## 1 Introduction

Connectivity is one of
the most basic concepts in graph theory^{1}^{1}1We refer the readers to [1, 4] for graph theoretical
notation and terminology not given here., both in combinatorial
and algorithmic senses. The classical connectivity has
two equivalent definitions. The connectivity of an undirected graph , written
, is the minimum size of a vertex set
such that is disconnected or has only one vertex. This
definition is called the cut-version definition.
The well-known theorem of Menger provides an equivalent
definition, which can be called the path-version definition.
For two distinct vertices and in ,
the local connectivity is the maximum number
of internally disjoint paths connecting and . Then is defined
to be the connectivity of .

The generalized -connectivity of a graph which was introduced by Hager [9] in 1985, is a natural generalization of the path-version definition of the connectivity. For a graph and a set of at least two vertices, an -Steiner tree or, simply, an -tree is a subgraph of which is a tree with . Two -trees and are said to be internally disjoint if and . The generalized local connectivity is the maximum number of internally disjoint -trees in . For an integer with , the generalized -connectivity is defined as

Observe that . If is disconnected and vertices of are placed in different connectivity components, we have . Thus, for a disconnected graph .

Both extremes for in relate to fundamental theorems in combinatorics. For , internally disjoint -trees are internally disjoint paths between the two vertices, and so the parameter is relevant to the well-known Menger theorem. For , internally disjoint -trees are edge-disjoint spanning trees of the graph, and so this parameter is relevant to the spanning tree packing problem [18, 19] and the classical Nash-Williams-Tutte theorem [17, 25]. Generalized connectivity of graphs has become an established area in graph theory, see a recent monograph [15] by Li and Mao on generalized connectivity of undirected graphs, see also a survey paper [14] of the area.

To extend generalized -connectivity to directed graphs, note that an -tree is a connected subgraph of containing . In fact, in the definition of we could replace “an -tree” by “a connected subgraph of containing .” Therefore, we define strong subgraph -connectivity by replacing “connected” with “strongly connected” (or, simply, “strong”) as follows. Let be a digraph of order , a -subset of and . Strong subgraphs containing are said to be -internally disjoint or, simply, internally disjoint if and for all .

Let be the maximum number of internally disjoint strong digraphs containing in . The strong subgraph -connectivity is defined as

By definition, if is not strong. Note that we define a digraph with one vertex to be strongly connected. Strong subgraph -connectivity allows us to extend applications of generalized -connectivity described in [14, 15] from undirected to directed graphs.

We will now overview results and conjectures on generalized -connectivity related to results and open problems of our paper. Li, Li and Zhou [13] showed that given a fixed positive integer , for any graph the problem of deciding whether can be solved in polynomial time. This was generalized by Li and Li [12] who proved that given two fixed positive integers and , for any graph the problem of deciding whether can be solved in polynomial time. For a fixed integer , but an arbitrary (i.e. part of input) integer , Li and Li [12] showed that the complexity changes provided PNP: Let be a fixed integer. For a graph a -subset of and an integer (), it is NP-complete to decide whether . Solving a conjecture of S. Li [11], Chen, Li, Liu and Mao [5] proved that in the above result, the bound 4 on can be replaced by 3 (which is the best possible provided PNP). Note that another conjecture of S. Li [11] remains open [14]: for a fixed integer , given a graph and an integer , it is NP-complete to decide whether Thus, the “global” analog of the generalized local connectivity intractability result still remains open. Li and Li [12] proved an intractability result similar to that of [5] given above when is fixed but is arbitrary: For a graph and a subset of , it is NP-complete to decide whether , where is a fixed integer.

It turns out that computing strong subgraph -connectivity becomes intractable much
earlier with respect to and above.
Let and be fixed integers.
In Theorem 2.1 by reduction from the Directed 2-Linkage
problem^{2}^{2}2The Directed -Linkage problem is formulated in the next section. we prove that deciding whether
is NP-complete for a -subset of .
Similarly to generalized -connectivity,
we do not know whether the problem of deciding is NP-complete for fixed and ,
but we conjecture that it is the case.

