Four Edge-Independent Spanning Trees
Alexander Hoyer, Robin Thomas

TL;DR
This paper proves a theorem about 4-edge-connected graphs, showing they contain four spanning trees with edge-disjoint paths to a root, and provides a polynomial-time method to construct such trees.
Contribution
It introduces an ear-decomposition theorem for 4-edge-connected graphs and demonstrates the existence and construction of four edge-independent spanning trees.
Findings
Existence of four edge-independent spanning trees in 4-edge-connected graphs
Polynomial-time algorithm for constructing these trees
Extension of ear-decomposition techniques to spanning tree construction
Abstract
We prove an ear-decomposition theorem for -edge-connected graphs and use it to prove that for every -edge-connected graph and every , there is a set of four spanning trees of with the following property. For every vertex in , the unique paths back to in each tree are edge-disjoint. Our proof implies a polynomial-time algorithm for constructing the trees.
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a
FOUR EDGE-INDEPENDENT SPANNING TREES 111Partially supported by NSF under Grant No. DMS-1202640. 222Presented at the 29th Cumberland Conference on Combinatorics, Graph Theory and Computing at Vanderbilt University.
Alexander Hoyer
and
Robin Thomas
School of Mathematics
Georgia Institute of Technology
Atlanta, Georgia 30332-0160, USA
ABSTRACT
We prove an ear-decomposition theorem for -edge-connected graphs and use it to prove that for every -edge-connected graph and every , there is a set of four spanning trees of with the following property. For every vertex in , the unique paths back to in each tree are edge-disjoint. Our proof implies a polynomial-time algorithm for constructing the trees.
March 2017. Revised October 2017.
1 Introduction
If is a vertex of a graph , two subtrees of are edge-independent with root if each tree contains , and for each , the unique path in between and is edge-disjoint from the unique path in between and . Larger sets of trees are called edge-independent with root if they are pairwise edge-independent with root .
Itai and Rodeh [6] posed the Edge-Independent Tree Conjecture, that for every -edge-connected graph and every , there is a set of edge-independent spanning trees of rooted at . Here, we prove the case of the Edge-Independent Tree Conjecture. That is, we prove the following:
Theorem 1**.**
If is a -edge-connected graph and , then there exists a set of four edge-independent spanning trees of rooted at .
There is a similar conjecture which has been studied in parallel, concerning vertices rather than edges. If is a vertex of , two subtrees of are independent with root if each tree contains , and for each , the unique path in between and is internally vertex-disjoint from the unique path in between and . Larger sets of trees are called independent with root if they are pairwise independent with root .
Itai and Rodeh [6] also posed the Independent Tree Conjecture, that for every -connected graph and for every , there is a set of independent spanning trees of rooted at .
The case of each conjecture was proven by Itai and Rodeh [6]. The case of the Independent Tree Conjecture was proven by Cheriyan and Maheshwari [1], and then independently by Zehavi and Itai [11]. Huck [5] proved the Independent Tree Conjecture for planar graphs (with any ). Building on this work and that of Kawarabayashi, Lee, and Yu [7], the case of the Independent Tree Conjecture was proven by Curran, Lee, and Yu across two papers [2, 3]. The Independent Tree Conjecture is open for nonplanar graphs with .
In 1992, Khuller and Schieber [8] published a later-disproven argument that the Independent Tree Conjecture implies the Edge-Independent Tree Conjecture. Gopalan and Ramasubramanian [4] demonstrated that Khuller and Schieber’s proof fails, but salvaged the technique, and proved the case of the Edge-Independent Tree Conjecture by reducing it to the case of the Independent Tree Conjecture. Schlipf and Schmidt [10] provided an alternate proof of the case of the Edge-Independent Tree Conjecture, which does not rely on the Independent Tree Conjecture. The case of the Edge-Independent Tree Conjecture is proven here, while the case remains open.
By adapting the technique of Schlipf and Schmidt [10], we prove an edge analog of the planar chain decomposition of Curran, Lee, and Yu [2]. We then use this decomposition to create two edge numberings which define the required trees.
