1. Introduction
Cactus graph is a connected graph in which every block is a cycle or an edge, in other words, no edge belongs to more than one cycle. Cactus graphs have been extensively studied and used as models for many real-world problems. This graph is one of the most useful discrete mathematical structures for modelling problems arising in the real-world. It has many applications in various fields, like computer scheduling, radio communication system, etc. Cactus graphs have been studied from both theoretical and algorithmic points of view. This graph is a subclass of planar graph and superclass of tree.
An 
-labelling of a graph 
is a function of 
from its vertex set 
to the set of nonnegative integers such that 
if 
and 
if
, where 
is the distance between the vertices 
and
, i.e., the number of edges between 
and
. The span of an 
-labelling 
of 
is max
. The 
-labelling 
of 
is the smallest 
such that 
has a 
- labelling of span
.
An interesting graph-labelling problem comes from the radio frequency assignment problem, as well as code assignment in computer networks. One version of the radio channel assignment problem [1] is to assign integer channels to a network of transmitters with distance restrictions, such that the several labels of interference between nearby transmitters are avoided and the span of the label used is minimized. A variation of the problem is code assignment in computer networks, i.e., to assign integer control codes to a network of computer stations with distance restrictions.
Bertossi and Bonuccelli [2] introduced a kind of code assignment to avoid hidden terminal interference; this is as follows. Since some modern computer networks consist of including mobile computers or computer displaced in wide areas, they need to use broadcast communication media such as busses (only in local area networks) or radio frequencies. The computer network which communicates by radio frequencies is called Packet Radio Network. It consists of computer stations (computers and transceivers), in which the transceivers broadcast outgoing message packets and listen for incoming message packets. Unconstrained transmission in broadcast media may lead to collision on interference, i.e., there is the time-overlap of two or more incoming message packets received at the destination station. This results in damaged useless packets at the destination. Collided message packets must be retransmitted. That increases the time delay of the transmission, and hence lowers the system throughput. Several protocols have been devised to reduce or eliminate the collisions. They form the medium access control sublayer. For example, under Code Division Multiple Access protocol, the collision-free property is guaranteed by the use of proper assignment of orthogonal control codes to stations and the spread of spectrum communication techniques (e.g., hopping over different time slots or frequency bands).
We represent the network by a graph, such that all stations are vertices and two vertices are adjacent if the corresponding stations can hear each other. Hence, two stations are at distance two, if they are outside the hearing range of each other but can be received by the same destination station. There are two types of collisions-interferences: direct collision, due to transmission of adjacent station, and hidden terminal collision, when stations at distance two transmit to the same receiving station at the same time.
To avoid hidden terminal interference, we assign a control codes to each station in the software as follows. For one station, to avoid hidden terminal interference from its adjacent stations (which hear each other) sending packets to it, we require distinct codes for its immediate adjacent station, i.e.,
. Here we suppose that there is a little direct interference in the system, i.e., direct interference is so week that we can ignore it. Apparently in the model of [2] there are some special hardware designs, which can avoid direct interference in the system. Hence, we allow the same code for two adjacent stations (which can hear each other), meaning
. Therefore, we have the 
-labelling case.
It is important to note that the 
-labelling problem is just a special case of ordinary graph labelling. Each feasible 
-labelling of a graph 
yields a feasible labelling of the graph
, where 
contains edge 
whenever 
and 
are distance two apart in
. Conversely, a labelling of 
becomes a feasible labelling of 
by calling the labels
, where 
represents the maximum colour number of the graph.
In this paper, we label the vertices of a cactus graph 
by 
-labelling and it is shown that 
, where 
is the degree of the graph
, i.e., 
max{deg(vi):
, deg(vi)) is the degree of the vertex vi)}.
2. Review of Previous Works
Some results are available on 
-labelling problem. Here we discuss some particular cases. When 
and 
then we get 
-labelling problem. Several results are known for 
-labelling of graphs, but, to the best of our knowledge no result is known for cactus graph. In this section, the known result for general graphs and some related graphs of cactus graph are presented.
The upper bound for 
of any graph 
is 
[3], where 
is the degree of the graph.
The problem is simple for paths 
of 
vertices. It can easily be verified that
, 
for 
[4].
When the first and the last vertices of 
are merged then 
becomes
. In [2], Bertossi and Bonuccelli showed that 
is equal to 1 if n is multiple of 4 and 2 otherwise.
For complete graph
, it is easy to check that
.