Thomassen [23] showed that for every positive integer there are digraphs which are strongly -connected, but which contain a pair of vertices not belonging to the same directed cycle. This implies that for every positive integer there are strongly -connected digraphs such that . Indeed, let and be vertices in a strongly -connected digraph such that no cycle contains both and . Suppose . Then there are -internally disjoint subgraphs and containing and . But then a path from to in and a path from to in form a cycle in , a contradiction.

The above negative results motivate
studying strong subgraph -connectivity for special classes of digraphs.
Arguably the most studied of them is the class of
tournaments, see, e.g., a recent informative account
[2] on tournaments and semicomplete digraphs by Bang-Jensen and Havet.
A digraph is semicomplete if there is at least one arc between
any pair of vertices. We show that the problem of deciding whether
for every semicomplete digraphs is
polynomial-time solvable for fixed and (Theorem 2.4).
This result can be viewed as an analog of the corresponding result of Li and Li [12] for
. The main tool used in our proof is a recent Directed
-Linkage theorem^{3}^{3}3Another interesting recent result on Directed
-Linkage was published in [6]. of Chudnovsky, Scott and Seymour
[7].

A digraph is called symmetric if for every arc there is an opposite arc . Thus, a symmetric digraph can be obtained from its underlying undirected graph by replacing each edge of with the corresponding arcs of both directions. We will say that is the complete biorientation of and denote this by We will show that for any connected graph , the parameter can be computed in polynomial time (Theorem 2.5). This result is best possible in the following sense, unless PNP. Let be a symmetric digraph and a fixed integer. Then it is NP-complete to decide whether for with (Theorem 2.8). To prove Theorem 2.8, we use an NP-complete problem from [5]. If we fix not only but also , the complexity changes again (unless PNP): in Theorem 2.10, we show that one can decide in polynomial time whether . To prove Theorem 2.10, we use the celebrated result of Robertson and Seymour [20] on the Undirected -Linkage problem.

Some inequalities concerning parameter were obtained in the literature, see e.g. [16, 22]. For a connected graph of order , Li, Mao and Sun [16] obtained the following inequality for : , where . Moreover, the upper and lower bounds are sharp. In the same paper, they also characterized graphs with .

Let be a strong digraph with order . For , we prove that (Theorem 3.5). The bounds are sharp; we also characterize those digraphs for which attains the upper bound. The main tool used in the proof of Theorem 3.5 is a Hamiltonian cycle decomposition theorem of Tillson [24].

For a positive integer , let

The paper is organized as follows. The next section is devoted NP-completeness results and polynomial algorithms discussed above. In Section 3 we prove sharp lower and upper bounds on strong subgraph -connectivity also discussed above. We conclude the paper with Section 4, where we discuss further direction of research on strong subgraph -connectivity and state some open problems.

## 2 Algorithms and Complexity

It is easy to decide whether for a digraph : it holds if and only if is strong. Unfortunately, deciding whether is already NP-complete for with , where is a fixed integer.

The well-known Directed -Linkage problem [1] is of interest in the next three theorems. The problem is formulated as follows: for a fixed integer , given a digraph and a (terminal) sequence of distinct vertices of decide whether has vertex-disjoint paths , where starts at and ends at for all

Let us prove our main intractability result.

###### Theorem 2.1

Let and be fixed integers. Let be a digraph and with . The problem of deciding whether is NP-complete.

Proof: Clearly, the problem is in NP. To show it is NP-hard, we reduce from the Directed 2-Linkage problem, which is NP-complete [8].

Let us first consider the case of and . Let be an instance of Directed 2-Linkage. Let us construct a new digraph (see Figure 1) by adding to vertices and arcs

Let It remains to show that is a positive instance of Directed 2-Linkage if and only if .

Let be a positive instance of Directed 2-Linkage with vertex-disjoint paths from to and from to , respectively. Then there are two internally disjoint strong subgraphs containing of , one induced by the arcs of and and the other by the arcs of and .

