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Article

# Graphs with Minimal Strength

by 2,* and
1
College of Mathematical Sciences, Harbin Engineering University, Harbin 150001, China
2
Faculty of Computer & Mathematical Sciences, Universiti Teknologi MARA (Segamat Campus), Johor 85000, Malaysia
3
College of Global Talents, Beijing Institute of Technology, Zhuhai 519088, China
4
Department of Mathematics, The Chinese University of Hong Kong, Hong Kong, China
*
Author to whom correspondence should be addressed.
Symmetry 2021, 13(3), 513; https://doi.org/10.3390/sym13030513
Received: 1 March 2021 / Revised: 15 March 2021 / Accepted: 16 March 2021 / Published: 21 March 2021
(This article belongs to the Special Issue Graph Labelings and Their Applications)

## Abstract

:
For any graph G of order p, a bijection $f : V ( G ) → { 1 , 2 , … , p }$ is called a numbering of G. The strength $s t r f ( G )$ of a numbering f of G is defined by $s t r f ( G ) = max { f ( u ) + f ( v ) | u v ∈ E ( G ) } ,$ and the strength $s t r ( G )$ of a graph G is In this paper, many open problems are solved, and the strengths of new families of graphs are determined.

## 1. Introduction

We only consider simple and loopless graph $G = ( V , E )$ with order $| V ( G ) | = p$ and size $| E ( G ) | = q$. If $u v ∈ E ( G )$, we say that v is a neighbor of u. The degree of a vertex v in a graph G is the number of neighbors of v in G, denoted as $deg G ( v )$. The minimum degree (and maximum degree) of G is the minimum (and maximum) degree among the vertices of G, denoted as $δ ( G )$ (and $Δ ( G )$). A vertex of degree 0 is called an isolated vertex and a vertex of degree 1 is called a pendant vertex; its incident edge is called a pendant edge. The set of all neighbors of u is denoted as $N G ( u )$. For $S ⊂ V ( G )$, let $N G ( S )$ be the set of all neighbors of the vertices in S. We shall drop the subscript G if there is no ambiguity. For $a < b$, the set of integers from a to b is denoted as $[ a , b ] Z$. For notations not defined in this paper, we refer to [1].
The magic square [2] is one of the oldest problems on symmetry [3]. It was later extended to the idea of magic graphs [4], super magic graphs [5], and super edge-magic graphs [6]. The notion of the strength of a graph G was then introduced by Ichishima et al. [7] as a generalization of super magic strength [8], which is effectively defined only for super edge-magic graphs (also called strong vertex-graceful [9] and strongly indexable [10]), to any nonempty graphs as follows.
A bijection $f : V ( G ) → [ 1 , p ] Z$ is called a numbering of the graph G of order p.
Definition 1.
The strength $s t r f ( G )$ of a numbering $f : V ( G ) → [ 1 , p ] Z$ of G is defined by
$s t r f ( G ) = max { f ( u ) + f ( v ) | u v ∈ E ( G ) } ,$
and the strength $s t r ( G )$ of a graph G itself is
A numbering f is called a strength labeling of G if $s t r f ( G ) = s t r ( G )$.
The super magic strength, $s m ( G )$, of a graph G is defined as the minimum of all magic constants over all super edge-magic labelings of G. A necessary and sufficient condition for a graph to be super edge-magic [11] then gives rise to the concept of the consecutive strength labeling of a graph, which is equivalent to super edge-magic labeling. In [7] (Section 4), Ichishima et al. showed that a graph G of order p has $s m ( G ) ≥ s t r ( G ) + p + 1$. Several new lower bounds of $s m ( G )$ in terms of other graph parameters were also obtained. Moreover, all of the bounds are sharp. Thus, one may use the strength of graphs to establish the super magic strength of the corresponding graphs.
Several lower and upper bounds for $s t r ( G )$ were obtained in [7]. The following two are needed in what follows.
Lemma 1.
If H is a subgraph of a graph G, then
$s t r ( H ) ≤ s t r ( G ) .$
Lemma 2.
