Recognizing Members of the Tournament Equilibrium Set is NP-hard

0.08 MB·English

by  Felix Brandt

Recognizing Members of the Tournament Equilibrium Set is
NP-hard
Felix Brandt Felix Fischer Paul Harrenstein
Institutfu¨rInformatik,Universita¨tMu¨nchen,Germany
{brandtf,fischerf,harrenst}@tcs.ifi.lmu.de
8
0
0
2 Abstract

n
Arecurringthemeinthemathematicalsocialsciencesishowtoselectthe“mostdesirable”elements
a
J givenabinarydominancerelationonasetofalternatives.Schwartz’stournamentequilibriumset(TEQ)

7 ranks among the most intriguing, but also among the most enigmatic, tournamentsolutions that have
been proposed so far in this context. Due to its unwieldy recursive definition, little is known about

] TEQ. In particular, its monotonicityremainsan open problemup to date. Yet, if TEQ were to satisfy
C
monotonicity,itwouldbeaveryattractivetournamentsolutionconceptrefiningboththeBankssetand
C
Dutta’s minimal covering set. We show that the problem of deciding whether a given alternative is
.
s containedinTEQisNP-hard.
c
[

2 1 Introduction
v
1
The central problem of the literature on tournament solutions is as appealing as it is simple: Given an ir-
6
9 reflexive, asymmetric, and complete binary relation over a set, findthe mostattractive elements of this set.
2
Asthestandardnotionofmaximalityisnotwell-definedinthepresenceofcycles,numerousalternativeso-
.
1 lutionconceptshavebeendevisedandaxiomatized(see,e.g.,Moulin,1986;Laslier,1997). Insocialchoice
1
theory, the base relation, which we call dominance relation, is commonly defined via pairwise majority
7
0 voting, andmanywell-knowntournament solutions yieldattractivesocialchoicecorrespondences.
: Over the years, the computational complexity of almost all common solution concepts has been com-
v
i pletely characterized (see, e.g., Woeginger, 2003; Conitzer, 2006; Brandt et al., 2007; Brandt and Fischer,
X
2007). Onenotableexceptionisthetournamentequilibriumset(TEQ)proposedbySchwartz(1990). Dueto
r
a itsunwieldyrecursivedefinition, littleisknownaboutTEQ(Dutta,1990;Laffondetal.,1993). However,if
acertaintechnicalconjecturestatedalmosttwodecadesagoturnedouttobetrue,itwouldconstituteoneof
themostattractive tournament solutions, refining both theminimalcovering setandtheBanksset (Laslier,
1997;Laffondetal.,1993). Laslierstatesthat“Unfortunately, noalgorithmhasyetbeenpublishedforfind-
ing the Minimal Covering set or the tournament equilibrium set of large tournaments. For tournaments of
order10ormore,itisalmostimpossibletofind(inthegeneralcase)thesesetsathand”(Laslier,1997,p.8).
While it has recently been shown that computing the minimal covering set is feasible in polynomial time
(Brandt and Fischer, 2007), itturns out that this is not the case for TEQ unless P equals NP.Weprove this
by first providing an alternative hardness proof for the Banks set which is then modified so as to apply to
TEQas well. Incontrast tothe Banks set, there isno obvious reason tosuppose that the TEQmembership
problem isinNP;itmayverywellbeevenharder.
1
2 Preliminaries
Atournament T isapair (A,≻),where Aisafiniteset ofalternatives and ≻anirreflexive, anti-symmetric,
and complete binary relation on A, also referred to as the dominance relation. Intuitively, a ≻ b signifies
thatalternativeabeatsbinapairwisecomparison. WewriteT fortheclassofalltournaments andwehave
T(A)denotethesetofalltournaments onafixedsetAofalternatives. IfT isatournamentonA,thenevery
subset X ofAinducesatournament T| definedas(X,≻| ),where≻| = {(x,y) ∈ X×X: x ≻ y}.
