In this subsection we will show that in any digraph G = ( V , E ) such that the outdegree and the indegree of any node is at least three, there are at least three edge-disjoint directed cycles of size ≥3.

To obtain the first cycle, start a forward trip T 1 = ( v 0 , v 1 , v 2 , ⋯ ) from any node v 0 . The strategy for building T 1 is to avoid returning to the node visited immediately before the current node (i.e., v i + 1 ≠ v i − 1 ). This is always possible because every node has an outdegree of at least 3. Because the total number of nodes in G is finite, T 1 cannot visit new nodes indefinitely and must eventually revisit a previously visited node. Let v k = v j (with j < k ) be the first repeated node. The subtrip ( v j , v j + 1 , ⋯ , v k ) forms a simple directed cycle C 1 . Since immediate backtracking was prohibited, k − j ≥ 3 , ensuring C 1 has size at least 3.

To obtain the second cycle, delete the arcs E ( C 1 ) from G . Because C 1 is a simple cycle, removing its arcs decreases both the indegree and outdegree of each node on C 1 by exactly 1. Thus, every node in G \ E ( C 1 ) retains an indegree and outdegree of at least 2. We then start a second trip T 2 in G \ E ( C 1 ) using the same non-backtracking rule ( v i + 1 ≠ v i − 1 ), which remains possible since the outdegree is at least 2. By finiteness, T 2 must close a loop to form a simple directed cycle C 2 . Avoiding immediate backtracking again guarantees that C 2 has size at least 3, and by construction, C 2 is arc-disjoint from C 1 .

To get a third cycle start a trip from any node using the same strategy as before: if possible do not return to the node visited right before the current node. This time it is not always possible since C 1 and C 2 might have used two outgoing arcs of some nodes. Let D ⊆ V be the set of those nodes.

Suppose at some point of our trip we do not yet have the third cycle and we are stuck at some node u , i.e., all the outgoing arcs of u have been used before.

Partition the whole trip T into two parts T 1 and T 2 where T 1 is the subtrip up to the first visit to u and T 2 is the subtrip after that point on. Note that T 2 is a strongly-connected balanced subgraph. Since T 2 does not contain any directed cycles of size ≥3 then it is a digon-tree, based on Lemma 1 (see Figure 3).

Figure 3. Partitioning the trip to T 1 and T 2 .

Note that the only common node of T 1 and T 2 is u ; otherwise, T would contain a cycle of size ≥3. Then any leaf v of the digon-tree T 2 is from D ; otherwise, based on our strategy we wouldn’t return from v to the node visited right before it (if u is a leaf of T 2 then u ∈ D , otherwise we would not be stuck at u ). Thus, T 2 contains at least two nodes from D , and so at least two nodes

from C 1 . Let w 1 , w 2 ∈ V ( C 1 ) ∩ V ( T 2 ) such that i n t ( w 1 ⇝ T 2 w 2 ) ∩ V ( C 1 ) = ∅ . Then Q 1 = w 1 ⇝ C 1 w 2 ⇝ T 2 w 1 and Q 2 = w 1 ⇝ T 2 w 2 ⇝ C 1 w 1 are edge-disjoint cycles of size

≥3 (see Figure 4).

Figure 4. Getting the third cycle.

So we have three edge-disjoint directed cycles of size ≥3: Q 1 , Q 2 and C 2 .

Let C 1 , C 2 and C 3 be the first three edge-disjoint cycles and p be the number of their common nodes. We show the existence of the fourth cycle by induction on p .

p = 0 . The arguments to get the third cycle in subsection 3.1 are still valid in this case to get a fourth cycle. Here D is defined as the set of the nodes shared by exactly two of the three cycles. Note that any two nodes from D are on the same cycle C i for some i ∈ { 1 , 2 , 3 } . Suppose p = 1 , and x is the only common node. This time D is defined as the set of the nodes shared by two or three cycles. Three cases are possible:

◇ if x is balanced, i.e., outdegree ( x ) = indegree ( x ) , then we can get a fourth cycle by starting a trip from any node v ≠ x and applying the arguments of subsection 3.1.

◇ if outdegree ( x ) > indegree ( x ) then taking a forward trip from u will give a fourth cycle.

