A new axiomatic approach to diversity

0.18 MB·

by  Chris Dowden

A new axiomatic approach to diversity
Chris Dowden
LIX, E´cole Polytechnique, 91128 Palaiseau Cedex, France
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a Abstract
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The topic of diversity is an interesting subject, both as a purely mathematical
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concept and also for its applications to important real-life situations. Unfor-
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tunately, although the meaning of diversity seems intuitively clear, no precise

] mathematical definition exists. In this paper, we adopt an axiomatic approach
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to the problem, and attempt to produce a satisfactory measure.
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Keywords: weighted graphs, diversity, axioms, metric spaces, biomathematics
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2000 MSC: 05C90, 92B05,91B14
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[ 1. Introduction

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Over the last twenty years, an important problem in conservation biology
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has been howbest to measurethe ‘diversity’of a setof species. This is because
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0 diversityhasemergedasaleadingcriterionwhenprioritisingspeciestobesaved
3 from extinction. The topic also has applications in a wide number of other
5 fields, such as linguistics and economics, but in this paper we examine it as a
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1 mathematical concept.
0 There are two distinct challenges. The first is how to accurately evaluate
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the diversity of any two elements (e.g. species), and the second is how to then
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use these pairwise-diversities, or ‘distances’, to produce scores for sets of size
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v greater than two. It is the latter, more mathematically interesting problem,
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X that we address here.
Biologists and economists have produced numerous papers (see [1]–[13] and
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a the references therein) investigating various different measures. Some of these
are very simple ‘rule of thumb’ methods (e.g. minimum distance [3], maximum
distance [2], average distance [12], total distance [2]), while others are more
elaborate (e.g. phylogenetic diversity [4], which we shall shortly discuss, or p-
median [5]). However, each of these is known to be imperfect, in that they
sometimes rank sets in a counter-intuitive order.
The most popular method seems to be phylogenetic diversity ([4]). Given
the tree-like structure of evolutionary relationships, phylogenetic diversity was
Email address: [email protected] (ChrisDowden)
Preprint submitted toElsevier January 28, 2011
developed for the specialised case when the pairwise-diversities induce a tree-
metric, with the score of a set of organisms being defined to be the length of
the minimal subtree connecting them. For example, given the tree shown in
Figure 1, the sets {u,v,w} and {u,v,x} score 14 and 22, respectively, and so
(cid:0)@
5(cid:0) 2
@
(cid:0)
@
4(cid:0)(cid:0)@3@ w(cid:0)r0(cid:0)@7@
(cid:0)r @r @
u v (cid:0) @
1(cid:0) 1@
(cid:0)r @r
x y
Figure1: Anexampleofphylogenetic diversity.
the latter would be considered as the more diverse.
One problem with phylogenetic diversity is that, in practice, the pairwise-
diversitieswilloftennotinduceatree-metric. Moreover,evenwithatree-metric,
sets can still sometimes get ranked in an undesirable order. For example, in
Figure 1 the set {u,x,y} wouldscore more than the set {u,w,y} (20 compared
to19),eventhoughwisverydifferentfrombothuandy,whilexisverysimilar
to y. Indeed, adding w to the set {u,y} does not increase the phylogenetic
diversity score at all, which seems strongly counter-intuitive!
It is the object of this paper to introduce a new axiomatic approach to di-
versity,in an attempt to produce a measure that never disagrees with intuition
(furthermore,weshallonlyassumethatthepairwise-diversitiessatisfytheprop-
erties of a metric, and not necessarily a tree-metric). Such an idea has already
been discussed in [9] and [2], but the axioms suggested were not completely
satisfactory, and hence resulted in some undesirable measures being accepted
(for example, it is shown that the unique way to satisfy the axioms of [2] is to
rank sets according to their maximum distance).
To avoid confusion, we should mention in passing that there are a number
of papers (see, for example, [13] or [7]) in which the term ‘diversity’ is used to
meanthetotalnumberoffeaturescontainedbyaset(e.g.thenumberofdifferent
books possessedby a set of libraries). Also, there is an unrelated section of the
literature (see, for example, [6] or [1]) which uses the term to refer to a type of
entropy. Inmost situations,these definitions arenot consistentwith the notion
of diversity explored in this paper (although there have been some attempts to
unify the different approaches,see e.g. [8] or [10]).
