Towards a Computational Proof of Vizing's Conjecture using Semidefinite Programming and Sums-of-Squares

Vizing's conjecture (open since 1968) relates the product of the domination numbers of two graphs to the domination number of their Cartesian product graph. In this paper, we formulate Vizing's conjecture as a Positivstellensatz existence question. In particular, we select classes of graphs according to their number of vertices and their domination number and encode the conjecture as an ideal/polynomial pair such that the polynomial is non-negative on the variety associated with the ideal if and only if the conjecture is true for this graph class. Using semidefinite programming we obtain numeric sum-of-squares certificates, which we then manage to transform into symbolic certificates confirming non-negativity of our polynomials. Specifically, we obtain exact low-degree sparse sum-of-squares certificates for particular classes of graphs. The obtained certificates allow generalizations for larger graph classes. Besides computational verification of these more general certificates, we also present theoretical proofs as well as conjectures and questions for further investigations.


Introduction
Sum-of-squares and its relationship to semidefinite programming is a cutting-edge tool at the forefront of polynomial optimization [5]. Activity in this area has exploded over the past two decades to span areas as diverse as real and convex algebraic geometry [19], control theory [17], proof complexity [14], theoretical computer science [2] and even quantum computation [3]. Systems of polynomial equations and other non-linear models are similarly widely known for their compact and elegant representations of combinatorial problems. Prior work on polynomial encodings includes colorings [1,16], stable sets [20,21], matchings [9], and flows [24]. In this project, we combine the modeling strength of systems of polynomial equations with the computational power of semidefinite programming and devise an optimization-based framework for a computational proof of an old, open problem in graph theory, namely Vizing's conjecture.
Vizing's conjecture was first proposed in 1968, and relates the sizes of minimum dominating sets in graphs G and H to the size of a minimum dominating set in the Cartesian product graph G H; a precise formulation follows as Conjecture 2.1. Prior algebraic work on this conjecture [22] expressed the problem as the union of a certain set of varieties and thus the intersection of a certain set of ideals. However, algebraic computational results have remained largely untouched. In this project, we present an algebraic model of Vizing's conjecture that equates the validity of the conjecture to the existence of a Positivstellensatz, or a sum-of-squares certificate of non-negativity modulo a carefully constructed ideal.
By exploiting the relationship between the Positivstellensatz and semidefinite programming, we are able to produce sum-of-squares certificates for certain classes of graphs where Vizing's conjecture holds. Thus, not only are we demonstrating an optimization-based approach towards a computational proof of Vizing's conjecture, but we are presenting actual minimum degree non-negativity certificates that are algebraic proofs of instances of this combinatorial problem. Although the underlying graphs do not further what is known about Vizing's conjecture at this time (indeed the combinatorics of the underlying graphs is fairly easy), such a construction of "combinatorial" Positivstellensätze is successfully executed for the first time here. The construction process is an elegant combination of computation, guesswork, computer algebra and proofs relying on clever definitions of certain polynomials as well as tricky manipulations.
Our paper is structured as follows. In Section 2, we present the necessary background and definitions from graph theory and commutative algebra. In Section 3, we begin the heart of the paper: we describe the ideal/polynomial pair that models Vizing's conjecture as a sum-of-squares problem. This pair is parametrized by the sizes n G and n H of the graphs G and H respectively, and on the sizes k G and k H of a minimum dominating set in these graphs. In Section 4, we describe our precise process for finding the sum-of-squares certificates along with an example. In Sections 5 and 6, we present our computational results and the Positivstellensätze, i.e., the theorems that arise for various generalizations. In particular, in Section 5, we introduce certain symmetric polynomials that not only allow for a compact notation, but also are vital in proving correctness of the certificates. With the help of the developed calculus, we investigate the graph classes where k G = n G and k H = n H − d and present certificates for d ∈ {0, . . . , 4} (all other parameters arbitrary). We provide formal proofs for d ≤ 2 and computational proofs using SageMath [8] for d ≤ 4. Moreover, for fixed integer d, we explain an algorithm for computing a certificate or proving that there is none of the conjectured form. Then, in Section 6, the case k G = n G − 1 and k H = n H − 1 is considered. For this class, we obtain certificates for n H ∈ {2, 3} (n G arbitrary) and prove their correctness. Finally, in Section 7, we summarize our project, state some concluding remarks and present our ideas for future work. For the sake of completeness, an appendix provides certificates (along with proofs) that arose during the application of our method but were dismissed after we obtained certificates with simpler forms.

Backgrounds and Definitions
In this section, we recall all necessary definitions from graph theory, polynomial ideals and commutative algebra.

Definitions from Graph Theory
Given a graph G with vertex set V (G), a set D ⊆ V (G) is a dominating set in G if for each v ∈ V (G) \ D, there is a u ∈ D such that v is adjacent to u (i.e., there is an edge between u and v) in G. A dominating set is called minimum if there is no dominating set of smaller size (i.e., cardinality). The domination number of G, denoted by γ(G), is the size of a minimum dominating set in G. The decision problem of determining whether a given graph has a dominating set of size k is NP-complete [11].
Given graphs G and H with edge sets E(G) and E(H) respectively, the Cartesian product graph G H has vertex set 2 V (G) × V (H) and edge set In 1968, Vadim G. Vizing conjectured the following beautiful relationship between domination numbers and Cartesian product graphs: Conjecture 2.1 (Vizing [28], 1968). Given graphs G and H, then the inequality
Example 2.2. In this example, we demonstrate the Cartesian product graph of two C 4 cycle graphs: In these graphs, represents a vertex in a minimum dominating set, and Vizing's conjecture holds with equality: γ(G) γ(H) = 2 · 2 = 4 = γ(G H). However, observe that some copies of G in G H do not contain any vertices of the dominating set, i.e., they are dominated entirely by vertices in other "layers" of the graph. This example highlights the difficulty of Vizing's conjecture.

Historical Notes
Vizing's conjecture is an active area of research spanning over fifty years. Early results have focused on proving the conjecture for certain classes of graphs. For example, in 1979, Barcalkin and German [4] proved that Vizing's conjecture holds for graphs satisfying a certain "partitioning condition" on the vertex set. The idea of a "partitioning condition" inspired work for the next several decades, as Vizing's conjecture was shown to hold on paths, trees, cycles, chordal graphs, graphs satisfying certain coloring properties, and graphs with γ(G) ≤ 2. These results are clearly outlined in the 1998 survey paper by Hartnell and Rall [15]. In 2004, Sun [26] showed that Vizing's conjecture holds on graphs with γ(G) ≤ 3. There are also results proving a weaker version of the conjecture, for example, the recent result of Zerbib [30] showing that γ(G) γ(H) + max{γ(G), γ(H)} ≤ 2 γ(G H). The 2009 survey paper [6] summarizes the work from 1968 to 2008, contains new results, new proofs of existing results, and comments about minimal counterexamples.

