Fooling Sets and the Spanning Tree Polytope

0.09 MB·

by  Kaveh Khoshkhah

Fooling Sets and the Spanning Tree Polytope
Kaveh Khoshkhah, Dirk Oliver Theis∗
InstituteofComputerScience UniversityofTartu
ofthe
U¨likooli17,51014Tartu,Estonia
{kaveh.khoshkhah,dotheis}@ut.ee
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Mon Jan 2 11:58:01 EET 2017
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n
a Abstract
J
Inthestudyofextensionsofpolytopesofcombinatorialoptimizationproblems,anotori-
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ousopenquestionisthatforthesizeofthesmallestextendedformulationoftheMinimum

SpanningTreeproblemonacompletegraphwithnnodes. Thebestknownlowerboundis
]
M Ω(n2),thebestknownupperboundisO(n3).
Inthisnoteweshowthatthevenerablefoolingsetmethod cannotbeusedtoimprove
D thelowerbound:everyfoolingsetfortheSpanningTreepolytopehassizeO(n2).
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s Keywords: Polyhedral Combinatorial Optimization, Extended Formulations, Lower
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Bounds.
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v 1 Introduction
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5 ([n])
3 The Spanning Tree polytope Pn has as its vertices the characteristic vectors R 2 of edge-
setsoftreeswithnodeset[n]:={1,...,n}. Acompletesystemofinequalitiesandequations
0
0 isthefollowing:
.
1
0 xe =n−1
7 e∈X([n])
1 2
: x ≤|S|−1 ∀S ⊂[n], |S|>1
v e
i eX∈(S)
X 2
[n]
r x ≥0 ∀e∈ .
a e (cid:18) 2 (cid:19)
Thissystemhasexponentiallymanyfacet-defininginequalities. Thereisaclassicalextended
formulation by Wong/Martin [10, 13] with O(n3) inequalities (and variables). A notorious
open problem in polyhedral combinatorial optimization, highlighted by M. Goemans at the
2010 Carge`se Workshop on Combinatorial Optimization, asks whether or not an extended
formulationwith o(n3)inequalitiesexists. TheonlyknownlowerboundisΩ(n2)—this isa
“trivial”lowerbound(itisthedimensionofthespanningtreepolytope).
One method for proving lower bounds to sizes of extended formulations is the so-called
fooling-set method (see, e.g., [9]). For that method, one looks at the function f which maps
every pair of an inequality and a solution to 0, if the solution satisfies the inequality with
equation,andotherwiseto1. A foolingsetofsize r isa list ofpairsF = (s1,t1),...,(sr,tr)
(cid:0) (cid:1)
∗SupportedbytheEstonianResearchCouncil,ETAG(EestiTeadusagentuur),throughPUTExploratoryGrant
#620,andbytheEuropeanRegionalDevelopmentFundthroughtheEstonianCenterofExcellenceinComputer
Science,EXCS.
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such that f(s ,t ) = 1 for j = 1,...,r, and f(s ,t )f(s ,t ) = 0 for i 6= j. If a fooling set of
j j i j j i
sizer exists,theneveryextendedformulationmustcontainatleastr inequalities.
Itisaneasyfolklorefactthateverypolytopehasafoolingsetofwhosesizethedimension
ofthepolytope. Butfoolingsetlowerboundshavetheirlimitations. Mostimportantly,there
can never be a fooling set larger than the square of the dimension of the polytope ([5]; see
also[2,11]). However,sincethatboundisknowntobetight[7,6],andthedimensionofthe
spanningtreepolytopeisΩ(n2),afoolingsetboundofΩ(n3)fortheSpanningTreepolytope
is possible. A small improvement was made in [12], where a O(n8/3logn) upper bound for
thelargestpossiblefoolingsetfortheSpanningTreepolytopeisshown.
Inthisnote,weprovethatthelargestsizeofafoolingsetintheSpanningTreepolytope
isO(n2).
Theorem1. FoolingsetsforthespanningtreepolytopehavesizeO(n2).
Webelievethatourtoolsmaybe usefulforotherattemptstogiveupperboundson sizes
offoolingsets.
2 Proof of Theorem 1
We start by making precise the function f. Since the number of nonnegativity inequalities
“x ≥0”isO(n2),andweaimforanO(n2)upperbound,wecanomitthem. Wedefine
e
S := S ([n] |S|>1
n (cid:12) o
[n(cid:12)]
T := T ⊂ (cid:12) T tree .
n (cid:18) 2 (cid:19)(cid:12) o
(cid:12)
(cid:12)
Thefunctionf isnowthefollowing:
0, ifthesub-forestofT inducedbyS isatree,
f: S ×T : (S,T)7→ i.e.,T ∩ S isconnected;
 2
1, otherwis(cid:0)e.(cid:1)