Let have two internally disjoint strong subgraphs containing . Since the in-degree of in is , we may without loss of generality assume that and . As has in-degree and we must have . As the out-degree of is , we analogously have (as ). So, for , both and are in . Therefore, there must be a path from to in and by definition of , will not have vertices outside of . As and are internally disjoint, the paths are disjoint.

Now let us consider the case of and . Add to copies of the 2-cycle and subdivide the arcs of every copy to avoid parallel arcs. Let us denote the new digraph by . Assume that there are internally disjoint strong subgraphs, , containing in . As the out-degree of in is we can without loss of generality assume that , and the (subdivided) arcs from to belong to , respectively. As and the in-degree of is no (subdivided) arc from to belongs to . Analogously, since and the out-degree of is , no (subdivided) arc from to belongs to . Therefore the (subdivided) arcs from to belong to , respectively. As in the case when we now note that and and that there therefore exists disjoint paths from to and to in , respectively.

Conversely if there exists disjoint paths from to and to in , then it is not difficult to create internally disjoint strong subgraphs containing in using the same approach as when as each (subdivided) -cycle also gives rise to a strong subgraph containing . Thus, we have proved the theorem in the case of and .

It remains to consider the case of and . Add to (where for ) new vertices and arcs of 2-cycles for each .
Subdivide the new arcs to avoid parallel arcs. Let . It is not hard to see that the resulting digraph has internally disjoint strong subgraphs if and only if is a positive instance of Directed 2-Linkage.

Recently, Chudnovsky, Scott and Seymour [7] proved the following powerful result, which was already used in [3].

###### Theorem 2.2

[7] Let and be fixed positive integers. Then the Directed -Linkage problem on a digraph whose vertex set can be partitioned into sets each inducing a semicomplete digraph and a terminal sequence of distinct vertices of , can be solved in polynomial time.

Now we will consider the problem of deciding whether for a semicomplete digraph . We will first prove the following:

###### Lemma 2.3

Let and be fixed positive integers. Let be a digraph and let be vertex disjoint subsets of , such that for all . Let and assume that every vertex in is adjacent to every other vertex in . Then we can in polynomial time decide if there exists vertex disjoint subsets of , such that and is strongly connected for each .

Proof: Let be the strongly connected components in , such that there is no arc from to for . We consider the following two cases.

Case 1: has a unique initial and a unique terminal component (which can be the same component) for all .

Let . For each , do the following. If is strongly connected then set and delete from . Otherwise, contract every strong component to a vertex and look at all possible subsequences, of which start with and end with . Let be such a subsequence, where , and . Now duplicate every vertex to and , for all and remove every that does not appear in the subsequence. We now add the sequence to our terminal sequence .

After having done the above for all we use Theorem 2.2 in order to determine if there are vertex disjoint paths satisfying our terminal sequence (that is, for every there is a path from to ). If such a linkage exists (for the terminal sequence of some subsequences above) then let include all internal vertices on paths between the pairs of vertices in as well as itself. Now observe that is strongly connected and all are vertex disjoint, as desired.

We will now show that if there exists , such that is strongly connected and all are vertex disjoint, then there exists a desired linkage. So, assume that such exist. As is strong, we note that it remains strong after contracting all strong components of to vertices. Thereafter there exists a shortest path from the terminal strong component of to the initial strong component of . Let the vertices on which correspond to (contracted) strong components of be (in the order they appear on ) and using this as the subsequence in our algorithm the subpaths of gives us the desired linkage between the ’s. Doing the above for all we see that our algorithm will indeed find the desired linkage (when considering the subsequences constructed above).

As and are constants, we note that there are at most a constant number of subsequences to consider, so the algorithm runs in polynomial time. This completes Case 1.

Case 2: Case 1 does not hold.

We will in this case transform the problem, such that we can solve it using Case 1. For all proceed as follows. Initiate a set as an empty set. If there is a unique initial strong component in and a unique strong terminal component in then let . If this is not the case, then let denote the set of initial strong components in and let denote the set of terminal strong components in . For every choose a vertex, such that has at least one arc into the component . We allow repetition of vertices in the sequence . (Such vertices must exist if there is a set containing such that is strong.) Analogously, for each choose a vertex, such that has at least one arc into it from the component . Again we allow to be not necessarily distinct. Now add vertices of and to .