Definitions around Polynomial Ideals and Sum-of-Squares
Our goal is to model Vizing's conjecture as a semidefinite program. In particular, we will create an ideal/polynomial pair such that the polynomial is non-negative on the variety associated with the ideal if and only if Vizing's conjecture is true.
In this subsection, we present a brief introduction to polynomial ideals, and the relationship between non-negativity and sum-of-squares. This material is necessary for understanding our polynomial ideal model of Vizing's conjecture. For a more thorough introduction to this material see [5] and [7].
Throughout this section, let I be an ideal in a polynomial ring P = K[z 1 , . . . , z n ] over a field K ⊆ R. The variety of the ideal I is defined as the set We say that the ideal I is radical if whenever f m ∈ I for some polynomial f ∈ P and integer m ≥ 1, then f ∈ I. It should be mentioned that radical ideals and varieties are closely connected. 3 The concrete ideals that we are using later on are all radical. This is a consequence of the following lemma. The notion square-free implies that when a polynomial is decomposed into its unique factorization, there are no repeated factors. For example, (z 2 in each variable (i.e., boolean ideals) are radical. In this work, we will make heavy use of one of the most prominent theorems of algebra, namely Hilbert's Nullstellensatz.
Theorem 2.4 (Hilbert's Nullstellensatz). Let K be a field (not necessarily real, as assumed everywhere else), P = K[z 1 , . . . , z n ] a polynomial ring, I ⊆ P an ideal and f ∈ P . If f (z * ) = 0 for all z * ∈ V(I), then there is a non-negative integer r with f r ∈ I. Remark 2.5. Our set-up implies the following: • If the ideal I is radical, then f (z * ) = 0 for all z * ∈ V(I) implies f ∈ I.
• If I = f 1 , . . . , f q for some f 1 , . . . , f q ∈ P , then it suffices to check all z * that are common zeros of f 1 , . . . , f q (over the algebraic closure K) instead of all z * ∈ V(I).
Therefore, if both assumptions are satisfied, then f (z * ) = 0 for all z * that are common zeros of f 1 , . . . , f q (over the algebraic closure K) implies f ∈ I.
We continue with our background by recalling the necessary notation for sum-of-squares for the ideal I of the polynomial ring P . As usual, we write f ≡ h mod I whenever f = h + g for some g ∈ I.
Definition 2.6. Let be a non-negative integer. A polynomial f ∈ P is called -sumof-squares modulo I (or -sos modulo I), if there exist polynomials s 1 , . . . , s t ∈ P with degrees deg s i ≤ for all i ∈ {1, . . . , t} and In the context of real-valued polynomials as we have it, algebraic identities like f = t i=1 s 2 i + g for some g ∈ I, are often referred to as Positivstellensatz certificates of non-negativity, and these identities can be found via semidefinite programming, which is at the heart of this project. We present a first example now and will describe precisely how these certificates are obtained in Section 4.
and the polynomial z 1 + z 2 − 2 is said to be 1-sos modulo I. The certificate consists of the single polynomial It is well-known that not all non-negative polynomials can be expressed as a sum-ofsquares. However, in the particular case when the ideal is radical and the variety is finite, we can state the following. Proof. Let f be a polynomial that can be expressed as a sum-of-squares modulo I, Since all polynomials in the ideal I vanish on the variety by definition and since the real-valued t i=1 s 2 i is clearly non-negative, f is non-negative on the variety V(I).
To prove the other direction, we recall a well-known argument. Suppose we have a polynomial f with f (z * ) ≥ 0 for all z * ∈ V(I). Suppose further that the finite variety V(I) equals {z * 1 , . . . , z * t } for a suitable t. We now construct t interpolation polynomials f i for i ∈ {1, . . . , t} (see [12]) such that Observe that the square of an interpolating polynomial is again an interpolating polynomial. Therefore, the difference polynomial vanishes on every point {z * 1 , . . . , z * t } in the variety. We now use Hilbert's Nullstellensatz (Theorem 2.4): Since the ideal I is radical, we apply Remark 2.5 on the difference polynomial (2.1) and get that it is in I. Therefore, if we let We observe that the in this case is quite large, since it is the degree of the interpolating polynomial f i , which depends on the number of points in the variety. However, we will rely on the fact that the sum-of-squares representation is not unique, and there may exist Positivstellensatz certificates of much lower degree, within reach of computation. As we will see in Section 5 and 6, this does indeed turn out to be the case.

Vizing's Conjecture as a Sum-of-Squares Problem
In this section, we describe Vizing's conjecture as a sum-of-squares problem. Towards that end, we will first define ideals associated with graphs G, H and G H, and then finally describe an ideal/polynomial pair where the polynomial is non-negative on the variety of the ideal if and only if Vizing's conjecture is true. We begin by creating an ideal where the variety of solutions corresponds to graphs with a given number of vertices and size of a minimum dominating set.
The notation underlying all of the definitions in this section-we will use it also through the whole article-is as follows. Let n G and k G ≤ n G be fixed positive integers, and let G be the class of graphs on n G vertices with a fixed 4 minimum dominating set D G of size k G . We then turn the various edges "on" or "off" (by controlling a boolean variable e gg for each possible edge {g, g }) such that each point in the variety corresponds to a specific graph G ∈ G.
(3.1c) 4 We fix the vertices of the dominating set without loss of generality as this corresponds to a simple renaming of the vertices. Doing this avoids the introduction of additional boolean variables for the vertices and reduces the size of the corresponding isomorphism group of the variety. It is therefore algorithmically favorable. 5 Being precise, the ideal IG is defined by the polynomials on the left-hand side of the equations (3.1a), (3.1b) and (3.1c). However, we think that the current phrasing provides the better insight and is closer to the intended way of thinking for this work. If one would like to write equations in a formally correct way, one first needs to evaluate the polynomial on the left-hand side at some suitable point, meaning to substitute the variables of the polynomial ring by real values. For example, the variable e gg ∈ PG is substituted by some (possibly a priori unknown) e * gg ∈ R, therefore the polynomial on the left-hand side of (3.1a) becomes the equation (e * gg ) 2 − e * gg = 0. Notation 3.2 is also related to this issue and brings the connection to the points in the associated variety.
We will, however, always be precise when the distinction between variable and evaluated (starred) form matters. Notation 3.2. Throughout this paper, we will use the following notations: We will use z for the tuple of variables of the polynomial ring P , so P = K[z]. When considering the variety V(I) associated to an ideal I ⊆ P , we use the notation z * ∈ V(I) for the elements in this variety.
Note that the polynomial ring variables (which are the components of z) correspond bijectively to the components of z * . In particular we will use e * gg for the component of z * = e * G ∈ V(I G ) corresponding to the polynomial ring variable e gg ∈ P G . Remark 3.3. Definition 3.1 is meaningful even in the case that n G = 1. The only vertex must be in the dominating set, so k G = 1. Pairs {g, g } cannot be chosen from the one-element set V (G), thus the set of variables e G is empty. This implies P G = K[e G ] is the polynomial ring over K in no variables (i.e., isomorphic to K). The polynomials defining the ideal I G disappear: There are no polynomials coming from (3.1a) again because of non-existing pairs {g, g }. Also, there are no polynomials coming from (3.1b) as V (G) \ D G is empty because both V (G) and D G consist exactly of the same vertex. There is a contribution from (3.1c) for S being the empty set, however this is the equation 0 = 0, so again no true contribution. Thus, the ideal I G ⊆ P G only consists of 0.
This, in turn, means that the variety V(I G ) is "full" meaning in our particular situation being the set containing the empty tuple only. Proof. For n G = k G = 1 this is clearly true, as there is exactly one element in both V(I G ) and G.
For n G 2 consider any point z * ∈ V(I G ). We use Notation 3.2. Since equations (3.1a) turn the edges "on" (e * gg = 1) or "off" (e * gg = 0), the point z * defines a graph G in n G vertices. Equations (3.1b) iterate over all the vertices inside the set D G , and ensure that for each vertex outside the set at least one edge from a vertex inside the set to this vertex is "on". Therefore, D G is a dominating set. Finally, equations (3.1c) iterate over all sets S of size k G − 1 and ensure that at least one vertex outside S is not incident to any vertex inside S for any S. Therefore, no set S of size k G − 1 is a dominating set. Thus, every point z * ∈ V(I G ) corresponds to a graph G on n G vertices with a minimum dominating set of size k G .
With the intuition given above it is straight forward to construct a point in V(I G ) for a graph on n G vertices with a minimum dominating set of size k G .
Similarly, for fixed positive integers n H and k H ≤ n H , let H be the class of graphs on n H vertices and a minimum dominating set of size k H . Again, we fix the dominating set to some D H to reduce isomorphisms within the variety. Next, we define the graph class G H and the ideal I G H . For the above classes G and H, the graph class G H is the set of product graphs G H for G ∈ G and H ∈ H.
The new variables needed for the ideal are the variables corresponding to the vertices in the product graph and indicate if such a vertex is in the dominating set or not. Let Definition 3.5. The ideal I G H ⊆ P G H is defined by the system of polynomial equations for g ∈ V (G) and h ∈ V (H).
Observe that we have no restrictions on the edge variables in this definition. It is only used as a stepping stone to the final and most important ideal in our polynomial model. Note that our definition of I viz depends on the specific parameters n G , n H , k G and k H . Notation 3.7. Analogously to Notation 3.2 we will write z * ∈ V(I viz ) for the elements of the variety of I viz . We will use e * gg , e * hh and x * gh for the component of z * corresponding to the polynomial ring variables e gg ∈ P G , e hh ∈ P H and x gh ∈ P G H respectively. Proof. We have already demonstrated in Theorem 3.4 that V(I G ), V(I H ) are in bijection to the graphs in n G , n H vertices with minimum dominating sets of size k G , k H respectively. It remains to investigate the restrictions placed on the x gh variables, which denote whether or not the vertex gh ∈ V (G H) appears in the dominating set of the product graph.
Let z * ∈ V(I viz ) be a point in the variety. We use Notation 3.2. With the arguments above this point induces a graph G ∈ G and a graph H ∈ H. Furthermore equations (3.2a) force the vertex variables x gh to evaluate to "on" (x * gh = 1) or "off" (x * gh = 0). We define D such that the vertex gh is in D if x * gh = 1 and is outside D otherwise. Equations (3.2b) ensure that D is a dominating set, because every vertex gh is dominated. Indeed, it is either in the set itself (i.e., 1 − x * gh = 0), or it is adjacent to a vertex in the dominating set D via an edge from the underlying graph in G or the underlying graph in H. In particular, the edge {g, g } is "on" and the vertex g h is in the dominating set (e * gg = 1 and x * g h = 1), or the edge {h, h } is "on" and the vertex gh is in the dominating set (e * hh = 1 and x * gh = 1). In either of these cases, the vertex gh of the box graph is dominated. Therefore, the points in the variety V(I viz ) are in bijection to the graphs in n G , n H vertices with minimum dominating sets of size k G , k H respectively, and their corresponding product graph with a dominating set D of any size.
With the intuition given above it is straight forward to construct a point in V(I G ) for graphs G, H and a dominating set D in G H.
Observe that there are no polynomials in I viz enforcing minimality on the dominating set in the product graph. This is essential when we tie all of the definitions together and model Vizing's conjecture as a sum-of-squares problem. In particular, we model Vizing's conjecture as an ideal/polynomial pair, where the polynomial must be non-negative on the variety associated with the ideal if and only if Vizing's conjecture is true. Definition 3.9. Given the graph classes G and H, define Proof. Assume that Vizing's conjecture is true, and fix the values of n G , k G , n H and k H . Therefore, for all graphs G ∈ G and H ∈ H, we have gh coming from f viz equals exactly the size of the dominating set in the box graph G H. Therefore, we have Similarly, if f viz (z * ) ≥ 0 for all z * ∈ V(I viz ), every dominating set in G H has size at least k G k H . In particular, the minimum dominating set in G H has size at least k G k H and Vizing's conjecture is true.