For every graph G of order p with $δ ( G ) ≥ 1$,
$s t r ( G ) ≥ p + δ ( G ) .$
Let $G + H$ be the disjoint union of G and H with $V ( G + H ) = V ( G ) ∪ V ( H )$ and $E ( G + H ) = E ( G ) ∪ E ( H )$. The disjoint union of m copies of G is denoted as $m G$. We first extend Lemma 2 to graphs with isolated vertices.
Lemma 3.
Let G be a graph with $δ ( G ) ≥ 1$. If $m ≥ 1$, then
$s t r ( G + m K 1 ) = s t r ( G ) .$
Proof.
Let f be a strength labeling of the graph G. We extend f to a numbering of $G + m K 1$ by assigning all m isolated vertices by labels in $[ p + 1 , p + m ] Z$. Clearly, $s t r f ( G + m K 1 ) = s t r f ( G ) = s t r ( G )$. Hence $s t r ( G + m K 1 ) ≤ s t r ( G )$. Combining with Lemma 1, we have the lemma. □
Thus, from now on, we only consider graphs without isolated vertices. In [7,12], the authors showed that $s t r ( G ) = p + δ ( G )$ if G is a path, cycle, complete graph, complete bipartite graph, ladder graph, prism graph, Möbius ladder, book graph, or $K m , n × K 2$, each of which has order p. Moreover, if $s t r ( G ) = p + δ ( G )$ for a graph G of order p with $δ ( G ) ≥ 1$, then $s t r ( G ⊙ n K 1 ) = ( n + 1 ) p + 1$, where $G ⊙ H$ is the corona product of G and H. The following problems are posed.
Problem 1.
Find sufficient conditions for a graph G of order p with $δ ( G ) ≥ 1$ to ensure $s t r ( G ) = p + δ ( G )$.
Problem 2.
Find good bounds for the strength of a graph.
Problem 3.
For every lobster T, determine the exact value of $s t r ( T )$.
Problem 4.
For every integer $n ≥ 3$, determine the strength of $Q n$, the n-dimensional hypercube.
In this paper, we obtained a sufficient condition for a graph to have $s t r ( G ) = | V ( G ) | + δ ( G )$. Moreover, we showed that every graph G either has $s t r ( G ) = | V ( G ) | + δ ( G )$ or is a proper subgraph of a graph H that has $s t r ( H ) = | V ( H ) | + δ ( H )$ with $δ ( H ) = δ ( G )$. Further, new good lower bounds of $s t r ( G )$ are obtained. Consequently, Problems 1 to 3 are solved. Moreover, we completely determine the strength of 2-regular graphs and give a partial solution to Problem 4.

## 2. Sufficient Condition

Let $G 1 = G 1 = G$ be a graph of order p with $p − 2 ≥ δ ( G ) = δ 1 ≥ 1$. Suppose that $G i$, $i ≥ 1$, is not $m i K 1$ for $m i ≥ 1$ nor $m i K 1 + K r$ with $m i ≥ 0$ for some $r ≥ 2$. We may denote $G i$ by $m i K 1 + G i$, where $m i ≥ 0$ and $δ i = δ ( G i ) ≥ 1$. Let $G i + 1$ be a graph obtained from $G i$ by deleting the $m i ≥ 1$ isolated vertices that exist in $G i$, and a vertex of degree $δ i$ together with all its neighbors in $G i$. Continue the procedure until $G s$ is either $m s K 1$ with $m s ≥ 1$ or $m s K 1 + K r$ with $m s ≥ 0$ for some $r , s ≥ 2$. This sequence ${ G i } i = 1 s$ of subgraphs is called a $δ$-sequence of G. When $G s = m s K 1$, we let $δ s = 0$ by convention. Let $y ˜ j ( G ) = m j + 1 − δ j$ for $1 ≤ j ≤ s$.
Example 1.
The following are two examples to illustrate the above construction. The black vertex is the chosen vertex that will be deleted at each stage.
