**2. Preliminaries**

The terminology and basic definitions in this section are summarized from [30] where the authors present the results on the efficient closed domination among standard products of digraphs.

Let *D* be a digraph with the vertex set *V*(*D*) and the arc set *<sup>A</sup>*(*D*). For any two vertices *u*, *v* ∈ *<sup>V</sup>*(*D*), we write (*<sup>u</sup>*, *v*) as the *arc* with *direction* or *orientation* from *u* to *v*, and say *u* is *adjacent* to *v*, or *v* is *adjacent* from *u*. For an arc (*<sup>u</sup>*, *v*) we also say that *u* is the *in-neighbor* of *v* and that *v* is the *out-neighbor* of *u*. For a vertex *v* ∈ *<sup>V</sup>*(*D*), the *open out-neighborhood* of *v* (*open in-neighborhood* of *v*) is *N*<sup>+</sup> *D* (*v*) = {*u* ∈ *V*(*D*) : (*<sup>v</sup>*, *u*) ∈ *A*(*D*)} (*N*− *D* (*v*) = {*u* ∈ *V*(*D*) : (*<sup>u</sup>*, *v*) ∈ *A*(*D*)}). The *in-degree* of *v* is *δ*− *D*(*v*) = |*<sup>N</sup>*− *D* (*v*)|, the *out-degree* of *v* is *δ*<sup>+</sup> *D*(*v*) = |*N*<sup>+</sup> *D* (*v*)| and the degree of *v* is *<sup>δ</sup>D*(*v*) = *δ*− *D*(*v*) + *δ*<sup>+</sup> *<sup>D</sup>*(*v*). Moreover, *N*− *D* [*v*] = *N*− *D* (*v*) ∪ {*v*} is the *closed in-neighborhood* of *v* (*N*<sup>+</sup> *D* [*v*] = *N*<sup>+</sup> *D* (*v*) ∪ {*v*} is the *closed out-neighborhood* of *v*). In the above notation we omit *D* if there is no ambiguity with respect to the digraph *D*. We similarly proceed with any other notation which uses such a style of subscripts. Throughout the paper we use [*k*] = {1, . . . , *k*}.

A vertex *v* of *D* with *<sup>δ</sup>*+(*v*) = |*V*(*D*)| − 1 is called an *out-universal vertex*, and if *<sup>δ</sup>*−(*v*) = |*V*(*D*)| − 1, then *v* is called an *in-universal vertex*. A vertex *v* of *D* with *<sup>δ</sup>*+(*v*) = 0 is called a *sink*, and if *<sup>δ</sup>*−(*v*) = 0, then *v* is called a *source*. If *<sup>δ</sup>*(*v*) = 0, then *v* is an *isolated vertex* or a *singleton*. An arc of the form (*<sup>v</sup>*, *v*) is called a *loop* and can be considered as a directed cycle of length one. A vertex *v* with *<sup>δ</sup>*(*v*) = 1 is called

a *leaf* and is either a sink (if *<sup>δ</sup>*+(*v*) = 0) or a source (if *<sup>δ</sup>*−(*v*) = 0). Clearly, any vertex *u* with *<sup>δ</sup>*(*u*) = 2 is either a sink, or a source, or *<sup>δ</sup>*−(*u*) = 1 = *<sup>δ</sup>*+(*u*).

The *underlying graph* of a digraph *D* is a graph *GD* with *V*(*GD*) = *V*(*D*) and for every arc (*<sup>u</sup>*, *v*) from *D* we have an edge *uv* in *<sup>E</sup>*(*GD*). If (*<sup>u</sup>*, *v*) and (*<sup>v</sup>*, *u*) are both arcs, then we have two edges between *u* and *v* in the underlying graph. A *directed path* is a digraph *D* ∼= *Pn* with one source and one sink where its underlying graph is isomorphic to a path *Pn*. Similarly, a *directed cycle* is a digraph *D* ∼= *Cn* without sinks and sources with a cycle *Cn* as its underlying graph. We also consider a loop as a directed cycle *C*1 of length one and double arc with different orientation as a directed cycle *C*2 of length two. The distance *dD*(*<sup>u</sup>*, *v*) between two vertices *u* and *v* is the minimum number of arcs on a directed path from *u* to *v* or ∞ if such a directed path does not exist. For *A* ⊆ *V*(*D*) we denote by *D* − *A* a digraph obtained from *D* by deleting all vertices from *A*. By *D*[*A*] we denote the subdigraph of *D* that is induced on the vertices from *A*.

Let *D* be a digraph and let *S* ⊆ *<sup>V</sup>*(*D*). The set *S* is called a *total dominating set* of *D* if the open out-neighborhoods centered in vertices of *S* cover *<sup>V</sup>*(*D*), that is *V*(*D*) = *v*∈*<sup>S</sup> <sup>N</sup>*+*D* (*v*). Let *S* be a total dominating set of *D*. If *<sup>N</sup>*+*D* (*v*) ∩ *<sup>N</sup>*+*D* (*u*) = ∅ for every two different vertices *u*, *v* ∈ *S*, then the set {*N*+*D* (*v*) : *v* ∈ *S*} not only covers *V*(*D*) but also partitions *<sup>V</sup>*(*D*). In this case we say that *S* is an *efficient open dominating set* (or an EOD set for short) of *D*. If there exists an EOD set *S* for the digraph *D*, then *D* is called an *efficient open domination digraph* (or an EOD digraph for short). For *A* ⊆ *V*(*D*) we say that *SA* ⊆ *V*(*D*) is *efficient open domination set only* (or *an EOD set only* for short) for a digraph *D* − *A* if every vertex from *V*(*D*) − *A* has exactly one in-neighbor in *SA* and in addition *A* ∩ *<sup>N</sup>*+*D* (*SA*) = ∅.

Let *D* and *F* be digraphs. Different products of digraphs *D* and *F* have, similarly as in graphs, their set of vertices equal to *V*(*D*) × *<sup>V</sup>*(*F*). We roughly and briefly discuss the four standard products of digraphs: the *Cartesian product D*✷*F*, the *direct product D* × *F* , the *strong product D F* and the *lexicographic product D* ◦ *F* (sometimes also denoted *D*[*F*]). Adjacency in different products is defined as follows.


Some examples of the above mentioned products appear in Figure 1.

**Figure 1.** The digraphs *D* and *E*, and their Cartesian, direct, strong and lexicographic products.

Let ∗∈{✷, ×, , ◦}. The map *pD* : *V*(*D* ∗ *F*) → *V*(*D*) defined by *pD*((*d*, *f*)) = *d* is called the *projection map onto D*. Similarly, we define *pF* as the *projection map onto F*. Projections are defined as maps between vertices, but frequently it is more convenient to see them as maps between digraphs. In this case we observe the subdigraphs induced by *B* ⊆ *V*(*D* ◦ *F*) and *pX*(*B*) for *X* ∈ {*<sup>D</sup>*, *<sup>F</sup>*}. Notice that in the Cartesian and in the strong product the arcs project either to arcs (with the same orientation) or to a vertex. In the case of the direct product arcs always project to arcs (with the same orientation). In the lexicographic product *D* ◦ *F* the projection *pD* maps arcs into arcs (with the same orientation) or into vertices. In the same product the projection *pF* maps arcs into vertices, into arcs with the same orientation, into arcs with different orientation or into two vertices without an arc between them.