Corollary 3.11. Vizing's conjecture is true if and only if for all positive integer values
Proof. The ideal I viz contains the univariate polynomial z 2 −z for each variable. Therefore, by Lemma 2.3, I viz is radical. Due to Lemma 3.8, the variety V(I viz ) is finite and obviously it is real. Therefore, by Lemma 2.8, we know that a polynomial is non-negative on V(I viz ), if and only if there exists an integer such that the polynomial is -sos modulo I viz . Hence the result follows from Theorem 3.10.
In this section, we have drawn a parallel between Vizing's conjecture and a sum-ofsquares problem. We defined the ideal/polynomial pair (I viz , f viz ) such that f viz (z * ) ≥ 0 for all z * ∈ V(I viz ) if and only if Vizing's conjecture is true. In the next section, we describe how to find these Positivstellensatz certificates of non-negativity, or equivalently, these Positivstellensatz certificates that Vizing's conjecture is true.

Overview of the Methodology
In our approach to Vizing's conjecture we "partition" the graphs G, H and G H by their sizes (number of vertices) n G and n H and by the sizes of their dominating sets k G and k H . Note that we aim for certificates for all partitions as this would prove the conjecture. However in the following we present our method which works for a fixed partition (i.e., for fixed values of n G , k G , n H and k H ), and only later relax this and generalize to parametrized partitions.
The outline is as follows: • Step 1: Model the graph classes as ideals For fixed values of n G , k G , n H and k H the first step is to create the ideal I viz as described in Section 3, in particular Definition 3.6. To summarize, we create the ideal I viz in a suitable polynomial ring in such a way that the points in the variety V(I viz ) are in bijection to the triples (G, H, D) where G is a graph in G, H is a graph in H and D is a dominating set of any size of G H; see Theorem 3.8. In this polynomial ring there is a variable for each possible edge of G and H (indicating whether this edge is present or not in the particular graphs G and H) and for each vertex of G H (indicating whether this vertex is in the dominating set of G H or not).
The second step is to use the polynomial ring variables mentioned above to reformulate Vizing's conjecture: It is true for a fixed partition if the polynomial f viz (Definition 3.9) is non-negative if evaluated at all points in the variety V(I viz ) of the constructed ideal. For showing that the polynomial is non-negative, we aim for rewriting it as a finite sum of squares of polynomials (modulo the ideal I viz ). If we find such polynomials, then these polynomials form a certificate for Vizing's conjecture for the fixed partition. To be more precise and as already described in Section 3, Vizing's conjecture is true for these fixed values of n G , k G , n H and k H if and only if f viz is -sos modulo I viz .
In the subsequent Section 4.2 we describe how to perform step three and do another reformulation, namely as a semidefinite program. Note that in order to do so, we need to have specified , the degree of the certificate. Note also that in order to prepare the semidefinite program, we use basis polynomials (i.e., special generators) of the ideals. These are obtained by computing a Gröbner basis of the ideal; see [7] for more information on Gröbner bases.
The fourth step (Section 4.3) is now to solve the semidefinite program. If the program is infeasible (i.e., there exists no feasible solution), we increase . On the other hand, if the program is feasible, then we can construct a numeric sum-of-squares certificate. As the underlying system of equations-therefore the future certificate-is quite large, we iterate the following tasks: Find a numeric solution to the semidefinite program, find or guess some structure in the solution, use these new relations to reduce the size of the semidefinite program, and begin again with solving the new program. This reduces the solution space and therefore potentially also the size (number t of summands) of the certificate and the number of monomials of the s i from Definition 2.6. The procedure above goes hand-in-hand with our next step (Section 4.4), namely obtaining (one might call it guessing) an exact certificate out of the numeric certificate.
Once we have a candidate for an exact certificate, we can check its validity computationally by summing up the squares and reducing modulo the ideal; see our step six described in Section 4.5.
We want to point out, that we still consider Vizing's conjecture for a particular partition of graphs. However, having such certificates for some partitions, one can go for generalizing them by introducing parametrized partitions of graphs. Our seventh step in Section 4.6 provides more information.
The final step is to prove that the newly obtained, generalized certificate candidate is indeed a certificate; see as well Section 4.6. Further certificates and different generalizations together with their proofs can be found in Sections 5 and 6.

