Rational agents are the quickest
Michael Mandlery
Royal Holloway College, University of London
This version: January 2011
Abstract
We consider agents who choose by proceeding through a checklist of criteria (for
any pair of alternatives the …rst criterion that ranks the pair determines the agent’s
choice).
Regardless of the discriminating capacity of the criteria, choices that
maximize complete and transitive preferences can always be the outcome of a ‘quick’
checklist that uses the minimum number of criteria. For any irrational preference
on the other hand there is always a discriminatory capacity for criteria such that
the preference is not the outcome of a quick checklist.
Due to their cognitive
limitations, agents are likely to use criteria that discriminate coarsely, but coarse
criteria turn out to be the e¢ cient way to generate preference rankings.
JEL codes: D01
Keywords: checklists, rationality, decision-making e¢ ciency.
I am grateful to Jose Apesteguia, Simon Blackburn, Michel De Lara, Daniel Ellsberg, Mordecai Kurz, Mark Machina, Paola Manzini, Marco Mariotti, Klaus Ritzberger, Herbert Scarf, Phil
Steitz, and particularly Claire Blackman for helpful discussions and advice. An extended abstract
of an earlier version of this paper appears in the Proceedings of TARK XII and is available at
http://personal.rhul.ac.uk/uhte/035/Mandler.TARK.2009.pdf
y
Address: Department of Economics, Royal Holloway College, University of London, Egham, Surrey
TW20 0EX, United Kingdom. Email: [email protected]
1
Introduction
Consider an agent who makes decisions by proceeding through a sequence of criteria. For
any pair of alternatives, the agent checks whether the …rst criterion C1 recommends one
over the other; if not, the agent proceeds to the second criterion C2 ; and so on.
For
example, the …eld of alternatives might be a set of possible meals, C1 might categorize
meals by their meat content (vegetarian, red or white meat, …sh, all other cases), C2 by
their cuisine (French, Italian, Chinese, all other countries), C3 by ‡avor (spicy, sweet,
or neither).
Each criterion will order some of the categories it de…nes but a criterion
need not be complete or transitive, e.g., it might cycle. We call a sequence of criteria a
checklist.
Checklists bring several advantages. If an agent has to form preferences by making
a separate judgment for each pair of alternatives a lot of decisions would be required.
Criteria in contrast make many rankings simultaneously. Checklists are also easy to manipulate: keeping an entire preference relation in mind is demanding, while remembering
a handful of criterion rankings is not.
Drawing on the psychological literature on the
limitations of working memory, we argue that agents will be able to handle only those
criteria that employ a few categories, as in our opening example; a criterion with even
…ve categories would require an agent to keep an unwieldy number of binary comparisons
(ten) in mind. But unlike the psychological literature we will stress the e¢ ciency advantage of criteria that employ only a few categories: the smaller the number of categories,
the more e¢ cient the agent’s decision-making will turn out to be.
Our main topic will be to compare the decision-making speed of checklists that lead to
rational and irrational preferences. A checklist C always implicitly maximizes a preference
relation
that we call the outcome of C. Conversely, given any preference
under mild restrictions be a checklist that leads to
.
, there will
Which preferences can be the
outcome of the quick checklists that employ the theoretical minimum number of criteria?
To ensure a fair contest, we compare preferences with the same number of indi¤erence
classes and let criteria share the same capacity to discriminate among alternatives. We
show that if
is complete and transitive –the preference is rational –there is always a
1
quick checklist that leads to
. In contrast, for any irrational preference there is a dis-
criminatory capacity for criteria such that the preference cannot be the outcome of a quick
checklist. Thus the set of agents who use e¢ cient decision-making procedures always includes the rational agents but, as we vary the discriminatory capacity, each irrational
agent will drop out of the set at some point. Even a preference whose only irrationality is
incompleteness, which might seem to require fewer decision-making discriminations, will
sometimes fail to be the outcome of a quick checklist.
The quick checklists that lead to rational preferences arise in natural ways: when each
criterion stems from a one-dimensional or linear scale then the preferences that result
will be rational, and if each criterion is ‘balanced’–its categories discriminate with equal
e¢ ciency across sets that have not yet been ranked by earlier criteria –then the checklist
will be quick.
The symmetry concept that underlies quick sequential decision-making is as important
as the speci…c results.
An agent’s …rst criterion discriminates among alternatives and
thus partitions the domain of alternatives into a set of categories or equivalence classes.
The equivalence classes of the agent’s second criterion then partition each of the …rst
criterion’s equivalence classes into smaller subsets, and so on.
For decision-making to
be quick, each criterion must be balanced and it must partition each set of not-yetdiscriminated alternatives as …nely as its discriminatory capacity permits.
A criterion
can accomplish this only if it orders every set of not-yet-discriminated alternatives in the
‘same way’ – formally, these sets must be order-isomorphic.
Criteria built from linear
scales provide the characteristic example of criteria that enjoy this feature.
There are irrational preferences that happen to have the required order-isomorphic
subsets and so there will be quick checklists that lead to these preferences.
But the
examples are unusual. The proportion of irrational preferences that are the outcome of
quick checklists shrinks to 0 at a super-exponential rate as the size of the domain of the
alternatives increases.
And the examples are fragile: there is no irrational preference
that is the outcome of a quick checklist regardless of the discriminatory capacities of the
decision-making criteria.
Rational preferences in contrast can always be the outcome of a quick checklist for
2
the simple reason that rational preferences are endowed with the richest possible supply
of subsets that are ‘ordered the same way’: for a rational preference, any two sets of
indi¤erence classes of the same size share the same ordering, which is why it is so easy for
a rational agent to employ balanced criteria. No irrational preference has this ‡exibility.
This study pursues an agenda that Mandler, Manzini, and Mariotti (2010) (henceforth
MMM) lays out but could not fully address.
MMM considered checklists with binary
criteria that partition alternatives into just two equivalence classes. Any preference that
is the outcome of a binary-criteria checklist must be rational and conversely any rational
preference will be the outcome of some quick checklist that uses binary criteria.
But
because binary criteria can never lead to irrational preferences, MMM could not stage a
race between rational and irrational agents.
Notice that each criterion in our opening
‘meals’example divides meals into at least three categories, not two; that example can
therefore admit irrational preferences.
Our results contrast with Herbert Simon’s (1990) view that utility-maximization places
‘a heavy (often unbearable) computational burden on the decision maker’. If we measure
computational burden by the number of checklist criteria, we come to the conclusion that
it is the irrational agents who bear the heavier burden. The di¢ culty with Simon’s and
kindred psychological views is that they take utility to exist independently of an agent’s
choices rather than being a mere representation of choice behavior; rational decisionmaking is di¢ cult from this perspective because an agent’s true utility can be hard to
discover. But if instead we take an agent’s checklist as primitive then rational preferences
can simply be the outcome of criteria that are themselves rational. Indeed the most crude
but also the most procedurally plausible criteria are the binary ones that rank some subset
of the alternatives over the remainder, and binary criteria necessarily generate rational
preferences.
Binary criteria in fact play a special role in the theory of sequential decision-making.
As mentioned at the outset, if we judge a checklist by the number of decisions an agent
has to make to build the checklist then the criteria that employ fewer categories always
more e¢ cient. Binary criteria are therefore the most e¢ cient of all. Since binary criteria
must lead to rational preferences, rational preferences are vindicated by this measure of
3
decision-making burden as well. The e¢ ciency of binary distinctions may help explain
their prevalence in everyday decision-making.
Choice behavior that arises from a set of criteria has been the subject of much recent
research.
See Kalai, Rubinstein, and Spiegler (2002) (KRS), Apesteguia and Ballester
(2009, 2010) (AB), and Manzini and Mariotti (2007). Besides MMM (2010), KRS and
AB (2010) are closest to the present paper in that they assess a set of criteria by the
number of criteria in the set (what we call ‘length’).
But the number of equivalence
classes in criteria is irrelevant for the goals of the KRS and AB papers, whereas for our
aim of making e¢ ciency comparisons it is imperative to give every agent access to criteria
of the same discriminatory power. Also, agents in our model proceed through an ordered
sequence of criteria where it is the …rst criterion that ranks x and y that determines
the ultimate ranking of x and y, as in Manzini and Mariotti (2007), AB (2009), MMM
(2010) and the much earlier lexicographic utility theory of Chipman (1960, 1971) and
Fishburn (1974).
Chipman’s framework bears close resemblance to ours but, since his
criteria are allowed to make uncountably many discriminations, the framework will often
fail to make tight e¢ ciency comparisons. Outside of the sequential-criteria setting of this
paper, others have found that rationality has appealing e¢ ciency properties.
Tversky
and Simonson (1993) adopt a view somewhere between Simon’s and ours and acknowledge
(grudgingly) the dexterity of rational decision-making. Salant (2009) provides computerscience grounds for the e¢ ciency of rational choice.
2
Criteria and Checklists
When an agent is not endowed with preference judgments and has to produce them, it
would be a daunting task to make a separate judgment for every pair of alternatives.
Even remembering such a long list of judgments would be di¢ cult. To produce or store
preference judgments, an agent can instead proceed through a sequence –a checklist –of
criteria.
Fix a domain of alternatives X.
An agent with a checklist goes through through
a sequence of criteria C1 ; :::; CT , where each Ci is an asymmetric binary relation on X.
4
If there is a criterion in the sequence that ranks a pair in X then the …rst criterion
to do so determines the agent’s choice from that pair.
preference
which is also an asymmetric binary relation
These binary choices de…ne a
on X. Criteria will typically
be coarser than preferences: they divide X into a smaller number of equivalence classes
than preferences do.
For a variation on the example in the introduction, X could be
the set of all movie downloads and an agent might make choices by …rst examining a
criterion that orders movies by language categories (movies in French, movies in English,
all others), then by an ordering of the genre categories (thriller, comedy, science …ction,
all others), and so on. Criteria need not be transitive and a criterion Ci need not rank
every pair of the categories or equivalence classes that Ci de…nes.
De…nition 1 A checklist is a …nite sequence of criteria C = (C1 ; :::; CT ) where each Ci
is an asymmetric relation on X. The preference
x
y , 9 i with 1
i
is the outcome of C if and only if
T such that x Ci y and not y Cj x for all j < i.
The integer T , which bounds the time the checklist can take to rank a pair in X, is called
the length of C.