For a fixed *f* ∈ *V*(*F*) we call set *Df* = {(*d*, *f*) ∈ *V*(*D* ∗ *F*) : *d* ∈ *V*(*D*)} a *<sup>D</sup>*-*layer through f* in *D* ∗ *F*, where ∗∈{✷, ×, , ◦}. Symmetrically, an *F*-layer *F<sup>d</sup>* through *d* is defined for a fixed *d* ∈ *<sup>V</sup>*(*D*). Notice that for the Cartesian product, for the strong product and for the lexicographic product, (*D* ∗ *F*)[*Df* ] is isomorphic to *D* and (*D* ∗ *F*)[*F<sup>d</sup>*] is isomorphic to *F*, respectively. In the case of the direct product loops play an important role. If there are no loops in *f* and in *d*, then the subdigraphs (*D* ∗ *F*)[*F<sup>d</sup>*] and (*D* ∗ *F*)[*Df* ] are isomorphic to an empty digraph on |*V*(*F*)| and |*V*(*D*)| vertices, respectively. If we have (*d*, *d*) ∈ *A*(*D*) and (*f* , *f*) ∈ *<sup>A</sup>*(*F*), then (*D* ∗ *F*)[*F<sup>d</sup>*] and (*D* ∗ *F*)[*Df* ] are isomorphic to *F* and *D*, respectively.

It is easy to see that open out-neighborhoods in the direct product of digraphs satisfy

$$N\_{D \times F}^{+}((d, f)) = N\_D^{+}(d) \times N\_F^{+}(f) \tag{1}$$

and for the lexicographic product of digraphs it holds that

$$N\_{D \circ F}^{+}((d\_\prime f)) = \left(N\_D^{+}(d) \times V(F)\right) \cup \left(\{d\} \times N\_F^{+}(f)\right). \tag{2}$$

Using these two equalities a complete characterization of the EOD digraphs among the direct and the lexicographic product is presented in the last section.

### **3. The Cartesian Product**

**Definition 1.** *Let F be a digraph and let S*1, ... , *Sk* ⊆ *<sup>V</sup>*(*F*)*. If Si is an EOD set only for F* − *Si*−1*, i* ∈ [*k*]*, where S*0 = ∅*, then we say that F is a k-EOD path divisible. Similarly, if Si is an EOD set only for F* − *Si*−1*, i* ∈ [*k*]*, where S*0 = *Sk, then we say that F is a k-EOD cycle divisible. We say that sets S*1, ... , *Sk are k-EOD path or k-EOD cycle divisible sets of F.*

Notice that every *k*-EOD path divisible digraph is also an EOD digraph, because *S*1 is an EOD set only for *F* − *S*0 = *F*. Therefore, an EOD digraph *F* with an EOD set *S*1 is 1-EOD path divisible. Also if *F* is *n*-EOD path divisible, then it is also *m*-EOD path divisible for every *m* ≤ *n*. In particular, let *F* be a directed cycle, that is *F* is an EOD digraph with the EOD set *S* = *<sup>V</sup>*(*F*). If we set *S*2*i*−<sup>1</sup> = *V*(*F*) and *S*2*i* = ∅, then *F* is *k*-EOD path divisible for every positive integer *k*. If *F* is *k*-EOD path divisible, then it can happen that *Si* ∩ *Sj* = ∅. See an example of this on Figure 2.

With the following example we underline the rich structure of *n*-EOD path (or cycle) divisible digraphs. We will show that every digraph can be an induced digraph of an *n*-EOD path (or cycle) divisible digraph. A complete digraph *Kn* contains an arc in both directions between all different vertices of *Kn*. Let *V*(*Kn*) = {*<sup>v</sup>*1, ... , *vn*}. Digraph *K*<sup>−</sup>*n* is obtained from *Kn* by deleting all arcs (*vi*+1, *vi*), *i* ∈ [*n* − 1]. For a digraph *F* we construct an *n*-EOD path divisible digraph *F*<sup>+</sup> in the folowing way. We take one copy of *F* and two copies of *K*<sup>−</sup>*n* , the first copy containing the vertices *V*1 = {*<sup>v</sup>*1, ... , *vn*} and the second copy containing the vertices *V*2 = {*v*1, ... , *<sup>v</sup>n*}. The arc set of *F*<sup>+</sup> contains *<sup>A</sup>*(*F*), all arcs from both copies of *K*<sup>−</sup>*n* , the set {(*vi*, *<sup>v</sup>i*),(*vi*, *vi*) : *i* ∈ [*n*]}, all arcs from the set {(*vi*, *f*) : *vi* ∈ *V*1, *f* ∈ *V*(*F*)} and an arbitrary subset of {(*f* , *vi*),(*f* , *vi*) : *f* ∈ *<sup>V</sup>*(*F*), *vi* ∈ *V*1, *vi* ∈ *<sup>V</sup>*2}. It is not hard to see that *F*<sup>+</sup> is an *n*-EOD path divisible digraph with *n*-EOD path divisible sets *Si*= {*vi*, *<sup>v</sup>i*} for every *i* ∈ [*n*].

 Similar construction can be done to ge<sup>t</sup> an *n*-EOD cycle divisible digraph. We only need to delete arcs (*<sup>v</sup>*1, *vn*) and (*v*1, *<sup>v</sup>n*) from *F*+. Also if *F* is an *n*-EOD cycle divisible digraph, then *F* is also an *kn*-EOD cycle divisible digraph, where *kn*-EOD cycle divisible sets are repeated cyclically *k* times.

**Figure 2.** 4-EOD path divisible digraph with *S*1 = {*<sup>v</sup>*3, *<sup>v</sup>*4}, *S*2 = {*<sup>v</sup>*6, *<sup>v</sup>*7}, *S*3 = {*<sup>v</sup>*1, *<sup>v</sup>*8}, *S*4 = {*<sup>v</sup>*2, *<sup>v</sup>*3} and with *S*1 ∩ *S*4 = {*<sup>v</sup>*3}.

Next we show that *n*-EOD path divisibility of *F* is essential for the Cartesian product *Pn*✷*F* to be an EOD digraph, where *Pn* is a directed path.

**Theorem 1.** *Let Pn be a directed path and let F be a digraph. The Cartesian product Pn*✷*F is an EOD digraph if and only if F is an n-EOD path divisible digraph.*

**Proof.** Let *Pn* = *v*1 ... *vn* be a directed path where *v*1 is the source and *vn* is the sink and let *F* be an arbitrary digraph.

First assume that *F* is an *n*-EOD path divisible digraph. Denote by *S*1, ... , *Sn* the subsets of *V*(*F*) that correspond with *n*-EOD path divisibility. We will show that *S* = <sup>∪</sup>*ni*=<sup>1</sup>{*vi*} × *Si* is an EOD set of *Pn*✷*F*, meaning that |*<sup>N</sup>*−(*vi*, *u*) ∩ *S*| = 1 for every *i* ∈ [*n*] and *u* ∈ *<sup>V</sup>*(*F*). For every (*<sup>v</sup>*1, *u*) it holds that |*<sup>N</sup>*−(*<sup>v</sup>*1, *u*) ∩ *S*| ≥ 1 because *S*1 is an EOD set of *F* and therefore {*<sup>v</sup>*1} × *S*1 is an EOD set for (*Pn*✷*<sup>F</sup>*)[*Fv*1 ] ∼= *F*. Since *v*1 is a source of *Pn*, there do not exist any other in-neighbors of vertices in *Fv*1 except those already in *Fv*1 , so |*<sup>N</sup>*−(*<sup>v</sup>*1, *u*) ∩ *S*| = 1. Next we observe (*vi*, *u*) for 2 ≤ *i* ≤ *n* and *u* ∈ *<sup>V</sup>*(*F*). If *u* ∈ *Si*−1, then (*vi*, *u*) has an in-neighbor in {*vi*−<sup>1</sup>} × *Si*−<sup>1</sup> ⊂ *S* and if *u* ∈ *V*(*F*) − *Si*−1, then (*vi*, *u*) has an in-neighbor in {*vi*} × *Si* ⊂ *S*. On the other hand these neighbors are unique in *S*, because *<sup>N</sup>*+*F* (*Si*) ∩ *Si*−<sup>1</sup> = ∅ and *Si* is an EOD set only for *F* − *Si*−1. Hence, *S* is an EOD set of *Pn*✷*F*, which is therefore an EOD digraph.