 1. → → → → $G 1 = G 1$ $G 2 = G 2$ $G 3 = G 3$ $G 4 = K 1 + G 4$ $G 5 = K 1$ $δ 1 = 2$ $δ 2 = 1 , m 2 = 0$ $δ 3 = 2 , m 3 = 0$ $δ 4 = 1 , m 4 = 1$ $δ 5 = 0 , m 5 = 1$
Now, $( m 2 + 1 − δ 2 ) + ( m 3 + 1 − δ 3 ) = ( 0 ) + ( − 1 ) = − 1$. So, this δ-sequence of the graph $G 1$ does not satisfy the condition of (1) mentioned below.
 2. → → → $G 1 = G 1$ $G 2 = K 1 + G 2$ $G 3 = K 1 + G 3$ $G 4 = K 2$ $δ 1 = 2$ $δ 2 = 2 , m 2 = 1$ $δ 3 = 1 , m 3 = 1$ $δ 4 = 1 , m 4 = 0$
This δ-sequence of the graph $G 1$ satisfies the condition of (1).
Example 2.
Consider the following graph G:
(1). Suppose that we choose a as the first vertex (similarly, we can choose b, c, or d). So, we have
 → → K1 $G 1 = G$ $G 2 = 3 K 1 + G 2$ $G 3 = G 3$ $G 4 = K 1$ $δ 1 = 2$ $δ 2 = 2 , m 2 = 3$ $δ 3 = 4 , m 3 = 0$ $δ 4 = 0 , m 4 = 1$
Now, $( m 2 + 1 − δ 2 ) + ( m 3 + 1 − δ 3 ) = ( 2 ) + ( − 3 ) = − 1$.
(2). Suppose that we choose x as the first vertex. So, we have
 → → K1 $G 1 = G$ $G 2 = G 2$ $G 3 = 3 K 3 G 3$ $G 4 = K 1$ $δ 1 = 2$ $δ 2 = 2 , m 2 = 0$ $δ 3 = 4 , m 3 = 3$ $δ 4 = 0 , m 4 = 1$
Now, $( m 2 + 1 − δ 2 ) = − 1$.
So, every δ-sequence of G does not satisfy condition (1).
The following theorem gives a sufficient condition for a graph to have minimal strength.
Theorem 1.
For a graph G of order p with $p − 2 ≥ δ ( G ) = δ 1 ≥ 1$, if there is a δ-sequence ${ G i } i = 1 s$ of G such that
then $s t r ( G ) = p + δ ( G )$.
Note that a sum with an empty term is treated as zero, as usual. If there is no ambiguity, we will write $y ˜ j ( G )$ as $y ˜ j$ and $z ˜ i ( G )$ as $z ˜ i$.
Proof.
Let ${ G i } i = 1 s$ be a $δ$-sequence of G satisfying condition (1). Let $u i$ be a vertex in $G i$ of degree $δ i$, which is deleted from $G i$ to obtain $G i + 1$, $1 ≤ i ≤ s − 1$. Now, $G 1 = G 1 = G$. We shall construct a numbering f of G such that $s t r f ( G ) = p + δ 1$.
Label $u 1$ by p and all its neighbors by 1 to $δ 1$ in arbitrary order. This guarantees that the largest induced edge label is $p + δ 1$ at this stage.
Suppose that we have labeled vertices in $V ( G ) ∖ V ( G i + 1 )$ by using the labels in $[ 1 , ∑ j = 1 i δ j ] ∪ [ p + 1 − ∑ j = 1 i ( m j + 1 ) , p ] Z$, where $1 ≤ i ≤ s − 1$. Moreover, the neighbors of $u i$ are labeled with labels in $[ 1 + ∑ j = 1 i − 1 δ j , ∑ j = 1 i δ j ] Z$, and all induced edge labels are at most $p + δ 1$, up to now. Note that a sum with an empty term is treated as zero.
Now, we consider the graph $G i + 1$.
(a) Suppose that $2 ≤ i + 1 < s$. We label the $m i + 1$ isolated vertices of $G i + 1$ with labels in $[ p + 2 − ∑ j = 1 i + 1 ( m j + 1 ) , p − ∑ j = 1 i ( m j + 1 ) ] Z$, respectively (if $m i + 1 = 0$, then this process does not exist), and $u i + 1$ by $p + 1 − ∑ j = 1 i + 1 ( m j + 1 )$ and its neighbors with labels in $[ 1 + ∑ j = 1 i δ j , ∑ j = 1 i + 1 δ j ] Z$, respectively.
