2194-2610-1-PB

8/12/2019 2194-2610-1-PB

1/30

An Algebraic Exploration of Dominating Sets

and Vizings Conjecture

S. MarguliesDepartment of Mathematics

Pennsylvania State UniversityState College, PA

[email protected]

I. V. HicksComputational and Applied Mathematics

Rice UniversityHouston, TX

[email protected]

Submitted: Oct 7, 2011; Accepted: Mar 29, 2012; Published: Apr 7, 2012

Mathematics Subject Classifications: 05C69, 13P10

Abstract

Systems of polynomial equations are commonly used to model combinatorialproblems such as independent set, graph coloring, Hamiltonian path, and others.We formulate the dominating set problem as a system of polynomial equations intwo different ways: first, as a single, high-degree polynomial, and second as a col-lection of polynomials based on the complements of domination-critical graphs. Wethen provide a sufficient criterion for demonstrating that a particular ideal represen-tation is already the universal Grobner bases of an ideal, and show that the secondrepresentation of the dominating set ideal in terms of domination-critical graphs

is the universal Grobner basis for that ideal. We also present the first algebraicformulation of Vizings conjecture, and discuss the theoretical and computationalramifications to this conjecture when using either of the two dominating set repre-sentations described above.

Keywords: dominating sets, Vizings conjecture, universal Grobner bases

1 Introduction

The combination of non-linear models and techniques from computer algebra is gaining

acceptance as a tool for exposing combinatorial properties of graph-theoretic problems.For example, prior work on polynomial encodings includes colorings [3, 13, 18, 21, 23,24, 25, 27, 28, 30], stable sets [20, 21, 25, 31], matchings [14], and flows [3, 29, 30].These non-linear encodings have been used to prove combinatorial results [2,22,24], andGrobner bases have been the building blocks of algorithms for solving integer programs[6, 10,34], but techniques from computer algebra are not widely used by combinatoristsand graph theorists, and thus have not yet been deeply explored. In this paper, we move

the electronic journal of combinatorics 19(2) (2012), #P1 1

8/12/2019 2194-2610-1-PB

2/30

beyond simply formulating a graph-theoretic problem as a system of polynomial equations,and instead formulate an entire graph-theoretic conjecture using systems of polynomialequations and algebraic techniques. Although the method we introduce here can probablybe applied to other open questions involving inequalities, we focus on the dominating setproblem and an algebraic approach to a famous open question from domination theory:

Vizings conjecture.Given a graphG, a setDV(G) is adominating setif for allv /D, there is auD

such that v is adjacent to u in G. The domination number ofG, denoted by(G), is thesize of a minimum dominating set in G. The decision problem of determining whether agiven graph has a dominating set of size k is NP-complete [15]. Given graphs G and H,the Cartesian product graph GHhas vertex set V(G) V(H), and edge set

E(GH) =

(gh,gh) : g= g and (h, h)E(H), or h = h and (g, g)E(G),where g, g V(G) and h, h V(H). In 1968, V. Vizing conjectured a beautiful rela-tionship between domination numbers and Cartesian product graphs:

Conjecture 1 (Vizing[35], 1968). Given graphsGand H, (G)(H) (GH).

Vizings conjecture is an active area of research spanning over forty years. Early resultshave focused on proving the conjecture holds for a certain classes of graphs. For example,in 1979, Barcalkin and German [5]proved that Vizings conjecture holds for graphs sat-isfying a certain partitioning condition on the vertex set. The idea of a partitioningcondition inspired work for the next several decades, as Vizings conjecture was shown tohold on paths, trees, cycles, chordal graphs, graphs satisfying certain coloring properties,and graphs with(G) 2. These results are clearly outlined in the 1998 survey paper byHartnell and Rall[17]. In 2000, Clark and Suen[9] showed that (G)(H) 2(GH),and in 2004, L. Sun [33]showed that Vizings conjecture holds on graphs with (G) 3.Finally, in 2009, Hartnell and Rall, et al. [7] contributed another thorough survey pa-per summarizing the work from 1968 to 2008, which contains new results, new proofs ofexisting results, and comments about minimal counter-examples.

Vizings conjecture is of great interest to graph theorists for two primary reasons. First,it is an easily stated, easily understood combinatorial problem whose proof has eludeddemonstration for decades. Second, it is of algorithmic interest, since domination ingraphs has applications ranging from optimal fire station placement to wireless networks(see survey [7]), and additionally, product graphs are a common algorithmic tool forstudying the scaling of NP-hard graph-theoretic invariants. In short, an elegant, elusiveconjecture with algorithmic consequences will always excite exploration.

We begin by reviewing basic ideas from algebraic geometry: unions of varieties, inter-sections of ideals, notions of radical ideals, and universal Grobner bases (Sections2and3).In Section3, for certain ideals, we develop a criterion for identifying a particular basis ofthe ideal as the universal Grobner basis. In Section4.1, we represent both the problem offinding graphsG with dominating sets of size k, and the problem of finding graphs G andH with dominating sets of size k, l respectively such that the Cartesian product graphGHhas a dominating set of size r, as systems of polynomial equations. In Section 4.2,


8/12/2019 2194-2610-1-PB

3/30

we develop the idea of k-domination-covers, or k-covers, which are the complements ofk-dominating graphs. We identify specific properties ofk-covers, which translate to newproperties of domination-critical graphs.

In Section4.3, we unify the seemingly disparate results that have appeared thus farin our paper: we prove that the sameideals described in Section4.1can be represented

by a set of polynomials based on k-covers, that is, by a set of polynomials based on thecomplements of domination-critical graphs. By the results in Section3, we demonstratethat this representation is the universal Grobner basis of the ideal.

Our paper culminates in Section5with an algebraic representation of Vizings conjec-ture. This representation is built upon the union of certain varieties and the intersectionof certain ideals. We initially present the algebraic version of Vizings conjecture withoutrespect to a particular representation. We then discuss the consequences of using eitherof the two representations presented in Sections4.1and4.3. We include comments aboutcomputational results and future computational directions, as well as approaches froma purely graph theory perspective. We conclude by clarifying the relationship betweenuniversal Grobner bases and Vizings conjecture viak-covers, and reclaim a known resultwhere Vizings conjecture holds.

2 Algebraic Definitions and Background

In this section, we outline the basic definitions and results concerning ideals and varietiesthat are utilized throughout the paper, and that are particularly necessary for the alge-braic approach to Vizings conjecture described in Section5. The results presented in thissection are well-known in the field of algebraic geometry and are presented in detail in [ 11],Chapters 2 and 4. The only new contribution is Lemma3, which is a small applicationof well-known results. The ideals referenced throughout are always ideals in polynomialrings, i.e. I K[x1, . . . , xn], where K is an algebraically-closed field. In our case, K = C.

Given a system of polynomial equations, f1 =f2 = = fs = 0, the ideal associatedwith the system is I=f1, . . . , f s, and the variety associated I (i.e. V(I)) is the set ofcommon zeros of{f1, . . . , f s}. In other words,V(I) ={x Cn :f1(x) = = fs(x) = 0}.An ideal iszero-dimensional ifV(I) is finite. Throughout the paper, we will often write apolynomial f I as()fi. In this case, () represents the coefficients of the generatorsfi, but because we do not refer to the coefficients explicitly, we do not need to give themindividual and precise labels such as ai.

Given two ideals, I =f1, . . . , f s and J =g1, . . . , gt, the product ideal IJ is theideal generated by all polynomials f g with f I and gJ. It can be shown ([11], pg183, Prop. 6) that I J (often denoted by I J) is the ideal generated byfigj : 1 i s, 1 j t. We summarize the results concerning the varieties ofI and Jas follows:

V(I) V(J) =V(figj : 1 i s, 1 j t) =V(I J) =V(I J).

An ideal I is radical iffm I, for some integer m 1, implies that f I. Given anidealI, the radical ofI, denoted

I, is the set{f : fm I for some integer m 1}. It


8/12/2019 2194-2610-1-PB

4/30

is easy to see that an ideal Iis radical if and only ifI=

I. We recall the following factconcerning product ideals and radical ideals (left as an exercise in [11]): Given radicalidealsI , J K[x1, . . . , xn],

IJ=I J. We note that in one particular case, it is trivial

to determine whether or not an ideal is radical.

Lemma 2. ([19], Section 3.7.B, pg. 246) Given a zero-dimensional idealI, ifIcontainsa univariate square-free polynomial in each variable, thenI is radical.

In this case, square-free implies that when a polynomial is decomposed into its uniquefactorization, there are no repeated factors. For example, (x2 + y)(x4 + 2z+ 3) is square-free, but (x2+y)(x4+2z+3)3 is not. In particular, Lemma2implies that ideals containingx2ixi = xi(xi 1) in each variable (i.e., the boolean ideals) are radical. In Section4, wewill see that all ideals associated with dominating sets (and therefore Vizings conjecture)are radical for this reason.

In general, it may be quite difficult to determine a basis for

IJ. However, ifIandJare radical ideals via Lemma2, and additionally, if the univariate, square-free polynomials

in each variable contained inI, Jareidentical, we can explicitly determine a basis for IJ.Lemma 3. Let I and J be ideals such that I =f1, . . . f s and J =g1, . . . , gt. Fur-thermore, for 1 i n, let fi = gi be square-free univariate polynomials in xi. Then

IJ=figj : 1 i s, 1 j t + fi : 1 i n.Proof. We prove the inclusion in both directions. For convenience, letM :=figj : 1 i s, 1 j t+fi : 1 i n. First, we show that

IJ M. Let h IJ.

Then, there exists an integer m 1 such that hm IJ. Thus, hm =()f g. Thus,hm M. But, by Lemma 2, the ideal M is radical since M contains a square-freeunivariate polynomial fi in each variable xi. Thus, h

m M implies that hM.Conversely, to show thatM

IJ, lethM. Then,h =()f g +()f. Consider

h2 =

()f g+

()f

()f g+

()f

.

