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 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 value of an optimal solution)? The answer leads to the concept of the Grundy covering number of 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 .
Let be a simple graph. For any , let be the open neighborhood of , i.e. (the symbol “” will be used recurrently to make definitions) and be the closed neighborhood of , i.e. .
A particular case of the Grundy covering problem occurs when is the hypergraph of the closed neighborhoods of vertices in a graph : where . 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 BRESAR2016 .
In order to illustrate these concepts, consider the graph of Figure 1 a) and the hypergraph of the closed neighborhoods of . A possible execution of the heuristic would be to pick , then and finally . In Figure 1 b), it is displayed from left to right what happens at each step. A vertex inside a box means that is the chosen set at that step. Filled circles denote those vertices that are already covered (set of the heuristic). Note that, at any step, at least one new vertex is covered: in step 2 vertex 5 is covered while in step 3 vertex 1 is covered. The Grundy domination number of this graph is 3 since it is not possible for the heuristic to perform 4 steps.
The study of these domination parameters was originally motivated by a domination game P1 ; P2 ; P3 , and their associated problems can model some applications where there are two players with opposing interests. For instance, consider a city divided into districts where the municipal government intends to offer a concession per year of a given service (e.g. Internet connection) to companies. Each year, a company is located in a district, which is committed to providing its service to both the district and its neighbors. In return, it requires having at least, within its domain (the district where it is installed and its neighbors), a captive district: only that company offers the service to it, for (at least) a year. The goal of the government is to plan which district should be chosen (for a company to be settled) each year so that, after a while, the city is entirely covered and to maximize the number of companies providing the service (and, thus, to foster longterm competitiveness). The given problem can be modeled as a Grundy domination problem where each vertex represents a district and two vertices are adjacent if and only if they represent neighboring districts. Indeed, the Grundy domination number gives the maximum number of companies. For instance, if the city is modeled as the graph of Figure 1 a), an optimal schedule is to settle a company on district 3 in year 1, another on district 4 in year 2 and the last one on district 5 in year 3. During the 2nd. year, district 5 only receives the service from the company on district 4. The same happens for district 1 and the company on district 5 during the last year. In Section 6, the city of Buenos Aires is considered.
The problems associated to these parameters are hard for general graphs. In BRESAR2014 , it is proven that the Grundy domination problem is hard for chordal graphs (it is also proven that this problem is polynomial for trees, cographs or split graphs). Regarding the total version of this problem, it is hard when is bipartite BRESAR2016 (but it is polynomial on trees, tidy and distancehereditary bipartites NASINI2017 ).
It is known that one of the most powerful tools to solve hard problems are branchandcut algorithms, which are based on Integer Programming. In this work, we introduce a problem that generalizes the Grundy domination and the Grundy total domination, and we propose integer programming formulations for this new problem. In order to obtain good upper and lower bounds, we also design a heuristic algorithm that combines a greedy strategy with a tabu search. Our approach is exact for instances ranging from 20 to 50 vertices (depending on the edge density of the graph) and, in particular, by taking only the heuristic algorithm, one can achieve good solutions on large instances.
In Section 2, we introduce a general version of the problem and show some useful properties. In particular, we calculate the exact value of the corresponding parameters for two families of graphs (paths and web graphs). In Sections 3 and 4, we introduce an integer programming model together with several families of valid inequalities that strengthen its linear relaxation. Besides, we present constraints that remove unnecessary integer points from the solution space and whose addition results in several formulations. In Section 5, we propose a heuristic for obtaining an initial upper bound and an initial feasible solution of our problem, and a tabu search for improving that initial solution. In Section 6, we perform extensive computational experiments to compare the formulations as well as to test two families of valid inequalities as cuts in a branchandcut framework. We also evaluate the performance of the tabu search. The test instances include randomly generated graphs, structured graphs, classical benchmark instances and two real instances from the aforementioned application in the city of Buenos Aires. Another experiment allows us to formulate a conjecture about the Grundy domination number on Kneser graphs. Finally, in Section 7 some conclusions are drawn.
Some results contained in this work appeared without proof in the extended abstract LAGOS2017 .
1.1 Definitions and notation
Let be a simple graph and be a subset of vertices of . Define the function , called neighborhood of , as follows:
Here, denotes the powerset of . Assume that no vertex from is isolated in so that, for all , and for some .
Some definitions given in BRESAR2014 ; BRESAR2016 (which only depends on ) are rewritten below in terms of the pair “”. These definitions abstract the behavior of the heuristic mentioned at the beginning of this work.
Let be a sequence of distinct vertices of . The sequence is called a legal sequence of if
holds for every . By convention, which is trivially non empty.