Weusetheterminology“S isconnectedinT”forthecasef(S,T)=0.
NowletF = (S1,T1)...,(Sr,Tr) beafoolingsetofsizerforf. Wedefinean(r×r)-matrix
H byHi,j :=f(S(cid:0)i,Tj). Wewillprove(cid:1)thefollowing.
Lemma2(MainLemma). EverycolumnofH hasatmostn(n−1)zeroentries.
TheMainLemmaimmediatelyimpliesTheorem1.
ProofofTheorem1fromLemma2. ThatF isafoolingsetmeansthatH (has1sonthediag-
onal, and)in each pair ofdiagonallyoppositeoff-diagonalentries, atleast oneis zero. So H
hasatleastΩ(r2)zeroentries. ButtheMainLemmaimpliesthatH hasatmostO(rn2)zero
entries.
Intheremainderofthesection,wewillprovetheMainLemma.
Westartwiththefundamentaldefinitionoftheproof.
Definition 3 (Witness). Let S ∈ S, T ∈ T. A witness for f(S,T) = 1 is a triple (a,x,b) of
nodeswhichsatisfies
(a) a,b∈S,x∈/ S;and
(b) xliesonthe(unique)pathinT betweenaandb.
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Let us justify the terminology. Clearly, if a witness exists, S cannot be connected in T,
andhencef(S,T)=1. Ontheotherhand,iff(S,T)=1,i.e.,thesub-forestofT inducedbyS
has multiple connected components, then there is a node x ∈/ S with the property that for
everya ∈ S thereis ab ∈ S such thatx lieson thepath in T betweena andb. (To findsuch
anxshrinktheconnectedcomponentsto“supernodes”,andletxbe any(non-shrunk)node
on a path between two “super nodes”.) Hence, a witness for f(S,T) = 1 exists if and only if
f(S,T)=1.
Clearly,if(a,x,b)isanwitnessforf(S,T)=1,thensois(b,x,a).
Lemma4. If(a,x,b)isawitnessforbothf(S ,T )=1andf(S ,T )=1theni=j.
i i j j
Proof. If(a,x,b)isawitnessforbothf(S ,T )=1andf(S ,T )=1,thenitisalsoawitness
i i j j
for both f(S ,T ) = 1 and f(S ,T ) = 1. But since F is a foolingset, this can only happen if
i j j i
i=j.
Lemma 5. Let S ∈ S, T ∈ T, and a,b,c ∈ S. If at least one of (a,x,b), (b,x,c), (c,x,a) is a
witnessforf(S,T)=1,thenatleasttwoofthemare.
Proof. Condition (a) of Definition 3 is satisfied by either none or all of the triples. As for
Condition (b), if say, x were on the path between a and b in T, but x was neither on the
pathbetweenbandcnoronthepathbetweencanda,thenT wouldhavetwodistinctpaths
betweenaandb,onethroughxandoneavoidingx—acontradiction.
Lemma 6. LetS ∈ S, T,T′ ∈ T,andx ∈ [n] suchthatW 6= ∅andS is connectedinT′.
x,S,T
Thenthereisanedge{a,b}∈T′∩ S with(a,x,b)∈W .
2 x,S,T
(cid:0) (cid:1)
Proof. ConsiderthetreeQwithnodesetS andedgesetT′∩ S . Amongallpairs{a,b}with
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(a,x,b) ∈ Wx,S,T take oneforwhichthedistance betweena a(cid:0)nd(cid:1)bin Qisminimal. Weclaim
that{a,b}∈ Q. Indeed,byLemma5,iftherewasaconthepath inQbetweenaandbwith
c∈/ {a,b},thenforatleastoneof{a,c}or{c,b}thecorrespondingtripletwithxinthemiddle
isinW ,andthedistanceinQisshorter.
x,S,T
Foreveryx∈[n](anode)andk ∈[r](acolumnofH),denotebyZ thesetofi∈[r](rows
x,k
ofH)forwhich
• S isconnectedinT ;and
i k
• thereexista,b∈S suchthat(a,x,b)isawitnessforf(S ,T )=1.
i i i
Lemma7. Foreveryx∈[n]andk ∈[r],wehave|Z |≤n−1.
x,k
Proof. For each i ∈ Z , by Lemma 6, there is an edge {a ,b } ∈ T such that (a,x,b) ∈
x,k i i k
W .
x,Si,Ti
ByLemma4,theedges{a ,b },i∈Z ,aredistinct. Hence|Z |≤n−1.
i i x,k x,k
TheproofoftheMainLemmaisnowalmostcomplete.
ProofofLemma2. Letk ∈[r]betheindexofacolumnofH. Foreveryi∈[r]withf(S ,T )=
i k
0, there is an x such that i ∈ Z . Hence the number of zero-entriesin that column of H is
x,k
atmost
n
|Z |,
x,k
xX=1
which,byLemma7isatmostn(n−1).
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3 Conclusion and Outlook
We have settled the question whether the fooling set method can be used to prove a non-
trivial lower bound for the extension complexity (i.e., smallest number of inequality in an
extended formulation) of the Spanning Tree polytope. There are two more, stronger, com-
binatorial lower bounds which are frequently used: the rectangle covering number and the
fractionalrectanglecoveringnumber. Thenextstepwouldbetofindoutifanyofthesegrow
more quickly than n2. In summary, though, despite the interest in the problem (e.g., [1, 8])
andpartialresultsandapproaches(e.g.,[3,4]),theproblemisnowmoreopenthanever.
Acknowledgments
This research was supported by the Estonian Research Council, ETAG (Eesti Teadusagen-
tuur),throughPUTExploratoryGrant#620. Wealsogratefullyacknowledgefundingbythe
European Regional DevelopmentFund through the Estonian Center of Excellence in Com-
puterScience,EXCS.
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