◇ if outdegree ( x ) < indegree ( x ) then taking a backward trip from x and applying arguments dual to those in subsection 3.1 will give a fourth cycle.

p > 1 . Suppose a fourth cycle exists whenever C 1 , C 2 and C 3 have less than p common nodes. Let’s show it for the case of p common nodes. Let U = { u 1 , ⋯ , u p } be the set of the common nodes.

Two cases are possible.

1) While traversing C 2 and C 3 starting from u 1 we meet the common nodes in the same order as in C 1 .

2) We meet them in a different order.

Case 1:

Subcase1.1: i n t ( u i ⇝ C k u i + 1 ) does not have any intersection with i n t ( u j ⇝ C l u j + 1 ) for i ≠ j , k ≠ l . Then let i n t ( u i ⇝ u i + 1 ) denote the set of internal nodes of u i ⇝ C k u i + 1 , k = 1 , 2 , 3 .

Start a forward trip T i from a node w ∈ i n t ( u i ⇝ u i + 1 ) .

Suppose we get stuck in i n t ( u i ⇝ u i + 1 ) before hitting any node outside it. If we get a cycle of size ≥3 inside i n t ( u i ⇝ u i + 1 ) , then we have a fourth cycle. Suppose we do not have a cycle and get stuck in node v ∈ i n t ( u i ⇝ u i + 1 ) , by getting a digon at the end of the trip. Then v is visited more than once, and the subtrip of T i after visiting v the first time is a digon-tree D (could be just a digon like in Figure 5). Note that each leaf of the digon-tree belongs to two of the three cycles, based on our strategy of avoiding digons in the trips. Take two nodes w 1 and w 2 on D such that (i) each of them belongs to two of the three cycles, (ii) no other node on the D -path connecting them belongs to two cycles. Then one of the three cycles, say C 1 , contains both w 1 and w 2 . If there are other nodes on the D -path connecting w 1 and w 2 that belong to C 1 , then we can replace w 1 and w 2 with other nodes on D that belong to C 1 and the D -path connecting them does not have nodes from C 1 . Suppose we encounter w 1 before w 2 while traversing cycle C 1 from u i to u i + 1 . Then we can replace C 1 with cycle Q 1 = w 1 ⇝ C 1 w 2 ⇝ D w 1 which has size ≥3 (see Figure 5).

Next, we show that Q 1 , C 2 , C 3 share no common nodes. Any node on the path w 1 ⇝ C 1 w 2 belongs to C 1 and to at most one of C 2 and C 3 ; thus, replacing C 1

with Q 1 cannot create a node common to all three cycles Q 1 , C 2 , C 3 . Furthermore, by the construction of D (part (ii) above), no interior node of D belongs to more than one of C 1 , C 2 , C 3 . More specifically, each interior node of D (if any exist) belongs to exactly one of C 2 and C 3 . Consequently, after replacing C 1 with Q 1 , each interior node of D will belong to Q 1 and to exactly one of C 2 and C 3 . In summary, Q 1 , C 2 , C 3 have no common nodes, allowing us to apply the induction hypothesis.

Figure 5. Example of Subcase 1.1.

Suppose we hit a node outside i n t ( u i ⇝ u i + 1 ) , and the first node of V ( C 1 ) ∪ V ( C 2 ) ∪ V ( C 3 ) \ i n t ( u i ⇝ u i + 1 ) hit by the trip is v ≠ u i + 1 . This is a situation similar to subcase 1.2 of Section 2: trip T i will give a shortcut for one of the three cycles. Let w ′ i be the last node from i n t ( u i ⇝ u i + 1 ) visited before v in trip T i . Then the path w ′ i ⇝ T i v can serve as a shortcut for one of the original

cycles, say C 1 . The new cycle C ′ 1 = w ′ i ⇝ T i v ⇝ C 1 w ′ i does not contain u i + 1 .

By construction, the path v ⇝ C 1 w ′ i contains the exact same set of common nodes

as before, with the exception of u i + 1 , which is excluded because u i + 1 ∉ V ( C ′ 1 ) .

Furthermore, the interior of the path w ′ i ⇝ T i v contains no nodes from

V ( C 1 ) ∪ V ( C 2 ) ∪ V ( C 3 ) . Consequently, the number of nodes common to C ′ 1 , C 2 , and C 3 is strictly less than p (specifically, p − 1 ), allowing us to obtain a fourth cycle by induction.