Now assume that *Pn*✷*F* is an EOD digraph and let *S* be its EOD set. Let *Si* = *pF*(*<sup>S</sup>* ∩ *Fvi*) for *i* ∈ [*n*] and let *S*0 = ∅. Clearly, every vertex from *Fv*1 must have exactly one in-neighbor in {*<sup>v</sup>*1} × *S*1 because *v*1 is a source of *Pn*. Therefore, *S*1 is an EOD set of *F* − *S*0 = *F* and *<sup>N</sup>*+*F* (*<sup>S</sup>*1) ∩ *S*0 = ∅. Now let *i* > 1. Vertices from {*vi*} × *Si*−<sup>1</sup> have in-neighbors in {*vi*−<sup>1</sup>} × *Si*−<sup>1</sup> and therefore do not have in-neighbors in {*vi*} × *Si*, meaning that *<sup>N</sup>*+*F* (*Si*) ∩ *Si*−<sup>1</sup> = ∅. On the other hand all other vertices in *Fvi* must have an in-neighbor in {*vi*} × *Si*, because *S* is an EOD set of *Pn*✷*F*. Thus, *Si* is an EOD set only for *F* − *Si*−1. Therefore, *S*1,..., *Sn* yield that *F* is an *n*-EOD path divisible digraph.

With *n*-EOD cycle divisibility one can describe all EOD digraphs among *Cn*✷*F* where *Cn* is a directed cycle. The proof is very similar to the proof of Theorem 1 and is therefore omitted. The main difference is that we do not need to treat layer *Fv*1 separately since everything follows from the general step.

**Theorem 2.** *Let Cn be a directed cycle and let F be a digraph. The Cartesian product Cn*✷*F is an EOD digraph if and only if F is an n-EOD cycle divisible digraph.*

We continue with a path that is oriented in such a way, that it has exactly one source of degree two.

**Theorem 3.** *Let D be a digraph with an underlying graph Pn* = *v*1 ... *vn with such an orientation that vk,* 1 < *k* < *n, is the only source, let m* = max{*k*, *n* − *k* + 1} *and let F be a digraph. The Cartesian product D* ✷*F is an EOD digraph if and only if F is an m-EOD path divisible digraph.*

**Proof.** Let *vk* be the only source of *D*. Thus *P* = *vkvk*−1 ... *v*1 is a directed path on *k* vertices, where *vk* is the source and *v*1 is the sink, and *P* = *vkvk*+<sup>1</sup> ... *vn* is a directed path on *n* − *k* + 1 vertices, where *vk* is the source and *vn* is the sink. Let *m* = max{*k*, *n* − *k* + <sup>1</sup>}. If *F* is *m*-EOD path divisible with sets *Si*, *i* ∈ [*m*], then *F* is *k*-EOD path divisible with sets *Si*, *i* ∈ [*k*] and also (*n* − *k* + 1)-EOD path divisible with sets *Si*, *i* ∈ [*n* − *k* + 1]. As shown in the proof of Theorem 1, sets *S*(*k*)=( ∪*<sup>k</sup> <sup>i</sup>*=<sup>1</sup>{*vi*} × *Sk*−*i*+<sup>1</sup>) and *<sup>S</sup>*(*n* − *k* + 1)=( ∪*<sup>n</sup> <sup>i</sup>*=*<sup>k</sup>*{*vi*} × *Si*) are EOD sets for *P* ✷*F* and *P*✷*F*, respectively. Clearly, *S* = *S*(*k*) ∪ *<sup>S</sup>*(*n* − *k* + 1) is an EOD set of *D* ✷*F* because *Fvk* ∩ *S*(*k*) = *Fvk* ∩ *<sup>S</sup>*(*n* − *k* + 1) and *D* ✷*F* is an EOD digraph.

Now assume that *D* ✷*F* is an EOD digraph and let *S* be its EOD set. Since *vk* is a source, *P* ✷*F* and *P*✷*F* are also EOD digraphs. By Theorem 1 *F* is a *k*-EOD path divisible digraph and an (*n* − *k* +<sup>1</sup>)-EOD path divisible digraph. Hence, *F* is also *m*-EOD path divisible for *m* = max{*k*, *n* − *k* + <sup>1</sup>}.

Before we deal with a path that is oriented in such a way, that it has exactly one sink of degree two, we need the following definition.

**Definition 2.** *Let F be a k-EOD and an -EOD path divisible digraph. We say that F is k*, *-*sink friendly *if there exist k-EOD path divisible sets S*1, ... , *Sk and -EOD path divisible sets S* 1, ... , *S such that Sk* ∩ *S* = ∅ *and there exists a set S*0 ⊆ *<sup>V</sup>*(*F*)*, which is an EOD set only for F* − (*Sk* ∪ *S* )*.*

**Theorem 4.** *Let D be a digraph with an underlying graph Pn* = *v*1 ... *vn with such an orientation that vk,* 1 < *k* < *n, is the only sink and let F be a digraph. The Cartesian product D* ✷*F is an EOD digraph if and only if F is* (*k* − <sup>1</sup>),(*<sup>n</sup>* − *k*)*-sink friendly.*

**Proof.** Let *F* be digraph and let *D* be a digraph with an underlying graph *Pn* = *v*1 ... *vn* where *vk*, 1 < *k* < *n*, is the only sink. This means that *v*1 and *vn* are the only sources of *D*.

