Kernels by properly colored paths in arc-colored digraphs
Yandong Bai, Shinya Fujita, Shenggui Zhang

TL;DR
This paper investigates the existence of kernels by properly colored paths in arc-colored digraphs, proposing a conjecture and verifying it for specific classes like unicyclic digraphs, semi-complete digraphs, and bipartite tournaments.
Contribution
It introduces a conjecture on kernels by properly colored paths in all such digraphs and confirms it for several important classes, providing weaker conditions for some cases.
Findings
Conjecture that all cycles properly colored imply such kernels.
Verified the conjecture for unicyclic digraphs.
Confirmed the conjecture for semi-complete digraphs and bipartite tournaments.
Abstract
A {\em kernel by properly colored paths} of an arc-colored digraph is a set of vertices of such that (i) no two vertices of are connected by a properly colored directed path in , and (ii) every vertex outside can reach by a properly colored directed path in . In this paper, we conjecture that every arc-colored digraph with all cycles properly colored has such a kernel and verify the conjecture for unicyclic digraphs, semi-complete digraphs and bipartite tournaments, respectively. Moreover, weaker conditions for the latter two classes of digraphs are given.
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Taxonomy
TopicsAdvanced Graph Theory Research · graph theory and CDMA systems · Graph Labeling and Dimension Problems
Kernels by properly colored paths in arc-colored digraphs
††thanks: The first author is supported by NSFC (Nos. 11601430 and 11671320), NPU (No. 2016KY0101) and China Postdoctoral Science Foundation (No. 2016M590969); the second author is supported by JSPS KAKENHI (No. 15K04979); and the third author is supported by NSFC (Nos. 11571135 and 11671320). Part of this work was done while the first and the third author were visiting Yokohama City University and the hospitality was appreciated.
Yandong Bai a,, Shinya Fujita b, Shenggui Zhang a
a Department of Applied Mathematics, Northwestern Polytechnical University,
Xi’an 710129, China
b International College of Arts and Sciences, Yokohama City University,
Yokohama 236-0027, Japan Corresponding author. E-mail addresses: [email protected] (Y. Bai), [email protected] (S. Fujita), [email protected] (S. Zhang).
Abstract
A kernel by properly colored paths of an arc-colored digraph is a set of vertices of such that (i) no two vertices of are connected by a properly colored directed path in , and (ii) every vertex outside can reach by a properly colored directed path in . In this paper, we conjecture that every arc-colored digraph with all cycles properly colored has such a kernel and verify the conjecture for unicyclic digraphs, semi-complete digraphs and bipartite tournaments, respectively. Moreover, weaker conditions for the latter two classes of digraphs are given.
Keywords: kernel; kernel by monochromatic (properly colored, rainbow) paths
1 Introduction
All graphs (digraphs) considered in this paper are finite and simple, i.e., without loops or multiple edges (arcs). For terminology and notation not defined here, we refer the reader to Bang-Jensen and Gutin [1].
A path (cycle) in a digraph always means a directed path (cycle) and a -cycle means a cycle of length , where is an integer. For a digraph , define its kernel to be a set of vertices of such that (i) no two vertices of are connected by an arc in , and (ii) every vertex outside can reach by an arc in . This notion was originally introduced by von Neumann and Morgenster [21] in 1944. Since it has many applications in both cooperative games and logic (see [2, 3]), its existence has been the focus of extensive study, both from the algorithmic perspective and the sufficient condition perspective. Among them, the following results are of special importance. For more results on kernels, we refer the reader to the survey paper [4] by Boros and Gurvich.
Theorem 1** (Chvátal [6]).**
It is NP-complete to recognize whether a digraph has a kernel or not.
Theorem 2** (Richardson [18], von Neumann and Morgenster [21]).**
*Let be a digraph. Then the following statements hold:
if has no cycle, then has a unique kernel;
if has no odd cycle, then has at least one kernel;
if has no even cycle, then has at most one kernel.*
An arc is called symmetrical if . For a cycle , we call two arcs and crossing consecutive, where addition is modulo . The following theorem has been proved.