The conjectures are related to network communication with redundancy. If represents a communication network, one can wonder if information can be broadcast through the entire network with resistance to edge failures (i.e. it would require simultaneous edge failures to disconnect a client from every broadcast). The Edge-Independent Tree Conjecture implies that the absence of edge bottlenecks of size less than is necessary and sufficient for a redundant broadcast to be possible from any source . The Independent Tree Conjecture answers the analogous problem where vertex failures are the concern, rather than edge failures.
2 The Chain Decomposition
In this paper, a graph will refer to what is commonly called a multigraph. That is, there may be multiple edges between the same pair of vertices (“parallel edges”) and an edge may connect a vertex to itself (a “loop”). All paths and cycles are simple, meaning they have no repeated vertices or edges. We consider a loop to induce a cycle of length one and a pair of parallel edges to induce a cycle of length two. Also, the presence of a loop increases the degree of a vertex by two. We will use the overline notation to name specific subgraphs, rather than for the graph complement.
Throughout this section, fix a graph with and a vertex . We begin by defining a decomposition analogous to the planar chain decomposition in [2].
Definition**.**
An up chain of with respect to a pair of edge-disjoint subgraphs (, ) is a subgraph of , edge-disjoint from and , which is either:
- i
A path with at least one edge such that every vertex is either or has degree at least two in , and the ends are either or are in , OR 2. ii
A cycle such that every vertex is either or has degree at least two in , and some vertex is either or has degree at least two in . We will consider to be both ends of the chain, and all other vertices in the chain to be internal vertices.
Chains which are paths will be called open and chains which are cycles will be called closed, analogous to the standard ear decomposition.
Definition**.**
A down chain of with respect to a pair of edge-disjoint subgraphs (, ) is an up chain with respect to (, ).
Definition**.**
A one-way chain of with respect to the pair of edge-disjoint subgraphs (, ) is a subgraph of , induced by an edge with ends and , such that is either or has degree at least two in , and is either or has degree at least two in . We call the tail of the chain and the head.
Definition**.**
Let be a sequence of subgraphs of . Denote and , so that and are the null graph. We say that the sequence is a chain decomposition of rooted at if:
The sets partition , AND 2. 2.
For , the subgraph is either an up chain, a down chain, or a one-way chain with respect to the subgraphs .
Definition**.**
The chain index of , denoted , is the index of the chain containing .
Definition**.**
An up chain is minimal if no internal vertex of is in .
Definition**.**
A down chain is minimal if no internal vertex of is in .
Definition**.**
A chain decomposition is minimal if all of its up chains and down chains are minimal.
Remarks**.**
A minimal up chain is a special case of an ear in the standard ear decomposition. 2. 2.
The chain decomposition is symmetric in the following sense. If is a chain decomposition rooted at , then is a chain decomposition rooted at , with the up and down chains switched and the heads and tails of one-way chains switched. Throughout this paper, we will refer to this fact as “symmetry”. 3. 3.
is either a closed up chain ending at or a one-way chain with as the tail, and is either a closed down chain ending at or a one-way chain with as the head. 4. 4.
In the planar chain decomposition in [2], up chains and down chains are analogous to the corresponding open chains. The elementary chain is analogous to a one-way chain.
Remark 2**.**
An up chain or down chain may be subdivided into several minimal chains by breaking at the offending internal vertices. These minimal chains may then be inserted consecutively to the decomposition at the index of the old chain. In this way, one can easily obtain a minimal chain decomposition from any chain decomposition.
We will prove Theorem 1 by combining the following results:
Theorem 3**.**
If is a -edge-connected graph and , then has a chain decomposition rooted at .
Theorem 4**.**
Suppose is a graph with no isolated vertices. If has a chain decomposition rooted at some , then there exists a set of four edge-independent spanning trees of rooted at .
3 Preliminary Results
While not needed for our main results, the following proposition demonstrates how the chain decomposition fits in with the various decompositions used in other cases of the Independent Tree Conjecture and Edge-Independent Tree Conjecture. A partial chain decomposition and its complement are “almost -edge-connected” in the following sense.
Proposition 5**.**
Suppose is a chain decomposition of a graph rooted at . Then for , and are connected. Further, if is a cut edge of (resp. ), then induces a one-way chain and one component of (resp. ) contains one vertex and no edges.
Proof.