Finally, it is worth remarking that, even without the biological motivation,
the question under discussion in this paper seems very natural — given two
collections of points in a metric space, which is the more spread out? It seems
surprising that the topic has never previously been investigated by mathemati-
cians.
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The remainder of the paper is divided into four main sections. In the first,
Section 2, we propose four basic axioms that any sensible diversity measure
shouldsatisfy. In Section3, we then presentsome measuresthatfulfill allthese
requirements (the first such measures ever to be produced). In Section 4, we
discussonefurtherpropertythataperfectdiversitymeasureshouldbeexpected
tosatisfy,andfinally,inSection5,wepresentanewmeasurethatseemstohave
all these properties.
2. Axioms
Throughout the rest of this paper, we shall assume that we are given a
completeweightedgraph,wheretheedge-weightsdenotethepairwise-diversities
ofthe vertices(this is slightlydifferentfromthe tree-likestructureofFigure1).
Wewillhenceforthrefertothesepairwise-diversitiesas‘distances’,andweshall
assume that they satisfy the properties of a metric. Our aim is to construct a
way to use these distances to give a score for the overalldiversity of any subset
of our collection of vertices. To that end, we will spend this section proposing
four axioms that any satisfactory diversity measure, D, should satisfy.
We start with three properties that are hoped to be intuitively natural:
Axiom 1. For any non-empty set of vertices S, we have D(S ∪{x}) ≥ D(S)
for all x, with equality if and only if x∈S.
Axiom 2. For any two vertex-sets S ={s ,s ,...,s } and T ={t ,t ,...,t }
1 2 n 1 2 n
satisfying D({t ,t })≥D({s ,s }) for all i and j, we have D(T)≥D(S), with
i j i j
equality if and only if D({t ,t })=D({s ,s }) for all i and j.
i j i j
Axiom 3. Continuity. Given any set of vertices S = {s ,s ,...,s } and
1 2 n
any ǫ > 0, there exists a δ(S,ǫ) > 0 such that, for any set of vertices T =
{t ,t ,...,t } satisfying |D({t ,t })−D({s ,s })|<δ for all i and j, we have
1 2 n i j i j
|D(T)−D(S)|<ǫ.
Itisworthobservingthattwootherdesirablepropertiesfollowautomatically
from these axioms. First, note that Axiom 1 implies D({x})= D({x,x}), and
hence:
Corollary 1. D({x})=0 for all x.
Secondly, it follows from applying Axiom 3 to the sets S = {s ,s ,...,s ,s }
1 2 n n
and T = {s ,s ,...,s ,x} (and using the triangle inequality) that we have
1 2 n
continuity when adding a new vertex:
Corollary 2. Given a set of vertices S = {s ,s ,...,s } and an ǫ > 0, there
1 2 n
exists a δ(S,ǫ) > 0 such that, for any vertex x satisfying D({x,s }) < δ, we
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have D(S∪{x})<D(S)+ǫ.
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Ourfourthaxiomismotivatedbythe principlethatconsistentresultsought
tobe obtainedregardlessofdifferencesinthe scaleusedtomeasurethe original
distances. For example, if we wish to compare the diversity of the locations
of stars in two different galaxies, then the resultant ranking should not depend
on whether the distances were measured in light-years or kilometres. In other
words,multiplyingallouroriginaldistancesbysomeconstantcshouldnotaffect
whether or not D(S)>D(T) for any sets S and T:
Axiom 4. Scaling. Given four sets of vertices S = {s ,s ,...,s }, S′ =
1 2 n
{s′,s′,...,s′ }, T = {t ,t ,...,t } and T′ = {t′,t′,...,t′}, if D({s′,s′}) =
1 2 n 1 2 k 1 2 k i j
cD({s ,s }) for all i and j and D({t′,t′}) = cD({t ,t }) for all i and j, for
i j i j i j
some constant c>0, then D(S′)>D(T′) if and only if D(S)>D(T).