Suppose we hit a node outside i n t ( u i ⇝ u i + 1 ) , and the first node of V ( C 1 ) ∪ V ( C 2 ) ∪ V ( C 3 ) \ i n t ( u i ⇝ u i + 1 ) hit by the trip is u i + 1 . This case is handled like subcase 1.3 of Section 2. If the forward trip T i hits u i + 1 and the backward trip S i hits u i for all i ∈ 1 , ⋯ , p , then the union of all forward and backward trips will give a fourth cycle.

Figure 6. Example of Subcase 1.2 in the proof for the fourth cycle.

Subcase1.2: i n t ( u i ⇝ C k u i + 1 ) has intersection with i n t ( u j ⇝ C l u j + 1 ) for i ≠ j ,

k ≠ l . Let k = 1 , l = 2 , and w is an intersection node. Then define two new

cycles: Q 1 = w ⇝ C 1 u i + 1 ⇝ C 1 u j ⇝ C 2 w , Q 2 = w ⇝ C 2 u j + 1 ⇝ C 2 u i ⇝ C 1 w (see Figure 6).

Number of common nodes of cycles Q 1 , Q 2 , C 3 is less than the number of common nodes of cycles C 1 , C 2 , C 3 . Particularly, u i , u i + 1 , u j , u j + 1 are not common nodes now.

On the other hand, no new common nodes are created. For instance, if the path u i ⇝ C 1 w in Q 2 intersects the path u j ⇝ C 2 w in Q 1 as well as C 3 , then any such intersection would also be a common node of C 1 , C 2 , and C 3 . Furthermore, since C 1 is a simple directed cycle, the path u i ⇝ C 1 w in Q 2 cannot share any vertices with the paths w ⇝ C 1 u i + 1 (except for w ) or with u i + 1 ⇝ C 1 u j in Q 1 . The arguments for the remaining pairs of paths, one from Q 1 and the other from Q 2 , are identical.

Even in the case when u j coincides with u i + 1 and u i coincides with u j + 1 , the number of common nodes will go down by two.

Thus, we can apply induction in this case.

Case2: u i ’s are encountered in different order while traversing the three cycles. Note that in this case the number of common nodes is at least three.

Subcase2.1: The order of u i ’s is σ or σ − 1 (the opposite order of σ ) for the cycles. Let C 1 and C 2 have the same order σ while C 3 has order σ − 1 .

If i n t ( u i ⇝ C 1 u i + 1 ) has intersection with i n t ( u j ⇝ C 2 u j + 1 ) for i ≠ j , then we can apply the idea of Subcase 1.2.Otherwise we can interchange u i ⇝ C 1 u i + 1 and u i ⇝ C 2 u i + 1 for any i . Let B i 1 = u i ⇝ C 1 u i + 1 ⇝ C 3 u i and B i 2 = u i ⇝ C 2 u i + 1 ⇝ C 3 u i . B i 1 and B i 2 are strongly-connected balanced digraphs. For given i , B i 1 and B i 2 cannot be both digon-trees since u i ⇝ C 1 u i + 1 and u i ⇝ C 2 u i + 1 are different. For every i , interchange u i ⇝ C 1 u i + 1 and u i ⇝ C 2 u i + 1 so that B i 1 contains a directed cycle Q i of size ≥3. We have at least three edge-disjoint Q i ’s obtained this way because the number of common nodes of C 1 , C 2 , C 3 is at least three in this case. Since we also have C 2 , then the number of directed cycles of size ≥3 is at least four.

Subcase 2.2: Two of the three cycles have neither the same nor the opposite directions. Let C 1 and C 2 be those two cycles. Let u 1 , u 2 , ⋯ , u p be the order we encounter the common nodes on cycle C 1 . Then for some i , the node following u i on cycle C 2 is neither u i + 1 nor u i − 1 . Suppose that node is u j (see Figure 7).

Figure 7. Example of Subcase 2.2.