By symmetry, we need only prove the result for the ’s. The connectivity follows from the fact that every type of chain is connected and contains at least one vertex in an earlier chain.
Suppose is a cut edge of some . Since is an edge in , we have and . We also know that is connected by the previous paragraph. Then cannot be part of an up chain, or else would be part of a cycle formed by the chain and a path in between the ends of (if is open; else the chain itself is a cycle). Also, cannot be part of a down chain, or else would be part of a cycle formed by and a path in between the ends of . Therefore, induces a one-way chain.
Let be the component of not containing , and suppose for the sake of contradiction that contains an edge. Let be an edge of with minimal chain index. Consider , the chain containing . Regardless of the chain type, some vertices in are incident to at least two edges in since , so one of these edges is not . This contradicts the minimality of . ∎
The next lemma and its corollary will allow us to ignore the possibility of loops in the graph when convenient.
Lemma 6**.**
Suppose is a chain decomposition of rooted at . If is in (resp ), then is incident to a non-loop edge in (resp ). If has degree at least two in (resp. ), then is incident to two distinct non-loop edges in (resp. ).
Proof.
Note that the second claim in the lemma implies the first, since a loop increases the degree of a vertex by 2, so it suffices to prove the second claim in the lemma.
Suppose is incident to a loop, which by symmetry we may assume is in . Of all loops incident to , choose the one with minimal chain index . Consider the chain classification of . The chain definitions all coincide for a loop, and require that () has degree at least two in . By the minimality of , is not incident to any loops in . It follows that is incident to two distinct non-loop edges in . ∎
Corollary 7**.**
Suppose is a chain decomposition of rooted at , and is a loop. Then is a chain decomposition of rooted at . Further, if has no isolated vertices, then has no isolated vertices.
Proof.
The first claim follows from the preceding lemma. For the second, observe that if is the only edge incident to its end, then it fails the conditions for every chain definition. ∎
Next, we prove the following useful fact about minimal chain decompositions.
Lemma 8**.**
Suppose is a graph with no isolated vertices, is a minimal chain decomposition of rooted at , and with . Then there are indices so that has degree exactly two in and .
Proof.
By symmetry, we need only find . Since has no isolated vertices, is in some chain. Consider the chain containing so that is minimal. Note that .
If is an up chain, then is an internal vertex of since , so has degree two in and degree at least two in . Therefore is not null, so . Then completes the proof.
The chain is not a down chain since .
So we may assume that is a one-way chain, and must be the head since . Therefore has degree at least two in , so we may consider the next chain to contain , say . Note that has degree one in by the definition of .
If is an up chain, then it is open and is an end of the chain, since the chain decomposition is minimal and has degree one in . The chain is not a down chain since has degree one in . If is a one-way chain, then is the head since () does not have degree at least two in . In all cases, has degree one in and degree at least two in . Therefore is not null, so . Then completes the proof. ∎
Finally, we show that the chain decomposition implies a minimum degree result.
Lemma 9**.**
Suppose is a graph with no isolated vertices, is a chain decomposition of rooted at , and with . Then has degree at least .
Proof.
By Corollary 7, we may assume that there are no loops in . If is in an up chain , then has degree at least in , and either degree in (if is internal) or degree at least in and degree at least in (if is an end). Either way, has degree at least in . By symmetry, the same is true if is in a down chain.
So we may assume that the only chains containing are one-way chains. Since has no isolated vertices, there is at least one such chain . Then has degree in and degree at least in (if is the tail) or (if is the head). We conclude that has degree at least in .
Assume for the sake of contradiction that does not have degree at least . Then has degree and is in exactly three one-way chains, say , , with . Consider . Since we know all of the chains containing , we can say that has degree in and degree in . This contradicts the definition of a one-way chain, as can be neither the head nor the tail of the chain . We conclude that has degree at least as desired. ∎
Remark**.**
If in addition to having a chain decomposition and no isolated vertices, then is -edge-connected so has degree at least as well. However, we will not need this result, and it will follow from Corollary 12.
4 The Mader Construction
We will adapt the strategy of Schlipf and Schmidt [10] in order to construct a chain decomposition. In particular, we will use a construction method for -edge-connected graphs due to Mader [9]. We limit our description of the construction to the needed case , since the method is more complicated for odd .