Theorem 3** (Duchet [8], Duchet and Meyniel [9], Galeana-Sánchez and Neumann-Lara[12]).**
*A digraph has a kernel if one of the following conditions holds:
each cycle has a symmetrical arc;
each odd cycle has two crossing consecutive arcs;
each odd cycle has two chords whose heads are adjacent vertices.*
It is worth noting that if we replace Condition (ii) in the definition of kernels by every vertex outside can reach by an arc or a path of length 2, then such a vertex subset, named quasi-kernel, always exists. This was proved by Chvátal and Lovász [7] in 1974. Jacob and Meyniel [17] furthermore showed in 1996 that every digraph has either a kernel or three quasi-kernels. For more results on quasi-kernels, see [5, 13, 16].
Let be a digraph and a positive integer. Call an -colored digraph if its arcs are colored with at most colors. Denote by the color assigned to the arc . A subdigraph of an arc-colored digraph is called monochromatic if all arcs of receive the same color, and is called rainbow if any two arcs of receive two distinct colors. Define a kernel by monochromatic paths (or an MP-kernel for short) of an arc-colored digraph to be a set of vertices of such that (i) no two vertices of are connected by a monochromatic path in , and (ii) each vertex outside can reach by a monochromatic path in .
The concept of MP-kernels in an arc-colored digraph was introduced by Sands, Sauer and Woodrow [19] in 1982. They showed that every 2-colored digraph has an MP-kernel. In particular, as a corollary, they showed that every 2-colored tournament has a one-vertex MP-kernel. Here note that each MP-kernel of an arc-colored tournament consists of one vertex. They also proposed the problem that whether a 3-colored tournament with no rainbow triangles has a one-vertex MP-kernel. This problem still remains open and has attracted many authors to investigate sufficient conditions for the existence of MP-kernels in arc-colored tournaments. Shen [20] showed in 1988 that for every -colored tournament with no rainbow triangles and no rainbow transitive triangles has a one-vertex MP-kernel, and also showed that the condition “with no rainbow triangles and no rainbow transitive triangles” cannot be improved for . In 2004, Galeana-Sánchez and Rojas-Monroythe [14] showed, by constructing a family of counterexamples, that the condition of Shen cannot be improved for , either. Galeana-Sánchez [10] showed in 1996 that every arc-colored tournament such that the arcs, with at most one exception, of each cycle of length at most four are assigned the same color has a one-vertex MP-kernel. Besides, Galeana-Sánchez and Rojas-Monroythe [15] showed in 2004 that every arc-colored bipartite tournament with all 4-cycles monochromatic has an MP-kernel. For more results on MP-kernels, we refer to the survey paper [11] by Galeana-Sánchez.
A subdigraph of an arc-colored digraph is called properly colored if any two consecutive arcs of receive distinct colors. Define a kernel by properly colored paths (or a PCP-kernel for short) of an arc-colored digraph to be a set of vertices of such that (i) no two vertices of are connected by a properly colored path in , and (ii) each vertex outside can reach by a properly colored path in .
By the definitions of kernels, MP-kernels and PCP-kernels, one can see in some sense that both MP-kernels and PCP-kernels generalize the concept of kernels in digraphs.
Observation 1**.**
*Let be a digraph. Then the following three statements are equivalent.
has a kernel;
-colored has an MP-kernel;
-colored has a PCP-kernel.*
In this paper we concentrate on providing some sufficient conditions for the existence PCP-kernels in arc-colored digraphs. For convenience, we write “PC path” for “properly colored path” in the following. Define the closure of an arc-colored digraph to be a digraph with vertex set and arc set there is a PC -path in . It is not difficult to see that the following simple (but useful) result holds.
Observation 2**.**
An arc-colored digraph has a PCP-kernel if and only if has a kernel.
