1 Introduction
Covering problems are some of the most studied problems in Graph Theory and Combinatorial Optimization due to the large number of applications. Consider a hypergraph
without isolated vertices. An edge cover of is a set of hyperedges that cover all vertices of , i.e. . The general covering problem consists in finding the covering number of , denoted by , which is the minimum number of hyperedges in an edge cover of BERGE .The most natural constructive heuristic for obtaining an edge cover of
is as follows. Start from empty sets and (the latter one keeps the already covered vertices). At each step , pick a hyperedge , add to and add all the elements of to . The process is repeated until . In addition, can only be chosen if at least one of its elements has not been previously included in , i.e. .How bad can a solution given by this heuristic be (compared to the optimal one) ? The answer leads to the concept of Grundy covering number of , denoted by , which computes the largest number of steps performed by such a constructive heuristic, or equivalently the largest number of hyperedges used in the resulting covering BRESAR2014 . In general, “Grundy numbers” relate the worst case of greedy heuristics. They were initially addressed in the context of coloring problems GRUNDYCOL .
A particular case of the covering problem, motivated by a domination game P1 ; P2 ; P3 , occurs when is the hypergraph of the closed neighborhoods of vertices in a simple graph , i.e. where and . Here, the Grundy covering number of is called Grundy domination number of BRESAR2014 . Analogously, the Grundy total domination number of is the Grundy covering number of the hypergraph of the open neighborhoods of vertices in (where the hyperedge related to is ) BRESAR2016 .
The problems associated to these parameters are hard even when belongs to a restricted family of graphs. For example, the Grundy domination problem is hard when is a chordal graph, although it is polynomial when is a tree, a cograph or a split graph BRESAR2014 . The total version of this problem is hard when is bipartite BRESAR2016 but it is polynomial on trees, tidy and distancehereditary bipartites NASINI2017 .
In this work, we introduce the General Grundy Domination Problem (GGDP), where the hyperedges can be either closed or open neighborhoods. We obtain the parameter associated with GGDP for paths and web graphs. We propose integer programming formulations for GGDP, which in particular can obtain the Grundy (total) domination number of a graph. We study the polytope associated to one of these formulations. We find families of valid inequalities and derive conditions under which they are facetdefining. Besides, we perform computational experiments to compare the formulations as well as to test some of the proposed valid inequalities as cuts in a B&C framework. Some results contained in this work appeared without proof in the extended abstract LAGOS2017 .
1.1 Definitions and notation
Let be a simple graph on vertices and be a subset of vertices of . Define the function as follows: if and if . Assume that no vertex from is isolated in , so that for all . will be simply called the neighborhood of .
Now, define as the hypergraph with . The Grundy covering number of will be called Grundy domination number of , to be denoted . The combinatorial problem of our interest is presented below:
GENERAL GRUNDY DOMINATION PROBLEM (GGDP)
INSTANCE: a graph and a set such that no isolated vertex is in .
OBJECTIVE: obtain .
Since the Grundy domination problems mentioned in the introduction are particular cases of our problem (note that is the Grundy domination number of while is the Grundy total domination number of ), they can be simultaneously addressed by GGDP.
We refer to the pair “” as an instance and we present the same definitions given in BRESAR2014 ; BRESAR2016 in terms of instances (of the GGDP).
Let be a sequence of distinct vertices of . The sequence is called a legal sequence of an instance if
holds for every . If is a legal sequence, then we say that footprints the vertices from , and that is the footprinter of every vertex . That is, footprints a vertex if does not belong to the neighborhood of , for .
If then is called a dominating sequence of . The legal dominating sequences of are in onetoone correspondence with the solutions of the constructive heuristic for the covering number of .
Let be any legal sequence of with maximal length, i.e. it is not possible to append a vertex at the end of . It is easy to see that is also a dominating sequence and, in particular, if its length is maximum then it is equal to . Therefore, when determining , we do not need to restrict ourselves to legal dominating sequences. Instead, we can only explore the structure of legal sequences, which is simpler.
For given vertices , we denote the expression , which basically means that the sequence is legal. Naturally, means . An instance is called clutter if is a clutter too, i.e. if for all distinct vertices . It is called a strong clutter if , for all distinct vertices .
We say that distinct vertices are twins if . If does not have twin vertices, then is called twin free. Clutters are twin free instances.
2 Properties of the GGDP and examples
In order to reduce the size of the input graph, two simple rules can be applied. In first place, if is the disjoint union of graphs and , then:
Therefore, we can restrict ourselves to connected graphs.
In addition, if there exist twin vertices , then
where is the graph obtained from by deleting . Hence, we can suppose that the instance is twin free.
Let be the minimum cardinality of for all . In BRESAR2014 ; BRESAR2016 , upper bounds of Grundy domination numbers are presented. They can easily be adapted to our problem:
Proposition 2.1.
BRESAR2014 ; BRESAR2016 .
Proof.