Transform to a Semidefinite Program
Semidefinite programming refers to the class of optimization problems where a linear function with a symmetric matrix variable is optimized subject to linear constraints and the constraint that the matrix variable must be positive semidefinite. A semidefinite program (SDP) can be solved to arbitrary precision in polynomial time [27]. In practice the most prominent methods for solving an SDP efficiently are interior-point algorithms. We use the solver MOSEK [23] within MATLAB. For more details on solving SDPs and on interior-point algorithms see [29].
It is possible to check whether a polynomial f is -sos modulo an ideal with semidefinite programming. We refer to [5, pg. 298] for detailed information and examples. We will now present how to do so for our setting only.
Let us first fix (for example, by computing) a reduced Gröbner basis B of I viz and fix a non-negative integer . Denote by v the vector of all monomials in our polynomial ring P of degree at most which can not be reduced 6 modulo I viz by the Gröbner basis B. Let p be the length of the vector v. Then f viz (of Definition 3.9) is -sos modulo I viz if and only if there is a positive semidefinite matrix X ∈ R p×p such that f viz is equal to v T Xv when reduced over B. Hence the SDP we end up with optimizes the matrix variable X ∈ R p×p subject to linear constraints that guarantee f viz being v T Xv as above. The objective function can be chosen arbitrarily because any matrix satisfying the constraints is sufficient for our purpose. More will be said on this later.
If the SDP is feasible, then due to the positive semidefiniteness we can decompose the solution X into X = S T S for some S ∈ R t×p and t ≤ p. Subsequently, we define the polynomial s i by the ith row of Sv and obtain Note that the last congruence holds due to the constraints in the SDP. Equation (4.1) then certifies that f viz can be written as a sum of squares of the s i , and hence, f viz is -sos modulo I viz according to Definition 2.6. If the SDP is infeasible, we have an indication that there is no certificate of degree . We increase to + 1, because f viz could still be ( + 1)-sos modulo I viz or posses a certificate of even higher degree. However, if no new reduced monomials appear in this increment, then by Lemma 2.8 and Theorem 3.10 Vizing's conjecture does not hold.
Example 4.1. We consider the graph classes G and H with n G = 3, k G = 2, n H = 3 and k H = 2. Using SageMath [8] we construct the ideal I viz , generated by 32 polynomials in 15 variables. Again using SageMath, we find a Gröbner basis of size 95.
First, we check the existence of a 1-sos certificate. The vector v for = 1 has length 12, i.e., we set up an SDP with a matrix variable X ∈ R 12×12 . Imposing the necessary constraints to guarantee 12 i=1 s 2 i ≡ f viz mod I viz leads to 67 linear equality constraints. Interior-point algorithms detect infeasibility of this SDP in less than half a second, this indicates that there is no 1-sos certificate.
Setting up the SDP for checking the existence of a 2-sos certificate results in a problem with a matrix variable of dimension 67 and 359 linear constraints. Interior-point algorithms find a solution X of this SDP in 0.72 seconds, this guarantees the existence of a numeric 2-sos certificate for these graph classes.

Obtain a Numeric Certificate
As described in Section 4.2 above, after solving the SDP we decompose the solution X. We do so by computing the eigenvalue decomposition X = V T DV and then setting S = D 1/2 V , where D is the diagonal matrix having the eigenvalues on the main diagonal. Since X is positive semidefinite, all eigenvalues are non-negative and we can compute D 1/2 by taking the square root of each of the diagonal entries. The matrices X, V and D are obtained through numeric computations, hence there might be entries in D that are rather close to zero but not considered as zero. We may try setting these almost-zero eigenvalues to zero, which reduces the number of polynomials of the sum-of-squares certificate.
Furthermore, a zero column in S means that the corresponding monomial is not needed in the certificate. Hence, we may try to compute a certificate where we remove all monomials corresponding to almost-zero columns. This can decrease the size of the SDP considerably and a smaller size of the matrix and fewer constraints is favorable for solving the SDP. Of course, if removing these monomials leads to infeasibility of the SDP, then removing these monomials was not correct. Every row of S corresponds to one polynomial s i of the numeric sum-of-squares certificate and every column of S corresponds to one monomial in v.
As already mentioned we can choose the objective function arbitrarily. Our experiments show that different objective functions lead to (significantly) different solutions. Therefore, we carefully choose a suitable objective function leading to a "nice" solution for each instance.
Example 4.2. We look again at the case we considered in Example 4.1, that is G and H with n G = 3, k G = 2, n H = 3 and k H = 2, for which we already obtained an optimal solution X and a numeric 2-sos certificate.
After computing (numerically) the eigenvalue decomposition X = V T DV , we set all almost-zero eigenvalues to zero and compute S = D 1/2 V , which results in a 12 × 67 matrix, i.e., 55 eigenvalues are considered as zero. In Figure 1 a heat map of matrix S is displayed. It seems unattainable to convert this obtained solution to an exact certificate (see Section 4.4), so we take a different path.
Using different objective functions and aiming for a certificate where only certain monomials appear can lead to results with a clearer structure. If the ith monomial should not be included, we can set the ith row and column of X equal to zero and obtain another SDP, where we have fewer variables and modified constraints. We now try to use only the 19 monomials 1, x gh and x gh x gh for all g ∈ V (G) and all h, This results in an SDP with a matrix variable of dimension 19 and 99 constraints. The SDP can be solved in 0.48 seconds, and again, we obtain matrix S (after setting almost-zero eigenvalues to zero), which now is of dimension 4 × 19. A heat map is given in Figure 2.
As one sees in Figure 2, S has a certain block structure, suggesting that in each s i the coefficients of the monomials depend only on the index g ∈ V (G) and there is no dependence on the indices h ∈ V (H). Therefore, we aim for a 2-sos certificate of the form x gh x gh (4.2a) Figure 2: Plotting the entries of matrix S as in Figure 1, but now we only allow the coefficients of 19 monomials to be non-zero. The numeric sum-of-squares certificate consists of 4 (number of rows) polynomials in 19 (number of columns) monomials. In particular the first three rows correspond to s 1 , s 2 and s 3 and the last row corresponds to s 0 as given in (4.2).
for i ∈ {1, . . . , n G } and it seems almost obvious which simple algebraic numbers the entries of X could be, for example 0.667 could be 2/3. We will use that in the following section.

Guess an Exact Certificate
We now have a guess for the structure of the certificate, but coefficients that are simple algebraic numbers are hard to determine from the numbers in S. On the other hand, the exact numbers in X seem to be rather obvious so we go back to the relation X = S T S. It implies that if we fix two monomials then the inner product of the vectors of the coefficients of these monomials in all the s i has to be equal to the corresponding number in X.
Setting up a system of equations using all possible inner products, we may obtain a solution to this system. This solution determines the coefficients in the certificate (and the certificate might be simplified even further). Example 4.3. We continue Example 4.1, that is we consider the graph classes G and H with n G = 3, k G = 2, n H = 3 and k H = 2.
The exact numbers in X given in Example 4.2 can be guessed easily. In fact, if this guess for X is correct, every choice of S such that S T S = X gives a certificate. Using the relation S T S = X we set up a system of equations on the parameters of (4.2). To be more precise, ...,n G . Then we can define the vectors a = ν α , b g = λg β and c g = µg γ , and S T S = X (together with the guessed values for X) implies that a, a = 2(n G − 1) 2 , has to hold for each g ∈ V (G), where ·, · denotes the standard inner product. Under the assumption that our guess for X was correct, each solution to this system of equations leads to a valid sum-of-squares certificate (4.2). We want a sparse certificate and the numeric solution suggests that ν 2 = ν 3 = 0 holds, so we try to obtain a solution with also ν 1 = 0 (even though the numeric solution does not fit into that setting). Using these values, the equations involving the vector a determine the exact values for α, β and γ as α With that, the system of equations simplifies to λ g , λ g = 1 9 , λ g , λ g = 0, µ g , µ g = 4 9 , µ g , µ g = 0, λ g , µ g = − 2 9 , λ g , µ g = 0.
Calculating n G i=1 s 2 i we find out that, due to the system of equations, the sum-of-squares simplifies to Hence, if (4.2) is a sum-of-squares certificate then also x gh x gh , (4.3a) To close this section let us highlight once more that we use the SDP and its solution to find out which monomials are used in the certificate and to obtain a structure of their coefficients. In particular we do not need a solution of the SDP of high precision, so solving the SDP is not a bottleneck in this example. It will turn out that this is also true for all other examples we consider.

Computationally Verify the Certificate
When a certificate is conjectured, it is straightforward to verify it computationally via SageMath [8]. To do so, it is necessary to compute the Gröbner basis of I viz . Observe that at this point, semidefinite programming is no longer needed.

Generalize the Certificate and Prove Correctness
In Sections 4.2 to 4.5, we presented a methodology for obtaining a sum-of-squares certificate for graph classes G and H with fixed parameters n G , k G , n H and k H . Assuming that the previously found pattern generalizes, one can iterate the steps outlined above to obtain certificates for larger classes of graphs.
Example 4.5. We want to generalize the certificate for the graph classes G and H with n G = 3, k G = 2, n H = 3 and k H = 2 to the case k G = n G − 1 ≥ 1, n H = 3 and k H = 2 for any n G ≥ 2.
Solving the SDP for the cases n G = 4 and n G = 5 again yields nicely structured matrices and in fact, all the calculations done for the case n G = 3 (which we already wrote down parametrized by n G above) go through.
Hence, we are able to generalize the sum-of-squares certificate (4.3) in the following way.
The proof will be given later on after introducing some more auxiliary results; see Section 6. Of course, once having the theorem above, it can be verified computationally for particular parameter values, for example for k G = 4 and n G = 5, where the computation of a Gröbner basis is feasible.