If is a legal sequence of , then we say that footprints the vertices from . That is, footprints a vertex if does not belong to the neighborhood of , for .
If then is called a dominating sequence.
For a given sequence , let . We say that a legal sequence of is maximal if there is no such that , i.e. the sequence with appended at the end, is legal. We say that a legal sequence of is maximum if there is no legal sequence of such that is longer than . A maximum legal sequence is also maximal.
Consider again the graph of Figure 1 and let . Part b) actually shows that the sequence is legal and dominating: (2, 3 and 4 are footprinted by 3), (5 is footprinted by 4), (1 is footprinted by 5) and . The sequence is also maximum. On the other hand, the sequence is a maximal legal sequence that is not maximum.
2 The General Grundy Domination Problem
Let be a simple graph on vertices and such that no vertex from is isolated in . Let denote the hypergraph with .
We define the general Grundy domination number of , denoted by , as the Grundy covering number of .
It gives rise to the following problem:
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 ( is indeed the Grundy
domination number of while is the Grundy total domination number of ), they can be addressed by a tool that
just solves the GGDP.
The following result will be useful to get the general Grundy domination number. It shows that every optimal solution is a maximum legal sequence and the “dominating” property comes for free.
Proposition 2.1.
Let be an instance of the GGDP and a maximal legal sequence of . Then, is dominating. Moreover, if is maximum, then is the length of .
Proof.
If were not dominating, there would exists a vertex not footprinted by any element of . As is an instance of the GGDP, there exists a vertex such that , and so . Thus, the sequence is legal since footprints , which contradicts the maximality of . Therefore, is dominating.
The second part of the statement is derived by the fact that legal dominating sequences are (by definition) in onetoone correspondence with the solutions provided by the constructive heuristic for . ∎
2.1 Properties on the GGDP
As we will see later, knowing a good upper bound will be fundamental for the performance of the exact algorithm that will be proposed (the worse the bound, the larger the integer linear program that must be solved). The following result gives a bound on the length of a sequence where the first elements are known.
Proposition 2.2.
Let be positive integers such that , and be legal sequences. Then,
Proof.
Let . Since is legal, vertices must footprint at least one different vertex from , implying that . Hence, . ∎
Corollary 2.3.
Let be a positive integer such that and be the minimum cardinality of for all legal sequences . Then, .
Let . For the case , finding is linear and provides an upper bound which can be tight sometimes. For instance, let be the graph of Figure 1 and . Here, and, by the corollary, . As the sequence is legal, . The bound was previously derived in BRESAR2014 for the Grundy domination number (i.e. ) and in BRESAR2016 for the total case (i.e. ), although their proofs are different from ours, which is valid for any .
Better bounds can be computed by increasing . For instance, let be the graph of Figure 2 a) and . Here, , , and, by the corollary, we get the upper bounds 7, 6 and 5 respectively. None of them is tight since (the sequence (1,2,4,6) is optimal). A similar example is given in Figure 2 b) for the case . Here, , and , thus obtaining the upper bounds 7, 6 and 5 respectively, but (the sequence (5,2,6,1) is optimal).
Another way to shorten the time of optimization is by reducing the size of the input graph.
We say that distinct vertices are twins if . If does not have twin vertices, then is called twin free. If there exist twin vertices , then
where is the graph obtained from by deleting . This rule can be applied recursively until the instance is twin free.
Now, if is the disjoint union of graphs and , then
This rule allows us to solve each component separately as the overall time is lower than solving directly.
2.2 The GGDP on paths and web graphs
When a new hard optimization problem on graphs is introduced, a natural question is what happens on simple structures such as paths and circuits. Below, we give for two families of graphs (paths and web graphs) with any . Recall that GGDP is already hard on bipartite or chordal graphs.
Let denote an induced path on vertices where . Note that, for any integer and any , when and otherwise. Therefore,
Observation 2.4.
Let and . If , then ; otherwise, .
We say that is a good configuration for if
(i) and ,
(ii) and ,
(iii) and either
(iii.1) and is a good configuration for the subpath
induced by or
(iii.2) and is a good configuration for the subpath
induced by .
For the sake of simplicity, when we say that is a good configuration for a subpath of , we are actually referring to the set . Moreover, in contrast with the definition of GGDP, we allow the subgraph induced by to have isolated vertices in the good configuration definition.
From the previous observation, we trivially have
Observation 2.5.
Let and . If is a good configuration for , then .
The following results give the general Grundy domination number of a path and, at the same time, show where the upper bound is tight.
Proposition 2.6.
Let , and . If or is a good configuration for , then ; otherwise, .
Proof.
If , we must have , because . So is a good configuration and . Now, consider the case . If then . And, if ( is a good configuration for ) then .
For , note that is a legal dominating sequence. Indeed, footprints (and itself, if ), footprints (and , if ), and footprints . Thus, . If , by Observation 2.4 we have . It remains to consider the case . By Observation 2.4, . Then, , and it is enough to prove that is a good configuration for if and only if there exists a legal dominating sequence of size .
First, assume that is a good configuration for . We use induction on . Recall that we have already obtained for , and is trivial for . For , (iii.1) or (iii.2) holds. Assume without loss of generality 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 good configuration 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 good configuration for . Since , is also a good configuration for . ∎