Any term in the expandedh2 can be viewed as ()figj fkgl = ()figj , or ()figjfk = ()figj,or ()fifj = ()figj, etc. Thus, any term in the expanded h2 can be written as ()f g, withany extra multiplicities in f or g simply folded into the coefficient. Therefore,

h2 =

()f g,

and there exists an integer m = 2 1 such that hm

I J, implyingh

IJ.

This brings us to the following critical fact: ifIandJare boolean, radical ideals, thenby Lemma3,

IJ=figj : 1 i s, 1 j t + xi(xi 1) : 1 i n= I J.

In Section4, when we represent the dominating set problem as a system of polynomialequations, the representations will be boolean, radical ideals as described above. Thus,the basis of their product ideals can be described via Lemma 3. This fact will be vital inSection5 when we present an algebraic representation of Vizings conjecture.


8/12/2019 2194-2610-1-PB

5/30

3 Universal Grobner Bases and Linear Factors

In this section, we provide a brief overview of the terminology (from [11], Chap. 2) per-taining to Grobner bases, and build off the ideas of De Loera in[21]to show that a specificset of linear factor polynomials is a universal Grobner basis. These linear factor ideals

allow us to provide a combinatorial interpretation of the universal Grobner basis of thedominating set ideal defined in Section4, and will be used in our algebraic exploration ofVizings conjecture.

Amonomial orderfor the monomials in the polynomial ring K[x1, . . . , xn] is a well-ordering which is multiplicative, and for which the constant polynomial is the smallest.The leading term lt(f) of a polynomial f K[x1, . . . , xn] is the largest monomial in f(and its corresponding coefficient) with respect to the monomial order. Given an idealI, a finite subsetG={g1, . . . , gr}ofI is a Grobner basis ofI(with respect to) if

lt(g1), . . . , lt(gr)=lt(I)=lt(f) :fI.

A finite subset GofIis a universalGrobner basis ofIif it is Grobner basis ofIwith respectto any monomial order. Although there are infinitely many monomial orderings, theuniversal Grobner basis ofIis finite and unique (see [11]for further details). BuchbergersS-pair criterion [8] is a specific criterion for determining if a given subset of polynomials isa Grobner basis ofI. In order to define Buchbergers S-pair criterion, we fix a monomialorderand recall the following notation:

fF denotes the remainder of division of the polynomial f by the ordered s-tupleF = (f1, . . . , f s) (see[11], Chap. 2, Sec. 3 for details.)

Example: Note that (x2y+xy2 +y2){xy1,y21}

=x+y+ 1, since

x2y+xy2 +y2 = (x+y)(xy 1) + (y2 1) +x+y+ 1,

but when the order ofF is changed, (x2y+xy2 +y2){y21,xy1,}

= 2x+ 1, since

(x2y+xy2 +y2) = (x+ 1)(y2 1) +x(xy 1) + 2x+ 1.

x is the least common multiple of the leading monomial lm(f) and the leadingmonomiallm(g), written asx =lcm

lm(f), lm(g)

.

TheS-pair off and g is the combination:

S(f, g) = x

lt(f) f x

lt(g) g.

We will now characterize Grobner bases in terms of S-pairs.

Theorem 4 ([8] (also see [11], Chap. 2, Sec. 6 for details)). Given an idealI, a basisG={g1, . . . , gr}ofIis a Grobner basis forI, if and only if, for all pairsi=j ,S(gi, gj )G = 0.


8/12/2019 2194-2610-1-PB

6/30

We now construct a set of polynomials, where each polynomial is a product of linearfactors, and show that if the set satisfies the linear factor criterion, then the set is auniversal Grobner basis. Let S={i1, i2, . . . , ik} {1, . . . , n}, and

P(S) := (xi1 1)(xi2 1) (xik 1), and x(S) =xi1xi2 xik .

Definition 5. Let{S1, . . . , S D}be a set such that eachSi {1, . . . , n}and|Si|= di. Wesay that the set{S1, . . . , S D}satisfies the linear factor criterionif, for any integers i, j, k,Sk is anota proper subset ofSi \ Si Sj , andSk is anota proper subset ofSj\ Si Sj.Theorem 6. Let{S1, . . . , S D} with Si {1, . . . , n} be a set that satisfies the linear

factor criterion, and let gi = P(Si). Then{g1, . . . , gD} is a universal Grobner basis ofg1, . . . , gD.

Before we prove this theorem, we present an example.

Example 7. Consider C[x1, . . . , x12], and let S1 =

{1, 2, 3, 5, 6

}, S2 =

{1, 2, 3, 7, 8

}, S3 =

{9, 10, 11, 12}, S4={4, 8, 9}. LetG={g1, g2, g3, g4}with gi = P(Si).g1 := (x1 1)(x2 1)(x3 1)(x5 1)(x6 1),g2 := (x1 1)(x2 1)(x3 1)(x7 1)(x8 1),g3 := (x9 1)(x10 1)(x11 1)(x12 1),g4 := (x4 1)(x8 1)(x9 1).

Since S1 S2 ={1, 2, 3}, S1\ (S1 S2) ={5, 6}, and S2\ (S1 S2) ={7, 8}. Neither S3nor S4 is a propersubset ofS1\ (S1 S2) or S2\ (S1 S2). This is true for all i, j pairs,thus{S1, S2, S3, S4} satisfies the linear factor criterion. Additionally, note that lt(gi) isalwaysx(Si), regardless of the specified monomial order. Thus, we see

S(g2, g4) =x1x2x3x4x7x8x9

x1x2x3x7x8g2x1x2x3x4x7x8x9

x4x8x9g4= x4x9g2 x1x2x3x7g4

= (x4+x9 1)g2 (x1x2x3+x1x2x7+x1x3x7+x2x3x7 x1x2 x1x3 x1x7 x2x3 x2x7 x3x7+x1+x2+x3+x7 1)g4.

Regardless of the monomial order, the leading terms of the coefficients ofg2 and g4 are

not divisible by the leading terms of g1, g2, g3 or g4, Thus, S(g2, g4)G

= 0. Performingsimilar calculations for all other pairs can quickly show that{g1, . . . , g4} is the universalGrobner basis of

g1, . . . , g4

.

For simplicity of notation, let Sionly= Si \ (Si Sj) andSjonly= Sj\ (Si Sj). Further-more, letSij be a subset ofSi Sj,Si only be a subset ofSionly, andSj only be a subset ofSjonly. Note that

|Sjonly| +di=|Sionly| +dj. (1)


8/12/2019 2194-2610-1-PB

7/30

Proof. We will show that the set of generatorsG ={g1, . . . , gD} satisfies BuchbergersS-pair criterion. Consider any two polynomialsgi, gj. We must show S(gi, gj )

G= 0. We

claim

S(gi, gj) =x(Si Sj)

x(Si)

gi

x(Si Sj)x(Sj)

gj =x(Sjonly)gi

x(Sionly)gj (2)

=

lt

P(Sjonly) P(Sjonly)gi ltP(Sionly) P(Sionly)gj. (3)

Before we prove the equality between lines 2 and 3, we note that, if true, we have

already shown that S(gi, gj)G

= 0. This can be seen by noting that, since{S1, . . . , S D}satisfies the linear factor criterion, given integers i, j, k, Sk is not a proper subset ofS

ionly

andSk is not a proper subset ofSjonly. Additionally, regardless of the monomial order, the

leading term ofgi isx(Si), and the leading term of

lt

P(Sjonly) P(Sjonly)is aproper

subset ofSjonly (similarly, ltP(Sionly)P(Sionly) is a proper subset ofSionly). Thus,

lt(gk) does not divide either

lt

P(Sjonly)P(Sjonly) or ltP(Sionly)P(Sionly),

and we can see that S(gi, gj )G

= 0.In order to prove the equality between lines2and3, we must show that every monomial

in line2either cancels within line 2, or appears in line 3with the same coefficient, andvice versa.

Let x(M) be a monomial appearing in line 2. Either

M=Sjonly Sij Si only, or M=Sionly Sij Sj only.

We note that x(M) appears in both x(Sjonly)gi and x(Sionly)gj only when S

i only = S

ionly

and Sj only = Sjonly. In this case, the coefficient for x(M) is (1)C (1)D where

C=diSij + Sionly, and D= dj Sij + Sjonly.

This formula can be explained by noting that, in any polynomial that is the productof linear factors, the coefficient for a given monomial is (1)C where C is the degree ofthe polynomial minus the degree of the given monomial. For example, consider f :=(x11)(x21)(x31)(x71). The coefficient for the monomialx1x2x3is (1)43 =1,and the coefficient for x2x7 is (1)42 = 1.

Using the identity given in Eq. 1, we note that di Sionly = dj S

jonly. Thus,C=D, and the monomial cancels.We will now consider the case where M=Sjonly Sij Si only andSi only=Sionly. In

this case, x(M) appears in line 2in the product x(Sjonly)gi, and the coefficient for x(M)

is (1)C where

C=diSij + Si only.


8/12/2019 2194-2610-1-PB

8/30

In line3,x(M) appears only in the productltP(Sionly) P(Sionly)gj , and the coef-ficient for x(M) is (1)C where

C =Sionly

Si only

+djSij

+Sjonly

+ 2.

But using the identity from Eq. 1,and substituting for dj, we see

C =Sionly Si only + Sjonly+di Sionly Sij + Sjonly + 2

=diSi only Sij + 2.

Thus, C is the same parity as C, which implies that the coefficient for x(M) in line 2,and the coefficient for x(M) in line3 are the same.

The case where M = SionlySijSj only and Sj only= Sjonly is the same. Thus, wehave shown that every monomial in line 2 either cancels, or appears in line 3 with thesame coefficient.

