Algorithmic complexity of proper labeling problems

TL;DR

This paper investigates the computational complexity of various proper labeling problems in graphs, presenting polynomial algorithms for some cases and NP-completeness results for others.

Contribution

It provides new insights into the algorithmic complexity of proper labeling variants, including polynomial algorithms and NP-completeness proofs.

Findings

Polynomial time algorithms for certain proper labeling variants

NP-completeness results for other labeling problems

Enhanced understanding of the computational boundaries of proper labeling

Abstract

A proper labeling of a graph is an assignment of integers to some elements of a graph, which may be the vertices, the edges, or both of them, such that we obtain a proper vertex coloring via the labeling subject to some conditions. The problem of proper labeling offers many variants and received a great interest during recent years. We consider the algorithmic complexity of some variants of the proper labeling problems, we present some polynomial time algorithms and $ \mathbf{NP} $-completeness results for them.

Tables

Table 1. Table 1: Recent results on proper labeling of graphs

Edge	$\mathrm{𝐍𝐏}$-h	$𝐏$	Upper Bound	Conjecture
Sum	3-regular	-	${\mathbb{N}}_5$	${\mathbb{N}}_3$
Product	3-regular	-	${\mathbb{N}}_4$	${\mathbb{N}}_3$
	Planar 3-col
Gap	Planar bipartite	Planar bipartite with $\delta >1$	-	${\mathbb{N}}_{\chi +1}$
Vertex
Sum	3-regular	-	${\mathbb{N}}_{{\Delta }^2-\Delta +1}$	${\mathbb{N}}_{\chi }$
	Planar 3-col
Product	3-regular	-	${\mathbb{N}}_{\chi \ln \chi +\chi \ln \ln \chi +2}$	${\mathbb{N}}_{\chi }$
	Planar 3-col
Degree	4-col	From ${\mathbb{N}}_2$	${\mathbb{N}}_{2\chi }$	${\mathbb{N}}_{\chi }$
Maximum	3-regular	Bipartite	-	-
Gap	Bipartite	Planar Bipartite	-	-
	Planar 3-col

Equations

$$Pr({\lambda }_v^i={\lambda }_u^i)$$

$$Pr(\forall i{\lambda }_v^i={\lambda }_u^i)$$

$$Pr(\exists vu\forall i{\lambda }_v^i={\lambda }_u^i)$$

Full text

Key words: Proper labeling; Computational complexity; Multiplicative vertex-coloring weightings; Gap vertex-distinguishing edge colorings ; Fictional coloring; Vertex-labeling by maximum; 1, 2, 3-Conjecture; Multiplicative 1, 2, 3-Conjecture.

Subject classification: 05C15, 05C20, 68Q25

1 Introduction

A proper labeling of a graph is an assignment of integers to some elements of a graph, which may be the vertices, the edges, or both of them, such that we obtain a proper vertex coloring via the labeling subject to some conditions. Karoński, Łuczak and Thomason initiated the study of proper labeling [19]. They introduced an edge-labeling which is additive vertex-coloring that means for every edge $uv$, the sum of labels of the edges incident to $u$ is different from the sum of labels of the edges incident to $v$ [19]. The problem of proper labeling offers many variants and received a great interest during recent years, for instance see [1, 9, 10, 18, 19, 24].

There is a concept which has a close relationship with proper labeling problem. An adjacent vertex-distinguishing edge coloring, or avd-coloring, of a graph $G$ is a proper edge coloring of $G$ such that no pair of adjacent vertices meets the same set of colors. Avd-coloring has been widely studied, for example see [6, 8, 16].

In this paper, we consider the algorithmic complexity of the following proper labeling problems. We will present some polynomial time algorithms and $\mathbf{NP}$-completeness results for them. Throughout the paper we denote $\{1,2,\cdots,k\}$ by $\mathbb{N}_{k}$.

Edge-labeling by sum.

An edge-labeling $f$ is an edge-labeling by sum if $c(v)=\sum_{e\ni v}f(e)$ is a proper vertex coloring. This parameter was introduced by Karoński et al. and it is conjectured that three labels $\{1,2,3\}$ are sufficient for every connected graph, except $K_{2}$ (1, 2, 3-Conjecture, see [19]). This labeling has been studied extensively by several authors, for instance see [1, 2, 7, 20, 21, 24]. Currently, we know that every connected graph has an edge-labeling by sum, using the labels from $\mathbb{N}_{5}$ [18]. Furthermore, Addario-Berry, Dalal, and Reed showed that almost all graphs have an edge-labeling by sum from $\mathbb{N}_{2}$ [2].

Dudek and Wajc showed that determining whether a given graph has an edge-labeling by sum from $\mathbb{N}_{2}$ is $\mathbf{NP}$-complete [14]. They showed that determining whether a graph has an edge-labeling by sum from either $\{0,1\}$ or $\{1,2\}$ is $\mathbf{NP}$-complete. They conjectured that deciding whether a given graph has an edge-labeling by sum from $\{a,b\}$ for any rational $a$ and $b$ is $\mathbf{NP}$-complete. By a short proof we improve the current results and show that for a given 3-regular graph $G$, determining whether $G$ has an edge-labeling by sum from $\{a,b\}$ is $\mathbf{NP}$-complete.

Theorem 1

For a given 3-regular graph $G$, determining whether $G$ has an edge-labeling by sum from $\mathbb{N}_{2}$ is $\mathbf{NP}$-complete.

Vertex-labeling by sum (Lucky labeling and sigma coloring).

A vertex-labeling $f$ is a vertex-labeling by sum if $c(v)=\sum_{u\sim v}f(u)$ is a proper vertex coloring. Vertex-labeling by sum is a vertex version of the above problem, which was introduced recently by Czerwiński et al. [10]. It was conjectured that every graph $G$ has a vertex-labeling by sum, using the labels $\{1,2,\ldots,\chi(G)\}$ [10] and it was shown that every graph $G$ with $\Delta(G)\geq 2$, has a vertex-labeling by sum, using the labels $\{1,2,\ldots,\Delta^{2}-\Delta+1\}$ [4]. A similar version of this labeling was introduced by Chartrand et al. [9].

It was shown that, it is $\mathbf{NP}$-complete to decide for a given planar 3-colorable graph $G$, whether $G$ has a vertex-labeling by sum from $\mathbb{N}_{2}$ [3]. Furthermore, it is $\mathbf{NP}$-complete to determine for a 3-regular graph $G$, whether $G$ has a vertex-labeling by sum from $\mathbb{N}_{2}$ [12].

Edge-labeling by product. (Multiplicative vertex-coloring)

An edge-labeling $f$ is an edge-labeling by product if $c(v)=\prod_{e\ni v}f(e)$ is a proper vertex coloring. This variant was introduced by Skowronek-Kaziów and it is conjectured that every non-trivial graph $G$ has an edge-labeling by product, using the labels from $\mathbb{N}_{3}$ (Multiplicative 1, 2, 3-Conjecture, see [25]). Currently, we know that every non-trivial graph has an edge-labeling by product, using the labels from $\mathbb{N}_{4}$ [25]. Also, every non-trivial, 3-colorable graph $G$ permits an edge-labeling by product from $\mathbb{N}_{3}$ [25]. We will prove the following for planar 3-colorable graphs and 3-regular graphs.

Theorem 2

*We have the following:

(i) For a given planar 3-colorable graph $G$, determining whether $G$ has an edge-labeling by product from $\mathbb{N}_{2}$ is $\mathbf{NP}$-complete.

(ii) For a given 3-regular graph $G$, determining whether $G$ has an edge-labeling by product from $\mathbb{N}_{2}$ is $\mathbf{NP}$-complete.*

**Vertex-labeling by product.

**A vertex-labeling $f$ is a vertex-labeling by product if $c(v)=\prod_{u\sim v}f(u)$ is a proper vertex coloring. Vertex-labeling by product is a vertex version of the above problem.