First assume that *F* is (*k* − <sup>1</sup>),(*<sup>n</sup>* − *k*)-sink friendly. This means that there exist sets *S*1, ... , *Sk*−<sup>1</sup> that yield (*k* − 1)-EOD path divisibility of *F* and sets *Sn*, *Sn*−<sup>1</sup> ... , *Sk*+<sup>1</sup> that yield (*n* − *k*)-EOD path divisibility of *F*. In addition, *Sk*−<sup>1</sup> ∩ *Sk*+<sup>1</sup> = ∅ and there exists a set *Sk* ⊆ *V*(*F*) which is an EOD set only for *F* − (*Sk*−<sup>1</sup> ∪ *Sk*+<sup>1</sup>). Let *A* = ∪*k*−<sup>1</sup> *i*=1 ({*vi*} × *Si*) and *B* = ∪*<sup>n</sup> <sup>j</sup>*=*k*+<sup>1</sup>({*vj*} × *Sj*). We will show that *S* = *A* ∪ *B* ∪ ({*vk*} × *Sk*) is an EOD set of *D* ✷*F*. Let *Q* = *v*1 ... *vk*−1 and *R* = *vnvn*−1 ... *vk*+<sup>1</sup> be directed subpaths of *D*. By Theorem 1, *Q* ✷*F* and *R* ✷*F* are EOD digraphs with EOD sets *A* and *B*, respectively. No vertex from *Fvk* is an in-neighbor of vertices from *Q* ✷*F* and *R* ✷*F*, because *vk* is a sink. Therefore, there exists exactly one in-neighbor in *S* for every vertex from *Q* ✷*F* and *R* ✷*F*. So, we only need to check (*vk*, *f*) for every *f* ∈ *<sup>V</sup>*(*F*). If *f* ∈ *Sk*−1, then (*vk*−1, *f*) ∈ *S* is the in-neighbor of (*vk*, *f*). On the other hand this is the only neighbor of (*vk*, *f*) from *S* because (*vk*+1, *f*) ∈/ *S* as *Sk*−<sup>1</sup> ∩ *Sk*+<sup>1</sup> = ∅ and because *Sk* is an EOD set only for *F* − (*Sk*−<sup>1</sup> ∪ *Sk*+<sup>1</sup>). By symmetry we can see that (*vk*, *f*) has also exactly one in-neighbor in *S* whenever *f* ∈ *Sk*+1. So let *f* ∈ *V*(*F*) − (*Sk*−<sup>1</sup> ∩ *Sk*+<sup>1</sup>). Clearly (*vk*, *f*) has

no in-neighbor from *S* in *Fvk*−<sup>1</sup> and in *Fvk*+<sup>1</sup> . Since *Sk* is an EOD set only for *F* − (*Sk*−<sup>1</sup> ∪ *Sk*+<sup>1</sup>), there exists exactly one in-neighbor *x* of *f* in *Sk* and (*vk*, *x*) is therefore the only neighbor of (*vk*, *f*) from *S*. Hence, *S* is an EOD set of *D*✷*F* which is therefore an EOD digraph.

Now assume that *D*✷*F* is an EOD digraph and let *S* be its EOD set. Again, for directed paths *Q* = *v*1 ... *vk*−1 and *R* = *vnvn*−1 ... *vk*+1, *Q*✷*F* and *R*✷*F* are EOD digraphs with no influence from *Fvk* in a product *D*✷*F*. Sets *Si* = *pF*(*<sup>S</sup>* ∩ *Fvi*) for *i* ∈ [*k* − 1] are (*k* − 1)-EOD path divisible sets by Theorem 1 and sets *Sj* = *pF*(*<sup>S</sup>* ∩ *Fvj*) for *j* ∈ {*k* + 1, ... , *n*} (in reversed order) are (*n* − *k*)-EOD path divisible sets by the same theorem. If *f* ∈ *Sk*−<sup>1</sup> ∩ *Sk*+1, then (*vk*, *f*) has two in-neighbors (*vk*−1, *f*) and (*vk*+1, *f*) in *S*, a contradiction. Therefore, we have *Sk*−<sup>1</sup> ∩ *Sk*+<sup>1</sup> = ∅. Let *Sk* = *pF*(*<sup>S</sup>* ∩ *Fvk* ) and let *f* be an arbitrary vertex from ∈ *V*(*F*) − (*Sk*−<sup>1</sup> ∪ *Sk*+<sup>1</sup>). Clearly, (*vk*, *f*) has exactly one in-neighbor in {*vk*} × *Sk* because *S* is an EOD set. Also (*vk*, *f* ), *f* ∈ *Sk*−<sup>1</sup> ∪ *Sk*+1, has no in-neighbor in {*vk*} × *Sk* as it has its unique in-neighbor either in *S*, in {*vk*−<sup>1</sup>} × *Sk*−<sup>1</sup> or in {*vk*+<sup>1</sup>} × *Sk*+1. Hence, *Sk* is an EOD set only for *V*(*F*) − (*Sk*−<sup>1</sup> ∪ *Sk*+<sup>1</sup>) and *F* is (*k* − <sup>1</sup>),(*<sup>n</sup>* − *k*)-sink friendly.

The next challenge considering digraphs with an underlying graph isomorphic to a path or to a cycle is when we have more sinks and sources of degree two. Clearly, after every sink there comes a source and after each source there is a sink. In the case of a sink *v* of degree two digraph *F* must be *k*, -sink friendly by Theorem 4, where *k* + 1 and + 1 are the distances to the sources that are closest to *v*. However this is not always enough. Let *x* and *y* be two sinks and let *u* be a source between them. By Theorem 4 digraph *F* must be *k*1, 1-sink friendly and *k*2, 2-sink friendly where *k*1 + 1 and *k*2 + 1 are the distances between *u* and *x* and *u* and *y*, respectively. We say that *F* is *k*1, *k*2-*source friendly* if *S*1 = *S*1. Here, sets *S*1, ... , *Sk*1 and *<sup>S</sup>*1, ... , *Sk*2 are appropriate *k*1- and *k*2-EOD path divisible sets from *k*1, 1-sink friendly and *k*2, 2-sink friendly constellation, respectively. Now, if we can assure sink friendliness for each sink, and also source friendliness for each source of a digraph *F*, then this is characteristic for *D*✷*F* to be an EOD digraph. Here, the underlying graph of *D* is either *Pk* or *Ck* with more than one source or sink of degree two. Because the proof is very similar and the formal statement is problematic (it depends on the status of vertices of degree one in *D*), we omit the proof of this.

We end this section with another fixed factor which this time has a star *<sup>K</sup>*1,*n* as its underlying graph. Vertex of degree *n* is the source and all the others are sinks.

**Theorem 5.** *Let D be a digraph with an underlying graph <sup>K</sup>*1,*n with the set of vertices* {*<sup>v</sup>*0, *v*1, ... , *vn*}*, where <sup>δ</sup>*+*D*(*<sup>v</sup>*0) = *n and let F be an arbitrary digraph. The Cartesian product D*✷*F is an EOD digraph if and only if F is a* 2*-EOD path divisible digraph.*

**Proof.** Let *D* be a digraph with an underlying graph *<sup>K</sup>*1,*n* with the set of vertices {*<sup>v</sup>*0, *v*1, ... *vn*}, where *<sup>δ</sup>*+*D*(*<sup>v</sup>*0) = *n*, which means that *v*0 is the source. Clearly, *<sup>δ</sup>*<sup>−</sup>*D*(*vi*) = 1 and *vi* is a sink for every *i* ∈ [*n*]. Let *F* be an arbitrary digraph. Denote by *A* the set of vertices of *D*✷*F*.

First assume that *F* is 2-EOD path divisible with sets *S*1 and *S*2. We will show that *S* = ({*<sup>v</sup>*0} × *<sup>S</sup>*1) ∪ (∪*ni*=<sup>1</sup>{*vi*} × *<sup>S</sup>*2) is an EOD set for *D*✷*F*, meaning that |*<sup>N</sup>*−((*<sup>e</sup>*, *f*)) ∩ *S*| = 1 for every (*e*, *f*) ∈ *A*. For every (*<sup>v</sup>*0, *f*) ∈ *A* it holds that |*<sup>N</sup>*−((*<sup>v</sup>*0, *f*)) ∩ *S*| ≥ 1 since *S*1 is an EOD set for *F* and therefore {*<sup>v</sup>*0} × *S*1 is an EOD set for (*<sup>D</sup>*✷*<sup>F</sup>*)[*Fv*0 ] ∼= *F*. Since *v*0 is a source, there do not exist any other in-neighbors of vertices *Fv*0 except those from *Fv*0 , so |*<sup>N</sup>*−((*<sup>v</sup>*0, *f*)) ∩ *S*| = 1. Now let (*vi*, *f*) ∈ *A*, *i* ∈ [*n*]. Vertices from {*<sup>v</sup>*0} × *S*1 are the in-neighbors of all of the vertices from {*vi*} × *S*1 and, since *F* is 2-EOD path divisible, vertices from {*vi*} × *S*2 are the in-neighbors of all of the vertices from {*vi*} × (*V*(*F*) − *<sup>S</sup>*1). So |*<sup>N</sup>*−((*vi*, *f*)) ∩ *S*| ≥ 1. By the definition of 2-EOD path divisibility it holds that |*N*+({*vi*} × *<sup>S</sup>*2) ∩ ({*vi*} × *<sup>S</sup>*1)| = 0, meaning that |*<sup>N</sup>*−((*vi*, *f*)) ∩ *S*| = 1.