X X X
As the dominance relation is not assumed to be transitive in general, it may contain cycles. Moreover,
thereneednotbeaso-calledCondorcetwinner,i.e.,analternativethatdominatesallothers. Thismakesthat
theusual notions ofmaximumandmaximalelementsarenolongerfeasible inthiscontext. Otherconcepts
havebeensuggested totakeovertheroleofsingling outthosealternatives thatcansomehowbeconsidered
the“winners” ofthetournament. Formally, atournament solution S isdefined asafunction thatassociates
witheachtournamentT onAasubsetS(T)ofA. Thedefinitionofatournamentsolutioncommonlyincludes
the requirement that S(T)benon-empty ifT isdefined onanon-empty setof alternatives andthat itselect
the Condorcet winner if there is one (Laslier, 1997, p.37). For X a subset of A, we also write S(X) for
the more cumbersome S(T| ), provided that the tournament T is known from the context. In this paper
X
we will be concerned with two particular tournament solutions, the Banks set and Schwartz’s tournament
equilibrium set(TEQ).Foraproperdefinition, however,weneedsomeauxiliary notionsandnotations.
Let R be a binary relation on a set A. We write R∗ for the transitive reflexive closure of R. By the top
cycle TC (R) we understand the maximal elements of the asymmetric part of R∗. A subset X of A is said
A
to be transitive ifR istransitive on X. For X ⊆ Y ⊆ A, X iscalled maximal in Y ifno proper superset of X
inY istransitive, i.e.,ifthereisnotransitive Z ⊆ Y with X ⊂ Z. Clearly, everytransitive setiscontained in
a maximal transitive set. Given a set Z = {Z} of pairwise disjoint subsets of A, a subset X of A will be
i i∈I
calledachoicesetforZ ifitcontains precisely oneelementfromeachsubsetZ ∈ Z.
i
Intournaments, maximal transitive setsare also referred toasBankstrajectories. TheBanksset BA(T)
ofatournamentT thencollectsthemaximalelementsoftheBankstrajectories.
Definition1(Banksset) LetT beatournament on A. Analternative a ∈ AisintheBanksset BA(T)ofT
ifaisamaximalelementofsomemaximaltransitive setinT.
The tournament equilibrium set of a tournament T on A is defined as the top cycle of a particular
subrelation of the dominance relation, the TEQ relation. The underlying idea is that an alternative is only
“properly” dominated, i.e., dominated according to the subrelation, if it is dominated by an element that is
selected by some tournament solution concept S. To make this idea precise, for X ⊆ A, wewrite D (a) =
X
{b ∈ X: b ≻ a}forthedominatorsofainX,omittingthesubscriptwhenX = A. Thus,foreachalternativea
one examines the dominator set D(a), and solves the subtournament T| by means of the solution S. In
D(a)
the subrelation a is then only dominated by the alternatives in S(D(a)). This of course, still leaves open
the question as to the choice of the solution concept S. Now, in the case of TEQ, S is taken to be TEQ
itself! The reason why this is a proper recursive definition is that if a ∈ X for some X ⊆ A the set D (a)
X
of dominators of a in X is a proper subset of X. I.e., in order to determine the TEQ relation in a particular
subtournament T,onehastocalculate theTEQofapropersubtournament ofT.
Definition2(Tournamentequilibriumset) LetT beatournamentonA. ForeachsubsetXofAwedefine
thetournament equilibrium setTEQ(X)for X asfollows:
TEQ(X) = TC (→ ),
X X
where→ isdefinedasthebinaryrelation onX suchthatforall x,y ∈ X,
X
x→ yifandonlyif x∈ TEQ(D (y)).
X X
2
a
b c
d e
Figure1: ExampleduetoSchwartz,1990,whereBA(T)= {a,b,c,d}andTEQ(T) = {a,b,c}. Therelation→
isindicated bythickedges.
ObservethattheTEQrelation → isinvariably asubset ofthedominance relation ≻andthatif D (x) , ∅,
X X
thenthereissomey∈ D (x)withy → x.
X X
ItcaneasilybeestablishedthattheBankssetandTEQbothselecttheCondorcetwinnerinatournament
ifthere is one. Moreover, in acyclic tournament onthree alternatives, the Banks set and TEQboth consist
ofallalternatives. Yet,theBankssetandTEQdonotcoincideforalltournaments. Foranexampleconsider
the tournament depicted in Figure 1. First we calculate the TEQ relation →. Observe that D(a) = {c}.