2 Main results
We first consider the computational complexity of finding a PCP-kernel in an arc-colored digraph.
Proposition 1**.**
It is NP-hard to recognize whether an arc-colored digraph has a PCP-kernel or not.
Proof.
Let be a digraph and a set of vertices with . Let be the digraph with and , i.e., adding a set of new vertices to together with all possible arcs from to . We can always choose a with . Color by using colors in such a way that the subdigraph is monochromatic and the arc set is -colored. It is not difficult to see that the -colored has a PCP-kernel if and only if has a kernel. By Theorem 1 the computational complexity of the latter problem is NP-complete. The desired result then follows directly. ∎
Now we present the following result.
Proposition 2**.**
*An arc-colored digraph has a PCP-kernel if one of the following conditions holds:
has no cycle;
the coloring of is proper (consecutive arcs receive distinct colors);
is properly-connected (each vertex can reach all other vertices by a PC path).*
Proof.
Note that has a cycle if and only if has a cycle. The statement (i) therefore follows directly from Theorem 2 (i) and Observation 2. Assume that the coloring of is proper. If is strongly connected, then each vertex forms a PCP-kernel. If is not strongly connected, then the set of sinks is a PCP-kernel. If is properly-connected, then by the definition of PCP-kernels each vertex forms a PCP-kernel. ∎
By Proposition 2 (i), every arc-colored digraph containing no cycle has a PCP-kernel. It is natural to ask what is the analogous answer for a digraph containing cycles. For the simplest case, i.e., is a cycle, we get the following result.
Theorem 4**.**
An arc-colored cycle has a PCP-kernel if and only if it is not a monochromatic odd cycle.
Call a digraph unicyclic if it contains exactly one cycle. Note that every cycle is unicyclic. For general arc-colored unicyclic digraphs, furthermore, for general digraphs containing cycles, a number of examples (see for example the arc-colored digraphs in Figures 1 and 3) show that additional conditions are needed to guarantee the existence of PCP-kernels. But what kind of conditions do we need? By Proposition 2 (ii) and (iii), if the coloring is proper or “close” to proper (roughly speaking), then it has a PCP-kernel. By Proposition 2 (i), the existence of cycles influences the existence of PCP-kernels. This yields a natural question to ask whether the condition “all cycles are properly colored” suffices or not. Based on this consideration, we propose the following conjecture.
Conjecture 1**.**
Every arc-colored digraph with all cycles properly colored has a PCP-kernel.
Remark 1. If Conjecture 1 is true, then it is best possible in view of the two arc-colored digraphs in Figure 1, in which solid arcs, dotted arcs and dashed arcs represent arcs colored by three distinct colors respectively. It is not difficult to check that neither of them has a PCP-kernel. For any even integer (resp. odd integer ), the sharpness of Conjecture 1 can be shown by replacing the path (resp. ) of the left digraph (resp. the right digraph) by a monochromatic path of length (resp. length ) using the color assigned to the previous short path. One can check that neither of the two new constructed digraphs has a PCP-kernel.
A digraph is semi-complete if for every two vertices there exists at least one arc between them. A tournament (bipartite tournament) is an orientation of a complete graph (complete bipartite graph). Note that each tournament is semi-complete. Theorem 4 shows that Conjecture 1 holds for cycles. We will also show that Conjecture 1 holds for general unicyclic digraphs, semi-complete digraphs and bipartite tournaments. In fact, for the latter two classes of digraphs, weaker conditions have been obtained, respectively.
Theorem 5**.**
Every arc-colored unicyclic digraph with the unique cycle properly colored has a PCP-kernel.
Remark 2. We see from the two unicyclic arc-colored digraphs in Figure 1 that the condition “the unique cycle is properly colored” cannot be dropped in Theorem 5.
Note that every two vertices in a semi-complete digraph are adjacent and thus every PCP-kernel in such a digraph consists of one vertex. We obtain the following result whose proof idea is similar to that in [20].