For a given graph $G$, let $g:V(G)\rightarrow\{1,\ldots,\chi(G)\}$ be a proper vertex coloring of $G$. Label the set of vertices $g^{-1}(1)$ by 1; also, for each $i$, $1<i\leq\chi(G)$ label the set of vertices $g^{-1}(i)$ by the $(i-1)$-th prime number; this labeling is a vertex-labeling by product. In number theory, Prime Number Theorem describes the asymptotic distribution of the prime numbers. As a consequence of Prime Number Theorem we have the following bound for the size of the $n$-th prime number $p_{n}$ (i.e., $p_{1}=2$, $p_{2}=3$, etc. ): $p_{n}<n\ln n+n\ln\ln n$, for $n\geq 6$ (see [5] p. 233). So, every graph $G$ has a vertex-labeling by product, from $\{1,2,\ldots,\chi\ln\chi+\chi\ln\ln\chi+2\}$. Here, we ask the following question.

Problem 1

.* Does every graph $G$ have a vertex-labeling by product, using the labels $\{1,2,\ldots,\chi(G)\}$?*

Every bipartite graph and 3-colorable graph has a vertex-labeling by product from $\mathbb{N}_{2}$ and $\mathbb{N}_{3}$, respectively. We will prove the following about the planar 3-colorable graphs.

Theorem 3

*We have the following:

(i) For a given planar 3-colorable graph $G$, determining whether $G$ has a vertex-labeling by product from $\mathbb{N}_{2}$ is $\mathbf{NP}$-complete.

(ii) For every $k$, $k\geq 3$, it is $\mathbf{NP}$-complete to determine whether a given graph has a vertex-labeling by product from $\mathbb{N}_{k}$.*

Also, we prove the following about random graphs.

Theorem 4

For every constant $p$, $p\in(0,1)$, almost all graphs in $G(n,p)$ have a vertex-labeling by product from $\mathbb{N}_{11}$.

Let $G$ be a 3-regular graph. $G$ has a vertex-labeling by product from $\mathbb{N}_{2}$ if and only if $G$ has a vertex-labeling by sum from $\mathbb{N}_{2}$. It was proved that it is $\mathbf{NP}$-complete to determine for a given 3-regular graph $G$, whether $G$ has a vertex-labeling by sum from $\mathbb{N}_{2}$ [12]. So we have the following:

Theorem A

It is $\mathbf{NP}$-complete to determine for a given 3-regular graph $G$, whether $G$ has a vertex-labeling by product from $\mathbb{N}_{2}$.

**Edge-labeling by gap.

**An edge-labeling $f$ is an edge-labeling by gap if

c(v)=\begin{cases}1&if\,\,d(v)=0,\\ f(e)_{e\ni v}&if\,\,d(v)=1,\\ \max_{e\ni v}f(e)-\min_{e\ni v}f(e)&otherwise,\end{cases}

is a proper vertex coloring. Every graph $G$ has an edge-labeling by gap if and only if it has no connected component isomorphic to $K_{2}$; for instance, put the different powers of two on the edges of $G$.

A similar definition was introduced by Tahraoui et al. [26]. They introduced the following variant: Let $G$ be a graph, $k$ be a positive integer and $f$ be a mapping from $E(G)$ to the set $\mathbb{N}_{k}$. For each vertex $v$ of $G$, the label of $v$ is defined as

c(v)=\begin{cases}f(e)_{e\ni v}&if\,\,d(v)=1,\\ \max_{e\ni v}f(e)-\min_{e\ni v}f(e)&otherwise,\end{cases}

The mapping $f$ is called gap vertex-distinguishing labeling if distinct vertices have distinct labels. Such a coloring is called a gap-k-coloring and is denoted by $gap(G)$ [26]. It was conjectured that for a connected graph $G$ of order $n$ with $n>2$, $gap(G)\in\{n-1,n,n+1\}$ [26]. They propose study of the variant of the gap coloring problem that distinguishes the adjacent vertices only.

Now, consider the following example.

Remark 1

Every complete graph $K_{n}$ of order $n$ with $n>2$, has an edge-labeling $f_{n}$ by gap from $\{1,2,\cdots,\chi(K_{n})+1\}$. Suppose that $K_{3}=v_{1}v_{2}v_{3}$ and let $f_{3}$ be the following function: $f_{3}(v_{1}v_{2})=4$, $f_{3}(v_{1}v_{3})=1$ and $f_{3}(v_{2}v_{3})=2$. Define $f_{n}$ recursively:

f_{n}(v_{i}v_{j})=\begin{cases}f_{n-1}(v_{i}v_{j})+1&if i,j<n,\\ 1&if $(i=n$ and $j\neq 2)$ or $(j=n$ and i\neq 2),\\ 2&otherwise.\end{cases}

Let $f$ be an edge-labeling by gap from $\mathbb{N}_{k}$, for a graph $G$ we have $k\geq\chi(G)-1$. Now, we state the following problem:

Problem 2

.* Does every connected graph $G$ of order $n$ with $n>2$, have an edge-labeling by gap from $\{1,2,\ldots,\chi(G)+1\}$?*

We will prove the following:

Theorem 5

*We have the following:

(i) For a given planar bipartite graph $G$ with minimum degree two, determining whether $G$ has an edge-labeling by gap from $\mathbb{N}_{2}$ is in $\mathbf{P}$.

(ii) For a given planar bipartite graph $G$, determining whether $G$ has an edge-labeling by gap from $\mathbb{N}_{2}$ is $\mathbf{NP}$-complete.

(iii) For every $k$, $k\geq 3$, it is $\mathbf{NP}$-complete to determine whether a given graph has an edge-labeling by gap from $\mathbb{N}_{k}$.*

**Vertex-labeling by gap.

**A vertex-labeling $f$ is a vertex-labeling by gap if

c(v)=\begin{cases}1&if\,\,d(v)=0,\\ f(u)_{u\sim v}&if\,\,d(v)=1,\\ \max_{u\sim v}f(u)-\min_{u\sim v}f(u)&otherwise,\end{cases}

is a proper vertex coloring. A graph may lack any vertex-labeling by gap. Here we ask the following:

Problem 3

.* Does there exist a polynomial time algorithm to determine whether a given graph has a vertex-labeling by gap?*

Every tree $T$ has a vertex-labeling by gap from $\mathbb{N}_{2}$. Let $T$ be a tree and $v\in V(T)$ be an arbitrary vertex, define:

f(u)=\begin{cases}1&if\,\,d(u,v)\equiv 0\ (\mod 4),\\ 2&otherwise.\end{cases}

It is easy to see that this labeling is a vertex-labeling by gap from $\mathbb{N}_{2}$.

We will show that there is dichotomy for the problem determining whether a given graph has a vertex-labeling by gap from $\mathbb{N}_{2}$: it is polynomially solvable for planar bipartite graphs but $\mathbf{NP}$-complete for bipartite graphs.

Theorem 6

*We have the following:

(i) For a given planar bipartite graph $G$, determining whether $G$ has a vertex-labeling by gap from $\mathbb{N}_{2}$ is in $\mathbf{P}$.

(ii) For a given bipartite graph $G$, determining whether $G$ has a vertex-labeling by gap from $\mathbb{N}_{2}$ is $\mathbf{NP}$-complete.

(iii) For every $k$, $k\geq 3$, it is $\mathbf{NP}$-complete to determine whether a given graph has a vertex-labeling by gap from $\mathbb{N}_{k}$.

(iv) It is $\mathbf{NP}$-complete to decide whether a given planar 3-colorable graph $G$ has a vertex-labeling by gap from $\mathbb{N}_{2}$.*

Note that, every bipartite graph $G=[X,Y]$ has a vertex-labeling by gap, label the set of vertices $X$ by 1 and label the set of vertices $Y$ by different even numbers.