We must now show that every monomial in line 3either cancels, or appears in line 2

with the same coefficient. Let x(M) be a monomial appearing in line 3. Either

M=Sj only Sij Si only, or M=Si only Sij Sj only.We note that in each case,Mcontains apropersubset ofSjonlyor apropersubset ofS

ionly.

Additionally, the two cases when Si only = Sionly, or S

j only = S

jonly have already been

explicated above. Thus, in both cases, we have already shown that x(M) appears in line2with the same coefficient.

We will now consider monomials of the formx(M) whereM=Sj onlySijSi only, andshow that these monomials cancel within line3. The coefficient forx(M) is (1)C(1)Dwhere

C=Sjonly Sj only +di Sij + Si only, and

D=Sionly Si only +djSij + Si only + 2.

As before, using the identify from Eq. 1 and substituting for dj , we see C is the sameparity asD. Thus, (1)C (1)D = 0, and the monomial cancels.

Thus, we have shown that every monomial in line 2either cancels within line 2, orappears in line 3 with the same coefficient, and vice versa. Additionally, we have notutilized the properties of any particular monomial order to do so. Thus, we have shownthat{g1, . . . , gD} is a universal Grobner basis forg1, . . . , gD.

The following corollary extends the theorem above to include boolean polynomialsof the form x2ixi. This is particularly interesting because boolean polynomials are acommon ingredient in non-linear models of combinatorial problems.

Corollary 8. Let{S1, . . . , S D} withSi {1, . . . , n} (and|Si|>1) be a set that satisfiesthe linear factor criterion. Letgi=P(Si), and letbi=x

2i xi. ThenG ={b1, . . . , bn}

{g1, . . . , gD} is the universal Grobner basis forb1, . . . , bn + g1, . . . , gD.


8/12/2019 2194-2610-1-PB

9/30

We observe that xi 1 andx2i xi are redundant equations, which explains the extracondition|Si|> 1, i.Proof. By Theorem6, we have already seen that{g1, . . . , gD}is a universal Grobner basis.Therefore, in order to show thatG satisfies Buchbergers criterion, it remains to show thatS(bi, bj)

G

=S(bi, gj)

G

= 0, without relying on properties of a particular monomial order,.Because monomial orders are by definition well-orders, lt(bi) is alwaysx2i , andlt(gi)

is always x(Si), regardless of the specific monomial order.

We will first show that S(bi, bj)G

= 0 for i=j . Note that

S(bi, bj) =x2i x

2j

x2i(x2i xi)

x2i x2

j

x2j(x2j xj ) =x2i xj xix2j =xj(x2i xi) xi(x2j xj).

Since |Si| > 1, the coefficientsxi and xj are not divisible by lt(gk), k. Thus,S(bi, bj)

G= 0.

We will now show that S(bi, gj)

G

= 0, for all i, j pairs. Case 1: Choose an i such that i /Sj, and write gj :=x(Sj ) +Prest. Note

S(bi, gj) =x2i x(Sj)

x2i(x2i xi)

x2i x(Sj )

x(Sj ) (x(Sj ) +Prest) =xix(Sj) x2i (Prest)

=xi(x(Sj ) +Prest) Prest(x2i xi) =xigj Prest(x2i xi).By the linear factor criterion, lt(Prest) is not divisible by lt(gk) for any k. Thus,

S(bi, gj)G

= 0.

Case 2: Choose an i such that i

Sj , and write gj := x(Sj)

x(Sj

\i) + Prest.

SinceiSj, this implies that lcmlt(bi), lt(gj)= xix(Sj ) =x2i x(Sj\ i). ThenS(bi, gj) =

x2i x(Sj\ i)x2i

(x2i xi) xix(Sj)

x(Sj)

x(Sj) x(Sj\ i) +Prest

=xiPrest

(4)

=

x(Sj\ i) P(Sj\ i)

(x2i xi), (5)

andS(bi, gj )G

= 0 by the linear factor criterion. The equality of lines 4and5can beseen as follows. The polynomialx(Sj \ i)P(Sj \ i) is the polynomial P(Sj \ i) withthe signs changed and the leading term removed. When multiplied by a positive

x2i and a negative xi, every monomial appearing in the product also appears inxi(Prest). To see this clearly, consider a monomial x(M) where M is a propersubset ofSj (this is the only kind of monomial appearing in Prest on line4). Thereare two cases:

Case 2a: i M. When x(M) is multiplied byxi (line 4), the sign ofthe leading coefficient changes, and the degree increases by one. In other


8/12/2019 2194-2610-1-PB

10/30

words,xix(M) appears in expanded productxi(Prest) with leading coefficient(1) (1)dj|M|. However,x(M\ i) appears in P(Sj\ i) withthe same paritysignas x(M) in Prest(i.e., (1)(dj1)(|M|1)). Thus, the productP(Sj \i)(x2i )produces the same monomial with the same leading coefficient as xi(Prest)(line5) and equality between lines4 and5 is preserved.

Case 2b: i / M. In this case, x(M) appears in P(Sj\i), but with oppositesignasx(M) inPrest. In particular,x(M) inPrest has coefficient (1)dj|M| asbefore, butx(M) inP(Sj\ i) has coefficient (1)(dj1)|M|. Thus, the productP(Sj \ i)(xi) produces the same monomial with the same leading coefficientasxi(Prest) and equality between lines 4 and 5is preserved

In cases 2a and 2b, we demonstrated that the monomials produced byP(Sj \i)(x2i )andP(Sj\i)(xi) respectively, appear in line 4. Thus, we have accounted forevery monomial in line5, and we have shown that line 4 is equal to line5.

We have shown that S(bi, bj)G

= 0 for i= j, and that S(bi, gj)G = 0, for all i, j pairs,without relying on any properties of a monomial order. Since we have already shownthat{g1, . . . , gD} is a universal Grobner basis, this means that we have shown that{b1, . . . , bn} {g1, . . . , gD}is a universal Grobner basis ofb1, . . . , bn + g1, . . . , gD. Thisconcludes our proof.

In Section4.3, we will represent the dominating set problem as a system of polyno-mial equations in such a way that the representation is already a universal basis. Thisrepresentation follows the work of De Loera [21], where the graph coloring problem isalso represented in such a way that it is a universal Grobner basis. These kinds of repre-sentations may prove useful in the context of combinatorial ideal membership questions,and also with the advance of algorithms specifically tailored for finding universal Grobner

basis, such as [4].

4 Dominating Sets, Ideals and Grobner Bases

In this section, we begin by formulating the dominating set problem as a system ofpolynomial equations. We extend this formulation to include a dominating set in a graphG, a dominating set in a graph H, and a dominating set in the product graph GH.We refer to this formulation (linking graphs G, H, and GH) in Section 5during thealgebraic formulation of Vizings conjecture. In Section4.2,we introduce the idea of a k-domination coveror a k-cover. We explain the relation betweenk-covers and domination-

critical graphs, and provide various examples ofk-covers. We also prove several propertiesofk-covers, and provide a conjecture for future work. We unify these ideas in Section4.3 by showing that k-covers provide the combinatorial interpretation of the universalGrobner basis of the ideals described in Section 4.1. Thus, the purpose of this sectionis to explore two different non-linear models of the dominating set problem based ontwo different combinatorial properties, and surprisingly, the second representation is theuniversal Grobner basis of the first.


8/12/2019 2194-2610-1-PB

11/30

4.1 Dominating Sets and Systems of Polynomial Equations

Let Snk represent the set of k-subsets of{1, 2, . . . , n}. Thus,|Snk | =

nk

, and S Snk

implies thatS {1, 2, . . . , n}is a particular subset with|S|= k. In the following systemof polynomial equations, there is one binary decision variable for every possible edge in agraph with n vertices. Thus, the variable e

ijis 1 if the edge between vertex i and vertex

j exists in the graph, and 0 otherwise. Since our graphs are undirected, we implicitlyassume the substitution eji = eij whenever j > i.

Theorem 9. There is a bijection between the set of solutions of the following system ofequations and the set of labeled graphsG inn vertices with a dominating set of sizek.

e2ij eij = 0, for1 i < j n,SSn

k

i /S

jS

(eij 1)

= 0. (6)

Example 10. Letn = 3 andk = 1. The variables in the system of polynomial equationsdefined by Theorem9 are e12, e13 and e23. Furthermore, S31 =

{1}, {2}, {3}.For the second equation, we have:

(e21 1) + (e31 1)

(e12 1) + (e32 1)

(e13 1) + (e23 1)

= 0.

However, as noted above, since the graph is undirected, we implicitly assume the substi-tutione21= e12, e31= e13 ande32 = e23. Thus, the system of equations is as follows:

e212 e12 = 0, e213 e13= 0, e223 e23 = 0

(e12 1) + (e13 1)(e12 1) + (e23 1)(e13 1) + (e23 1)= 0.The solutions to the system of equations are in bijection with the labeled graphs on threevertices with a dominating set of size one:

Proof. We refer to the system of polynomial equations defined by Theorem9asthrough-out this proof. We will define a map between the solutions of and the labeled graphs

G in n vertices with a dominating set of size k, and show that is a bijection betweenthe two sets.To deal with the problem of graph isomorphisms, we consider a set ofn vertices with

a fixed labeling 1 through n. Thus, two graphs G and G on this labeled set ofn verticesare equal if and only if (i, j) E(G) implies that (i, j) E(G). In other words, ifGand G are isomorphic, they are not necessarily equal under this definition. Then, ourmap simply takes a solution to , and converts it to a graph G on the labeled set ofn


8/12/2019 2194-2610-1-PB

12/30

vertices by adding the edges (i, j) toE(G) if and only if the variable eij = 1. Then, isone-to-one, since given any two solutions s ands, ifG = (s) =(s) =G, then clearlys= s.