The wheel
, is obtained by joining 
and
, i.e.,
. It is also easy to check that 
.
Bertossi and Bonuccelli [2] investigated the 
- labelling problem on complete binary trees, proving that 3 labels suffice. An optimum labelling as follows can be found. Assign first labels 0, 1 and 2, respectively, to the root, its left child and its right child. Then, consider the nodes by increasing levels: if a node has been assigned label
, then assign the remaining labels to its grandchildren, but giving different to brother grandchildren. The above procedure can be generalized to find an optimum 
-labelling for complete (
)-ary trees, requiring span
. It is straight forward to see that when 
and 
this result gives the 
- number for complete binary trees and paths respectively.
It is shown in [5] that for any tree
, 
is equal to
, implying that
. An optimal 
-labelling can be also determined by exploiting the algorithm provided in [6] for optimally 
- labelling trees.
Bodlaender et al. [7] compute upper bounds for graphs of treewidth bounded by 
proving that
. They give also approximation algorithms for the 
- labelling running in 
time.
In [2], the NP-completeness result for the decision version of the 
-labelling problem is derived when the graph is planar by means of a reduction from 3-VERTEX COLORING of straight-line planar graph. An exhausted survey on 
-labelling is available in [8].
In [7], an approximation algorithm is designed for 
-labelling a permutation graph in 
time; it guarantees the bound
.
The n-dimensional hypercube 
is an 
-regular graph with 
nodes. Then 
and there exists a labelling scheme using such a number of labels. This labelling is optimal when 
for some 
and it is a 2-approximation otherwise [9]. For a bipartite graph, 
[7]. Later this lower bound has been improved by a constant factor of 
[10]. A study on 
-labelling of cartesian product of a cycle and path is done by Chiang and Yan [11].
When 
and 
then we get 
-labelling problem. This problem was introduced by Grrigs and Yeh [12,13] in connection with the problem of assigning frequencies in a multihop radio network. Some results of 
-labelling problem are given below.
Kral and Skrekovski [14] improve the upper bound for any graph
,
. The best known result till date is 
due to Goncalves [15].
Heuvel and Mc Guinness showed that 
[16] for planar graphs. Molloy and Salavatipour [17] reduced this upper bound to
. Wang and Lih [18] proved that if 
is a planar graph of girth (girth is defined to be the length of a shortest cycle in
) at least 5, then 
In [19], we have showed that the upper and the lower bounds for 
of a cactus graph 
is 
Adams et al. [20], give different bounds for certain generalized petersen graphs. A study on 
-labelling of cartesian product of a cycle and a path is done by Chiang and Yan [11].
For further studies on the 
-labelling, see [21- 30].
When 
and 
then we get another special case which is called 
-labelling problem. Some results of 
-labelling problem are given below.
For path, 
and 
for each
, and 
is 2 if 
is a multiple of 3 and it is 3 otherwise [31].
3. The L(0,1)-Labelling of Induce Sub-Graphs of Cactus Graphs
Let 
be a given graph and subset 
of
. The induced subgraph by
, denoted by 
is the graph given by
, where 
. Some induced subgraphs of cactus graph are shown in Figure 1.
The cactus graphs have many interesting subgraphs, those are illustrated below. An edge is nothing but
,
(a) (b) (c) (d)
Figure 1. Some induce subgraphs of cactus graph.
so
. The star graph 
is a subgraph of cactus graph, therefore, one can conclude the following result.
Lemma 1. For any star graph 
(1)
4. L(0,1)-Labelling of Cycles
In [2], Bertossi and Bonuccelli have labeled 
by 
-labelling and they have obtained the following result. Here we have given a constructive prove of this result.
4.1. L(0,1)-Labelling of One Cycle
Lemma 2. [2] For any cycle 
of length
,
(2)
Proof. Let 
be the vertices of the cycle
. We classify 
into five groups, viz., 
, 
, 
, 
and
.Then the 
-labelling of the vertices of a cycle are as follows.
Case 1. Let
.
Case 2. Let 
(mod 4), i.e.,
.
Case 3. Let 
(mod 4), i.e.,
.
The label of first 
vertices 
are same as in Case 2. For the last vertex
, 
is define as
Case 4. Let 
(mod 4), i.e.,
.
Here the label of first 4k + 1 vertices 
are same as in Case 3. For the last vertex
, 
is define as
Case 5. Let 
(mod 4), i.e.,
.
