Applied Mathematics
p-ISSN: 2163-1409 e-ISSN: 2163-1425
2012; 2(1): 8-16
doi:10.5923/j.am.20120201.02
Tzvetalin S. Vassilev, Laura J. Huntington
Department of Computer Science and Mathematics, Nipissing University, North Bay, Ontario, P1B 8L7, Canada
Correspondence to: Tzvetalin S. Vassilev, Department of Computer Science and Mathematics, Nipissing University, North Bay, Ontario, P1B 8L7, Canada.
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Copyright © 2012 Scientific & Academic Publishing. All Rights Reserved.
The Atom Bond Connectivity index, also known as ABC index was defined by Estrada[4] with relation to the energy of formation of alkanes. It was quickly recognized that this index reflects important structural properties of graphs in general. The ABC index was extensively studied in the last three years, from the point of view of chemical graph theory[5,6], and in general graphs[1]. It was also compared to other structural indices of graphs[2]. Das derives multiple results with implications to the minimum/maximum ABC index on graphs. With relation to trees, it is known that among all the trees of the same number of vertices, the maximum ABC index is attained for the star graph. However, it is not known which tree(s) minimize(s) the ABC index. The problem seems to be hard. It is partially addressed in many sources[5,1,6], but remains open. In this paper we further investigate the trees that minimize the ABC index. Our investigations are limited to chemical trees, i.e. trees in which the maximum vertex degree is 4. The chemical trees were introduced to reflect the structure of the carbon chains and the molecules based on them. Our approach is algorithmic. We identify certain types of edges (chemical bonds) that are important and occur frequently in chemical trees. Further, we study how the removal of a certain edge, the introduction of certain edge or the contraction of certain edge affects the ABC-index of the tree. We pay particular attention to the examples of minimal ABC index chemical trees provided by Dimitrov[3].
Keywords: Chemical graph theory, Structural indices of graphs, Chemical trees
Cite this paper: Tzvetalin S. Vassilev, Laura J. Huntington, On the Minimum ABC Index of Chemical Trees, Applied Mathematics, Vol. 2 No. 1, 2012, pp. 8-16. doi: 10.5923/j.am.20120201.02.
where the summation goes over all the edges of G, du and dv are the degrees of the terminal vertices u and v of edge uv and E(G) is the set of edges of G with cardinality m = |E(G)|.

Trees:
The following are not chemical trees:Graphs of vertex degree greater than 4 and graphs with cycles:
Contribution of an edge uv whose terminal vertices are of degree d and 2:
Contribution of an edge uv whose terminal vertices are of degree 1 and 3:
Contribution of an edge uv whose terminal vertices are of degree 1 and 4:
Contribution of an edge uv whose terminal vertices are of degree d and 3:
Contribution of an edge uvwhose terminal vertices are of degree d and 4:

edge be an edge with a terminal vertex of degree
and the other terminal vertex of degree
. For example, a 3-4 edge is an edge between vertices of degree 3 and 4 respectively. Also let
denote a transformation of an
edge to a
edge. For example,
means that a 2-2 edge was transformed into a 2-1 edge by adding or removing certain vertices and edges as specified in each unique case presented.
This is the only possible arrangement for a chemical tree with 2 vertices.n=3
The chain is once again the only option.n=4


n=5




n=6






Thus for n≤6, the chain has the minimal ABC index.
And the second is the following tree:
Note that all edges in this tree have at least one terminal vertex of order 2; this means that all the edges contribute the same amount to the ABC index
Thus, the ABC index of this formation is equal to the ABC index of the chain with 7 vertices.
Therefore, there are two different chemical trees with 7 vertices of minimal ABC index.Similarly for chemical trees with n=8, there are two different trees with 8 vertices of minimal ABC index.The chain:
And another tree:
Note that similar to the tree of minimal ABC index for 7 vertices, all of this tree’s edges have at least one vertex of degree 2.For
, there are 4 trees of minimal ABC index including the chain.