Summary
In this section we saw by an example how to use our machinery combined with clever guessing in order to obtain sum-of-squares certificates for proving that Vizing's conjecture holds for fixed values of n G , k G , n H and k H , and how this can be used to obtain certificates for a less restricted set of parameters. We will use the next sections in order to present further certificates and generalizations that we found using our method and for which we were able to prove correctness.

Exact
Certificates for k G = n G and k H = n H − d In this section we deal with certificates for the case k G = n G and k H = n H − d. Towards this end we will first prove several auxiliary results in Section 5.1. Next we present and prove certificates for d = 0, d = 1 and d = 2 in Sections 5.2, 5.3 and 5.4. Then in Section 5.5 we will see how this brings insight on the structure of the certificates. We are therefore able to formulate a conjecture on the structure of the certificate for general d and also present a strategy for proving it. This will be complemented by a more computational approach for checking the conjecture for a given value d; in particular we will prove the conjecture for d = 3 and d = 4 with the help of SageMath [8].

Auxiliary Results: Sigma Calculus
In this section we will develop the machinery needed to prove the correctness of our (exact) certificates. It will turn out that the key is to be able to do operations with certain symmetric polynomials, which will be introduced in Definition 5.4. Another important tool will be again Theorem 2.4, Hilbert's Nullstellensatz. It's implications formulated as Remark 2.5 will be used repeatedly, for example in the proof of the following first lemma. Translating this lemma in terms of congruence relations, we have e gg ≡ 0 mod I G and e gg ≡ 0 mod I viz for all {g, g } ⊆ V (G).
Let us briefly consider Lemma 5.1 from a graph theoretic point of view. Due to Theorem 3.4 the points in the variety of I G are in bijection to the graphs in G, which are the graphs on n G vertices with domination number k G = n G . It is easy to see that such graphs can not have any edges, because otherwise the domination number would be strictly less than n G . Hence e * gg = 0 holds for all points z * in the variety of I viz , when we use Notation 3.2. This intuitively justifies Lemma 5.1 by graph theoretical considerations.
Let us consider the second factor of (3.2b). If n G = 1, then there is no g = g ∈ V (G), so this product is empty and equals 1. If n G ≥ 2, then e * gg = 0 (the component of z * corresponding to e gg ) for all g ∈ V (G) because of Lemma 5.1, and the product equals 1 again. Hence (3.2b) implies Assume that z * is not a zero of f . Then clearly x * gh = 0 for all h ∈ T . In particular, Therefore, for each h ∈ T , there is a h ∈ V (H) such that e * hh = 1 and x * gh = 1. As For each h ∈ T , the h (defined above) is in S, so the summand e * h h = e * hh = 1 for h = h . All other summands are either 0 or 1, hence each sum h ∈S e * h h is at least one, so in particular non-zero. This is again a contradiction.
Hence for all n G ≥ 1 our assumption was wrong, so z * is a zero of f . Now, Hilbert's Nullstellensatz (Theorem 2.4 and Remark 2.5) implies f ∈ I viz , so (5.1) is satisfied.
Furthermore, (5.1) above can be rewritten as Therefore, the congruence (5.2) follows from the fact that Note that from a high-level point of view, if k H = n H − d, then Lemma 5.2 allows us to rewrite particular products of d + 1 terms as a sum of products of at most d terms and therefore to reduce the degree of polynomials.
To continue and in order to apply the above findings, we introduce the following polynomials.

Definition 5.4. Let g ∈ V (G) and i be a non-negative integer. We define
In a classical setting the polynomial σ g,i is the elementary symmetric polynomial of degree i in n H variables. In the following we will investigate the arithmetic of the σ g,i over the ideal I viz and aim for getting nice expressions for products of σ g,i .
Lemma 5.5. Let k G , n G , k H , n H ≥ 1 and let i ≥ j. Then Note that we can extend the summation range to 0 ≤ r ≤ j as σ g,i = 0 for all i > n H . This makes the formulation of the lemma completely independent of the parameters k G , n G , k H and n H . Moreover, we will see in the proof that we actually only need generators x 2 − x in the ideal, making the lemma valid in a more general setting. Remark 5.6. As needed later, we state Lemma 5.5 for some particular values of i and j. We have σ 2 g,1 ≡ σ g,1 + 2σ g,2 mod I viz , σ g,2 σ g,1 ≡ 2σ g,2 + 3σ g,3 mod I viz , σ 2 g,2 ≡ σ g,2 + 6σ g,3 + 6σ g,4 mod I viz .
Now we come back to the proof of Lemma 5.5. In the following, we use the phrase power products to refer to products of powers of variables with non-negative exponent, or in other words, to the summands of a polynomial without their coefficient.
Proof of Lemma 5.5. We start with a couple of remarks. All summands of σ g,i and σ g,j have degree i and j respectively. Hence all summands in the product σ g,i σ g,j are summands of degree i + j. Furthermore, whenever two summands in σ g,i and σ g,j contain the same factor x gh , a resulting factor x 2 gh can be reduced to x gh over I viz because of (3.2a). Therefore, all summands in σ g,i σ g,j are square-free and will have degree at least i and at most i + j. Clearly the degree is also bounded by n H . Moreover σ g,i and σ g,j are symmetric in h ∈ V (H). By all these considerations, it is therefore possible to write for some coefficients δ r ∈ Z. In fact, these coefficients are non-negative.
For the following considerations, we always reduce modulo I viz , therefore reducing exponents of monomials larger than one to exponents exactly one. So let us fix a power product x i of σ g,i of degree i (i.e., x i = h∈S x gh for some S with |S| = i) and count power products x j of σ g,j so that the product x i x j is of degree i + r (as said, after reducing the power product over I viz ). Apparently, there have to be r factors in x j which are not factors of x i ; there are n H −i r possible such choices. The remaining j − r factors of x j have to be among the factors of x i , hence there are i j−r possible choices. Finally, we note that there are n H i choices for the fixed power product x i above. In total, expanding the product σ g,i σ g,j results in a sum of n H −i r i j−r n H i powerproducts of degree i + r for each r. We now collect these power products to determine the coefficients δ r of the corresponding summand. Each sum σ g,i+r consists of n H i+r power products of degree i + r. Hence and due to the representation (5.4), we have which completes the proof.
Lemma 5.5 allows us to replace products of our symmetric polynomials σ g,i by sums. This will become very handy in proving certificates.
We now go back to our particular set-up with k H = n H − d. The next important ingredient is the following lemma, which allows us to reduce some σ g,d+j+1 of "high" degree. Proof. For each h∈S x gh in the definition of σ g,d+j+1 , we fix an arbitrary partition S = T ∪ W with T and W disjoint in a way that |T | = d + 1 and |W | = j. Therefore, we obtain With Lemma 5.2, we can reformulate this to Due to symmetry in h ∈ V (H), minimum degree j and maximum degree d + j of the right-hand side, we can rewrite (5.5) to a representation σ g,d+j+1 ≡ d r=0 (−1) d+r β r σ g,j+r mod I viz for some coefficients β r ∈ Z.
In order to determine these β r , we count the number of power products of degree j + r on the right-hand side of (5.5). There are n H d+j+1 possible choices for S, only the one particular fixed partition S = T ∪ W and d+1 r possible choices for U out of T . Hence there are n H d+j+1 d+1 r power products of degree j + r appearing in (5.5), all of which have the same sign. Due to the fact that σ g,j+r contains n H j+r monomials, this implies that As mentioned, the lemmata above provide "reduction rules" for some quantities σ g,i or products of such quantities. We now derive explicit formulas for particular instances.
Remark 5.10. Let us fix d, i.e., our full set-up is k G = n G ≥ 1 and k H = n H − d ≥ 1, and let us fix g ∈ V (G). Then, more systematically speaking, whenever f is a finite K-linear combination of terms of the form σ g,i and σ g,i σ g,j for non-negative integers i and j, we can reduce f to a form The idea is to use Lemma 5.5 for σ g,i σ g,j in order to get rid of these products and replace them by terms of the form σ g,i . After this step, one can repeatedly use Lemma 5.2 in order to replace all σ g,i for i > d by linear combinations of σ g,i with i ≤ d. All these operations are efficient; the coefficients of the individual steps are given directly in Lemmata 5.5 and 5.2.
Finally let us mention that an implementation of the arithmetic described in the previous remark, for example with SageMath [8], is handy: It makes it easily possible to verify the results of Remarks 5.8 and 5.9.
This completes the section on our auxiliary results which we need in the following to prove our certificates.