We will also say that the checklist C leads to the preference
if
is the outcome of
C. De…nition 1 is a variant of both Chipman’s (1960) lexicographic utility representation
and a Manzini-Mariotti (2007) sequential choice procedure.1 An agent with a checklist C
need not think about the preferences he or she is implicitly maximizing though of course
we can use De…nition 1 to deduce
from C. As we will see, we can also think of a choice
function as the outcome of a checklist.
How many categories are criteria likely to have? The psychological and informationprocessing literatures suggest that the number will be small. For most decisions, an agent
using a criterion Ci will have to hold in consciousness the binary comparisons of the categories that Ci de…nes. Miller (1956) famously concluded that the number of ‘chunks’that
an agent can hold in mind is roughly seven. But since Miller the number has been steadily
whittled down. Simon (1974) argued that …ve is more accurate. A binary comparison
1
There are also related concepts in set theory, e.g., Cuesta Dutari (1943, 1947).
5
quali…es as an object-…le in the model of Kahneman et al. (1992), and Treisman (2006)
judges that subjects can hold only three or four object-…les in memory. An encyclopedic
overview of the evidence, Cowan (2000), concludes that the ‘magic number’that bounds
working memory is four.2 Now if di is the number of categories de…ned by a criterion Ci
then the number of binary comparisons of categories is given by the binomial coe¢ cient
di
2
=
di (di 1)
.
2
So a bound of four on the number of binary comparisons would give a
bound of three on di . On the other hand, for complete and transitive criteria, an agent
would presumably need to keep only di itself in mind; if the agent remembers the ranking
of each category over the next best category then he or she would presumably know the
remaining rankings implicitly. In summary, the psychological literature suggests that a
criterion that needs to be manipulated in working memory could have at most three or
four categories. Consider ‘meals’: if an agent wants to order meals on a menu using a
‘cuisine’criterion that divides meals into 5 categories then he or she would have to keep
10 category comparisons in mind, which indeed seems unwieldy.
As we will see in section 6, the cognitive constraints that limit the number of categories
are not simply an unfortunate fact of mental life.
The constraints have an e¢ ciency
advantage: as criteria become more coarse, the number of decisions needed to make a
given number of preference discriminations will fall. Coarse criteria also ensure that an
agent makes trade-o¤s: if an agent’s …rst criterion is …ne then it would by itself determine
the agent’s preference between most pairs of alternatives but if the criterion is coarse then
the criteria that come later in the checklist will have a chance to be decisive.
We will use the equivalence classes of criteria to formalize the concept of category.
The same de…nition of equivalence classes will serve for preferences as well.
De…nition 2 Given an asymmetric relation > on X, the binary relation
on X is
de…ned by
x
y , fz 2 X : z > xg = fz 2 X : z > yg and fz 2 X : x > zg = fz 2 X : y > zg.
We call
2
the equivalence (or indi¤ erence) relation of >.
See Luck and Vogel (1997) for a characteristic example of the research surveyed.
6
It is easy to con…rm that
is in fact an equivalence relation and thus furnishes an
appropriate de…nition of indi¤erence for preferences.
specialized to the case where
For a textbook discussion of
,
is transitive, see Fishburn (1970).
De…nition 3 Given an asymmetric relation > and its equivalence relation
equivalence class is a nonempty I
X such that, for all x 2 I, x
, a >-
y , y 2 I.
Given an asymmetric > on X, the >-equivalence classes form a partition of X. When we
apply De…nitions 2 and 3 to a preference
we use
and call it an indi¤erence relation.3
of
to denote the equivalence relation
To keep our terminology clean, we reserve
equivalence for equivalence relations of criteria. We assume throughout the paper and
has …nitely many indi¤erence classes.4
without further remark that any preference
The number of indi¤erence classes in a preference
does not say how ‘di¢ cult’
to construct, it just measures the number of subsets of X that
is
treats in a behaviorally
distinguishable way: each indi¤erence class identi…es a distinct set of preferred items
or a distinct set of dispreferred items (or both).
We discuss alternative measures of
discrimination in section 3.
To see how the count of criterion equivalence classes works, consider again the ‘meals’
example that opened the paper. Two of the criteria partitioned meals into four categories
and one criterion partitioned meals into three categories. If, for example, the categories of
the ‘cuisine’criterion C2 (French, Italian, Chinese, and a remainder of all other countries)
are to form equivalence classes then for any two meals in the same category, say two
3
The reader may wonder why we use a strict preference
to de…ne indi¤erence classes rather than
using a weak preference % and letting the symmetric part of % be our de…nition of indi¤erence. The
di¢ culty with the latter tactic is that a % might generate two or more indi¤erence classes that are
unranked by % but that have identical upper and lower contour sets; that is, we might have x and y such
that neither x % y nor y % x holds but where
fz 2 X : z > xg = fz 2 X : z > yg and fz 2 X : x > zg = fz 2 X : y > zg
where > denotes the strict part of %. In such a case, x and y are fully interchangeable from a behavioral
point of view. In particular, any checklist that leads to % would have to place x and y in the same
>-indi¤erence class. So if we nevertheless labeled x and y as being in separate indi¤erence classes, we
would give a checklist that leads to % unwarranted credit for tackling more indi¤erence classes than show
up in the agent’s behavior. Conversely, % might classify two elements as indi¤erent when their upper
and lower contour sets do di¤er and hence would have to be distinguished by some criterion.
4
Our framework extends to preferences with an arbitrary number of indi¤erence classes and to checklists that consist of an in…nite sequence of criteria but both Observation 2 and Theorem 3 would no longer
obtain.
7
French meals x and y, x C2 y must not obtain and C2 must rank x and y the same way
relative to any non-French meal z. The criterion C2 will then have no more than four
equivalence classes; it may have fewer if two categories have identical upper and lower
contour sets but this will not occur when C2 actually ranks every pair of categories.
Checklists can generate preference rankings e¢ ciently for two reasons. First, a single
ranking of a pair of criterion categories rank many pairs of alternatives.
Second, the
criteria in a checklist can ‘cut across’one other and therefore rank diverse sets of alternatives; speci…cally, a criterion that comes later in a checklist can discriminate within the
categories demarcated by earlier criteria, e.g., C2 can discriminate among those alternatives that C1 fails to rank. With meals, for example, if C1 is a ‘meat content’criterion
and C2 is a ‘cuisine’ criterion then C2 can generate preference judgements within each
C1 -equivalence class, assuming that each meat category has dishes from more than one
cuisine.
If a criterion Cj partitions each set of alternatives not yet discriminated by Ci , i < j,
into the same number of equivalence classes then we call the criterion balanced.5
we will see, for a rational preference
checklist that leads to
As
there are always balanced criteria that form a
, which is the reason why rational preferences can always be the
outcome of a quick checklist. Agents who want to decide quickly will seek out balanced
criteria since they discriminate e¢ ciently, performing the same number of discriminations
on each set of as-yet-unranked alternatives. An unbalanced criterion in contrast wastes
an opportunity to make discriminations on some sets of alternatives.
A rich source of balanced criteria are one-dimensional scales, where some attribute of
the alternatives is scaled by a distinct real number: for each x 2 X the attribute speci…es
a number s(x) and, when x 6= y, s(x) 6= s(y). If X were a domain of cars, examples of such
scales might be price, horsepower, or even qualitative properties, like prestige, that can
be mapped into a one-dimensional scale. Suppose an agent uses one-dimensional scales
to build criteria; for each scale, the agent’s criterion either ranks more of the attribute
5
Formally, the criteria in a checklist (C1 ; :::; CT ) are balanced if for each Cj , j < T , and each maximal
A
X such that for all i < j there is a Ci -equivalence class Ei with A
Ei , we have A \ Ej 6= ?
for each Cj -equivalence class Ej . We exempt CT from this requirement due to the fact that to make
a given number of preference discriminations not all of the …nal A sets need to be partitioned into the
same number of equivalence classes.
8
as better than less or vice versa (as with price). As we will see in Theorem 2, the agent
can then build balanced criteria; and balanced criteria, since they discriminate e¢ ciently,
produce quick checklists.
Choice functions as the outcome of checklists.
We have presented checklists as
a way to choose from two-element sets: given fx; yg, the agent proceeds through the
checklist until he comes to the …rst criterion Ci that ranks x and y and then eliminates
the Ci -inferior item and if no criterion ranks x and y then the agent deems both to
be acceptable.
But checklists can also be applied to larger choice sets: if we feed an
arbitrary choice set A to the checklist C = (C1 ; :::; CT ) we then eliminate in each round
i any element of A that has survived so far and that Ci ranks as inferior to some other
survivor. We thus de…ne ‘survivor sets’Si (A) recursively by
S0 (A) = A
Si (A) = fx 2 Si 1 (A) : @ y 2 Si 1 (A) with y Ci xg for i = 1; :::; T ,
as in Manzini and Mariotti (2007), AB (2009), MMM (2010).
Suppose c is a choice
function on a domain D of nonempty subsets of X that includes the two-element sets; for
every A 2 D, c(A) is a nonempty subset of A. We can then de…ne a choice function c
to ‘be the outcome of the checklist C’if ST (A) = c(A) for all A 2 D. This setting is in
fact the principal interpretation we have in mind for checklists, but the extra generality
of A’s with more than two elements would be pointless for our purposes. As we will see,
a rational preference
can be the outcome of a minimum-length checklist; and it will be
clear that a choice function de…ned on arbitrary sets that maximizes
can be the outcome
of exactly the same checklist. Thus rational choice functions can also be the outcome of
minimum-length checklists. We will point out in section 5 that the converse that holds
for preferences also holds for choice functions: only the choice functions that maximize
a rational preference can be the outcome of a minimum-length checklist regardless of the
discriminatory capacity of the criteria.
9
3
The number of discriminations
A fair test of decision-making speed must compare checklists whose criteria employ the
same discriminatory capacity, which we will measure by the number of criterion equivalence classes.
De…nition 4 A discriminatory capacity is a sequence of integers d = (d1 ; d2 ; :::) where
each di
2. The checklist (C1 ; :::; CT ) is a d-checklist if and only if each Ci has di or
fewer Ci -equivalence classes.
We exclude di = 1 since then Ci can make no discriminations (if di = 1 then Ci = ?).
We use special labels for the following checklists: a (2; 2; :::)-checklist is a binary checklist
!
and a (3; 3; :::)-checklist is a 3 -checklist.