Definition**.**
A Mader operation is one of the following operations:
Add an edge between two (not necessarily distinct) vertices. 2. 2.
Consider two distinct edges, say with ends , and with ends , , and “pinch” them as follows. Delete the edges and , add a new vertex , then add the new edges with one end and the other end respectively. While and must be distinct, the ends need not be. In this case, will have parallel edges to any repeated vertex.
Theorem 10** ([9, Corollary 14]).**
A graph is -edge-connected if and only if, for any , one can construct in the following way. Begin with a graph consisting of and one other vertex of , connected by four parallel edges. Then, repeatedly perform Mader operations to obtain .
Remark**.**
Mader does not explicitly state that one can include a fixed vertex in , but it follows from his work. His proof starts with , and then reverses one of the Mader operations while maintaining -edge-connectivity. An edge can be deleted unless is minimally -edge-connected, in which case he finds two vertices of degree in his Lemma 13. He then shows that any degree vertex can be “split off” (the reverse of a pinch) in his Lemma 9, so we can always split off a vertex not equal to .
5 Proof of Theorem 3
Due to Theorem 10, it suffices to prove that a chain decomposition can be maintained through a Mader operation. The decomposition in the starting graph is as follows. Two of the edges form a closed up chain. The remaining two edges form a closed down chain.
Suppose the graph is obtained from the graph by a Mader operation, with both graphs -edge-connected. Assume that we have a chain decomposition of . By Remark 2, we may assume that we have a minimal chain decomposition. We wish to create a new chain decomposition of .
5.1 Adding an Edge
Suppose is obtained from by adding an edge with ends , . If one of the ends is the root , we can classify the new edge as a one-way chain with tail at, say, the very beginning of the chain decomposition. The head must have at least two incident edges in later chains, since all chains are later.
If neither end is , choose the minimal index such that or has degree exactly two in , guaranteed to exist by Lemma 8. Note that since is null. Without loss of generality, has degree exactly two in . By the definition of , has degree at most two in , and therefore degree at least two in . We classify the new edge as a one-way chain with tail and head , between the chains and .
We consider the impact of these changes on other chains in the graph. A new chain was added, but none of the other chains changed index relative to each other. Vertices may have increased degree in the ’s or the ’s due to the new edge, but increasing degree does not invalidate any chain types. Note that some chains may no longer be minimal, so the new chain decomposition in is not necessarily minimal.
5.2 Pinching Edges
Suppose is obtained from by pinching the edges with ends , and with ends , , replacing them with edges . We will use the notation for the chain containing , where is the subpath between and an end of so that , and is defined similarly. Note that (resp. ) may have no edges if (resp. ) is an end of . In the same way, we will use the notation for the chain containing .
We now prove several claims to deal with all possible chain classification and chain index combinations for and .
Claim 1**.**
If , then has a chain decomposition rooted at .
Proof.
If , then . Without loss of generality, and , so that the chain can be written as (where may have no edges if ). Recall that and are distinct, so is not a one-way chain.
By symmetry, we may assume is an up chain. In , we replace the chain with the following chains (in the listed order); see Figure 2 for an illustration:
. This is an up chain. Since the edges and have not yet been used, the new vertex is incident to two edges in later chains. 2. 2.
. This is a one-way chain with tail and head . The tail is incident to two edges in earlier chains, namely and . The head is incident to two edges in later chains since it was an internal vertex in the old up chain . 3. 3.
. This is a one-way chain with tail and head . The tail is incident to two edges in earlier chains, namely and . The head is incident to two edges in later chains since it was an internal vertex in the old up chain . 4. 4.
. Only add this chain if contains an edge. This is an up chain. The new ends are each incident to an edge in an earlier chain ( and , respectively) and are each incident to two edges in later chains since they were interior vertices of the old up chain .
We consider the impact of these replacements on other chains in the graph. We inserted most of the edges of the old chain at the same chain index , preventing any changes. The exception is the pinched edges and which were deleted, but the ends each received new incident edges inserted at the same chain index . Thus, we have maintained the chain decomposition. This proves Claim 1. ∎
Without loss of generality, we assume the following for the remainder of the proof:
- •
.
- •
If is a one-way chain, then is the tail and the head.