Let be a legal dominating sequence of of maximum size and let be a vertex footprinted in the last step. Since was not dominated in the previous steps, . Then, , implying . ∎
Now, we consider two families of graphs (paths and web graphs) that will serve as examples. First, let denote an induced path on vertices where . Note that, for any integer and any , when and otherwise. Therefore,
Observation 2.2.
Let and . If then ; otherwise, .
We say that is a good configuration (or “g.conf. ” for short) for if
(i) and ,
(ii) and ,
(iii) and either
(iii.1) and is a g.conf. for the subpath induced by or
(iii.2) and is a g.conf. for the subpath induced by .
For the sake of simplicity, when we say that is a g.conf. for a subpath of , we are actually referring to the set . Moreover, differently from the definition of GGDP, we allow the subgraph induced by to have isolated vertices in the g.conf. definition.
From the previous observation, we trivially have
Observation 2.3.
Let and . If is a g.conf. for , then .
The following results give the Grundy domination number of a path and show where the upper bound is tight.
Proposition 2.4.
Let , and . If or is a good configuration for then ; otherwise, .
Proof.
If , we must have , because . So is a g.conf. and . Now, consider the case . If then . And, if ( is a g.conf. for ) then .
For , note that is a legal dominating sequence. Indeed, footprints (and itself, if ), footprints (and , if ), and footprints . Thus, . If , by Observation 2.2 we have . It remains to consider the case . By Observation 2.2, . Then, , and it is enough to prove that is a g.conf. for if and only if there exists a legal dominating sequence of size .
First, assume that is a g.conf. for . We use induction on . Recall that we have already obtained for , and is trivial for . For , (iii.1) or (iii.2) holds. Assume w.l.o.g. that (iii.1) holds (the other case is symmetric). The induction hypothesis ensures the existence of a legal dominating sequence for , with . Consider the extended sequence . is legal and dominating for with .
Now, assume that there exists a legal dominating sequence such that . We use again induction on . is trivially a g.conf. for when . For , any dominating legal sequence of length must start with an endpoint of , say , and such a vertex must belong to . Moreover, must be the last vertex in the sequence, and the vertices between and define a legal dominating sequence for . By the induction hypothesis, is a g.conf. for . Since , is also a g.conf. for . ∎
Corollary 2.5.
Let , and . If is a good configuration for , then ; otherwise, .
The second nontrivial example is related to web graphs ANNEGRET . Let be positive integers such that . A web is a graph with and . Note that , where and stand for the addition and subtraction modulo . Therefore, if , and if .
Proposition 2.6.
Let be the web graph and ( possibly empty). is a clutter if and only if or . Moreover, in the following cases: (i) , or (ii) there is such that induces a path , , and is a good configuration for . Otherwise .
Proof.
Suppose that . Then, for all . If , let . Then, and is not a clutter. If , for all . In addition, for all . Therefore, is a clutter.
Now, suppose that . Note that, for all such that , there is such that . Therefore, and is a clutter.
In order to determine , consider the sequence of length . It is a legal dominating sequence since vertex footprints and (and itself if ), vertex footprints (and vertex if ) and, if , vertex footprints for all . The last footprinted vertex is . Therefore, we obtain . If , we are done since .
Assume now that . Let , and . Note that . It suffices to show that if and only if induces a path and is a good configuration for it. Recall that . The unique way of getting a dominating legal sequence of length is by starting with and then choosing a vertex that will footprint only one more vertex at each step. This means that, at any step but the last one, we cannot choose a vertex from because it would footprint and at least one vertex from . So, after , we must choose all the vertices in , and finally a vertex from . In addition, must induce a path, otherwise some of its vertices would footprint at least 2 vertices. Consider the sequence where and is a maximum legal sequence of . In virtue of Corollary 2.5, if is a g.conf. for and otherwise. Therefore, if and only if induces a path and is a g.conf. for it. ∎
For example, consider the web graphs depicted in Figure 1. Filled Circles denote the vertices of , i.e. . Upper bounds for are (if ) and (if ). In (a), neither is a g.conf. for nor for , so a legal sequence of maximum length is . In (b), since is a g.conf. for , a legal sequence of maximum length is . Note also that this latter instance is not a clutter since .
3 Integer programming formulations
Legal sequences can be modeled as binary vectors as follows. For every
and , let be a binary value such that if is chosen in step . Also, for every and , let be a binary value such that if is available to be footprinted in step (i.e. not footprinted by any of the chosen vertices in previous steps). The following formulation computes the parameter :subject to  
(1)  
(2)  
(3)  
(4)  
(5)  
Constraints (1) ensure that at most one vertex is chosen in each step. Constraints (2) guarantee that each vertex can be chosen no more than once. Constraints (3) specify that can be chosen in the next step only if there is at least one nonfootprinted vertex of in the current one (that will become footprinted). Constraints (4) force to footprint if some vertex of is chosen. Finally, constraints (5) say that footprinted vertices will remain footprinted for the next steps.
Although this formulation works fine, there exist several integer solutions for each legal sequence. Below, we present families of constraints that remove unnecessary solutions.