The label of first 
vertices 
are same as in Case 2. For the last three vertices
, 
, 
, 
is define as
Thus, from all above cases, we conclude that
4.2. L(0,1)-Labelling of Two Cycles
Lemma 3. Let G be a graph which contains two cycles and they have a common cutvertex. If 
be the degree of G, then
(3)
Proof. Let 
contains two cycles 
and 
of lengths 
and 
respectively. Let 
be the cutverx and 
be the degree of
. Let 
and 
be the vertices of 
and 
respecvely. The labelling procedure of
’s of 
of same as given in Lemma 2. Now we label the cycle 
as follows.
Case 1. Let 
and
.
The label of the cutvertex 
is 0, i.e.,
. The label of other vertices of 
are as follows:
Case 2. For 
(mod 4) and
.
Case 3. For 
(mod 4) and
.
Case 4. For 
(mod 4) and
.
The label of the vertices of 
are same as given in Case 3 of that lemma.
Case 5. For 
(mod 4) and 
In this case, we label the of 
as given in Case 3.
Case 6. For 
(mod 4) and
.
Here we label the adjacent vertices of 
by 
and
. Now we label the other vertices 
of 
as follows.
The above 
is redefine for the vertex 
as
.
In particular when
, then we label the vertices of 
as follows.
The label of the cutvertex 
and two adjacent vertices 
and 
are same as above. And we label the remaining vertex 
by
.
Case 7. For 
(mod 4) and 
(mod 4).
Here the label of two vertices
, 
of 
are same as given in the above case. Now we label the vertices 
as
For the vertices 
and
, 
is defined as 
and 
In particular when
, then we label the vertices of 
as follows.
The label of the cutvertex 
and two adjacent vertices of 
of 
are same as above. Now we label the remaining vertices of 
as
Case 8. For 
(mod 4) and 
(mod 4).
The label of 
of 
are same as in above case. For the vertex
, we label it as
.
When
, then the label of the vertices
, 
, 
, 
and 
are same as in the above case, and
.
Case 9. For 
(mod 4) and 
(mod 4).
We label the vertices of 
and the vertices
, 
, 
, 
and 
of 
according to the Case 8. For the vertex
, 
is
In particular, when
, the the label of the vertices of 
are same as the label of the vertices of 
as in the above case except the vertex
. The label of the vertex 
is 
Case 10. For 
(mod 4) and 
(mod 4).
We label first 
vertices 
of 
as
For the last four vertices
, 
, 
and
, the above 
is define as
In particular when
, the label of the vertices
, 
, 
and 
are same as the label of the vertices 
shown above.
Case 11. For 
(mod 4) and 
(mod 4).
We label first 
vertices 
of 
as per Case 10. And for the last five vertices, 
is define as
In particular when 
then the label of the vertices 
are same as the label of the last five vertices of 
of the above case.
Case 12. For 
(mod 4) and 
(mod 4).
Now we label first 
vertices of 
as
For the last two vertices 
and
, the 
is
Case 13. For 
(mod 4) and 
(mod 4).
Here we label the other vertices of 
as in Case 11.
Case 14. For 
(mod 4) and 
(mod 4).
In this case, the label of 
are same as in Case 12.
Case 15. For 
(mod 4) and 
(mod 4).
We label the vertices of 
as per Case 12.
Thus from the above cases, it follow that
4.3. L(0,1)-Labelling of Three Cycles
Lemma 4. Let G be a graph, contains three cycles and they have a common cutvertex
. If 
be the degree of
, then,
(4)
Proof. Let
, 
and 
be three cycles join by a common cutvertex
, of lengths
, 
and 
respectively. Let 
be the degree of the cutvertex, i.e.,
. Let
, 
, 
,
;
, 
, 
,
;
, 
, 
, 
be the vertices of the cycles respectively.
Now we label the graph as follows.
Case 1. Let
, 
and
.
In Case 1 of Lemma 3, we label a graph which contains two cycles of length three and they have a common cutvertex
. According to the previous lemma we label the vertices of the third cycle of length 3 as follows:
Case 2. For
, 
, 
, where 
.
All the subcases of this case, the label of two vertices 
and 
of the cycle 
are
and the label of other two cycles are of different types, they are discussed below.
When 
(mod 4) and
.
Here the label of the vertices of the cycles 
and
, joined with a common cutvertex 
are same as in Case 2 of Lemma 3.
When 
(mod 4) and
.
The label of the vertices of 
and 
are same as in Case 3 of Lemma 3.
When 
(mod 4) and
.