- •
If is a one-way chain, then is the tail and the head.
Claim 2**.**
Suppose that either is a one-way chain whose head has degree one in , or is a one-way chain whose tail has degree one in . Then has a chain decomposition rooted at .
Proof.
By symmetry, we may assume is a one-way chain whose head has degree one in .
First, we replace with . This is a one-way chain with tail and head . The tail was the tail of the old one-way chain . The head has two (in fact three) incident edges in later chains, namely , , .
- •
Case 1: is an up chain. Since has degree one in , if is closed then is not the end of . By swapping and if necessary, we may assume that is not the end of in . Thus, the end of in is still either or incident to an edge in an earlier chain, despite having not placed yet. We use the edges of and , , to construct chains at the index as follows:
. This is an up chain. The new end, , has one incident edge in an earlier chain () and two incident edges in later chains (, ). By assumption, the old end in is still either or incident to an edge in an earlier chain. 2. 2.
. This is a one-way chain with tail and head . The tail is incident two edges in earlier chains (, ). The head is either or incident to two edges in later chains, since has degree one in by assumption. 3. 3.
. This is a one-way chain with tail and head . The tail has two (in fact three) incident edges in earlier chains (, , ). The head is either or incident to two edges in later chains, since it was part of the old up chain . 4. 4.
. Only add this if contains an edge. This is an up chain. The new end, , has one incident edge in an earlier chain () and two incident edges in later chains since it was an internal vertex of the old up chain . Since we placed above, the end of in has is either or incident to an end in an earlier chain, even if the end is .
- •
Case 2: is a down chain. Since has degree one in , , so each vertex of is still either or incident to two edges in earlier chains, despite having not placed yet. We use the edges of and , , to construct chains at the index as follows:
. Only add this if contains an edge. This is a down chain. The new end, , has one incident edge in a later chain () and two incident edges in earlier chains since it was an internal vertex of the old down chain . 2. 2.
. This is a one-way chain with tail and head . The tail is either or incident to two edges in earlier chains since it was part of the old down chain . The head is incident to two edges in later chains (, ). 3. 3.
. This is a down chain. The new end, , has one incident edge in a later chain () and two incident edges in earlier chains (, ). 4. 4.
. This is a one-way chain with tail and head . The tail has two (in fact three) incident edges in earlier chains (, , ). The head is either or incident to two edges in later chains since has degree one in and by assumption, so has degree at least three in unless it is .
- •
Case 3: is a one-way chain. Since has degree one in , so the tail is still either or incident to two edges in earlier chains, despite having not placed yet. We use the edges , , to construct chains at the index as follows:
. This is a one-way chain with tail and head . The tail is either or incident to two edges in earlier chains as discussed above. The head is incident to two edges in later chains (, ). 2. 2.
. This is a one-way chain with tail and head . The tail is incident to two edges in earlier chains (, ). The head is either or incident to two edges in later chains since it was the head of . 3. 3.
. This is a one-way chain with tail and head . The tail has two (in fact three) incident edges in earlier chains (, , ). The head is either or incident to two edges in later chains since has degree one in and by assumption, so has degree at least three in unless it is .
We consider the impact of these replacements on other chains in the graph. As before, most of the edges of the old chains and were inserted at the same chain indices and respectively, preventing any changes. The pinched edges and were deleted, but the ends , , each received new incident edges , , inserted at the same chain indices (, , and respectively). However, was inserted at a different chain index than the deleted edge since was at while is at . By the claim assumptions, has degree one in , so there are no chains containing between and , and so no chains were affected by the change. Thus, we have maintained the chain decomposition. This proves Claim 2. ∎
We may now assume the following for the remaining cases:
- •
If is a one-way chain, then has degree at least two in .
- •
If is a one-way chain, then has degree at least two in .
We also make the following conditional definitions, which will aid in distinguishing the remaining cases:
- •
If is a one-way chain and is not in , then define the minimal index such that and . Since is minimal, has degree one in (incident only to the pinched edge ). From this and the fact that is a minimal chain, it follows that either is one of two distinct ends of the up chain , or is the head of the one-way chain which is not a loop.
