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Balanced matrices
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Conforti, M, Cornuejols, G and Vuskovic, K (2006) Balanced matrices. Discrete Applied
Mathematics, 306 (19-20). 2411 - 2437 . ISSN 0166-218X
https://doi.org/10.1016/j.disc.2005.12.033
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Mihele Conforti
Gerard Cornuejols y
Kristina Vuskovi z
dediated to the memory of Claude Berge
April 2003, revised April 2004, January 2005
DipartimentodiMatematiaPuraedAppliata,UniversitadiPadova,ViaBelzoni7, 35131Padova,Italy. onfortimath.unipd.it
y
Tepper Shool of Business, Carnegie Mellon University, Pittsburgh, PA 15213, USA; and LIF, Universite de Marseille, 163 Av de Luminy, 13288 Marseille, Frane. g0vandrew.mu.edu
z
Shool of Computing, University of Leeds, Leeds LS2 9JT, UK. vuskoviomp.leeds.a.uk
A0;1matrixisbalaned if,ineverysubmatrixwithtwononzero entriesperrowandolumn,thesumoftheentriesisamultipleoffour. ThisdenitionwasintroduedbyTruemperandgeneralizesthenotion ofbalaned0;1matrixintroduedbyBerge. Inthistutorial,wesurvey what is urrently known about these matries: polyhedral results, ombinatorial and struturaltheorems, and reognitionalgorithms.
1 Introdution
A 0;1 matrix H is a hole matrix if H ontains two nonzero entries per row and per olumn and no proper submatrix of H has this property. A hole matrix H is square, say of order n, and its rows and olumns an be permutedsothat itsnonzeroentries are h
i;i
,1in,h i;i+1
,1in 1, h
n;1
and noother. Note thatn 2 and the sum of the entries of H is even. A holematrix isodd ifthe sum of itsentries isongruent to2mod4and even if the sum of its entries is ongruent to 0mod4.
Apolytopeisintegral if allitsvertieshaveonlyinteger-valuedomponents. The set paking polytope,dened by an nm 0;1 matrix A, is
P(A)=fx2R n
:Ax1; 0x1g;
where1denotes a olumnvetor ofappropriatedimension whoseentries are allequal to 1.
The next theorem haraterizes abalaned 0;1 matrix A interms of the setpakingpolytopeP(A)aswellasthe setoveringpolytopeQ(A)andthe set partitioning polytopeR (A):
Q(A)=fx:Ax1; 0x1g;
R (A)=fx:Ax=1; 0x1g:
Theorem 2.1 (Berge [3℄,Fulkerson,Homan, Oppenheim[41℄) Let M bea 0;1 matrix. Then the following statements are equivalent:
(i) M isbalaned.
(ii) For eah submatrixA of M, theset overing polytope Q(A)is integral.
(iii) For eah submatrix A of M, the set pakingpolytope P(A) isintegral.
(iv) For eah submatrixA of M, the set partitioningpolytope R (A) is inte-gral.
Given a 0;1 matrix A, let p(A), n(A) denote respetively the olumn vetors whose i
th
omponents p i
(A), n i
(A) are the number of +1s and the number of 1s in the i
th
row of matrix A. Theorem 2.1 extends to 0;1 matriesas follows.
Theorem 2.2 (Conforti, Cornuejols [17℄) Let M be a 0;1 matrix. Then the followingstatements are equivalent:
(i) M isbalaned.
P(A)=fx:Ax 1 n(A); 0x1g isintegral.
(iv) For eah submatrix A of M, the generalized set partitioning polytope R (A)=fx:Ax=1 n(A); 0x1g is integral.
Toprovethis theorem,weneed thefollowingtwo results. Therst one is aneasy appliationof omputation of determinants by ofator expansion.
Remark 2.3 Let H be a 0;1 holematrix. If H isan even holematrix, H is singular, and if H is an odd hole matrix, det(H)=2.
Lemma 2.4 If A is a balaned 0;1 matrix, then the generalized set parti-tioning polytope R (A) is integral.
Proof: Assume that A ontradits the theorem and has the smallest size (number of rows plus number of olumns). Then R (A) is nonempty. Let x be a frational vertex of R (A). By the minimality of A, 0 < x
j
< 1 for all j. It follows that A is square and nonsingular. Sox is the unique vetor in R (A).
Let a 1
;:::;a n
denotethe row vetors of A and letA i
be the (n 1)n submatrixofAobtainedbyremovingrowa
i
. BytheminimalityofA,theset partitioningpolytope R (A
i
)=fx2R n
:A i
x=1 n(A i
); 0x1g is an integralpolytope. SineAissquare andnonsingular,thepolytopeR (A
i )has exatlytwoverties,sayx
S ;x
T
. SinexisinR (A i
),thenx=x S
+(1 )x T
. Sine 0 < x
j
< 1 for all j and x S
;x T
have 0,1 omponents, it follows that x
S +x
T
= 1. Let k be any row of A i
. Sine both x S
and x T
satisfy a k
x = 1 n(a
k
), this implies that a k
1 = 2(1 n(a k
)), i.e. row k ontains exatly two nonzero entries. Applyingthis argument totwo dierent matriesA
i , it follows that every rowof A ontains exatly two nonzero entries.
IfAhasaolumnj withonlyonenonzeroentrya kj
Theorem 2.5 Let A be a balaned 0;1 matrix with rows a;i2S, and let S 1 ;S 2 ;S 3
be a partition of S. Then
T(A)=fx2 R n
: a i
x1 n(a i
) for i2S 1
;
a i
x=1 n(a i
) for i2S 2
;
a i
x1 n(a i
) for i2S 3
;
0x1g
is an integral polytope.
Proof: If x is a vertex of T(A), it is a vertex of the polytope obtained fromT(A)by deletingthe inequalitiesthat arenot satisedwith equalityby
x. By Theorem 2.4, every vertex of this polytope has 0;1omponents. 2
Theorem 2.5does not hold when the upperboundx1 is removed. To see this, onsider the matrix
A= 0 B B B B B B B B B B B
1 1 1 1 1 0 0
1 1 0 0 0 0 0
1 0 1 0 0 0 0
1 0 0 1 0 0 0
0 0 0 0 1 1 1
0 0 0 0 0 1 0
0 0 0 0 0 0 1
1 C C C C C C C C C C C A : Then( 1 2 ; 1 2 ; 1 2 ; 1 2
;2;1;1)istheunique solutionofAx=1 n(A)and there-foreitisafrationalvertex ofthe polyhedronT(A)with x1removed, for any partitionof the rows of A intoS
1 , S 2 and S 3 .
Proof of Theorem 2.2: Sine balaned matries are losed under taking submatries,Theorem 2.5 shows that (i) implies(ii),(iii)and (iv).
Assume that A ontains anodd hole submatrix H. By Remark 2.3, the vetor x = (
1 2
;:::; 1 2
) is the unique solution of the system Hx = 1 n(H). This proves allthree reverse impliations. 2
2.1 Total Dual Integrality
if a linear system Ax b is TDI and b is integral, then fx :Ax bg is an integralpolyhedron.
Theorem 2.6 (Fulkerson,Homan, Oppenheim[41℄) Let A= 0 B A 1 A 2 A 3 1 C A bea
balaned 0;1 matrix. Thenthe following linear systemisTDI:
A 1
x1 (1)
A 2
x1
A 3
x=1
x0:
Theorem2.6and theEdmonds-Gilestheorem implyTheorem2.1. Inthis setion,we prove the following more generalresult.
Theorem 2.7 (Conforti, Cornuejols [17℄) Let A = 0 B A 1 A 2 A 3 1 C A
be a balaned
0;1 matrix. Thenthe following linear system is TDI:
A 1
x1 n(A 1
) (2)
A 2
x1 n(A 2
)
A 3
x=1 n(A 3
)
0x1:
The followingtransformation of a0;1 matrix A intoa 0;1matrix B is oftenseen inthe literature: to every olumn a
j
of A, j = 1;:::;p,assoiate two olumns of B, say b
P j
and b N j
, where b P ij
= 1if a ij
=1, 0 otherwise, and b
N ij
= 1 if a ij
= 1, 0 otherwise. Let D be the 0;1 matrix with p rows and 2polumnsd
P j
and d N j
suh that d P jj
=d N jj
=1 and d P ij
=d N ij
Given a 0;1 matrix A = B A 1 A 2 A 3 C
A and the assoiated 0;1 matrix B =
0 B B 1 B 2 B 3 1 C
A, dene the following two linear systems:
A 1
x1 n(A 1
) (3)
A 2
x1 n(A 2
)
A 3
x=1 n(A 3
)
0x 1;
B 1
y1 (4)
B 2
y1
B 3
y=1
Dy=1
y0:
Avetorx2R p
satises(3)if andonlyifthe vetor (y P
;y N
)=(x;1 x) satises(4)andthistransformationmapsintegervetorsintointegervetors. Hene the polytope dened by (3) is integral if and only if the polytope dened by (4) is integral. We show that, if A is a balaned 0;1 matrix, then both (3) and (4) are TDI.
Lemma 2.8 If A = 0 B A 1 A 2 A 3 1 C A
is a balaned 0;1 matrix, the orresponding
system (4) isTDI.