Remark 2

A hypergraph $H$ is a pair $(X,Y)$, where $X$ is the set of vertices and $Y$ is a set of non-empty subsets of $X$, called edges. The $k$-coloring of $H$ is a coloring $\ell:X\rightarrow\mathbb{N}_{k}$ such that, for every edge $e$ with $\mid e\mid>1$, there exist $v,u\in X$ such that $\ell(u)\neq\ell(v)$. It was shown by Thomassen [27] that, for any $k$-uniform and $k$-regular hypergraph $H$, if $k\geq 4$, then $H$ is 2-colorable. For a given $r$-regular bipartite graph $G=[X,Y]$ with $r>3$, consider the hypergraph $H$ with the vertex set $X$ and edge set $Y$ such that such that $v\in e$ in $H$ if and only if $v\in X$ is adjacent to $e\in Y$ in $G$. Let $\ell$ be a $2$-coloring for $H$. Define

c(v)=\begin{cases}\ell(v)&if\,\,v\in X,\\ 2&otherwise,\end{cases}

this labeling is vertex-labeling by gap from $\mathbb{N}_{2}$. So every $r$-regular bipartite graph $G=[X,Y]$ with $r\geq 4$, has a vertex-labeling by gap from $\mathbb{N}_{2}$. But, there are $3$-uniform, $3$-regular hypergraphs that are not $2$-colorable. For instance, consider the Fano Plane. The Fano Plane is a hypergraph with seven vertices $\mathbb{Z}_{7}$ and seven edges $\{\{i,i+1,i+3\}:1\leq i\leq 7\}$.

Problem 4

.* Determine the computational complexity of deciding whether a given 3-regular bipartite graph $G$ have a vertex-labeling by gap from $\mathbb{N}_{2}$.*

Vertex-labeling by degree. (Fictional coloring)

A vertex-labeling $f$ is a vertex-labeling by degree if $c(v)=f(v)d(v)$, where $d(v)$ is the degree of vertex $v$, is a proper vertex coloring. This parameter was introduced by Bosek, Grytczuk, Matecki and Żelazny [30]. They conjecture that every graph $G$ has a vertex-labeling by degree from $\{1,2,\cdots,\chi(G)\}$. Let $p$ be a prime number and let $G$ be a graph such that $\chi(G)\leq p-1$, they proved that $G$ has a vertex-labeling by degree from $\mathbb{N}_{p-1}$, so every graph $G$ has a vertex-labeling by degree from $\{1,2,\cdots,2\chi(G)\}$ [30].

Since determining the chromatic number of regular graph is $\mathbf{NP}$-hard, hence for a given regular graph $G$, determining the minimum $k$ such that $G$ has a vertex-labeling by degree from $\mathbb{N}_{k}$ is $\mathbf{NP}$-hard. We will prove the following:

Theorem 7

(i)* Determining whether a given graph has a vertex-labeling by degree from $\mathbb{N}_{2}$ is in $\mathbf{P}$.

(ii) For every $k$, $k\geq 3$, for a given graph $G$ with $\chi(G)=k+1$ determining wether $G$ has vertex-labeling by degree from $\mathbb{N}_{k}$ is $\mathbf{NP}$-complete.*

**Vertex-labeling by maximum.

**A vertex-labeling $f$ is a vertex-labeling by maximum if $c(v)=\max_{u\sim v}f(u)$ is a proper vertex coloring. A graph $G$ may lack any vertex-labeling by maximum and it has a vertex-labeling by maximum from $\mathbb{N}_{2}$ if and only if $G$ is bipartite.

Remark 3

Let $k$ be the minimum number such that $G$ has a vertex-labeling by maximum from the set $\mathbb{N}_{k}$, then $k-\chi(G)$ can be arbitrarily large. For instance, for a given $t$, $t>3$ consider the graph $G$ with vertex set $V(G)=\{a_{i}:1\leq i\leq t\}\cup\{b_{j}:1\leq j\leq t-2\}$ and edge set $E(G)=\{a_{i}a_{i+1}:1\leq i\leq t-1\}\cup\{a_{j}b_{j},b_{j}a_{j+1}:1\leq j\leq t-2\}$. $G$ is 3-colorable but $k=t$.

Every triangle-free graph has a vertex-labeling by maximum (put different numbers on vertices) and if $G$ is a graph such that every vertex appears in some triangles, then $G$ does not have any vertex-labeling by maximum. Here, we present a nontrivial necessary condition that can be checked in polynomial time for a graph to have a vertex-labeling by maximum. First consider the following definition.

Definition 1

For a given graph $G$ the subset $S$ of vertices is called triangular-structured-vertices (TSV) if every $v\in S$ appears in a triangle in $G[S]$ and for every two adjacent vertices $v$ and $u$, where $v\in S$ and $u\notin S$, there exists a vertex $z\in S$ such that $z$ is adjacent to $v$ and $u$.

Let $S$ be a TSV for $G$. By way of contradiction, assume that $f$ is a vertex-labeling by maximum for $G$ and $v\in S\cup N(S)$ is a vertex such that $f(v)=\max_{u\in S\cup N(S)}f(u)$. Then $v$ has two neighbors $x$ and $y$ in $S$ with $\max_{u\sim x}f(u)=\max_{u\sim y}f(u)=f(v)$. This is a contradiction. Therefore, if $G$ has a TSV, then $G$ does not have a vertex-labeling by maximum.

Theorem 8

*We have the following:

(i) For a given 3-regular graph $G$, determining whether $G$ has a vertex-labeling by maximum from $\mathbb{N}_{3}$ is $\mathbf{NP}$-complete.

(ii) For every $k$, $k\geq 3$, it is NP-complete to decide whether $G$ has a vertex-labeling by maximum from $\mathbb{N}_{k}$ for a given $k$-colorable graph $G$.

(iii) There is a polynomial time algorithm to determine whether $G$ has a TSV.*

Here, we ask the following questions:

Problem 5

.* Is the necessary condition, sufficient for a given graph to have a vertex-labeling by maximum?*

Problem 6

.* Does there exist a polynomial time algorithm to determine whether a given graph has a vertex-labeling by maximum?*

Notation

Throughout this paper all graphs are finite and simple. We follow [15, 29] for terminology and notations are not defined here, and we consider finite undirected simple graphs $G=(V(G),E(G))$. We denote a 3-Sat formula by $\Phi=(X,C)$, where $X$ is the set of variables and $C$ is a set of clauses over $X$ such that each clause $c\in C$ has $\mid c\mid=3$.

A proper vertex coloring of $G$ is a function $c:V(G)\longrightarrow L$, such that if $u,v\in V(G)$ are adjacent, then $c(u)$ and $c(v)$ are different. A proper vertex $k$-coloring is a proper vertex coloring with $|L|=k$. The smallest integer $k$ such that $G$ has a proper vertex $k$-coloring is called the chromatic number of $G$ and denoted by $\chi(G)$. Similarly, for $k\in\mathbb{N}$, a proper edge $k$-coloring of $G$ is a function $c:E(G)\longrightarrow\mathbb{N}_{k}$, such that if $e,e^{\prime}\in E(G)$ share a common endpoint, then $c(e)$ and $c(e^{\prime})$ are different. The smallest integer $k$ such that $G$ has a proper edge $k$-coloring is called the edge chromatic number of $G$ and is denoted by $\chi^{\prime}(G)$. By Vizing’s theorem [28], the edge chromatic number of a graph $G$ is equal to either $\Delta(G)$ or $\Delta(G)+1$. Those graphs $G$ for which $\chi^{\prime}(G)=\Delta(G)$ are said to belong to $Class$ $1$, and the others to $Class$ $2$.

We denote the induced subgraph $G$ on $S$ by $G[S]$. Also, for every $v\in V(G)$ and $S\subseteq V(G)$, $N(v)$ and $N(S)$ denote the neighbor set of $v$ and the set of vertices of $G$ which has a neighbor in $S$, respectively. Furthermore, let $H$ be a subgraph of $G$, for every vertex $v$, $v\in V(H)$, we denote $\mid\{u:vu\in E(H),u\in V(H)\}\mid$ by $d_{H}(v)$.

Summery of results

2 Proof of Theorem 1

We reduce Monotone Not-All-Equal (NAE) 3-Sat to our problem in polynomial time. It is shown that the following problem is $\mathbf{NP}$-complete [15].