Corollary 2.7.
Let , and . If is a good configuration for , then . Otherwise, .
The second (less trivial) family are web graphs ANNEGRET . Let be positive integers such that . A web is a graph with and . See examples in Figure 3. Note that , where and stand for the addition and subtraction modulo . Therefore, if , and if .
Proposition 2.8.
Let be the web graph and ( possibly empty). Then, in the following cases: (i) , or (ii) there is such that induces a path , , and is a good configuration for . Otherwise, .
Proof.
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 2.7, if is a good configuration for and otherwise. Therefore, if and only if induces a path and is a good configuration for it. ∎
For example, consider the web graphs depicted in Figure 3. Filled circles denote the vertices of , i.e. . Upper bounds for are (if ) and (if ). In a), is neither a good configuration for nor for , so a legal sequence of maximum length is . In b), since is a good configuration for , a legal sequence of maximum length is .
3 Integer programming formulation
Let be an instance of the GGDP, and and be a lower bound and upper bound on . We present an ILP formulation for the problem by modeling the iterative process performed by the constructive heuristic described in the introduction. For the sake of readability, through this section, we also present examples based on the graph of Figure 2 a) with and .
Legal sequences will be modeled as binary vectors: a given sequence is represented by an array of
consecutive slots where each slot can be empty or store a vertex. An empty slot is skipped when forming the sequence from its corresponding array. For instance, the array , where “” denotes the empty slot, represents the sequence . Observe that a sequence can be represented by different arrays. An empty slot can be interpreted as an innocuous step (without performing any action) in the process, just to attain steps in total.Now, for a given array , let be a binary value such that if and only if is in the th. slot, i.e. . In order these variables to reflect an array composed of distinct vertices, the following constraints must hold:
(1)  
(2) 
Constraints (1) ensure that at most one vertex is chosen in each slot.
Constraints (2) guarantee that each vertex can be chosen no more than once.
In order to model the legality of sequences, we consider another set of binary variables that keeps a record of footprinted vertices. First, let
, if is empty, and otherwise, for . In other words, set contains those vertices available to be footprinted in step and subsequent ones. In case the slot is empty, no vertex is footprinted at step . Now, observe that the sequence represented by an array is legal if, for every nonempty , .Let be a binary value such that if and only if . In particular, for all meaning that can be treated as a constant.
The next constraints model the chain of inclusions :
(3) 
(the inclusion naturally holds because ).
Now, we model when is empty and when . This is equivalent to ask that , if is empty or . It can be ensured by the following constraints:
(4) 
Indeed, suppose that is empty or . Then, the l.h.s. of (4) is zero, implying , as desired. On the other hand, if , then (4) becomes redundant.
In order to model when is not empty, every such that must not belong to . Constraints (5), presented below, remove those elements from :
(5) 
Therefore, constraints (3)(5) guarantee the correct valuation of the variables.
Finally, in order the sequence (represented by the array) to be legal, for any such that is not empty, there must be at least one available vertex of at step , which becomes footprinted at :
(6) 
Indeed, if (i.e. ), then the r.h.s. of (6) is positive implying that some must satisfy ,
that is and .
The previous families of constraints are enough to model legal sequences but not all of them are necessary. In fact, we can dispense with constraints (4) (as proved below). Omitting them allows marking a vertex as nonavailable to be footprinted at step even if it was not indeed footprinted, that is and but for all . This does not affect the sequence legality. It leads to the following formulation, which we call , that obtains the array that maximizes the amount of nonempty slots:
subject to  
(1)  
(2)  
(3)  
(5)  
(6)  
Theorem 3.9.
Formulation gives the value of .
Proof.
Let be the array defined by an optimal solution of . Precisely, for each ,
Note that is uniquely defined by (1). Consider the sequence represented by that array. We have to show that is legal and has maximum length. Since the nonempty slots have distinct vertices by (2), so do . Now, consider for some . There is some such that (i.e. ) in the array. By constraints (3) and (6), there exists such that . Besides, by constraints (3) and (5), we have and , for all . It follows that for all . Therefore,
which implies that is legal.
Now, suppose that is a legal sequence longer than , e.g. . As , fits in an array. Consider and the sets , for all and for all (if ). Clearly, the array and the sets correspond to a feasible integer point with objective value of , which leads to a contradiction. ∎
4 Strengthening the formulation
In this section, we explore several ways of strengthening the linear relaxation of by the addition of inequalities.
4.1 Symmetrybreaking inequalities
In an ILP model, two integer solutions are called symmetric if they share the same objective value and one can be obtained from the other by swapping its components MARGOT . In a broader sense, symmetric solutions have a common characteristic (for instance, both represent the same solution of the problem) so that one can be eliminated without losing the correctness of the model. Sometimes, ILP models with several symmetric solutions may have poor performance and one way to reduce the number of symmetric solutions is to incorporate symmetricbreaking constraints to the formulation. Certainly, constraints (4) defined above remove some symmetric solutions. Below, we present two families of constraints that also do this job.