Proof: The proof is by indution on the number m of rows of B. Let =(
P ;
N )2Z
2p
denote anintegralvetor and R 1
;R 2
;R 3
the index sets of the rows of B
1 ;B
2 ;B
3
max m X i=1 u i + X j=1 v j (5)
uB +vD
u i
0; i2R 1
u i
0; i2R 2
:
Sine v j
only appears in two of the onstraints uB +vD and no onstraint ontains v
j
and v k
, it follows that any optimal solution to (5) satises
v j
= min( P j m X i=1 b P ij u i ; N j m X i=1 b N ij u i ): (6)
Let(u;v) be an optimalsolution of (5). If u is integral, then sois vby (6), andwearedone. Soassumethatu
`
isfrational. Letb `
bethe orresponding rowofB,andletB
`
be thematrix obtained fromB by removingrowb `
. By indution on the number of rows of B, the system (4)assoiated with B
` is TDI. Hene the system
max X i6=` u i + p X j=1 v j u ` B `
+vD bu ` b ` (7) u i
0;i2R 1
nf`g
u i
0;i2R 2
nf`g
has an integral optimalsolution(~u;v).~ Sine (u
1
;:::;u ` 1
;u `+1
;:::;u m
;v 1
;:::;v p
) is a feasible solution to (7) and Theorem 2.5shows that
P m i=1 u i + P p j=1 v j
is aninteger number,
X i6=` ~ u i + p X j=1 ~ v j d X i6=` u i + p X j=1 v j e= m X i=1 u i + p X j=1 v j bu ` :
Thereforethevetor(u
;v
)=(~u 1
;:::;u~ ` 1
;bu `
;u~ `+1
;:::;u~ m
;v~ 1
;:::;v~ p
1 2 3 A 1 ;A 2 ;A 3
. ByLemma 2.8, the linearsystem (4)assoiated with (3)isTDI. Letd 2R
p
be any integral vetor. The dual of minfdx : x satises (3)g is the linear program
max w(1 n(A)) t1
wA t d (8)
w i
0;i2R 1
w i
0;i2R 2
t0:
For every feasible solution (u;v) of (5) with = ( P
; N
) = (d;0), we onstrut a feasible solution (w;
t) of (8) with the same objetive funtion value as follows:
w = u
t
j =
(
0 if v
j = P i b N ij u i P i b P ij u i P i b N ij u i d j if v j =d j P i b P ij u i : (9)
Whenthevetor(u;v) isintegral,theabovetransformationyieldsanintegral vetor (w;
t). Therefore (8) has an integral optimal solution and the linear system (3) is TDI. 2
Thistheoremdoesnotholdwhentheupperboundx1isdropped from the linear system asshown by the example given afterTheorem 2.5.
3 Colorings and Hypergraphs
3.1 Biolorings
Ak-oloringofamatrixAisapartitionofolumnsofAintoksetsor\olors" (someof them may beempty). In this setionwe onsider 2-olorings.
submatrix of A isbiolorable.
Ghouila-Houri[43℄introduedthenotionofequitablebioloringfora0;1 matrix A as follows. The olumns of A are 2-olored intoblue olumnsand redolumnsinsuhaway that,foreveryrowof A,the sum ofthe entries in the blue olumns diers from the sum of the entries in the red olumns by atmost one.
Theorem 3.2 (Ghouila-Houri[43℄) A 0;1 matrix A is totally unimodular if and only if every submatrix of A has an equitable bioloring.
ThistheoremgeneralizesaresultofHellerandTompkins[50℄formatries with atmost two nonzero entries per row.
A 0;1 matrix A is biolorable if itsolumns an be 2-olored into blue olumns and red olumns in suh a way that every row with two or more nonzeroentries eitherontains twoentries of opposite sign inolumnsofthe same olor, or ontains two entries of the same sign in olumns of dierent olors. Equivalently, for no row with at least two nonzero entries all the 1s have the same olor, say blue, and all the 1's are red. For a 0;1 matrix, this denition oinides withBerge's notionof bioloring. Clearly, if a0;1 matrix has an equitable bioloring as dened by Ghouila-Houri, then it is biolorable. So the theorem below implies that every totally unimodular matrix isbalaned.
Theorem 3.3 (Conforti, Cornuejols [17℄) A 0;1 matrix A is balaned if and only if every submatrix of A is biolorable.
Proof: Assume rst that A is balaned and let B be any submatrix of A. Remove from B any row with fewer than two nonzero entries. Sine B is balaned, so is the matrix (B; B). It follows from Theorem 2.5 that the inequalities
Bx 1 n(B) (10)
Bx 1 n( B)
dene an integral polytope. Sine it is nonempty (the vetor ( 2
;:::; 2
) is a solution),itontains a0,1vetor x . Color aolumnj of B red if x
j
=1and blue otherwise. By(10), this is a validbioloringof B.
Conversely, assume that A ontains an odd hole matrix H. We laim that H is not biolorable. Suppose otherwise. Sine H ontains exatly 2 nonzeroentries perrow, the bioloringonditionshows that thevetorof all zeroes an be obtained by adding the blue olumnsand subtrating the red olumns. SoH is singular,a ontradition to Remark 2.3. 2
In Setion 4.1, we prove abioloring theorem that extends all the above results(Theorem 4.3).
Cameron and Edmonds [10℄ showed that the following simple algorithm nds abioloringof a balaned matrix.
Algorithm(Cameron and Edmonds [10℄)
Input: A 0;1 matrix A.
Output: A bioloringof A ora proof that the matrix A is not balaned.
Stopif allolumns are olored or if some row isinorretlyolored. Oth-erwise, olor a newolumn red or blue as follows.
If some row of A fores the olor of a olumn, olor this olumn aord-ingly.
If no row of A fores the olor of a olumn, arbitrarily olor one of the unolored olumns.
Intheabovealgorithm,arowa i
forestheolor ofaolumnwhenallthe olumnsorrespondingtothe nonzero entries of a
i
have been oloredexept one, say olumn k, and row a
i
, restrited to the olored olumns, violates the bioloringondition. In this ase, the bioloring rule ditates the olor of olumnk.
When the algorithm fails to nd a bioloring, the sequene of forings that resulted in an inorretlyolored row identies an odd hole submatrix of A.
Note that a matrix A may be biolorable even if A is not balaned. In fat, the algorithm may nd a bioloring of A even if A is not balaned.
For example, if A = 0
B
1 1 1 0 1 1 0 1 1 0 1 1
1
C
the algorithmannotbeused as a reognitionof balanedness.
3.2 k-Colorings
A 0;1 matrix A isk-olorable if thereexists ak-oloringof itsolumns suh thatfor every rowi that has atleast two 1sinolors J[L thereare entries a
ij =a
il
=1 where j 2J and l 2L. This is equivalentto saying that every pairofolorsJ;Lonstitutesabioloring(asdenedintheprevioussetion) of the submatrixA
JL
of A, indued by olumns J[L.
Theorem 3.4 (Berge [4℄) A 0;1 matrix A is balaned if and only if every submatrix of A isk-olorablefor every k.
Proof: The \if" part follows from Theorem 3.1. We now show that if every olumnsubmatrix of A isbiolorable, then A is k-olorablefor every k. By Theorem 3.1 this proves the result. For a given k-oloringof A, let r(i) be the number of olorsthat are represented inrow i,i.e. the number ofolors J for whih a
ij
=1for somej 2J. Considera k-oloringof Asuhthat the sum of r(i) over all rows i of A is maximized. Suppose that this k-oloring of A does not satisfy the above denition. Then there are olors J;L that donotgiveabioloringof the matrixA
JL
. LetJ 0
;L 0
beabioloringof A JL
, and onsider a new k-oloring of A where J and L are replaed by J
0 and L
0
and all the other olors stay the same. In this new oloring the sum of r(i) over all rows i of A has inreased, in omparison to the original one, a
ontradition. 2
The above proof shows that if A is abalaned matrix one an eÆiently onstrut a k-oloring of A, that satises the above ondition, using the algorithmof Cameron and Edmonds.
Similarly the notion of equitable bioloringis extended to the notion of equitablek-oloring. Ak-oloringofa0;1matrixAisequitableifeverypair ofolors J;Lonstitutes anequitablebioloringof A
JL
. Asimilar argument asin the proof above,givesthe following result.
sothat every pair of olorsJ;L onstitutesa bioloringof A JL
.
Conjeture 3.6 (Conforti and Zambelli) A 0;1 matrix A is balaned if and only if every submatrix of A is k-olorable for every k.
Fork =2the onjeture isequivalenttoTheorem3.3. This onjeture is open for k =3. Notethat the onjeture holds for every totallyunimodular matrix A sine every equitable k-oloring of A is a k-oloring that satises theabove ondition. De Werra [37℄ givesaweakernotionofk-olorabilityof a0;1 matrix and proves that balaned matriessatisfy it.