*Monotone Not-All-Equal 3-Sat .

Instance*: A 3-Sat formula $(X,C)$ such that there is no negation in the formula.

Question: Is there a truth assignment for $X$ such that each clause in $C$ has at least one true literal and at least one false literal?

Without loss of generality suppose that $\Phi$ is a formula such that every variable $x$, $x\in X$ appears in more than one clause. For every variable $x$, $x\in X$, we denote the number of clauses which contains $x$ by $\gamma(x)$. For a given formula $\Phi$ we construct a 3-regular graph $G$, such that $\Phi$ has a NAE truth assignment if and only if $G$ has an edge-labeling by sum from $\mathbb{N}_{2}$. For every variable $x$ if $\gamma(x)>2$, let $A_{x}$ be a cycle of length $\gamma(x)$. Otherwise let $A_{x}$ be two vertices $a$ and $b$ with two parallel edges between $a$ and $b$. Construct the graph $B_{x}$ by replacing every edge $ab$ of $A_{x}$ by $I(a,b)$ which is shown in Figure 1. Let $c_{i_{1}},\cdots,c_{i_{\gamma(x)}}$ be the clauses which contains $x$. $B_{x}$ has $\gamma(x)$ vertices of degree two, call these vertices $x_{i_{1}},\cdots,x_{i_{\gamma(x)}}$, in an arbitrary order. Now, for every variable $x$, $x\in X$, put a copy of $B_{x}$ and for every clause $c_{i}$, $c_{i}\in C$, put a vertex $c_{i}$. For every clause $c_{i}$ and every variable $x$, if $x$ appears in $c_{i}$, then join $c_{i}$ to $x_{i}$. Clearly, $G$ is a simple 3-regular graph (see Figure 2 for more details).

First assume that $G$ has an edge-labeling $f$ by sum from $\mathbb{N}_{2}$ and let $\ell$ be the coloring which is induced by $f$. By the structures of $A_{x}$ and $B_{x}$, for a variable $x$ which appears in $c_{i_{1}},\cdots,c_{i_{\gamma(x)}}$ the labels of edges $c_{i_{1}}x_{i_{1}},\cdots,c_{i_{\gamma(x)}}x_{i_{\gamma(x)}}$ are the same. Now, for every variable $x$, which appears in $c_{i_{1}},\cdots,c_{i_{\gamma(x)}}$ put $\Gamma(x)=true$ if and only if the label of edge $c_{i_{1}}x_{i_{1}}$ are 2. Let $c_{i}$ be a clause with the variable $x$, $y$, $z$. First, suppose that $\{f(x_{i}c_{i}),f(y_{i}c_{i}),f(z_{i}c_{i})\}\neq\{1,2\}$, then $\ell(x)=\ell(c_{i})$, but this is a contradiction. So, we have $\{f(x_{i}c_{i}),f(y_{i}c_{i}),f(z_{i}c_{i})\}=\{1,2\}$. Consequently, $\Gamma$ is a NAE satisfying assignment. Next, suppose that $\Phi$ has a NAE satisfying assignment $\Gamma:X\rightarrow\{true,false\}$, for every variable $x$, which appears in $c_{i_{1}},\cdots,c_{i_{\gamma(x)}}$ label all the edges $c_{i_{1}}x_{i_{1}},\cdots,c_{i_{\gamma(x)}}x_{i_{\gamma(x)}}$ by 2 if and only if $\Gamma(x)=true$. It is easy to extent this labeling to an edge-labeling by sum from $\mathbb{N}_{2}$. The proof is complete. $\clubsuit$

3 Proof of Theorem 2

(i) Clearly, the problem is in $\mathbf{NP}$. We reduce Cubic Planar 1-In-3 3-Sat to our problem. Moore and Robson [22] proved that the following problem is $\mathbf{NP}$-complete.

*Cubic Planar 1-In-3 3-Sat.

Instance*: A 3-Sat formula $\Phi=(X,C)$ such that every variable appears in exactly three clauses, there is no negation in the formula, and the bipartite graph obtained by linking a variable and a clause if and only if the variable appears in the clause, is planar.

Question: Is there a truth assignment for $X$ such that each clause in $C$ has exactly one true literal?

Consider an instance $\Phi$ of Cubic Planar 1-In-3 3-Sat. We transform this into a planar 3-colorable graph $G_{\Phi}$ such that $G_{\Phi}$ has an edge-labeling by product from $\mathbb{N}_{2}$ if and only if $\Phi$ has a 1-in-3 assignment. We use two gadgets $H_{x}$ and $I_{c}$ which are shown in Figure 3. The graph $G_{\Phi}$ has a copy of $H_{x}$ for each variable $x\in X$ and a copy of $I_{c}$ for each clause $c\in C$. Also, for each clause $c=y\vee z\vee w$ add the edges $cy$, $cz$ and $cw$. First, suppose that $G_{\Phi}$ has an edge-labeling by product from $\mathbb{N}_{2}$. In every copy of $H_{x}$ and $I_{c}$ the label of every edge is determined. In Figure 3, the label of each edge is written on the edge and the color of each vertex induced by edge labels is written on the vertex. Every variable $x$ appears in exactly three clauses, suppose that $x$ appears in $c_{i}$, $c_{j}$ and $c_{k}$. By attention to the induced colors of neighbors of $x$ in $H_{x}$, the labels of edges $c_{i}x$, $c_{j}x$ and $c_{k}x$ are equal. Furthermore, by attention to the $H_{x}$ and $I_{c}$, for every clause $c=x\vee y\vee z$, the label of exactly one of the edges $cx$, $cy$ and $cz$ is 2 (note that if the labels of edges $cx$, $cy$ and $cz$ are 2, then the induced colors of $x$ and $c$ are same). Now, for every variable $x$, which appears in $c_{i}$, $c_{j}$ and $c_{k}$ put $\Gamma(x)=true$ if and only if the labels of edges $c_{i}x$, $c_{j}x$ and $c_{k}x$ are 2. Clearly, $\Gamma$ is a 1-in-3 satisfying assignment. Next, suppose that $\Phi$ has a 1-in-3 satisfying assignment $\Gamma:X\rightarrow\{true,false\}$, for every variable $x$, which appears in $c_{i}$, $c_{j}$ and $c_{k}$, label $c_{i}x$, $c_{j}x$ and $c_{k}x$ by 2 if and only if $\Gamma(x)=true$. It is easy to extent this labeling to an edge-labeling by product from $\mathbb{N}_{2}$.

(ii) Let $G$ be a 3-regular graph. $G$ has an edge-labeling by product from $\mathbb{N}_{2}$ if and only if $G$ has an edge-labeling by sum from $\mathbb{N}_{2}$, so by Theorem 1, we can prove the theorem. $\clubsuit$

4 Proof of Theorem 3

(i) Clearly, the problem is in $\mathbf{NP}$. We reduced Cubic Planar 1-In-3 3-Sat to our problem. In [22] it was proved that Cubic Planar 1-In-3 3-Sat is $\mathbf{NP}$-complete. First, we construct an auxiliary graph $\mathcal{H}^{c}_{i}$. Put a copy of triangle $K_{3}=z_{1}^{c}z_{2}^{c}z_{3}^{c}$. For every vertex $z_{j}^{c}$, $1\leq j\leq 2$, put $2i$ new isolated vertices $t_{1}^{j},t_{2}^{j},\ldots,t_{2i}^{j}$ and join $z_{j}^{c}$ to all of them. Also, add the edges $t_{1}^{j}t_{2}^{j}$, $t_{3}^{j}t_{4}^{j},\ldots,t_{2i-1}^{j}t_{2i}^{j}$. Next, put $2i-2$ new isolated vertices $t_{1}^{3},t_{2}^{3},\ldots,t_{2i-2}^{3}$ and join $z_{3}^{c}$ to all of them. Finally, add the edges $t_{1}^{3}t_{2}^{3}$, $t_{3}^{3}t_{4}^{3},\ldots,t_{2i-3}^{3}t_{2i-2}^{3}$. Call the resulting graph $\mathcal{H}^{c}_{i}$. Now, consider an instance $\Psi$, we transform this into a planar 3-colorable graph $G_{\Psi}$ such that $G_{\Psi}$ has a vertex-labeling by product from $\mathbb{N}_{2}$ if and only if $\Psi$ has a 1-in-3 assignment. For each clause $c\in C$ put a vertex $c$ and a copy of $\mathcal{H}^{c}_{3}$, $\mathcal{H}^{c}_{5}$ and $\mathcal{H}^{c}_{6}$. Also, join the vertex $c$ to $z_{3}^{c}\in\mathcal{H}^{c}_{3}$, $z_{3}^{c}\in\mathcal{H}^{c}_{5}$ and $z_{3}^{c}n\in\mathcal{H}_{6}^{c}$. Next, for each variable $x\in X$ put a vertex $x$. Finally, for each clause $c=x\vee y\vee w$ add the edges $cx$, $cy$ and $cw$.