By considering the case when |T|=2, this implies the following:
Corollary 3. Given two sets S = {s ,s ,...,s } and S′ = {s′,s′,...,s′ },
1 2 n 1 2 n
if D({s′,s′}) = cD({s ,s }) for all i and j, for some constant c > 0, then
i j i j
D(S′)=cD(S).
There is also an ‘equidistance’ axiom that we will discuss (extensively) in
Section 4.
Note that it is implicit in our whole approachthat the diversityscore ought
to be a function only of the distances. While this is certainly sensible if the
vertices represent points in Euclidean space, since the set of distances uniquely
defines the set of points (up to translations, rotations and reflections), it is
perhaps less clear for some other scenarios.
For example, consider the points in {0,1}4 shown in Table 1 and suppose
Points Co-ordinates
u 1 0 0 0
v 0 1 0 0
w 0 0 1 0
x 0 0 0 1
y 1 1 1 0
Table1: Aselectionofpointsonafour-dimensionalcube.
thatwewishtochooseonenewvertex,eitherxory,toaddtotheset{u,v,w}.
Sincexandybothhave(Hamming)distanceexactly2fromalltheotherpoints,
it is automatic with our approach that they would be considered as equally
good, i.e. the diversity of the set {u,v,w,x} would equal the diversity of the
set {u,v,w,y}. However, since these two sets have fundamental differences
(one is four-dimensional and one is three-dimensional), it could be argued that
forcingtheirdiversityscorestobeequalis,whilehardlyridiculous,ratherheavy-
handed. Nevertheless, we shall turn a blind eye to such objections, as it is
difficult to know how to proceed otherwise.
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3. New diversity measures
Intheprevioussection,weproposedfouraxiomsthatanydiversitymeasure
should satisfy. Although the problem seems fairly natural, it is surprisingly
difficulttoconstructameasurethatfulfillsalltheserequirements(indeed,every
method suggested in the existing literature fails either Axiom 1 or Axiom 2).
However, in this section we shall now present a simple system for obtaining
measuresthatdosatisfyallfouraxioms. Thisisnottheendofthestory,though,
and in Section 4 we shall argue that these measures are still not satisfactory.
Given a real-valued function f, let us define the measure D recursively
f
(from our given distances) by using the equation
D (S)=f(S)+ max {D (T)} (1)
f f
T⊂S,|T|=|S|−1
for all vertex-sets S of size greater than two. Let us call the function f suit-
able if: (a) f is a continuous function of the distances; (b) if any of the
distances are 0, then f = 0; (c) if none of the distances are 0, then f is
strictly positive and is a monotonically strictly increasing function of the dis-
tances; and (d) f is ‘scale-invariant’ in the sense of Axiom 4. For exam-
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ple, we could choose f({s1,s2,...,sn}) to be 1≤i<j≤nD({si,sj}) (n2) or
−1 (cid:16) (cid:17)
1 , or any linear combinatiQon of these. We will now see
1≤i<j≤n D({si,sj})
(cid:16)thPat Df satisfies the a(cid:17)xioms if f is suitable:
Theorem 4. The diversity measure D defined in equation (1) satisfies Ax-
f
ioms 1–4 if the function f is suitable.
Proof It is immediately clear by induction that D satisfies Axioms 2–4, so it
f
only remains to show that Axiom 1 is satisfied. To do this, we need to prove
that (i) adding a vertex already in the set does not alter the score, and (ii)
adding a vertex not already in the set strictly increases the score.
We shall proceed by induction. Suppose that (i) and (ii) both hold when
addinga vertexto anysetofsizeless thank (note thatthe basestepis adirect
consequence of the fact that the distances satisfy the properties of a metric),
and let us now consider the case when we are adding a vertex s to a set
k+1
S ={s ,s ,...,s } of size exactly k.
