Table Of ContentSYMMETRY IN EXTENDED FORMULATIONS OF THE
PERMUTAHEDRON
1
1 KANSTANTSINPASHKOVICH
0
2
ABSTRACT. It is well known that the permutahedron Πn has 2n −2 facets.
n TheBirkhoffpolytopeprovidesasymmetricextendedformulationofΠnofsize
a Θ(n2).Recently,Goemansdescribedanon-symmetricextendedformulationof
J
ΠnofsizeΘ(nlogn). Inthispaper,weprovethatΩ(n2)isalowerboundfor
4
2 thesizeofsymmetricextendedformulationsofΠn.
]
O
C 1. INTRODUCTION
.
h Extended formulations of polyhedra have gained importance in the recent past,
t
a becausethisconceptallowstorepresentapolyhedronbyahigher-dimensional one
m
withasimplerdescription.
[ Toillustratethepowerofextendedformulationswetakealookatthepermutahe-
2 dronΠn ⊆ Rn,whichistheconvexhullofallpointsobtainedfrom(1,2,··· ,n) ∈
v Rn by coordinate permutations. The minimal description of Π in the space Rn
n
6
looksasfollows[1]:
4
4 n(n+1)
3 X xv =
2
2. v∈[n]
1 |S|(|S|+1)
9 Xxv ≥ forall∅ 6= S ⊂ [n]
2
0
v∈S
:
v Thus the permutahedron Π has n! vertices and 2n −2 facets. At the same time
n
i
X itiseasy toderive anextended formulation ofsizeΘ(n2)fromthe Birkhoffpoly-
r tope[1]:
a
X izi,v = xv forallv ∈ [n]
i∈[n]
X zi,v = 1 foralli ∈ [n]
v∈[n] (1)
X zi,v = 1 forallv ∈ [n]
i∈[n]
z ≥ 0 foralli,v ∈[n]
i,v
Theprojection ofthepolyhedron described by(1)tothex-variables givestheper-
mutahedron Π . Clearly, every coordinate permutation of Rn maps Π to itself.
n n
Date:January25,2011.
1
2 KANSTANTSINPASHKOVICH
The extended formulation (1) respects this symmetry in the sense that every such
permutation of the x-variables can be extended by some permutation of the z-
variables such that these two permutations leave (1)invariant (up to reordering of
theconstraints).
Also there exists a non-symmetric extended formulation of the permutahedron
of size Θ(nlog(n)) [2]. This is the best one can achieve [2] due to the fact that
every face of Π (including the n! vertices) is a projection of some face of the
n
extension. And since the number of faces of a polyhedron is bounded by 2 to the
number of its facets, we can conclude that every extension of the permutahedron
hasatleastlog (n!) = Θ(nlog(n))facets.
2
As we show in this paper the size of the extended formulation (1) is asymp-
totically optimal for symmetric formulations of the permutahedron. Thus there
existsagapinsize betweensymmetric and non-symmetric extended formulations
ofΠ . Thissituation appears insomeothercasesaswell,e.g. thecardinality con-
n
strained matching polytopes and the cardinality constrained cycle polytopes [3].
But even if the gaps observed in those cases are more substantial, the permutahe-
dronisinteresting because ofthepossibility todeterminetightasymptotical lower
bounds Ω(n2) and Ω(nlog(n)) on the sizes of symmetric and non-symmetric ex-
tendedformulations.
Thepaper isorganised asfollows. Section 2contains definitions ofextensions,
thecrucialnotionofsectionandsomeauxilaryresults. Insection3weexploitsome
knowntechniques [3],[4]andsomenewapproaches toprovealowerboundonthe
numberofvariablesandfacetsinsymmetricextensions ofthepermutahedron.
Acknowledgements. I thank Volker Kaibel for valuable comments, which led to
simplificationofLemma8andLemma9,andforhisveryusefulrecommendations
onwording.
2. EXTENSIONS, SECTIONS AND SYMMETRY
Here we list some known definitions and results, which will be used later. For
abroader discussion of symmetry in extended formulations of polyhedra werefer
thereaderto[3].