We will now show that the image of is a subset of the set of labeled graphs G witha dominating set of size k . The boolean equations e2ij

eij = 0 force every variableeij to

be zero or one; thus, the boolean equations turn edges on or off when applying themap, and every solution corresponds to a particular graph G. We must now show thatsince the solution satisfies Eq. 6, the particular graph G formed from the solution has adominating set of size k . Let A and B denote different pieces of Eq. 6as follows.

SSnk

i /S

jS

(eij 1)

A

B

= 0.

The equation is only satisfied if one of the inner summations B (corresponding to some

setS) is zero. The value of an individual summandA in B is either zero or1. However,two different summands A, A in the summation B can never mutually cancel, since ifthey are both non-zero, they are both1|S|. Thus, they are either both1 or both +1.Therefore, a summationB is only zero if every individual summand A is also zero, and Ais only zero if at least one edge variable has value one. Therefore, every i /S is adjacentto a jS. In other words,Ghas a dominating set of size k.

Having shown that the image of is a subset of the set of labeled graphs G with adominating set of size k, we must now show that is onto, or that given any graph Gwith a dominating set of size k, 1(G) is a solution to . The map 1(G) would beapplied as follows: if the edge eij is present in G, turn it on by setting the variableeij = 1. If the edge eij is not present in G, turn the variable eij off by setting thevariable eij = 0. Clearly, the boolean equations e2ijeij = 0 are satisfied. Since thegraph G has a dominating set of size k, let S ={i1, . . . , ik} be such a dominating set.Thus, every vertex i not in the dominating set Smust be adjacent to a vertex j that isin the dominating set S. In other words, an edge eij is on from i / S to j S, andthe (eij 1) term in the product corresponding to the dominating set Sis equal to zero.Since every vertex i /Smust satisfy this condition, this implies that every summand inthe summation (corresponding to the dominating set S)

i /S

jS

(eij 1)

is equal to zero, and thus the entire summation is equal to zero. Since Eq. 6is a productof summations, the equation is satisfied.

Thus, is one-to-one and onto, and is a bijection between solutions of and the setof labeled graphsGwith a dominating set of size k .

We will now use the system of polynomial equations defined in Theorem9as a buildingblock to model dominating sets in G, H andGH.


8/12/2019 2194-2610-1-PB

13/30

Theorem 11. There is a bijection between the set of solutions of the following system ofpolynomial equations and the set of labeled graphsG, H inn, n vertices with dominatingsets of sizek, l respectively such that their Cartesian product graphGHhas a dominat-ing set of sizer.

Representing a graphG inn vertices with a dominating set of sizek:

e2ij eij = 0, for1 i < j n,SSn

k

i /S

jS

(eij 1)

PkG

= 0. (PkG)

Representing the graphH inn vertices with a dominating set of sizel:

e2ij eij = 0, for1 i < j n,

SSn

l

i /S

jS

(eij 1)

PlH

= 0. (PlH)

Representing the Cartesian product graphGHwith a dominating set of sizer:

SSnnr

gh /S

ghS

(egg 1)

ghS

(ehh 1)

PrGH= 0. (PrGH)

Proof. It is clear from Theorem 9that there is a bijection between the solutions of theequations representing graphs G and Hand the set of graphs in n, n vertices with domi-nating sets of size k, l respectively. The equation representing GH is of the same form,except that this equation takes into account the unique structure of the product graph.For example, if a vertex not inSis adjacent to a vertex in S, the adjacency is either dueto an edge in G, or an edge in H. In particular, gh / S is either adjacent to a vertexghS if the edge (g, g) is on in G, or gh / S is adjacent to a vertex gh S if theedge (h, h) is on in H. Thus, the proof of the bijection follows the logic of the proof ofTheorem9.

Since the system of polynomial equations described in Theorem 11 depends on thevariables n, k,n, l and r, we define the ideal I(n,k,n, l , r) as

I(n,k,n, l , r) :=PkG, PlH, PrGH, e2 e,wheree2 edenotes the entire set of boolean edge equations

{e2ij eij : 1 i < j n, e2ij eij : 1 i < j n, }.


8/12/2019 2194-2610-1-PB

14/30

We note that I(n,k,n, l , r) is radical by Lemma2, since the boolean equations e2 eare univariate and square-free. We now definek-covers, which provide a combinatorialinterpretation of the universal Grobner basis ofI(n,k,n, l , r) in Section4.3. We note thatwhat is important about the next section is not the uniqueness or complexity of the defi-nition ofk-covers, but rather the interesting relationship between the ideal I(n,k,n, l , r),

the combinatorial idea of a k-cover, and the universal Grobner basis of the ideal.

4.2 k-Covers and k-Dominating Sets

We begin by recalling that a graph G is domination-critical if, for any two non-adjacentverticesu, v, the graphG :=G +(u, v) has(G) =(G)1. In other words, a graphGisdomination-critical if, whenever any edge is added to G, the domination number decreasesby one. Thus, k-domination-critical graphs are created from k-dominating graphs byadding edges. In this section, we study the complements of domination-critical graphs.We define a k-domination-cover (or k-cover) as analogous to a k-dominating graph, anda minimal k-cover as analogous to a k-domination-critical graph. Just as a domination-critical graph is obtained from ak-dominating graph by adding edges, a minimal k-coveris obtained from a k-cover byremovingedges.

Before we begin, we recall a few definitions. LetKn denote the complete graph in nvertices (ann-clique). ForSV(G), theopen neighborhood ofS(denoted byNG(S), orsimplyN(S) when the context of the graph is clear) is defined to be

NG(S) :={v : (u, v)E(G) anduSand v /S}.

Thecommon neighborhood ofSV(G) (denoted bycmNG(S), or simplycmN(S) whenthe context of the graph is clear) is defined to be

cmNG(S) :=uS

NG({u}).

Example 12. In the following example, let S={0, 1, 3}. Then cmN(S) ={6}.

Definition 13. A graphCis ak-cover ifcmNC(S)=for allSV(C) with|S|= k1.We say that a graph C is a k-cover if every set of size k 1 has a common neighbor.

We note that k-covers are only defined for k 2.

Definition 14. Ak-coverC isminimal if for all eE(C),C :=C eis not a k-cover.


8/12/2019 2194-2610-1-PB

15/30

Example 15. Here we see a minimal 3-cover (left) and its complement (right).

We refer to graphs of the type described by Definition 13 as k-covers, because weare taking n vertices and then covering those n vertices with k-cliques such that thek-cliques intersect in a very particular way. Notice that the 3-cover illustrated in Ex. 15takes six vertices, and then covers those six vertices with triangles such that any subsetof size two has a common neighbor.

Proposition 16. Given ak-coverC, everyvV(C) appears in a clique of sizek.Proof. Since every set of size k

1 inV(C) has a common neighbor, every vertex has at

least one outgoing edge. Thus, for anyv1V(C),v1 is adjacent to some v2. If 2 k 1,bothv1 andv2 are each adjacent to a third vertex v3. In other words, v1, v2andv3 form a3-clique. By repeating this processk 1 times, we form a k-clique containingv1. Sincev1was an arbitrary starting point, this algorithm can be repeated for any vertex, and everyvV(C) appears in a clique of size k .

We also observe that while 2-covers can be disconnected, k-covers with k 3 areconnected. Additionally, since the diameter of a graph is the longest shortest path betweenany two vertices, and every two vertices has a common neighbor,k-covers withk 3 havediameter at most two.

Proposition 17. A graph G is k-domination-critical if and only if G is a minimal k-cover.

Proof. IfG is k-domination-critical, then G has a dominating set of size k, but no dom-inating set of size k1. Thus, every (k 1)-subset of vertices in G has a commonneighbor, and G is a k-cover. Furthermore, since G is k-domination-critical, for any twonon-adjacent vertices u, v,

G+ (u, v)

= k1. In other words, if any edge (u, v) is

removed fromG, then there is at least one (k 1)-subset of vertices that no longer has acommon neighbor. Thus, G is a minimal k-cover.

Conversely, assume that G is a minimal k-cover. Then, for all eE(G),G eis notak-cover. This implies that there is some set D

V(G) of sizek

1 that does not have

a common neighbor. In other words, for all v V(G)\D, v is notadjacent to somevertex in D. In other words, D is a dominating set of size k1 in G. Thus, we haveshown that when any edge is added toG (or removed fromG), there is a dominating setof size k 1 in G. Thus, in order to prove that G is k-domination-critical, it remains toshow that there exists a dominating set of size k in G, and that there does not exist adominating set of size k 1 inG.


8/12/2019 2194-2610-1-PB

16/30

Clearly, there is no dominating set of size k 1 inG, sinceG is ak-cover. We will nowshow that G contains a dominating set of size k. Consider any two non-adjacent verticesu, v V(G). Without loss of generality, let D ={i1, . . . , ik2, u} be the dominating setof sizek 1 inG + (u, v). SinceD is a dominating set inG + (u, v), but not a dominatingset inG, the only vertex thatD doesnotdominate inG is v. Thus,D + vis a dominating

set inG, and since|D+v|= k, G isk-domination-critical.We note that in[32], Sumner and Blitch extensively explore properties of 3-domination-

critical graphs. They explicitly categorize 2-domination-critical graphs as follows:

Theorem 18 (Sumner and Blitch, 1983). A graphG is 2-domination-critical if and onlyifG=ni=1K1,ni withn 1.

The Sumner-Blitch categorization of 2-domination-critical graphs is equivalent to Def-inition13, since for any 2-cover, every vertex is adjacent to at least one other vertex.

Example 19.Here we see a 2-cover on 16 vertices. We note that this 2-cover is the union

of two K1,3 graphs, two K1,2 graphs, and a K1,1 graph. Additionally, every single vertex(every set Swith|S|= 1) has a common neighbor.