1 2 k
First, let us work towards (i) by supposing (without loss of generality)
that s = s . By part (b) of the definition of suitability, we then have
k+1 k
f S∪{s } =0andsoD S∪{s } =max D S∪{s } \s .
k+1 f k+1 i≤k+1 f k+1 i
H(cid:0)ence, it suffi(cid:1) ces to prove t(cid:0)hat D S(cid:1)∪{s } \sn is(cid:16)m(cid:0)aximised by(cid:1)taki(cid:17)nog
f k+1 i
i ∈ {k,k + 1}. But note that, fo(cid:16)r(cid:0)i < k, the(cid:1)indu(cid:17)ction hypothesis implies
D S∪{s } \s =D (S\s )≤D (S), and so we are done.
f k+1 i f i f
(cid:16)N(cid:0)ow let us w(cid:1)ork t(cid:17)owards (ii) by supposing that sk+1 ∈/ S. If the vertices of
S are all distinct, then f(S ∪{s }) > 0 and the result is clear. If not, then
k+1
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let S′ denote a maximally sized subset of S with vertices that are all distinct.
By the induction hypothesis, we have D (S′ ∪{s }) > D (S′). But note
f k+1 f
that, by a combinationofthe induction hypothesisand(i), the left-handside is
D (S∪{s }) and the right-hand side is D (S).
f k+1 f
One particular choice for a suitable function would be to take
(n)
f({s ,s ,...,s })= 2 , (2)
1 2 n 1
1≤i<j≤n D({si,sj})
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as this has the aesthetically pleasing property that it will always be equal to
1 for the ‘regular’ case when the distances are all 1 (and hence D (S) will be
f
equalto|S|−1forthiscase). However,thisisapersonalchoiceandiscertainly
not an axiom!
4. The equidistance axiom
In the previous section, we saw a scheme for generating diversity measures
that satisfy Axioms 1–4. However, as briefly mentioned earlier, there is also a
fifth axiom that is necessary — that of equidistance. In this section, we shall
explainwhysuchanaxiomisdesirableanddiscusshowtodefineit. Wewillstart
with the case when we just have graphs of order three, which we shall see only
requiresa smallmodification to ourpreviousmeasures;then we willinvestigate
a seemingly natural way to extend the concept to graphs of arbitrary order,
whichwe shallsee actually produces an intriguingcontradiction;and finally we
will propose a more careful definition. In Section 5, we shall then describe a
pretty measure that appears to always give nice results.
Let us imagine that we have two vertices x and y that are distance 1 apart,
and that we wish to add one more vertex to this set. Suppose that we are free
to choose any element from {z : D({x,z})+D({y,z})=2}. It seems intuitive
that the overall diversity score ought to be greater the more equidistant the
new vertex is between x and y. Unfortunately, this is actually not true for
the diversity measure defined at the end of the last section, where we use the
suitablefunctionofequation(2)inrecursion(1),sinceweknowthattheregular
unit triangle scores 2, whereas the triangle with lengths 1, 1 and 3 scores
2 2
3 +max 1,1,3 = 9 + 3 > 2. This example establishes the need for a
1+2+2 2 2 11 2
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new axiom:
(cid:8) (cid:9)
Axiom 5. Three-vertex equidistance. Given two sets S = {s ,s ,s } and
1 2 3
T = {t ,t ,t }, if D(t ,t ) = D(s ,s ) and D(t ,t )+D(t ,t ) = D(s ,s )+
1 2 3 1 2 1 2 1 3 2 3 1 3
D(s ,s ) = λ, but D(t ,t )− λ < D(s ,s )− λ and D(t ,t )− λ <
2 3 1 3 2 1 3 2 2 3 2
D(s ,s )− λ , then D(T)>D(S).