- •
If is a one-way chain and is not in , then define the maximal index such that and . Since is maximal, has degree one in (incident only to the pinched edge ). From this and the fact that is a minimal chain, it follows that either is one of two distinct ends of the down chain , or is the tail of the one-way chain which is not a loop.
Claim 3**.**
Suppose that either one of is not defined, or . Then has a chain decomposition rooted at .
Proof.
The chains replacing will have indices adjacent to and (if it is defined). Likewise, the chains replacing will have indices adjacent to and (if it is defined). Thus, by the assumptions of this claim, the chains replacing will have lower chain index than the chains replacing . This fact will be needed when confirming that the new chains are valid. We begin by replacing as follows:
- •
Case 1: is an up chain. We replace it with . This is an up chain. The new vertex has two incident edges in later chains, namely and .
- •
Case 2: is a down chain. We replace it with the following chains (in the listed order):
. Only add this chain if contains an edge. This is a down chain. The new end has an incident edge in a later chain, namely . 2. 2.
. Only add this chain if contains an edge. This is a down chain. The new end has an incident edge in a later chain, namely . 3. 3.
. This is a one-way chain with tail and head . The tail is either or incident to two edges in earlier chains since it was in the old down chain . The head has two incident edges in later chains, namely and . 4. 4.
. This is a one-way chain with tail and head . The tail is either or incident to two edges in earlier chains since it was in the old down chain . The head has two incident edges in later chains, namely and .
- •
Case 3: is a one-way chain whose head is in . We replace it with the following chains (in the listed order):
. This is a one-way chain with tail and head . The tail was the tail of the old one-way chain . The head has two (in fact three) incident edges in later chains, namely . 2. 2.
. This is an up chain. The vertex is either or incident to two edges in later chains since it was the head of the old one-way chain , and it has an incident edge in an earlier chain by assumption. The vertex has two incident edges in later chains, namely and , and is incident to from the previous chain.
- •
Case 4: is a one-way chain whose head is not in . Then is defined as above.
First, we replace with . This is a one-way chain with tail and head . The tail was the tail of the old one-way chain . The head has two (in fact three) incident edges in later chains, namely .
- –
Subcase 1: is one of two distinct ends of the up chain . Replace with . This is an up chain. Since was a path and is a new vertex, this new chain is a path. The new end is adjacent to one edge in an earlier chain () and two edges in later chains ( and ).
- –
Subcase 2: is the head of the one-way chain which is not a loop. Then is not required to be in for to be a valid chain. In fact, is not required to be in any of by the definition of and the assumptions of this case. Thus, we can leave as is and insert the chain immediately after . This is an up chain. The vertex is incident to an edge in the previous chain , and is either or incident to two edges in later chains since it is the head of . The vertex is adjacent to one edge in an earlier chain () and two edges in later chains ( and ).
The procedure for replacing is symmetric, by following the above steps in the reversed chain decomposition.
We consider the impact of these replacements on other chains in the graph. In most cases, we replaced the old chain with new chains inserted at the same chain index , preventing any changes. The pinched edge was deleted, but the end received a new incident edge at the same chain index . In Cases 1-3, the same is true for . In Case 4, received a new incident edge either at or immediately after the chain index . However, by the definition of and the claim assumptions, no chains were affected by the new chain index except , which was specifically considered and shown to be valid in Case 4. By similar arguments, the changes caused by replacing also did not invalidate any chains. Thus, we have maintained the chain decomposition. This proves Claim 3. ∎
Claim 4**.**
Suppose that both of are defined and . Then has a chain decomposition rooted at .
Proof.
Recall that is either an up chain or a one-way chain with head , and is either a down chain or a one-way chain with tail . Since , we conclude that must be a one-way chain with tail and head , and since and are defined. We can replace and with the following chains, in the listed order. The first two will be placed immediately before index , and the last two immediately after index ; see Figure 3 for an illustration:
. This is a one-way chain with tail and head . The tail was the tail of the old one-way chain and we are placing this chain after index . The head has two (in fact three) incident edges in later chains, namely . 2. 2.
. This is a one-way chain with tail and head . By the definition of , the tail is either or incident to two edges in earlier chains than , and we are placing this chain immediately before index . The head has two incident edges in later chains, namely and . 3. 3.