Now assume that *D*✷*F* is an EOD digraph and let *S* be its EOD set. Let *vi* be an arbitrary vertex of the star *D* different from *v*0. Vertices from *S* ∩ *Fv*0 are in-neighbors of some (or all) of the vertices of *Fvi* . Denote the set of those vertices by *Fvi* . Vertices from *Fvi* − *Fvi* have to have in-neighbors in *S* ∩ *Fvi* , since they do not have in-neighbors in *S* ∩ *Fv*0 . Vertices from *S* ∩ *Fvi* are not in-neighbors of any of the vertices from *Fvi* , since that would mean that there exists (*vi*, *f*) ∈ *Fvi* for which |*<sup>N</sup>*−(*vi*, *f*) ∩ *S*| > 1, a contradiction with *S* being an EOD set of *D*✷*F*. Let *S*1 := *pH*(*<sup>S</sup>* ∩ *Fs*0 ) and *S*2 := *pH*(*<sup>S</sup>* ∩ *Fs*1 ). Clearly, *S*1 is an EOD set of *F* and *S*2 is an EOD set only for *F* − *S*1. Hence, *F* is a 2-EOD path divisible digraph.

### **4. The Strong Product**

In this section we first characterize all EOD strong product digraphs *D F*, such that the underlying graphs of *D* and *F* are cycles *Cm* and *Cn*, respectively. Then we extend this result to a characterization of all EOD digraphs *D F* where *D* and *F* have uni-cyclic graphs as their underlying graphs. We also conjecture that there are no more EOD digraphs among the strong product of digraphs. We start with several lemmas that come in handy later.

**Lemma 1.** *Let D and F be two digraphs without isolated vertices. If one of them has a source, then D F is not an EOD digraph.*

**Proof.** Let *D* and *F* be two digraphs. If *D* has a source *u* and *F* has a source *v*, then vertex (*<sup>u</sup>*, *v*) is a source in *D F* and it has no in-neighbor. Hence, there does not exist an EOD set for *D F*. Without loss of generality let *D* have a source *u* and let *F* be an arbitrary digraph without a source. We will try to construct an EOD set *S* for *D F*. Since *u* is a source, vertex (*<sup>u</sup>*, *y*), *y* ∈ *<sup>V</sup>*(*F*), has in-neighbors only in *Du*, and since *F* does not have a source at least one in-neighbor of (*<sup>u</sup>*, *y*) exists. Let (*<sup>u</sup>*, *y*) ∈ *S* be the in-neighbor of (*<sup>u</sup>*, *y*). Again, since *u* is a source and *F* has no source, there exists an in-neighbor (*<sup>u</sup>*, *y*) ∈ *S* of (*<sup>u</sup>*, *y*) ∈ *Du*. Denote by *u* an out-neighbor of *u* in *D*. It exists since *D* contains no isolated vertices. By the definition of the strong product of two digraphs, both (*<sup>u</sup>*, *y*) and (*<sup>u</sup>*, *y*) are in-neighbors of (*u*, *y*) and since (*<sup>u</sup>*, *y*),(*<sup>u</sup>*, *y*) ∈ *S* vertex (*u*, *y*) has two different in-neighbors in *S*. Meaning that *S* is not an EOD-set, so *D F* is not an EOD digraph.

In the rest of this section we use the following notation and orientation for directed cycles *Cm* = *c*1*c*2 ... *cm* and *Cn* = *dndn*−<sup>1</sup> ... *d*1 on *m* and *n* vertices, respectively, see Figure 3, and with (*ci*, *dj*) we denote a vertex of a strong product of those two cycles. All operations on the first index *i* are via (mod m) and on the second index *j* are via (mod <sup>n</sup>). We also partition *V*(*Cm Cn*) into sets

$$\begin{aligned} A &= \{ (c\_i, d\_j); i + j = 3q + 1, q \in \mathbb{N} \}, \\ B &= \{ (c\_i, d\_j); i + j = 3q + 2, q \in \mathbb{N} \} \quad \text{and} \\ C &= \{ (c\_i, d\_j); i + j = 3q, q \in \mathbb{N} \}. \end{aligned} \tag{3}$$

**Lemma 2.** *If there exists an EOD set S for Cm Cn, m*, *n* ≥ 3*, and* (*ci*, *dj*) ∈ *S, then* (*ci*−1, *dj*+<sup>1</sup>) ∈ *S.*

**Proof.** Let *Cm Cn*, *m*, *n* ≥ 3, be an EOD digraph, let *S* be its EOD set and let (*ci*, *dj*) ∈ *S*. The vertex (*ci*, *dj*) is an in-neighbor of (*ci*+1, *dj*),(*ci*, *dj*−<sup>1</sup>) and (*ci*+1, *dj*−<sup>1</sup>). On the other hand (*ci*, *dj*) must also have an in-neighbor in *S*. The only in-neighbors of (*ci*, *dj*) are (*ci*−1, *dj*),(*ci*, *dj*+<sup>1</sup>) and (*ci*−1, *dj*+<sup>1</sup>). If (*ci*−1, *dj*) ∈ *S*, then (*ci*, *dj*) and (*ci*−1, *dj*) are both in-neighbors of (*ci*, *dj*−<sup>1</sup>), a contradiction. Similarly, if (*ci*, *dj*+<sup>1</sup>) ∈ *S*, then (*ci*, *dj*) and (*ci*, *dj*+<sup>1</sup>) are both in-neighbors of (*ci*+1, *dj*), a contradiction again. Hence, (*ci*−1, *dj*+<sup>1</sup>) must be in *S*.

**Lemma 3.** *If there exists an EOD set S for Cm Cn, m*, *n* ≥ 3*, and* (*ci*, *dj*) ∈ *S, then* (*ci*, *dj*+<sup>3</sup>),(*ci*−3, *dj*) ∈ *S.*

**Proof.** Let *Cm Cn*, *m*, *n* ≥ 3, be an EOD digraph and let *S* be its EOD set. With possible change of notation let (*cm*, *d*1) ∈ *S*. A vertex (*cm*, *d*1) is an in-neighbor of the vertices (*<sup>c</sup>*1, *d*1), (*cm*, *dn*) and (*<sup>c</sup>*1, *dn*). Vertex (*cm*−1, *d*2) belongs to *S* by Lemma 2. Clearly, (*cm*−1, *d*2) is also the in-neighbor of vertices (*cm*−1, *d*1) and (*cm*, *d*2). So (*cm*, *d*2) ∈/ *S* because otherwise (*cm*, *d*1) has two in-neighbors in *S*. If (*cm*, *d*3) ∈ *S*, then (*cm*, *d*2) has two in-neighbors in *S* again. So (*cm*, *d*3) ∈/ *S*.