Let Q 1 = u i ⇝ C 2 u j ⇝ C 1 u i and Q 2 = u i ⇝ C 1 u j ⇝ C 2 u i . Q 1 and Q 2 are strongly-connected balanced graphs. Since u i − 1 ∈ { u j ⇝ C 1 u i } \ { u i ⇝ C 2 u j } , Q 1 cannot be a digon-tree and contains a directed cycle Q ′ 1 of size ≥3. Similarly, since u i − 1 ∈ { u j ⇝ C 2 u i } \ { u i ⇝ C 1 u j } , Q 2 cannot be a digon-tree and contains a directed cycle Q ′ 2 of size ≥ 3.

Note that the set U ′ of common nodes of Q ′ 1 , Q ′ 2 , C 3 is a subset of U . But u i + 1 ∉ V ( Q ′ 1 ) . Thus, u i + 1 ∉ U ′ , | U ′ | < | U | = p and taking Q ′ 1 , Q ′ 2 and C 3 as the first three cycles we will get a fourth cycle by induction.

Note that in a worst-case scenario four cycles is the best we can achieve. To illustrate that, consider the following example. Let G = ( V , E ) be a complete digraph on a 4-node set V = { x , y , z , u } . In this graph we can find at most four edge-disjoint directed cycles of size ≥3: C 1 = x → y → u → x , C 2 = y → x → z → y , C 3 = u → z → x → u , C 4 = z → u → y → z . This is not surprising since the total number of edges is only 12 in this example.

The main algorithm given in this subsection is similar to the one given in [8] for the undirected case.

First consider the following related problem. The input is a directed graph G with nonnegative weights on the edges, a root vertex r and an integer p . The goal is to find a minimum-weight directed subgraph H of G , such that for each vertex v there are p openly disjoint paths from r to v (alternately, from v to r for each v ) in the subgraph H . Frank and Tardos [10] gave an algorithm which solves this problem in polynomial time. We will use that algorithm as a subroutine for solving our problem.

In the algorithm for solving the k -vertex-connectivity problem, using a set of root vertices R plays an important role. Below we will discuss two subroutines that use root vertices. Those are called ROOT-VERTICES-IN ( G , p , R ) and ROOT-VERTICES-OUT ( G , p , R ). The input of both subroutines is a k -vertex-connected directed graph G = ( V , E ) , a nonnegative weight function w defined on the edges, and a set R of p vertices. The output of ROOT-VERTICES-IN ( G , p , R ) (ROOT-VERTICES-OUT ( G , p , R )) is an approximate minimum-weight subgraph of G in which there are p openly disjoint paths from any vertex v to R (from R to any vertex v ). Note that v could be in R , in which case we find p − 1 paths from v to the remaining vertices in R (from remaining vertices of R to v ).

The algorithm ROOT-VERTICES-OUT ( G , p , R ) works as follows. Augment G by adding a new vertex r and add p new directed edges of weight 0 from r to each vertex in R . Use the algorithm of [10] on this graph with r as the root and find a minimum-weight subgraph H o u t with p openly disjoint paths from r to any vertex of G . Since these disjoint paths go through the vertices of R , for any vertex v in G , there are p openly disjoint paths from R to v in H o u t .

The algorithm ROOT-VERTICES-IN ( G , p , R ) works similarly. Augment G by adding a new vertex r and add p new directed edges of weight 0 from each vertex in R to r . Use the algorithm of [10] on this graph with r as the root and find a minimum-weight subgraph H i n with p openly disjoint paths from any vertex of G to r . Since these disjoint paths go through the vertices of R , for any vertex v in G , there are p openly disjoint paths from v to R in H i n .

Let S i n and S o u t be the subgraphs of G obtained from correspondingly H i n and H o u t by deleting r and its adjacent edges.

Lemma 2There are p openly disjoint paths from R to v in S o u t .For any v ∈ R , there are p − 1 openly disjoint paths from R − v to v in S o u t .There are p openly disjoint paths from v to R in S i n .For any v ∈ R , there are p − 1 openly disjoint paths from v to R − v in S i n .

Lemma 3The weight of S i n ( S o u t )is not more than the weight of a minimum-weight p -vertex-connected spanning subgraph of G .