There are other measures of the number of discriminations that we could use instead,
both for preferences and criteria. For example, when there is no strict ranking between
a pair of indi¤erence or equivalence classes then one might want to declare that fewer
discriminations are present. From this point of view, the best measure of the number of
discriminations in a preference or criterion would be the number of pairs of indi¤erence or
equivalence classes that are strictly ranked. Both Theorems 1 and 3 below, appropriately
rephrased, would continue to hold; in particular if we are given a bound on the number
of pairs of equivalence classes that each criterion is permitted to rank then any rational
preference will be the outcome of a checklist with the theoretical minimum number of
criteria.
Since even unranked equivalence classes are behaviorally distinguishable, we
will take the simpler path of using the number of indi¤erence or equivalence classes as
our measure of discrimination.
4
Checklists for rational and irrational preferences
To de…ne rationality, we use a version of completeness that applies to asymmetric relations.
De…nition 5 An asymmetric relation > on X with equivalence relation
if and only if for every x; y 2 X either x (> [
10
) y or y (> [
) x or both.
is complete
Equivalently, > is complete if and only if for all >-equivalence classes I 6= J, either I > J
or J > I.6
We de…ne an asymmetric relation
transitive.
to be rational if and only if
It is easy to con…rm that an asymmetric
and only if it is negatively transitive (where x
y
is complete and
is complete and transitive if
z implies x
z), the traditional
de…nition of rationality for asymmetric relations.
One way that a rational
can arise is when each criterion Ci in a d-checklist C =
(C1 ; :::; CT ) is itself rational.
Observation 1 If each criterion Ci in a d-checklist C is rational and C leads to
then
,
is rational.
The proof (in the appendix along with all proofs omitted from the text) in fact shows
that the conclusion above still holds when each Ci is only required to be rational on the
items that have not yet been discriminated by any preceding criterion.
Contra Herbert Simon, it need not be di¢ cult for an agent to be rational; rational
preferences can merely indicate that an agent uses rational criteria. Indeed, the crudest
criteria are the binary ones where the discriminatory capacity d is (2; 2; :::). Since binary
criteria are necessarily both complete and transitive, the preferences that are the outcome
of binary checklists are always rational; binary checklists may look primitive, but they
lead to choice behavior that Simon took to be demanding.
Since only rational preferences can be the outcome of binary checklists, a checklist
that leads to an irrational preference must have a discriminatory capacity d with some
coordinates di > 2. The following observation notes that for an arbitrary
go to a d where di
we need only
3 occurs su¢ ciently often.
Observation 2 Any preference
with n indi¤erence classes can be the outcome of some
d-checklist if at least n coordinates of d have di
3.
If there is no restriction on the number of di ’s at least as big as 3, a simple way to
build a d-checklist that leads to
is just to list the preference rankings in
: for any two
6
For any binary relation R on X, any subsets A; B X, and any c 2 X, we use A R B to mean a R b
for all a 2 A and b 2 B, c R A to mean c R a for all a 2 A, and A R c to mean a R c for all a 2 A.
11
-indi¤erence classes I and J with I
J, de…ne a criterion CI;J by x CI;J y if and only
if x 2 I and y 2 J, and then form a d-checklist C that assigns the CI;J in any order to
the Ci with di
3. (For the Cj with dj = 2 that come before the …nal CI;J , set Cj = ?.)
This C evidently leads to
and C is a d-checklist since each CI;J has three equivalence
classes: I, J, and Xn(I [ J). An alternative and usually shorter construction is to de…ne
for each
-indi¤erence class I a criterion CI by y CI x if and only if x 2 I and y
form C by assigning the CI to the Ci with di
3, again arranged in any order (and again
setting the remaining Ci to be empty). This checklist leads to
2. If
x, and
is complete, our …rst construction in contrast uses
n
2
=
, proving Observation
n(n 1)
2
nonnull criteria.
The worst case – where the quickest d-checklist requires n nonnull criteria –can in fact
occur (see Example 3).
5
The comparison of decision-making speed
It may well be that an irrational agent, overwhelmed by the di¢ culty of making decisions,
makes only a few preference discriminations; such behavior, not surprisingly, can be the
outcome of a very short checklist. The length of a checklist therefore gauges decisionmaking speed only relative to the number of indi¤erence classes in the preference that the
checklist leads to. For the same reason, one should compare only those checklists that
share the same discriminatory capacity.
To determine the theoretical minimum length for a checklist, we generalize slightly the
standard logarithm: for any discriminatory capacity d, let dlogd ne denote the smallest
integer t such d1 d2
dt
n. When each di equals the same integer – as with binary
checklists –dlogd ne is indeed the ceiling of logdi n.7
Theorem 1 If a d-checklist C leads to a preference with n indi¤erence classes then C
has at least dlogd ne criteria.
Theorem 1 leads to the following de…nition of a quick checklist.
7
If, for n > 0 and d
(1; 1; : : :), we de…ne logd n = s + r where (s; r) satis…es d1
ds drs+1 = n and
r 2 [0; 1], then throughout the paper we can use the standard de…nitions of d e and b c. That is, for a
real number a, dae is the smallest integer b such that b
a and bac is the largest integer b such that
b a.
12
De…nition 6 A d-checklist C is quick if and only if it has dlogd ne criteria where n is
the number of indi¤erence classes in the outcome of C.
The label ‘quick’ is well-deserved: if we …x d, then, as the number of
-indi¤erence
classes n increases, dlogd ne increases at a less-than-polynomial rate. Since the ratio of
dlogd ne to n converges to 0 rapidly as n increases, quick checklists are highly e¢ cient
discriminators.8
There are two ways to view decision-making speed. One can look at the d-checklists
that arise from speci…c decision-making procedures and ask whether or not these checklists
are quick.
Alternatively one can remain agnostic about which procedures arise and
instead ask if speci…c preference relations could in principle be the outcome of a quick
checklist for various values of d.
The two perspectives are related since there is a particular decision-making procedure,
based on one-dimensional or linear scales, that will always generate a quick checklist that
leads to a rational preference.
We discussed one-dimensional scales in section 2; since
only the ordering that a scale represents matters, we now generalize slightly to linear
scales, that is, transitive asymmetric relations >i on X such that for each distinct pair
x; y either x >i y or y >i x.
An agent can always use linear scales to form a quick
checklist, as Theorem 2 below indicates. The construction used for the theorem employs
the ‘balanced’criteria discussed in section 2 that will appeal to agents who want to decide
quickly. Recall that balanced criteria do not waste opportunities to discriminate; they
e¢ ciently perform the same number of discriminations on each set of alternatives that
have not yet been ranked. Thus the message behind Theorem 2 is that one can always
build balanced criteria out of linear scales. Before stating the result, we need to de…ne
exactly what it means for a criterion to ‘use’a linear scale: a criterion Ci in a checklist
C uses >i to order alternatives if, for any pair x; y where Ci is the …rst criterion in C to
place x and y in di¤erent equivalence classes, x Ci y , x >i y.
Theorem 2 For any family of linear scales (>i )i 1 , discriminatory capacity d, and n
jXj, there is a quick d-checklist that leads to a rational preference with n indi¤erence
8
An earlier version of this paper de…ned a checklist to be quick if it minimizes the expected number
of criteria an agent has to examine before making a choice; this de…nition leads to similar results.
13
classes such that each Ci in the checklist uses >i to order alternatives.
We turn to the main contest: which preferences can always be the outcome of a quick
decision-making checklist, the rational or the irrational preferences? This way of posing
the question judges preferences by their potential decision-making speed.
While any
preference could be the outcome of a checklist with arbitrarily many criteria, whether or
not a preference is the outcome of some quick checklist is a clear-cut marker.
Theorem 3 will show that, for any complete and transitive
capacity d,
and any discriminatory
is the outcome of a quick d-checklist: rational agents can potentially make
choice discriminations at least as quickly as any other type of agent. An example will be
highly instructive.
Example 1 For X = f1; 2; 3; 4; 5; 6; 7; 8; 9g, let be the usual ordering of the integers.
!
!
The unique quick 3 -checklist (where 3 = (3; 3; :::)) that leads to is then the (C1 ; C2 )
given by
f9; 8; 7g C1 f6; 5; 4g C1 f3; 2; 1g and f9; 8; 7g C1 f3; 2; 1g
f9; 6; 3g C2 f8; 5; 2g C2 f7; 4; 1g and f9; 6; 3g C2 f7; 4; 1g,
which is pictured in Figure 1. Criterion C1 divides X into a top third, a middle third,
and a bottom third, and then ranks these thirds as
does, while C2 groups together
the top items from the three C1 -equivalence classes, the middle items from the three C1 equivalence classes, and the bottom three items from the three C1 -equivalence classes,
and then linearly ranks the tops over the middles over the bottoms. Since C2 makes the
same number of discriminations within each C1 -equivalence class, it is balanced.
Example 1 provides a model, for any rational
that leads to
, of how to de…ne a quick checklist
. One begins by partitioning X into d1 C1 -equivalence classes that are
ordered by
. Then at each subsequent stage i build a balanced Ci as follows: for each
maximal A
X that has not yet been divided into distinct criterion equivalence classes,
partition A into di cells that are ordered by
that have the same rank according to
and group together all the cells across A’s
to form the Ci -equivalence classes.
14
C1:
C1 and C2:
C2:
9
9
8
9
7
6
8
3
7
6
6
8
5
5
5
2
4
4
3
3
2
7
1
4
2
1
1
G
Figure 1: A quick 3 -checklist for a rational ;
C1 :
C1 and C2:
C2 :
a b
c
a
d e
f
g h
i
c
b e
h
a d
g
b
e
d
f
c
f
i
g
h
i
G
Figure 2: A quick 3 -checklist for an irrational ;
Although for analytical purposes we have begun with a rational preference
, the quick
checklists we have described should not be seen as decision-making procedures adopted
by agents who know their preferences in advance.
If an agent uses a family of linear
scales to order alternatives and builds balanced criteria, then the checklist that results
will, up to a labeling of alternatives, exactly follow the Example 1 template (see the proof
of Theorem 2).
The critical feature of a rational preference
, and the key to its decision-making
speed, is that each subset of not-yet-discriminated alternatives is ‘ordered the same way’
by
.
sets of
Rational preferences always display this symmetry: for a rational
any two
-indi¤erence classes with the same cardinality will be order isomorphic.9 This
property in fact characterizes the rational preferences with at least four indi¤erence classes
(Mandler (2009)). The fact that a rational
enjoys this property means that a criterion
Ci can simultaneously order every set of items that has not yet been discriminated prior
to Ci in a way that respects
.
Nevertheless, in some carefully constructed cases an irrational preference can be the
outcome of a quick checklist.