If for some we cannot choose as above, we stop and consider other choices for the previous values of . Analogously, for .

If we have succeeded in choosing and for every , then for each we add the corresponding vertices and to and call the resulting set . Note that .

If is a terminal component in then must contain a vertex not in , as otherwise would be a terminal component of a contradiction to containing a vertex (not in and therefore not in ) that has an arc into it from . However, as all vertices not in are adjacent, this implies that there is a unique terminal strong component in . Analogously, there is a unique initial strong component in .

We now use the approach in Case 1, for all possible choices of vertices and for all . As there are at most possible choices of vertices and for each observe that we have to use the approach in Case 1 at most times, which is a polynomial as and are constants.

If the above algorithm finds the sets, , then clearly they exist.
Conversely, if the sets do exist then when is not strong, observe that each initial strong component in must
have an arc into it from a vertex in and each terminal strong component in must
have an arc into out of it to a vertex in . Picking these vertices as our vertices and , observe that
our algorithm will indeed find sets , as desired.

###### Theorem 2.4

For any fixed integers , we can decide whether for a semicomplete digraph in polynomial time.

Proof: Let be fixed and let be a set of vertices of a semicomplete digraph . To prove this theorem it suffices to show that deciding whether can be done in polynomial time.

Let be obtained from by replacing every by copies, i.e. replacing with for all . Let for all . Assume for now that is an independent set in (this will change later), but if is an arc from to , then is in for all . An analogously, if is an arc from to , then is in for all .

Let be a partition of the arcs in , where some sets may be empty. That is, every arc in belongs to exactly one . For each add the arcs of to . That is, if then add the arc to . This completes the construction of (for a given partition ).

We can now decide if there exist disjoint vertex sets in such that and is strongly connected for all in polynomial time by Lemma 2.3. If, for some partition, , such ’s exist then we will show that and if this is not the case then we will show that . As there are only a polynomial number of partitions (as and are constants), this gives us a polynomial algorithm.

First assume that such ’s exist for some partition, . Then the subgraph in on vertex set and with the arcs is strongly connected and as all ’s are vertex disjoint (and the arc sets ’s are disjoint) observe that , as desired.

Conversely if , then there exists strongly connected subgraphs
such that for all . Without loss of generality, we may assume that every arc of
belongs to some (as otherwise just add it to some ). Letting and observe that our
algorithm does find the desired ’s and we are done.

Now we turn our attention to symmetric graphs. We start with the following structural result.

###### Theorem 2.5

For every graph we have .

Proof: We may assume that is a connected graph. Let be a digraph whose underlying undirected graph is and let , where are distinct vertices of . Observe that . Indeed, let and let be -internally disjoint strong subgraphs of . Thus, by choosing a path from to in each , we obtain internally disjoint paths from to , which correspond to internally disjoint paths between and in . Thus, and it suffices to show that .

Let
for some . We know that
there are at least internally disjoint paths connecting
and in , say . For each
, we can obtain a strong subgraph containing
, say , in by replacing each edge
of with the corresponding arcs of both directions. Clearly,
any two such subgraphs are internally disjoint, so we have
and we are done.

Theorem 2.5 immediatly implies the following positive result, which follows from the fact that can be computed in polynomial time.

###### Corollary 2.6

For a graph , can be computed in polynomial time.

Theorem 2.5 states that when . However when , then is not always equal to , as can be seen by . Chen, Li, Liu and Mao [5] introduced the following problem, which turned out to be NP-complete.

CLLM Problem: Given a tripartite graph with a 3-partition such that , decide whether there is a partition of into disjoint 3-sets such that for every and is connected.

###### Lemma 2.7

[5] The CLLM Problem is NP-complete.

Now restricted to symmetric digraphs , for any fixed integer , the problem of deciding whether is NP-complete for with .

###### Theorem 2.8

For any fixed integer , given a symmetric digraph , a -subset of and an integer , deciding whether , is NP-complete.

