. Note that
c + i = n since every alternative in B(v 0 ) is either comparable or incomparable to x but
not both. Thus, we have
n+1
X
0
GBai (v ) =
i=1
=
n
X
i=1
n
X
GBai (v) + c + i
GBai (v) + n
i=1
=
=
=
=
=
Hence,
Pn+1
i=1
GBai (v 0 ) =
n(n − 1)
+n
2
n2 − n 2n
+
2
2
n2 + n
2
n(n + 1)
2
(n + 1)[(n + 1) − 1]
2
(n + 1)[(n + 1) − 1]
. Thus, assuming that our result holds for
2
any graded ballot with n alternatives, we have shown that it then holds for any graded
ballot with n+1 alternatives. It follows from induction that the theorem holds and we con-
5. GRADED/POSET BORDA COUNT
clude that
Pn
i=1 GBai (v)
=
109
n(n − 1)
for any ballot in a Graded Borda election consisting
2
of n alternatives.
Since the previous two proofs were both conducted without any reference to the graded
aspect of our graded posets, we can see that these results apply not only to graded posets
but to any type of possible poset. Hence, we can extend our Graded Borda election method
to a Poset Borda election method by simply keeping the same formula for allocating ballot
Borda Scores seen in Definition 5.2.1 but extending the domain of our function to allow
voters to use any kind of partially ordered ballot. By Theorems 5.2.3, 5.2.4 and 5.2.5,
we see that this Poset Borda election method retains the two mathematical properties
satisfied by Bucket Borda and Graded Borda which we hoped for any valid and desirable
extension of Linear Borda to satisfy.
6
Open Questions
We have successfully developed an extension of Linear Borda which allows voters to use
partially ordered ballots while still allocating ballot Borda Scores in a manner that is perfectly compatible with linear, bucket and graded poset ballots. We have also shown that
the Poset Borda (Graded Borda) election method from Section 5.2 satisfies the two desirable mathematical properties any non-linear Borda Count voting sytem can and should
satisfy (we would, of course, like for Poset Borda to satisfy all three of the desirable properties of Linear Borda defined in Section 4.1 but as was shown in the impossibility proof
of the same section this is not possible for non-linear Borda Count election methods).
However there are still a fair number of questions I have on the topic, and I present them
here for any who may be interested in continuing certain aspects of the research begun in
this project. Most of these are not questions I had when I began the project but rather
questions that developed as I worked through different examples and proved certain results. Some of these are questions which, given more time, I feel I could have answered
myself and others are problems that I spent time working on and was unable to sufficiently
solve.
6. OPEN QUESTIONS
111
• The most obvious extension of this project is to do a similar analysis of Poset Borda
as was provided for Bucket Borda in Chapter 4. While it seems likely that changing
the domain of our Borda Count function from all possible bucket ordered sets to
all possible partially ordered sets would have little change on the results provided
in Chapter 4, I think it would still be interesting to see how each of these voting
system criteria apply to an election utilizing partially ordered ballots. Since posets
allow such a wide variety of orderings for the alternatives in a ballot it is possible
that the introduction of these ballots may cause some criteria which were satisfied
by Bucket Borda to now be unsatisfied. I do not think these proofs would be very
difficult to compute, and given more time I would have provided them in this paper.
• The discussion of inverse profiles and resolvability in Section 4.3 could certainly be
explored further. In particular, I would like to know if there are certain profiles for
which the inverse of the profile is unique, and if so, I would like to try and figure
out what aspect of these profiles causes them to have a unique inverse. Now that we
have developed a formula for allocating ballot Borda Scores to the alternatives in
a partially ordered ballot, I would also like to prove or disprove that the inverse of
a partially ordered ballot is unique. I am fairly sure that a partially ordered ballot
will not always have a unique inverse, and if this is true, then I would like to try and
characterize when a partially ordered ballot will have a unique inverse and when it
will have more than one inverse. Finally, although it seems quite obvious, I would
like to have come up with a rigorous proof that shows Borda Count is a resolvable
voting system.
• Some other open questions initially raised by Professor Cullinan refer to the ability of
a given Borda Count election to satisfy a certain property given a restriction on the
number of voters or number of alternatives in the election. For example, in Theorem
6. OPEN QUESTIONS
112
4.2.13 we prove that if a Bucket Borda election satisfies the Condorcet criterion,
then the election is clone independent. In Figure 4.2.4 we see that given a profile
consisting of 6 alternatives and 5 voters, it is possible for a Bucket Borda election
to fail the Condorcet criterion. It would be interesting to now figure out and prove
what is the smallest profile in which it is possible for a Bucket Borda election to
fail the Condorcet criterion? From the examples I’ve computed it seems to me that
in elections with few voters and few alternatives the Condorcet winner (if it exists)
must always be the Bucket Borda social choice. Hence, it would be interesting to
see how large we must make our profiles in order for the Bucket Borda social choice
to differ from the Condorcet winner. Another example of an unanswered question
of this type is what is the smallest profile and number of alternatives in an election
such that Bucket Borda fails to satisfy the plurality criterion? In Theorem 4.1.11 we
use a counterexample to show that a Bucket Borda election consisting of 19 voters
and 3 alternatives can fail the plurality criterion. Is this the smallest profile in which
Bucket Borda can fail plurality and if so, how do we prove that this is so? These are
interesting questions which Professor Cullinan introduced rather late in the project
and I did not have time to thoroughly explore.
• Finally, I would have liked to include more analysis and discussion of Arrow’s impossibility theorem which was only briefly mentioned in Section 2.2. Since we introduce our own impossibility theorem in Section 4.1, I would’ve liked to discuss
in more depth why voting systems cannot simultaneously satisfy the three criteria
mentioned in Arrow’s proof, and then explore Borda Count or other voting systems
for further examples of voting properties or criterion that cannot all hold in a single
voting system.
References
[1] M. Ackerman et al. Elections with Partially Ordered Preferences. 2007.
[2] R. Fagin et al. Comparing and Aggregating Rankings with Ties. 2004.
[3] F. Roberts and B. Tesman. Applied Combinatorics. 2nd Ed. Pearson Prentice
Hall, New Jersey: 2005.
[4] R. Stahl. Flag h-vectors of Distributive Lattices. 2008.
[5] A. Taylor and A. Pacelli. Mathematics and Politics: Strategy, Voting, Power,
and Proof. 2nd ed. Springer-Verlag, New York: 2008.
[6] N. Tideman. Independence of Clones as a Criterion for Voting Rules. Social
Choice and Welfare, 4 (1987), 185-206.
[7] D. Woodall. Properties of Preferential Election Rules. Voting Matters, 3 (1994),
8-15.