Hence, c is the Condorcet winner in D(a) and we have TEQ(D(a)) = {c} and c → a. For alternative b we
have D(b) = {a,e}. Since a ≻ e, alternative a is the Condorcet winner in D(b), and we may conclude that
TEQ(D(b)) = {a}. Hence, a → b whereas e 6→ b. In an analogous fashion we find for all x ∈ {a,b,c,d,e}
that x → c if and only if x = b as well as that x → d if and only if x = a. For alternative e, however,
D(e) = {a,c,d}. Since a ≻ d ≻ c ≻ a, a three-cycle, we have that TEQ(D(e)) = {a,c,d} and hence a → e,
c → e as well as d → e. The top cycle of the relation → thus found can then be seen to coincide with
the set {a,b,c}, which then also constitutes the TEQ of this tournament. By contrast, the Banks set in this
example consists of the four elements a, b, c and d. Maximal transitive sets of which a, b, c and d are the
maximal elements are, e.g., {a,b,d}, {b,c}, {c,a,e} and {d,c,e}, respectively. Alternative e is not included
intheBanksset. Theonlytransitive subsets ofwhicheisthemaximalelementare{e}and{e,b}. However,
both these sets are included in maximal transitive sets of which e is not the maximal element, e.g., the set
{a,e,b}. Thus TEQ and the Banks set may differ. However, the former is known to always be included in
thelatter.
Proposition 1(Schwartz,1990) LetT = (A,≻)beatournament. Then,TEQ(T) ⊆ BA(T).
Proof: Weprovebystructural induction on X thatTEQ(X) ⊆ BA(X)forallsubsets X of A. Thecase X = ∅
istrivial, asthenTEQ(X) = BA(X) = ∅. So,assume thatTEQ(X′) ⊆ BA(X′),forall X′ ( X. Weprove that
TEQ(X) ⊆ BA(X)as well. Tothis end, consider an arbitrary a ∈ TEQ(X). Either D (a) = ∅ or D (a) , ∅.
X X
In the former case, ais the Condorcet winner in X and therefore a ∈ BA(X). Inthe latter case, x → a for
X
some x ∈ X. Having assumed that a ∈ TEQ(X), i.e., a ∈ TC(→ ), there is also an x′ ∈ X with a → x′.
X X
Accordingly, a ∈ TEQ(D (x′)). Bytheinduction hypothesis, alsoa∈ BA(D (x′)). Therefore,thereissome
X X
maximaltransitivesetY inD (x′)ofwhichaisthemaximalelement. Then,Y∪{x′}isatransitivesetinX.
X
NowletY′ ⊆ XbeamaximaltransitivesetinXcontainingY∪{x′}witha′asmaximalelement. Observethat
a′ ∈ BA(X). Then,a′ ≻ x′andsoa′ ∈ D (x′). NowconsiderY′∩D (x′). Clearly,Y∩D (x′)isatransitive
X X X
setin D (x′)whichcontains a′ asitsmaximalelement. Moreover, Y ⊆ Y′∩D (x′). Bymaximality ofY it
X X
thenfollowsthatY = Y′∩D (x′)andthata = a′. Wemayconclude thata ∈ BA(X). (cid:3)
X
In the remainder of this paper, we assume the reader to be familiar with the well-known complexity
classesPandNPandthenotionofpolynomial-time reducibility (see,e.g.,Papadimitriou, 1994).
3
Algorithm1TournamentEquilibrium Set
procedureTEQ(X)
R ← ∅
B←C ← argmin |D(a)|
a∈X
loop
R ← R∪{(b,a): a∈C∧b ∈ TEQ(D(a))}
D ← S TEQ(D(a))
a∈C
if D ⊆ BthenreturnTC (R)endif
B
C ← D
B← B∪C
endloop
3 A Heuristic for Computing the Tournament Equilibrium Set
ItisnotveryhardtoseethatthenaivealgorithmforcomputingTEQ,whichsimplyrecursivelycomputes→,
requires exponential time in the worst case. Running time can be greatly reduced by using a heuristic that
relies on the conjecture that the top cycle of any TEQ relation consists of only one strongly connected
component. ThisconjecturewasalreadymadebySchwartz(1990)andhaslaterbeenshowntobeequivalent
to the (conjectured) monotonicity of TEQ (Laffond et al., 1993). Algorithm 1 computes TEQ by starting
with the set B of all alternatives that have dominator sets of minimal size. These alternatives are likely to
be included in TEQ and the small size of their dominator sets speeds up the determination of their TEQ-
dominators. In the following, all alternatives that TEQ-dominate any alternative in B are iteratively added
to B until no more such alternatives can be found, in which case the algorithm returns the top cycle of
B. Clearly, the worst-case running time of this algorithm is still exponential, and it will be shown in the
remainderofthepaperthatthishastobethecaseforeveryalgorithm computing TEQunless P = NP.