A polytope Q ⊆ Rd together with a linear map p : Rd → Rm is called an
extension of a polytope P ⊆ Rm if the equality p(Q) = P holds. Moreover,
ifQistheintersectionofanaffinesubspaceofRd andthenonnegativeorthantRd
+
then Q is called a subspaceextension. A (finite) system of linear equations and
inequalities whose solutions are the points in an extension Q of P is anextended
formulationforP.
Through the proof we mostly deal with such objects as sections s : X → Q,
which are maps that assign to every vertex x ∈ X of P some point s(x) ∈ Q∩
p−1(x). Such a section induces a bijection between X and its image s(X) ⊆ Q,
whoseinverseisgivenbyp.
For an arbitrary group G ⊆ S(m) acting on the set X of vertices of P, an
extension as above is symmetric (with respect to the action of G on X), if for
SYMMETRYINEXTENDEDFORMULATIONSOFTHEPERMUTAHEDRON 3
everyπ ∈ Gthereisapermutation κ ∈ S(d)withκ .Q = Qand
π π
p(κ .y) = π.p(y) forally ∈ p−1(X). (2)
π
Wedefineanextended formulation A=y = b=,A≤y ≤ b≤ describing thepoly-
hedron
Q = {y ∈Rd : A=y = b=,A≤y ≤ b≤}
extending P ⊆ Rm as above to besymmetric(with respect to the action of G on
the set X of vertices of P), if for every π ∈ G there is a permutation κ ∈ S(d)
π
satisfying (2) and there are two permutations ̺= and ̺≤ of the rows of (A=,b=)
π π
and(A≤,b≤),respectively,suchthatthecorrespondingsimultaneouspermutations
of the columns and the rows of the matrices (A=,b=) and (A≤,b≤) leaves them
unchanged. Clearly,inthissituationthepermutationsκ satisfyκ .Q = Q,which
π π
impliesthefollowing.
Lemma 1. Every symmetric extended formulation describes a symmetric exten-
sion.
Wecallanextensionweaklysymmetric(withrespecttotheactionofGonX)if
thereisasections :X → Qsuchthatforeveryπ ∈Gthereisapermutationκ ∈
π
S(d)withs(π.x) = κ .s(x)forallx ∈ X. Itcanbeshownthateverysymmetric
π
extensionisweaklysymmetric.
Dealing with weakly symmetric extension we can define an action of G on the
set S = {s ,...,s } of the component functions of the section s : X → Q
1 d
with π.sj = sκπ−−11(j) ∈ S for each j ∈ [d]. In order to see that this definition
indeedyieldsagroupaction,observethat,foreachj ∈ [d],wehave
(π.sj)(x) = sκπ−−11(j)(x) = (κπ−1.s(x))j = sj(π−1.x) forallx ∈ X, (3)
from which one deduces id .s = s for the identity element id in G as well
m j j m
as(ππ′).s = π.(π′.s )forallπ,π′ ∈ G. Theisotropygroupofs ∈ S underthis
j j j
actionis
iso(s )= {π ∈ G : π.s = s }.
j j j
From (3) one sees that, π is in iso(s ) if and only if we have s (x) = s (π−1.x)
j j j
for all x ∈ X or, equivalently, s (π.x) = s (x) for all x ∈ X. In general, it will
j j
be impossible to identify the isotropy groups iso(s ) without more knowledge on
j
thesection s. However,foreach isotropy group iso(s ),one canatleastbound its
j
index(G : iso(s ))bythenumberofvariables:
j
(G :iso(s )) ≤ d
j
Thenextlemmacouldbeusedtotransformanarbitrarysymmetricextension to
somesubspace symmetricextension.
Lemma2. Ifthereisasymmetricextension inRd˜withf facetsforapolytope P ,
thenthereisalsoasymmetricsubspace extension inRd withd ≤ 2d˜+f forP.
Usingthislemmaandhavingalowerboundonthenumberofvariablesinsym-
metricsubspace extensions ofthegivenpolytope P onegetsalowerbound onthe
totalnumberoffacetsandvariablesinanysymmetricestension ofP.
4 KANSTANTSINPASHKOVICH
3. BOUND ON SYMMETRIC SUBSPACE EXTENSION OF THE PERMUTAHEDRON
Now we would like to establish a lower bound on the number of variables in
symmetricsubspace extensions ofthepermutahedron.