3.3 Balaned Hypergraphs
A 0;1 matrix A an be represented by ahypergraph. Then the denition of balanedness for 0;1 matries is a natural extension of the property of not ontainingoddylesforgraphs. Infat,thisisthemotivationthatledBerge [2℄tointroduethenotionofbalanedness: AhypergraphH=(V;E),where V represents the olumn set and E represents the row set of A, is balaned if every odd yle C of H has anedge ontaining at least three nodes of C. Equivalently, H is balaned if the assoiated 0;1 matrix A is balaned. We refertoBerge[6℄foranintrodutiontothetheoryofhypergraphs. Several re-sultsonbipartitegraphsgeneralizetobalanedhypergraphs, suhasKonig's bipartitemathingtheorem,asstatedinthe nexttheorem. Inahypergraph, a mathing is a set of pairwise noninterseting edges and a transversal is a node set interseting allthe edges.
Theorem 3.7 (Berge, LasVergnas[7℄) In abalaned hypergraph, the maxi-mumardinalityof a mathingequalstheminimum ardinalityof a transver-sal.
Proof: Follows fromTheorem 2.6applied with A 1
=A 3
=; and the primal objetive funtion max
P j
x j
. 2
mumnumber of nodes inan edge equalsthe maximumardinality of afamily of disjoint transversals.
Proof: Let min
be the minimum ardinality of an edge in H , and let A be the inidene matrix of H . Sine A is balaned, by Theorem 3.4, A is
min -olorableand this oloringindues a partitionof V in
min
olors. Let J be a olor. We show that J is a transversal of H . Assume not; then there is anedge e that does not ontain any node oloredJ. Sine jej
min
, there exists a olor, say L, that ontains at least two nodes of e. This shows that the submatrixA
JL
is not biolored, aontradition. 2
The hromati number of ahypergraphis theminimum numberofolors needed to olor its nodes so that no edge ontains two nodes of the same olor.
Theorem 3.9 (Berge [5℄) In a balaned hypergraph H = (V;E), the maxi-mum number of nodes in an edge equals the hromati number of H .
Proof: Let max
be the maximum number of nodes in anedge of H , and let A be the inidene matrix of H . Sine A isbalaned, it is
max
-olorableby Theorem 3.4. By the same argument as before, suh a oloring provides a
oloringof H . 2
Oneof therst resultsonmathingsingraphsisthe followingelebrated theorem of Hall.
Theorem 3.10 (Hall[49℄) A bipartite graph has no perfet mathing if and only if there exist disjoint node sets R and B suh that jBj>jR j and every edge having one endnode in B has the other in R .
The following resultgeneralizes Hall'stheorem tobalaned hypergraphs.
Huk and Triesh [55℄ give aombinatorialproof.
Proof: Assume H admits a perfet mathing M. Then for every disjoint subsetsR , B of V suh that jB\ej jR\ej for every e2E, we have:
jBj= X
e2M
jB\ej X
e2M
jR\ej=jR j:
Sothe ondition isneessary.
We prove suÆieny: Assume H admits no perfet mathing and let A be the node-edge inidene matrix of H . Then by Theorem 2.1, the system Ay=1; y0denesanintegralpolytope. Therefore,sineHhasnoperfet mathing, this system has no solution. Hene, by Farkas' lemma,there is a vetor x suh that A
T
x0 and 1 T
x<0. We an assume 1 x1. Let z = 1 x. Then 0 z 2, A
T
z A T
1 and 1 T
z > 1 T
1 = jVj. Consider the linear program:
min (A T
1) T
u+2 T
v Au+Iv 1 u;v 0:
By Theorem 2.6 its onstraints form a TDI system. Sine the system satisedby z orresponds tothe dual of the abovelinearprogram, it follows that it has an integral solution z. So there is anintegral vetor x suh that A
T
x 0; 1 T
x < 0; 1 x 1. Now set B = fv 2 Vjx v
= 1g and R=fv 2Vjx
v
=1g. Then B, R satisfy the onditions of the theorem. 2
Itis wellknown thata bipartitegraphwith maximumdegreeontains edge-disjoint mathings. The same property holds for balaned hyper-graphs. The following result is equivalent to Theorem 3.9. We provide a proofbased on Theorem 3.11.
Corollary 3.12 Theedges ofabalanedhypergraphHwithmaximumdegree an be partitioned into mathings.
jB \ej for every edge e of H . Adding these inequalities over all edges, we get jR jjBjsine H is -regular, aontradition. SoH ontains a perfet mathingM. Removingtheedgesof M, theresultnowfollowsbyindution. 2
3.4 2-Setion Graphs and Clique-Hypergraphs
The main resultof this setionwas found by Prisner[63℄.
The 2-setion graph of a hypergraph H=(V;E)is the simpleundireted graph G = (V;E) having the same node set as H ; two of its nodes are adjaent if and only if they belong to the same edge of H .
A hypergraph H = (V;E) is a lique-hypergraph if E is the family of all the maximal liques of its 2-setion graph G. Obviously, if H is a lique-hypergraph, H does not ontain any repeated or dominated edges. In [48℄ an algorithm is given, to list the set K of all maximal liques of a graph G=(V;E). ItsrunningtimeisO(jVjjEjjK j). Sothe lique-hypergraph of agraph G an be eÆiently onstruted.
Lemma 3.13 A hypergraph H=(V;E) isa lique-hypergraph if and only if H ontains no dominated or repeated edge,and for every triple of edges, say e 1 ;e 2 ;e 3
, the set of nodes V 123
=(e 1
\e 2
)[(e 2
\e 3
)[(e 1
\e 3
) is ontained in some edge of H .
Proof: LetG bethe 2-setiongraphof H . Sine V 123
is ontainedinalique ofG, the ondition isobviously neessary. We nowprovesuÆieny. If H is notalique-hypergraph, thensomeset ofnodespairwiseadjaentinGisnot ontained in and edge of H ; letV
0
be a minimalsuh set. Clearly jV 0
j 3. BytheminimalityofV
0
,foreveryv 2V 0
,thesetV 0
nv isontainedinanedge e
v
ofH . Assumefv 1
;v 2
;v 3
gV 0
. NowV 0 (e v1 \e v2 )[(e v1 \e v3 )[(e v2 \e v3 ) and e v 1 ;e v 2 ;e v 3
satisfy the above onditions. 2
Let us dene a hypergraph to be semi-balaned if its inidene matrix ontains no 33 hole matrix. Balaned hypergraphs are obviously semi-balaned.
Given hypergraph H = (V;E), let E max
be the subset of E onsisting of one opy of every maximal edge of H , and letH
max
=(V;E max
max (V;E
max
) is a lique-hypergraph.
Proof: By onstrution, H max
ontains no dominated or repeated edge. So assume H
max
is not a lique-hypergraph. By Lemma 3.13, H max
ontains edges e
1 ;e
2 ;e
3
suh that the set V 123
is not ontained in any other edge of H
max
. Inpartiular,thereexistnodesv 12
2(e 1
\e 2
)ne 3
andv 13
;v 23
similarly dened. Let A be the inidene matrix of H . Nowthe rows and olumnsof A orrespondingto v
12 ;v
13 ;v
23 and e
1 ;e
2 ;e
3
induea 33hole matrix. 2
Lemma 3.15 Let H =(V;E) be a semi-balaned hypergraph not ontaining any repeated edges. Then every edge of H
max
ontains two verties that do not belongto any other edge of H .
Proof: Obviously H and H max
have the same 2-setion graph G. F urther-more,sine H is semi-balaned,sois H
max
. Soby Lemma 3.14, H max
isthe lique-hypergraph of G. Assume the lemma is false, and let e 2 E
max be an edge violating the above ondition. Obviously, e ontains at least three nodes. Sine every pair of nodes in e belong to some other edge of E
max , G is also the 2-setion graph of the hypergraph H
max
ne. However, sine e is missing, H
max
ne is not the lique-hypergraph of G. By Lemma 3.14, H
max
neisnotsemi-balanedandhenebothH max
,H ,arenotsemi-balaned,
aontradition. 2
Corollary 3.16 (Prisner [63℄) Let H be a balaned hypergraph that is the lique-hypergraph of G. Then the number of edges of H is bounded by the number of edges of G.
Proof: ByLemma 3.15, every edgeof H ontains anedge of G that belongs
tonoother edge of H . 2
4 Related Integer Polytopes
4.1 k-Balaned Matries
matries are all totally unimodular is said to be almost totally unimodular. Camion[12℄ proved the following:
Theorem 4.1 (Camion[12℄and Gomory[ited in[12℄℄) Let A be an almost totallyunimodular 0;1matrix. ThenA issquare,detA=2 andA
1 has only
1 2
entries. Furthermore, eah row and eah olumn of A has an even number of nonzero entries and the sum of all entries in A equals 2 mod 4.
Proof: Clearly A is square, say nn. If n = 2, then indeed, det A = 2. Now assumen 3. Sine A is nonsingular,it ontains an (n 2)(n 2)
nonsingular submatrix B. Let A =
B C
D E
!
and U =
B 1 0 DB 1 I ! :
Thendet U =1and UA=
I B
1 C
0 E DB
1 C
!
:We laimthat the 22
matrix E DB 1
C has all entries equal to 0;1. Suppose to the ontrary that E DB
1
C has an entry dierent from 0;1 in row i and olumn j. Denoting the orresponding entry of E by e
ij
, the orresponding olumn of C by
j
and row of Dby d i , B 1 0 d i B 1 1 ! B j d i e ij ! = I B 1 j 0 e ij d i B 1 j !
and onsequently A has an (n 1)(n 1) submatrixwith a determinant dierent from0;1,a ontradition.