2 3 2 (cid:12) (cid:12) (cid:12) (cid:12) (cid:0) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:1)
(cid:12) One way to(cid:12)approach the problem of trying to satisfy Axiom 5 would be to
find a suitable function f for which the partial derivative with respect to the
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maximum distance (when the total distance is fixed) is always less than −1,
thus offsetting the contribution to D of the max {D(T)} term.
f T⊂S:|T|=2
However,aneatersolutionistoinsteaddevelopaseparatediversitymeasure
for sets of size three that does satisfy Axiom 5 (and also Axioms 1–4) and
then simply use the recursion of equation (1) on this (it can be checked that
Theorem 4 will still hold). For example, we could set
1
D({s ,s ,s })=g({s ,s ,s })+ D({s ,s }),
1 2 3 1 2 3 i j
2
1≤i<j≤3
X
where g is any suitable function that also satisfies Axiom 5. It is simple to see
that D({s ,s ,s }) satisfies Axioms 2–5, and Axiom 1 (for the case when we
1 2 3
are comparing sets of size three with sets of size two) follows from using the
triangle inequality on the second term. An aesthetically pleasing choice, if we
then use recursion (1) with the suitable function defined in equation (2), is to
−1
take g({s ,s ,s })= 3 1 , so that
1 2 3 2 1≤i<j≤3 D({si,sj})
(cid:16) (cid:17)
P
−1
3 1 1
D({s ,s ,s })= + D({s ,s }). (3)
1 2 3 i j
2 D({s ,s }) 2
i j
1≤i<j≤3 1≤i<j≤3
X X
 
Ofcourse,wereallyneedsomething moregeneralthanjustthe three-vertex
rule of Axiom 5. For example, consider the two four-vertex sets, S and S′,
depicted in Figure 2. It seems intuitive that S′ should be considered as more
sr4 sr′4
(cid:20)T (cid:20)T
(cid:20) (cid:20)
T T
(cid:20) (cid:20)
1 T 1 T
S (cid:20) 3 S′ (cid:20)
4 T 4 1 T1
(cid:20)3(cid:20) (cid:17)(cid:17)rQQs3 T3T (cid:20)(cid:20) (cid:17)(cid:17)rQQs3 TT
(cid:20) (cid:17) Q (cid:20) (cid:17) Q
(cid:20) (cid:17)1 1Q T (cid:20) (cid:17)1 1Q T
(cid:17) QT (cid:17) QT
r(cid:20)(cid:17) QTr r(cid:20)(cid:17) QTr
s1 1 s2 s1 1 s2
Figure2: Twofour-vertexsets,S andS′.
diverse than S, as the position of s′ is equidistant in relation to s , s and s .
4 1 2 3
Unfortunately, the diversity measure that we have just defined gives D(S′)=3
and D(S) = 6 +D({s ,s ,s }) = 4 + 3 1 + 1 4 + 4 +1 =
3+3+3+1+1+1 1 2 4 5 2 3+3+1 2 3 3
4 4 4 4
97 >3.
30 (cid:0) (cid:1)
One naturalwayto extend Axiom 5 to any number ofvertices is the follow-
ing:
‘Axiom’ 5′ Strong equidistance. Given two sets S = {s ,s ,...,s } and
1 2 n
T ={t ,t ,...,t }, if D(t ,t )=D(s ,s ) for all i,j <n and D(t ,t )=
1 2 n i j i j i<n i n
P
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(∗)
D(s ,s ) = λ, but D(t ,t )− λ ≤ D(s ,s )− λ for all i < n,
i<n i n i n n−1 i n n−1
then D(T) ≥ D(S), with(cid:12)equality if and o(cid:12)nly i(cid:12)f there is equalit(cid:12)y in (*) for all
P (cid:12) (cid:12) (cid:12) (cid:12)
i<n. (cid:12) (cid:12) (cid:12) (cid:12)
Unfortunately, as we shall now see, it turns out that this strong version
actually leads to inconsistencies with our earlier axioms!:
Theorem 5. ‘Axiom’ 5′ is inconsistent with Axioms 1–3.