Theorem 3. For every n ≥ 6 there exists no weakly symmetric subspace exten-
sionofthepermutahedron Π withless than n(n−1) variables (withrespect tothe
n 2
groupG = S(n)actingviapermuting thecorresponding elements).
Fortheproof, weassumethat Q ⊆ Rd withd < n(n−1) isaweakly symmetric
2
subspace extension of Π . Weak symmetry is meant with respect to the action
n
of G = S(n) on the set X of vertices of Π and we assume s : X → Q to be a
n
sectionasrequired inthedefinitionofweaksymmetry.
The operator Λ(ζ) maps any permutation ζ to the vector (ζ−1(1), ζ−1(2), ···,
ζ−1(n)). Thus,wehave
X = {Λ(ζ) : ζ ∈ S(n)}
whereS(n)isthesetofallpermutations ontheset[n],and
(π.Λ(ζ))v = Λ(ζ)π−1(v)
holdsforallπ ∈ S(n),ζ ∈ S(n).
In order to identify suitable subgroups of the isometry groups iso(s ), we use
j
the following result on subgroups of the symmetric group S(n) [4]. Here A(n)
denotesthealternating group,i.e. thegroupconsisting ofallevenpermutations on
theset[n].
Lemma4. ForeachsubgroupU ofS(n)with(S(n) :U) ≤ n fork < n,there
(cid:0)k(cid:1) 4
issomeW ⊆ [n]with|W| ≤ ksuchthat
{π ∈ A(n) : π(v) = v forallv ∈ W} ⊆ U
holds.
Asweassumedd < n ,Lemma4impliesthatforallj ∈ [d]
(cid:0)2(cid:1)
{π ∈ A(n) : π(v) = v forallv ∈V } ⊆ iso(s )
j j
for some set V ⊂ [n], |V | ≤ 2. Wecan prove that V can be choosen to contain
j j j
notmorethanoneelement, whichwedenotebyv .
j
Lemma5. Foreachj ∈ [d]thereissomev ∈ [n]suchthat
j
{π ∈ A(n) : π(v )= v }⊆ iso(s )
j j j
Thiselementv isuniquely determined unlessA(n)⊆ iso(s )
j j
Proof. The statement of lemma is automatically true if the set V is empty or
j
contains just one element. So let us assume the set V to consist of two ele-
j
ments{v,w}. Ifthegroupiso(s )hastwofixedblocksV and[n]\V ,thefollow-
j j j
inginequality
n(n−1)
d < ≤ (S(n) :iso(s ))
j
2
SYMMETRYINEXTENDEDFORMULATIONSOFTHEPERMUTAHEDRON 5
holds. Thus we can find some permutation τ ∈ iso(s ) such that w.l.o.g τ(v) 6∈
j
{v,w}.
Later it would be convenient to have τ(w) = w and τ ∈ A(n). If τ(w) 6= w
or τ 6∈ A(n) we regard a new permutation τ′ = τ−1βτ ∈ A(n), where β ∈
A(n),β(v) = v, β(w) = w, βτ(w) = τ(w) and βτ(v) 6= τ(v). Forevery n ≥ 6
such permutation β can be found since τ(v) 6∈ {v,w,τ(w)}. The construction
of τ′ guarantees that τ′(w) = w, τ′(v) 6= v and τ′ ∈ iso(s ). Hence we can
j
assumethatτ(w) = wandτ ∈ A(n).
Toprovethatthedescribedinthelemmaelementv existswecanshowthatthe
j
elementwisoneofsuchpossibilities
{π ∈ A(n) : π(w) = w}⊆ iso(s ) (4)
j
Anypermutation π ∈ A(n),π(w) = w withtheproperty π(v) 6= v canberep-
resented as(π(τα)−1)ταforanyα ∈ S(n). Wechoose apermutation α ∈ A(n)
suchthatα(v) = v,α(w) = wandαπ−1(v) = τ−1(v). Theexistenceofαcanbe
trivially proved for n ≥ 6. Thus the permutation π belongs to iso(s ) because all
j
three permutations τ, αand π(τα)−1 belong toiso(s )( note that π(τα)−1 and α
j
areevenpermutations andfixelementsv,w).