Theorem 6**.**
Every arc-colored semi-complete digraph with no monochromatic triangles has a vertex such that all other vertices can reach by a PC path of length at most 3.
Corollary 1**.**
Every arc-colored semi-complete digraph with no monochromatic triangles has a PCP-kernel.
Remark 3. The condition “with no monochromatic triangles” in Theorem 6 and Corollary 1 cannot be dropped. Recall that every tournament is semi-complete and one can verify that the 2-colored tournament shown in Figure 2 has no PCP-kernels and no vertex defined in Theorem 6, in which solid arcs and dotted arcs represent arcs colored by two distinct colors, respectively. Larger -colored tournaments containing no PCP-kernel for general can be constructed by adding new vertices together with new colors to the new added arcs such that has no outneighbors in the set of new added vertices.
Theorem 7**.**
Every arc-colored bipartite tournament with all -cycles and -cycles properly colored, or , has a PCP-kernel.
Remark 4. The conditions in Theorem 7 cannot be dropped in view of the 3-colored bipartite tournament shown in Figure 3, in which solid arcs, dotted arcs and dashed arcs represent arcs colored by three distinct colors, respectively. One can see that and contains neither PC 4-cycles nor PC 6-cycles. One can also see that the closure of is semi-complete, in which the new added arcs are represented by thick dashed lines. Note that a semi-complete digraph has a kernel if and only if it has a vertex such that all other vertices can reach by an arc. One can see that does not contain such a vertex, so by Observation 2 we get that has no PCP-kernels. Furthermore, we can construct infinite family of bipartite tournaments which can show that the conditions in Theorem 7 cannot be dropped. Let be an arbitrary -colored bipartite tournament with . Define to be the union of and as follows: take all possible arcs between and going from to and denote this set of arcs by , let and , let the colors on and remain the same and let the coloring of be arbitrary. Then has no PCP-kernel since the proposition that has a PCP-kernel implies that has a PCP-kernel.
In the rest of the paper, we always use to denote for two digraphs and ; if consists of a single vertex , then we denote by . For two vertices and , if is an arc then we say dominates and sometimes write .
3 Proofs of Theorem 4 and Theorem 5
Proof of Theorem 4.
The necessity of Theorem 4 follows from the fact that each odd cycle has no kernel. For the sufficiency, it is equivalent to show that (i) every arc-colored odd cycle with at least two colors has a PCP-kernel and (ii) every arc-colored even cycle has a PCP-kernel. We prove the result by constructing such a kernel .
Let be an arc-colored cycle and assume w.l.o.g. that the vertices are located in a clockwise direction. If is an monochromatic even cycle, then we can let . Now assume that is an arc-colored cycle with at least two colors. If the coloring is proper, then clearly each vertex forms a PCP-kernel. Now assume that the coloring is not proper and assume w.l.o.g. that is a monochromatic path of maximum length (which is at least two). Put into , where is the largest integer such that . Here, note that since is neither a monochromatic odd cycle nor a PC cycle, we have and or . Afterwards, we consider, in a counter-clockwise direction, the first appeared maximal monochromatic path of length at least two in , say . Now put into , where is the largest integer such that . Continue this procedure until there is no monochromatic path of length at least two and let be the last appeared maximal monochromatic path of length at least two. It follows that
[TABLE]
where
[TABLE]
[TABLE]
, and addition is modulo . It is not difficult to check that no two vertices of are connected by a PC path in . For each , one can also verify that each vertex in can reach some vertex in by a PC path of length one, and each vertex in , can reach by a PC path; in other words, every vertex outside can reach by a PC path in . Therefore, the set is a PCP-kernel of . ∎
Proof of Theorem 5.