Our categorization is an extension of the Sumner-Blitch categorization since Definition13is generalized for the complements ofk-dominating graphs, although further charac-terizations are needed for minimal k-covers. We now link minimal k + 1-covers withk-dominating graphs.

Theorem 20. A graphG = (V, E) has a dominating set of sizek if and only if for allminimalk+ 1-coversC= (V, Ecov), E(G) Ecov=.

Via this theorem, we characterize graphs with dominating sets of size one in termsof 2-covers. Since one is the smallest dominating set, this explains why we only definek-covers for k 2.

Proof. Assume G = (V, E) has a dominating set D V(G) of size k. Consider anyminimal k+ 1-cover C= (V, Ecov). Since D is also a subset ofV(C), and since|D|= k,by Definition13,D has a common neighbor v inC. Thus, for everyuD, (u, v)Ecov.SinceD is a dominating set in G and v /D, there must exist an edge (u, v)E(G) withu

D. Since (u, v)

E(G) and (u, v)

Ecov,E(G)

Ecov

=

.

Conversely, assume that for all minimal k + 1-covers C= (V, Ecov),E(G) Ecov=.We must show that G contains a dominating set of size k . We proceed by contradiction.Assume that G does not contain a dominating set of size k, and form a new graphGcrit by adding edges to G until the resulting graph is k+ 1-domination-critical. Thus,E(G)E(Gcrit), and Gcrit has a dominating set of size k+ 1, but no dominating set ofsizek.


8/12/2019 2194-2610-1-PB

17/30

Next, we consider the complement of Gcrit (denoted by Gcrit). Since Gcrit is k+ 1-critical,Gcrit is a minimal k + 1-cover by Proposition17. We note that E(Gcrit)E(G),andE(Gcrit) E(Gcrit) =. However, E(G)E(Gcrit). Therefore,E(Gcrit) E(G) =.But this is a contradiction, sinceGcrit is a minimal k + 1-cover, and E(G) Ecov=, forall minimalk + 1-covers C= (V, Ecov). Thus,Gcontains a dominating set of size k .

We note that, according to Thm. 20, in order to demonstrate that a graphG doesnothave dominating set of size k, it is sufficient to produce ak +1-coverCthat is a subgraphofG. Thisk+ 1-cover then becomes acertificateof the non-existence of ak-dominatingset: it is a coNP certificate. However, even though the cover itself is polynomial-size in|G|, there are no known polynomial-time algorithms for verifying that every k-subset ofChas a common neighbor. Therefore, there is no conflict between the use ofk-covers ascertificates, and the conjectured non-equality of NP and coNP.

We now explore properties ofk-covers.

Theorem 21. A graphCis ak-cover if and only if, for allSV(C)with1 |S| k1,there exists a cliqueQincmNC(S)such that|Q|= k|S|. Moreover, everyvcmNC(S)appears in a cliqueQof sizek |S|.Proof. Assume that a graphC is ak-cover. GivenSV(C), if|S|= k 1, we know bydefinition of ak-cover thatcmN(S) is non-empty. Therefore, a clique Q of size one existsin cmN(S), and moreover, all vertices in cmN(S) trivially form cliques of size one.

Now consider S V(C) with 1 |S| k 2. As before, by definition, since|S| < k1, cmN(S) is non-empty. Let v1 cmN(S). Since|Sv1| k1, the setSv1 has a common neighbor v2. Thus, the set{v1, v2} forms a clique of size two incmN(S). If |Sv1v2| k1, we repeat this operation, and we note that the setSv1v2 has a common neighbor v3, and that{v1, v2, v3} forms a clique of size threeincmN(S). We repeat this operation exactly k |S| 1 times until we have formed thek |S|-cliqueQ ={v1, v2, . . . , vk|S|}in cmN(S).

Therefore, for all 1 |S| k 1, there exists a clique Q in cmN(S) such that|Q| = k |S|. Moreover, every vcmN(S) appears in a clique Q of size k |S|, sincethis operation can be repeated with reference to any vertex v .

Conversely, let Cbe a graph such that for all SV(C) with 1 |S| k 1, thereexists a clique Q in cmN(S) such that|Q| = k |S|. In particular, let Sbe a subset ofV(C) with|S|= k 1. Then, there exists a clique Qof size one in cmN(S). Therefore,by definition,C is ak-cover.

We have already shown that every vertex in a k-cover appears in a k-clique, but we

have not discussed how these k-cliques intersect. We present the following conjecture.

Conjecture 22. A graphC is ak-cover if and only if for all SV(C) with|S|= k 1,there exists a mapqfromvSto k-cliques inCsuch that for anyu, vS,|q(u)q(v)| k 2.Example 23. Here is an example of Conjecture 22 on a 3-cover of 7 vertices.


8/12/2019 2194-2610-1-PB

18/30

ConsiderS={c, g}. Then, ifq(c) ={c,d,f}and q(g) ={a,f,g},|q(c) q(g)|=|{f}|=1 =k

2. Additionally, consider S=

{e, g

}. Then, ifq(e) =

{b,d,e

}and q(g) =

{a,b,g

},

|q(e) q(g)|=|{b}|= 1 =k 2. Upon inspection, we can see that given any set Swith|S|= k 1 = 2, a similar map qcan be constructed and the conjecture holds.

4.3 k-Covers, k-Dominating Sets and a Universal Grobner BasisofI(n,k,n, l , r)

In Section4.1, we outlined a representation of the k-dominating set problem as a system ofpolynomial equations, and defined the ideal I(n,k,n, l , r). In Theorem9, I(n,k,n, l , r)was defined by three, high-degree polynomials PkG, P

lH and P

rGH, and their associated

boolean edge equations. Each of these polynomials was a series of products, with each

product a sum of products. Thus, the polynomialsPkG, PlH and PrGH were a brute-forceenumeration of every possible vertex set of sizek , dominating or otherwise, in the graph.

In this section, we present another representation of the ideal I(n,k,n, l , r). Thisrepresentation is based on k + 1-covers, with each polynomial equation correspondingto a minimal k+ 1-cover ofV(G). Thus,I(n,k,n, l , r) can also be generated by manydifferent polynomials of comparatively low degree, as opposed to only three polynomialsof high degree. However, there is currently no known algorithm for enumerating minimalk+ 1-covers, which is equivalent to the problem of enumeratingk+ 1-domination-criticalgraphs. Thus, another point of comparison between the representation expressed hereand Theorem9 is that the polynomials described in Theorem9 can be explicitly writtendown, whereas polynomials based on k+ 1-covers, while combinatorially explicit, maybe difficult to write down explicitly enough for the purpose of computation. However,the potential advantage of the cover representation is that k+ 1-covers are collections ofedges that satisfy the linear factor criterion (Definition5). Therefore, by Theorem6 andCorollary8, the representation described below is a universal Grobner basis.

In presenting a combinatorial interpretation of any Grobner basis, we must be partic-ularly careful of input instances where the system of polynomial equations is infeasible.In these cases, the associated variety is empty and the associated ideal is the entire ring.This implies that the universal Grobner basis is simply the number one. In our case, wewill exclude input instances where the system of polynomial equations is infeasible byrestricting the values of n, n, k and l. However, in Theorem 38, we will show that the

instances lost by this restriction correspond exactly to instances where Vizings conjectureis known to be true.

Before we begin, we introduce some notation. Let Cnk+1 be the set of all minimalk+ 1-covers on n vertices. For C Cnk+1, let

P(C) :=

eE(C)

(e 1).


8/12/2019 2194-2610-1-PB

19/30

Theorem 24. There is a bijection between the solutions of the following system of equa-tions and the labeled graphsG inn vertices with a dominating set of sizek.

e2ij eij = 0, for1 i < j n,P(C) = 0, for eachC

C

nk+1. (7)

We note that ifk =n, there are no n+ 1 covers on n vertices. Thus, Cnn+1 =, andthe only equations that appear are of the form e2 e= 0. Thus, any graph is a solutionto this system, which is reasonable since every graph has a dominating set of size n.

Proof. As in the proof of Theorem9, we must define a map and show that provides abijection between the set of solutions and the set of graphs with a dominating set of sizek. We will use the same map defined in the proof of Theorem9, and thus, we alreadyknow that the map is one-to-one, and we only need show that the image of is the set ofgraphs with a dominating set of size k, and that the map is onto.

To consider the image of the map , we note that, as in Theorem 9, every solution

corresponds to a particular graph G (the assignment of the variables eij to zero or onesimply turns the edges on or off inG). Since the cover equations (Eqs. 7) are satisfied,this implies that for all minimal k+ 1-covers, E(G) E(C)=. Then, by Theorem20,G has a dominating set of size k .

To show that is onto, consider a graph G in n vertices that has a dominating set ofsize k. As before, we must show that 1(G) (defined by setting variables eij to zero orone, depending on whether or not eij E(G)) maps to a solution. Clearly, the booleanequationse2ij eij = 0 are satisfied. Since the graph G has a dominating set of size k, forany minimal k+ 1-cover C, E(G)E(C)= by Theorem20. Therefore, at least oneedge in every cover is on and the cover equations (Eqs. 7) are satisfied.

Let I(n, k) denote the ideal generated by polynomials described in Theorem 24. Wewill now prove that this representation is a universal Grobner basis. We note that we dis-covered the connection between covers and Grobner bases by experimental investigationsusing CoCoA Lib [1].

Corollary 25. The basisI(n, k) described in Theorem24is a universal Grobner basis.

Proof. We must show that{P(C) : C Cnk+1} satisfies the linear factor criterion (Def.5). LetCi, Cj, Ck be covers in C

nk+1. Since each are minimal covers, it is clear thatCk can

never be a proper subset ofCi\CiCj or Ci\CiCj. From Thm. 6and Cor. 8, wecan see that the basis ofI(n, k) described in Thm. 24is a universal Grobner basis.