4 An Alternative NP-Hardness Proof for Membership in the Banks Set
Theproblem ofdeciding whetheraparticular alternative isincluded intheBankssetofagiventournament
T isknown to beNP-complete. Thiswasfirstdemonstrated by Woeginger (2003) bymeans of areduction
from graph three-colorability. Here we will give an alternative proof of this result. Our proof works by a
reduction from 3SAT, the NP-complete satisfiability problem for Boolean formulas in conjunctive normal
form with exactly three literals per clause (see, e.g., Papadimitriou, 1994). The construction used in this
paperisarguably simplerthanWoeginger’s. Moreover,amuchsimilarconstruction willbeusedinthenext
section to prove NP-hardness of the analogous decision problem for TEQ. The tournaments used in these
reductions willbothbetakenfromaspecialclassT∗,whichweintroduce next.
Definition3(TheclassT∗) A tournament (A,≻) is in the class T∗ if it satisfies the following properties.
There is some odd integer n ≥ 1, the size of the tournament, such that A = C ∪ U ∪ ··· ∪ U , where
1 n
C,U ,...,U are pairwise disjoint and C = {c ,...,c }. Each U is a singleton if i is even, and U =
1 n 0 n i i
{u1,u2,u3} if i is odd. The complete and asymmetric dominance relation ≻ is such that the following five
i i i
properties holdforallc ∈C ,c ∈C ,u ∈ U ,u ∈ U (0 ≤ i, j≤ n):
i i j j i i j j
(i) c ≻ c , ifi < j,
i j
(ii) u ≻ c , ifi= j,
i j
(iii) c ≻ u, ifi, j,
j i
4
(iv) u ≻ u , ifi< jandatleastoneofiand jiseven (i.e.,u oru isaseparating node),
i j i j
(v) uk ≻ ul, ifiisoddandk ≡ l−1 (mod 3) (i.e.,u1 ≻ u2 ≻ u3 ≻ u1).
i i i i i i
Wealsorefertoc byd,for“decisionnode”andtoS U byU. Fori = 2k,wehaveasanotational
0 1≤i≤n n
convention U = Y = {y }andsetY = S Y . Thesenodesarecalled separating nodes.
i k k 1≤2k≤n k
Observethatthisdefinitionfixesthedominancerelationbetweenanytwoalternativesexceptforsomepairs
ofalternatives thatarebothinU.
As a next step in the argument, we associate with each instance of 3SAT atournament in the class T∗.
An instance of 3SAT is given by a formula ϕ in 3-conjunctive normal form (3CNF), i.e., ϕ = (x1 ∨ x2 ∨
1 1
x3)∧···∧(x1 ∨ x2 ∨ x3), whereeach x ∈ {x1,x2,x3: 1 ≤ i ≤ m}isaliteral. Foreachclause x1∨ x2 ∨ x3
1 m m m i i i i i i
weassume x1, x2 and x3 tobedistinct literals. Wemoreoverassumetheliterals tobeindexedandby X we
i i i i
denote the set {x1,x2,x3}. For literals x we have x¯ = ¬p if x = p, and x¯ = p if x = ¬p, where p is some
i i i
propositional variable. We may also assume that if x and y are literals in the same clause, then x , y¯. We
say a 3CNF ϕ = (x1 ∨ x2 ∨ x3)∧···∧(x1 ∨ x2 ∨ x3) is satisfiable if there is a choice set V for {X}
1 1 1 m m m i 1≤i≤m
suchthatv′ = v¯ fornov,v′ ∈ V. Nextwedefineforeach3SAT formulaϕthetournament TBA.