Let be an arc-colored unicyclic digraph with a PC cycle . Note that the cycle must be an induced cycle since otherwise two cycles will appear. Note also that each vertex of forms a PCP-kernel of . If is strongly connected, then is a cycle and the desired result follows directly. Now assume that is not strongly connected. Then there exist strongly connected components , , of such that there is no arc from to for any . Let be the component containing the cycle . One can see that . One can also see that each is a single vertex, since otherwise another cycle will appear. We distinguish two cases and show the result by constructing a PCP-kernel .
If , then let be an arbitrary vertex of and we put into . Let be the largest integer such that there is no PC -path. Put into . Let be the largest integer such that there is no PC -path. Put into . Continue this procedure until all the remaining vertices in can reach by a PC path. Let be the last vertex putting into . The terminal vertex set is clearly a PCP-kernel.
If , then contains at least one sink and we put all sinks, say , into . By similar procedure above we can put, step by step, the vertices with into . Let be the set of vertices which cannot reach the current by a PC path. If , then put an arbitrary vertex of (instead of all vertices of ) into and continue the procedure. If , then and we can use the same procedure above to get a PCP-kernel . ∎
4 Proof of Theorem 6
For convenience, in this proof, call a vertex good if all other vertices can reach by a PC path of length at most 3. One can see that it suffices to consider the tournament case. Let be an -colored tournament, where is a positive integer. For , note that each monochromatic tournament with no monochromatic triangles is transitive, then the unique sink is a good vertex. So we may assume that and is an arc-colored tournament with at least two colors. We prove the result by induction on .
Since each arc-colored transitive triangle and each non-monochromatic triangle has a good vertex, the result holds for . Now assume that is a minimum counterexample with . It follows that each -colored tournament with no monochromatic triangles and with order less than has a good vertex. So for each vertex of the subtournament has a good vertex. Denote by the good vertex of corresponding to the given coloring of . Then , since otherwise is a good vertex of . For two distinct vertices and , we claim that . If not, then by the definition of there exist a PC -path in and a PC -path in . It follows immediately that there exist a PC -path and a PC -path in . Thus, is a good vertex of , a contradiction.
Now consider the subdigraph induced on the arc set . Since each vertex of has both indegree and outdegree one, then consists of vertex-disjoint cycles. If has at least two cycles, then by induction hypothesis the induced subtournament on each cycle has a good vertex, which is obviously a good vertex of , a contradiction. So consists of one cycle.
Let . By the choices of the arcs, there exists no PC -path of length at most 3 in , addition is modulo in this proof. Consider the three vertices , if , then since there exists no monochromatic triangle we have either is PC -path of length 2 or is a PC -path of length 2, a contradiction. So . In fact, one can see from the simple proof that for any .
Let be the minimum integer such that and for any . Such an integer exists by the fact that . We may assume that for any . Since there exists no PC -path of length 2, we have , say . By assumption, there exists no monochromatic triangle, we have . Since there exists no PC -path of length 3, we have . Since is not a monochromatic triangle, we have . Similarly, we can show that and for any . This implies that and a PC -path of length 2 appears, a contradiction.
5 Proof of Theorem 7
Proof of Theorem 7 (i).
For the 1-colored case, by Observation 1, it suffices to consider the existence of a kernel. We claim that either or is a kernel. If is not a kernel, then there exists such that each vertex of is an inneighbor of , implying that is a kernel. So every 1-colored bipartite tournament has a PCP-kernel (not necessary to satisfy the required condition). In the following we assume and consider PCP-kernels in -colored bipartite tournaments with at least two colors.
We write if or . It is not difficult to verify, see also in [15], that the following lemma holds. We need to keep in mind of this lemma in the forthcoming proof.
Lemma 1** (Galeana-Sánchez and Rojas-Monroy [15]).**
*Let be an arc-colored bipartite tournament. Then
for each directed walk in we have iff (mod 2);
every closed directed walk of length at most 6 is a cycle in .*
For two vertices and in , denote by the distance from to . The following lemma will play a key role in the proof.
Lemma 2**.**
If there exists a PC -path but exists no PC -path in , then .
Proof.