Anypermutation π ∈ A(n),π(w) = w belongs toiso(s )whenever π(v) = v.
j
Thereforewecanconclude that(4)holds.
Havingsomeotherelementu ∈ [n],u6= w suchthat
{π ∈A(n) : π(u) = u} ⊆ iso(s ) (5)
j
wecanprovethatA(n) ⊆ iso(s ),since everypermutation π ∈ A(n)isacompo-
j
sitionofnotmorethanfourpermutations described by(4)and(5). (cid:3)
The next theorem has an important role in the proof because it describes the
actionofA(n)onthecomponents s .
j
Theorem 6. There exists a partition of the set [d] into sets A and B , such that
i j
each set B consists just of one element b and each set A consists of n ele-
j j i
mentsai,ai,...,ai elementswith
1 2 n
s (π.x) = s (x) s (π.x) = s (x) (6)
ait aiπ−1(t) bj bj
foranyvertexx ∈X andallπ ∈ A(n).
Let us consider, for v ∈ [n−2], permutations ρ consisting of just one cy-
v
cle (v,v +1,v +2). Wewould like to show the existence of apartition A , {b }
i j
whichsatisfiesallcardinality assumptions ofTheorem 6andsatisfies equation (6)
forallx ∈ X andallpermutationsρ . SuchapartitionsatisfiesTheorem6because
v
everypermutation π ∈ A(n)isaproductofpermutations ρ .
v
Through the whole proof the action of S(d) is restricted to the action on vec-
tors s(x) for x ∈ X. It means that two permutation κ′ and κ from S(d) are
equivalent for us if sκ′−1(j)(x) = sκ−1(j)(x) for all x from X and for all j
from [d]. For example, we can take the identity permutation id instead of κ
d
ifsκ−1(j)(x) = sj(x)forallx ∈ X andallj ∈ [d].
6 KANSTANTSINPASHKOVICH
Lemma7. Foreachπ = (w ,w ,w ) ∈ A(n)thereexistsapermutationinS(d),
1 2 3
which is equivalent to κ such that all cycles of this permutation are of the form
π
(j ,j ,j )withv = w andA(n) 6⊆ iso(s )forallt ∈ [3].
1 2 3 jt t jt
Proof. The permutation κ3 is equivalent to the identity permutation id since the
π d
permutation π3 istheidentity permutation id .
n
Thus any cycle C of the permutation κ permutes indices of identical compo-
π
nentfunctions ofsifthecyclelength |C|isnotdivisable bythree. Hence, wecan
assumethateverycycleC ofκ haslength|C|= 0 mod 3.
π
ThesameargumentallowsustotransformeachcycleC = (j ,j ,···j )ofthe
1 2 3l
permutation κ into the following cycles (j ,j ,j ),..., (j ,j ,j ), offer-
π 1 2 3 3l−2 3l−1 3l
inganequivalentpermutationtoκ . Thuswemayassumethatκ containscycles
π π
oflengththreeonly.
Letusconsider oneofthecycles(j ,j ,j )ofthepermutation κ . Weinvesti-
1 2 3 π
gatetwopossiblecases.
Iftheelement v doesnot belong to{w ,w ,w }orA(n) ⊆ iso(s )thenwe
j1 1 2 3 j1
haveπ ∈ iso(s )andthusπ,π2 ∈ iso(s ),whichyields
j1 j1
sj1(x) = sj1(π.x) = sκπ−1(j1)(x) = sj3(x)
sj1(x) = sj1(π2.x) = sκπ−2(j1)(x) = sj2(x)
This shows that the component functions s , s , s are identical. Thus the cy-
j1 j2 j3
cle(j ,j ,j )canbedeleted.