In Section4.1, we used the system of polynomial equations defined in Theorem 9tomodel the larger question of dominating sets in graphs G, HandGH(Theorem11). Wewill repeat the process here, using the cover-based model from Theorem24 as a buildingblock. However, a system of polynomial equations based onk+1-covers is also intrinsicallybased on graph intersections. Thus, we must define the edge variables that appear in theintersection of an arbitrary r-cover onnn vertices and an arbitrary product graph.


8/12/2019 2194-2610-1-PB

20/30

Definition 26. Given graphs Gand H, let Cbe a graph on the vertices V(G) V(H).Then the setC

E(G) E(H) is defined to be the following set of edges:

C=

(g, g) : (gh,gh)E(C) (h, h) : (gh,gh)E(C).Now we present the cover-based model that links dominating sets inG, HandG

H.As before, this system of polynomial equations will be used in the algebraic representation

of Vizings conjecture in Section5, and the restriction on the values ofn, k, n, land r willbe more meaningful in that context.

Theorem 27. Let n,k,n, l and r be integers such that r min(n, n). There is abijection between the set of solutions of the following system of polynomial equations andthe set of labeled graphsG, H inn, n vertices with dominating sets of sizek, lrespectivelysuch that their Cartesian product graphGHhas a dominating set of sizer.

Representing a graphG inn vertices with a dominating set of sizek:

e2ij eij = 0, for1 i < j n,P(C) = 0, for eachC Cnk+1.

Representing the graphH inn vertices with a dominating set of sizel:

e2ij eij = 0, for1 i < j n,

P(C) = 0, for eachC Cnl+1.

Representing the Cartesian product graphGHwith a dominating set of sizer:

P C= 0, for eachC Cnnr+1.Proof. It is clear from Theorem24that there is a bijection between the solutions of theequations representing graphsG and Hand the set of labeled graphs inn, n vertices withdominating sets of size k, l respectively. The equation representingGH is of the sameform, except that it takes into account the unique structure of the product graph.

Assume that GHhas a dominating set DV(GH) of size r . We will show thatthe cover equations associated withGHare satisfied. LetCnn

r+1be a minimalr +1-coveronV(GH). SinceCnn

r+1 is anr + 1-cover, every subset of size r has a common neighborin V(Cnn

r+1) = V(GH). In particular, D has a common neighbor ghV(Cnnr+1). SinceD is a dominating set ofGH and gh is not in D, the vertex gh must be dominated

by a vertex in D. Therefore, either gh D with (h, h) E(H), or gh D with(g, g)E(G). In first case, the edge (gh, gh)E(Cnnr+1) E(GH), and in the secondcase, the edge (gh,gh)E(Cnnr+1)E(GH). In either case, E(Cnnr+1)E(GH)=,and each of the cover equations associated with GHare satisfied.

Conversely, assume that each of the cover equations associated withGHare satisfied.We must show that GHhas a dominating set of size r. We proceed by contradiction.


8/12/2019 2194-2610-1-PB

21/30

Assume thatGHdoes not have a dominating set of size r, and consider the complementofGH(denoted by GH).

Since GH does not contain a dominating set of size r, every subset of size r inV(GH) has a common neighbor. Thus, by Definition13, GH is a r+ 1-cover. Fur-thermore,E(GH)

E(GH) =

. We remove edges fromGHuntil we find a subgraph

C that is a minimal r+ 1-cover. SinceCis a subgraph ofGH, E(GH) E(C) =.However, edges in Cof the form (gh,gh) where g=g and h=h do not correspond tovariables in our system of polynomial equations, and thus, we have not yet shown thatthe setC isnon-empty and the equation P(C) = 0 is not satisfied. We must showthat Ccontains at least one edge of the form (gh,gh) or (gh,gh).

By assumption, r min(n, n). Without loss of generality, let n n and let SV(C) = V(GH) be a subset of size r such that{g : gh S} = V(G). In otherwords, choose S such that there is at least one vertex in S per G-level. Since C isan r + 1-cover, S has a common neighbor gh V(C), and since gh is connected toevery vertex in Sand there is at least one vertex per G-level in S, there exists a vertexgh

S. Thus, (gh,gh)

E(C), implying the set

C is non-empty. But recall that

E(GH)E(C) =, which implies that the equation P(C) = 0 is not satisfied. Thisis a contradiction with our assumption that each of the cover equations associated withminimalr+ 1-covers are satisfied. Thus, GHcontains a dominating set of size r.

The ideal described by Theorem27is the set of polynomials vanishing on graphsG, Hwith dominating sets k, l respectively, such that GH has a dominating set of size r.Thus, the ideal described by Theorem 27is the same as the ideal described by Theorem11: both are I(n,k,n, l , r). However, the question of whether the basis described byTheorem11 is a universal Grobner basis is not immediately resolvable. Given two r + 1-coversCand C onnn vertices, it does seem possible that for large values ofrand smallervalues ofk and l, a k+ 1-cover on n vertices (or an l+ 1-cover on n vertices) might bea proper subgraph ofCC. In such a scenario, the collection of linear factors wouldnotsatisfy the linear factor criterion described by Definition5, and the basis would notbe a universal Grobner basis. The natural question to then pose is the following: forwhat values ofr is the cover representation ofI(n,k,n, l , r) a universal Grobner basis?In Section5, we will show that this question is equivalent to Vizings conjecture.

5 An Algebraic Formulation of Vizings Conjecture

We now express Vizings conjecture in terms of the ideals described in Section 4. Bydefining two particular zero-dimensional, radical ideals and then intersecting them, wetransform Vizings conjecture from a question about dominating sets and product graphsto an ideal membership question involving the product of two polynomials. This algebraicrepresentation suggests a variety of computational approaches, from large-scale, sparselinear algebra computations such as those demonstrated in [24, 23, 26]to Grobner basiscalculations customized for the linear factor criterion (Def. 5). Although Grobner basiscomputations and the ideal membership question itself are both known to be EXPSPACE-


8/12/2019 2194-2610-1-PB

22/30

complete [16], algebraic algorithms customized for specific NP-complete combinatorialproperties (such as graph 3-colorability) have yielded surprisingly practical computationapproaches on some large examples [23]. However, refinements to known algorithmsbased on the algebraic and combinatorial properties of these particular dominating setrepresentations have not yet been explored. We leave this question for future work.

After describing the algebraic representation of Vizings conjecture, we transfer theideal membership question involving the product of two cover polynomials back to therealm of graph theory. This transformation yields a conjecture involving k-covers that isthe complement of Vizings conjecture. We conclude by reclaiming a known result with acover-based proof, and linking universal Grobner bases to Vizings conjecture.

5.1 Ideals, Varieties and Vizings Conjecture

We begin by defining three different varieties, and then relating the vanishing idealsassociated with these varieties to the ideals described in Section 4. Each of the followingvarieties are a set of points such that each point represents a pair of graphs G, H inn, n

vertices such that GHhas a dominating set of size kl 1. Additionally,1. LetVk1l be a variety such that every point represents a pair of labeled graphsG, H

with dominating sets of size k 1, l respectively.2. LetVkl1 be a variety such that every point represents a pair of labeled graphs G, H

with dominating sets of size k, l 1 respectively.3. Let Vkl be a variety such that every point represents a pair of labeled graphsG, H

with dominating sets of size k, l respectively.

Throughout this section, graphs are always labeled graphs. We define the following ideals.

Ik1l :=I(n, k l, n, l , k l 1), Ikl1:= I(n,k,n, l l,kl 1), Ikl :=I(n,k,n, l , k l 1).

Notice that in each of these ideals, we have set r = kl1. According to the definitionsof I(n,k,n, l , r) given in Section 4, Vkl = V(I

kl), V

k1l = V(I

k1l ) and V

kl1 = V(I

kl1).

Recall that if a given ideal I is radical, then I(V(I)) = I. Since the ideals Ikl, Ik1l and

Ikl1 are radical (by Lemma 2), regardless of whether we choose the representation ofIkl from Section 4.1, or the representation of I

kl from Section 4.3, I

kl = I(V

kl ), I

k1l =

I(Vk1l ) andIkl1 = I(V

kl1). In the following lemmas, we will relate Vizings conjecture

to these ideals and varieties. It is important to note that the algebraic representation ofVizings conjecture described below is independentof the internal representation of the

idealI(n,k,n, l , r). In other words, the following lemmas will hold foranyrepresentationofIkl, I

k1l andI

kl1 as long as the ideals are radical, and I

kl =I(V

kl ), I

k1l =I(V

k1l ) and

Ikl1= I(Vkl1). Thus, if another representation ofI(n,k,n

, l , r) is found in the future witha differing set of computational or combinatorial properties, this formulation of Vizingsconjecture remains valid.

Lemma 28. Vk1l Vkl1= Vkl if and only if Vizings conjecture is true.


8/12/2019 2194-2610-1-PB

23/30

Proof. Every point in Vkl corresponds to a graph G with (G) k and a graph H with(H) l such that (GH) kl 1.

Assume that Vizings conjecture is true. Thus, (G)(H) (GH) for all graphsG, H. In particular, if (G) = k and (H) = l, then kl (GH). In particular,(GH) can never equal kl

1. Thus, if there is a pair of graphs G, H such that

(GH) kl 1, then either (G) k 1 or (H) l 1. Thus, any point appearingin Vkl also appears inV

k1l Vkl1, and Vk1l Vkl1= Vkl .

Conversely, assume that Vk1l Vkl1 = Vkl . Then, for every pair of graphs G, H suchthat (GH) kl1, then either (G) k1 or (H) l1. In other words, if(G) =k and (H) =l, then(GH)> kl 1. This implies that(G)(H) (GH),or that Vizings conjecture is true.