A variable can be set to zero even if it is not footprinted by any vertex. In order to forbid this situation, we consider these constraints:
(6) (7) 
It is allowed to choose no vertex in a step, leading to several symmetric solutions. They can be avoided by replacing (1) by
(8) (9) where is a lower bound of . These constraints force to assign vertices to the first steps when the solution represents a legal sequence of length . In addition, sequences of length smaller than are removed.

We can impose that every integer solution represents a dominating sequence:
(10)
Here is a table of possible formulations. All of them include constraints (2)(5).
Form.  (1)  (6)(7)  (8)(9)  (10)  Solutions 

16253  
205  
463  
43  
1668  
124  
68  
28 
There is a onetoone correspondence between legal sequences of and solutions of , and a onetoone correspondence between dominating legal sequences and solutions of .
The last column shows the number of integer solutions for the bull graph, i.e. and , with , and .
Although one would expect that the fewer the number of solutions is, the smaller the size of the B&B tree will be, it is known that the addition of symmetrybreaking inequalities not always help to improve the optimization MARGOT . Therefore, we carried out computational experiments in order to determine which formulation performs better. They are reported in Section 5.
4 The polytope of legal sequences
From now on, we suppose that has at least 3 vertices. We recall that is connected and is twin free.
The following easy result will be useful throughout the section:
Lemma 4.7.
Let .
(i) There exists such that .
(ii) If is a clutter, .
Proof.
In first place, note that and , for every vertex , since is a connected graph with at least 3 vertices.
(i) Suppose that for all . Since does not have twin vertices, .
Therefore, for some , and so is a , which is absurd.
(ii) Suppose that such that , and so is an edge.
Since is a clutter, for all .
Then, is a connected component of , which is also absurd.
∎
Let be the convex hull of the set of binary feasible solutions in formulation . In this section, we study the facial structure of . We choose it for two reasons. On the one hand, every valid inequality of this polytope is still valid for with . On the other hand, the dimension of just depends on the size of the instance, which is an interesting feature for polyhedral studies. The same does not happen with the other polytopes.
For instance, if we consider (recall that in this polytope and hold only if someone is footprinting in step ), then the equality is valid for all , due to (4) and (6). In addition, the equality is valid for all and all pairs such that . Indeed, in this case, by (7) and (4), respectively. If , the equality is clear. If , then by (5) and because is the only possible footprinter of and must be chosen at step or before.
To analyse (in this polytope, sequences of size always use the first steps), we introduce a new parameter. Let be the index of the largest step where can occur in some legal sequence of . Clearly, for all , is at most and, in particular, coincides with it for some . Thus, obtaining these parameters is as hard as determining the Grundy domination number itself. In , the equality is valid for all and all . In addition, if and , then for all and all . Indeed, no matter are the first chosen vertices, some of them will footprint any vertex in . In Prop. 4.9, we give the dimension of for the case .
Regarding (in this polytope, legal sequences are also dominating), the inequality is valid, where is the classical domination number of . If such an inequality becomes an equality. In BRESAR2014 , the authors partially characterize the family of graphs where . Full characterization of this family is still an open question for , as well as for BRESAR2016 .
Polytopes associated with the remaining formulations are even harder to study.
4.1 Dimension
For the sake of readability, write and assume . Also, define as the solution where and for all and . Note that it always belongs to .
Below, we define some operations in order to manipulate solutions. Let be an integer solution of .