Consequently, det A =det UA =det(E DB 1
C) =2. So, every entryof A
1
isequalto0; 1 2
. Suppose A 1
has anentryequal to 0, say in row i and olumn j. Let
A be the matrix obtained from A by removing olumni and leth
j
bethe j th
olumn of A 1
with row i removed. Then Ah j =u j
,whereu j
denotes thej th
unitvetor. Sine
Ahasrankn 1, this linear system of equations has a unique solution h
j
. Sine
A is totally unimodular and u
j
is integral, this solutionh j
is integral. Sine h j
6=0,this ontraditsthe fatthat everyentry ofh
j
isequalto0; 1 2
. SoA 1 has only 1 2 entries.
This property and the fatthat AA 1
and A 1
A are integral,implythat A has aneven numberof nonzero entries ineah rowand olumn.
Finally, let denote a olumn of A 1
and S = fi : i = + 1 2 g and
S =fi : i
= 1 2
indexedbyS. SineAisaunitvetor,thesum ofallentriesintheolumns ofAindexedby Sequals k+2. SineeveryolumnofAhas anevennumber of nonzero entries, k is even, say k = 2p for some integer p. Therefore, the
sum of allentries in A equals 4p+2. 2
For any positive integer k, we say that a 0;1 matrix A is k-balaned if A does not ontain any almost totally unimodular submatrix with at most 2k nonzero entries ineah row. Truemper[70℄ givesaonstrutionof allthe minimalmatriesthat are not k-balaned.
Note that every almost totally unimodular matrix ontains at least 2 nonzero entries per row and per olumn. So the odd hole matries are the almosttotally unimodular matrieswith at most 2 nonzero entries per row. Therefore the balaned matries are the 1-balaned matries and the to-tally unimodular matries with n olumns are the k-balaned matries for k bn=2. The lass of k-balaned matries was introdued by Truemper and Chandrasekaran [72℄ for 0,1 matries and by Conforti, Cornuejols and Truemper[24℄for0;1matries. Letkdenoteaolumnvetorwhoseentries are allequal tok.
Theorem 4.2 (Conforti,Cornuejolsand Truemper[24℄) Let Abe anmn k-balaned 0;1 matrix with rows a
i
, i 2 [m℄, b be a vetor with entries b i
, i2[m℄, and let S
1 ;S
2 ;S
3
be a partitionof [m℄. Then
P(A;b)=fx2R n
: a i
xb i
for i2S 1
a i
x=b i
for i2S 2
a i
xb i
for i2S 3
0x1g
is an integral polytope for all integral vetors b suh that n(A) b k n(A).
Proof: Assume the ontrary and let A be a k-balaned matrix of smallest ordersuhthat P(A;b) has afrationalvertex xfor some vetor b suhthat
n(A) b k n(A) and some partition S 1
;S 2
;S 3
of [m℄. Then by the minimalityofA,xsatisesalltheonstraintsinS
1 [S
2 [S
3
atequality. Sowe mayassumeS
1 =S
3
otherwiseletA betheolumnsubmatrixofAorrespondingtothefrational omponentsofx andA
p
bethe olumnsubmatrixofA orrespondingtothe omponents of x that are equal to 1. Let b
f
= b p(A p
)+n(A p
). Then n(A
f )b
f
k n(A
f
)sine b f
=b p(A p
)+n(A p
)=A f
xn(A f
)and beauseb
f
=b p(A p
)+n(A p
)b+n(A p
)k n(A)+n(A p
)k n(A f
). SinetherestritionofxtoisfrationalomponentsisavertexofP(A
f ;b f ) withS 1 =S 3
=;,the minimalityofAisontradited. SoAisasquare non-singularmatrix whihis not totallyunimodular. Let Gbe analmosttotally unimodular submatrix of A. Sine A is not k-balaned, G ontains a row i suh that p
i
(G)+ n i
(G) > 2k. Let A i
be the submatrix of A obtained by removing row i and let b
i
be the orresponding subvetor of b. By the minimalityof A,P(A
i ;b
i
)withS 1
=S 3
=;isanintegralpolytopeandsine A is nonsingular, P(A
i ;b
i
) has exatly two verties, say z 1
and z 2
. Sine
x is a vetor whose omponents are all frational and x an be written as the onvex ombinationof the 0;1 vetors z
1 and z
2
, then z 1
+z 2
=1. For `=1;2, dene
L(`)=fj; either g ij
=1andz ` i
=1or g ij
= 1andz ` i =0g: Sine z 1 +z 2
= 1, it follows that jL(1)j+jL(2)j = p i
(G)+n i
(G) > 2k. Assume w.l.o.g. that jL(1)j >k. Now this ontradits
jL(1)j= X j g ij z 1 j +n i
(G)b i
+n i
(A)k
where the rst inequality follows from A i z 1 =b i . 2
This theorem generalizes previous results by Homan and Kruskal [51℄ for totally unimodular matries, Berge [3℄ for 0;1 balaned matries, Con-forti and Cornuejols [17℄ for 0;1 balaned matries, and Truemper and Chandrasekaran [72℄ for k-balaned 0,1matries.
A 0;1matrix A has ak-equitable bioloringif itsolumnsan be parti-tionedinto blue olumnsand red olumnsso that:
thebioloringis equitableforthe row submatrixA 0
determinedby the rows of A with atmost 2k nonzero entries,
inolumnsof dierent olors.
Obviously, anmn 0;1matrix A isbiolorable if and onlyif A has a 1-equitable bioloring, while A has an equitable bioloring if and only if A has a k-equitable bioloringfor k bn=2. The following theorem provides anewharaterizationof the lass ofk-balaned matries,whihgeneralizes the bioloring results of Setion 3.1 for balaned and totally unimodular matries.
Theorem 4.3 (Conforti,CornuejolsandZambelli[26℄) A0;1matrix A is k-balaned if and only if every submatrix of A has a k-equitable bioloring.
Proof: Assume rst that A is k-balaned and let B be any submatrix of A. Assume, up torowpermutation,that
B = B
0
B 00
!
whereB 0
isthe rowsubmatrixof B determinedby the rows of B with 2k or fewer nonzero entries. Consider the system
B 0
x
$ B
0 1
2 %
B 0
x
& B
0 1
2 '
B 00
x k n(B
00
) (11)
B 00
x k n( B
00 )
0x 1
Sine B is k-balaned, also B
B !
is k-balaned. Therefore the
on-straint matrix of system (11) above is k-balaned. One an readily verify
that n(B 0
) j
B 0
1 2
k
k n(B
0
) and n( B 0
) l
B 0
1 2
m
k n( B
0 ).
Therefore, by Theorem 4.2 applied with S 1
= S 2
= ;, system (11) denes an integral polytope. Sine the vetor (
1 2
;:::; 1 2
i
is,in fat,k-equitable.
Conversely, assumethatAisnot k-balaned. ThenA ontains analmost totally unimodular matrix B with at most 2k nonzero elements per row. Suppose that B has ak-equitablebioloring,then suha bioloringmust be equitablesineeahrowhas,atmost,2knonzeroelements. ByTheorem4.1, Bhas aneven numberofnonzeroelementsineahrow. Thereforethesumof theolumnsoloredblueequalsthesumoftheolumnsoloredred,therefore
B isa singularmatrix, a ontradition. 2
Given a0;1matrixAandpositiveintegerk,oneanndinpolynomial time a k-equitable bioloring of A or a ertiate that A is not k-balaned asfollows:
Find a basi feasible solution of (11). If the solution is not integral, A isnot k-balaned by Theorem 4.2. If the solutionis a 0,1 vetor, it yields a k-equitable bioloringasin the proof of Theorem 4.3.
Notethat,aswiththealgorithmofCameronandEdmonds [10℄disussed inSetion 3.1, a0,1 vetor may befound even when the matrix A is not k-balaned.
Using the fat that the vetor ( 1 2
;:::; 1 2
) is a feasible solution of (11), a basifeasible solutionof(11) an atuallybederived instronglypolynomial time usingan algorithmof Megiddo [59℄.
4.2 Perfet and Ideal 0;1 Matries
A 0;1 matrix A is said to be perfet if the set paking polytope P(A) is integral. A0;1matrixAisideal ifthesetoveringpolytopeQ(A)isintegral. The study of perfet and ideal 0;1 matries is a entral topi in polyhedral ombinatoris. Theorem 2.1 shows that every balaned 0;1 matrix is both perfet and ideal.
inidene vetors of the maximalliques of G.
Theorem 4.4 (Lovasz [57℄, Fulkerson [40℄, Chvatal [14℄) Let A be a 0,1 matrix. The set paking polytope P(A) is integral if and only if the rows of A of maximalsupport form the lique-node matrix of a perfet graph.
Now we extend the denition of perfet and ideal 0;1 matries to 0;1 matries. A 0;1 matrix A is ideal if the generalized set overing polytope Q(A) = fx : Ax 1 n(A); 0 x 1g is integral. A 0;1 matrix A is perfet if the generalized set paking polytope P(A) = fx : Ax 1 n(A); 0x 1g is integral. By Theorem 2.2, balaned 0;1 matries are both perfet and ideal.