ϕ
Definition4(Banksconstruction) Letϕ bea3CNF(x1 ∨ x2 ∨ x3)∧···∧(x1 ∨ x2 ∨ x3). Define TBA =
1 1 1 m m m ϕ
(C∪U,≻)asthetournamentintheclassT∗ ofsize2m−1suchthatforall1≤ j< 2m,
X if j= 2i−1,
U =  i
j {yi} if j= 2i
andsuchthatforall x∈ X and x′ ∈ X (1 ≤ i, j ≤ m),
i j
x≻ x′ ifboth j< iand x′ = x¯orbothi< jand x′ , x¯.
Observe that in conjunction withthe other requirements on the dominance relation ofa tournament inT∗,
thiscompletely fixesthedominancerelation ≻ofTBA.
ϕ
AnexampleforsuchatournamentisshowninFigure2. WearenowinapositiontoproveNP-completeness
ofdeciding whetheraparticular alternative isintheBanksset.
Theorem1 Theproblemofdecidingwhetheraparticular alternative isintheBankssetofatournamentis
NP-complete.
Proof: Membership in NP is obvious. For a fixed alternative d, we can simply guess a transitive subset of
alternatives V withdasmaximalelementandverifythatV isalsomaximalw.r.t.setinclusion.
For NP-hardness, we show that TBA contains a maximal transitive set with maximal element d if and
ϕ
only if ϕ is satisfiable. First observe that V is a maximal transitive subset with maximal element d in TBA
ϕ
onlyif
(i) forall1≤ i < 2mthereisau∈ U suchthatu ∈ V,and
i
(ii) thereareno1 ≤ i< j < 2m,u ∈ U,u′ ∈ U withu,u′ ∈ V suchthatu ≻ u.
i j j i
Regarding (i), ifthere is an 1 ≤ i < 2m such that noelement of U iscontained inV, wecan always add c
i i
to V in order to obtain a larger transitive set. If (ii) were not to hold, both i and j have to be odd for u to
j
dominate u. However, in light of (i), there has to be k with i < k < j and u′′ ∈ U such that u′′ ∈ V. It
i k
follows that V isnot transitive because u, u′′, and u′ form acycle. Ifthere ismaximal transitive set V with
5
c
5
c
4
c
3
c
2
c
1
d
¬p s q
y
1
p s r
y
2
p q ¬r
Figure2: TournamentTBA forthe3-CNFformulaϕ = (¬p∨s∨q)∧(p∨s∨r)∧(p∨q∨¬r).
ϕ
maximalelementdcomplying withboth(i)and(ii),asatisfying assignmentofϕcanbeobtainedbyletting
allliteralscontained inX∩V betrue.
For the opposite direction, assume that ϕ is satisfiable. Then there is a choice set W for {X} such
i 1≤i≤m
that x′ = x¯ for no x,x′ ∈ W. Obviously V = W ∪ {y ,...,y } ∪ {d} does not contain any cycles and
1 m−1
thusistransitive withmaximalelementd. Inordertoobtain alarger transitive setwithadifferent maximal
element, weneed toadd c forsome 1 ≤ i ≤ m to V. However, V ∪{c} always contains acycle consisting
i i
of c, d, and u for some u ∈ U , contradicting the transitivity of V ∪{c}. We have thus shown that d is the
i i i
maximalelementofsomemaximaltransitive setinTBA containing V asasubset. (cid:3)
ϕ
5 NP-hardness of Membership in the Tournament Equilibrium Set
Asthemainresultofthispaper,wedemonstratethattheproblemofdecidingwhetheraparticularalternative
isintheTEQofatournamentisNP-hard. Tothisend,werefinetheconstructionusedintheprevioussection
toproveNP-completeness ofmembershipintheBanksset.
Definition5 Let ϕ a 3-conjunctive normal form (x1 ∨ x2 ∨ x3)∧···∧(x1 ∨ x2 ∨ x3). Further for each
1 1 1 m m m
1 ≤ i < m,letthere beasetZ = {z1,z2,z3}.DefineTTEQ asthetournament (A,≻)inT∗ ofsize4n−3such
i i i i ϕ
that A =C∪U ∪···∪U andforall1≤ i ≤ m,
1 4n−3
X if j= 4i−3,
Uj = Zii if j= 4i−1,
{yi} otherwise.