Now, the vertices of $V ( G ) ∖ V ( G i + 2 )$ are labeled by using the labels in $[ 1 , ∑ j = 1 i + 1 δ j ] ∪ [ p + 1 − ∑ j = 1 i + 1 ( m j + 1 ) , p ] Z$.
Since each isolated vertex of $G i + 1$ is only adjacent to some neighbors of $u i$, and $u i + 1$ may be adjacent with some neighbors of $u i$, the largest new induced edge label related to these vertices is
The largest new induced edge label related to $u i + 1$ and its neighbors in $G i + 1$ is
$p + 1 − ∑ j = 1 i + 1 ( m j + 1 ) + ∑ j = 1 i + 1 δ j = p + 1 − ∑ j = 2 i + 1 ( m j + 1 − δ j ) − ( m 1 + 1 − δ 1 ) = p + 1 − z ˜ i + 1 − ( m 1 + 1 − δ 1 ) ≤ p + 1 − ( 1 − δ 1 ) = p + δ 1 .$
Repeat this process until $i + 1 = s$.
(b) Suppose that $i + 1 = s$. Now, $G s = m s K 1$ with $m s ≥ 1$ or $m s K 1 + K r$ for some $r ≥ 2$ and $m s ≥ 0$. In this case, the set of unused labels is $[ 1 + ∑ j = 1 s − 1 δ j , p − ∑ j = 1 s − 1 ( m j + 1 ) ] Z$. That is, $m s = p − ∑ j = 1 s − 1 ( m j + 1 + δ j )$ or $m s + r = p − ∑ j = 1 s − 1 ( m j + 1 + δ j )$.
When $G s = m s K 1$, the process is the same as in the above case. Hence, we have a numbering for G with the strength $p + δ 1$.
When $G s = m s K 1 + K r$, where $δ s + 1 = r = p − m s − ∑ j = 1 s − 1 ( m j + 1 + δ j )$, we label the $m s$ isolated vertices of $G s$ with labels in $[ p − m s + 1 − ∑ j = 1 s − 1 ( m j + 1 ) , p − ∑ j = 1 s − 1 ( m j + 1 ) ] Z$, respectively (if $m s = 0$, then this process is not performed). Finally, label the vertices of $K r$ with labels in $[ 1 + ∑ j = 1 s − 1 δ j , p − m s − ∑ j = 1 s − 1 ( m j + 1 ) ] Z$, respectively. Then, the largest new induced edge labels related to the neighbors of $u s − 1$ are
$p − ∑ j = 1 s − 1 ( m j + 1 ) + ∑ j = 1 s − 1 δ j = p − ( m 1 + 1 − δ 1 ) − z ˜ s − 1 ≤ p − 1 + δ 1 < p + δ 1 .$
The largest new induced edge labels in $K r$ are
$[ p − m s − ∑ j = 1 s − 1 ( m j + 1 ) ] + [ p − 1 − m s − ∑ j = 1 s − 1 ( m j + 1 ) ] = [ p − m s − ∑ j = 1 s − 1 ( m j + 1 ) ] + [ ∑ j = 1 s δ j ] = p + 1 − ∑ j = 1 s ( m j + 1 − δ j ) ≤ p + 1 − m 1 − 1 + δ 1 = p + δ 1 .$
Hence, we have a numbering f such that $s t r f ( G ) = p + δ 1$. Therefore, $s t r ( G ) ≤ p + δ 1$. By Lemma 2, $s t r ( G ) = p + δ 1$. □
Example 3.
Consider the graph $G 1$ described in Example 1. Using the second δ-sequence of $G 1$ and following the construction in the proof of Theorem 1, we have the following strength labeling f of $G 1$ such that $s t r f ( G 1 ) = 14$.
From Example 2, G does not satisfy the hypothesis of Theorem 1, but there is a strength labeling f for it with $s t r f ( G ) = p + δ ( G ) = 17$ as follows.