1 2 3
Hence, we may w.l.o.g. assume v = w . Foreach τ′ ∈ A(n) with τ′(w ) =
j1 1 3
w and τ := πτ′π−1 ∈ A(n) we have τ(w ) = πτ′π−1(w ) = πτ′(w ) =
3 1 1 3
π(w ) = w . Since τ ∈ A(n)and τ(w ) = w wehave τ ∈ isos , andthus(see
3 1 1 1 j1
thefifthequation inthefollowingchain)forallx ∈ X
sj3(π−1τπ.x) = sκπ−1(j1)(π−1τπ.x) = sj1(ππ−1τπ.x)
= sj1(τπ.x) = sj1(τ.(π.x)) = sj1(π.x) = sκπ−1(j1)(x) = sj3(x)
Hence τ′ ∈ iso(s ) holds. Thus we have v = w , unless A(n) ⊆ iso(s ),
j3 j3 3 j3
which as in the treatment of the case A(n) ⊆ iso(s ) would allow us to remove
j1
thecycle(j ,j ,j ). Similarly,onecanestablishv = w . (cid:3)
1 2 3 j2 2
Now we want to prove the next useful lemma, which will help us to construct
thedesired partition.
Lemma8. Lettwopermutationsπ = (w ,w ,w )andσ = (w ,w ,w )begiven
1 2 3 2 3 4
withw 6= w andsupposethatthecorresponding permutationsκ andκ satisfy
1 4 π σ
the conditions from Lemma 7. If the permutation κ contains a cycle (j ,j ,j )
π 1 2 3
withv = w forallt ∈ [3]thenoneofthefollowingistrue:
jt t
a) Thepermutation κ contains acycle(j ,j ,j )withv = w .
σ 2 3 4 j4 4
b) Thepermutation κσ contains twocycles(j2,j3′,j4′)and(j2′′,j3,j4′′)withvj′′ =
2
w2,vj′ = w3 andvj′ = vj′′ = w4.
3 4 4
Proof. Assume that permutation κ does not contain any cycle involving the in-
σ
dex j . Every permutation µ ∈ A(n) can be represented as a combination τ′στ,
2
SYMMETRYINEXTENDEDFORMULATIONSOFTHEPERMUTAHEDRON 7
where τ′, τ are even permutations with τ′(w ) = τ(w ) = w . Thus for any
2 2 2
permutation µ ∈ A(n)wehave
sj2(µ.x) = sj2(τ′στ.x) = sj2(στ.x) = sκσ−1(j2)(τ.x) = sj2(τ.x) = sj2(x)
This contradicts conditions on κ from Lemma 7. We proceed in a similar way
π
whennocycleinκ involves j .
σ 3
Iftherearetwodifferentcycles(j ,j′,j′)and(j′′,j ,j′′)inthepermutationκ
2 3 4 2 3 4 σ
wewould like to prove that the component functions mentioned in the lemma are
identical. Forthis let us consider the permutation πσ which could be written as a
combination of two disjoint cycles (w ,w )(w ,w ). From this we can conclude
1 2 3 4
that(πσ)2 istheidentitypermutationid ,whatimpliesthat(κ κ )2 isequivalent
n π σ
toid .
d
Forallx ∈ X wehave
sj3(x) = sj3((πσ)2.x) = sκπ−1(j3)(σπσ.x) = sj2(σπσ.x) = sκσ−1(j2)(πσ.x)
andwecancontinuethischainofequationsusingthatvj′ = w4isnotequaltoany
4
oftheelementsw ,w andw inthefollowingway:
1 2 3
sκσ−1(j2)(πσ.x) = sj4′(πσ.x) = sj4′(σ.x) = sκσ−1(j4′)(x) = sj3′(x)
Thus we proved that sj3 and sj3′ are identical component functions. Considering
expression sj2((πσ)2.x)wegetthatsj2 andsj2′′ areidentical aswell. (cid:3)
Lemma 9. For every cycle (j ,j ,j ) in the permutation κ we can find a set
1 2 3 ρ1
S = {j ,j ,··· ,j } such that, for every v ∈ [n−2], there is a per-
(j1,j2,j3) 1 2 n
mutation equivalent to ρ , which contains the cycle (j ,j ,j ) and has the
v v v+1 v+2
properties requiredinLemma7.
Proof. LetusconstructthesetS inseveralsteps. WestartwithS =
(j1,j2,j3) (j1,j2,j3)
{j ,j ,j },whichsatisfiesthecondition ofthelemmaforv = 1.