AsintheBanksconstruction, weletforall x∈ X and x′ ∈ X (1 ≤ i, j≤ m)
i j
x≻ x′ ifboth j< iand x′ = x¯orbothi< jand x′ , x¯.
6
c
9
c
8
c
7
c
6
c
5
c
4
c
3
c
2
c
1
d
¬p s q
y
1
z1 z2 z3
1 1 1
y
2
p s r
y
3
z1 z2 z3
2 2 2
y
4
p q ¬r
Figure3: TournamentTTEQ forthe3-CNFformulaϕ = (¬p∨ s∨q)∧(p∨ s∨r)∧(p∨q∨¬r). Omitted
ϕ
edgesareagainassumedtopointdownwards.
Finally,forall1≤ i, j ≤ m, xk ∈ X andzl ∈ Z ,
i i j j
xk ≻ zl ifandonlyifi < jorbothi = jandk = l.
i j
Anexampleforsuchatournament isshowninFigure3.
Wenowproceed toshowthata3SAT formulaϕissatisfiableifandonlyifthedecision noded isinthe
tournament equilibrium setofTTEQ.
ϕ
Lemma1 Let T = (C ∪U,≻)be a tournament in T∗ and let B ⊆ C ∪U such that d ∈ B. Then, for each
u ∈ U ∩Bthereexistsc ∈C∩Bsuchthatc→∗ u
B
Proof: Let c ∈ C ∩ B be such that D (c)∩C = ∅, i.e., c is the alternative in C with the highest index
i B i i
amongthoseincluded in B. Then,
c → cforallc∈ B∩C withc, c. (1)
i B i
7
Forthis, merelyobserve thatbyconstruction c istheCondorcet winner in D (c). Hence, c ∈ TEQ(D (c))
i B i B
andc → c.
i B
Thelemmaitself thenfollows fromthestronger claim thatforeach u ∈ U ∩Bthereissomec ∈ C ∩B
withbothc→∗ uandc∈ TEQ(B). Thisclaimweprovebystructural induction onsupersets Bof{d}.
B
If B = {d}, U ∩ B = ∅ and the claim is satisfied trivially. So let {d} be a proper subset of B. Again, if
U ∩ B = ∅, the claim holds trivially. So we may assume there be some u ∈ U ∩ B. Then, d ∈ D (u) by
B
constructionofT. IfD (u)∩U = ∅,D (u)isanon-emptysubsetofC∩B,andsoisTEQ(D (u)). Itfollows
B B B
that for some c ∈ TEQ(D (u))∩C we have c → u. If, on the other hand, D (u)∩U , ∅, the induction
B B B
hypothesis isapplicable and wehave c ∈ TEQ(D (u)) forsome c ∈ C ∩B. Hence, c → u. Withuhaving
B B
been chosen arbitrarily, we actually have that for all u ∈ U ∩ B, there is some c ∈ C ∩ B with c → u. It
B
remainstobeshownthatthereissomec ∈C∩TEQ(B)withc →∗ u.
B
Tothisend,againconsider c ∈ C∩Bsuchthat D (c)∩C = ∅. Itsuffices toshowthatc →∗ bforall
i B i i B
b ∈ B, as then both c ∈ TEQ(B)∩C and c →∗ u. So, consider an arbitrary b ∈ B. If b = c, the case is
i i B i
trivial. Ifb ∈ C ∩Bbutb , c,wearedone by(1). Ifinstead b ∈ U ∩B,then c →∗ bforsomec ∈ C ∩B,
i B
aswehaveshowninthefirstpartoftheproof. Ifc= c,wearedone. Otherwise,wecanapply(1)toobtain
i
c → c′ →∗ bandhencec →∗ b. (cid:3)
i B B i B
Theorem2 Decidingwhetheraparticularalternative isinthetournamentequilibriumsetofatournament
isNP-hard.