Hooker [54℄ was the rst to relate idealness of a 0;1 matrix to that of afamilyof 0,1matries. A similarresult forperfetion wasobtained in[19℄. Theseresults were strengthened by Guenin[46℄ and by Boros,
Cepek [8℄for perfetion, and by Nobili, Sassano [61℄ for idealness. The key tool for these resultsis the following:
Given a 0;1matrix A, letP and Rbe0;1matriesofthe same dimen-sionasA,withentries p
ij
=1if andonlyifa ij
=1,andr ij
=1ifandonlyif
a ij
= 1. The matrix D A
=
P R
I I !
is the 0;1 extensionof A. Notethat
thetransformationx +
=xand x =1 xmaps everyvetorxinP(A)into avetor inf(x
+
;x )0: Px +
+R x 1; x +
+x =1g and every vetor x inQ(A) into a vetor in f(x
+
;x )0: Px +
+R x 1; x +
+x =1g. SoP(A)and Q(A)are respetivelythefaesof P(D
A
)and Q(D A
),obtained by setting the inequalitesx
+
+x 1 and x +
+x 1 at equality. Thus, if P(D
A
) is anintegralpolytope,then so is P(A). Similarly Q(D A
)integral impliesQ(A) integral. Toget a onverse, weintroduethe following notion.
Consider a 0;1 matrix A with two rows a 1
and a 2
suh that there is one index k suh that a
1 k a
2 k
= 1 and, for all j 6= k, a 1 j a
2 j
= 0. A disjoint impliationof Aisthe0;1vetor a
1 +a
2
. Fora0;1matrixA, thematrix A
+
obtained by reursively addingall disjointimpliationsand removingall dominated rows (those whose support is not maximal in the paking ase; thosewhosesupportisnotminimalintheoveringase)isalledthe disjoint ompletionof A. Note that P(A)=P(A
+
) and Q(A)=Q(A +
ideal if and only if the 0;1 matrix D A
+ isideal.
Furthermore Ais ideal if and only if minfx:x2Q(A)g has anintegral optimum x for every vetor 2f0;1;1g
n .
Theorem 4.6 (Guenin[46℄) Let A be a 0;1 matrix suh that P(A) is not ontained in any of the hyperplanes fx:x
j
=0g or fx:x j
=1g. ThenA is perfet if and only if the 0;1 matrix D
A
+ is perfet.
Note that this result does not hold when the assumption on the hyper-planes fx : x
j
= 0g and fx : x j
= 1g is dropped. For example,
on-sider A = 0
B
1 1 1
1 1 1
1 1 1
1
C
A. Then P(A) is an integral polytope sine it
onlyontains the point 0,whereas P(D A
+)
isnot an integral polytopesine A
+
= A and P(D A
) has the frational vertex (x +
;x ) where x +
=( 1 2 ;
1 2 ;
1 2 ) and x =0.
Theorem 4.7 (Guenin[46℄) Let A be a 0;1 matrix suh that P(A) is not ontained in any of the hyperplanes fx : x
j
= 0g or fx : x j
= 1g. Then A is perfet if and only if maxfx : x 2 P(A)g admits an integral optimal solution for every 2 f0;1g
n
. Moreover, if A is perfet, the linear system Ax1 n(A), 0x1 isTDI.
This is the natural extension of the Lovasz's theorem for perfet 0;1 matries. The next theorem haraterizes perfet 0;1 matries in terms of exluded submatries. A row of a 0;1 matrix A is trivial if it ontains at most one nonzero entry. Note that trivial rows an be removed without hangingP(A).
Theorem 4.8 (Guenin[46℄) Let A be a 0;1 matrix suh that P(A) is not ontained in any of the hyperplanes fx:x
j
=0g or fx:x j
=1g. ThenA is perfet if and only if A
+
does not ontain
1)
1 1 1 1
!
or
1 1
1 1
!
minimally imperfet 0,1 matrix B by swithingsigns of all entriesin a subset of the olumns of B.
Forideal0;1matries, asimilarharaterizationwasobtained interms of exluded\weak minors"by Nobiliand Sassano [61℄.
4.3 Propositional Logi
In propositional logi, atomi propositions x 1
;:::;x j
;:::;x n
an be either trueorfalse. Atruthassignmentisanassignmentof"true"or"false"toevery atomiproposition. Aliteral isanatomipropositionx
j
oritsnegation:x j
. Alauseisadisjuntionofliteralsandissatised byagiventruthassignment if atleast one of its literals istrue.
A survey of the onnetions between propositionallogi and integer pro-grammingan be found in[53℄.
A truth assignmentsatises a set of m lauses
_
j2P i
x j
_( _
j2N i
:x j
) for i=1:::;m
ifandonlyif theorresponding0;1vetor satisesthesystem ofinequalities X
j2Pi x
j X
j2Ni x
j
1 jN
i
j for i=1:::;m:
The above system of inequalitiesis of the form
Ax 1 n(A); (12)
where A isan mn 0;1matrix.
weight of the satised lauses. MAXSAT an be formulated as the integer program:
min P
m i=1
w i
s i
Ax+s1 n(A) x2f0;1g
n
;s2f0;1g m
:
Logial inferene inpropositionallogi is assoiated toa set S of lauses (the premises) and a lause C (the onlusion), and onsists of deiding whether every truth assignment that satisesall the premises in S also sat-ises the onlusion C.
Let Ax1 n(A) be the system of inequalitiesassoiated with the set S of premises. The onlusion C = (
W j2P(C)
x j
)_( W
j2N(C) :x
j
) annot be deduedfromSifandonlyifthereexists a0;1vetorsatisfyingthefollowing system:
Ax1 n(A); x
j
=0 for all j 2P(C); x
j
=1 for all j 2N(C):
Equivalently, the onlusion C an be represented by the inequality X
j2P(C) x
j
X
j2N(C) x
j
1 jN(C)j;
or, more ompatly, x 1 jN(C)j where denotes the n-vetor with omponents
j
=1forj 2P(C), j
= 1forj 2N(C)and j
=0otherwise. Then C annot be dedued fromS if and only if the integer program
minfx:Ax1 n(A); x2f0;1g n
g (13)
has a solutionwith value jN(C)j.
programinthat ase:
Theorem 4.9 Let S be an ideal set of lauses. Then SAT, MAXSAT and logial inferene an be solved in polynomial time by linear programming.
This has onsequenes for probabilisti logi as dened by Nilsson [60℄. Being able to solve MAXSAT in polynomial time provides a polynomial time separation algorithm for probabilisti logi via the ellipsoid method, as observed by Georgakopoulos, Kavvadias and Papadimitriou [42℄. Hene probabilistilogi is solvable inpolynomialtime for ideal sets of lauses.
Lemma 4.10 Let S be an ideal set of lauses. If every lause of S ontains more than one literal then, for every atomi proposition x
j
, there exist at least two truth assignments satisfying S, one in whih x
j
is true and one in whih x
j
isfalse.
Proof: Sine the point x j
= 1=2; j = 1;:::;n belongs to the polytope Q(A)=fx:Ax 1 n(A); 0x 1g and Q(A) is an integral polytope, thentheabovepointanbeexpressedasaonvexombinationof0;1vetors inQ(A). Clearly,foreveryindex j, thereexists inthe onvex ombinationa 0;1 vetorwith x
j
=0and anotherwith x j
=1. 2
AonsequeneofLemma4.10isthat,foranidealsetoflauses,SATan be solved more eÆiently than by generallinear programming.
Theorem 4.11 (Conforti,Cornuejols [16℄) Let S be an ideal set of lauses. Then S is satisable if and only if a reursive appliation of the following proedure stops with an empty set of lauses.
Reursive Step
If S =; then S is satisable.
If S ontains a lause C with a single literal (unit lause), set the orre-sponding atomi proposition x
j
so that C is satised. Eliminate from S all lauses that beome satised and remove x
j
from all the other lauses. If a lause beomesempty, then S is not satisable (unit resolution).
IfeverylauseinS ontainsatleasttwoliterals,hooseanyatomi propo-sition x
j
appearing in a lause of S and add to S one of the lauses x j
and :x
problem when S is an ideal set of lauses, see [16℄. For balaned (or ideal) sets of lauses, it is anopen problemto solve MAXSAT inpolynomialtime by a diret method, without appealing to polynomial time algorithms for generallinear programming.
4.4 Nonlinear 0;1 Optimization
Consider the nonlinear 0;1maximizationproblem
max
X
k a
k Y
j2T k
x j
Y
j2R k
(1 x j
)
x 2 f0;1g n
where,w.l.o.g.,allorderedpairs(T k
;R k
)aredistintandT k
\R k
=;. Thisis anNP-hardproblem. A standardlinearizationofthis problemwasproposed by Fortet [39℄:
max X
a k
y k
y k
x j
0for all k s:t:a k
>0; for allj 2T k
y k
+x j
1for all k s:t:a k
>0; for allj 2R k
y k
X
j2T k
x j
+ X
j2R k
x j
1 jT
k
jfor all k s:t:a k
<0
y k
; x j
2 f0;1gfor all k andj:
When the onstraint matrix is balaned, this integer program an be solved as a linear program, as a onsequene of Theorem 2.7. Therefore, in this ase, the nonlinear 0;1 maximizationproblem an be solved in polyno-mialtime. The relevaneof balanedness inthis ontextwaspointed out by Crama[33℄.