. This is a one-way chain with tail and head . The tail has two incident edges in earlier chains, namely and . By the definition of , the head is either or incident to two edges in later chains than , and we are placing this chain immediately after index . 4. 4.
. This is a one-way chain with tail and head . The tail has two (in fact three) incident edges in earlier chains, namely . The head was the head of the old one-way chain , and we are placing this chain before .
We consider the impact of these replacements on other chains in the graph. The deleted edge was replaced by two edges with chain index greater than , so we must be careful. The edge was inserted before index , but had degree at least two in , so losing a degree in later subgraphs will not invalidate any chains. The edge was inserted immediately after index , so by the definition of , the only chain affected is . Since has as a head, losing a degree in will not invalidate the chain. By a symmetric argument, the changes caused by and do not invalidate any chains. This proves Claim 4. ∎
Claim 5**.**
Suppose that both of are defined, and . Then has a chain decomposition rooted at .
Proof.
We can replace and with the following chains, at the indicated chain indices; see Figure 4 for an illustration:
. Add this chain at index . This is a one-way chain with tail and head . The tail was the tail of the old one-way chain and we are placing this chain at index . The head has two (in fact three) incident edges in later chains, namely . 2. 2.
. Add this chain immediately after . This is a one-way chain with tail and head . By the definition of , the tail is either or incident to two edges in earlier chains than , and we are placing this chain after index . The head has two incident edges in later chains, namely and . 3. 3.
. Add this chain immediately before . This is a one-way chain with tail and head . The tail has two incident edges in earlier chains, namely and . By the definition of , the head is either or incident to two edges in later chains than , and we are placing this chain before index . 4. 4.
. Add this chain at index . This is a one-way chain with tail and head . The tail has two (in fact three) incident edges in earlier chains, namely . The head was the head of the old one-way chain , and we are placing this chain at index .
We consider the impact of these replacements on other chains in the graph. The edge was deleted, but received a new incident edge at the same chain index . The edge was inserted before index , but the index is still smaller than , so by the definition of , no chains are affected. By a symmetric argument, the changes caused by and also do not invalidate any chains. This proves Claim 5. ∎
The claims cover all possibilities of pinching edges. The proof of Theorem 3 is complete. The proof also implies a polynomial-time algorithm to construct a chain decomposition.∎
6 Proof of Theorem 4
Assume that we have a chain decomposition of . By Remark 2, we may assume that the chain decomposition is minimal. We will adapt the strategy of Curran, Lee, and Yu [3] to prove Theorem 4. In particular, we will construct two partial numberings of the edges of using the chain decomposition. We will then construct four spanning trees in two pairs, with one pair associated with each numbering. Within each pair, paths back to the root will be monotonic in the associated numbering to ensure independence. Between pairs, paths back to the root will be monotonic in chain index to ensure independence.
Using Corollary 7, we may assume that there are no loops in . By Lemma 8, for each vertex , there are two distinct non-loop edges incident to whose chain indices are strictly smaller than the chain index of any other edge incident to . Likewise there are two distinct edges whose chain indices are strictly larger than the chain index of any other edge adjacent to . We will name these edges as follows:
Definition**.**
For each vertex , the two -edges of are the two incident edges with the lowest chain index. Similarly, the two -edges of are the two incident edges with the highest chain index.
Remark 11**.**
By the definition of a down chain, the edges of down chains are never -edges. Likewise, by the definition of an up chain, the edges of up chains are never -edges.
Next, we will iteratively define a numbering , which will assign distinct values in to all edges in up chains and one-way chains. Here, two “consecutive” edges in a chain will refer to two edges in the chain which are incident to an internal vertex of the chain, so the two edges incident to the end of a closed chain are not consecutive, despite being adjacent.
We begin by numbering the edges in , and then number the edges of each up chain and one-way chain in order of chain index. When we reach a chain , we may assume that all edges in belonging to up chains and one-way chains have been numbered, which includes all -edges in by Remark 11. We use the following procedure to number the edges in :
- •
If is a closed up chain containing , then number the edges in so that the values change monotonically between consecutive edges in the chain. The particular numbers used are arbitrary.
- •
If is a closed up chain not containing , then both -edges of the common end have already been numbered. Call these two -edges numbering edges of . Say the numbering edges of have -values and . Number the edges in so that the values change monotonically between consecutive edges in the chain, and all values are between and .