Vertex (*cm*−2, *d*3) ∈ *S* by Lemma 2 since (*cm*−1, *d*2) ∈ *S* and (*cm*−2, *d*3) is an in-neighbor of (*cm*−2, *d*2) and (*cm*−1, *d*3). One of the in-neighbors (*cm*−1, *d*3), (*cm*, *d*4) or (*cm*−1, *d*4) of the vertex (*cm*, *d*3) must be in *S*. If (*cm*−1, *d*3) ∈ *S*, then (*cm*, *d*2) has two in-neighbors in *S*. Similarly, if (*cm*−1, *d*4) ∈ *S*, then (*cm*−1, *d*3) has two in-neighbors in *S*. Hence, (*cm*, *d*4) ∈ *S*.

We can exchange the role of factors and by symmetric arguments ge<sup>t</sup> that (*ci*−3, *dj*) also belongs to *S*.

**Figure 3.** The strong product of two directed cycles *Cm* and *Cn*.

Now we can characterize all EOD digraphs among strong product digraphs of two cycles.

**Theorem 6.** *Let D and F be digraphs with underlying graphs Cm and Cn, respectively. The strong product D F is an EOD digraph if and only if both D and F are directed cycles, m* = 3 *and n* = 3*k for some k*, ∈ N*.*

**Proof.** First, let *m* = 3 and *n* = 3*k*, *k*, ∈ N, and let *D* ∼= *Cm* and *F* ∼= *Cn* be two directed cycles. Recall the sets *A*, *B* and *C* from (3). We will show that *A* is an EOD set of *Cm Cn*. The in-neighbor of a vertex (*ci*, *dj*) ∈ *B*, *i* + *j* = 3*q* + 2, that is in *A* is (*ci*−1, *dj*), since (*i* − 1) + *j* = (*i* + *j*) − 1 = (<sup>3</sup>*q* + 2) − 1 = 3*q* + 1. The in-neighbor of a vertex (*ci*, *dj*) ∈ *C*, *i* + *j* = 3*q*, that is in *A* is (*ci*, *dj*+<sup>1</sup>), since *i* + (*j* + 1)=(*i* + *j*) + 1 = 3*q* + 1. The in-neighbor of a vertex (*ci*, *dj*) ∈ *A*, *i* + *j* = 3*q* + 1, that is in *A* is (*ci*−1, *dj*+<sup>1</sup>), since (*i* − 1)+(*j* + 1) = *i* + *j* = 3*q* + 1. So every vertex *v* from *V*(*Cm Cn*) is efficiently dominated by *A*. Moreover *v* has exactly one in-neighbor in *A* since exactly one in-neighbor of *v* has the sum of indices equal to 3*q* + 1.

To prove the contrary let *D* and *F* be two digraphs with underlying graphs *Cm* and *Cn*, respectively, such that *D F* is an EOD digraph with an EOD set *S*. If one of the cycles is not directed, then it has a source. By Lemma 1 the strong product *D F* is not an EOD digraph. So we may assume that both *D* and *F* are directed cycles. With possible change of notation we may assume that (*cm*, *d*1) ∈ *S*. By consecutive use of Lemma 3 we ge<sup>t</sup> that {(*cm*, *d*3*k*+<sup>1</sup>) : *k* ∈ [*n*/3]} ⊆ *S* and that {(*cm*−3, *d*1) : ∈ [*m*/3]} ⊆ *S*. If *n* = 3*k*, then (*cm*, *dn*−<sup>2</sup>) ∈ *S* and (*cm*, *dn*+<sup>1</sup>) ∈ *S* by Lemma 3 again where (*cm*, *dn*+<sup>1</sup>)=(*cm*, *d*1). If *n* = 3*k* + 1, then (*cm*, *dn*) ∈ *S* and (*cm*, *dn*+<sup>3</sup>) ∈ *S* by Lemma 3 again where (*cm*, *dn*+<sup>3</sup>)=(*cm*, *d*3). Hence, (*cm*, *d*3),(*cm*, *d*4) ∈ *S* and they are both the in-neighbors of (*<sup>c</sup>*1, *d*3), a contradiction. If *n* = 3*k* + 2, then (*cm*, *dn*−<sup>1</sup>) ∈ *S* and (*cm*, *dn*+<sup>2</sup>) ∈ *S* by Lemma 3 again where (*cm*, *dn*+<sup>2</sup>)=(*cm*, *d*2). Hence, (*cm*, *d*1),(*cm*, *d*2) ∈ *S* and they are both the in-neighbors of (*<sup>c</sup>*1, *d*1), a

contradiction again. Therefore, *n* = 3*k*. By symmetric arguments we also ge<sup>t</sup> that *m* = 3 and the proof is completed.

Next we expand Theorem 6 and present a bigger class of EOD strong product digraphs. For this let *T*1, ... , *Tm* be arbitrary trees with roots *r*1, ... ,*rm*, respectively. We define an underlying graph *C*<sup>+</sup> *m* such that we identify root *ri* with vertex *ci* of a cycle *Cm* for every *i* ∈ [*m*]. Clearly, *C*<sup>+</sup> *m* is exactly a uni-cyclic graph, but we need the before mentioned structure. Notice that *C*<sup>+</sup> *m* ∼= *Cm* if every tree *Ti* is a one vertex tree. We say that a digraph with the underlying graph *C*<sup>+</sup> *m* is *well oriented* if *Cm* is a directed cycle and every edge from *Ti* is oriented away from the root *ri* for every *i* ∈ [*m*]. We use the same notation *C*<sup>+</sup> *m* for a digraph with the underlying graph *C*<sup>+</sup> *m* .

**Theorem 7.** *Let m*, *n* ≥ 3 *be two positive integers. The strong product C*<sup>+</sup> *m C*<sup>+</sup> *n is an EOD digraph if and only if both C*<sup>+</sup> *m and C*<sup>+</sup> *nare well oriented, m* = 3 *and n* = 3*k for some k*, ∈ N*.*

**Proof.** First, let *m* = 3 and *n* = 3*k*, *k*, ∈ N, and let *C*<sup>+</sup> *m* and *C*<sup>+</sup> *n* be well oriented. We show this direction in two steps. First let *C*<sup>+</sup> *n* ∼= *Cn* and we show that *C*<sup>+</sup> *m Cn* is an EOD digraph. By Theorem 6 *Cm Cn* is an EOD digraph with an EOD set *A* from (3). We extend set *A* to set *A*<sup>+</sup> for which we then show that it is an EOD set of *C*<sup>+</sup> *m Cn*. First we choose the notation for all the vertices from (*C*<sup>+</sup> *m Cn*) − *V*(*Cm Cn*). With *vi* we denote all the vertices from *C*<sup>+</sup> *m* − *V*(*Cm*) with *dC*+ *m* (*cm*, *vi*) = *i*. Notice that different vertices from *C*<sup>+</sup> *m* can have the same notation. Vertices from (*C*<sup>+</sup> *m Cn*) − *V*(*Cm Cn*) are then denoted as usual by (*vi*, *dj*). Furthermore, we denote sets *A* = {(*vi*, *dj*) : *i* + *j* = 3*q* + 1, *q* ∈ N}, *B* = {(*vi*, *dj*) : *i* + *j* = 3*q* + 2, *q* ∈ N} and *C* = {(*vi*, *dj*) : *i* + *j* = 3*q*, *q* ∈ N}. Now we partition *V*(*C*<sup>+</sup> *m Cn*) into sets *A*<sup>+</sup> = *A* ∪ *A* , *B*<sup>+</sup> = *B* ∪ *B* and *C*<sup>+</sup> = *C* ∪ *C*, where *A*, *B* and *C* are from (3). We will show that *A*<sup>+</sup> is an EOD set of *C*<sup>+</sup> *m Cn*.