Let be a shortest PC -path, where and . The result holds clearly for . Now let and assume the opposite that each -path has length at least 3.
Claim 1**.**
There exists no arc from to .
Proof.
The statement holds directly for . Assume that . Let . Then and . Now , is a 4-cycle and by assumption it is properly colored. So and is a PC -path of length less than , a contradiction. ∎
Claim 2**.**
There exists such that .
Proof.
Assume the opposite that for each . If is odd, then and either there exists a -path of length 1 or there exists a PC -path of length 1, a contradiction. So is even. Recall that . If , then and are PC 4-cycles. So is a PC -path, a contradiction. If , then since otherwise is a -path of length 2. Now is a PC 4-cycle and is a PC 6-cycle. Thus, is a PC -path, a contradiction. If , then and since otherwise either or is a -path of length 2. Besides, we have since otherwise and are PC 4-cycles and is a PC -path. We also can show that . If not, then is a PC 4-cycle and is a PC -path of length less than , a contradiction. Then there exist two PC 6-cycles and . It follows that is a PC -path, a contradiction. So from now on assume that .
We claim first that . If not, then is a PC 4-cycle and thus is a PC -path of length less than , a contradiction. We also claim that . If not, then since and are PC cycles we have and . It follows that is a PC -path of length less than , a contradiction.
Recall that for each and all 4-cycles and 6-cycles are properly colored. Thus,
[TABLE]
is a PC -path, contradicting the assumption in Lemma 2. ∎
Let be the minimum integer in such that and let . By Claim 1, we have . If , then ; otherwise, is a -path of length 2. By Claim 1, we also have . Since is a PC 4-cycle, we get that is a PC -path of length less than , a contradiction. So we have .
By the choice of , we have and is a PC 4-cycle. Hence . If , then is a PC -path of length less than , a contradiction. So .
By the minimality of we have . Since , we have . If , then is a PC 6-cycle and is a PC -path of length less than , a contradiction. So .
If , then by the choice of we have . Now , is a PC 6-cycle and is a PC -path of length less than , a contradiction. So . If , then is a PC 4-cycle and is a PC -path of length less than , a contradiction. So and .
Now we claim that . If not, then is a PC 4-cycle and is a PC -path of length less than , a contradiction. We may also claim that . If not, then is a PC 4-cycle and is a PC -path of length less than , a contradiction. Similarly, we can show that for any odd with . Clearly, there will be a -path of length at most 2.
The proof of Lemma 2 is complete. ∎
In view of Theorem 3 (i), it suffices to show that every cycle of has a symmetrical arc. Assume the opposite that there exists a cycle in containing no symmetrical arc and denote it by
[TABLE]
We will get a contradiction by showing that has a symmetrical arc. Here we distinguish two cases.
Case 1**.**
.
Since a bipartite tournament contains no odd cycle, there exists an arc of in , say . By Lemma 2, there exists a -path of length 2 in , say .
If , then is a (properly colored) 4-cycle in and . This implies that has a symmetrical arc , a contradiction.
If , then by Lemma 2 and Lemma 1 (ii) there will be a 5-cycle which contradicts the well-known fact that a bipartite tournament contains no odd cycle.
Now let . Then by Lemma 2, there exist a -path of length 2 and a -path of length 2 in , say and . By Lemma 1 (ii) and our assumption we get that is a PC 6-cycle. This implies that each arc in is a symmetrical arc, a contradiction.
Case 2**.**
.
In view of Lemma 2, there exists a -path of length at most 2 for each in , where . Let be the shortest -path in and let . Then is a closed directed walk in . For convenience, denote this closed walk by
[TABLE]
where and .
If , then is a PC 4-cycle and . Note that either or . This implies that has a symmetrical arc, a contradiction. Similarly, if , then we can show that either or is a symmetrical arc of , a contradiction.