Certificates for k G = n G and k H = n H
The easiest and almost trivial case is the one with k G = n G and k H = n H , so d = 0. We get the following certificate and therefore have proven with our method that Vizing's conjecture holds in this case. Note that we can simplify this 0-sos certificate of Theorem 5.11 to an empty sum using no polynomial, but we give the formulation of Theorem 5.11 to highlight the similarity to the other certificates we will present in this section.
Proof of Theorem 5.11. We have x gh ≡ 1 mod I viz for all g ∈ V (G) and h ∈ V (H) as already mentioned in Remark 5.3 due to Lemma 5.2. Hence we obtain so the s g form indeed a 0-sos certificate for f viz .
Note that the certificate of Theorem 5.11 has the lowest degree possible.

Certificates for
The easiest non-trivial case is the one with k G = n G and k H = n H − 1, so d = 1. Using the above machinery (explained in the methodology Section 4) we first found a complicated sum-of-squares certificate, which is presented in Appendix A.1. We were eventually able to transform this complicated certificate to the following much easier certificate and therefore have proven with our method that Vizing's conjecture holds in this case.
Theorem 5.12. For k G = n G ≥ 1 and k H = n H − 1 ≥ 1, Vizing's conjecture is true as the polynomials are a 1-sos certificate of f viz .
Proof. The polynomials s g can alternatively be written as s g = σ g,1 − (n H − 1). Using Remark 5.8 yields Consequently, this evaluates to g∈V (G) so the s g form indeed a 1-sos certificate for f viz .
Note that the certificate of Theorem 5.12 has the lowest positive degree possible and furthermore only uses very particular monomials of degree at most 1.

Certificates for k G = n G and k H = n H − 2
The next slightly more difficult case is the one for k G = n G and k H = n H − 2, so d = 2. Also in this case we first found a more complicated certificate (see Appendix A.2) which we were able to transform to the following simple certificate.
Theorem 5.13. For k G = n G ≥ 1 and k H = n H − 2 ≥ 1, Vizing's conjecture is true as the polynomials are a 2-sos certificate of f viz .
Remark 5.14. We want to point out that Theorem 5.13 is true whenever α, β, γ are solutions to the system of equations and that in Theorem 5.13 one particular easy solution is stated.
Proof of Theorem 5.13. The polynomials s g can alternatively be written as s g = ασ g,0 + βσ g,1 + γσ g,2 . (Note that σ g,0 = 1.) Using Remark 5.8 yields s 2 g = (α + βσ g,1 + γσ g,2 ) 2 = α 2 + β 2 σ 2 g,1 + γ 2 σ 2 g,2 + 2αβσ g,1 + 2αγσ g,2 + 2βγσ g,1 σ g,2 ≡ α 2 + β 2 (σ g,1 + 2σ g,2 ) + 2αβσ g,1 + 2αγσ g,2 Due to the particular values of α, β and γ this simplifies to Note that for all computationally considered instances of the form k G = n G and k H = n H − 2, the SDP for = 1 was infeasible, so for all of those instances there seems to be no 1-sos certificate and one really needs monomials of degree 2 in the s i in order to obtain a certificate. Nevertheless, degree 2 is still very low. Furthermore also in this sum-of-squares certificate only very particular monomials are used; it can be considered sparse therefore. This is confirmed by the following example.

Computational
Certificates for k G = n G and k H = n H − d When taking a closer look at the certificates in Theorem 5.11, Theorem 5.12 and Theorem 5.13, one can guess a structure from the certificates found so far. In particular there seems to be a d-sos certificate for the case k G = n G and k H = n H − d. Hence, at this point, we can formulate a conjecture which intuitively seems to be the "correct" generalization.
where α i are the solutions to a certain system of polynomial equations, are a d-sos certificate of f viz .
Moreover, the proofs of Theorems 5.12 and 5.13 give rise to an algorithmic approach for finding certificates. We formulate this as the following proposition.
Proposition 5.17. Let d be a non-negative integer. Then there is an algorithm that either finds a certificate of the form as given in Conjecture 5.16 or outputs that there is no certificate of that form.
Proof. We first describe our algorithm by following the proofs of Theorems 5.12 and 5.13.
Suppose s g is of the form as given in Conjecture 5.16, so where simply the definition of σ g,i (see Definition 5.4) was used. We now use binomial expansion for s 2 g . In the result there will be terms of the form σ g,i σ g,j for i, j ≤ d. We can now use the arithmetic described in Remark 5.10 to end up with Note that all coefficients of this polynomial additionally depend on the variable n H . The formal summation over all g ∈ V (G) is trivial; we obtain g∈V (G) In order to obtain a d-sos certificate, this has to be equal to Comparing the coefficients of n G = g∈V (G) σ g,0 and the other g∈V (G) σ g,i (for 1 ≤ i ≤ d) yields the system of equations We want to point out that the existence of a real-valued solution α 0 , . . . , α d (as functions in n H ) is equivalent to the fact that to the s g being as in Conjecture 5.16 form a d-sos certificate. Therefore computing the variety associated to the system of equations (5.7), i.e., finding all solutions of this system, is the last step of an algorithm that has the properties stated in Proposition 5.17, and the proof is completed.
Remark 5.18. The system of equations (5.7) does not depend on n G , but only on d and n H . Hence, whenever we find a solution to (5.7)-this might be for a fixed value of n H or parametrized in n H -then this gives rise to a certificate for those values and all possible values of n G = k G .
Before we exploit the algorithm provided by Proposition 5.17 which finds a certificate of the form as given in Conjecture 5.16, let us mention that it consists of two main steps: The first step is to construct the system of equations (5.7) and the second is to find a solution to this system of equations.
Let us reconsider the proofs of Theorems 5.12 and 5.13. There, we already have a particular certificate at hand, and we prove that it is in fact a certificate by performing essentially the first main step of the algorithm. In fact the system of equations (5.6) corresponds to (5.7) for d = 2 as the variables (α, β, γ) (of (5.6)) equal (α 0 , α 1 , α 2 ) (of (5.7)). Even though the computations for proving the theorems above are tedious, they are straight forward.
So, let us come back to the algorithm of Proposition 5.17. For finding a certificate for general d ≥ 3 the situation is more difficult: The computations get very messy, so it seems infeasible to get the system of equations (5.7) in closed form depending on the parameter d. Moreover, even for the case d = 2 it is not obvious that the system of equations (5.6) even has a solution. Still, we want to use Proposition 5.17 for obtaining more certificates, so let us consider the cases d = 1 and d = 2 once more, but this time with the help of SageMath [8].
Using the algorithm provided in the proof of Proposition 5.17 allows to reprove Theorems 5.12 and 5.13 computationally with SageMath. It turns out that the variety of the system of equations (5.7), whose points are the solutions (α 0 , . . . , α d ) of (5.7), is of dimension 1 which means that the dependency on n H is the only dependency on a free parameter. For d = 1, the solution is essentially unique (except for the obvious replacement of s g by −s g ). For d = 2, in the solution presented in Theorem 5.13 we can additionally replace each occurrence of √ 2 in any of (α, β, γ) by − √ 2 and obtain another solution. In other words we can choose the signs of ± √ 1 and ± √ 2. This observation will be revisited in Remark 5.22.
In the same manner and with a lot of patience, we can let the algorithm run for d = 3 and get the result presented as Theorem 5.20 below. However, the computation can be speeded up in the following way. This will allow to also also cover the case d = 4.
Remark 5.19. Suppose that the coefficients α 0 , . . . , α d are polynomials in n H with degrees bounded by d. (This is the case for Theorems 5.12 and 5.13, so this assumption is reasonable.) Then, by fixing a particular value for n H , the time of the computation of the α 0 , . . . , α d is now dramatically reduced. Doing this for d + 1 different values n H allows to compute the coefficients α 0 , . . . , α d as interpolation polynomials in n H .
It should be noted that this interpolation trick is technically/computationally not as innocent as one might think: One has to carefully choose the values n H in order to "keep track" of the branch of one particular solution, as the solution of (5.7) is not unique.
By using the strategy explained in the previous remark, we are able to show the following.
Remark 5.22. Let us consider the set-up and certificate as presented in Conjecture 5.16 again. In particular, let us have a look at the coefficient α d for various d. By using the certificates obtained in this Section 5, we may rewrite this coefficient as We therefore ask the following question: Is it true that d = 5 : To summarize, in this section we have obtained certificates for the cases with k G = n G ≥ 1 and k H = n H − d for d ∈ {0, 1, 2, 3, 4} by our method. For d ∈ {0, 1, 2} we have proven these results by hand, for d ∈ {3, 4} we have proven them computationally. We will continue to prove the correctness of certain certificates in the next section.