By Theorem 6 each vertex from *A*, *B* and *C* has exactly one in-neighbor in *A*. The in-neighbor of a vertex (*vi*, *dj*) ∈ *B* , *i* + *j* = 3*q* + 2, that is in *A*<sup>+</sup> is either (*vi*−1, *dj*) or (*ci*−1, *dj*), since (*i* − 1) + *j* = (*i* + *j*) − 1 = (<sup>3</sup>*q* + 2) − 1 = 3*q* + 1. The in-neighbor of a vertex (*vi*, *dj*) ∈ *C* , *i* + *j* = 3*q*, that is in *A*<sup>+</sup> is (*vi*, *dj*+<sup>1</sup>), since *i* + (*j* + 1)=(*i* + *j*) + 1 = 3*q* + 1. The in-neighbor of a vertex (*vi*, *dj*) ∈ *A* , *i* + *j* = 3*q* + 1, that is in *A*<sup>+</sup> is either (*vi*−1, *dj*+<sup>1</sup>) or (*ci*−1, *dj*+<sup>1</sup>), since (*i* − 1)+(*j* + 1) = *i* + *j* = 3*q* + 1. So every vertex *x* from *V*(*C*<sup>+</sup> *m Cn*) has an in-neighbor in *A*+. Moreover *x* has exactly one in-neighbor in *A*<sup>+</sup> since exactly one in-neighbor of *x* has the sum of indices equal to 3*q* + 1.

By symmetric arguments we can show that *C*<sup>+</sup> *m C*<sup>+</sup> *n* is an EOD digraph whenever *C*<sup>+</sup> *m* ∼= *Cm*. So, we can assume that *C*<sup>+</sup> *m* - *Cm* and *C*<sup>+</sup> *n* - *Cn*. We know by the above arguments that *C*<sup>+</sup> *m Cn* is an EOD digraph with an EOD set *A*+. Since there is no arc from vertices of *D* = ( *C*<sup>+</sup> *m C*<sup>+</sup> *n* ) − *V*(*C*<sup>+</sup> *m Cn*) to vertices of (*C*<sup>+</sup> *m Cn*) we will use the set *A*<sup>+</sup> for *C*<sup>+</sup> *m Cn* and enlarge it to *A*∗ that will be an EOD set of *C*<sup>+</sup> *m C*<sup>+</sup> *n* . For this we first need to present the following notation for vertices of *D*. By (*ci*, *uj k*) we denote all the vertices from *D* that belong to layers (*C*<sup>+</sup> *m* ) *dj* and (*C*<sup>+</sup> *n* )*ci* and are at the distance *k* from (*ci*, *dj*). Similarly, we use (*vi*, *uj k*) for all the vertices from *D* that belong to layers (*C*<sup>+</sup> *m* ) *dj* and (*C*<sup>+</sup> *n* )*vi* and are at the distance *k* from (*vi*, *dj*). Notice that different vertices from *D* can have the same notation. Beside *A*<sup>+</sup> we put (*ci*, *uj k*) and (*vi*, *uj k*) in *A*∗ if (*i* + *j* = 3*q* + 1 and *k* = 3*p*) or (*i* + *j* = 3*q* + 2 and *k* = 3*p* − 2) or (*i* + *j* = 3*q* and *k* = 3*p* − 1) for some *p*, *q* ∈ N.

We will show that *A*∗ is an EOD set of *C*<sup>+</sup> *m C*<sup>+</sup> *n* . We already know that *A*<sup>+</sup> ⊆ *A*∗ is an EOD set of *C*<sup>+</sup> *m Cn* and we need to show that every vertex from *D* has exactly one in-neighbor in *A*<sup>∗</sup>. Notice that every (*xi*, *uj k*), where *x* ∈ {*<sup>c</sup>*, *<sup>v</sup>*}, has exactly three in-neighbors (*xi*−1, *uj <sup>k</sup>*),(*xi*−1, *uj <sup>k</sup>*−<sup>1</sup>) and (*xi*, *uj <sup>k</sup>*−<sup>1</sup>). (If *k* = 1, then we put *uj* 0 = *dj*.) We need to consider nine cases. They are presented in the following Table 1.

In the first two columns we present all nine options. The middle column contains the in-neighbor of (*xi*, *uj k*) from *A*∗ and the last two columns show why this is the in-neighbor of (*xi*, *uj k*) in *A*<sup>∗</sup>. Finally, we show that only one of the three in-neighbors of (*xi*, *uj k*) is in *A*<sup>∗</sup>. If (*xi*−1, *uj <sup>k</sup>*−<sup>1</sup>) ∈ *A*<sup>∗</sup>, then exactly one index of the other two in-neighbors differs by 1 from the same index of (*xi*−1, *uj <sup>k</sup>*−<sup>1</sup>) and they are

therefore not in *A*<sup>∗</sup>. By symmetry (*xi*−1, *<sup>u</sup>jk*−<sup>1</sup>) is also not in *A*∗ whenever either (*xi*−1, *ujk*) or (*xi*, *<sup>u</sup>jk*−<sup>1</sup>) is in *A*<sup>∗</sup>. So let (*xi*, *<sup>u</sup>jk*−<sup>1</sup>) ∈ *A*<sup>∗</sup>. In this case we can build a similar table as before, only that this table shows that (*xi*−1, *ujk*) is not in *A*<sup>∗</sup>. Similarly, also (*xi*, *<sup>u</sup>jk*−<sup>1</sup>) ∈/ *A*∗ when (*xi*−1, *ujk*) ∈ *A*<sup>∗</sup>.


**Table 1.** Nine cases considered that show that every vertex from *D* has exactly one in-neighbor in *A*<sup>∗</sup>.

To prove the contrary let *<sup>C</sup>*+*m <sup>C</sup>*+*n* be an EOD digraph with an EOD set *S*. If there exists an arc in a tree *Ti* that is not oriented away from the root, then we have a source in *Ti* and with that a contradiction with Lemma 1. Hence, all arcs of trees from *<sup>C</sup>*+*m* are oriented away from the root. If cycle *Cm* is not a directed cycle, then we have a source *cj* on *Cm* for some *j* ∈ [*m*]. Because all arcs of *Tj* are oriented away from the root *rj* = *cj*, we have a source *cj* in *<sup>C</sup>*+*m* as well, a contradiction with Lemma 1 again. Hence, *<sup>C</sup>*+*m* is well oriented. Also, *<sup>C</sup>*+*n* must be well oriented by the same arguments. Next we observe a subdigraph *Cm Cn* of *<sup>C</sup>*+*m <sup>C</sup>*+*n* . By the orientation of all arcs of all the trees, we see that there does not exist an arc from vertices of (*C*+*m <sup>C</sup>*+*n* ) − *V*(*Cm Cn*) to vertices of *Cm Cn*. Therefore, *Cm Cn* is an EOD digraph as well and by Theorem 6 we ge<sup>t</sup> *m* = 3 and *n* = 3*k* for some positive integers and *k*.