As we have seen, if a checklist is quick then after the
application of the …rst k criteria each set of alternatives A that has not yet been discriminated by the criteria must be
-order-isomorphic.10 While an irrational preference does
not have as rich a supply of order-isomorphic subsets as a rational preference does, it
still might have enough –given a particular discriminatory capacity d –to form a quick
d-checklist.
Example 2 Let X = fa; b; c; d; e; f; g; h; ig and de…ne
fa; b; cg
a
b
fd; e; f g
c
a and d
by
fg; h; ig and fa; b; cg
e
f
9
d and g
fg; h; ig
h
i
g.
Two sets, where each is endowed with a binary relation, (A; A ) and (B; B ), are order-isomorphic
if and only if there is a bijection f : A ! B such that, for all x; y 2 A, x A y , f (x) B f (y).
10
More precisely, if C1 ; :::; Ck are the …rst k criteria then these not-yet-discriminated sets A are the
cells of the coarsest common re…nement of the partitions C1 ; :::; Ck where each Ci is formed by the Ci equivalence classes. Also, a quick checklist requires only that there is a set B, endowed with a binary
relation and containing no more than dk+1
dT elements, such that the -indi¤erence classes in each
not-yet-discriminated set A can be homomorphically embedded into B.
15
Thus there are three triples that
orders linearly but within each triple
orders the
!
items as a cycle. The 3 -checklist (C1 ; C2 ) de…ned below and pictured in Figure 2 both
leads to
and is quick:
fa; b; cg C1 fd; e; f g C1 fg; h; ig and fa; b; cg C1 fg; h; ig
fa; d; gg C2 fb; e; hg C2 fc; f; ig C2 fa; d; gg.
As in Example 1, the criteria are balanced.
The only important di¤erence between Examples 1 and 2 is that C2 in Example 2 is
a cycle rather a line. Both cases share the indispensable feature that the C1 -equivalence
classes are ordered the same way by
, which allows C2 to be balanced. There are only
a handful of preferences with nine singleton indi¤erence classes that enjoy this property
whereas the total number of irrational preferences with nine indi¤erence classes is vast: it
is therefore an exceedingly unusual event for a nine-indi¤erence-class to be the outcome
!
of a quick 3 -checklist. We further consider the likelihood that an irrational
has a
quick checklist in section 7.
Here we pin down the unique ‡exibility of rational preferences by showing that for any
irrational
there is always a discriminatory capacity d such that
is not the outcome of a
quick d-checklist. Since any rational preference can be the outcome of a quick d-checklist
for any d, the property of being the outcome of a quick d-checklist for all d’s characterizes
rationality.
One proviso is necessary however: since only rational preferences can be
the outcome of binary checklists, the characterization should not use a binary checklist
to show that an irrational preference cannot always be quick – otherwise we would just
be repeating the substance of Observation 1. Given n, we say that the discriminatory
capacity d is admissible if and only if di > 2 for some i 2 f1; :::; dlogd neg. As always,
any discriminatory capacity d, admissible or not, satis…es di
Theorem 3 A preference
any admissible d,
with n
2 for all i.
5 indi¤erence classes is rational if and only if, for
is the outcome of a quick d-checklist.
The delicacy of the checklist in Example 2 illustrates the logic of Theorem 3, which is
the most important result in this paper. To achieve quickness, the items in each of the
three C1 -equivalence classes must be ordered by
16
in the same way. But the very fact
that
has three
-order-isomorphic subsets can make it impossible to partition X into
a di¤erent number of
-order-isomorphic subsets, as a quick d-checklist with d1 6= 3 will
require; an irrational preference may have a particular combination of order-isomorphic
subsets but it cannot have all the possible combinations. Indeed the
in Example 2 is
not the outcome of either a (5; 2)-checklist –that is, a d checklist with (d1 ; d2 ) = (5; 2) –
or a (2; 5)-checklist.
So far we have focussed on complete preferences.
The following example resolves
the puzzle, raised in the Introduction, that while it would seem that incomplete preferences should accelerate decision-making (the criteria that lead to
need to make fewer
rankings), our main result asserts the opposite.
Example 3 For X = fw; x; y; zg, let
y
z (so
is both
!
!
incomplete and intransitive). Set d = 3 . If there were a quick 3 -checklist that leads
to
be de…ned by w
x
, it must have length 2 = dlog3 4e. So then either C1 makes two out of the three
-rankings –that is, C1 has w C1 x C1 y or x C1 y C1 z or (w C1 x and y C1 z) –or C1 makes
one
-ranking and C2 makes the other two. If the criterion Ci that makes two rankings
has w Ci x Ci y then the omitted item z must be placed in the same Ci -equivalence class
as one of fw; x; yg and thus be ranked by Ci vis-à-vis x if joined with w or y or ranked
vis-à-vis w if joined with x, contrary to the rankings given by
. The Ci that makes two
rankings similarly cannot have x Ci y Ci z. And …nally if Ci has (w Ci x and y Ci z) then
(since di = 3) either fw; yg Ci fx; zg or w Ci fx; yg Ci fzg, which again imposes rankings
not given by
.
!
The minimum-length 3 -checklists that lead to
use three criteria, for example the
C de…ned by
w C1 x, x C2 y, y C3 z.
The above
!
is not quite the worst case for 3 -checklists.
last in the race are the
The
’s that come in
!
’s with n indi¤erence classes that require a 3 -checklist of
length n. The smallest n where such a
a
b
c
d
e
5 that leads to this
arises is 5: let on fa; b; c; d; eg be the cycle
!
a. The argument that there is no 3 -checklist shorter than length
is similar to the case above and we leave the details to the reader
(see also Mandler (2009)).
17
The moral of Example 3 is that while it may seem that incomplete preferences should
enjoy a speed advantage, incompleteness in fact forces any -unranked pair to be unranked
by every criterion. This restriction burns up criterion equivalence classes and hence slows
down the quickest checklist. The conventional description of incomplete preferences as
‘indecisive’turns out to be apt.
Extension to choice functions. We saw in section 2 that any checklist de…nes a choice
function c that selects from arbitrary choice sets. If c maximizes a rational preference
then c can be the outcome of the quick checklist that the proof of Theorem 3 uses. For a
converse, let the domain of choice functions include all of the two-element subsets of X.
And let us say that c ‘can always be the outcome of a quick checklist’if, for any admissible
d, c can be the outcome of a d-checklist with dlogd ne criteria, where n is the number of
indi¤erence classes in the base relation R of c.11
Theorem 3 then implies that c can
always be the outcome of a quick checklist if and only if its base relation is rational (and
5). It is easy to see that any choice function c that is the outcome of a checklist has a
n
rational base relation if and only if c maximizes a rational preference. A full equivalence
therefore obtains between choice functions that maximize rational preferences and choice
functions that can always be the outcome of a quick checklist.
6
Rational preferences are the easiest to build
So far we have taken the discriminatory capacity d to be exogenous; we had no choice
since a fair comparison of two preferences requires us to see if both can be the outcome
of criteria of the same power.
Here we reverse perspective and ask: for an agent who
uses criteria to build preferences, what is the best d to use?
Speci…cally, which d lets
preferences be constructed at the lowest cost as measured by the number of criterion
rankings the agent has to make? The answer is that it is always cheaper to use the coarser
criteria that distinguish fewer categories. The minimum cost is therefore achieved only
by binary checklists, where d = (2; 2; :::). So any agent who uses the minimum number of
binary criteria –minimum for the number of preference discriminations the agent ends up
11
R is de…ned by a R b , (c(fa; bg) = fag).
18
distinguishing – will make preference discriminations at the lowest possible cost. Since
binary criteria necessarily lead to rational preferences, we have another advantage of
rationality: only those agents who construct rational preferences can make the minimum
decision-making e¤ort.
The cost advantage of coarse criteria casts an economic light on the empirical observation of psychologists, discussed in section 2, that agents have only a limited ability to
retain and manipulate concepts in working memory. These limitations might seem to be
an unambiguous defect of our cognitive capacities; what we show here is that our limited
mental powers force us into making preference discriminations e¢ ciently.
We do not measure the cost of building a checklist that makes n preference discriminations by its length; by that standard it would always be cheapest to set d1 = n since
then an agent could make do with a single criterion.
Length moreover does not take
into account the cost of making the judgments that de…ne a checklist’s criteria. Instead
we measure the cost of a criterion Ci by the number of pairwise comparisons of Ci ’s
equivalence classes: for any two Ci -equivalence classes the agent must decide whether one
or the other is superior or whether they are unranked. The cost of a criterion with di
equivalence classes is therefore given by the binomial coe¢ cient
di
2
=
di (di 1)
2
and the
cost of a checklist as a whole is given by the sum of the costs of its criteria.
Increasing some of the di in a discriminatory capacity d seems to present a trade-o¤:
while the a¤ected criteria become more expensive to form one can make do with fewer of
them. The …rst e¤ect turns out to dominate the second: the criterion-building cost of a
larger di always outweighs the advantage of using fewer criteria. It follows that binary
criteria end up being the cheapest.
To illustrate suppose there are n = 9 indi¤erence
A quick binary checklist will require dlog2 9e = 4 criteria and hence 4 22 = 4
!
criterion decisions whereas a quick 3 -checklist, seemingly a better …t with n = 32 , requires
classes.
dlog3 9e = 2 criteria and 2
notably smaller than
9
2
3
2
= 6 decisions.
Observe that both of these numbers are
= 36, the number of pairwise comparisons that a preference with
9 indi¤erence classes makes: building preference distinctions from a checklist, even when
the checklist is not the most e¢ cient possible, can require much less decision-making e¤ort
than building the distinctions directly.
19
Let e(Ci ) denote the number of equivalence classes in a criterion Ci . We then de…ne
P
the cost of a checklist C = (C1 ; :::; CT ) to be Ti=1 e(C2 i ) (where 12 = 0).
Theorem 4 If the checklist C = (C1 ; :::; CT ) and discriminatory capacity d satisfy
(d1 ; :::; dT ) < (e(C1 ); :::; e(CT ))12
then there is a quick d-checklist that leads to a preference with the same number of indifference classes as the outcome of C but with lower cost than C.
Let a more-than-binary checklist be a (C1 ; :::; CT ) such that some Cj , j 2 f1; :::; T g, has
e(Cj ) > 2. Since we can drop any Ci with e(Ci ) = 1 from a checklist C without changing
C’s cost or the number of indi¤erence classes that C leads to, repeated application of
Theorem 4 to any more-than-binary checklist gives:
Corollary 1 Any more-than-binary checklist C has a larger cost than any quick binary
checklist that leads to a preference with the same number of indi¤erence classes as the
outcome of C.