For a given legal sequence , consider an array such that the first elements are those from the sequence and the remaining slots are empty, i.e. . We call it canonical. For instance, the sequence (1,2,4,6) is represented by the canonical array . Those solutions associated to noncanonical arrays can be removed by
(7) which means that, if for some , is not empty, then is not empty too.

As optimal sequences are dominating, we can impose that every integer solution represents a dominating sequence:
(8)
Different combinations of the symmetrybreaking constraints presented above give rise to 8 formulations listed in the following table. For instance, is the same as plus constraints (4):
Form.  (1)  (2)  (3)  (4)  (5)  (6)  (7)  (8) 

(only )  
(only )  
(only )  
(only ) 
In some rows, constraints (1) are omitted (except for the case ) since they are dominated by (7).
All of the enumerated formulations are correct (in the sense that they deliver ) since is correct and the other ones just delete unnecessary solutions. In particular, two of them deserve a special attention:
Observation 4.10.
There is a onetoone correspondence between legal sequences of and integer solutions of .
Observation 4.11.
There is a onetoone correspondence between dominating legal sequences of and integer solutions of .
Although one would expect that the fewer the number of solutions is, the smaller the size of the branchandbound tree will be, the addition of symmetrybreaking inequalities not always help to improve the optimization. Therefore, we carried out computational experiments in order to determine which formulation performs better. Results are reported in Section 6.
4.2 Removing nonoptimal solutions
If a lower bound is known, solutions corresponding to sequences of size less than are not optimal and, thus, are not needed. They can be removed by forcing to be at least . Even better, we can impose that the first slots must be used:
(9) 
These constraints can be incorporated to any of the 8 formulations. In that case, some constraints may become dominated and can be omitted: (1) for and (7) for .
4.3 Valid inequalities
It is known that a cuttingplane algorithm is one of the most efficient tools to deal with an integer linear programming problem WOLSEY . The main idea is to consider the linear relaxation and try to strengthen it by adding violated valid inequalities. The algorithm can use general cuts that do not take advantage of the structure (such as Gomory cuts), or specially developed ones that exploit the properties of the problem. In this section, we present some inequalities that are valid for the formulations given above, which will be the basis for a cuttingplane algorithm.
Let be the convex hull of the set of integer feasible solutions in .
Given a nonempty set of vertices and a positive integer , let denote the subset of vertices with exactly neighbors in , i.e.
Observe that can be partitioned in sets for . For example, consider the graph of Figure 4 where . Filled circles represent vertices of , i.e. . Then, , where , (vertices inside a rhombus) and (vertices inside a square).
Now, we present a general family of valid inequalities. Consider:

an integer ,

an integer ,

a nonempty set of vertices,

a set (possibly empty) such that

a nonempty set of vertices, , such that

,

for all , and

for all .


integers such that , and for all .
We define the inequality as:
(10) 
Before formally proving its validity, we present an example that can make it easier to understand. Consider again the instance of Figure 4 with and . Let , , (thus, ), , (thus, ), , and . As and are disjoint subsets of and , the inequality, i.e.
is well defined. Below, there is a representation of variables “” from the inequality as a matrix having a row per vertex and a column per step.
Step 1  Step 2  Step 3  Step 4  Step 5  Step 6  Step 7  
Vertex  
8  
3  
2  
1  
4  
5  
6  
7 
Variables “” associated to vertices from form a ladder (, , , , , and ). Due to