5 The Struture of Balaned Matries
5.1 Bipartite Representation of a 0;1 Matrix
bipartite.
The bipartite representation of a 0;1 matrix A is the bipartite graph G(A)=(V
r [V
;E) having a node in V r
for every row of A, a node in V
forevery olumnofA and anedgeij joiningnodes i2V r
and j 2V
ifand onlyif the entry a
ij
of A equals 1.
Notethat a0;1matrix isbalaned if and onlyif itsbipartite representa-tion isa balaned graph.
The bipartite representation of a 0;1 matrix A is the signed bipartite graphG(A)=(V
r [V
;E)havinganodeinV r
foreveryrowof A,anode in V
for every olumn of A and anedge ij joiningnodes i2V r
and j 2V
if andonlyiftheentrya
ij
isnonzero. Furthermorea ij
isthesignoftheedgeij. This onept extends the one introdued above. Conversely, for a bipartite graphG=(V
r [V
;E),withsigns 1onitsedges, thereisauniquematrix A for whih G = G(A) (up to transposition of the matrix, permutation of rows and permutation of olumns).
5.2 Signing 0,1 Matries: Camion's Algorithm and
Truemper's Theorem
A0;1matrixisbalaneableifitsnonzeroentriesanbesigned+1or-1sothat the resulting 0;1 matrix is balaned. A bipartitegraph G is balaneableif G=G(A) and A isa balaneable matrix.
Camion [12℄ observed that the signing of a balaneable matrix into a balaned matrixis unique up tomultiplying rows orolumnsby 1,and he gave a simple algorithmto obtain this signing. We present Camion's result next.
LetA bea 0;1 matrix and letA 0
beobtained from A by multiplying a set S of rows and olumns by 1. A is balaned if and only if A
0
is. Note that,inthebipartiterepresentationof A,thisorresponds toswithingsigns onalledgesoftheutÆ(S). NowletR bea0,1matrixandG(R )itsbipartite representation. Sine every edgeof amaximal forest F of G(R )is ontained in a ut that does not ontain any other edge of F, it follows that if R is balaneable, there exists a balaned signing of R in whih the edges of F have any speied (arbitrary)signing.
CAMION'S SIGNING ALGORITHM
Input: A 0,1 matrix A and its bipartite representation G, a maximal forest F of G and an arbitrary signing of the edges of F.
Output: A signing of G in whih the edges of F are signed as speied in the input, and if A is balaneablethen the signing isbalaned.
Index the edges of G e 1
;:::;e n
, so that the edges of F ome rst, and every edge e
j
, j jFj+1, together with edges havingsmaller indies, loses aholeH
j
of G. For j =jFj+1;:::;n,signe j
so thatthesum of theweights of H
j
isongruentto 0mod4.
Notethat the rows and olumnsorresponding tothe nodes of H j
dene ahole submatrix of A.
Thefatthatthereexists anindexingofthe edgesofGasrequiredinthe signingalgorithmfollows fromthe followingobservation. Forj jFj+1,we anselet e
j
sothat the path onnetingthe endnodes of e j
inthe subgraph (V(G);fe
1 ;:::;e
j 1
g) isshortest possible. The holeH j
identiedthisway is alsoa holeinG. Thisfores the signingof e
j
, sineallthe otheredges ofH j are signed already. So, one the (arbitrary) signing of F has been hosen, the signingof G is unique. Therefore we have the following result.
Theorem 5.1 If the input matrix A is a balaneable 0,1 matrix, Camion's signing algorithm produes a balaned 0;1 matrix B. Furthermore every balaned0;1matrixthatarises from A bysigningits nonzeroentries either +1or 1, an be obtained byswithing signs on rows and olumns of B.
If one applies Camion's algorithm to the bipartite representation of the following matrix, the signing produed would leave one of the four holes unbalaned, proving that the matrix is not balaneable.
0
B
1 1 0 1 1 0 1 1 0 1 1 1
1
C A
asfollows. Let G bea opy of Gthat is not signed. Test whether G is bal-aneable. Ifitisnot,then Gisnot balaned. Otherwise,letF beamaximal forest of G
0
. Run the signingalgorithmon G 0
with the edges of F signed as they are inG. Then G isbalaned if and only if the signing of G
0
oinides with the signingof G.
We now give a haraterization due to Truemper [71℄ of the bipartite graphsthat are balaneable.
In a bipartite graph, a wheel (H;v) onsists of a hole H and a node v havingatleast threeneighborsinH. The wheel(H;v)isoddifv has anodd number of neighbors in H. A 3-path onguration is an indued subgraph onsistingofthreeinternallynode-disjointpathsonnetingtwononadjaent nodesuandv andontainingnoedgeotherthanthoseofthepaths. Ifuand v are in opposite sides of the bipartition, i.e. the three paths have an odd numberofedges,the3-pathongurationisalleda3-odd-pathonguration. InFigure1,solid linesrepresent edgesand dottedlines represent pathswith atleast one edge.
u H
v
v
Figure1: Anodd wheel and a 3-odd-path onguration
bal-balaneable. These are in fat the only minimal bipartite graphs that are not balaneable,as shown by the following theorem.
Theorem 5.2 (Truemper[71℄)Abipartitegraphisbalaneableifand onlyif it does not ontain an odd wheelor a 3-odd-path onguration as an indued subgraph.
We proveTheorem 5.2following Conforti, Gerardsand Kapoor [27℄. For a onneted bipartitegraph G that ontains a lique utset K
t with t nodes, let G
0 1
;:::;G 0 n
be the onneted omponents of GnK t
. The bloks of Gare the subgraphs G
i
indued by V(G 0 i
)[K t
for i=1;:::;n.
Lemma 5.3 If a onneted bipartite graph G ontains a K 1
or K 2
utset, then G is balaneableif and only if eah blok is balaneable.
Proof: IfG isbalaneable,thensoare the bloks. Thereforeweonlyhave to prove the onverse. Assume that all the bloks are balaneable. Give eah blok a balaned signing. If the utset is aK
1
utset, this yields abalaned signingofG. Ifthe utsetisaK
2
utset,re-sign eah blok sothatthe edge ofthatK
2
hasthe sign+1. Nowtaketheunionofthesesignings. Thisyields
abalaned signing of G again. 2
Thus, intheremainder oftheproof,weanassumethatGisaonneted bipartitegraphwith noK
1 orK
2
utset.
Lemma 5.4 Let H be a hole of G. If G 6= H, then H is ontained in a 3-path onguration or a wheel of G.
Proof: Choose two nonadjaent nodes u and w in H and a uw-path P = u;x;:::;z;w whose intermediate nodes are inGnH suh that P is asshort as possible. Suh a pair of nodes u;w exists sine G6=H and G has noK
1 or K
2
utset. If x = z, then H is ontained in a 3-path onguration or a wheel. So assume x6=z. Byour hoie of P, u is the only neighborof x in H and wis the only neighborof z inH.
and that y is adjaent to u and w. But then V(H)[V(P) indues awheel
with enter y. 2
For e 2 E(G), let G e
denote the graph with a node v H
for eah hole H ofG ontaininge and anedge v
H i
v H
j
if and onlyif there exists a wheelor a 3-pathonguration ontaining both holes H
i
and H j
.
Lemma 5.5 G e
is a onneted graph.