So binary criteria have the potential to generate preference discriminations more e¢ ciently than any other type of criteria. It is therefore only the agents who end up being
rational who have preferences that can be built with minimum e¤ort.
There is practical advice in Corollary 1.
Returning to the restaurant meals that
opened the paper, suppose you want a method for ordering meals that discriminates
su¢ ciently many preference classes but that requires the fewest decisions. Corollary 1
instructs you to use binary criteria that form a quick checklist. To be quick, the binary
criteria must be balanced: as you proceed through your criteria, each binary distinction
must divide each set of not-yet-discriminated meals into two parts. As Theorem 2 showed,
criteria built from linear scales can accomplish this goal. So, in the case of meals, if an
agent uses linear attributes such as calorie count, cost, spiciness, etc., to build criteria
then each set of not-yet-discriminated alternatives can be divided into two parts with
more and less respectively of the attribute. Some attributes that need not be linear –
12
For T -vectors x and y, x < y means xi
yi for all i and x 6= y.
20
meat versus …sh –may also happen to divide sets of not-yet-discriminated meals into two
parts. The upshot of Corollary 1 is that a binary structure such as this, although it seems
crude, is the most e¢ cient way to partition meals into n indi¤erence classes. Notice that
if one can …nd the required binary distinctions, the e¢ ciency advantage will remain even
if on each day you go out to eat your ranking of some of the binary categories changes –
one day you prefer …sh and the next you prefer meat.
The binary checklists we have described determine a preference on all possible meals
but they will also serve as a way to search the smaller choice set –the menu –of a particular
restaurant. When a binary checklist does not generate enough indi¤erence classes to pick
out a single meal from a menu, the meals that survive undominated through all of the
criteria would then be declared indi¤erent and the agent would pick an arbitrary survivor.
This binary method may o¤er a good description of how people sometimes decide. Our
analysis pushes Rubinstein (1996) one step further: not only do linear binary relations
stand out in their usefulness but those relations that stem from binary categories end up
being the cheapest way to produce preference judgments.
7
The proportion of irrational preferences that are
the outcome of quick checklists
While a rational preference is always the outcome of a quick d-checklist, for any d, only
carefully constructed irrational preferences can be quick and then only for particular d’s.
Since on any domain except the very smallest the number of irrational preferences is
vast, the proportion of irrational preferences with a quick checklist must be quite small.
To illustrate, consider again the preferences in Examples 1 and 2 in section 5, which
!
both had nine singleton indi¤erence classes and were the outcome of quick 3 -checklists.
We can mimic these examples to generate all of the ’s with nine singleton indi¤erence
!
classes that are the outcome of a quick 3 -checklist. In any such (C1 ; C2 ), each Ci
must have three Ci -equivalence classes and each Ci -equivalence class must consist of three
-indi¤erence classes.
Moreover, we must put the three
-indi¤erence classes in each
!
C1 -equivalence class into separate C2 -equivalence classes. A 3 -checklist that does not
21
meet these requirements would have to put some nonindi¤erent pair x; y in both the same
C1 -equivalence class and the same C2 -equivalence class, and then since x and y would be
ranked the same way relative to every other z 2 X we would have x
y. Now up to
an order isomorphism there are only four asymmetric orderings R with three equivalence
classes. Labeling the equivalence classes a, b, and c, these are de…ned by: (1) a R b R c and
a R c, (2) a R b R c R a, (3) a R b and b R c, and (4) a R b.13 By letting C1 order the C1 equivalence classes according to one of the orderings (1) through (4) and letting C2 order
the items in each C1 -equivalence class according to one of the same orderings (the same
ordering for every C1 -equivalence class), we can generate, up to an order isomorphism, all
of the preferences with nine singleton indi¤erence classes that are the outcome of quick
!
3 -checklists. Evidently there are 16 of which we have seen two: the in Example 2 is
the outcome of criterion orderings (1) and (2) and the
in Example 1 is the outcome of
two copies of ordering (1). Since 15 of these 16 preferences are not rational and since,
up to an order isomorphism, there are approximately 400 billion irrational
’s with nine
singleton indi¤erence classes it is apparently exceedingly unusual event for a
!
class to be the outcome of a quick 3 -checklist.
in this
The restriction to singleton indi¤erence classes slightly exaggerates the unlikeliness of
an irrational
being the outcome of a quick checklist, but only slightly. The following
table reports for 3
n
9 the number of preference relations on f1; :::; ng that fail
to be complete and transitive (regardless of the size of indi¤erence classes), the number
!
of these preferences that are the outcome of a quick 3 -checklist, and the ratio of these
two numbers, i.e., the proportion of irrational preferences on Xn that are the outcome of
!
quick 3 -checklists. In all cases, to keep the count under control we give the number of
preferences up to an order isomorphism.14
13
The orderings de…ned by (a R b; c R b), (a R b; a R c), and the empty ordering do not qualify since they
all have fewer than three equivalence classes.
!
14
Since there are only …ve relevant Ci orderings of Ci -indi¤erence classes when d = 3 , it is feasible
!
to enumerate the irrational preferences on Xn that are outcomes of quick 3 -checklists and then weed
out the preferences that are order isomorphic. I am indebted to Claire Blackman for writing code that
performs this enumeration (it is available on request). The total number of preferences on Xn up to order
isomorphism is given in Harary (1969); for the number of irrational preferences we subtract the number
of rational preferences, which equals 2n 1 , from the Harary total.
22
n
!
# irrational preferences # with quick 3 -checklists
ratio
3
3
3 1
4
34
31 :91176471
5
566
226 :39929329
6
21; 448
1; 084 :05054084
7
2; 142; 224
3; 847 :00179580
8
575; 016; 091
11; 143 :00001938
9
415; 939; 242; 776
217; 902 :00000052
To give a general result on the proportion of irrational preferences that are the outcome
of a quick checklist as the size of the domain increases, …x for the remainder of the section
the discriminatory capacity at an arbitrary d such that there is a
i 2 N.
with di
For any n, let Xn denote a generic domain of n objects, f1; :::; ng, let
for all
ir (n)
denote the set of irrational preferences on Xn ,
ir (n)
and let j
ir (n)j
=f
Xn
Xn :
denote the cardinality of
is asymmetric and not rationalg,
ir (n).
As for quick checklists, let q ir (n) denote
the set of irrational preferences on Xn that are the outcome of a quick checklist,
q ir (n) = f 2
ir (n)
:
is the outcome of a quick d-checklistg,
and let jq ir (n)j denote the cardinality of q ir (n).
Theorem 5 The proportion of irrational preferences on Xn that are the outcome of a
quick d-checklist converges to 0 at a super-exponential rate as n ! 1. That is, for any
k > 0, ekn jjq irir(n)j
converges to 0.
(n)j
8
Conclusion
Rationality may not be as di¢ cult as Herbert Simon supposed. If we identify a preference
relation with a decision procedure and not with an agent who must uncover from his mind
23
a preexisting utility function, then there need be no intrinsic challenge to choosing rationally. For agents who use checklists, rational preferences will emerge if the constituent
building blocks of a checklist – the criteria – are themselves rational; moreover criteria
are necessarily rational when agents use simple binary criteria. Irrationality can still occur, of course, but in the present setting there is no computational argument in its favor.
Indeed, it is the rational agents who hold the advantage. The agents who use rational
criteria can potentially make choice discriminations with strictly greater e¢ ciency than
the agents who do not.
While any irrational agent is slow compared to the potential speed of a rational agent,
the principle that a preference that is well-endowed with order-isomorphic subsets will
have a speed advantage can be used to compare the relative slowness of di¤erent irrational
preferences. Suppose that an agent with preference
knows that in a coming stretch of
time, say a month, the only decisions he will face involve alternatives drawn from some
subdomain Y
X. Suppose Y can be partitioned into a pair of sets A and B and that A
and B are ranked by
. If the
-indi¤erence classes in A and B are
-order isomorphic
then there is a quick checklist on the subdomain Y : the restriction of
to A [ B will be
the outcome of a d-checklist where d1 = 2 and d2 is the number of
-indi¤erence classes
in A (or B). Of course, analogous checklists arise for all k-tuples (rather than pairs) of
sets that partition a subdomain as long as the partitioning sets are ordered by
and are
order-isomorphic. A preference’s endowment of order-isomorphic subsets thus measures
its decision-making speed across domains and discriminatory capacities. Of course since
rational agents have the maximum possible endowment of such subsets, by this measure
too they are the quickest.
A
Appendix: Remaining results and proofs
Proof of Observation 1. We prove a slightly stronger result, that if each Ci in C is
rational when restricted to any A
X such that x; y 2 A ) (not x Cj y and not y Cj x
for j < i) then the outcome
of C is rational. Given the d-checklist (C1 ; :::; CT ), set
d = maxfd1 ; :::; dT g and, for any x 2 X and criterion Ci , de…ne Xxi = fy 2 X : not
y Cj x and not x Cj y for 1 j < ig, Jxi = fJ Xxi : 9 a Ci -equivalence class E such that
J = E \ Xxi 6= ?g, and xi = jfJ 2 Jxi : x Ci Jgj. Since the Ci -equivalence classes form
a partition of X, Jxi is a partition of Xxi . Given our assumption that Ci is rational on
24
Xxi , Ci de…nes a rational ordering on the elements of Jxi . We identify x with the integer
Int(x) whose expansion in base d has xi as its ith digit (counting from the left). If C
leads to , then x y if and only if the …rst criterion Ci that ranks x and y has x Ci y,
which (since Ci is rational on Jxi ) obtains if and only if the …rst digit i where Int(x) and
Int(y) di¤er in base d has xi > yi , i.e., Int(x) > Int(y). It is also easy to con…rm that
x y if and only if Int(x) = Int(y). Since Int(x) may therefore serve as a utility number
for any x 2 X, has a utility representation and is therefore complete and transitive.
Singleton Indi¤erence Class Lemma. If a preference with n indi¤erence classes is
the outcome of a length T checklist then there is a preference with n singleton indi¤erence
classes that is also the outcome of a length T checklist.
Proof.