First, suppose that $G_{\Psi}$ has a vertex-labeling $f$ by product from $\mathbb{N}_{2}$ and let $\ell$ be the induced coloring by $f$. In every copy of $\mathcal{H}_{3}^{c}$ the label of vertex $z_{3}^{c}$ is $2$. We have the similar property for $\mathcal{H}_{5}^{c}$ and $\mathcal{H}_{6}^{c}$. By the structure of $\mathcal{H}_{3}^{c}$, we have $f(c)=1$ and in the subgraph $\mathcal{H}_{3}^{c}$, $\ell(z_{3}^{c})=8$; similarly for every subgraph $\mathcal{H}_{5}^{c}$, $\ell(z_{3}^{c})=32$ and for each subgraph $\mathcal{H}_{6}^{c}$, we have $\ell(z_{3}^{c})=64$. So for every clause vertex $c$ we have $\ell(c)=16$. Now, for every variable $x$, put $\Gamma(x)=true$ if and only if $f(x)=2$. Since for every clause $c$, $\ell(c)=16$, $\Gamma$ is a 1-in-3 satisfying assignment. On the other hand, suppose that $\Psi$ is 1-in-3 satisfiable with the satisfying assignment $\Gamma:X\rightarrow\{true,false\}$, for every variable $x$, label the vertex $x$ by 2 if and only if $\Gamma(x)=true$. The labels of other vertices are determined and it is clear that this labeling is a vertex-labeling by product from $\mathbb{N}_{2}$.

(ii) For every $k$, $k\geq 3$, we present a polynomial time reduction from 3-colorability to our problem.

*3-Colorability

Instance*: A graph $G$.

Question: Is $\chi(G)\leq 3$?

First define the following sets: $\mathcal{A}_{k}=\{mn:m,n\in\mathbb{N}_{k}\}$, $\mathcal{B}_{k}=\{\frac{m}{n}:m,n\in\mathbb{N}_{k}\}$. Also, define $\alpha(k)=\max_{\mathcal{D}_{k}\in\mathcal{C}_{k}}\mid\mathcal{D}_{k}\mid$, where $\mathcal{C}_{k}$ is the set of sets such that for every set $\mathcal{D}_{k}\in\mathcal{C}_{k}$, we have $\mathcal{D}_{k}\subseteq\mathcal{A}_{k}$ and $\{\frac{d}{d^{\prime}}:d,d^{\prime}\in\mathcal{D}_{k}\}\cap\mathcal{B}_{k}=\emptyset$. Since $k$ is constant, so we can compute $\alpha(k)$ in $O(1)$. Note that for $k$, $k\geq 3$, $\alpha(k)\geq 3$. Now, for a given graph $G$ with $n$ vertices $v_{1},v_{2},\ldots,v_{n}$, join all vertices of $G$ to the all vertices of complete graph $K_{\alpha(k)-3}$ with the vertices $v_{n+1},\ldots,v_{n+\alpha(k)-3}$. Call the resulting graph $G^{*}$. Now consider the graph $G^{**}$ with the vertex set $\{v_{i}^{j}:i\in\mathbb{N}_{n+\alpha(k)-3},j\in\mathbb{N}_{k}\}$ such that $v_{x}^{y}$ is joined to $v_{z}^{w}$ if and only if $x=z$ or $v_{x}v_{z}\in E(G^{*})$. Finally, consider a copy of graph $G^{**}$, for every $i$, $1\leq i\leq n+\alpha(k)-3$, put two new isolated vertices $v_{i}^{\prime}$ and $v_{i}^{\prime\prime}$ and join them to the set of vertices $\{v_{i}^{1},\ldots,v_{i}^{k}\}$. Call the resulting graph $\widetilde{G}$.

We show that $\widetilde{G}$ has a vertex-labeling by product from $\mathbb{N}_{k}$ if and only if $G$ is 3-colorable. Let $f$ be a vertex-labeling by product for $\widetilde{G}$. Clearly, $f(v_{1}^{1}),\ldots,f(v_{1}^{k})$ should be different integers. For every $i$, $i\in\mathbb{N}_{n+\alpha(k)-3}$, we have: $\{f(v_{i}^{j}):j\in\mathbb{N}_{k}\}=\mathbb{N}_{k}$. Furthermore, for every $i_{1}$, $i_{2}$, $1\leq i_{1}<i_{2}\leq n+\alpha(k)-3$, we have: $f(v_{i_{1}}^{\prime})f(v_{i_{1}}^{\prime\prime}),f(v_{i_{2}}^{\prime})f(v_{i_{2}}^{\prime\prime})\in\mathcal{A}_{k}$. Also, for every $i_{1}$ and $i_{2}$, if $v_{i_{1}}v_{i_{2}}\in E(G^{*})$, then

$\dfrac{f(v_{i_{1}}^{\prime})f(v_{i_{1}}^{\prime\prime})}{f(v_{i_{2}}^{\prime})f(v_{i_{2}}^{\prime\prime})}\notin\mathcal{B}_{k}$.

Therefore, $\mid\{f(v_{i}^{\prime})f(v_{i}^{\prime\prime}):1\leq i\leq n+\alpha(k)-3\}\mid\geq\alpha(k)-3+\chi(G)$. So, $\widetilde{G}$ has a vertex-labeling by product from $\mathbb{N}_{k}$ if and only if $\chi(G)\leq 3$. The proof is complete. $\clubsuit$

5 Proof of Theorem 4

For a given $G(n,p)$, let $\{\mathcal{A}_{1},\ldots,\mathcal{A}_{5}\}$ be a partition of vertices to five parts such that those parts are of equal or almost equal sizes (that is, $\lfloor n/5\rfloor$ or $\lceil n/5\rceil$). For every vertex $v$ denote the number of neighbors of $v$ in $\mathcal{A}_{i}$ by $\lambda_{v}^{i}$. Label the vertices of $\mathcal{A}_{1},\ldots,\mathcal{A}_{5}$ by $2,3,5,7,11$, respectively. This labeling is a vertex-labeling by product if for every two adjacent vertices $v$ and $u$, there is an index $i$, $1\leq i\leq 5$ such that $\lambda_{v}^{i}\neq\lambda_{u}^{i}$.

[TABLE]

By Stirling’s approximation $\sqrt{2\pi n}(\frac{n}{e})^{n}\leq n!\leq\sqrt{e^{2}n}(\frac{n}{e})^{n}$, we have:

[TABLE]

The proof is complete. $\clubsuit$

6 Proof of Theorem 5

(i) Moret proved in [23] that Planar Not-All-Equal 3-Sat is in $\mathbf{P}$ by an interesting reduction to a known problem in $\mathbf{P}$, namely Planar(Simple) MaxCut.

*Planar Not-All-Equal 3-Sat.

Instance*: A 3-Sat formula $(X,C)$ such that the following graph obtained from 3-Sat is planar. The graph has one vertex for each variable, one vertex for each clause; all variable vertices are connected in a simple cycle and each clause vertex is connected by an edge to variable vertices corresponding to the literals present in the clause (note that positive and negative literals are treated exactly alike).