- •
If is an open up chain containing , then is an end and the other end is some . At least one -edge of has already been numbered. Choose an -edge which has already been numbered and call it a numbering edge of . Say that is the -value of the numbering edge. Number the edges in so that the values increase between consecutive edges in the chain when moving from to , and all values are larger than .
- •
If is an open up chain not containing , then at least one -edge of each end has been numbered. If the ends are and , we can choose two distinct edges so that is an -edge of and is an -edge of . We can choose these two distinct edges because otherwise, the only -edge of or in would be a single edge between and , and then would not be connected. Call the edges numbering edges of . Without loss of generality, . Number the edges in so that the values increase between consecutive edges in the chain when moving from to , and all values are between and .
- •
If is a one-way chain whose tail is , then number the edge of arbitrarily.
- •
If is a one-way chain whose tail is not , then both -edges of the tail are already numbered, say with -values and . Number the edge of between and .
We symmetrically define a numbering , which assigns distinct values in to the edges of down chains and one-way chains, by using the above procedure in the reversed chain decomposition.
We are finally ready to construct the trees. Define the subgraphs as follows. For each , consider the two -edges of . Assign the edge with the lower -value to and the edge with the higher -value to . Similarly, consider the two -edges of . Assign the edge with the lower -value to and the edge with the higher -value to .
Several properties of will follow from the following claim.
Claim**.**
For any , consider the edge assigned to at . Let be the other end of . If , let be the edge assigned to at . Then and .
Proof.
Let be the edge assigned to at . The edge is not in a down chain by Remark 11. We break into two cases.
- •
Suppose is in an up chain . Since the chain decomposition is minimal and , its -edges are either in , or else have chain index less than . In either case, as desired.
Note that is either in , or else is the numbering edge of at the end . By the numbering procedure, we know that is between and the -value of one of the -edges of , say . By the definition of , , so it follows that . Again by the definition of , , so as desired.
- •
Suppose induces a one-way chain . Since is an -edge, has degree at most one in , so must be the head of . Then is the tail of , so the -edges of have chain indices smaller than , which means and as desired.
From the numbering procedure, we know that is between the -values of the two -edges of , with being the smaller by the definition of . So, as desired.
In both cases we have and . This proves the claim. ∎
With the claim proven, it follows that the edges assigned to are all distinct, there are no cycles in , and following consecutive edges assigned to produces a path which is decreasing in chain index, strictly decreasing in -value, and can only end at . Thus, is connected and is a spanning tree of . A similar argument shows that is a spanning tree of where paths to are decreasing in chain index and strictly increasing in -value. Due to the opposite trends in -values, and are edge-independent with root .
By symmetry, we obtain analogous results for and . It remains to show that a tree from and a tree from are edge-independent. The paths back to from a vertex are decreasing in chain index in one tree and increasing in chain index in the other tree, but not strictly. The first edges in these paths are an -edge and a -edge of , respectively. By Lemmas 8 and 9, there is a positive difference in chain index between these initial edges, so the paths are in fact edge-disjoint. The proof of Theorem 4 is complete. The proof also implies a polynomial-time algorithm to construct the edge-independent spanning trees.∎
7 Summary of Results
With Theorems 3 and 4 proven, we obtain Theorem 1. In fact, we can examine the argument more carefully to extract a stronger, summarizing result.
Corollary 12**.**
Suppose is a graph with no isolated vertices and . Then the following statements are equivalent.
* is -edge-connected.* 2. 2.
There exists so that has a chain decomposition rooted at . 3. 3.
For all , has a chain decomposition rooted at . 4. 4.
There exists so that has four edge-independent spanning trees rooted at . 5. 5.
For all , has four edge-independent spanning trees rooted at .
Proof.
Theorem 3 gives us . Theorem 4 gives us and . Trivially, we have and . Therefore, we need only show .
Assume for the sake of contradiction that has four edge-independent spanning trees rooted at some , but is not -edge-connected. Suppose is an edge cut with . Consider a vertex in a component of not containing . Using the paths in each of the edge-independent spanning trees, we find that there exist four edge-disjoint paths between and . This contradicts the existence of . ∎
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