Proof Letthevertex-setsS ={s ,s ,s },T ={t ,t ,t }andU ={u ,u ,u }
1 2 3 1 2 3 n 1 2 3
be as shown in Figure 3. By Axiom 2, we have D(S) > D(T) and so, by
s t u
r3 3r r3
S (cid:20)T T (cid:20)T U (cid:20)T
(cid:20) T (cid:20) T n (cid:20) T
n+2 n+2
2(cid:20) T1 1(cid:20) T1 n+1(cid:20) T n+1
(cid:20) T (cid:20) T (cid:20) T
r(cid:20) Tr r(cid:20) Tr r(cid:20) Tr
s1 1 s2 t1 1 t2 u1 1 u2
Figure3: ThesetsS,T andUn intheproofofTheorem5.
continuity (Axiom 3), there exists a k such that D(S)>D(U ).
k
Now consider the set U′ = {u ,u ,...,u } ⊃ U constructed from U
k 1 2 k+2 k k
by setting D({u ,u }) = 0 for all l ≥ 4 (i.e. adding in k −1 extra copies of
2 l
u )and, similarly, the setS′ ={s ,s ,...,s } constructedfromS by setting
2 1 2 k+2
D({s ,s })=0 for all l ≥4 (i.e. adding in k−1 extra copies of s ).
2 l 2
If we assume the strong equidistance of ‘Axiom’ 5′, then D(U′) > D(S′).
k
But, by Axiom 1, D(U′) = D(U ) and D(S′) = D(S). Hence, we find that
k k
D(U )>D(S), and so we have a contradiction.
k
Note that (with a bit of care to ensure that the triangle inequality is not
violated during the proof) a form of this example still produces a contradic-
tion even if we alter the strong equidistance ‘axiom’ to include the condition
D(t ,t ) = D(s ,s ) for all i ≥ 3 as well as D(t ,t ) = D(s ,s ).
i n i n i<n i n i<n i n
This seems very surprising!
P P
However,letusnowrecallouroriginalfour-vertexexampleofFigure2. The
intuitionthatit wasdesirablefor the fourthvertexto be equidistantinrelation
to s , s and s perhaps stems from the fact that these other three vertices
1 2 3
were all in symmetric positions to begin with. Hence, it seems sensible that a
satisfactoryequidistanceaxiomshouldhavetotakeintoaccountsuchsymmetry
considerations(note that this was not necessaryfor the three-vertexcase,since
every set of size two is automatically symmetric).
Another fault with ‘Axiom’ 5′ is that, by only consideringthe case whenwe
wishtoaddanewvertex,itperhapsdoesnotcoveralldesirablesituations. For
example, consider the sets S and T shown in Figure 4. It seems intuitive that
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sr4 tr4
(cid:20)T (cid:20)T
(cid:20) (cid:20)
T T
(cid:20) (cid:20)
2 T 1 T
S (cid:20) T (cid:20)
1 T1 1 T1
(cid:20) (cid:20)
(cid:20) (cid:17)(cid:17)rQQs3 TT (cid:20) (cid:17)(cid:17)rQQt3 TT
(cid:20) (cid:17) Q (cid:20) (cid:17) Q
(cid:20) (cid:17)1 1Q T (cid:20) (cid:17)1 1Q T
(cid:17) QT (cid:17) QT
r(cid:20)(cid:17) QTr r(cid:20)(cid:17) QTr
s1 0 s2 t1 1 t2
Figure4: TwosetsS andT.
T should be thought of as the more diverse,since the edges t t and t t are in
1 2 3 4
symmetric positions, but this would not have been coveredby ‘Axiom’ 5′.
Hence, itseemsdesirableto formulateexactlywhatwemeanby‘symmetry’
before we attempt to propose another equidistance axiom. With this in mind,
we now give two definitions:
Definition 6. Let G be called a partial graph if it can be formed from an un-
labelled edge-weighted complete graph by deleting some of the edge-weights and,
instead, giving (distinct) labels to the associated edges.
Forexample,the graphsG andG showninFigure5 aretwoofthe partial
1 2
graphsthatcanbeformedfrom(theunlabelledversionof)thegraphrepresent-
ing the set S in Figure 4.
r r
(cid:20)T (cid:20)T
(cid:20) (cid:20)
T T
(cid:20) (cid:20)
G e T G e T
(cid:20) 2 (cid:20) 2
1 1 T1 2 e Te
(cid:20) 4(cid:20) 6
(cid:20) (cid:17)(cid:17)rQQ TT (cid:20) (cid:17)(cid:17)QrQ TT
(cid:20) (cid:17) Q (cid:20) (cid:17) Q
(cid:20)(cid:17)(cid:17)e1 1QQTT (cid:20)(cid:17)(cid:17)e1 e3QQTT
r(cid:20)(cid:17) QTr r(cid:20)(cid:17) QTr
0 e
5
Figure5: PartialgraphsG1 andG2.
