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 [
Bertossi and Bonuccelli [
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 [
. 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)}.
-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.
of any graph 
is 
[
is the degree of the graph.
of 
vertices. It can easily be verified that
, 
for 
[
are merged then 
becomes
. In [
is equal to 1 if n is multiple of 4 and 2 otherwise.
, it is easy to check that
.
, is obtained by joining 
and
, i.e.,
. It is also easy to check that 
.
- 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.
, 
is equal to
, implying that
. An optimal 
-labelling can be also determined by exploiting the algorithm provided in [
- labelling trees.
proving that
. They give also approximation algorithms for the 
- labelling running in 
time.
-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 [
-labelling a permutation graph in 
time; it guarantees the bound
.
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 [
[
[
-labelling of cartesian product of a cycle and path is done by Chiang and Yan [
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.
,
. The best known result till date is 
due to Goncalves [
[
. Wang and Lih [
is a planar graph of girth (girth is defined to be the length of a shortest cycle in
) at least 5, then 
of a cactus graph 
is 
-labelling of cartesian product of a cycle and a path is done by Chiang and Yan [
-labelling, see [21- 30].
and 
then we get another special case which is called 
-labelling problem. Some results of 
-labelling problem are given below.
and 
for each
, and 
is 2 if 
is a multiple of 3 and it is 3 otherwise [
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 
,
. The star graph 
is a subgraph of cactus graph, therefore, one can conclude the following result.
by 
-labelling and they have obtained the following result. Here we have given a constructive prove of this result.
of length
,
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.
.
(mod 4), i.e.,
.
(mod 4), i.e.,
.
vertices 
are same as in Case 2. For the last vertex
, 
is define as
(mod 4), i.e.,
.
are same as in Case 3. For the last vertex
, 
is define as
(mod 4), i.e.,
.
vertices 
are same as in Case 2. For the last three vertices
, 
, 
, 
is define as
be the degree of G, then
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.
and
.
is 0, i.e.,
. The label of other vertices of 
are as follows:
(mod 4) and
.
(mod 4) and
.
(mod 4) and
.
are same as given in Case 3 of that lemma.
(mod 4) and 
as given in Case 3.
(mod 4) and
.
by 
and
. Now we label the other vertices 
of 
as follows.
is redefine for the vertex 
as
.
, then we label the vertices of 
as follows.
and two adjacent vertices 
and 
are same as above. And we label the remaining vertex 
by
.
(mod 4) and 
(mod 4).
, 
of 
are same as given in the above case. Now we label the vertices 
as
and
, 
is defined as 
and 
, then we label the vertices of 
as follows.
and two adjacent vertices of 
of 
are same as above. Now we label the remaining vertices of 
as
(mod 4) and 
(mod 4).
of 
are same as in above case. For the vertex
, we label it as
.
, then the label of the vertices
, 
, 
, 
and 
are same as in the above case, and
.
(mod 4) and 
(mod 4).
and the vertices
, 
, 
, 
and 
of 
according to the Case 8. For the vertex
, 
is
, 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 
(mod 4) and 
(mod 4).
vertices 
of 
as
, 
, 
and
, the above 
is define as
, the label of the vertices
, 
, 
and 
are same as the label of the vertices 
shown above.
(mod 4) and 
(mod 4).
vertices 
of 
as per Case 10. And for the last five vertices, 
is define as
then the label of the vertices 
are same as the label of the last five vertices of 
of the above case.
(mod 4) and 
(mod 4).
vertices of 
as
and
, the 
is
(mod 4) and 
(mod 4).
as in Case 11.
(mod 4) and 
(mod 4).
are same as in Case 12.
(mod 4) and 
(mod 4).
as per Case 12.
. If 
be the degree of
, then,
, 
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.
, 
and
.
. According to the previous lemma we label the vertices of the third cycle of length 3 as follows:
, 
, 
, where 
.
and 
of the cycle 
are
(mod 4) and
.
and
, joined with a common cutvertex 
are same as in Case 2 of Lemma 3.
(mod 4) and
.
and 
are same as in Case 3 of Lemma 3.
(mod 4) and
.
and 
are same as in Case 4 of Lemma 3.
(mod 4) and
.
and 
as in Case 5 of Lemma 3.
, 
, 
, where 
.
are
and 
are same as in Case 6, Case 7, ..., Case 15, respectively of Lemma 3.
, 
, 
(mod 4), where
.
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 
(joined by a common cutvertex
), as same as in Case 6, Case 7, ..., Case 15 of Lemma 3.
, 
, 
(mod 4), where
.
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 
(joined by a common cutvertex
), as same as in Case 10, Case 11, ..., Case 15 of Lemma 3.
, 
, 
(mod 4), where
.
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 
are same as in Case 13, Case 14 and Case 15 respectively of Lemma 3.
(mod 4), 
(mod 4), 
(mod 4).
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
be the degree of the cutvertex.
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.
be the degree of the cutvertex.
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 
, 
and 
be the vertices of
, where
. We label 
by 0.
’s are as follows.
,
of 
is 
.
, when 
contains 
number of cycles of length 3.
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
.
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.
’s are given by.
is
.
, then
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.
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,
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
, 
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
. Then the degree of 
is
. Let
, 
and 
be the vertices of 
th cycle. We label the two vertices 
and 
as
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.
will preserve for the graph which contains 
number of cycles and 
number of edges.
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 
,
be constructed from 
by adding an edge to each vertex. To label this graph we consider five cases.
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.
.
's are as
(mod 4).
are assigned as 
for
.
(mod 4).
’s are given by
; 
for 
and 
and
.
(mod 4).
's are given by
, for
;
for
;
, for
.
(mod 4).
's are as follows:
, for
;
;
, for
.
.
be a graph obtained from 
by adding an edge to each of the pendent vertex of
, then
to each of the pendent vertices 
s. So in the new graph 
are the pendent vertices.
.
, 
, for
.
(mod 4).
, for
.
(mod 4).
, 
and
, for 
.
(mod 4).
, for
.
(mod 4).
, for
, and
, for
.
.
contains a cycle of any length and each vertex of the cycle has another cycle of any length, then
is the degree of
.
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.
.
according to the following rule
(mod 4).
,
(mod 4).
)th vertex 
of 
contains a cycle of length 4 then we label the vertices of 
as
(mod 4).
's are as follows:
, 
, 
, 
contain combined 
's and 
's then the minimum span is 3.
(mod 4).
, 
, 
contains a cycle of length 4 then the label of the vertices adjacent to 
are
, which is equal to 
or
.
.
. If each vertex of the cycle contains two or more cycles of length more than 2, then
be the degree of G.
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.
be its degree, then
be a path of length 
of the caterpillar graph and 
be the vertices of
. We label the vertices of 
according to the following rule.
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
.
's as per the rule to label the vertices of
.
.
.
, where 
is an integer.
be the degree of the lobster graph, then
be a path of length 
of the lobster graph and 
be the vertices of
. First we label the vertices of 
according to Lemma 11.
. 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 (
)].
-value of a lobster is
.
-labelling. Now we have to prove that for any tree the value of 
is
, where 
is the degree of the tree.
of degree
,
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, 
,
.
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
.
.
-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.
is the degree of a cactus graph
, then
-labelling.
-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
. Currently we are engaged to find the bounds for 
-labelling for different values of
, 
on cactus graphs.