#### 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 well-studied 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 circular-arc 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 c-dominating 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 circular-arc 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 well-studied 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 .

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).

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).

 â€‰ Input: The nodes of the path , . â€‰ Output: The -label of the path . â€‰ Step 1: Rearrange the intervals as follows: â€‰ â€ƒCase I: is odd â€‰ â€ƒ; â€‰ â€ƒ, for ; â€‰ â€ƒ, for ; â€‰ â€ƒ remains same; â€‰ â€ƒCase II: is even â€‰ â€ƒ; â€‰ â€ƒ, for ; â€‰ â€ƒ, for ; â€‰ â€ƒ remain same; â€‰ Step 2: Label the node by , i.e., , for â€‰ end.

#### 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).

 â€‰ Input:, and , . â€‰ Output: for ; . â€‰ Step 1: for to , where â€‰ â€ƒfor to â€‰ â€ƒlet be the th element of â€‰ â€ƒif , then add to the set ; â€‰ â€ƒend for; â€‰ end for; â€‰ Step 2: for to â€‰ â€ƒfor to â€‰ â€ƒâ€ƒ and be the elements of and respectively; â€‰ â€ƒâ€ƒif , then add to the set ; â€‰ â€ƒend for; â€‰ end for; â€‰ end.

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).

 â€‰ Input: The intervals of an and for and where . â€‰ Output:, the -label of the interval , . â€‰ Step 1: Rearrange the intervals as follows: â€‰ â€ƒCase I: is odd â€‰ â€ƒ; â€‰ â€ƒ, for ; â€‰ â€ƒ, for ; â€‰ â€ƒ remains same; â€‰ â€ƒCase II: is even â€‰ â€ƒ; â€‰ â€ƒ, for ; â€‰ â€ƒ, for ; â€‰ â€ƒ is unchanged; â€‰ Step 2: (Initialization) â€‰ â€ƒ; â€‰ â€ƒ; â€‰ Step 3: for to â€‰ â€ƒ; â€‰ â€ƒ; â€‰ â€ƒ; â€‰ end for; â€‰ Step 4:; â€‰ â€ƒ; â€‰ Step 5:; â€‰ end.

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:

, , , , ,