Lemma 29. Ik1l Ikl1= Ikl if and only if Vizings conjecture is true.Proof. IfIk1l Ikl1 = Ikl, then Vk1l Vkl1 = Vkl (and vice versa). This can be seen viathe results in Section2,since ifIk1l Ikl1 = Ikl, thenV(Ik1l Ikl1) =V(Ik1l )V(Ikl1) =V

k1

l Vkl1= V(I

kl) =V

kl . Thus, Vizings conjecture is true via Lemma 28.

We have now equated Vizings conjecture with a question about the equality of twoideals. To prove this equality, we must prove the inclusion in both directions. Theinclusion in one direction is fairly easy to see; however, the inclusion in the other directionremains open.

Lemma 30. Vk1l Vkl1Vkl .Proof. Consider any point p in Vk1l Vkl1. This point either represents graphs G, Hwith dominating sets k1, l respectively, or graphs G, Hwith dominating sets k, l1respectively (in both cases,GHcontains a dominating set of size kl

1). However, ifG

contains a dominating set of sizek 1 (orl 1),G also contains a dominating set of sizek (or sizel). Therefore, any point representing graphs G, Hwith dominating setsk 1, lrespectively, or graphs G, H with dominating sets k, l1 respectively also representsgraphs G, Hwith dominating sets k, l respectively. Thus, any point p Vk1l Vkl1 isalso a point pVkl .Corollary 31. Ikl Ik1l Ikl1.Proof. Since Vk1l Vkl1Vkl , this implies that I(Vkl )I

Vk1l Vkl1

, which implies

that Ikl Ik1l Ikl1 (see [11], Chapter 4).

Thus, in order to conclude a proof Vizings conjecture, we must show that Ik1

l Ik

l1Ikl . However, recall from the conclusion of Section2, if the representations ofIkl, Ik1l andIkl1 satisfy the conditions specified in Lemma 3, we can explicitly describe a basis forIk1l Ikl1:

Ik1l Ikl1=

Ik1l Ikl1=

f g: fIk1l , gIkl1

+ e2 e.


8/12/2019 2194-2610-1-PB

24/30

Thus far in this section, without specifying the representation for Ikl , etc., we haveshown that proving Vizings conjecture is equivalent to proving the equality of two ideals.We will now show that for each of the representations specified in Section 4, proving theopposite inclusion reduces to a problem in ideal membership, which implies, far moresignificantly, that proving Vizings conjecture reduces to a computational problem in

Grobner bases or linear algebra.

Using the Representation from Theorem 11:

Let h be any polynomial in Ik1l Ikl1, where Ik1l and Ikl1 are represented using thepolynomials defined by Theorem9. ThenhIk1l Ikl1 can be written as:

h= ()Pk1G PkG+ ()Pk1G Pl1H + ()Pk1G PGH+ ()Pk1G (e2 e) + ()PlHPkG+ ()PlHPl1H+ ()PlHPGH+ ()PlH(e2 e) + ()PGHPkG+ ()PGHPl1H + ()PGHPGH+ ()PGH(e2 e) + ()(e2 e)PkG+ ()(e2 e)Pl1H

+ ()(e2

e)PGH+ ()(e2

e)(e2

e) + ()(e2

e), (8)wherePGH is equal to P

kl1GH. We must show that h is also in I

kl. Specifically, we must

show that there exists polynomial coefficients such that

h= ()PkG+ ()PlH+ ()PGH+ ()(e2 e). (9)

Comparing Eqs. 8 and 9, we see that the only term which is not alreadyexpressed interms of polynomials in Ikl is the product P

k1G P

l1H . Thus, we must find an algebraic

relationship or asyzygy such that

Pk1G Pl1

H = ()PkG+ (

)PlH+ (

)PGH+ (

)(e2

e). (10)

This is equivalent to asking whether or not Pk1G Pl1

H Ikl. Here is the link betweenVizings conjecture and ideal membership: a conjecture about dominating sets in productgraphs is an ideal membership question about the product of two polynomials.

Lemma 32. Pk1G Pl1

H Ikl if and only if Vizings conjecture is true.Proof. IfPk1G P

l1H Ikl, then any polynomialhIk1l Ikl1is also inIkl. In other words,

Ik1l Ikl1Ikl . Lemma30and Corollary31establish the other direction of the inclusion,and Ik1l Ikl1 = Ikl. Thus, by Lemma 29, Vizings conjecture is true. Conversely, ifVizings conjecture is true, thenIk1l

Ikl1= I

kl, which implies that P

k1G P

l1H

Ikl.

Unfortunately, performing these computations was quite difficult. We worked on ma-chines with dual Opteron nodes, 2 GHz clock speed, and 12 GB of RAM. We used thecustom C++ exact arithmetic linear algebra solver and the method described in [24],[23]and [26]. We first tested the smallest possible example ofn= k = n = l = 2. In orderto find the syzygy defined by Eq. 10, we constructed a 100 84 linear system, solved inunder a second, yielding a syzygy of degree 6. However, when we tested the next largest


8/12/2019 2194-2610-1-PB

25/30

example, with n = 3, k = n = l = 2, we were unable to find a syzygy, although wedetermined that the syzygy had degree 8 or larger. The degree 7 linear algebra systemhad size 158, 4122, 310 and took 6,862 seconds ( 2 hours) to solve. However, thedegree 8 linear algebra system remained unsolved even after days of computation, mostlikely due to the magnitude of the numbers involved. For example, one of the coefficients

in PkG was 2,106,048,060. It is easy to imagine the computationally intensive nature ofexact arithmetic on numbers of that order in a linear system with millions of rows andcolumns. However, we again note that the specific combinatorial properties of these rep-resentations have not yet been exploited. It is a question of future work to improve thesecomputations.

Using the Representation from Theorem 27:

In Lemma32, we showed that proving Vizings conjecture using the representation de-scribed by Theorem 11 was equivalent to proving that Pk1G P

l1H Ikl. We now show

that when the representation ofIkl corresponding to covers (Theorem 27) is used, Viz-

ings conjecture reduces to both an ideal membership question involving the product oftwo cover polynomials, and a graph theoretic question that is the complement of Vizingsconjecture.

In the cover representation of Ikl, we denote the various cover polynomials asP(Cnk ), P(C

n

l ) andP Cnnkl . As before, hIk1l Ikl1 can be written as follows:

h= ()P(Cnk )P(Cnk+1) + ()P(Cnk )P(Cn

l ) + ()P(Cnk )P Cnnkl + ()P(Cnk )(e2 e)

+ ()P(Cnl+1)P(Cnk+1) + ()P(Cn

l+1)P(Cn

l ) + ()P(Cn

l+1)P Cnnkl + ()P(Cnl+1)(e2 e)

+ ()P Cnnkl P(Cnk+1) + ()P Cnnkl P(Cnl ) + ()P Cnnkl P Cnnkl + (

)P C

nn

kl (e2 e) + ()(e2

e)P(Cnk+1) + (

)(e2

e)P(Cn

l )

+ ()(e2 e)P Cnnkl + ()(e2 e)(e2 e) + ()(e2 e). (11)Again, to conclude the proof of Vizings conjecture, we must show that h Ikl. Morespecifically, we must show that there exist coefficients such that

h= ()P(Cnk+1) + ()P(Cn

l+1) + ()P Cnnkl + ()(e2 e). (12)

Comparing Eqs. 11and12,we see that the only term which is not already expressed interms of polynomials inIkl is the productP(C

nk )P(C

n

l ). Thus, we must find an algebraicrelationship or asyzygy such that

P(Cnk )P(C

n

l ) = ()P(Cnk+1) + ()P(C

n

l+1) + ()P Cnnkl + ()(e2 e).This is equivalent to asking whether or not P(Cnk )P(C

n

l )Ikl. However, recall that thecover representations are not just a basis, but a universal Grobner basis. Our experimentalinvestigations using CoCoA Lib [1]have indicated that that not only isP(Cnk )P(C

n

l )Ikl,but that there exists aP

Cnnkl such thatP(Cnk )P(Cnl ) =PCnnkl . In other words,P(Cnk )P(C

n

l ) is itself an element of the Grobner basis ofIkl . We were not able to verify


8/12/2019 2194-2610-1-PB

26/30

this conjecture on large examples because of the exponential number of monomials inthe expanded cover polynomials. For example, P(Cnk ) may be compactly representedas a product ofL linear factors, but when P(Cnk ) is expanded during the calculation ofthe Grobner basis, 2L monomials are generated. Indeed, even very small examples oftentook days of computation. We are interested in exploring modifications to Grobner basis

algorithms to exploit the factored form of these input bases for future work.We will now prove several lemmas and extrapolate the conjecture that there exists a

polynomial P Cnnkl such that P(Cnk )P(Cnl ) =P Cnnkl to a very specific graph

theory conjecture. Recall thatCnnkl (Definition26) denotes a specific set of edges.Lemma 33. Given a minimal k-cover Cnk and a minimal l-cover C

n

l , if there exists a(kl)-coverCnn

kl such thatE

CnkCn

l

=Cnnkl , then Vizings conjecture is true.

Proof. If there exists a (kl)-cover Cnn

kl such that E

CnkCn

l

= Cnnkl , then

P(Cnk )P(Cn

l ) = P Cnnkl , and P(Cnk )P(Cnl ) Ikl. Thus, Ik1l Ikl1 = Ikl , and

Vizings conjecture is true.

We will now define a product graph that specifically relates to the intersection.Definition 34. Given graphs G and H, the star product GHhas vertex set V(G)V(H) and edge set E(GH) =E(GH) (gh,gh) : g=g and h=h.Proposition 35. Given a k-cover Cnk and a l-cover C

n

l , then E

CnkCn

l

=

CnkCn

l

.

Proof. This follows directly from Definitions26 and34.

We note that C = CnkCn

l contains the largest amount of edges such thatC =CnkC

n

l . The question that remains is the following.