The above results give rise to the following conjecture. We believe that it is true, but the proof is a challenge.

**Conjecture 1.** *The strong product D F is an EOD digraph if and only if D* ∼= *<sup>C</sup>*+*m and F* ∼= *<sup>C</sup>*+*n are well oriented, m* = 3 *and n* = 3*k for some k*, ∈ N*.*

### **5. The Direct and the Lexicographic Product**

We conclude this paper with characterizations of the EOD digraphs among the direct and the lexicographic product. They follow from (1) and (2), respectively, and are no surprise. The following result for the direct product is an analogue of the result for the EOD graphs from [8] (under the name of total perfect codes).

**Theorem 8.** *Let D and F be digraphs. The direct product D* × *F is an EOD digraph if and only if D and F are EOD digraphs.*

**Proof.** Let *D* and *F* be EOD digraphs with EOD sets *SD* and *SF*, respectively. We will show that *SD* × *SF* is an EOD set of *D* × *F*. By (1) it holds that

$$V(D \times F) \subseteq \bigcup\_{(d,f) \in S\_D \times S\_F} N^+\_{D \times F}((d,f)).$$

Suppose there exists a vertex (*d*0, *f*0) that has two different in-neighbors (*d*, *f*) and (*d*, *f* ) in *SD* × *SF*. If *d* = *d*, then *f* = *f* , and by (1) we have

$$N\_{\mathcal{D}\times F}^{+}((d,f))\cap N\_{\mathcal{D}\times F}^{+}((d',f')) = \left(N\_{\mathcal{D}}^{+}(d)\times N\_{\mathcal{F}}^{+}(f)\right)\cap \left(N\_{\mathcal{D}}^{+}(d)\times N\_{\mathcal{F}}^{+}(f')\right) = N\_{\mathcal{D}}^{+}(d)\times \left(N\_{\mathcal{F}}^{+}(f)\cap N\_{\mathcal{F}}^{+}(f')\right).$$

Thus, *f*0 has two different in-neighbors *f* and *f* in *SF*. That is a contradiction since *SF* in an EOD set of *F*. If *f* = *f* , then *d* = *d* and we obtain a contradiction by symmetric arguments. Meaning that *d* = *d* and *f* = *f* . Again by (1) the vertex *d*0 has two different in-neighbors *d* and *d* in *SD* and *f*0 has two different in-neighbors *f* and *f* in *SF*, a contradiction with *SD* and *SF* being EOD sets of *D* and *F*, respectively. Therefore, no two vertices from *SD* × *SF* have a common out-neighbor, meaning that *D* × *F* is an EOD digraph.

Now let *D* × *F* be an EOD digraph and *S* be its EOD set. Let *f* ∈ *F* be an arbitrary vertex. Every vertex from *Df* has exactly one in-neighbor in *S*. Denote with *Sf* the set of all those vertices. We will show that *pD*(*Sf*) is an EOD set of *D*. Let *d* and *d* be two different vertices from *pD*(*Sf*). Choose *f* , *f* ∈ *V*(*F*) such that (*d*, *f* ),(*d*, *f* ) ∈ *Sf* . If there exists *d*0 such that *d* and *d* are its in-neighbors, then (*d*0, *f*) has two in-neighbors (*d*, *f* ) and (*d*, *f* ) in *S*, a contradiction with *S* being an EOD set of *D* × *F*. By (1) and because *S* is an EOD set of *D* × *F* it also holds that *V*(*D*) ⊆ *<sup>d</sup>*∈*pD*(*Sf*) *<sup>N</sup>*+*D* (*d*). Therefore, *pD*(*Sf*) is an EOD set of *D*, meaning that *D* is an EOD digraph. By symmetric arguments *F* is also an EOD digraph and with that the proof is completed.

The result for EOD digraphs among the lexicographic product of digraphs is an analogue to the graph version from [9].

**Theorem 9.** *Let D and F be digraphs. The lexicographic product D* ◦ *F is an EOD digraph if and only if*


**Proof.** Let *D* be a digraph on *n* vertices without edges and *F* be an EOD digraph. Then *D* ◦ *F* is isomorphic to *n* copies of *F* and since *F* is an EOD digraph, *n* copies of *F* also form an EOD digraph.

Now, let *D* be an EOD digraph, let *SD* be its EOD set and let *f*0 be a sink in *F*. We will show that *SD* × { *f*0} is an EOD set of *D* ◦ *F*. By (2) it holds that *<sup>N</sup>*+*D*◦*<sup>F</sup>*((*d*, *f*0)) = *<sup>N</sup>*+*D* (*d*) × *V*(*F*) since *f*0 is a sink in *F*. So *<sup>d</sup>*∈*SD <sup>N</sup>*+*D*◦*<sup>F</sup>*((*d*, *f*0)) equals *V*(*D* × *<sup>F</sup>*). If for *d*, *d* ∈ *SD* and *d* = *d* there exists a vertex in *D* ◦ *F* which in-neighbors are both (*d*, *f*0) and (*d*, *f*0), then there also exists a vertex in *D* which in-neighbors are both *d* and *d*. A contradiction with *SD* being an EOD set of *D*. Therefore, *D* ◦ *F* is an EOD digraph.

Conversely, let *D* ◦ *F* be an EOD digraph, *S* its EOD set and (*d*, *f*) ∈ *S* an arbitrary vertex. If *f* is not a sink in *F*, then there exists a vertex *f* ∈ *<sup>F</sup>d*, such that (*d*, *f*) is an in-neighbor of (*d*, *f* ). Denote with (*d*1, *f*1) the unique in-neighbor of (*d*, *f*) from *S*. If *d*1 = *d*, then (*d*, *f* ) has both (*d*, *f*) and (*d*1, *f*1) as its in-neighbors, which is not possible. Hence, *d*1 = *d*. If *d* has any out-neighbors, then for every out-neighbor *d* of *d* a vertex (*d*, *f*) has both (*d*, *f*) and (*d*1, *f*1) as its in-neighbors, a contradiction. So no *d* such that (*d*, *f*) ∈ *S* has any out-neighbors. Since every vertex (*d*, *f* ) ∈ *V*(*D* ◦ *F*) has exactly one in-neighbor (*d*, *f*) ∈ *S*, we conclude that *d* = *d* yields that *d* has at least one out-neighbor, which is not possible. Therefore, *d* = *d* and no *d* ∈ *V*(*D*) has any out-neighbors. Meaning that *D* is a digraph without arcs. To prove that *F* is an EOD digraph choose an arbitrary *F*-layer *F<sup>d</sup>* (which always induces a digraph isomorphic to *F*). Clearly, the vertices in *F<sup>d</sup>* that are also in *S* form an EOD set of (*D* ◦ *F*)[*F<sup>d</sup>*] ∼= *F*. So *F* is an EOD digraph and (*i*) follows.

Now assume *f* is a sink. Notice that in this case *F<sup>d</sup>* is a subset of all out-neighbors of (*d*0, *f*), where *d*0 is an in-neighbor of *d*. We will prove that *pD*(*S*) is an EOD set of *D*. Suppose it is not. Then there exist *d*, *d* ∈ *pD*(*S*) with a common out-neighbor. With this and (2) we have a contradiction with *S* being an EOD set of *D* ◦ *F*. Meaning that *pD*(*S*) is an EOD set of *D* and (*ii*) follows.
