Edge Disjoint Spanning Trees in an Undirected
Graphs
Abhinav Garg, Nayan Deshmukh, Sunil Pandey
August 2016
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Introduction and background
Under the RTE program we were mentored by Prof. Ovidiu Daeseu. We addressed the following problem :Given an undirected, unweighted graph G with n vertices and 2n-2 edges, one
of which is a double edge, check if G admits two edge disjoint spanning trees,
and if yes output them
Algorithm for finding 2 edge-disjoint spanning trees in a directed graph was
first proposed by Prof. R. Tarjan Edge-disjoint spanning trees and depth-first
search with running time of O(n). This algorithm used dfs as the primary tool.
However, the case of undirected graphs was left open. Recently, an algorithm
with running time O(n1.5 ) was proposed for a general undirected graph. The
algorithm was presented for a general undirected graph. Also, the algorithm was
very complex and it’s application in real life was hard. We had an intuition that
it should be possible to give an O(n) algorithm even for the case of undirected
graph along the same lines as for directed graph. As of today, we don’t have
an O(n) time algorithm for undirected graph neither a proof for it’s absence.
Our setting however are much more restrictive( n vertex 2n-2 edges) making the
intution for existance of O(n) time algorithm much more stronger. So our aim
was primarily to propose an O(n) time algorithm for finding two edge disjoint
spanning trees.
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Approaches Explored
We have tried a lot of approaches to progress towards our aim. Following are
the discriptions of the same.
2.1
On the lines of Tarjan’s algorithm
We worked on extending the Tarjan’s algorithm for directed graph to undirected
graphs.
2.1.1
Implementation
For a better insight of Tarjan’s algorithm of directed graph on undirected graph,
We implemented the algorithm in C++ initially using LEDA and then N graphs
support libraries. The implementation can be found at implementation of tarjan’s Algorithm. This implemetation however is still not ready for use as the
important conclusions from the implementation were infered before it’s completition.
2.1.2
Inferences on Tarjan’s algorithm
To develop an algorithm based on Tarjan’s these were te possible directions that
we opted :2
• Convert Undirected To directed:- We thought to put 2 edges ont in
forward direction and other in backward direction and then run tarjan’s
algorithm on it.
Shortcomings:- Tarjan’s algorithm returned 2 dfs trees. One in forward
direction, Another in backward direction.
• Mapping between Tarjan’s trees:- We tried to see if making tree
directed by putting two edges for each one can produce four trees from
tarjan’s algorithm and then map these 4 tress to two trees for undirected
graph. Difficulties :- It was quite difficult to think of such mapping and
no significant prgress was made in this field.
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Lemmas Proved
Given an undirected,unweighted graph G with n vertices and 2n-2 edges, one
of which is a double edge and no bridges, decide if G admits two edge-disjoint
spanning trees and if so produce the trees. We will try to construct two edge
disjoint spanning trees namely T1 and T2. Since the number of edges < 2n, we
will have a vertex with degree 2 or 3.
Lemma 3.1. Given the original graph G, for each of it’s two degree vertex we
can always randomly put one of it’s edge in T1 and the other in T2.
Proof. As the vertex is having degree 2, We can always say that one of it’s edge
should lie in T1 and the other in T2. To show that any of these edge can lie
in any of the tree, consider the following vertex numbered 3, having degree two
with neighbour’s as 1 and 2. Now, vertex 3 should be the leaf of both T1 and
T2. Say it was connected to 1 in T1 and 2 in T2, So we can anytime find 2 in
T1 and 1 in T2 and exchange 3. Therefore, we can randomly put edges of a 2
degree vertex in the original graph.
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Corollary 3.1.1. The above theorem is also valid for the graph G induced by
removing the two degree vertex along with it’s edges.
Proof. When we remove the 2 degree vertex, The graph left would still have 2
Edge-disjoint spanning trees,So we can still apply lemma 1.1.
Lemma 3.2. We can randomly put two edges of a 3 degree vertex in G3 in one
of the tree and the other edge in other tree.
Now we are left with a graph s.t. it has all vertices with degree ≥ 3. Also
degree of these vertices in graph G3 was ≥ 4.
Lemma 3.3. We can always put edges of three degree vertices in T1 and T2
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such that E1 ⊂ E1 and E2 ⊂ E2 is ensured.
Note:- The above proof is much more stronger than it looks. This
not only ensures that we create no cycle while dealing with three
degree vertices but also that going along these lines with proper steps
in future we can find T1 and T2, more precisely E1 and E2 correctly.
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Lemma 3.4. Given the graph G3 , for each of its vertices having degree 3 in
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G3 and had degree 4 in G3 we can randomly put one of its edges in the tree in
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which this vertex is disconnected in G3 .
Excepted Progress :-As we know that in G3 we have n vertices and 2n2 edges, Therefore we must have at least 4 three-degree vertex. Processing
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them the way we mentioned in the previous paper, we would arrive at G3 with
maximum n-4 vertex and 2n-14 edges which mean atleast 12 3-degree vertex.We
could remove one edge for each 3-degree vertex.Therefore, 2n-26 edges and n-4
vertices.
Notice, that while doing the above algorithm we’ll always check for cycles.
Also in case removing an edge from a three degree vertex creates a two degree
vertex, We again will put one it’s two edges randomly in T2.
Algorithm Proposed :The algorithm proposed reduces atleast n edges from the original graph and
put them in the trees T1 and T2. Following are the list of steps of the algorithm:STEP 1:- Sort the vertices in ascending order according to their degree.
STEP 2:- Start removing two edges corresponding to every vertex starting from
the first one in the sorted list and put one of them in tree T1 and other in tree
T2. If you encounter a vertex that doesn’t have 2 edges left, ignore it and move
ahead to the next vertex.
Lemma 3.5. After executing above algorithm we’ll have less than n edges left.
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Validation of the proposed algorithm
4.1
Creating instances
To create the input set to check how the algorithm is responding, we have used
Laman graph that has 2n-3 edges with n vertices. Laman graph has the property
that if a DFS is constructed and then one of the remaining edge is doubled and
the same edge is changed in the parent graph so that it has 2n-2 edges now,
then that graph admits two edge disjoint trees.
We are currently implementing our algorithm and testing it on the generated
test cases.
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Proofs
The authors reserve the copyrights of the proof. For any query, Please contact
:• [email protected]
• [email protected]
• [email protected]
• [email protected]
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