Proof: By reduction from 3-SAT. Consider an arbitrary 3-DNF ϕ and construct the tournament TTEQ =
ϕ
(C∪U,≻). Thiscanbedoneinpolynomial time. Weshowthat
ϕissatisfiableifandonlyifd ∈ TEQ(TTEQ).
ϕ
For the direction from left to right, observe that by an argument analogous to the proof of Theorem 1 it
can be shown that ϕ is satisfiable if and only if d ∈ BA(TTEQ). So assuming that ϕ is not satisfiable yields
ϕ
d < BA(TTEQ). Bytheinclusion ofTEQintheBanksset(Proposition 1),itfollowsthatd < TEQ(TTEQ).
ϕ ϕ
For the opposite direction, assume that ϕ is satisfiable. Then there is a choice set W for {X} such
i 1≤i≤m
that x′ = x¯ for no x,x′ ∈ W. Obviously W ∪{y ,...,y }∪{zj ∈ Z: xj ∈ W} = {u ,...,u } contains no
1 m−1 i i 1 n
cycles and thus is transitive. Without loss of generality we may assume that u ∈ U for all 1 ≤ i ≤ n. For
i i
each1 ≤ i≤ n+1,defineasubsetDu1,...,un ofalternativesasfollows. SetDu1,...,un = AandletDu1,...,un denote
i n+1 i
Ti≤j≤n+1D(uj) for each 1 ≤ i ≤ n. Hence, D1u1,...,un ( ··· ( Dun1+,1...,un. In an effort to simplify notation, we
write →i and Di(x) for →Du1,...,un and DDu1,...,un(x), respectively. We will also assume u1,...,un to be fixed
andwrite D for Du1,...,un. Itithensufficestioprovethat
i i
d ∈ TEQ(D ),forall1 ≤ k ≤ n+1. (2)
k
Thetheorem then follows asthespecial case inwhich k = n+1. Wefirstmake thefollowing observations
concerning theTEQrelation→ ineach D,foreach1 ≤ i, j≤ n+1:
i i
(i) u ∈ D ifandonlyif j< i,
j i
(ii) c ∈ D ifandonlyif j< i,
j i
(iii) ci →i+1 cj if j < i≤ n,
(iv) ui →i+1 ci,ifi ≤ n.
8
For(i),observethatif j< i,u ∈ D(u)bytransitivity oftheset{u ,...,u }. Hence,u ∈ D. Ifontheother
j i 1 n j i
hand j ≥ i,then u < D(u )andthus u < D. For(ii), observe thatc ∈ D(u)foralli , jand thus c ∈ D
j j j i j i j i
if j < i. However, c < D(u ) and hence c < D if j ≥ i. For (iii), merely observe that c is the Condorcet
j j j i i
winner in Di+1(cj), if j < i ≤ n. Toappreciate (iv), observe that by construction Di+1(ci) has to be either a
singleton {u} for some u ∈ U , or U itself. The former is the case if U ⊆ Y, or if U ⊆ X and i , n. The
i i i i i i
latter holds if Ui = Un or if Ui ⊆ Z. In either case, TEQ(Di+1(ci)) = Di+1(ci) and ui →i+1 ci holds. For
the case in which Ui ⊆ X with i , n, let Ui = {ui,u′i,u′i′}. Byconstruction, Ui+2 ⊆ Z and u′i,u′i′ < D(ui+2).
Accordingly, u′i,u′i′ < Di+1. Formui ∈ Di+1 itthenfollowsthat Di+1 ∩Ui = {ui}.
Wearenowinapositiontoprove(2)byinductiononk. Fork = 1,observethatdisaCondorcetwinner
in D and thus d ∈ TEQ(D ). For the induction step, let k = i + 1. With observation (i) we know that
1 1
ui ∈ Di+1 and, in virtue of the induction hypothesis, also that d ∈ TEQ(Di). Hence, d →i+1 ui. Moreover,
by observations (iii) and (iv), ci →i+1 d →i+1 ui →i+1 ci, i.e., ci, d and ui constitute a →i+1-cycle. In
virtue of Lemma 1 and observation (ii), we may conclude that ci →∗i+1 a for all a ∈ Di+1. Accordingly,
{ci,d,ui} ⊆ TCDi+1(→i+1)andd ∈ TEQ(Di+1),whichconcludes theproof. (cid:3)
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