Question: Is there a Not-All-Equal (NAE) truth assignment for $X$?

By a simple argument it was shown that the following problem is in $\mathbf{P}$ (for more information see [12]).

*Planar NAE 3-Sat Type 2.

Instance*: A 3-Sat formula $(X,C)$ such that the following graph obtained from 3-Sat is planar. The graph has one vertex for each variable, one vertex for each clause and each clause vertex is connected by an edge to variable vertices corresponding to the literals present in the clause.

Question: Is there a NAE truth assignment for $X$?

Now, consider the following variant.

*Planar NAE Sat Type 2.

Instance*: Set $X$ of variables, collection $C$ of clauses over $X$ such that each clause $c\in C$ has $\mid c\mid\geq 2$ and the following graph obtained from sat is planar. The graph has one vertex for each variable, one vertex for each clause and each clause vertex is connected by an edge to variable vertices corresponding to the literals present in the clause.

Question: Is there a NAE truth assignment for $X$?

We can convert any instance $\Phi$ of Planar NAE Sat Type 2 to an instance $\Psi$ of Planar NAE 3-Sat Type 2 in polynomial time. For a given instance $\Phi$, for each clause with exactly two literals like $c=(x\vee y)$, put two clauses $x\vee y\vee t$ and $x\vee y\vee\neg t$ in $\Psi$, where $t$ is a new variable. Also, for each clause with exactly four literals $c=(x\vee y\vee w\vee z)$, put two clauses $x\vee y\vee t$ and $w\vee z\vee\neg t$ in $\Psi$, where $t$ is a new variable. For clauses with more than five variables we have a similar argument.

For a given graph $G$, we can use a depth-first search algorithm to identify the connected components of $G$. $G$ has an edge-labeling by gap from $\{1,2\}$ if and only if each connected component of $G$ has an edge-labeling by gap from $\{1,2\}$. So, without loss of generality we can assume that $G$ is connected. Let $G[X,Y]$ be a connected planar bipartite graph such that $G$ does not have a vertex of degree one. Let $f$ be an edge-labeling by gap from $\{1,2\}$ for $G$. The color which is induced by $f$ for the set of vertices $X$ is one and the induced color by $f$ for the set of vertices $Y$ is zero, or vice versa (Property 1).

For every vertex $v\in X$, consider a variable $v$ in $\Phi$ and for every vertex $u\in Y$ put a clause $(\bigvee_{v\sim u}v)$ in $\Phi$. Now determine whether $\Phi$ has a NAE truth assignment. If $\Phi$ has a NAE truth assignment $\Gamma$, for every vertex $v$, $v\in X$ label all edge incident with $v$ by $1$ if and only if $\Gamma(v)=false$. Label other edges of $G$ by $2$, call this labeling $f$. It is easy to see that $f$ is an edge-labeling by gap. If $\Phi$ does not have a NAE truth assignment. Then, for every vertex $v\in Y$, consider a variable $v$ in $\Psi$ and for every vertex $u\in X$ put a clause $(\bigvee_{v\sim u}v)$ in $\Psi$. Now determine whether $\Psi$ has a NAE truth assignment. If $\Psi$ has a NAE truth assignment $\Gamma$ by a similar method we can find an edge-labeling $f$ by gap from $\{1,2\}$ for $G$. If $\Psi$ does not have a NAE truth assignment, by Property 1, $G$ does not have any edge-labeling by gap from $\{1,2\}$.

(ii) Let $\Phi$ be a $3$-Sat formula with clauses $C=\{c_{1},\ldots,c_{k}\}$ and variables $X=\{x_{1},\ldots,x_{n}\}$. Let $G(\Phi)$ be a graph with the vertices $C\cup X\cup(\neg X)$, where $\neg X=\{\neg x_{1},\ldots,\neg x_{n}\}$, such that for each clause $c_{j}=y\vee z\vee w$, $c_{j}$ is adjacent to $y,z$ and $w$, also every $x_{i}\in X$ is adjacent to $\neg x_{i}$. $\Phi$ is called planar $3$-Sat type $2$ formula if $G(\Phi)$ is a planar graph. It was shown that the problem of satisfiability of planar $3$-Sat type $2$ is $\mathbf{NP}$-complete [13].

*Planar $3$-Sat type $2$.

Instance*: A $3$-Sat type $2$ formula $\Phi$.

Question: Is there a truth assignment for $\Phi$ that satisfies all the clauses?

In planar $3$-Sat type $2$, if we only consider the set of formulas $\mathcal{F}$ such that for every $\Phi$, $\Phi\in\mathcal{F}$, $G(\Phi)$ is connected, the problem remains $\mathbf{NP}$-complete. We reduce this version to our problem. Consider an instance $\Phi$, we transform this into a graph $G_{\Phi}$ such that $G_{\Phi}$ has an edge-labeling by gap from $\mathbb{N}_{2}$ if and only if $\Phi$ has a satisfying assignment. For each variable $x\in X$ put a a copy of cycle $C_{4}=xt_{x}\neg xt_{x}^{\prime}$. For each clause $c\in C$ put a copy of gadget $P_{4}=cc^{\prime}c^{\prime\prime}c^{\prime\prime\prime}$. Now, put a copy $C_{6}$. Also, for each clause $c=y\vee z\vee w$ add the edges $cy$, $cz$ and $cw$. Finally, let $x$ be an arbitrary variable, join $x$ to one of the vertices of $C_{6}$. Call the resulting graph $G_{\Phi}$. $G_{\Phi}$ is connected, bipartite and planar. First, suppose that $G_{\Phi}$ has an edge-labeling $f$ by gap from $\mathbb{N}_{2}$ and $l$ is the proper coloring which is induced by $f$. Since for every variable $x$ the degrees of vertices $x$ and $\neg x$ are greater than one, also for every clause $c$ the degree of vertex $c$ is 4 and $G_{\Phi}$ is connected, hence in the induced coloring $l$ by $f$, for the set of variables and the set of clauses, we have $l(x_{1})=l(\neg x_{1})=\ldots=l(\neg x_{n})=l(x_{n})\neq l(c_{1})=\ldots=l(c_{m})$ and $l(x_{1})\neq 2\neq l(c_{1})$.

First, suppose that $l(x)=1$, since $x$ is adjacent to one of the vertices of $C_{6}$, in this situation $G_{\Phi}$ does not have any edge-labeling $f$ by gap from $\mathbb{N}_{2}$. So for all $x\in X$ and $c\in C$, we have $l(x)=0$, $l(\neg x)=0$ and $l(c)=1$. Hence, the labels of all edges incident with $x$ are same. Also, for every variable $x$, because of $t_{x}$, the labels of all edges incident with $x$ are different from the labels of all edges incident with $\neg x$. Now, for every variable $x$, which appears in $c_{i},c_{j},\ldots,c_{k}$ put $\Gamma(x)=true$ if and only if the label of edge $c_{i}x$ is $2$. For every clause $c=x\vee y\vee w$, $l(c)=1$. If the labels of all edges $cx,cy,cw$ are 1, then $f(cc^{\prime})=2$. Thus $f(c^{\prime}c^{\prime\prime})=2$ and $f(c^{\prime\prime}c^{\prime\prime\prime})=1$, so $l(c^{\prime\prime})=l(c^{\prime\prime\prime})$, but this is a contradiction. Therefore, $2\in\{f(cx),f(cy),f(cw)\}$. Therefore, $\Gamma$ is an satisfying assignment. Now, let $\Gamma$ be a satisfying assignment for $\Phi$. For every variable $x$, label all the edges incident with $x$ by $2$ if and only if $\Gamma(x)=true$. It is easy to extend this labeling to an edge-labeling $f$ by gap from $\{1,2\}$. This completes the proof.

(iii) For every $k$, we present a polynomial time reduction from k-colorability, $k>2$, to our problem.

*k-Colorability

Instance*: A graph $G$.

Question: Is $\chi(G)\leq k$?