For each -indi¤erence class I, let eI be an arbitrary element of I and for
any x 2 X, let [x] denote the -indi¤erence class that contains x. Given a d-checklist
(C1 ; :::; CT ) that leads to , de…ne the checklist C 0 = (C10 ; :::; CT0 ) by
x Ci0 y , e[x] Ci e[y]
for i = 1; :::; T and all x; y 2 X. It is easy to con…rm that each Ci0 is asymmetric and has
no more than di equivalence classes. Letting 0 be the outcome of C 0 ,
x
y , e[x] e[y] , the …rst Ci that ranks e[x] ; e[y] has e[x] Ci e[y]
, the …rst Cj0 that ranks x; y has x Cj0 y , x 0 y,
where in fact the indices i and j coincide. Since any a; b 2 X with a b are always in
the same Ci0 -equivalence class for i = 1; :::; T , the binary relation B on fI X : I = [x]
for some x 2 Xg de…ned by [a] B [b] if and only if a b must also be the outcome of a
d-checklist of length T .
Proof of Theorem 1. If a preference with n indi¤erence classes is the outcome of
a d-checklist of length T , the Singleton Indi¤erence Class Lemma implies there is a dchecklist C of length T that leads to a with n singleton indi¤erence classes. We show
that T dlogd ne.
Given a d-checklist C 0 of length k, de…ne a C 0 -undecided set to be a A
X such that
x; y 2 A ) (for each i = 1; :::; k, 9 a Ci0 -equivalence class I such that fx; yg I), de…ne
a maximal C 0 -undecided set to be a C 0 -undecided set A such that no C 0 -undecided set B
has jBj > jAj, and …nally de…ne the maximal C 0 -undecided cardinality to equal jAj where
A is a maximal C 0 -undecided set.
We show, by induction on k, that if the C that leads to has length k then the maximal
C-undecided cardinality is greater than or equal to (d1
dk ) 1 n. This claim holds for k =
1 since if every Ci -equivalence class has strictly fewer than d1 1 n elements we would have
jXj < n. Assuming then that the claim holds for arbitrary k, let C = (C1 ; :::; Ck ; Ck+1 )
be a d-checklist of length k + 1. So there is a (C1 ; :::; Ck )-undecided set A with jAj
(d1
dk ) 1 n. Since the dk+1 Ck+1 -equivalence classes, I1 ; :::; Idk+1 , form a partition of
X, the sets I1 \ A; :::; Idk+1 \ A partition A; hence one of these sets, say Ij \ A, must have
at least (d1
dk+1 ) 1 n elements. And since Ij \ A is a (C1 ; :::; Ck ; Ck+1 )-undecided set,
the maximal (C1 ; :::; Ck ; Ck+1 )-undecided cardinality is at least (d1
dk+1 ) 1 n.
Since C leads to , the maximal C-undecided cardinality must equal 1. If not there
25
would be distinct x; y 2 X and a Ci -equivalence class Ii for each i = 1; :::; T such that
fx; yg
Ii for i = 1; :::; T ; and then, 8z 2 X, (x Ci z , y Ci z) and (z Ci x , z Ci y)
which implies that x and y are in the same -indi¤erence class. Hence the previous
paragraph gives 1 (d1
dT ) 1 n, that is, d1
dT n. So T dlogd ne.
Proof of Theorem 2. Set C0 = ?. Given a binary relation Ci on X, let Ii denote
the partition of X whose cells are the Ci -equivalence classes.
So in particular I0 =
W
fXg. Given the partitions I0 ; :::; Ir of X, we let Jr = j=0;:::;r Ij denote the coarsest
common re…nement of I0 ; :::; Ir . We proceed recursively, taking C0 ; :::; Ci 1 as given,
where i 1 < dlogd ne and where, if jXj is …nite, the cells of Ji 1 di¤er in cardinality
by at most 1. Suppose that jJi 1 j di
n. Then de…ne for each J 2 Ji 1 a partition
K
j J k= fJ(1);
l m:::; J(di )g of J such that (1) if jJj is …nite each cell of KJ contains either
jJj
or jJj
elements, (2) if jJj is in…nite each cell of KJ has at least n elements,
di
di
and (3) J(r) >i J(s) if r < s. For the case jJi 1 j di > n de…ne to be the solution
of d1
di 1 = n and de…ne
P ( J )J2Ji 1 so that each J equals either b c or d e and
1
min[di ; jJj] and J2Ji 1 J = n are satis…ed. Then, for each J 2 Ji 1 set the
J
partition
k J =l fJ(1);
m :::; J( J )g of J so that if jJj is …nite then each cell of KJ contains
j K
jJj
jJj
either J or J elements and (2) and (3) are satis…ed. In all cases, for any k = 1; :::; di
S
S
such that J2Ji 1 J(k) 6= ? de…ne a Ci -equivalence class by Ii (k) = J2Ji 1 J(k) and
then set Ci by Ii (r) Ci Ii (s) if and only if r < s and both Ii (r) and Ii (s) are nonempty.
Assuming the recursion is well-de…ned, each Ci must use >i to order alternatives since
the cells of Ji 1 are the maximal subsets of X that have not yet been discriminated
by Cj , j < i. The proof of Observation 1 then shows that C must lead to a rational
preference. Also, still assuming the recursion is well-de…ned, if jXj is
j …nite then
k repeated
j
k
bjXj=dk c
jXj
application of the nested division property of ceilings and ‡oors,
=
dk+1
dk dk+1
l
m l
m
djXj=dk e
jXj
and
= dk dk+1 , implies that the cells of Ji will di¤er in cardinality by at most
dk+1
1.
To ensure that the recursion is well-de…ned and that the checklist that results is quick,
we need to show that (i) if jJi 1 j di n then each J 2 Ji 1 has at least di elements, (ii)
if jJi 1 j di > n then thereP
is a J for each J 2 Ji 1 equal to either b c or d e such that
1
min[d
;
jJj]
and
J
i
J2Ji 1 J = n, and (iii) if i = dlogd ne then jJi j = n.
We suppose …rst that jXj is …nite. If jJi 1 j di n then each J 2 Ji 1 must have at
least di elements: if some Jb 2 Ji 1 has Jb < di then each J 2 Ji 1 has jJj di (since
P
the cardinality di¤erence is at most 1) and hence jXj = J2Ji 1 jJj < jJi 1 j di
n, a
contradiction. Hence for each i with jJi 1 j di
n each KJ consists of exactly di cells.
So, if jJi 1 j di = n for some i then d1
di 1 di = n and hence i = dlogd ne and thus
(C1 ; :::; Ci ) is a quick d-checklist.
Now assume there is no i such that jJi 1 j di = n and consider the smallest i such
that jJi 1 j di > n. Since in addition (from the previous paragraph) jJi 1 j = d1
di 1 ,
we have i = dlogd ne. Since d1
di 1 < n, we have = d1 ndi 1 > 1. And since
jJi 1 j = d1
di 1 , jJi 1 j di > n implies that = d1 ndi 1 < di . So, for each J 2 Ji 1 ,
j
k
l
m
jXj
jXj
1
d
.
Now
each
J
2
J
has
either
or
elements. So if
J
i
i 1
d1 di 1
d1 di 1
26
j
k
l
m
jXj
jXj = n then d1 jXj
=
b
c
and
= d e and so, by setting J = b c for J with
d1 di 1
j
k di 1
l
m
n
n
jJj = d1 di 1 and J = d e for J with jJj = d1 di 1 = d e, we have ( J )J2Ji 1 such
P
that J jJj for each J and J2Ji 1 J = n. For this jXj = n case, let fn and cn denote
the number of cells in Ji 1 with the ‡oor and ceiling cardinalities:
fn =
cn =
J 2 Ji
1
J 2 Ji
1
: jJj =
: jJj =
n
d1
,
di
1
di
1
n
d1
.
So fn + cn = jJi l1 j and fmn b c + cn d e = n. Now if jXj > n the number of cells in
Ji 1 with at least d1 ndi 1 elements must be larger than cn and the number of cells with
j
k
n
elements must be smaller than fn . Hence we can still set J = b c for exactly
d1 di 1
fn of the cells in Ji 1 and set J = d e for exactly cn of the cells in
PJi 1 while still obeying
the restriction that J jJj for each J. So in this case as well J2Ji 1 J = n.
Finally let jXj be in…nite. Letting (Jin )i=1;:::;dlogd ne denote the coarsest common re…nement partitions that occur when jXj = n, it is easy to see, for each i 2 f1; :::; dlogd neg and
each positive integer k, that jfJ 2 Ji 1 : jJj kgj
fJ 2 Jin 1 : jJj kg . It therefore
remains true that if jJi 1 j di n then each J 2 Ji 1 has jJj di and, for the smallest i
such that jJi 1 j di > n, we can set J = b c for fn cells in Ji 1 and J = d e for cn cells
in Ji 1 while still satisfying J jJj for each J.
Proof of Theorem 3.
‘Only if’. Let be a preference on X with n indi¤erence classes that is complete and
transitive. Fix some d which in this half of the proof we can allow to be not admissible.
For any x 2 X, let [x] be the -indi¤erence class that contains x. For any -indi¤erence
class I X, de…ne
r(I) = jfJ
X : J is a
-indi¤erence class such that I
Jgj ,
which can be viewed as the utility number of any element of I.
We can identify
each integer r in [0; n 1] with its expansion in the mixed-radix number system (see
Knuth (1997b)) with bases d1 ; :::; dT , where T = dlogd ne. That is, for each r there is
a unique sequence of T integers, k(r; 1); :::; k(r; T ), where 0
k(r; j) < dj , such that
PT 1
r = k(r; T ) + j=1 k(r; j)(dT dT 1
dj+1 ). For j = 1; :::; T , de…ne Cj by x Cj y if and
only if k(r([x]); j) > k(r([y]); j). For any x; y 2 X, suppose there is a smallest integer i in
[1; T ] such that k(r([x]); i) 6= k(r([y]); i). If say k(r([x]); i) > k(r([y]); i) then r[x] > r[y]
and so x y. Moreover, x Ci y and not y Cj x for all integers j < i, as desired. If on the
other hand k(r([x]); i) = k(r([y]); i) for all integers i in [1; T ], then r([x]) = r([y]) and so
x y. And then not x Ci y and not y Ci x for all i in [1; T ].
‘If’. Let have n 5 indi¤erence classes and be the outcome of a quick d-checklist
for each admissible d. We will show that is rational. Let ‘C is a (d1 ; :::; dt )-checklist’
mean that C is a checklist of length t where each Ci has di or fewer equivalence classes.