Exact Certificates for
In this section we will finally prove Theorem 4.6 and therefore obtain a certificate for the case k G = n G − 1 ≥ 1, n H = 3 and k H = 2. As a byproduct we will also obtain a certificate for the case k G = n G − 1 ≥ 1, n H = 2 and k H = 1. Towards that end we will use some of the results of Section 5.1 and derive further results of a similar nature.

Auxiliary Results
We start with the following lemma. This lemma is equivalent to e gg ≡ 0 mod I G and e gg ≡ 0 mod I viz for all {g, g } ⊆ D G . Lemma 6.1 is plausible from a graph theoretical point of view. Indeed, due to Theorem 3.4 the points in the variety of I G are in bijection to the graphs in G, which are the graphs on n G vertices with domination number k G = n G − 1 and a minimum dominating set D G . Clearly in such graphs there are no edges between any two vertices of D G , because if there would be such an edge, the domination number would decrease. Hence, for each point in the variety of I viz , we have that the component e * gg is zero for every {g, g } ⊆ D G .
Proof of Lemma 6.1. For k G = n G − 1 = 1 there is no {g, g } ⊆ D G , so there is nothing to prove.
We use Notation 3.2. Let z * ∈ V(I G ), so z * is a common zero of (3.1a), (3.1b) and (3.1c). We assume that z * is not a zero of f = e gg .
Let us set S = S g = { g ∈ D G : g = g}. Then, due to (3.1c) we have g∈Sg e * gg g∈Sg e * gĝ = 0 (6.1) and conclude that one of the two factors has to be zero. Due to (3.1a), all e * gg and e * gĝ appearing in (6.1) are in {0, 1}, and moreover e * gg ∈ {0, 1}. Then e * gg = 1 (as it is assumed to be non-zero), and, because g ∈ S g , the first factor of (6.1) is non-zero. Therefore the second factor of (6.1) must be zero and hence e * gĝ = 0 for allg ∈ S g .
By symmetry (switching the roles of g and g ), we obtain e * gĝ = 0 for allg ∈ S g and therefore get e * gĝ = 0 for allg ∈ S g ∪ S g = D G . Thus, g∈D G (1 − e * gĝ ) = 1, but due to (3.1b) this product should be zero; a contradiction. Hence z * is also a zero of f = e gg , and Hilbert's Nullstellensatz (Theorem 2.4 and Remark 2.5) implies that f = e gg ∈ I G .
In particular we will need the following consequence of Lemma 6.1.
Note that also Corollary 6.2 can be explained from a graph theoretic point of view. To be precise there can be only one edge in a graph on n G vertices with domination number n G − 1, because an additional edge would decrease the domination number. Therefore, for each point in the variety of I G , the product of components corresponding to two different edge variables has always to be equal to 0 due to Theorem 3.4.
Proof of Corollary 6.2. If {g 1 , g 2 } ⊆ D G or {g 3 , g 4 } ⊆ D G the result follows from Lemma 6.1 because then e g 1 g 2 ∈ I G or e g 3 g 4 ∈ I G . Hence we only have to consider the case {g 1 , g 2 } ⊆ D G and {g 3 , g 4 } ⊆ D G . Let {ĝ} = V (G) \ D G , then without loss of generality this case is equivalent to g 1 = g 3 =ĝ and g 2 = g 4 .
We use Hilbert's Nullstellensatz like in Lemma 6.1 to prove the statement. Let z * ∈ V(I G ) be a common zero of (3.1a), (3.1b) and (3.1c). If we can prove that z * is also a zero of f = eĝ g 2 eĝ g 4 we are done.
so one of these two factors has to be zero; without loss of generality (due to symmetry in g 2 and g 4 ), let us assume the first factor. As e * gg 2 ∈ {0, 1} for allg ∈ V (G) by (3.1a), we then have in fact e * gg 2 = 0 for allg ∈ V (G). In particular we have e * gg 2 = 0 becauseĝ ∈ S. This is what we wanted to show.
We need Corollary 6.2 in order to prove the next result.
Note that it would again be possible to justify Lemma 6.3 in terms of graph theory using Theorem 3.8. It would need a case distinction for n H = 2 and n H = 3 and several more case distinctions on whether g, g ∈ D G , whether h 1 , h 2 , h 3 , h 4 ∈ D H and on the cardinality of {h 1 , h 2 , h 3 , h 4 }. We refrain from presenting the details here.
Proof of Lemma 6.3. First observe that without loss of generality we can assume that g ∈ D G , as |D G | = n G − 1 and therefore not both of g and g can be in V (G) \ D G . For notational convenience let {ĝ} = V (G) \ D G , and note that g might or might not be equal toĝ.
Next observe that without loss of generality h 4 = h 1 , because n H ∈ {2, 3}, and h 1 = h 2 and h 3 = h 4 by assumption. We obtain that the sets {h 1 , h 2 } and {h 3 , h 4 } are both of cardinality 2 and not disjoint.
In order to prove Lemma 6.3 we will use Hilbert's Nullstellensatz for analogously as it has been done in the proofs of Lemma 6.1 and Corollary 6.2. Note that we use Notation 3.7. Towards that end let z * ∈ V(I viz ), i.e., z * is a common zero of (3.1a), (3.1b) and (3.1c) for both G and H and of (3.2a) and (3.2b). Due to (3.1a) and (3.2a) all of e * gg , e * hh and x * gh are either 0 or 1. Assume that z * is not a zero of f , then x * gh 1 = x * gh 2 = x * g h 1 = x * g h 3 = 0. Furthermore, as g ∈ D G we have e * gg = 0 for allg ∈ D G by Lemma 6.1.
Case g =ĝ. We have g ∈ D G and can deduce from (3.2b) for g h 1 and g h 3 analogously as for g that Due to Corollary 6.2 and e * gĝ = 1 we have e * g ĝ = 0. Therefore (6.3a) and (6.3b) imply that e * h 1 h = e * h 3 h = 1, which is a contradiction to Corollary 6.2. So in this case z * is also a zero of f and hence f ∈ I viz holds because of Hilbert's Nullstellensatz.
Case g =ĝ. Due to Corollary 6.2 and e * gĝ = 1 we can deduce that e * g g = 0 for all g ∈ V (G) \ {g}. Therefore (3.2b) for the vertices g h 1 and g h 3 of the box graph class become We have x * gh 1 = x * gh 1 = 0, so from (6.2a) and (6.4a) it follows that x * gh = x * gh = 1 and e * h 1 h = e * h 1 h = 1. Corollary 6.2 applied on the graph class H implies h = h and therefore also h 2 = h 3 holds. Furthermore, this corollary also implies that e * h 2 h = 0, and hence x * gh 2 = 1 because of (6.2b). But this is a contradiction because x * gh 2 = x * g h 3 = 0. So also in this case z * is a zero of f and therefore f ∈ I viz holds.
Remark 6.4. In particular for k G = n G − 1, n H ∈ {2, 3} and k H = n H − 1, Lemma 6.3 implies Next we will need some more polynomials in order to be able to cope with σ g,i in a better way. Definition 6.5. Let i and j be two non-negative integers. We define Observe that τ i,j = τ j,i holds. As a next step we will use Lemma 6.3 in order to determine τ 2,2 . Lemma 6.6. Let k G = n G − 1 ≥ 1, n H ∈ {2, 3} and k H = n H − 1. Then Proof. By definition holds. By using Lemma 6.3 as stated in Remark 6.4 we obtain (6.5)