1 2 3
By Lemma 8 with π = ρ and σ = ρ we have to consider two possible
1 2
cases concerning the cycle(j ,j ,j ). Incase (a)wecanextend theset S
1 2 3 (j1,j2,j3)
to{j ,j ,j ,j },suchthatS satisfiestheconditions inthelemmaforv =
1 2 3 4 (j1,j2,j3)
1,2. In case (b) we can update κ by changing cycles (j ,j′,j′), (j′′,j ,j′′)
ρ2 2 3 4 2 3 4
to(j ,j ,j′),(j′′,j′,j′′),whatwillproduceapermutationequivalenttoκ . And
2 3 4 2 3 4 ρ2
wecanchooseS beequalto{j ,j ,j ,j′}.
(j1,j2,j3) 1 2 3 4
In this manner we go through all v ∈ [n−2] setting π = ρ and σ = ρ to
v−1 v
extendthesetS and,ifnecessary, toupdatethepermutation κ . (cid:3)
(j1,j2,j3) ρv
Applying Lemma9 to all cycles of κ we get some disjoint sets S in-
ρ1 (j1,j2,j3)
dexedbysubsetsofcyclesofκ . Moreover,thereisnocyclesinκ ,··· ,κ ,
ρ1 ρ2 ρn−2
whichdoes notcontain anyindex from theconstructed setsS . Thisisdue
(j1,j2,j3)
to Lemma8applied to pairs π = ρ , σ = ρ for v ranging from 2 to n−2( in
v v−1
thisorder).
NowwecanchoosethesetsA tobethesetsS andthesingletones{b }
i (j1,j2,j3) j
accordingly. Lemma9guarantees equation (6).
8 KANSTANTSINPASHKOVICH
Now we have some understanding of how permutations A(n) act on the com-
ponentfunctions ofs. Havingthisknowledgewecancreateanaffinecombination
ofpointsinthesubspace extensionQofΠ ,whichhasnon-negative components,
n
butdoesnotprojecttothepermutahedron Π .
n
Let us introduce the following subgroup of A(n) defined by one element w of
theset[n-1]
H∗ = {π ∈ A(n) : π([w]) = [w]}
w
ThusH∗ isthesetofallevenpermutations of[n],whichmap[w]toitself.
w
Letusconsiderthefunction s∗ : X ×[n−1]→ Rdefinedvia
s(π.x)
s∗(x,w) = Pπ∈Hw∗
|H∗|
w
The value of s∗ is a convex combination of points from Q and thus lies in Q for
anyx ∈ X.
LetusconsiderapartitionA ,{b }asinTheorem6. First,wetakealookatthe
i j
setsB andanyvertexxfromX:
j
s (π.x) s (x)
s∗ (x,w) = Pπ∈Hw∗ bj = Pπ∈Hw∗ bj = s (x)
bj |H∗| |H∗| bj
w w
Hencethevalueofs∗ dependsjustonxandisequaltothevalueofs (x)forany
bj bj
vertexx ∈ X.
Forthecomponents corresponding toai fromthesetA ,Theorem6givesus
t i
s∗ (x,w) = Pπ∈Hw∗ sait(π.x) = Pπ∈Hw∗ saiπ−1(t)(x) (7)
ait |Hw∗| |Hw∗|
Whentbelongstotheset[w],wecancalculate thecomponent s∗ as
ai
t
s (x)
s∗ait(x,w) = Pπ∈Hw∗|Haw∗iπ|−1(t) = vX≤w |H1w∗|π−X1(t)=vsaiv(x)
π∈H∗
w
Moreover,wecanestimatethenumberofpermutations inthesecondsumby
(w−1)!(n−w)!
|{π ∈ H∗ : π−1(t)= v}| =
w 2
andthusconclude that
s∗ait(x,w) = vX≤w |H1w∗|π−X1(t)=vsaiv(x) = vX≤w saivw(x)
π∈H∗
w
Whentdoesnotbelongto[w],wecanderiveasimilarequality
s (x)
s∗ait(x,w) = X na−iv w
v>w
SYMMETRYINEXTENDEDFORMULATIONSOFTHEPERMUTAHEDRON 9
Now wewould like to outline the idea of the contradiction. Let us assume that
wefoundsomeelementw oftheset[n−1]suchthatthestatements
if sai (Λ(idn)) > 0 then Xsai(Λ(idn)) > 0 (8)
w v
v>w
and
if saiw+1(Λ(idn)) > 0 then Xsaiv(Λ(idn)) > 0 (9)
v≤w
hold for all sets A (we will establish the existence of such a w in Lemma 10
i
below).