Now assume that . Let be the minimum integer such that . Then is a PC 4-cycle in and . If , then and thus either or is a symmetrical arc in , a contradiction. So and . By the choice of , we have . Then is a PC 6-cycle and there exists a PC -path. So is a symmetrical arc in , a contradiction. ∎
Proof of Theorem 7 (ii).
If , then has no cycle and the result follows from Proposition 2 (i). So we can assume w.l.o.g. that . By contradiction, suppose the opposite that has no PCP-kernel. By Proposition 2 we can assume that has a cycle. It is not difficult to check that if is a source then has a PCP-kernel if and only if has a PCP-kernel. So we assume also that has no source in . Let and let
-path in D$$\},
.
If , then is a PCP-kernel. So we assume that . Two vertices and are called contractible if for any vertices and we have iff , iff , and whenever . Recall that all digraphs we consider here are simple, that is, contain no loops. So there exists no arc between any two contractible vertices. We now show the following claim.
Lemma 3**.**
Let be two contractible vertices in an arc-colored digraph . Then has a PCP-kernel if and only if has a PCP-kernel.
Proof.
For the necessity, let be a PCP-kernel of . If , then by the definition of contractible vertices is a PCP-kernel of . If and , then is a PCP-kernel of . If , then is also a PCP-kernel of . For the sufficiency, let be a PCP-kernel of . If , then is a PCP-kernel of . Now assume that . If there exists a PC -path, then is a PCP-kernel of . Otherwise, there exists no PC -path and is a PCP-kernel of . ∎
Now we assume that does not contain two contractible vertices and distinguish two cases in the following.
Case 1**.**
.
Since has no source in , each vertex in has one outneighbor and one inneighbor in . For a vertex which has at least one inneighbor in , since there exists no PC path from to , we have for any ; otherwise, for each with there exists with , which yields a PC path from to , a contradiction. For convenience, denote by the common color assigned for the arcs from to . By the definition of , the following claim holds.
Claim 1**.**
For two vertices , if for some , then .
Let be a maximal subset of such that no two vertices of are connected by a PC path. If , then is a PCP-kernel. Assume that . Let
: there exists no PC -path in D$$\}.
If , then is a PCP-kernel. So assume that . Let be an arbitrary vertex in . Then by the definitions of and , there exists a PC -path for some .
Claim 2**.**
Every PC -path has length 2.
Proof.
By contradiction, assume w.l.o.g. that there exists a PC -path of length 4, say , where . Since , we have and . We show that there exists a PC -path for any . If , then is a desired path. Now let . Since , we have either or . If , then since we have and is a desired path. If , then similarly and is a desired path. It follows that is a PCP-kernel, a contradiction. ∎
Now we can assume w.l.o.g. that is a PC -path. Remark that and, by Claim 1, each vertex with can reach by a PC path . Let
If , then is a PCP-kernel, a contradiction. So assume that .
Claim 3**.**
There exists no PC -path.
Proof.
Assume the opposite that there exists a PC -path, say , for some . Then since otherwise is a PC -path for each , contradicting that . Now we show that is a PCP-kernel. Since , we have for each . So is a PC path for each . For each , note that , since , we have . Then is a PC -path. It therefore follows that is a PCP-kernel. ∎
Claim 4**.**
There exists no PC path connecting two vertices of .
Proof.
By symmetry, assume that is a PC path for some two vertices . Note that , otherwise, there exists a PC -path , contradicting Claim 3. Since , we get that a PC -path, a contradiction. ∎
Let be the set of vertices which cannot reach by a PC path. By Claims 3 and 4, no two vertices of are connected by a PC path. It follows that is a PCP-kernel, a contradiction.
Case 2**.**
.