For a given set and , the operation footprint, denoted by , means that vertices from are footprinted in step . That is:

for all and

for all and

for all and

for all and
In the case that we need to footprint just one vertex , we write .


For a given vertex and , the operation choose, denoted by , means that is chosen in step , and simply set to the solution .
We remark that, each time an operation is applied, we must check if is an integer solution of , since it is generally not guaranteed. The same operations can be used to propose points of .
Proposition 4.8.
The polytope is full dimensional.
Proof.
Suppose that the following equation is valid for all points of :
We prove that all of its coefficients are zero by giving pairs of points and using the fact that since both terms are equal to .

Let and consider the points and . Since the difference between both points lies just in , we get .

Let and and consider and . We get .

At this point for any since . Let and . Consider and . We get .
Therefore, is full dimensional. ∎
Proposition 4.9.
Let and . The dimension of is and a minimal system of equalities is:
Proof.
We have seen that the equalities from are all valid for and it is straightforward to prove that they are mutually independent each other. Thus, the stated expression gives an upper bound for . Suppose that the following equality holds for all points of :
To complete the proof, it suffices to show that all the coefficients are zero, except for , , which have to be all equal to each other. We use an approach similar to the one applied in Prop. 4.8.
Note that, for a given , there exists a legal sequence such that is chosen in step . Moreover, for a given , there exists a legal sequence such that is chosen in step and is the last element. can be obtained from by dropping vertices.

Let and . There exists , and so does not footprint . Consider and . We obtain . By assigning in the order , we get for all .

At this point, . Let . We prove that for all by taking the points and .

Let and . Consider as the point describing and as the point representing without (its last element). Note that still satisfies (8) because . Using , we get .
∎
4.2 Facetdefining inequalities
We start by mentioning some basic properties about the general facial structure of .
Proposition 4.10.
Let be a valid inequality for . Then:
(i) ;
(ii) for all and .
Moreover, if is facetdefining for , then:
(iii) , if the inequality is different from , for all and .
Proof.
(i) Since , .
(ii) Consider the point such that:
for all and ;
for all (such that ) and ;
for all and .
Clearly . Then, .
(iii) Let be the facet defined by .
Since the inequality is different from , there is a point such that .
Let be the vector obtained from by setting .
Clearly . Then, .
Also, , which implies .
∎
We use the following technique for proving sufficient conditions for a given inequality to be facetdefining of . Let be the face defined by the inequality. Hence, for all . If is another equation valid for all points of , provide a real number and prove that this latter equality can always be obtained from the product between and the former one.
Necessary conditions for to be facetdefining are proved as follows: negate the condition, then give a stronger inequality that dominates it or provide an equality (not equivalent to ) that is satisfied by all the points of .
Proposition 4.11.
The following inequalities define a facet of :
(i) for all and
(ii) for all if and only if
(iii) Constraints (5) for each and if and only if for all ,
(i.e. every leaf of adjacent to must belong to ).
Proof.
(i) Here, . Let .
Then, we have to prove that , and for all .
Items 1 and 2 given in the proof of dimension (Prop. 4.8) show again that . At this point
for any .
Let and such that and consider and
. We get .
(ii) Suppose that . Here, . Let . Then, we have to prove these cases: 1) for all , and 2)