Proof: Supposenot. Lete=uw. Choosetwoholes H 1
and H 2
ofGwith v H
1 andv
H 2
indierentonnetedomponentsofG e
,withtheminimumdistane d(H
1 ;H
2
) in Gnfu;vg between V(H 1
) fu;wg and V(H 2
) fu;wg and, subjet tothis, with the smallest jV(H
1
)[V(H 2
)j. Let T be a shortest path from V(H
1
) fu;vg to V(H 2
) fu;vg in Gn fu;vg. Notethat T isjustanodeof V(H
1
)\V(H 2
)nfu;vgwhen this set is nonempty. The graph G
0
indued by the nodes in H 1
, H 2
and T has noK 1 or K
2
utset. By Lemma 5.4, H 1
is ontained in a 3-path onguration or a wheel of G
0
. Sine eah edge of a 3-path onguration ora wheel belongs to two holes, there exists a hole H
3 6= H
1
ontaining edge e in G 0 . Sine v H1 and v H3
are adjaent in G e
, it follows that v H2
and v H3
are in dierent omponents of G
e
. Sine H 1
and H 3
are distint holes, H 3
ontains a node inV(H
2
)[V(T)nV(H 1
). IfH 3
ontainsanode inV(T)n(V(H 1
)[V(H 2
)), then V(H
1
)\V(H 2
)=fu;vgand d(H 3
;H 2
)<d(H 1
;H 2
)a ontradition to the hoie of H
1 ,H 2 . Therefore H 3
ontains a node x in V(H 2
)nV(H 1
). Byour hoie of H 1
, H
2
, we have that V(H 1
)\ V(H 2
)n fu;vg is nonempty. Let P 1 = H 1 ne and P 2 = H 2
ne and let s, t be the nodes in V(H 1
)\V(H 2
) suh that the st-subpathP
st 2
ofP 2
ontainsx andisshortest. LetP st 1
bethe st-subpathof P
1
. SineH 2
isahole,P st 1
ontainsanintermediatenodez 2V(H 1
)nV(H 2
). Now V(H
3
)[V(H 2
) is ontained in V(H 1
)[V(H 2
)nz, a ontradition to our hoie of H
1 ,H
2
. 2
1 2
edge of G. Sine Gnfu;vg is onneted, there exists a spanning tree F of G where u and v are leaves. Arbitrarily sign F and use Camion's signing algorithminGnfugand Gnfvg. Bythe minimalityof G,these two graphs are balaneable and therefore Camion's algorithm yields a unique signing of all the edges exept e. Furthermore, all holes not going through edge e are balaned. Sine G is not balaneable, any signingof e yieldssome holes going through e that are balaned and some that are not. By Lemma 5.5, there exists a wheel or a 3-path onguration C ontaining an unbalaned hole H
1
and a balaned hole H 2
both going through edge e. Now we use the fat that eah edge of C belongs to exatly two holes of C. Sine the holes of C distint fromH
1
and H 2
do not gothrough e, they are balaned. Furthermore,applyingthe above fat toalledges of C, the sum of alllabels inC is 1mod2,whihimplies that C has an odd number ofedges. Thus C isan odd wheel ora 3-odd-path onguration, aontradition. 2
5.3 Deomposition Theorems
Inthis setion,wepresentdeompositiontheoremsforbalaned0;1matries duetoConforti,CornuejolsandRao[23℄andbalaneable0;1matriesdueto Conforti,Cornuejols,KapoorandVuskovi[21℄. Westatethe deomposition theoremsintermsof thebipartiterepresentationofsuhmatries,asdened inSetion 5.1.
5.3.1 Cutsets
AsetS ofnodes(edges)ofaonnetedgraphGisanode(edge) utset ifthe subgraphofGobtained byremovingthe nodes (edges)inS,isdisonneted.
Foranode x, letN(x)denote theset of allneighborsof x. In abipartite graph, an extended star is dened by disjoint subsets T, A, N of V(G) and anode x2T suh that
(i) N N(x),
(ii) every node of A is adjaent to every node of T,
T
x
A
N
Figure2: Extended star
Figure3: A 1-join,a 2-joinand a 6-join
This onept was introdued by Conforti, Cornuejols and Rao [23℄ and is illustrated inFigure 2. An extended star utset is one where T [A[N is a node utset. An extended star utset with N =; is alled a bilique utset. Anextended star utsethaving T =fxg isalled a star utset. Notethat a star utset isa speial ase of a bilique utset.
A graph G has a 1-join if its nodes an be partitioned into sets H 1
and H
2
, with jH 1
j 2 and jH 2
j 2, so that A 1
H
1 , A
2
H
2
are nonempty, allnodes of A
1
are adjaent to all nodes of A 2
and these are the only adja-enies between H
1
and H 2
. This onept was introdued by Cunningham and Edmonds [35℄.
A graph G has a 2-join if its nodes an be partitioned into sets H 1
and H
2
so that A 1
;B 1
H
1 , A
2 ;B
2
H
2
where A 1
, B 1
, A 2
, B 2
are nonempty and disjoint,all nodes of A
1
are adjaent to allnodes of A 2
10
are adjaent to all nodes of B 2
and these are the only adjaenies between H
1
and H 2
. Also, for i = 1;2, H i
has at least one path from A i
to B i
and if A
i
and B i
are both of ardinality 1, then the graph indued by H i
is not a hordless path. We also say that E(K
A 1
A 2
)[E(K B
1 B
2
) is a 2-join of G. This onept was introdued by Cornuejols and Cunningham [32℄.
In a onneted bipartite graph G, let A i
, i = 1;:::;6, be disjoint non-empty node sets suh that, for eah i, every node inA
i
is adjaent to every node in A
i 1 [ A
i+1
(indies are taken modulo 6), and these are the only edgesinthesubgraph Aindued by thenode set [
6 i=1
A i
. Assumethat E(A) is an edge utset but that no subset of its edges forms a 1-join or a 2-join. Furthermore assume that no onneted omponent of GnE(A) ontains a node in A
1 [A
3 [A
5
and a node in A 2
[A 4
[A 6
. Let G 135
be the union of the omponents of GnE(A) ontaining a node in A
1 [A
3 [A
5
and G 246 be the union of omponents ontaining a node in A
2 [ A
4 [A
6
. The set E(A) onstitutes a 6-join if the graphs G
135
and G 246
A graph is strongly balaneable if it is balaneable and ontains no yle with exatly one hord. This lass of bipartitegraphs is well studied in the literature,see[28℄. WedisussitinSetion5.5.2. Thefollowinggraph,whih is not strongly balaneable, plays an important role: R
10
is the bipartite graph on ten nodes dened by the yle C = x
1 ;:::;x
10 ;x
1
of length ten withhordsx
i x
i+5
,1i5,seeFigure4. Equivalently,R 10
isthebipartite
representationofthematrix 0
B B B B B B
1 1 0 1 0 0 1 1 0 1 1 0 1 1 0 0 1 0 1 1 1 0 1 0 1
1
C C C C C C A
,whihappears inSeymour's
deomposition of totallyunimodular matries [66℄. Note that the signing of R
10
that assigns +1 to the edges of C and 1 to all the other edges is a balaned signing of R
10
. The orresponding 0;1 matrix is atually totally unimodular.
Theorem 5.6 (Conforti,Cornuejols,KapoorandVuskovi[21℄)Abalan e-able bipartite graph that is not strongly balaneableis either R
10
or ontains a 2-join, a 6-join or an extended star utset.
Figure5exhibitsexamplesshowingthatnoneofthethreekindsofutsets an be dropped fromTheorem 5.6.
A triad onsists of three internally node-disjoint paths t;:::;u; t;:::;v and t;:::;w, where t, u, v, w are distintnodes and u, v, w belong to the same side of the bipartition. Furthermore,the graph indued by the nodes of the triad ontains no other edges than those of the three paths. Nodes u, v and ware alled the attahments ofthe triad.
Afanonsistsofahordlesspathx;:::;ytogetherwithanodezadjaent to at least one node of the path, where x, y and z are distint nodes all belonging to the same side of the bipartition. Nodes x, y and z are alled the attahments of the fan.
A onneted 6-hole is a graph indued by two disjoint node sets T() and B() suh that eah indueseither a triad ora fan,the attahmentsof T() and B() indue a 6-hole and there are no other adjaenies between the nodes of T() and B(). Figure 6 depits the four types of onneted 6-holes.
The following theorem onerns the lass ofbalaneable bipartitegraphs that donot ontain a onneted 6-holeor R
10
asindued subgraph.
Theorem 5.7 (Conforti, Cornuejols and Rao [23℄) A balaneable bipartite graph not ontaining R
10
or a onneted 6-hole as indued subgraph either is strongly balaneable or ontains a 2-join or an extended star utset.
Soitremainstondadeompositionofbalaneablebipartitegraphsthat ontain R
10
oronneted6-holes asindued subgraph. Thisis aomplished asfollows.
Theorem 5.8 (Conforti,Cornuejols,Kapoorand Vuskovi[21℄) A balan e-ablebipartitegraphontainingR
10
asaproperinduedsubgraphhasabilique utset.
Theorem 5.9 ([21℄)A balaneablebipartite graphthat ontains aonneted 6-hole as indued subgraph, has an extended star utset or a 6-join.
Conforti,Cornuejols,KapoorandVuskovi[21℄giveapolynomialtime algo-rithm to hek whether a 0;1 matrix A is balaned. The algorithmworks onthe bipartiterepresentation G(A)introdued. Sine eahedge ofG(A) is signed +1 or 1 aording to the orresponding entry in the matrix A, we allGa signed bipartite graph.
Let G be a onneted signed bipartite graph. The removal of a node utsetoredgeutsetdisonnetsG intotwoormoreonneted omponents. Fromtheseomponentsweonstrutbloks ofdeompositionbyaddingsome new nodes and signed edges. We say that a deomposition is balanedness preserving when it has the following property: all the bloks are balaned if and only if G itself is balaned. The entral idea in the algorithm is to de-ompose G using balanedness preserving deompositionsintoapolynomial number of basi bloks that an be heked for balanedness in polynomial time.
For the 2-join and 6-join, the bloks an be dened so that the deom-positions are balanedness preserving. For the extended star utset it is not known how to onstrut bloks of deomposition that are balanedness preserving and generate a polynomial deomposition tree. Tooverome this problem,thealgorithmusesthe ideaofleaning,rstintroduedby Conforti and Rao[29℄, [30℄. An inputgraph Gis rst transformed into alean graph G
0
(to be dened later), and then G 0
is deomposed, the deompositions in G
0
being balanedness preserving.
Reently Zambelli[74℄, based onanidea introduedby Chudnovsky and SeymourforreognizingBergegraphs[15℄,hasgiven apolynomialalgorithm to test balanedness in a signed bipartite graph that does not use the de-omposition theorem: it uses leaning and shortest paths tehniques. We summarizehere the ideas behind hisalgorithm.
