Abstract
Graph labelling problem has been broadly studied for a long period for its applications, especially in frequency assignment in (mobile) communication system, ray crystallography, circuit design, etc. Nowadays, surjective labelling is a wellstudied problem. Motivated from the labelling problem and the importance of surjective labelling problem, we consider surjective labelling (labelling) problems for paths and interval graphs. For any graph , an labelling is a mapping so that, for every pair of nodes and , if , then ; and if , then , and every label is used exactly once, where represents the distance between the nodes and , and is the number of nodes of graph . In the present article, it is proved that any path can be surjectively labelled if , and it is also proved that any interval graph having nodes and degree can be surjectively labelled if . Also, we have designed two efficient algorithms for surjective labelling of paths and interval graphs. The results regarding both paths and interval graphs are the first result for surjective labelling.
1. Introduction
The frequency assignment problem is bottomed from the problems of distance labelling of graph. In 1992, labelling was invented by Griggs and Yeh [1] in conjunction with channel assigning problem in a multihop radio network.
For any graph , an labelling is a mapping , so that if and if . The span of labelling of is . The labelling number of is the smallest natural number so that has an labelling of span .
A surjective labelling of is a mapping so that when and when , and it requires that each label, , be used only once, where is the number of nodes of .
In Figure 1, we have shown an labelling of a path with 5 nodes and Figure 2 shows labelling of the same graph. In Figure 1, identical label is used several times but in Figure 2 the labels 1 to 5 are used only once. So, in labelling, there is a more complex task compared to labelling.
In 1994, Sakai has proved some results regarding distance two labelling of chordal graph. Later, in 2007, Bertossi and Bonuccelli have studied approximate coloring of trees and interval graphs. Amanathulla and Pal have studied a lot of problems regarding labelling of graphs, like labelling problems on permutation graphs [2], labelling problems on interval graphs [3], labelling problems on circulararc graphs [4],  and labelling problems of square of paths [5], and labelling numbers of squares of paths, complete graphs, and complete bipartite graphs [6]. In 2019, Berhe and Wang have studied computation of certain topological coindices of graphene sheet and C4C8(S) nanotubes and nanotorus [7]; also, Goyal et al. [8] have studied new composition of graphs and their Wiener indices. Hosamani et al. [9] have studied graphs with equal dominating and cdominating energy. Very recently, B. M. Gurevich has published one paper regarding classes of infinite loaded graphs with randomly deleted edges [10] and Ranjini et al. have studied degree sequence of graph operator for some standard graphs [11]. For general graph, . In 1992, Griggs et al. showed that and have proposed a conjecture [1].
In 1993, Jonas [12] has shown that . Chang et al. [13] have showed . KrÃ¡lâ€™ and Skrekovski [14] proved that and they further improved it to [15].
Different bounds for were obtained for different classes of graphs. Some results regarding upper bound of labelling are shown in Table 1.
In [37], Lingscheit et al. investigated minimal and surjective labelling for path, cycle, complete graph, caterpillar, and complete bipartite graph. They have shown that can be surjectively labelled when . Very recently, Amanathulla and Pal have studied labelling of cycle and circulararc graph and obtained good results for it [38].
labelling of graphs is a rapidly studied problem for its applications in various fields, especially in channel assignment in radio network. In labelling, although there is a light chance to overlap the frequencies in radio network, it cannot be neglected, but in labelling there is no chance to overlap the frequencies, as in this case the labels are distinct. For this reason, in the recent year, labelling of graph has become a wellstudied problem due to its applications. This motivates us to consider labelling of paths and s. Recently, many researchers applied various related concepts on graphs in different aspects (see, e.g., [39â€“43]).
In the present article, it is shown that any path is surjectively labelled by labelling if and it also shown that any having nodes can be surjectively labelled if . Two polynomial time algorithms are also established to label a path and an by labelling.
The remainder of this article is organized as follows: in Section 2, some notations and preliminary definitions are given. In Section 3, labelling of path has been presented. In Section 4, labelling of is investigated. The last section presents concluding remarks.
2. Preliminaries and Notations
A path is a graph , where , for all , where , and it is denoted by . Here, we consider which is not a path, so , because if , then it may be a path.
Let the set of intervals in real line be , where , , , and are the left and right endpoints of . For any interval , we draw a node , and two nodes and have joined by a line segment that if the corresponding intervals have common portion, then we obtain an [44]. Throughout the paper, an interval and a node are the same. An and its interval representation are shown in Figure 3.
Notations. For any with nodes and corresponding set of intervals , we define the following:(1): the set of used labels which are used before labelling the interval , for every interval .(2): the set of used labels at distance from the interval , before labelling , for every .(3): the set of valid labels for the interval before labelling , satisfying the adjoining condition of labelling, for every interval .(4): the valid set of labels of the interval before labelling , satisfying labelling condition, for every interval .(5): the set of valid labels of the interval before labelling , for every interval .(6): labelling number of .(7): the label of the interval .(8): the set of labels after completion of labelling of .
3. Surjective Labelling of Paths
In this portion, we have presented labelling of path and have showed that any path is surjectively labelled if . Also, we have presented a greedy algorithm to label a path by labelling.
Theorem 1. For ,
Proof. Let be a path having nodes.â€‰Case 1: .â€‰This result holds trivially.â€‰Case 2: .â€‰The labels used are 1 and 3 and hence .â€‰Case 3: .â€‰There are two possible cases shown in Figures 4(a) and 4(b). The labelling sequences are and . From the above result, it is concluded that can be labelled for .
(a)
(b)
Theorem 2. The minimum path that can be labelled by labelling is .
Proof. From Theorem 1, we have and , so, for , a path cannot be labelled by labelling. The labelling pattern of path (see Figure 5) shows that can be labelled by labelling. Hence, is the minimum path that can be labelled by labelling (Figure 6).
(a)
(b)
For this path, the node . Here, , so this path can be surjectively labelled by labelling. is the label of the node , for .
According to Algorithm 1, we rearrange the nodes as follows:
, , , , , , , , , , , , , , , , , , , , , and remains unchanged.
Now, node is labelled by ; that is, , for each . After completion of surjective labelling of , the node and the label of the corresponding node are shown in Figure 7(b).