The label of the vertices of 
and 
are same as in Case 4 of Lemma 3.
When 
(mod 4) and
.
We label the vertices of 
and 
as in Case 5 of Lemma 3.
Case 3. For
, 
, 
, where 
.
In all subcases of this case, the label of three vertices of 
are
And the label of the vertices of first two cycles 
and 
are same as in Case 6, Case 7, ..., Case 15, respectively of Lemma 3.
Case 4. For
, 
, 
(mod 4), where
.
For all subcases of this case, we label the vertices of the third cycle 
as same as the labelling of the vertices of 
in Case 6 of Lemma 3 except two vertices 
and 
(the adjacent vertices of
). Then we label these vertices as 
and
.
And we label the first two cycles 
and 
(joined by a common cutvertex
), as same as in Case 6, Case 7, ..., Case 15 of Lemma 3.
Case 5. For
, 
, 
(mod 4), where
.
In all subcases of this case, we label the vertices of the third cycle 
using the same process to labelling the vertices of 
in Case 7 of Lemma 3 except two vertices 
and 
(the adjacent vertices of
). Now we label the adjacent vertices of 
of 
as 
and
.
And we label the first two cycles 
and 
(joined by a common cutvertex
), as same as in Case 10, Case 11, ..., Case 15 of Lemma 3.
Case 6. For
, 
, 
(mod 4), where
.
In all subcases of this case, we label the vertices of the third cycle 
using the same process of labelling of the vertices of 
in Case 7 of Lemma 3, except two vertices 
and
. The label of 
and 
are 
and
.
And the label of the vertices of the cycles 
and 
are same as in Case 13, Case 14 and Case 15 respectively of Lemma 3.
Case 7. For 
(mod 4), 
(mod 4), 
(mod 4).
The label of the vertices of 
and 
are same as in Case 15 of Lemma 3. Here
. Then we label the vertices of 
using the same process of 
in Case 9 of Lemma 3 except the vertices 
and
. We label these vertices as 
and
. Thus from all above cases, it follow that
4.4. L(0,1)-Labelling of Four Cycles
Using th results from Lemma 3 and Lemma 4 we can write the following statement.
Lemma 5. Let G be a graph which contains four cycles of any length and they have a common cutvertex. Then,
(5)
where 
be the degree of the cutvertex.
Corollary 1. Let G be a graph which contains finite number of cycles and they have a common cutvertex. If the vertices of the cycles (except the cutvertex) contain one or more edges then,
(6)
4.5. L(0,1)-Labelling of Finite Number of Cycles
Let 
be a graph which contains 
number of cycles of length 3. Sometimes a cycle of length three is called triangle. A triangle is a subgraph of a cactus graph. Also, a triangle shaped star, (i.e., all the triangles that have a common cutvertex) is a subgraph of a cactus graph. Now, we consider a triangle shaped star for 
-labelling. Let 
be the 
triangles meet at a common cutvertex 
and we denote this graph by
which is equivalent to
. The number of vertices and edges of 
are 
and 
respectively. Again the graph 
may also contains 
number of cycles of finite length.
Then from Lemmas 3-5 we conclude the general form of these lemmas which is given below.
Lemma 6. Let the graph G contains n number of cycles of any length and they joined at a cutvertex, then
(7)
where 
be the degree of the cutvertex.
Proof. At first we prove that when 
contains 
number of cycles of length 3 then the value of 
is
, where 
be the degree of the graph. Let 
be the 
number of triangles joined with a common cutvertex 
(shown in Figure 2).
Let
, 
and 
be the vertices of
, where
. We label 
by 0.
Then according to the previous lemmas the labels of
Figure 2. A graph contains n numbers of triangles.
’s are as follows.
For
,
Now, the label of the vertex 
of 
is 
.
Therefore, 
, when 
contains 
number of cycles of length 3.
Again, if we consider a graph 
contains 
number of 
and 
number of
, then, 
, where 
be the degree of cutvertex, then the general form can be proved by mathematical induction, that is, when a graph 
contains finite number of cycles of any length then
.
Let 
contains 
number of
’s 
and 
number of
’s
. Let 
be the common vertex and degree of 
is
. Again let
, 
, 
, 
be the vertices of 
and
, 
, 
be the vertices of
. We label 
as 0. Then we label the other vertices of
’s as follows.
For 
and then the label of the vertices of
’s are given by.
For j = 0, 1, ···, n − 1,
Now the label of third vertex of 
is
Therefore,
.
The general form can be proved by mathematical induction.