In step 1 we establish a simple consequence of the assumption that some (2; n2 )-checklist
27
leads to . We use this consequence in step 2 to show that any (3; n3 )-checklist that
leads to must have a linear C1 . We in turn use this fact in step 3 to show that if is
also the outcome of a quick (3; 2; :::; 2)-checklist then is rational. Given the proof of
the Singleton Indi¤erence Class Lemma, we can assume for each checklist C that each indi¤erence class is contained in exactly one Ci -equivalence class for each Ci in C. Also,
for two -indi¤erence classes x and y, we use x ? y to mean :(x y) and :(y x).
Step 1. Let L and U be the C1 -equivalence classes for some (C1 ; C2 ) that is a (2; n2 )checklist that leads to where U C1 L and therefore U L. (C1 must have two equivalence classes since otherwise the outcome of (C1 ; C2 ) could have at most n2 indi¤erence
classes.) Given the proof of Theorem 1, any pair of -indi¤erence classes must either
be in distinct C1 or C2 equivalence classes. Hence if n is even the C1 -equivalence classes
contain n2 -indi¤erence classes each and if n is odd the C1 -equivalence classes contain
n
and n2
-indi¤erence classes.
2
Step 2. Let I, J, K be the C1 -equivalence classes for some (C1 ; C2 ) that is a (3; n3 )checklist that leads to . Using the fact that the intersection of each C1 -equivalence
class and each C2 -equivalence class can contain at most one -indi¤erence class and
n (mod 3) then each
that C2 has n3 -equivalence classes, it is easy to see that: if 0
C1 -equivalence class contains n3 -indi¤erence classes, if 2
n (mod 3) then the C1 equivalence classes contain n3 , n3 , and n3
-indi¤erence classes, and if 1 n (mod 3)
1, n3 , and
then the C1 -equivalence classes contain either ( n3 , n3 , and n3 ) or ( n3
n
) -indi¤erence classes. It follows that the union of the two C1 -equivalence classes
3
with the fewest -indi¤erence classes has at least 2 n3
-indi¤erence classes. If n 6
n
n
n
n
then 2 n3
and
if
n
=
5
then
+
=
.
Thus
the same union always has
2
3
3
2
n
-indi¤erence classes, a fact we use below.
at least 2
We show that fI; J; Kg is linearly ordered by C1 and hence by . Since C1 must have
exactly 3 equivalence classes –if C1 had k < 3 equivalence classes then k n3 < n violating
Theorem 1 –the only other possibilities (up to order isomorphism) are (i) IC1 JC1 KC1 I,
(ii) IC1 JC1 K, and (iii) IC1 J, considered as (i)-(iii) below.
Given a (3; n3 )-checklist (C1 ; C2 ) that leads to , call a C2 -equivalence class full if it
consists of exactly 3 -indi¤erence classes.
Lemma. At most two C2 -equivalence classes are not full.
Proof. If 3 C2 -equivalence classes were not full then those equivalence classes could
contain at most 6 -equivalence classes. The remaining n3
3 C2 -equivalence classes
could together contain at most 3( n3
3) -indi¤erence classes. Thus the entire set of
n
C2 -equivalence classes could contain at most 3 n3
3 < n -indi¤erence classes, a
3
contradiction.
Case (i). We then have I
J
K
I. Choose the labels of the C1 -equivalence
classes so that I \ U 6= ?. Then, since K I and U L, K \ L = ?. Hence K \ U 6= ?
and, applying the same argument, J \ L = ?. Hence J \ U 6= ? and, again, I \ L = ?.
Thus L = ?, a contradiction.
Case (ii). We have I
J
K. For n
7 the Lemma implies there must be a
C2 -equivalence class E that is full. For 5 n 6, C2 has 2 equivalence classes, hence
again a full C2 -equivalence class E. First suppose that J \ L 6= ? and J \ U 6= ?.
Then I U and K L. Since the three -indi¤erence classes in E must be in separate
C1 -equivalence classes, x I \ E 6= ? and z
K \ E 6= ?. Given that I and K are
28
not ranked by C1 , we have x ? z, which contradicts the fact that I
U and K
L.
Next suppose that either J
L or J
U . If say J
U then I
U and, since I [ J
-indi¤erence classes, K = L. But then the presence of a full
must contain at least n2
C2 -equivalence class E again implies that there are two -indi¤erence classes x and z in
I and K respectively with x ? z, again contradicting the fact that I
U and K
L.
The subcase where J L is similar.
Case (iii). We have I J. Observe …rst that each full C2 -equivalence class E must
either be contained in L or U : letting the three -indi¤erence classes in I \ E, J \ E,
and K \ E be denoted, respectively by x, y, and z, we have x ? z and y ? z and so
x [ z and y [ z must both be contained in either L or U . In the following subcases there
must be a set E of full C2 -equivalence classes such that jEj > n2 : n 10 (since, by the
Lemma, at most 4 -indi¤erence classes are in C2 -equivalence classes that are not full),
n = 6; 9 (since 0 n (mod 3) and hence every C2 -equivalence class is full), n = 8 (since
n
= 3 and hence 2 C2 -equivalence classes must be full), and n = 7 and there are 2 full
3
C2 -equivalence classes. Since jEj > n2 , there must two full C2 -equivalence classes, E
and F , such that E
L and F
U . But for the indi¤erence classes x = E \ I and
y = E \ J we have y x, a contradiction.
Two tedious subcases remain. The …rst is where n = 7 and there is one full C2 equivalence class E. Call the two nonfull C2 -equivalence classes F and G. Assume for
concreteness that E U (the E L possibility is similar). Then I U since IC1 (J \E).
Hence (J [ K) \ (F [ G) must contain at least three -indi¤erence classes that belong to
L and consequently both J [ K and F [ G must each contain at least one -indi¤erence
class that belongs to L. Since E U and E intersects each C1 -equivalence class, we must
have EC2 (F [ G). Now suppose I \ (F [ G) 6= ?. Let us then choose the labels F
and G so that I \ F 6= ?. Then the two -indi¤erence classes in G are both in L: one
must be in J and the other in K. Consequently since (I \ F ) U we must have F C2 G.
Given that EC2 G, that F C2 G, and that J and K are not ranked by C1 , we conclude that
the items in G cannot form two -indi¤erence classes (each item in G is -worse than
the other 5 -indi¤erence classes). On the other hand suppose I \ (F [ G) = ?. Then,
if F and G are unranked by C2 , (F [ G) \ J cannot contain two -indi¤erence classes
(each item in (F [ G) \ J is -worse than the -indi¤erence classes in E and -unranked
relative to the -indi¤erence classes in KnE). While if F and G are ranked by C2 –say
F C2 G –then G cannot contain two -indi¤erence classes (each item in G is -worse than
the other 5 -indi¤erence classes). The second subcase is where n = 5. Since n3 = 2,
one of the C2 -equivalence classes, say E, must be full and C2 must rank E and the other
C2 -equivalence class, F . Since E is full and 3 = 52 , either E = L or E = U . Suppose
for concreteness that E = U (the E = L possibility is similar). If F \ I 6= ? then
F \I
U and hence U would contain at least 4 -indi¤erence classes. So F \ J 6= ?
and F \ K 6= ? but, since E F and C2 does not rank J and K, F \ (J [ K) cannot
form two distinct -indi¤erence classes.
Thus the set of C1 -equivalence classes above, which we now denote Z, is linearly
ordered by .
Step 3. We conclude by showing that if
is the outcome of a (3; 2; :::; 2)-checklist
n
b
C with length 1 + log2 3 in addition to the steps 1 and 2 checklists, then
must be
n
b equals
linear. The length is chosen so that log2 3 , the number of binary criteria in C,
29
b1 -equivalence class consists of no more
the minimum integer t such that 3 2t n. Each C
log2 n
d
e
bi , i 2, is linear these -indi¤erence
3
than 2
-indi¤erence classes and since each C
n
classes are linearly ordered by . So we may instead consider a (3; 2dlog2 3 e )-checklist
(C1 ; C2 ) that leads to where C2 is linear and show that is linear. As in step 2, C1
must have exactly 3 equivalence classes, which we again label I, J, and K. Since
is
linear if C1 is, it is su¢ cient to consider the nonlinear possibilities for C1 , which (up to
order isomorphism) are (i) IC1 JC1 KC1 I, (ii) IC1 JC1 K, and (iii) IC1 J. Case (i) repeats
case (i) in step 2.
Case (iii). We have I J. For any z 2 X or -indi¤erence class z, we use E(z) to
denote the C2 -equivalence class that contains z.
For any nonlinear , there exists a -indi¤erence class k
K such that (I) there
is a -indi¤erence class i I with :iC2 k and (II) there is a -indi¤erence class j 2 J
with :kC2 j. To see this, suppose to the contrary that for every -indi¤erence class
k
K either (I) or (II) is violated. For example, let (I) be violated, i.e., every indi¤erence class i I satis…es iC2 k . Then any -indi¤erence class x with x E(k) \ J
satis…es xC2 k and any -indi¤erence class x with E(k) \ J x satis…es kC2 x –so in fact
E(k) consists of only one -indi¤erence class. Consequently the checklist C 0 where k is
‘transferred’into J but is otherwise identical to C is a checklist that leads to ; formally
C20 = C2 and C10 = C1 [ f( ; ) : 2 I and 2 kg. Notice that either C 0 is not of
type (iii) or, if it is (with C10 -equivalence classes I 0 , J 0 , and K 0 that satisfy I 0 C1 J 0 ), then
k
J 0 . A similar construction applies when (II) is violated. Proceeding sequentially
through the -indi¤erence classes in K, we eliminate all of the -indi¤erence classes in
e1 ; C
e2 ) that leads to where each C
ei is linear.
K and conclude that there is a checklist (C
Hence is linear.
Given that any -indi¤erence class k K satis…es (I) and (II), (k i or k ? i), (j k
or k ? j), and i j. Step 2 then implies that i [ j [ k 2 Z for some Z 2 Z. Suppose
Y 2 Z and Y
Z. Any -indi¤erence class x J satis…es i x and so x cannot be
a subset of Y . For a -indi¤erence classes x (Y \ I) then we must have xC2 k (since
otherwise k x or k ? x). To pin down the features of -indi¤erence classes in Y \ K,
we use (z), for any -indi¤erence class z, to denote the set in Z that contains z. Let x
be a -indi¤erence classes contained by Y \K. To show that xC2 w for all w J, suppose
to the contrary that :xC2 w for some w
J. Then (x)
(w). Since furthermore
i w, we have (w) Z and, since is transitive on Z, we get the contradiction that
x cannot be a subset of Y . For x (Y \ K) it must in addition be the case that xC2 k.