For a given connected graph $G$ with more than two vertices, we construct a graph $G^{*}$ such that $\chi(G)\leq k$ if and only if $G^{*}$ has an edge-labeling by gap from $\mathbb{N}_{k}$. For every vertex $v$, $v\in V(G)$, put a copy of gadget $P_{3}=v^{\prime}v^{\prime\prime}v^{\prime\prime\prime}$ and join $v^{\prime}$ to $u^{\prime}$ if and only if $vu\in E(G)$. Call the resulting graph $G^{*}$. First, suppose that $G^{*}$ has an edge-labeling $f$ by gap from $\{1,2,\cdots,k\}$ and $\ell$ is the coloring which is induced by $f$. For every vertex $v^{\prime}$, we have $\ell(v^{\prime})\in\{0,1,\ldots,k-1\}$, so $\ell$ is a proper vertex coloring for $G$. Now, let $c$ be a proper vertex coloring for $G$. For every vertex $v^{\prime}$, $v^{\prime}\in V(G^{*})$, label all edges incident with $v^{\prime}$ except $v^{\prime}v^{\prime\prime}$ by $1$ and label $v^{\prime}v^{\prime\prime}$ by $c(v)$. Finally for every edge $v^{\prime\prime}v^{\prime\prime\prime}$, label $v^{\prime\prime}v^{\prime\prime\prime}$ by $c(v)$ if $c(v)\neq 1$, otherwise label $v^{\prime\prime}v^{\prime\prime\prime}$ by $k$. This labeling is an edge-labeling by gap from $\mathbb{N}_{k}$. $\clubsuit$

7 Proof of Theorem 6

(i) For a given graph $G$, we can use a depth-first search algorithm to identify the connected components of $G$. $G$ has a vertex-labeling by gap from $\{1,2\}$ if and only if each connected component of $G$ has a vertex-labeling by gap from $\{1,2\}$. So, without loss of generality we can assume that $G$ is connected. Let $G[X,Y]$ be a connected planar bipartite graph with more than two vertices. Consider a copy of $G$, remove all vertices of degree one from $G$ and call the resulting graph $G^{\prime}[X^{\prime},Y^{\prime}]$, where $X^{\prime}\subseteq X$. If $E(G^{\prime})=\emptyset$, then $G$ is a star, so $G$ has a vertex-labeling by gap from $\{1,2\}$. So suppose that $E(G^{\prime})\neq\emptyset$ and let $f$ be a vertex-labeling by gap from $\{1,2\}$ for $G$. Since $G^{\prime}[X^{\prime},Y^{\prime}]$ is connected and bipartite, so the color which is induced by $f$ for the set of vertices $X^{\prime}$ is one and the color which is induced by $f$ for the set of vertices $Y^{\prime}$ is zero, or vice versa (Property A).

For every vertex $v\in X$, consider a variable $v$ in $\Phi$ and for every vertex $u\in Y^{\prime}$ put a clause $(\bigvee_{vu\in E(G)}v)$ in $\Phi$. Now determine whether $\Phi$ has a NAE truth assignment. If $\Phi$ has a NAE truth assignment $\Gamma$, for every vertex $v$, $v\in X$ label $v$ by $1$ if and only if $\Gamma(v)=false$. Label other vertices of $G$ by $2$. It is easy to see that this labeling is a vertex-labeling by gap from $\{1,2\}$ for $G$. If $\Phi$ does not have a NAE truth assignment, then, for every vertex $v\in Y$, consider a variable $v$ in $\Psi$ and for every vertex $u\in X^{\prime}$ put a clause $(\bigvee_{vu\in E(G)}v)$ in $\Psi$. Now determine whether $\Psi$ has a NAE truth assignment. If $\Psi$ has a NAE truth assignment $\Gamma$ by a similar method we can find vertex-labeling by gap from $\{1,2\}$ for $G$. Otherwise, by Property $A$, $G$ does not have any vertex-labeling by gap from $\mathbb{N}_{2}$. See Algorithm 1.

(ii) We reduce NAE 3-Sat to our problem in polynomial time. For a given $\Phi$, we transform $\Phi$ into a graph $G_{\Phi}$ such that $G_{\Phi}$ has a vertex-labeling by gap from $\mathbb{N}_{2}$ if and only if $\Phi$ has a NAE satisfying assignment. Construction of $G_{\Phi}$ is similar to the construction of $G_{\Phi}$ in the proof of (iii) of Theorem 5, except the gadget $P_{4}=cc^{\prime}c^{\prime\prime}c^{\prime\prime\prime}$. For each clause $c\in C$ instead of $P_{4}=cc^{\prime}c^{\prime\prime}c^{\prime\prime\prime}$, put an isolated vertex $c$. First, suppose that $G_{\Phi}$ has a vertex-labeling $f$ by gap from $\mathbb{N}_{2}$ and $l$ is the induced proper coloring by $f$. By an argument similar to argument of proof of Theorem 5, for every clause $c=x\vee y\vee w$, $l(c)=1$. So $\{f(x),f(y),f(w)\}=\{1,2\}$, therefore $\Gamma$ is a NAE satisfying assignment. Now, let $\Gamma$ be an satisfying assignment for $\Phi$. For every variable $x$, label the vertex $x$ by $2$ if and only if $\Gamma(x)=true$. This completes the proof.

(iii) The proof is similar to the proof of part (iii) of Theorem 5.

(iv) It was shown that 3-colorability of planar 4-regular graphs is NP-complete [11]. By a proof similar to the proof of part (iii) of Theorem 5, we can prove the theorem. $\clubsuit$

8 Proof of Theorem 7

(i) We reduce our problem to 2-Sat problem in polynomial time.

*2-Sat.

Instance*: A 2-Sat formula $\Phi$.

Question: Is there a truth assignment for $\Phi$ that satisfies all the clauses?

For a given graph $G$ of order $n$ we construct a 2-Sat formula $\Phi$ with $n$ variables $v_{1},\ldots,v_{n}$ such that $G$ has a vertex-labeling by degree from $\{1,2\}$ if and only if there is a truth assignment for $\Phi$. For every edge $e=v_{i}v_{j}$ , if $d(v_{i})=d(v_{j})$, add the clauses $v_{i}\vee v_{j}$ and $\neg v_{i}\vee\neg v_{j}$ and if $d(v_{i})=2d(v_{j})$, add the clause $v_{i}\vee\neg v_{j}$ , otherwise if $2d(v_{i})=d(v_{j})$, add the clause $\neg v_{i}\vee v_{j}$. First, suppose that $\Gamma$ is a satisfying assignment for $\Phi$. For every vertex $v_{i}$, label $v_{i}$ by 2 if and only if $\Gamma(v_{i})=true$. It is easy to see that this labeling is a vertex-labeling by degree from $\{1,2\}$. Next, let $f$ be a vertex-labeling by degree from $\{1,2\}$, for every variable $v_{i}$, put $\Gamma(v_{i})=true$ if and only if $f(v_{i})=2$. As we know, 2-Sat problem is in P [15]. This completes the proof.

(ii) It was shown that 3-colorability of planar 4-regular graphs is NP-complete [11]. For a given 4-regular graph $G$ and a number $k$, $k\geq 3$, we construct a graph $\mathcal{H}^{*}$ with chromatic number $k+1$ in linear time such that $\mathcal{H}^{*}$ has a vertex-labeling by degree from $\mathbb{N}_{k}$ if and only if $G$ is 3-colorable. Let $\mathcal{P}_{k}=\{4,5,\ldots,k\}$.