So, the converse of Theorem 1 is not true.
Thus, Theorem 1 provides a solution to Problem 1. Note that every tree T has the property that $δ i = 1$ and $m i ≥ 0$ for each $i ≥ 2$. We immediately have $s t r ( T ) = | V ( T ) | + 1$ and the following corollary that answers more than what Problem 3 asks.
Corollary 2.
If G is a forest without an isolated vertex, then $s t r ( G ) = | V ( G ) | + 1$.
Corollary 3.
The one-point union of cycles G of order p has $s t r ( G ) = p + 2$.
Proof.
Remove a degree 2 vertex that is adjacent to the maximum degree vertex of G and its neighbors to obtain a subgraph $G 2$, which is a disjoint union of path(s). So, G admits a $δ$-sequence that satisfies (1). □
Corollary 4.
If G is a wheel or fan graph of order p, then $s t r ( G ) = p + δ ( G )$.
In constructing a $δ$-sequence of G, if we change the choice of choosing a vertex of degree $δ i$ to a vertex of degree $d i$, then we get another sequence of subgraphs of G. This sequence is called a d-sequence of G. Let $y j ( G ) = m j + 1 − d j$, $z i ( G ) = ∑ j = 2 i y j ( G )$ and denote $d 1$ as $d G$. By the same argument as that when proving Theorem 1, we have:
Theorem 5.
For a graph G of order p with $p − 2 ≥ δ ( G ) ≥ 1$, if there is a d-sequence ${ G i } i = 1 s$ of G such that
then $s t r ( G ) ≤ p + d G$. The equality holds if $d G = δ ( G )$.
Example 4.
Consider the graph G in Example 2. The following is a d-sequence of G.
 → → → $G 1 = G$ $G 2 = 3 K 1 + G 2$ $G 3 = G 3$ $G 4 = K 2$ $d G = d 1 = δ ( G ) = 2$ $d 2 = 3 , m 2 = 3$ $d 3 = 2 , m 3 = 0$ $d 4 = 1 , m 4 = 0$
Now, $z 2 = 1$, $z 3 = 0$, and $z 4 = 0$. By Theorem 5, $s t r ( G ) = 17$, as shown in Example 3.
Lemma 4.
Let T be a forest without an isolated vertex of an order at of least 3, and let $P T$ be the set of pendant vertices that are adjacent to a vertex of a degree of at least 2. There is a δ-sequence of T of length s such that $z ˜ s ( T ) ≥ | P T | − | N T ( P T ) |$, where $z ˜ i ( T )$ is defined in (1). Moreover, all $z ˜ i ( T )$ satisfy (1).
Proof.
Obviously, the lemma holds when T is of order 3. Suppose that the lemma holds when the order of T is k or less, where $k ≥ 3$.
Now, consider a forest T of order $k + 1$. Choose a vertex $u ∈ P T$. Let v be the vertex adjacent to u with degree d. We shall consider the forest $T − u − v$.
Suppose that $T = K 1 , k$, which is a star; then, $T − u − v = ( k − 1 ) K 1$ and ${ T , ( k − 1 ) K 1 }$ is a $δ$-sequence of T. Note that $| P T | = k$ and $| N T ( P T ) | = 1$. Clearly, $z ˜ 2 ( T ) = ( k − 1 ) + 1 − 0 = k > | P T | − | N T ( P T ) |$.
Now, we assume that T is not a star. Let $T − u − v = m K 1 + T ′$, where $m ≥ 0$ and $T ′$ is a forest without an isolated vertex.
Suppose that the order of $T ′$ is 2; then, $T ′ = K 2$ and ${ T , m K 1 + K 2 }$ is a $δ$-sequence of T, where $m = k − 3$. Now, $z ˜ 2 ( T ) = m = k − 3$. If $T = K 1 , k − 2 + K 2$, then $| P T | = k − 2$ and $| N T ( P T ) | = 1$. We get $z ˜ 2 ( T ) = | P T | − | N T ( P T ) |$. If T is a tree, then $| P T | = k − 1$ and $| N T ( P T ) | = 2$. So, we still get $z ˜ 2 ( T ) = | P T | − | N T ( P T ) |$.