Letζ besomeevenpermutation suchthattheconditions
ζ([w−1]) ⊂ [w] and ζ(w+1) ∈ [w]
arefulfiled. Sincethepermutation ζ iseven,by(6)weget
s (Λ(ζ)) = s (Λ(id )) and s (Λ(ζ)) = s (Λ(id ))
bj bj n ait aiζ−1(t) n
The point y = (1+ǫ)s∗(Λ(id ),w) −ǫs∗(Λ(ζ),w) is an affine combination
n
ofpoints from Q. Thecomponents b of the point y are equal tothe non-negative
j
values (Λ(id )). Thecomponentai isequaltothevalue
bj n t
1
w((1+ǫ)Xsaiv(Λ(idn))−ǫ X saiv(Λ(idn))−ǫsaiw+1(Λ(idn)))
v≤w v≤w−1
inthecaset ≤ wandtothevalue
1
n−w((1+ǫ)Xsaiv(Λ(idn))−ǫ X saiv(Λ(idn))−ǫsaiw(Λ(idn)))
v>w v>w+1
inthecaset > w. Aftersimplification thosevalueslooklike
1
w(Xsaiv(Λ(idn))+ǫsaiw(Λ(idn))−ǫsaiw+1(Λ(idn)))
v≤w
and
1
n−w(Xsaiv(Λ(idn))+ǫsaiw+1(Λ(idn))−ǫsaiw(Λ(idn)))
v>w
From (8) and (9) follows that there exists ǫ > 0 such that for all A all compo-
i
nentsai ofthepointy arenon-negative. Butthecombinationygivesapointinthe
t
projection totheoriginal variables, whichviolates theinequality
w(w+1)
X xv ≥
2
v∈[w]
This is true since the projection of s∗(Λ(id ),w) belongs to the corresponding
n
faceandtheprojection ofs∗(Λ(ζ),w)doesnot. Thusforanyǫ > 0thepoint(1+
ǫ)s∗(Λ(id ),w)−ǫs∗(Λ(ζ),w)cannotbelongtothepermutahedron Π .
n n
Thenextlemmafinishestheproofbecauseoftheabovementionedconstruction.
Lemma 10. There exists such element w from the set [n−1] which satisfies the
conditions (8)and(9).
10 KANSTANTSINPASHKOVICH
Proof. Since each setA consists ofncomponents, wecan conclude that number
i
ofsetsA islessthan n−1 (recalld< n(n−1)).
i 2 2
Foreach set A there can exist just one element ufrom [n−1], which violates
i
the statement (8) for the mentioned inex i (it can just be the maximal element
from [n−1]forwhich s (Λ(id )) > 0). Analogously, for each setA there can
ai n i
u
existjustoneelementufrom[n−1],whichviolatesthestatement(9).
Thus, for at least one element w ∈ [n−1] both (8) and (9) are satisfied for all
setsA . (cid:3)
i
CombiningLemma2andTheorem3wegetthefollowingtheorem,whichgives
usalowerboundonthenumberofvariablesandfacetsinsymmetricextensionsof
thepermutahedron.
Theorem 11. For every n ≥ 6 there exists no symmetric extended formulation
of the permutahedron Π with less than n(n−1) variables and constraints (with
n 4
respecttothegroupG =S(n)actingviapermutingthecorresponding elements).
REFERENCES
[1] MicheleConforti,GerardCornue´jols,andGiacomoZambelli.Extendedformulationsincombi-
natorialoptimization.Technicalreport,Universita`diPadova,2009.
[2] Michel Goemans. Smallest compact formulation for the permutahedron.
http://math.mit.edu/~goemans/PAPERS/permutahedron.pdf.
[3] VolkerKaibel,KanstantsinPashkovich,andDirkOliverTheis.Symmetrymattersforthesizes
ofextendedformulations.arXiv:0911.3712v1[math.CO].
[4] Mihalis Yannakakis. Expressing combinatorial optimization problems by linear programs. J.
Comput.SystemSci.,43(3):441–466,1991.
KANSTANTSINPASHKOVICH
E-mailaddress: [email protected]