Proof: It is easy to see that this problem is in NP. We divide our proof into two steps:

In the first step, let be a tripartite graph with 3-partition such that . We will construct a graph , a -subset and an integer such that there are internally disjoint -trees in if and only if is a positive instance of the CLLM Problem.

We define as follows: let and . Set and .

If there are internally disjoint -trees in , then each tree contains exactly a vertex from , a vertex from and a vertex from since for all . Furthermore, in each such tree, elements of have exactly one common neighbor in . Since these trees are internally disjoint, there is a partition of into disjoint sets each having three vertices, such that for every we have that , and is connected.

If there is a partition of into disjoint sets each having three vertices, such that for every we have , and is connected, then let be a spanning tree of together with the edge set , where . It is easy to see that are the desired internally disjoint -trees.

In the second step, we construct a symmetric digraph from by replacing each edge with the corresponding arcs of both directions. If there are internally disjoint -trees in , then for each such tree, we can get a strong subgraph containing in by replacing each edge with the corresponding arcs of both directions. Clearly, all these subgraphs of are internally disjoint and contain . If there are internally disjoint strong subgraphs containing , say , in , then each contains exactly a vertex from , a vertex from and a vertex from since . For every , let be a spanning tree of the underlying undirected graph of . Observe that are internally disjoint -trees in . We now have that there are internally disjoint -trees in if and only if there are internally disjoint strong subgraphs containing in .

Now, by Lemma 2.7 and the two steps above, we are done.

The last theorem assumes that is fixed but is a part of input. When both and are fixed, the problem of deciding whether for a symmetric digraph , is polynomial-time solvable. We will start with the following technical lemma.

###### Lemma 2.9

Let be fixed. Let be a graph and let be an independent set in with . For , let be any set of arcs with both end-vertices in . Let a forest in be called -acceptable if the digraph is strong and contains . In polynomial time, we can decide whether there exists edge-disjoint forests such that is -acceptable for all and for all .

Proof: Assume that there exists a set of required forests. Observe that if there is a leaf in a forest , can be deleted from and will remain the required set. Thus, we may assume that all leaves in are vertices of .

Below we will use the fact that in a tree without degree-2 vertices, the number of internal vertices is smaller than the number of leaves. This fact can be easily proved by induction by deleting a leaf. Let be a tree in a forest of and let be the tree obtained from by contracting all degree-2 vertices not belonging to . We will call the skeleton of . Note that may contain edges, that are not edges of (see Figure 2 for an example). By producing the skeleton of every tree of , we obtain the skeleton of the forest .

We will now bound the number of possible skeletons obtained from ’s. By Cayley’s formula, the number of distinct trees on labeled vertices is bounded by . By considering every tree on vertices, then assigning the vertices to vertices of and finally deleting a subset of edges in the tree, we obtain a forest in . Note that after deleting isolated vertices every skeleton obtained from an is created this way. Therefore the number of possible skeletons is bounded by

Note that the above number is a polynomial in as is considered constant, which implies that is constant. Thus, the number of distinct skeletons for the set is bounded by , which is still a polynomial in as is also considered to be a constant.

We now consider a set of skeletons where is -acceptable and for all . It remains to show that there is a polynomial-time algorithm for deciding whether there exists a set of required forests such that is the skeleton of . To obtain such an algorithm, we will use the celebrated result of Robertson and Seymour [20] that the Undirected -Linkage problem is polynomial-time solvable. For every forest and every make copies of in , such that all copies have the same neighbourhood as . If belongs to several forests in (which can happen if ) then do the above for every forest, implying that we increase the number of copies of several times. Let denote all copies of . Note that if then all vertices in are adjacent to all vertices in . For every edge in a forest in we now wish to find a path from a copy of to a copy of . We can decide if all such vertex-disjoint paths exist by the Undirected -Linkage problem.

One can show that at most one edge from to is used in such a collection of paths, however this property can also be guaranteed by for each edge adding a new vertex and adding all edges from to instead of the edges between and (then at most one path of length at most from to can be used, as it has to go through

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