Since xC2 k and xC2 w for all -indi¤erence classes w
J, E(x) must contain just one
-indi¤erence class: since i J the only other possible -indi¤erence class in E(x) must
be a subset of I and any -indi¤erence class u with u
E(x) \ I satis…es uC2 x and
any -indi¤erence class u with E(x) \ I u satis…es xC2 u. We conclude that if v is a
-indi¤erence class in Y then v is a subset of fa 2 I : aC2 kg or fa 2 K : aC2 k and E(a)
contains one -indi¤erence classg. Hence each -indi¤erence class in Y is in a distinct
C2 -equivalence class F such that F C2 k. A similar argument shows that if Y 2 Z and
Z
Y then any -indi¤erence class v
Y is a subset of fa 2 J : kC2 ag or fa 2 K :
kC2 a and E(a) contains one -indi¤erence classg, and so each -indi¤erence class in Y is
in a distinct C2 -equivalence class G such that kC2 G. Since the union of the two sets in Z
-indi¤erence
with the smallest number of -indi¤erence classes contains at most 2 n3
30
classes (see step 2), there must be at least 2
n
readily show that 2 n + 1 > 2dlog2 3 e .15
n
3
+ 1 C2 -equivalence classes. But one may
3
Case (ii). This case is similar to but simpler than (iii). We have I J K. If
is not linear then there must be -indi¤erence classes i I and k K such that :iC2 k;
otherwise IC2 K and hence I
K and so we could add IC10 K to the original C1 thus
creating a checklist of two linear orders. Similarly to case (iii), i, k, J, any -indi¤erence
class x I with iC2 x, and any -indi¤erence class y K with yC2 k must all be subsets of
some Z 2 Z. So if Y 2 Z and Y
Z and v is a -indi¤erence class, then v I and vC2 i
and therefore each -indi¤erence class in Y is in a distinct C2 -equivalence class F such
that F C2 i. Similarly, if Y 2 Z and Z Y and u is a -indi¤erence class, then u K
and kC2 u and therefore each -indi¤erence class in Y is in a distinct C2 -equivalence class
G such that kC2 G. Again the union of the two sets in Z with the smallest number of
-indi¤erence classes and so there must be
-indi¤erence classes contains at most 2 n3
n
at least 2 3 + 1 C2 -equivalence classes, a contradiction.
Proof of Theorem 4.
Lemma 1. If C = (C1 ; :::; Ct ) satis…es e(Ci ) 2 for all i 2 f1; :::; tg, e(Ck ) > 2 for some
k 2 f1; :::; tg,
(d1 ; :::; dt )
(e(C1 ); :::; e(Ck 1 ); e(Ck )
1; e(Ck+1 ); :::; e(Ct )),
and d0 is a discriminatory capacity with (d01 ; :::; d0t+1 ) = (d1 ; :::; dt ; 2), then any d0 -checklist
C 0 with t + 1 criteria has lower cost than C.
Proof. Since e(Ci0 )
e(Ci ) for i = 1; :::; t, and e(Ck0 )
e(Ck ) 1, and since 2b is
increasing in b, the di¤erence between the cost of C and C 0 is at least
t
X
e(Ci )
2
i=1
0
@
=
X
e(Ci )
e(Ck )
+
2
2
i2f1;:::;tgnfkg
e(Ck )(e(Ck )
2
1)
(e(Ck )
1
1
+ 1A
1)(e(Ck )
2
2)
1 = e(Ck )
2 > 0.
Lemma 2. If n d1
dt , di 1 for all i 2 f1; :::; tg, and dk > 1 for some k 2 f1; :::; tg,
then n 2d1
dk 1 (dk 1)dk+1
dt .
Proof. Since dk > 1 implies 2(dk 1) = dk + (dk 2) dk , we have
2(dk
1)d1
dk 1 dk+1
dt
d1
dt
n.
Lemma 3. Let d0 = (d1 ; :::; dk 1 ; (dk 1); dk+1 ; :::; dt ; 2; 2; :::). If n > d1
dk 1 (dk
0
1)dk+1
dt then there is a quick d -checklist C that leads to a preference with n indi¤erence classes such that (e(C1 ); :::; e(Ct )) = (d01 ; :::; d0t ).
Proof. Let the preference be > on X = f1; :::; ng and let C be the quick d0 -checklist given
0
in the proof of Theorem 2 where each >i is > (and taking
mithe d in the Theorem).
h l d to be
n
0
and since n > d01
d0t
Since, as the proof of Theorem 2 shows, e(Ci ) = min di ; d0 d0
1
15
For any x
1, including x =
n
3,
1+log2 bxc
dlog2 xe and so 2 bxc
31
i 1
2dlog2 xe and 2 bxc+1 > 2dlog2 xe .
m
l
d0i for i t, we have e(Ci ) = d0i for i t.
implies d0 nd0
1
i 1
To prove the Theorem, suppose that the given C leads to a preference with n indi¤erence classes. We proceed recursively beginning with the d1 -checklist C 1 = C
of length T where d1 = (e(C1 ); :::; e(CT ); 2; 2; :::). Suppose at step j we are given
dj = (dj1 ; :::; djT ; 2; 2; :::) and a dj -checklist C j of length T + tj that leads to a preference with n indi¤erence classes such that (e(C1j ); :::; e(CTj +tj )) = (dj1 ; :::; djT +tj ). Theorem 1 then gives dj1
djT +tj
n. If there is a k 2 f1; :::; T g such that djk > dk and
n > dj1
djk 1 (djk 1)djk+1
djT +tj then, setting tj+1 = tj +1 and dj+1 = (dj1 ; :::; djk 1 ; (djk
1); djk+1 ; :::; djT ; 2; 2; ::), Lemma 2 and the proof of Theorem 3 imply there is a quick dj+1 checklist C j+1 of length T + tj+1 that leads to a preference with n indi¤erence classes.
By Lemma 3 we can choose C j+1 so that e(Cij+1 ) = dj+1
for i = 1; :::; T + tj . But since
i
j+1
j+1
j+1
j+1
n > d1
dT +tj , we must have e(CT +tj+1 ) = dT +tj+1 as well. By Lemma 1, C j+1 has
j+1
lower cost than C j . If (dj+1
1 ; :::; dT ) = (d1 ; :::; dT ), let the recursion terminate.
j
If, for all k 2 f1; :::; T g, dk > dk implies n dj1
djk 1 (djk 1)djk+1
djT +tj then de…ne
dj+1 by reducing some djk with djk > dk by 1 and leaving the remaining dji unchanged and
j+1
iterating this operation until the …rst point where either (1) (dj+1
1 ; :::; dT ) = (d1 ; :::; dT )
j+1
In case (1), the proof of Theorem 3
and n
dj+1
dj+1
dj+1
1
T +tj or (2) n > d1
T +tj .
j+1
j+1
implies there is a quick d -checklist C
that leads to a preference with n indi¤erence
j
j+1
j+1
classes and, since (d1 ; :::; dT +tj ) < (d1 ; :::; djT +tj ), C j+1 has lower cost than C j . Given
j+1
that (dj+1
In case (2), let dbj+1 be
1 ; :::; dT ) = (d1 ; :::; dT ), let the recursion terminate.
the discriminatory capacity de…ned one stage prior to dj+1 in the iteration. So n
dbj+1
dbj+1
1
T +tj . We may now apply the three lemmas just as in the previous paragraph:
j+1
since, by Lemma 2, n dj+1
dj+1
-checklist
1
T +tj +1 , Theorem 3 implies there is a quick d
j+1
C
of length T + tj+1 , where tj+1 = tj + 1, that leads to a preference with n indi¤erence
classes, which by Lemma 3 can satisfy e(Cij+1 ) = dj+1
for i = 1; :::; T +tj+1 . By Lemma 1,
i
j+1
C j+1 has lower cost than C j . If (dj+1
;
:::;
d
)
=
(d
;
1 :::; dT ), let the recursion terminate.
1
T
j+1
j+1
j
j
Since (d1 ; :::; dT ) < (d1 ; :::; dT ) the recursion must terminate and since at terminaj+1
tion (dj+1
1 ; :::; dT ) = (d1 ; :::; dT ), we have the desired d-checklist.
Proof of Theorem 5.
There are n(n2 1) unordered pairs of elements in Xn and for any x; y 2 Xn an asymmetric relation must specify one of three possibilities (x y, y x, or (not x y and
n(n 1)
not y x)). Hence the number of asymmetric relation on Xn is given by 3 2 . Now
for each asymmetric binary relation n 1 on Xn 1 we may select a binary relation R on
Xn that ranks only n relative to the elements of Xn 1 (each ordered pair in R has n as
(n 1)(n 2)
2
one of its coordinates) such that n 1 [ R is in ir (n). Hence 3
can serve as a
lower bound for j ir (n)j.
The number of ways that Xn can be partitioned into cells is given by the Stirling
number of the second kind S(n; ) (see Goldberg et al. (1972) and Knuth (1997a) and for
the formula for S(n; ) we use below). The number of asymmetric binary relations on
(
1)
the domain f1; :::; g is 3 2 , a total that includes all of the asymmetric binary relations
(
1)
with m equivalence classes for any m
. Hence S(n; k)3 2 provides an upper bound
on the number of asymmetric binary relations Cj on Xn with or fewer Cj -equivalence
32
classes. Finally, since a preference on Xn has at most n indi¤erence classes, a quick
d-checklist has no more than dlog ne criteria (Theorem 1). Hence an upper bound U (n)
on jq ir (n)j is given by
U (n) = S(n; k)3
(
dlog ne
1)
2
.
Since (Goldberg et al. (1972)) we may write S(n; ) as
S(n; ) =
1X
( 1)
! i=0
we have
S(n; )
Hence there is a r > 0 such that S(n; k)3
U (n)
(
1X
! i=0 i
1)
2
(rn )1+log
n
in ,
i
i
n
.
is bounded above by rn . Hence
= e(1+log
n)n ln r
.
So, for any …xed k > 0, ekn jq ir (n)j is bounded above by e(1+log
(n 1)(n 2)
(n 1)(n 2)
ln 3
2
2
, equals e
and since
bound for j ir (n)j, 3
lim
(n
n!1
1)(n
2
2)
ln 3
n)n ln r+kn
((1 + log n)n ln r + kn)
. Since our lower
= 1,
ekn jjq irir (n)j
! 0 as n ! 1.
(n)j
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