Hence the result.
Lemma 7. If a graph G contains finite number of cycles of any length and finite number of edges and they have a common cutvertex of degree
, then
(8)
Proof. Suppose that the lemma is true for 
number of cycles of any length and 
number of edges. Now we have to prove that if we add a cycle of any length to the cutvertex then the value of 
for the new graph will be same, i.e., the value of 
will preserve for 
number of cycles of any length.
Now the graph 
contains 
number of cycles of any length and 
number of edges joined with a common cutvertex
. Then the degree of 
is
. In the previous lemma we proved that when a graph contains finite number of cycles of any length then,
At first we prove that if 
contains 
number of cycles of length 3 and 
number of edges then the value of 
for that graph remains same. When all cycles are of length 3 then according to the Lemma 5 the label of two vertices of 
th cycle of length 3 are as
where
, 
and 
are three vertices of 
th cycle. Let
,
;
,
;
;
, 
are the vertices of 
edges. Here the label of the cutvertex 
is 0. Then we label the other vertices of the edges as follows
Now we add another cycle of length 3 to the cutvertex
. Then the degree of 
is
. Let
, 
and 
be the vertices of 
th cycle. We label the two vertices 
and 
as
Here we see that the label of third vertex of 
th cycle is 
as
. That is, the value of 
of the graph which contains 
number of cycles of length 3 and 
number of edges is same.
Similarly, we can prove that the value of 
will preserve for the graph which contains 
number of cycles and 
number of edges.
Hence the result.
5. L(0,1)-Labelling of Sun
Let us consider the sun 
of 
vertices. This graph is obtained by adding an edge to each vertex of a cycle
. So 
is a subgraph of
. But, what is the value of 
Lemma 8. For any sun
,
(9)
Proof. Let 
be constructed from 
by adding an edge to each vertex. To label this graph we consider five cases.
Let 
be the vertices of 
and 
is adjacent to 
and
. To complete
, we add an edge (
,
) to the vertex
, i.e.,
’s are the pendent vertices. The labelling procedure of 
is same as given in Lemma 2. Now we label the pendent vertices as follows.
Case 1. Let
.
The label of 
's are as
Case 2. Let 
(mod 4).
The label of 
are assigned as 
for
.
Case 3. Let 
(mod 4).
The label of the vertices
’s are given by
; 
for 
and 
and
.
Case 4. Let 
(mod 4).
The label of the vertices 
's are given by
, for
;
for
;
, for
.
Case 5. Let 
(mod 4).
In this case the label of the pendent vertices 
's are as follows:
, for
;
;
, for
.
Hence
.
Lemma 9. Let 
be a graph obtained from 
by adding an edge to each of the pendent vertex of
, then
(10)
Proof. Let the graph is obtained by adding an edge 
to each of the pendent vertices 
s. So in the new graph 
are the pendent vertices.
Case 1. Let
.
, 
, for
.
Case 2. Let 
(mod 4).
, for
.
Case 3. Let 
(mod 4).
, 
and
, for 
.
Case 4. Let 
(mod 4).
and
, for
.
Case 5. Let 
(mod 4).
, for
, and
, for
.
Thus, from all above cases we have
.
Lemma 10. If the graph 
contains a cycle of any length and each vertex of the cycle has another cycle of any length, then
(11)
where 
is the degree of
.
Proof. At first we prove that if the graph 
contains a cycle 
of length 
and each vertex of 
contains another cycle of length 3, or length 3 and length 4, or length 4, then 
or
. Let 
be the vertices of 
and
;
;
; 
are the vertices of all
’s which are joined with each vertex of
. If the vertex 
of 
contains a cycle of length 4 then let the vertices of 
be
, 
and
. Again if the vertex 
contains a cycle of length 4 then let the vertices of 
be
, 
and
. Therefore, all the vertices of 
are cutvertices.
Now we label the graph as follows.
Case 1. For
.
Now, we label the vertices of other cycles joined with the vertices of 
according to the following rule
If there are three cycles of length 4 then we label the vertices as follows:
and 
Case 2. Let 
(mod 4).
For
,
If all other cycles are of length 4 then the label of the last vertex of the last cycle is 3.
Case 3. Let 
(mod 4).
and 
If the (
)th vertex 
of 
contains a cycle of length 4 then we label the vertices of 
as
Case 4. Let 
(mod 4).
Now the label of all 
's are as follows:
for
, 
for
, 
and for
, 
If 
contain combined 
's and 
's then the minimum span is 3.