Definition 7. GivenapartialgraphGwhoselabellededgesaree ,e ,...,e ,let
1 2 k
ussaythate is symmetrictoe inGifthereexistsapermutatione ,e ,...,e
1 l i1 i2 ik
ofthelabels for which i =l andrelabellinge as e for allj does notchangeG.
1 j ij
For example, it can be seen that e is symmetric to e in G (indeed, all
1 2 2
edges are symmetric here) by considering the permutation shown in Figure 6.
In G , however, e and e are not symmetric.
1 1 2
It can be checked that, for each partial graph, symmetry defines an equiva-
lence relation on the labelled edges.
9
r
(cid:20)T
(cid:20)
T
(cid:20)
G e T
(cid:20) 4
2 e Te
1(cid:20) 5
(cid:20) (cid:17)(cid:17)rQQ TT
(cid:20) (cid:17) Q
(cid:20)(cid:17)(cid:17)e2 e6QQTT
r(cid:20)(cid:17) QTr
e
3
Figure6: Apermutationoftheedges ofG2.
Nowthatwehavedefinedsymmetry,wewillreturntothe issueofformulat-
ing a successful equidistance axiom. Our new version comes in two parts, the
first of which we present immediately:
Axiom 5′′a Symmetric equidistance, part one. Let G be an unlabelled edge-
S
weighted complete graph representing the set S, and let G be a partial graph
p
that can be formed from G . Let us denote the equivalence classes of the la-
S
belled edges in G by E = {e ,e ,...,e }, E = {e ,e ,...,e },...,
p 1 11 12 1k1 2 21 22 2k2
andE ={e ,e ,...,e }. Now let G bethegraph formed from G bysetting
l l1 l2 lkl T p
w (e ) = 1 w (e) for all i and j (i.e. averaging out the weights of
GT ij ki e∈Ei GS
all theedges in each equivalence class) and let T bethe set representedbyG (it
T
P
can be checked that G will satisfy the triangle inequality). Then D(T)≥D(S),
T
with equality if and only if T =S.
For example, Axiom 5′′a could be applied to the graphs G , G and G
S p T
shown in Figure 7.
q q q
(cid:20)T (cid:20)T (cid:20)T
G (cid:20) G (cid:20) G (cid:20)
T T T
S 2(cid:20)(cid:20)(cid:20) 1q TT1 p e(cid:20)1(cid:20)(cid:20)e2q TTe3 T 34(cid:20)(cid:20)(cid:20) 34q TT43
(cid:17)Q T (cid:17)Q T (cid:17)Q T
(cid:20) (cid:17) Q (cid:20) (cid:17) Q (cid:20) (cid:17) Q
(cid:17) Q T (cid:17) Q T (cid:17) Q T
(cid:20)(cid:17) 1 1 QT (cid:20)(cid:17) 1 1 QT (cid:20)(cid:17) 1 1 QT
q(cid:20)(cid:17) QTq q(cid:20)(cid:17) QTq q(cid:20)(cid:17) QTq
1 1 1
Figure7: AnexampleofAxiom5′′a.
ItisadeliberatedecisiontoonlystateAxiom5′′aforthecasewhentheedges
ineachequivalenceclassarecompletelyevenlyweightedinG ,ratherthanjust
T
more evenly weighted than in G , in order to take care that the axiom is not
S
strongerthanourintuition. Forexample,considerthesetsS,T andT′ depicted
in Figure 8. It certainly seems reasonable to say that T should be considered
more diverse than S (and this can be deduced by applying Axiom 5′′a to the
relevant four edges), but, in the light of Theorem 5, it is perhaps going too far
to claim that T′ should also definitely be consideredmore diverse than S (and,
10