Certificates for
Now we are finally able to prove Theorem 4.6, which provides a sum-of-squares certificate of degree 2 for k G = n G − 1 ≥ 1, n H = 3 and k H = 2. In fact we will prove the existence of sum-of-squares certificates not only in this case, but also for the case k G = n G − 1 ≥ 1, n H = 2 and k H = 1 in the following theorem.
Theorem 6.7. For k G = n G − 1 ≥ 1, n H ∈ {2, 3} and k H = n H − 1 Vizing's conjecture is true as the polynomials x gh x gh and , are a sum-of-squares certificate with degree 2 of f viz .
The previous two identities together with (6.8) and the fact that σ g,4 = 0 trivially holds because n H ∈ {2, 3} yield In order to obtain a certificate s 2 0 + g∈V (G) s 2 g ≡ f viz = −k G k H + g∈V (G) σ g,1 mod I viz has to hold. For n H = 2 we have σ g,3 = 0, so in this case every solution to the system of equations yields a valid certificate. It is easy to check that α = n G − 1, β = −1, γ = 1, κ = 0 and λ = 1 is a solution.
For n H = 3 the above equations and also βγ + κλ + γ 2 + λ 2 = 0 has to be fulfilled in order to obtain a certificate. Also in this case it can be verified is a solution to the system of equations. Therefore the polynomials s 0 and s g for g ∈ V (G) form indeed a certificate.

Missing Certificates for
Previously we have seen that in many cases it is possible to obtain a certificate not only for particular values of n H and k H , but for general values, like it was done in Section 5. Therefore it is a natural question, whether we can generalize the certificate from Theorem 4.6 for the case k G = n G − 1 ≥ 1, n H = 3 and k H = 2 to a certificate for the case k G = n G − 1 ≥ 1 and k H = n H − 1 ≥ 1. We have successfully generalized the certificate for n H = 2 with Theorem 6.7. Unfortunately it turns out that this is not possible for n H ≥ 4. Example 6.8. There seems to be no 2-sum-of-squares certificate for the case n G = 4, k G = 3, n H = 4, k H = 3 that uses only the monomials of the form 1, x gh and x gh x gh for all g ∈ V (G) and all {h, h } ⊆ V (H), as the corresponding SDP is infeasible.
The SDP for the case n G = 4, k G = 3, n H = 4, k H = 3 which takes into account all monomials of degree at most 2 is feasible. Therefore we expect that there is an exact 2-sum-of-squares certificate using all monomials also for these parameter values.
When we take a closer look on the proofs of Section 6.1 and 6.2 we get some insight in why this is the case. First, Lemma 6.3 is not true anymore for n G ≥ 4, so we can not use the reduction of all products of 4 variables as presented in Remark 6.4. Furthermore σ g, 4 is not equal to 0 anymore for n G ≥ 4, so in the proof of Theorem 4.6 the coefficient of σ g,4 would have to be 0, which is not possible as the coefficient is 6γ 2 + 6λ 2 .
As a result we would have to search for a certificate with more monomials than just 1, x gh and x gh x gh for the case k G = n G − 1 and k H = n H − 1 for n H ≥ 4.

Conclusions
In this project, we modeled Vizing's conjecture as an ideal/polynomial pair such that the polynomial is non-negative on the variety of a particularly constructed ideal if and only if Vizing's conjecture is true. We were able to produce low-degree, sparse Positivstellensatz certificates of non-negativity for certain classes of graphs using an innovative collection of techniques ranging from semidefinite programming to clever guesswork to computer algebra.
In particular, Vizing's conjecture with parameters k G = n G − 1 ≥ 1, k H = n H − 1 and n H ∈ {2, 3} has a 2-sum-of-squares Positivstellensatz certificate. Furthermore Vizing's conjecture with parameters k G = n G and k H = n H − d has a d-sum-of-squares Positivstellensatz certificate for d ≤ 4. We have conjectured a broader combinatorial pattern based on these certificates, but proving validity is left to future work.
However, at this time, we have indeed proved Vizing's conjecture for several classes of graphs using sum-of-squares certificates. Although we have not advanced what is currently known about Vizing's conjecture, we have introduced a completely new technique (still to be thoroughly explored) to the literature of possible approaches.

Future Work
The most pressing matter that arises in this paper is the following. We have investigated the case k G = n G and k H = n H − d. In the future we want to prove Conjecture 5.16 or find other certificates for the cases d ≥ 5. In particular it would be interesting to know if there is an easy structure for the leading coefficient α d in such a certificate as mentioned in Remark 5. 22.
On a small scale, in order to obtain more insight on the structure of certificates a next step will be to investigate further specific parameter settings. In particular, finding a certificate for the case k G = 1 (and all other parameters arbitrary) is among our next candidates.
On a large scale, it is known that Vizing's conjecture holds if one of the graphs G or H has domination number at most three [6], therefore, to find new results we need to get certificates for k G ≥ 4 and k H ≥ 4. Furthermore, it suffices to consider graphs that contain no isolated vertices. For such graphs the number of vertices is at least twice the domination number [25]. Hence, parameters where we can obtain new results on Vizing's conjecture must satisfy n G ≥ 8, k G ≥ 4, n H ≥ 8, and k H ≥ 4. Therefore, in our future work we intend to continue pushing the computational aspect of this project. One way to do so is to exploit symmetries in order to simplify the computation of a Gröbner basis, as computing the Göbner basis is one of the computational bottlenecks. One alternative possibility to deal with this bottleneck is to avoid the computation of a Gröbner basis by increasing the number of variables in the SDP. Another question of interest is if one could use symmetry to reduce the complexity of the SDP.
Up to now we always used the solution of the SDP in order to obtain insight in the structure of the certificate and then algebraic manipulations yielded the actual certificate.
It would be interesting to solve the SDP exactly over the algebraic reals and not only to a high precision over the rationals, in order to obtain an exact certificate as soon as the SDP is solved. However, this is a highly non-trivial task and is left for future research.
Another line of research is to change the model from a Positivstellensatz certificate to a Hilbert's Nullstellensatz certificate, and thus change from numeric semidefinite programming to exact arithmetic linear algebra. This approach must also be thoroughly investigated.
Finally, it would be very interesting to conjecture a global relationship between the values of n G , n H , k G and k H , and the degree of the Positivstellensatz certificate, and perhaps even recast the conjecture in terms of the theta body hierarchy described in [13].

A. Appendix: More Complicated "Intermediate" Certificates
In Sections 5.2, 5.3 and 5.4 we presented simple sum-of-squares certificates for the case k G = n G and k H = n H −d with d ∈ {0, 1, 2}. In fact, these easy certificates where obtained only after some computational experiments, in which more complicated certificates were found. We present these intermediate results and certificates here in this appendix.
For obtaining such a certificate, we use the machinery presented in Section 4.3 to get a numerical certificate. From this, we can guess a structure of the occurring coefficients, like it was done in Example 4.2. We will see that these more complicated certificates-they were found by an SDP solver-have a geometric aspect. By studying this aspect it was possible to simplify the more complicated certificates to the certificates presented in Sections 5.3 and 5.4. Hence retrospectively, these more complicated certificates are formally not needed for the proofs of existence of sum-of-squares certificates. Nevertheless we include them here to give a more accurate and complete picture of the process of how to obtain certificates.
A.1. k G = n G and k H = n H − 1 In this case the certificates found by observing a structure and guessing the coefficients of the numerical certificate have the following form. We can prove this theorem directly, but go a different way here: This result is one intermediate step and useful and necessary for conjecturing Theorem 5.12. But once Theorem 5.12 is proved, Theorem A.1 is not a dependency anymore. Therefore, we can reuse the statement made in Theorem 5.12 in the proof here without falling into a cyclic argumentation.