Recall that has no source in . By the assumption we have that every vertex in has one outneighbor and one inneighbor in . Let
[TABLE]
[TABLE]
In the following proof we need to keep in mind that each vertex in can reach by a PC path and each vertex in can reach by a PC path. If , i.e., there exists no PC path connecting and , then clearly is a PCP-kernel. Now let and assume w.o.l.g. that . If there exist with , then since each vertex in can reach by a PC path passing through either or we get that is a PCP-kernel. So we can assume that for each . Let
[TABLE]
We now claim that . Assume the opposite that . If , then since each vertex in can reach by a PC path passing through and an arbitrary vertex in we get that is a PCP-kernel. If and , then since each vertex in can reach by a PC path passing through , and each vertex in can reach by a PC path passing through together with an arbitrary vertex in and , we can get that is a PCP-kernel. If and , then by a similar analysis and the observation that no two vertices of are connected by a PC path we have that is a PCP-kernel. So .
If there exist with , then similar to the analysis for we have that is a PCP-kernel. Thus, we can assume that for each . For the sake of a better presentation, define the following vertex sets, see also in Figure 4 in which a vertex encircled may represent a set of vertices, and solid arcs, dotted arcs, dashed arcs represent respectively the arcs colored by , and a color not in .
[TABLE]
[TABLE]
[TABLE]
[TABLE]
Since contains no two contractible vertices, we have . Note that no two vertices of are connected by a PC path in and also the following holds.
[TABLE]
[TABLE]
We distinguish two subcases.
Subcase 2.1**.**
.
It follows that and . If , then each vertex in can reach by a PC path passing through and an arbitrary vertex in , which implies that is a PCP-kernel. If , then since each vertex in can reach by a PC path passing through , and each vertex in can reach by a PC path passing through together with an arbitrary vertex in and , together with the observation that no two vertices in are connected by a PC path, we can get that is a PCP-kernel.
Subcase 2.2**.**
.
If , then since we get that each vertex in can reach by a PC path passing through and an arbitrary vertex in . It follows that is a PCP-kernel.
If and , then each vertex in can reach by a PC path passing through , and each vertex in can reach by a PC path passing through , an arbitrary vertex in and . Recall that no two vertices of are connected by a PC path in . So is a PCP-kernel.
If and , then we can show that either or is a PCP-kernel. If , then each vertex in can reach by a PC path passing through , and each vertex in can reach by a PC path passing through , an arbitrary vertex in and . It follows that is a PCP-kernel. If , noting that no two vertices of are connected by a PC path, then we can similarly show that is a PCP-kernel.
Now assume that and . If , then , each vertex in can reach by a PC path passing through , an arbitrary vertex in and , and clearly every vertex in can reach by a PC path passing through . It follows that is a PCP-kernel. If and , then each vertex in can reach by a PC path passing through and each vertex in can reach by a PC path passing through . Recall that no two vertices of are connected by a PC path. Then is a PCP-kernel. If and , noting that no two vertices of are connected by a PC path, then since each vertex in can reach by a PC path passing through and each vertex in can reach by a PC path passing through , and , we can obtain that is a PCP-kernel. Now let and . If , then since each vertex in can reach by a PC path passing through , and each vertex in can reach by a PC path passing through , we can get that is a PCP-kernel. If , then by observing that no two vertices of are connected by a PC path we can similarly show that is a PCP-kernel. ∎
6 An extension
Recall that an arc-colored digraph is rainbow if any two arcs receive two distinct colors. Another interesting topic deserving further consideration is the existence of a kernel by rainbow paths in an arc-colored digraph , which is defined, similar to the definition of MP-kernels or PCP-kernels, as a set of vertices of such that (i) no two vertices of are connected by a rainbow path in , and (ii) every vertex outside can reach by a rainbow path in . Similar to the proof of Proposition 1, we can get the computational complexity of finding a kernel by rainbow paths in an arc-colored digraph.
Proposition 3**.**
It is NP-hard to recognize whether an arc-colored digraph has a kernel by rainbow paths or not.
Proof.
Let and be defined as in Proposition 1. Color by using colors in such a way that the subdigraph is monochromatic and the arc set is -colored. Then one can see that the -colored has a kernel by rainbow paths if and only if has a kernel. By Theorem 1 the desired result holds. ∎
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