All four of these trees have edges which each have at least one terminal vertex of degree 2. Thus, all the edges in these trees contribute the same amount to the ABC index and since each of these graphs has 8 edges, all four of these trees have the same minimal ABC index as the chain of 9 vertices 
. This would be the case if all the edges of the tree had at least one terminal vertex of degree 2. But what if the tree could have an edge that would contribute less than one which has a vertex of degree 2 (namely less than
).Consider the following tree with n=10:
All of this tree’s edges have at least one vertex of degree 2 except for the one that is circled. This edge is between two vertices of degree 3. The contribution of the circled edge is as follows:
Compare the contribution of this edge to an edge that has at least one of its terminal vertices of degree 2.
Thus, the indicated edge with vertices of degree 3 contributes less than one of the edges of a chain would and so the ABC index of this tree is less than the ABC index of a chain of 10 vertices. In fact, it is the only tree with 10 vertices of minimal ABC index according to Furtula.
From the table above, these edges that contribute less than
are the following: a 3-3 edge, 3-4 edge and a 4-4 edge. (Note that here we do not consider edges of degree greater than 4 because we are currently focusing on chemical trees).
Note that this tree is very similar to the tree of minimal ABC index for 10 vertices; it simply has one more vertex and edge added to the horizontal chain.Let us explore the different ways one could form a tree with 11 vertices to try to determine why the new edge and vertex were placed where they were.
The original given by Furtula:Number of 2-1 edges:4Number of 2-2edges:1Number of 3-2edges:4Number of 3-3 edges:1The example above (Example A):Number of 2-1 edges:4Number of 2-2 edges:0Number of 3-2 edges:2Number of 3-3 edges:0Number of 3-4 edges:1Number of 4-1edges:1Number of 4-2 edges:2It seems that to go from the original given by Furtula to Example A the following transformations take place:
so in terms of contributions there in no change to the ABC Index
no change
no change
CHANGE: 
Since any edge with at least one terminal vertex of degree 2 will contribute
to the ABC index regardless of the degree of the other terminal vertex, no change occurs to the ABC index from the first three transformations. However, the last two transformations result in a greater contribution to the ABC index and thus Furtula’s tree with 11 vertices is still the minimal tree.Let us try another transformation (Example B):
We have now lost two 2-1 edges and gained two 3-1 edgesWe have also lost a 3-2 edge for a 3-3 edge.Since the contribution of a 2-1 edge is less than the contribution of a 3-1 edge, this transformation increases the ABC index.However, as mentioned above, the 3-3 edge contributes less than the 3-2 edge so we must calculate the decrease in the ABC index and ensure that it is less than the increase caused by the first transformation.Contribution of a 2-1 edge:
Contribution of a 3-1 edge:
Thus the total increase to the ABC index is:
Contribution of a 3-2 edge:
Contribution of a 3-3 edge:
Thus the total decrease to the ABC index is:
Therefore, this transformation ultimately results in an increase in the ABC index and Furtula’s tree is still the minimal example.What about the following tree, call it example C?
Here we see that there is no change to the ABC index from the transformation; however, this is not a new formation. In fact, this tree has the same ABC index as Furtula’s for the simple reason that it is isomorphic to Furtula’s tree.
Given what we now know about the contribution of certain edges, we would like to have as many 3-3, 3-4 and/or 4-4 edges as possibleTry starting with a vertex of degree 4
Now let us connect this tree with another tree to form a 4-4 edge.
Here we have used 7 of the 12 vertices.We know that 4-1 edges contribute more than 4-2 edges so let us add a pendant edge to each of the pendant vertices to develop the following tree:This tree has 12 vertices but is this tree of minimal ABC index?
Note that this tree still has two 4-1 edges. The question is: does it make sense to take one of these 4-1 edges and add it to one of the pendant edges? This would change one of the vertices of degree 4 to degree 3 so instead of having a 4-4 edge and two 4-2 edges we would now have a 3-4 edge and two 3-2 edges. The 3-2 edges will contribute the same as the 4-2 edges so they are not interesting. The question becomes: does a 3-4 edge contribute less than a 4-4 edge?The answer is no. A 3-4 edge contributes slightly more than a 4-4 edge. So the following tree is not the minimal tree:
What if we take this edge and move it as follows so that we form another 4-2 edge instead of a 4-1 edge?:
Now we have a slight increase in the ABC index because a 4-4 edge becomes a 3-4 edge but we also change a 4-1 edge to a 4-2 edge, which decreases the ABC index. Comparing the increase and decrease we see that the decrease is larger than the increase and thus this tree has a smaller ABC index than the previous trees.
Thus, we establish the tree of minimal ABC index with n=12 (Confirmed by Furtula) 
of graphs described by Das and some results regarding graphs that are not chemical trees (such as those in class
) will be introduced.
postulated by Kinkar Das. Let H be a graph in
and let
be the number of edges in H. Then:1. H is a connected graph2. H has minimum vertex degree
.3. H has q edges between vertices of maximum degree
where
4. H has
edges for which at least one of the terminal vertices has degree
.And q is given by:
Thus, an edge, say
, is of one of two possible types: either
has at least one terminal vertex of degree
or both the terminal vertices of h are of maximum degree
[1].
imply that the contribution of an edge h to the ABC(H) is either 
Therefore, we deduce the following equation for the ABC(H):
Therefore,
Note that any graph T which is tree cannot be in
for the simple reason that a tree has pendant vertices of degree 1 and thus the minimum vertex degree for any tree is 
. As mentioned above, these graphs will not have any pendant edges. Let us start with the most basic cycle, K3.This graph, K3 is connected and it has minimum vertex degree
but
since it has maximum vertex degree 
Is
?
K4 is connected, has maximum vertex degree
but now K4 has minimum vertex degree
What happens if we remove one of the diagonal edges?
Call this graph A. A
since A is connected, has
, has
and all of its edges are either between vertices of maximum degree or between vertices with at least on vertex of degree 2.Let us try adding triangular shaped graphs onto some of the vertices to see if we can generate another example of a graph from the class 
This graph, call it B, is not in
. It has
and it has a 4-3 edge. This edge is not between two vertices of maximum vertex degree, nor does it have at least one vertex of degree 
Let us try to remedy this by adding a second triangular graph to the right side of the graph B as follows:We still have
and 4-3 edges.Let us try making the core a complete graph K4 by adding an edge as follows:
Now check that this graph, call it C, meets all the criteria in order for it to belong to
1. It is connected.2. It has minimum vertex degree
3. It has
edge between vertices of maximum degree
4. It has
edges for which at least one of the terminal vertices is of degree
Thus,
and we can now easily calculate its ABC index because we know that for all graphs
,Consider the following graph D:

since D is connected, D has
D has
and all of the edges of D are either between two vertices of degree
or have at least one vertex of degree 
.
In fact, we can add 3 more edges to form a graph E with maximum degree
where 
And, we can add 6 edges to turn the core of the graph E into K6 and generate a new graph F as follows:
Now we can add 3 triangles to graph F to form the following graph G that is also in
In G, 


This formulation allows for the taking into account of the modified second Zagreb index
which is equal to the sum of the products of the reciprocal of the degrees of pairs of adjacent vertices, that is,
.This modified second Zagreb index is used by Das is the formulation of many theorems and their subsequent corollaries establishing upper and lower bounds on the ABC index of specific simple connected graphs with particular characteristics[1].
For a pendant edge uv, with du=1 and dv=v,
And obviously, to maximize this you would take the highest value for v possible, thus you need all the edges to be of the from
is the maximum vertex degree in the graph G.But why choose to have all pendant edges?Well, if we look at the original formulation of the ABC Index:
Note that in order to maximize this equation we need to make du+dv-2 as large as possible while keeping dudv as small as possible. And, as it turns out, this is achieved by making one of the degrees 1 and the other 