Construction of $\mathcal{H}^{*}$. First, consider a copy of graph $G$ and call its vertices old vertices . For every old vertex $v$ and for every $\alpha\in\mathcal{P}_{k}$, consider two copies of complete graphs $K_{k-1}$. Call this two copies by $K_{k-1}^{(v,\alpha)}$ and $K_{k-1}^{(v,\alpha)^{\prime}}$. Join $v$ to all vertices of $K_{k-1}^{(v,\alpha)}$ and $K_{k-1}^{(v,\alpha)^{\prime}}$, Next join all vertices of $K_{k-1}^{(v,\alpha)}$ to a new vertex $v_{\alpha}$ and also join all vertices of $K_{k-1}^{(v,\alpha)^{\prime}}$ to a new vertex $v_{\alpha}^{\prime}$. Now, join $v_{\alpha}$ to $v_{\alpha}^{\prime}$. Call the resulting graph $\mathcal{H}$. For a given graph $\mathcal{H}$, for every old vertex $v$ and for every $\alpha\in\mathcal{P}_{k}$, join every vertex of $u$, $u\in V(K_{k-1}^{(v,\alpha)})\cup V(K_{k-1}^{(v,\alpha)^{\prime}})\cup\{v_{\alpha},v_{\alpha}^{\prime}\}$ to $\alpha\times d_{\mathcal{H}}(v)-d_{\mathcal{H}}(u)$ new isolated vertices $l_{u}^{1},l_{u}^{2},\ldots,l_{u}^{\alpha\times d_{\mathcal{H}}(v)-d_{\mathcal{H}}(u)}$. Call the resulting $\mathcal{H}^{*}$.

The chromatic number of $\mathcal{H}^{*}$ is $k+1$. Because every two vertices in $V(K_{k-1}^{(v,\alpha)})\cup\{v_{\alpha}\}$ have the same degree, every vertex-labeling by degree needs at least $k$ distinct labels. Let $f$ be a vertex-labeling by degree from $\{1,\ldots,k\}$ for $\mathcal{H}^{*}$. We claim that for every old vertex $v$, $f(v)\in\{1,2,3\}$. To the contrary suppose that $f(v)\geq 4$, so there is a $\alpha\in\mathcal{P}_{k}$ such that $\alpha=f(v)$. Therefore, there is no vertex in $K_{k-1}^{(v,\alpha)}$ with the label $1$, hence $l(v_{\alpha})=1$. Similarly, $l(v_{\alpha}^{\prime})=1$. But $v_{\alpha}$ and $v_{\alpha}^{\prime}$ are adjacent and have the same degree. This is a contradiction. So, for every old vertex $v$, we have $f(v)\in\{1,2,3\}$. Since the degree of old vertices are equal, therefore $f$ is a proper vertex coloring for old vertices, so $G$ is 3-colorable. On the other hand, let $c$ be proper vertex coloring for $G$ with 3 colors. It is easy to find a vertex-labeling by degree from $\{1,\cdots,k\}$ for $\mathcal{H}^{*}$. $\clubsuit$

9 Proof of Theorem 8

(i) Clearly, the problem is in $\mathbf{NP}$. It was shown that it is $\mathbf{NP}$-hard to determine the edge chromatic number of a cubic graph [17]. Let $G$ be a $3$-regular graph. We construct a $3$-regular graph $G^{\prime}$ from $G$ such that $G^{\prime}$ has a vertex-labeling by maximum from $\mathbb{N}_{3}$ if and only if $G$ belongs to $Class$ $1$. In order to construct $G^{\prime}$, for every vertex $v\in V(G)$ with the neighbors $x$, $y$ and $z$ consider two disjoint triangles $v_{x}v_{y}v_{z}$ and $v_{x}^{\prime}v_{y}^{\prime}v_{z}^{\prime}$ in $G^{\prime}$. Also, for every edge $e\in E(G)$, consider two adjacent vertices $e$ and $e^{\prime}$ in $G^{\prime}$. Finally, for every edge $e=uv\in E(G)$, join $e$ to $v_{u}$ and $u_{v}$; also join $e^{\prime}$ to $v_{u}^{\prime}$ and $u_{v}^{\prime}$. Name the constructed graph $G^{\prime}$ (see Figure 4). Since $G^{\prime}$ has triangles, so every vertex-labeling by maximum needs at least $3$ distinct labels. First suppose that $G^{\prime}$ has a vertex-labeling $f$ by maximum from $\mathbb{N}_{3}$ and let $\ell$ be the induced vertex coloring by $f$. For every vertex $v\in V(G)$ with the neighbors $x$, $y$ and $z$ in $G$, we have $\{\ell(v_{x}),\ell(v_{y}),\ell(v_{z})\}=\{1,2,3\}=\{\ell(v_{x}^{\prime}),\ell(v_{y}^{\prime}),\ell(v_{z}^{\prime})\}$. Suppose that there are $u$ and $v$ such that $\ell(v_{u})=\ell(v_{u}^{\prime})=3$, since $f$ can not assign $3$ to a vertex in a triangle, so $f(vu)=f((vu)^{\prime})=3$. Hence $\ell(vu)=\ell((vu)^{\prime})=3$ and this is a contradiction. So we have the following fact:

There are no $u$ and $v$ such that $\ell(v_{u})=\ell(v_{u}^{\prime})=3$ (Fact 1).

Now, consider the following proper $3$-edge coloring for $G$.

$g:E(G)\longrightarrow\{1,2,3\}$,

g(uv)=\begin{cases}1&if\,\,f(uv)=3,\\ 2&if\,\,f((uv)^{\prime})=3,\\ 3&otherwise.\end{cases}

By Fact 1, $g$ is well-defined and $G$ belongs to $Class$ $1$. On the other hand, assume that $g:E(G)\longrightarrow\{1,2,3\}$ is a proper 3-edge coloring. Define $f:V(G^{\prime})\longrightarrow\{1,2,3\}$ such that for every edge $uv\in E(G)$, $f(v_{u})=f(v_{u}^{\prime})=1$, $f(uv)=g(uv)$ and $f((uv)^{\prime})\equiv g(uv)+1(\mod 3)$. It is easy to see that $f$ is a vertex-labeling by maximum.

(ii) First, consider the following simple procedure:

$\diamond$ For a given graph $G$, put a new vertex $v$ and join it to the all vertices of $G$, next put a new vertex $u$ and join it to $v$. Name the constructed graph $G^{\prime}$.

We can construct $G^{\prime}$ in polynomial time and $G$ has a vertex-labeling by maximum from $\mathbb{N}_{k}$ if and only if $G^{\prime}$ has a vertex-labeling by maximum from $\mathbb{N}_{k+1}$. In Part (i), we proved that, it is $\mathbf{NP}$-complete to decide whether a given $3$-colorable graph $G$ has a vertex-labeling by maximum from $\mathbb{N}_{3}$. So by repeating the above procedure, for every $k$, $k\geq 3$, we can prove that, it is $\mathbf{NP}$-complete to decide whether a given $k$-colorable graph $H$ has a vertex-labeling by maximum from $\mathbb{N}_{k}$. This completes the proof.

(iii) Consider Algorithm $2$. When Algorithm 2 terminates, if it returns ”$G$ has the TSV $S$”, then $S$ is a TSV, so $G$ does not have any vertex-labeling by maximum. Suppose that Algorithm $2$ returns ”$G$ does not have a TSV”, but $G$ has a TSV $S^{\prime}$. In the lines $2$ and $3$ of the algorithm, the set of vertices $S^{\prime}$ are added to $S$. Now, consider the line $5$ of the algorithm and let $v\in S^{\prime}$ be the first vertex from the set $S^{\prime}$ that is eliminated from $S$. When Algorithm $2$ chooses the vertex $v$, $v$ is in a triangle in $G[S^{\prime}]$, so it is in a triangle in $G[S]$. Therefore, there is a vertex $u$ such that $uv\in E(G)$, $v\in S^{\prime}$, $u\in N(S)\setminus S$ and there is no vertex $z\in S$ such that $z$ is adjacent to $v$ and $u$. So $S^{\prime}$ is not a TSV. This is a contradiction. So when Algorithm $2$ returns ”$G$ has no TSV”, $G$ does not have any TSV. $\clubsuit$

10 Acknowledgment

We would like to thank Wiktor Żelazny for his valuable answers to our questions about the definition of fictional coloring. Also, we would like to express our deep gratitude to the referees for their constructive and fruitful comments.