(a)
(b)
4. Surjective L(2, 1)Labelling of s
Here, some lemmas that we have used to develop the proposed algorithm are presented.
Lemma 1. For any , , for .
Proof. Let be an having nodes. The labelling of starts from the leftmost interval. Let node be corresponding to the interval of the . Suppose that in a stage the intervals (for some ) are previously labelled by labelling and the remaining intervals are unlabelled.
Let . This means that the number of distinct labels used for labelling distance two intervals from the interval before labelling is . Since the degree of the is , there exists an interval (see Figure 8) and those are adjoining to intervals at most. In Figure 8, is adjoining to . Among the intervals, some intervals ( in Figure 8) are of distance two apart from and among the intervals there is an interval ( in Figure 8) whose distance is not two from . Hence, ; that is, .
Observation 1. For any , , for any interval , .
Observation 2. For any , , for any interval .
Theorem 4. Any with nodes is surjectively labelled if .
Proof. Since has nodes, let . Since we want to label the intervals of an by labelling, every label is used exactly once and the labels must be in . So,Again, since has nodes, to label the whole graph by labelling, distinct labels must be required. Also, since , in the extreme unfavorable cases labels are required to label graph by labelling. Again, in labelling, the highest label is equal to . Hence, an is surjectively labelled using labelling if .
If , then the may or may not be labelled by labelling, because in the worst case labels are required to label the , which is not equal to . This contradicts the condition that the used label must belong to and the highest label must be equal to for labelling.
4.1. Algorithm for Surjective L(2, 1)Labelling of s
In this part, two algorithms are designed: one is to compute and the other is to compute label for an (Algorithm 2).

Lemma 2. for is correctly computed by Algorithm 2 and the time complexity of the above algorithm is .
Proof. According to Algorithm 2, each element differs from by at least 2 for each . Therefore, for all and for all . So, Algorithm 2 correctly computes the set for each , . Again, each element of differs from by at least 1 for each . Therefore, for all and for all , and for all and for all . Therefore, Algorithm 2 correctly computes for every . As is the cardinality of the set of labels , for and , and also , where . So, is computed by using at most times, that is, using times. Again, , so, is computed using at most times, that is, using times. Since , the iterative time for algorithm is .
Lemma 3. For each , is the nonempty largest set satisfying distance one condition of labelling, , for every , and , for any .
Proof. Since and (by Observation 1), for every . Therefore, , so, is a nonempty set. Also, let be an arbitrary set of labels, which satisfies distance one condition of labelling, , for all , and . Then, for , for any . Thus, . So, . Then, . Since is arbitrary, is the largest nonempty set of labels which satisfies distance one condition of labelling, , for every , and , for any .
Lemma 4. For any , is the nonempty largest set satisfying labelling condition, , for every , , and .
Proof. Since and , for (by Observation 1), for , ; that is, for all and for all . Hence, is the valid label of ; therefore, . This shows that is a nonempty set. Also, let be an arbitrary set of labels which satisfies labelling conditions, for every , and . Then, for , for and for any . Thus, . Thus, . So, . Since is arbitrary, is the largest nonempty set of labels which satisfies labelling, for every , and , for any (Algorithm 3).

Theorem 5. Algorithm 3 correctly labels an by labelling, where .
Proof. Let be an with nodes such that . We rearranged the nodes so that no two consecutive intervals are adjacent to each other. After rearrangement of the intervals, let and let and .
We consider circumstances in which the intervals are already labelled for and the remaining intervals are not labelled. In this stage, our aim is to label by labelling. Now, is the nonempty largest set of labels satisfying labelling, for any , and for every (by Lemma 4).
Again, , so is the nonempty largest set satisfying labelling, as the label in was not used previously to label any interval and also satisfies labelling. Therefore, , where . Since is the largest set of labels satisfying labelling, is the least surjective label of . Since , the label of must be less than or equal to . Again, since , the interval is labelled by using only the labels from which have not been used earlier to label any interval. Since is arbitrary, any is surjectively labelled by labelling by Algorithm 3.
Theorem 6. The running time of Algorithm 3 is , where .
Proof. According to Algorithm 3, the label of , that is, , is computed if the set is computed. Now by Lemma 2 we see that by algorithm one can compute the set , , by using time. Again, by Appendix, algorithm diff, have been computed in time, for any . Therefore, the time needed for computing is . Since we need to find for , the time complexity for Algorithm 3 is , that is, .
4.1.1. Illustration of Algorithm
We take an having 14 nodes (see Figure 9) and label that graph by Algorithm 3. The graph after completion of surjective labelling is given in Figure 10.
For the above graph, the set of intervals and . Here, , so this can be surjectively labelled by labelling. is the label of the interval , for .
According to Algorithm 3, at first, we rearrange the intervals as follows:
, , , , ,