Quasi-Kernels for Oriented Paths and Cycles

If D is a digraph, then   K V D  is a quasi-kernel of D if   D K is discrete and for each   y V D K   there is x K  such that the directed distance from y to x is less than three. We give formulae for the number of quasi-kernels and for the number of minimal quasi-kernels of oriented paths and cycles.


Introduction
Notation For a digraph D, and

 
A D denote its vertex set and arc set, respectively.If denotes the subdigraph of D induced by U and and denote, respectively, the inand out-neighborhoods of U. If , these latter sets may be written and .x K   such that the directed distance in D from y to x is either one or two;

 
2) Independence: there is no arc in   A D between vertices of K.
If K is a quasi-kernel for digraph D, x K  and   y V D K   and the directed distance from y to x in D is either one or two, then we will say that x quasi-absorbs y.
By [1], every digraph has at least one quasi-kernel.The number of quasi-kernels possessed by various classes of digraphs has been addressed in several papers.For example, in [2] Gutin et al. give necessary and sufficient conditions for a digraph to have at least three quasi-kernels.Clearly, any independent superset of a quasi-kernel is a quasi-kernel.In [3] we give sufficient conditions for a digraph to have at least three minimal quasi-kernels.In [4], Heard and Huang provide sufficient conditions for at least three disjoint quasi-kernels in certain classes of digraphs.In the present work, we address the problem of counting the number of quasi-kernels and the number of minimal quasi-kernels of oriented paths and cycles.
Notation For a digraph D, denotes the number of distinct quasi-kernels of D and denotes the number of minimal quasi-kernels of D.

 
Our approach is to recursively construct all quasikernels and all minimal quasi-kernels of various classes of digraphs in order to count them.The following general lemma will be useful for the examination of these special classes.
Lemma 1 If D 1 and D 2 are digraphs such that , where x is a sink in both and , then Proof.Using the notation from the statement of the lemma, if clearly satisfies the quasi-absorbing property for 1 2 and D has no arcs between a vertex of , then clearly both K 1 and K 2 are independent.Since x is a sink in D, a directed path in D lies completely in D 1 or D 2 .Thus the quasiabsorbing property is satisfied by K 1 in D 1 and by K 2 in D 2 .It follows that every quasi-kernel of D arises as the union of a quasi-kernel of D 1 and one of D 2 , so . It is clear that K 1 and K 2 are both minimal for D 1 and

Paths
By Lemma 1, to count the number of quasi-kernels in an arbitrary oriented path, it suffices to consider those with no internal sinks, i.e. the digraphs and defined below.
, , , , , , For , , m n denotes the result of identifying the source vertices of and n .We will denote the vertices where and , and 3) denotes the number of quasi-kernels of containing the vertex .
There are two quasi-kernels of containing 1 , viz.
and 

, thus
The results we obtain will be expressed in terms of the e function, so our first goal is to obtain a closed-form expression for its value.We start with a useful recursion relation.
then, by the quasi-absorbing property, K must contain either v 3 or v 4 and, by independence, K cannot contain both, nor can it contain the vertex v 2 .The equation follows.
For some of what follows, it is convenient to extend the domain of the function e using the above recursion relation backwards.Thus, since and , we set and the recursion relation as stated in Lemma 2 holds for .
It is straightforward to confirm that the characteristic equation of the above recursion relation, , has the following roots: where The solution of the recursion relation can be written , where x 1 , x 2 and x 3 are selected to satisfy the initial conditions It can also be shown that and that .Thus, for all , Since   e n is integer valued, the result follows.Theorem 2 For all integers     The only quasi-kernels of n which are not minimal are those that contain both v 1 and v 3 .There are clearly of these, so . Again, the recursion relation produces the desired result.
Theorem 3 For all integers , , 2 , respectively.The set of all quasi-kernels of , m n can be partitioned according to how the quasi-absorbing property is satisfied for the vertex s.Let K be a quasi-kernel of and consider cases.
is a quasi-kernel of containing

 
4) K contains one vertex from and one from but does not contain s.
Here, .Thus we have a final The table in Figure 1 depicts the relevant fragments of all possible quasi-kernels not containing the source.The possiblilities are sorted according to the subscripts on the v's and secondarily to the subscript on the u's.The leftmost column enumerates the cases for reference herein.The right-most column gives, in the case of a non-minimal quasi-kernel, an example of an extraneous vertex and, otherwise, the number of minimal quasi-kernels containing the given fragment.For example in case 1, v 2 could be removed from a quasi-kernel containing that fragment and leave a set which is still a quasi-kernel whereas in case 8, the given fragment could be extended by any of the   quasi-kernels of containing v 5 and, independently, any of the   , 5 , , containing u 3 to yield a minimal quasi-kernel of , m n .Certain of the vertices have been left off the table if their presence in the minimal quasi-kernel is neither mandatory nor forbidden in the case depicted.For example in row 7, u 5 is not depicted since, in the configuration shown, it is present in some minimal quasi-kernels and absent in others.
e m e m e n e n e m e n Summing the contributions over this mutually exclusive and exhaustive list of cases: The last two terms can be combined using Lemma 2 since


Combining this with the number already obtained for th s 2, 3

e m e n e m e m e n e m e n e m e n
ose that contain the source gives the desired result.As stated at the beginning of this section, Theorem and 4 used with Lemma 1 provide the number of quasikernels and the number of minimal quasi-kernels for arbitrary oriented paths.
Counting quasireduced to counting those of oriented paths.Of course any oriented cycle can be obtained by identifying the "leaf" vertices of an oriented path.The reverse process we shall call "splitting".
Theorem Proof.Observe first that every quasi-kernel of such a C and K is a quasi-kernel for C, then there are three such quasi-kernels.The quasi-absorbing erty requires that K contain at prop least one of the vertices v 1 , v 2 and v 3 .On the other hand, independence ensures that the cases above are mutually exclusive.

3 
provides the following solution.(The identity = 1 was used in obtaining these expressions.) omit the details, but this follows in a straightforward way, making use of the facts that As noted in its definition, the digraph , m n contains subdigraphs isomorphic to m and n .We will abuse the notation by using and

2 ) 3 )
K contains v 2 or v 3 , but not s, u 2 , nor u 3 .PNote that the quasi-absorbing property requires that 4 of m with first vertex either v 2 or v 3 and K contains u 2 or u 3 , but not s = v 1 , v 2 nor v 3 .This is the preceding case with m and n reversed.It contributes     3 e me n 1   additional quasi-kernels.

1 ls
are exactly the images under the is minimal, so     = m C k C .C is the result of iden- both v 4 and u 4 .It is easy to see that such a quasi-