The algorithm rst detets whether the input graph has a 3-odd-path onguration(asdened inSetion 5.2), based onthe followingresult:
Inabipartite graphG, onsidera3-odd-pathongurationwiththe small-est number of nodes, indued by paths P
1 ;P
2 ;P
3
onneting nodes u and v. Let m
i
be a middle node of path P i
. In a subgraph obtained from G by re-movingsomeneighborsof uand v, any shortest pathfrom m
i
to uand v an be substituted for P
i
tite graphontains a 3-odd-path onguartion.
A detetable 3-wheel is a wheel (H;v) where v has three neighbors inH andtwo ofthe neighbors of v inH havedistane twoinH. Byananalogous methodZambellishows the following:
There exists a polynomial time algorithm that heks whether a bipartite graph that does not ontain a 3-odd-pathonguration, ontains a detetable 3-wheel.
ByTheorem5.2, if abipartitegraphontains a3-odd-path onguration ora detetable 3-wheel, itis not balaneable.
A node v is major for a hole H if v has at least three neighbors in H. Thefollowingresultisproved by Conforti,Cornuejols,KapoorandVuskovi [21℄.
Theorem 5.10 Let H be a smallest unbalaned hole in a signed bipartite graph. Then H ontains two edges suh that every major node for H is adjaent to at least one of the endnodes of these two edges.
A signed bipartite graph is lean if it is either balaned or ontains a smallestunbalaned hole H with no major verties for H.
Based onthe above theorem a polynomial time algorithmis onstruted in [21℄, that takes as input a signed bipartite graph G and outputs a lean graphG
0
, suh that Gis balaned if and onlyif G 0
is balaned.
Let G be a signed bipartite graph that does not ontain a 3-odd path ongurationnoradetetable3-wheel. ThelaststepofZambelli'salgorithm isbased on the following:
Let G be a lean signed bipartite graph that does not ontain a 3-odd-path onguration or a detetable 3-wheel. There exists a polynomial time algorithm,based onshortestpathmethods,thathekswhetherGisbalaned.
Several sublasses of balaned matries have beautiful deomposition prop-ertiesof their own. Totallyunimodularmatriesfor example an be deom-posed using adeep theorem of Seymour [66℄. This result is surveyed in[64℄, [62℄ or [31℄ and we do not review it here. We review instead the struture and properties of several other lasses of balaned matries.
5.5.1 Totally Balaned 0;1 Matries
A 0;1 matrix A is totally balaned if every hole submatrix of A is the 22 submatrix of all 1s. Equivalently, a bipartite graph G is totally balaned if every hole of G has length 4. Totally balaned matries arise in loation theory. Several authors (Golumbi and Goss [45℄, Anstee and Farber [1℄, Homan, Kolen and Sakarovith [52℄ and Lubiw [58℄ among others) have given properties of these matries.
Abilique isaomplete bipartitegraphwithatleast one node fromeah sideof thebipartition. Foranode u,letN(u)denotetheset ofallneighbors of u. An edge uv is bisimpliial if the node set N(u)[ N(v) indues a bilique. The following theorem of Golumbi and Goss [45℄ haraterizes totallybalaned bipartitegraphs.
Theorem 5.11 (Golumbi,Goss,[45℄)Atotallybalanedbipartitegraphhas a bisimpliial edge.
Thistheoremyieldsapolynomialtime algorithmtotestwhether a bipar-tite graph G is totallybalaned: for if e is a bisimpliialedge of G, then G istotallybalaned if and onlyif Gne is totallybalaned.
A 0;1matrix A isin standard greedy formif it ontains no22
subma-trix of the form
1 1 1 0
!
, where the order of the rows and olumns in the
submatrix is the same as in the matrix A. This name omes from the fat that the linear program
max X
y i
yA (14)
1 k 1 P k 1 i=1 a ij y i j
; j = 1;:::;n and 0 y i
p i
; i = 1;:::;k 1, set y k
to the largest value suh that
P k i=1 a ij y i j
; j = 1;:::;n and 0 y k
p k
: Theresultinggreedy solutionisanoptimum solutiontothis linearprogram. What doesthis havetodo withtotallybalaned matries? Theansweris in the next theorem.
Theorem 5.12 (Anstee, Farber [1℄, Homan, Kolen, Sakarovith [52℄, Lu-biw[58℄) A 0;1matrix istotally balaned ifand only ifits rowsand olumns an be permuted into standard greedy form.
This transformation an be performed intime O(nm 2
)[52℄.
Totally balaned 0;1 matriesome up invariousways in the ontextof faility loation problems ontrees. Forexample, the overing problem
min n X 1 j x j + m X 1 p i z i X j a ij x j +z i
1; i=1;:::;m (15)
x j
;z i
2 f0;1g
an be interpreted as follows: j
is the set up ost of establishing a faility atsite j, p
i
isthe penalty if lienti isnot served by any faility,and a ij
=1 if afailityat site j an serve lient i,0 otherwise.
When the underlying network is a tree and the failitiesand lients are loatedatnodes ofthetree,itisustomarytoassumethatafailityatsitej anserve allthe lientsina neighborhoodsubtree of j,namely,allthe lients withindistane r
j
fromnode j.
AnintersetionmatrixofthesetfS 1
;:::;S m
gversusfR 1
;:::;R n
g,where S
i
, i=1;:::;m, andR j
,j =1;:::;n,are subsetsofagiven set,isdened to bethe mn 0;1 matrix A=(a
ij
)where a ij
=1if and only if S i
\R j
6=;.
Theorem 5.13 (Giles[44℄)Theintersetionmatrixofneighborhoodsubtrees versus nodes of a tree is totally balaned.
dual of (14), the greedy solution desribed earlier for (14) an be used, in onjuntion with omplementary slakness, to obtain an elegant solution of the overing problem. The above theorem of Giles has been generalized as follows.
Theorem 5.14 (Tamir [67℄) The intersetion matrix of neighborhood sub-trees versus neighborhood subtrees of a tree is totally balaned.
Otherlassesoftotallybalaned0;1matriesarisingfromloation prob-lems ontrees an befound in [68℄.
5.5.2 Restrited and Strongly Balaned Matries
Asigned bipartitegraph G isrestrited balaned if the weight of every yle ofGisongruentto0mod4. Asignedbipartitegraphisstrongly balanedif every yle of weight 2 mod4 has at least two hords. Restrited (strongly, resp.) balaned 0;1 matries are dened to be the matries whose bipar-tite representation is a restrited (strongly,resp.) balaned bipartite graph. It follows from the denition that restrited balaned 0;1 matries are strongly balaned, and it an be shown that strongly balaned 0;1 matri-es are totallyunimodular,see [28℄. Restrited (strongly, resp.) balaneable 0,1matries are those where the nonzero entries an be signed +1or 1 so that the resulting0;1matrix is restrited (strongly,resp.) balaned.
Theorem 5.15 (Conforti, Rao [28℄) A strongly balaneable bipartite graph either isrestrited balaneable or ontains a 1-join.
Crama, Hammer and Ibaraki [34℄ dene a 0;1matrix A to be strongly unimodular if every basis of (A;I) an be put in triangularform by permu-tationof rows and olumns.
Theorem 5.16 (Crama, Hammer, Ibaraki [34℄) A 0;1 matrix is strongly unimodular if and only if it is strongly balaned.
either is 2-bipartite or ontains a utnode or ontains a 2-join onsisting of two edges.
Based on this theorem, Yannakakis designed a linear time algorithmfor hekingwhether a0;1matrixisrestritedbalaned. Adierentalgorithm forthis reognitionproblem wasgiven by Confortiand Rao[28℄:
Construtaspanningforestin thebipartite graph andhekifthere exists a yle of weight 2 mod 4 whih is either fundamental or is the symmetri dierene of fundamental yles. If no suh yle exists, the signed bipartite graph is restrited balaned.
Abipartitegraphislinear if itdoesnot ontain ayle oflength4. Note thatanextendedstarutsetinalinearbipartitegraphisalwaysastarutset, duetoCondition (ii)inthedenition ofextended starutsets. Confortiand Rao[29℄ proved the following theorem for linearbalaned bipartitegraphs:
Theorem 5.18 (Conforti,Rao[29℄)A linear balaned bipartite graph either is restrited balaned or ontains a star utset.
A yle C in a signed bipartite graph G is unbalaned if the sum of the weights of the edges in C is ongruent to 2 mod 4. It is easy to see that a signed bipartite graph has a balaned yle if and only if it has a balaned hole. Itfollowsthatthefollowingtwolassesofgraphsareequivalent: signed bipartite graphs in whih all yles are unbalaned, and signed bipartite graphs in whih all holes are unbalaned. These graphs are haraterized by Conforti, Cornuejols and Vuskovi in [25℄, where a linear algorithm for testingmembership inthis lass isgiven.
5.6 Some Conjetures and Open Questions
5.6.1 Eliminating Edges
Conjeture 5.19 (Conforti, Cornuejols, Kapoor, Vuskovi [21℄) In a bal-aned signed bipartite graph G, either every edge belongs to some R