Case 5. Let 
(mod 4).
For
, 
and for
, 
If the vertex 
contains a cycle of length 4 then the label of the vertices adjacent to 
are
From all the above cases, we see that 3 or 4 are used to label
, which is equal to 
or
.
Therefore,
.
The proves of the other cases are similar.
Corollary 2. Let G be a graph which contains a cycle of length
. If each vertex of the cycle contains two or more cycles of length more than 2, then
(12)
where 
be the degree of G.
6. L(0,1)-Labelling of Caterpillar Graph
Now, we label another important subclass of cactus graphs called caterpillar graph.
Definition 1. A caterpillar 
is a tree where all vertices of degree 
lie on a path, called the backbone of
. The hairlength of a caterpillar graph 
is the maximum distance of a non-backbone vertex to the backbone.
Lemma 11. If G be a caterpillar graph and 
be its degree, then
(13)
Proof. Let 
be a path of length 
of the caterpillar graph and 
be the vertices of
. We label the vertices of 
according to the following rule.
Let us assume that 
be any vertex of 
and
, 
are the adjacent vertices of
. As we label of the vertices of 
by 0 or 1 so without loss of generality let us consider that the label of
, 
and 
are 0, 0 and 1 respectively. Again let us consider that 
number of paths
;
, of same lengths are joined to the vertex
. Let
; 
are the vertices of 
paths other than
. Now we label the vertices of these paths as in the following method:
for 
and
.
And we label other vertices of 
's as per the rule to label the vertices of
.
Now,
.
The result will be same when finite number of paths of different lengths are joined to one or more vertices of the path of the caterpillar graph.
So,
.
7. L(0,1)-Labelling of Lobster
Another subclass of cactus graphs is the lobster graph. The definition of lobster graph is given below.
Definition 2. A lobster is a tree having a path (of maximum length) from which every vertex has distance at most
, where 
is an integer.
The maximum distance of the vertex from the path is called the diameter of the lobster graph. There are many types of lobsters given in literature like diameter 2, diameter 4, diameter 5, etc.
Lemma 12. Let G be a lobster graph. If 
be the degree of the lobster graph, then
(14)
Proof. Let 
be a path of length 
of the lobster graph and 
be the vertices of
. First we label the vertices of 
according to Lemma 11.
Then we label the other vertices of that graph. Let us denote the other vertices of the graph by
. Here 
are adjacent to 
and
,
. The label of the vertices of 
are either 0 or 1. Then we label the vertices 
by 2, 3 and so on [it depends upon (
)].
Thus the 
-value of a lobster is
.
We know from [2] that 3 labels are sufficent to label a complete binary tree by 
-labelling. Now we have to prove that for any tree the value of 
is
, where 
is the degree of the tree.
Lemma 13. For any tree 
of degree
,
(15)
Proof. Let 
be a tree with degree
. We first label the root of the tree by 0. Now we know from the definition of 
-labelling that the label difference between any two adjacent vertices is at least 0 and the label difference between any two vertices which are at distance two is at least 1. Now we label the children of the root from left to right by 0, 1, 2, 
,
.
Let us consider the 
th vertex of the tree. Assume that the label of the parent of 
is known. Then the allowable label for the children of 
are 0, 1, 2, 
except the label of the parent of
. Now, we label the children of 
by 0, 1, 2, 
, 
, 
, where 
is the degree of the vertex
, except the label of the parent of
. This process is valid for any vertex 
of the tree. Thus the maximum label used to label the entire tree by 
-labelling is max{
: 
}, which is exactly equal to
.
Hence
.
The 
-labelling of all subgraphs of cactus graphs and their combinations are discussed in the previous lemmas. From these results we conclude that the 
-value of any cactus graph can not be more than
. Hence we have the following theorem.
Theorem 1. If 
is the degree of a cactus graph
, then
(16)
The graph of Figure 3 is an example of a cactus graph, contains all possible subgraphs and its 
-labelling.
8. Conclusion
The bounds of 
-labelling of a cactus graph and various subclass viz., cycle, sun, star, caterpillar, lobster and tree are investigated. The bounds of 
for these graphs are 
or 2, for sun, star, caterpillar, lobster and tree it is
. For the cactus graph the bound is
, where 
is the
Figure 3. L(0,1)-labelling of a cactus graph.
maximum degree of the cactus graph
. Currently we are engaged to find the bounds for 
-labelling for different values of
, 
on cactus graphs.