Conjecture 36. Given minimal k, l-coversCnk , C

n

l , CnkC

n

l is a kl-cover.This is the complement of Vizings conjecture. We observe that we can easily prove

two of the known properties of covers on CnkCn

l .

Proposition 37. Given minimalk, l-coversCnk , Cn

l , every vertex inV(CnkC

n

l ) is con-tained in akl-clique.

Proof. Let gh be a vertex in CnkCn

l . We must show that gh appears in a kl-clique.Since Cnk , C

n

l are a k, l-covers respectively, g appears in a k-clique in Cnk , and happears

in anl-clique inCn

l . Letg {g1, . . . , gk1}be thek-clique inCnk , and leth{h1, . . . , hl1}be the l-clique in Cn

l . We claim

Q= g {g1, . . . , gk1} h {h1, . . . , hl1}is akl-clique inCnkC

n

l . Letghandgh be two vertices inQ. Ifg = g , then (h, h)Cnl ,

since h, h appear in an l-clique. Ifh = h, then (g, g)Cnk , since g, g appear in an k-clique. In both cases, (gh,gh) is an edge in CnkC

n

l . Finally, if g= g and h= h,then (gh,gh) is an edge inCnkC

n

l by the definition of the star product. Thus, we haveshown that Qis a kl-clique.


8/12/2019 2194-2610-1-PB

27/30

Additionally, we can see that ifk and l are strictly greater than two, thenCnkCn

l hasdiameter at most two. This can be seen by choosing any two verticesgh, gh inCnkC

n

l ,and noting that there must exist a third vertex gh such that gh is adjacent to bothgh and gh. Thus, under these conditions, the diameter ofCnkC

n

l is at most two.When framing Vizings conjecture in terms of covers and the star product, we can

easily reclaim the complement of a known result by El-Zahar and Pareek [ 12].

Theorem 38. Given minimal k, l-covers Cnk , Cn

l , such that kl1 < min(n, n), thenCnkC

n

l is akl-cover.

Proof. Consider any set Sofkl 1 vertices inCnkCnl . LetPnk denoteCnk -projection ofS(the set ofg coordinates such that ghis a vertex in S), and let Pn

l denoteCn

l -projectionofS (the set ofh coordinates such that gh is a vertex in S). Sincekl1

8/12/2019 2194-2610-1-PB

28/30

approaching a combinatorial problem via algebraic geometry. In particular, given a prob-lem such as Vizings conjecture, creating an algebraic formulation is an opportunity for anentirely new set of tools to try their strength against an old open problem. For example,during the actual computation of the universal Grobner basis, given that both the initialideal representation and the final universal Grobner basis have known combinatorial in-

terpretations, is there a graph-theoretic significance to the intermediate polynomials? DoBuchbergers S-pair polynomials correspond to a particular subgraph operation? Sinceno graph-theoretic counterexample to Vizings conjecture has been found in 50 years, is itpossible that the search for a polynomial counterexample (or proof) is a less prohibitivechallenge?

Additionally, the computational significance of the linear factor criterion introducedin this paper is yet to be explored. In this case, the factored form of the both the inputandoutput ideal is known before the computation even begins. With this combinatorialknowledge in hand, could a Grobner basis algorithm be specifically tailored to avoidcertain types of polynomial expansions when computing on ideals of this form? Finally, auniversal Grobner basis that is the product of linear factors is not unique to the dominatingset problem; preliminary investigations indicate that other polynomial models of othercombinatorial problems (such as independent set and factoring) share this property.

In short, this paper explores algebraic representations of the dominating set problem,focusing on a combinatorial interpretation of the universal Grobner basis. The conse-quence of this result on computation is the subject of future work.

Acknowledgements

The authors would like to thank the anonymous referees for their comments. We alsoacknowledge the support of NSF DMS-0729251, NSF-CMMI-0926618, DMS-0240058 andthe Rice University VIGRE program.

References

[1] J. Abbott and A.M. Bigatti. CoCoALib: a C++ library for doing Computations inCommutative Algebra. Available at http://cocoa.dima.unige.it/cocoalib.

[2] N. Alon. Combinatorial Nullstellensatz. Combinatorics, Prob. and Computing, 8:729, 1999.

[3] N. Alon and M. Tarsi. Colorings and orientations of graphs. Combinatorica, 12:125

134, 1992.[4] E. Babson, S. Onn, and R. Thomas. The Hilbert zonotope and a polynomial time

algorithm for universal Gorbner bases. Advances in Applied Mathematics, 30(3):529544, 2003.

[5] A. Barcalkin and L. German. The external stability number of the cartesian productof graphs. Bul. Akad. Stiinte RSS Moldoven., 1(94):58, 1979.


8/12/2019 2194-2610-1-PB

29/30

[6] V. Blanco and J. Puerto. Partial Grobner bases for multiobjective integer linearoptimization. SIAM Journal Discret. Math., 23(2):571595, 2009.

[7] B.Stjan, B.Bresar, P.Dorbec, W.Goddard, B.Hartnell, M.Henning, S.Klavzar, andD.Rall. Vizings conjecture: A survey and recent results, 2009. preprint.

[8] B. Buchberger. An algorithm for finding the basis elements of the residue class ring ofa zero-dimensional polynomial ideal. Journal of Symbolic Computation, 4(3-4):475511, 2006.

[9] W. Clark and S. Suen. An inequality related to Vizings conjecture. ElectronicJournal of Combinatorics, 7(Note 4), 2000.

[10] P. Conti and C. Traverso. Buchberger algorithm and integer programming. InProceedings of the 9th International Symposium, on Applied Algebra, Algebraic Al-gorithms and Error-Correcting Codes, AAECC-9, pages 130139. Springer-Verlag,1991.

[11] D. Cox, J. Little, and D. OShea. Ideals, Varieties and Algorithms. Springer, New

York, 1998.[12] M. El-Zahar and C.M. Pareek. Domination number of products of graphs. Ars

Combin., 31:223227, 1991.

[13] S. Eliahou. An algebraic criterion for a graph to be four-colourable. AportacionesMatematicas, 6:327, 1992.

[14] K.G. Fischer. Symmetric polynomials and Halls theorem.Discrete Math, 69(3):225234, 1988.

[15] M. Garey and D. Johnson. Computers and Intractability: A Guide to the Theory ofNP-Completeness. W.H. Freeman and Company, 1979.

[16] J. Gathen and J. Gerhard. Modern Computer Algebra. Cambridge University Press,Cambridge, second edition, 1999.

[17] B. Hartnell and D. Rall. Domination in cartesian products: Vizings conjecture. InDomination in graphs, pages 163189, New York, 1998. Mono. Textbooks Pure andAppl. Math, Dekker.

[18] C.J. Hillar and T. Windfeldt. An algebraic characterization of uniquely vertex col-orable graphs. Journal of Combinatorial Theory Series B, 98:400414, 2008.

[19] M. Kreuzer and L. Robbiano. Computational Commutative Algebra I. Springer-Verlag, 2000.

[20] S. Li and W. Li. Independence number of graphs and generators of ideals. Combi-

natorica, 1:5561, 1981.

[21] J.A. De Loera. Grobner bases and graph colorings. Beitrage zur Algebra und Ge-ometrie, 36(1):8996, 1995.

[22] J.A. De Loera, C. Hillar, P.N. Malkin, and M. Omar. Recognizing graph theoreticproperties with polynomial ideals. Elect. J. of Combinatorics., 17(1):R114, 2010.


8/12/2019 2194-2610-1-PB

30/30

[23] J.A. De Loera, J. Lee, P.N. Malkin, and S. Margulies. Hilberts Nullstellensatz andan algorithm for proving combinatorial infeasibility. In Proceedings of the twenty-

first international symposium on Symbolic and algebraic computation, pages 197206.ISSAC, 2008.

[24] J.A. De Loera, J. Lee, S. Margulies, and S. Onn. Expressing combinatorial op-

timization problems by systems of polynomial equations and the Nullstellensatz.Combinatorics, Probability and Computing, 18(4):551582, 2009.

[25] L. Lovasz. Stable sets and polynomials. Discrete Mathematics, 124:137153, 1994.

[26] S. Margulies. Computer Algebra, Combinatorics, and Complexity: Hilberts Null-stellensatz and NP-Complete Problems. University of California at Davis, Ph.D.Thesis, 2008.

[27] Y. Matiyasevich. A criteria for colorability of vertices stated in terms of edge orien-tations (in russian). Discrete Analysis (Novosibirsk), 26:6571, 1974.

[28] Y. Matiyasevich. Some algebraic methods for calculation of the number of colorings

of a graph (in russian). Zapiski Nauchnykh Seminarov POMI, 293:193205, 2001.[29] K. Mizuno and S. Nishihara. Nowhere-zero flow polynomials. Journal of Combina-

torial Theory, Series A, 108(2):205215, 2004.

[30] M. Mnuk. Representing graph properties by polynomial ideals. InComputer Algebrain Scientific Computing, pages 431444. 4th Inter. Work. on Comp. Alg. in Sci.Comp., 2001.

[31] A. Simis, W. Vasconcelos, and R. Villarreal. On the ideal theory of graphs. Journalof Algebra, 167(2):389416, 1994.

[32] D. Sumner and P. Blitch. Domination critical graphs. Journal of CominatorialTheory, Series B, 34:6576, 1983.

[33] L. Sun. A result on Vizings conjecture.Discrete Mathematics, 275(1):363366, 2004.

[34] R. Thomas. A geometric buchberger algorithm for integer programming.Mathemat-ics of Operations Research, 20(4):864884, 1995.

[35] V. Vizing. Some unsolved problems in graph theory. Uspekhi Mat. Nauk, 23:117134,1968.

Date post:	03-Jun-2018
Category:	Documents
Upload:	luis-alberto-fuentes
View:	215 times
Download:	0 times

2194-2610-1-PB

Documents