Proof: We will give the proof for S o u t ; the proof for S i n is similar. Let T be a minimum-weight p -vertex-connected spanning subgraph of G . Add the zero-weight edges from r to the vertices of R to obtain a new subgraph T ′ . T ′ has p openly disjoint paths from r to each vertex of G . Since the algorithm of [10] returns a minimum-weight solution H o u t satisfying that property, we have that

The algorithm for the k -vertex-connectivity problem first finds a subset R of k vertices such that the complete graph induced on R is relatively low-weight. Then it calls ROOT-VERTICES-IN( G , p , R ) and ROOT-VERTICES-OUT( G , p , R ) to obtain subgraphs S i n and S o u t of G . The algorithm returns S i n ∪ S o u t ∪ K R as its output, where K R is the complete subgraph on R .

First, we show that the output of the algorithm is k -vertex-connected, regardless how we find R .

Lemma 4The subgraph S i n ∪ S o u t ∪ K R is k -vertex-connected.

Proof: The proof by contradiction. Assume that the graph contains a vertex cut C with | C | < k , i.e., the removal of C breaks G into strongly-connected components C 1 , C 2 , ⋯ , C l , where l ≥ 2 . Since K R is a complete subgraph of R , all vertices of R remaining after the removal of C belong to C i for some i . Consider a vertex v ∈ C j for some j ≠ i . By lemma 2 there must be k openly disjoint paths from v to R and k openly disjoint paths from R to v in S i n ∪ S o u t ∪ K R . Thus, the removal of less than k nodes will leave v strongly connected to the remaining nodes of R . This is a contradiction since v ∉ C i . □

The key of getting a good approximation factor is to find a low-weight K R . Next, we will show how to achieve that for k = 2 , 3 by using our results in sections 2 and 3.

For k = 2 , the subset R is chosen as the nodes of the minimum-weight digon in G . It can be found in O ( n 2 ) time since there are ( n 2 ) possible digons.

Theorem 1The total weight of S i n ∪ S o u t ∪ K R is at most 2 1 3 times the optimum.

Proof: Let T be a minimum-weight 2-vertex-connected spanning subgraph of G . Based on the result of section 2, T has at least three edge-disjoint directed cycles: C 1 , C 2 , C 3 . Since each digon is itself a cycle,

Based on lemma 3,

For k = 3 , the subset R is chosen as the nodes of the minimum-weight directed

cycle C of size 3 in G . It can be found in O ( n 3 ) time since there are 2 ⋅ ( n 3 )

possible directed cycles of size 3.

Theorem 2The total weight of S i n ∪ S o u t ∪ K R is at most 2 3 4 times the optimum.

Proof: Let T be a minimum-weight 3-vertex-connected spanning subgraph of G . Based on the result of section 3, T has at least four directed cycles of size ≥3: C 1 , C 2 , C 3 , C 4 . For any i = 1 , 2 , 3 , 4 , let C i ' be a directed cycle of size 3 obtained from C i by taking shortcuts if necessary. Based on the triangle inequality, weight ( C ′ i ) ≤ weight ( C i ) , for i = 1 , 2 , 3 , 4 . Therefore, since C is the minimum-weight directed cycle of size 3,

Thus,

Let C ′ be the cycle of the opposite orientation to C , that is, if C = v 1 → v 2 → v 3 → v 1 then C ′ = v 1 → v 3 → v 2 → v 1 . Based on the triangle inequality,

Adding up the left and right hand sides of the above three inequalities will result in

That is, the weight of the cycle of the opposite orientation to C is no more than twice the weight of C . Based on (1) and (2),

Using also the results of lemma 3,

The total running time of the algorithm is dominated by finding the set R and executing the two calls to the Frank–Tardos subroutine, ROOT-VERTICES-IN ( G , p , R ) and ROOT-VERTICES-OUT ( G , p , R ). For an input digraph G = ( V , E ) with n = | V | and m = | E | :

Root Set Identification: For k = 2 , finding the minimum-weight digon requires O ( n 2 ) time by inspecting all ( n 2 ) vertex pairs. For k = 3 , finding the minimum-weight directed 3-cycle takes O ( n 3 ) time by inspecting all 2 ( n 3 ) candidate 3-cycles. Subroutine Executions: For constant p = k ∈ { 2 , 3 } , the subroutines ROOT-VERTICES-IN or ROOT-VERTICES-OUT have O ( n 2 m ) running time [10].

Since m ≥ n in strongly connected digraphs, the O ( n 2 m ) subroutine phase bounds the total execution time. Thus, the main algorithm achieves an overall time complexity of O ( n 2 m ) for both k = 2 and k = 3 .