Suppose that the order of $T ′$ is greater than 2. By the induction assumption, there is a $δ$-sequence of $T ′$, say ${ T ′ = T 1 , T 2 , ⋯ , T s }$, such that $z ˜ s ( T ′ ) ≥ | P T ′ | − | N T ′ ( P T ′ ) |$ for some $s ≥ 2$.
Now, consider ${ T , T 1 , ⋯ , T s }$ of T. Note that $z ˜ 2 ( T ) = m + 1 − δ ( T ′ ) = m$. Let a be the number of vertices of degree 2 in T, but degree 1 in $T ′$; then, $| P T ′ | = | P T | − ( m + 1 ) + a$, where $a ≥ 0$. Let $b = | N T ′ ( P T ′ ) ∖ N T ( P T ) |$. Therefore, $| N T ′ ( P T ′ ) | ≤ | N T ( P T ) | − 1 + b$ (some vertices in $N T ( P T )$ may not be in $N T ′ ( P T ′ )$).
If a vertex w in T of degree 2 becomes of degree 1 in $T ′$, then w may be in $P T ′$ and may have at most one neighbor in $N T ′ ( P T ′ )$ that is not in $N T ( P T )$. So, $a ≥ b$.
Now, we have
$z ˜ s + 1 ( T ) = z ˜ 2 ( T ) + z ˜ s ( T ′ ) = m + z ˜ s ( T ′ ) ≥ m + | P T ′ | − | N T ′ ( P T ′ ) | ≥ m + [ | P T | − ( m + 1 ) + a ] − [ | N T ( P T ) | − 1 + b ] = | P T | − | N T ( P T ) | + a − b ≥ | P T | − | N T ( P T ) | .$
By induction, the lemma holds for any forest T of an order greater than 2. □
Remark 1.
In the proof of Lemma 4, we can see that $z ˜ 2 ( T ) = | P T | − | N T ( P T ) | + 1$ if T is a star.
Theorem 6.
Keep all notations defined in Theorem 1 and Lemma 4. Let H be a graph with $δ ( H ) ≥ 1$. Suppose that ${ H = H 1 , ⋯ , H s }$ is a d-sequence of H. Suppose that
$Z = min { z i ( H ) | 2 ≤ i ≤ s }$
is not positive. Suppose that T is a graph with a d-sequence ${ T = T 1 , ⋯ , T t }$ satisfying (2). Let $G = H + T$. If $z t ( T ) ≥ d H − Z$; then, $| V ( G ) | + δ ( G ) ≤ s t r ( G ) ≤ | V ( G ) | + d T$.
Proof.
Since $T t$ is either $m K 1$ for some $m ≥ 1$ or $m K 1 + K r$ for some $m ≥ 0$, we have the following two cases.
For the first case, $y t ( T ) = m + 1$ and ${ T 1 , ⋯ , T t − 1 , m K 1 + H , H 2 , ⋯ , H s }$ is a d-sequence of G. Then,
$z j ( G ) = z j ( T ) ≥ 0$, $2 ≤ j ≤ t − 1$;
$z t ( G ) = z t − 1 ( T ) + y t ( H + T ) = z t − 1 ( T ) + [ m + 1 − d H ] = z t ( T ) − d H ≥ z t ( T ) − d H + Z ≥ 0$;
$z t + 1 ( G ) = z t − 1 ( T ) + ( m + 1 − d H ) + z 2 ( H ) = z t ( T ) − d H + y 2 ( H ) = z t ( T ) − d H + z 2 ( H )$.
In general,
$z t + j ( G ) = z t − 1 ( T ) + ( m + 1 − d H ) + z j + 1 ( H ) = z t ( T ) − d H + z j + 1 ( H ) ≥ z t ( T ) − d H + Z ≥ 0 , 1 ≤ j ≤ s − 1 .$
For the last case, ${ T 1 , ⋯ , T t , H 1 , H 2 , ⋯ , H s }$ is a d-sequence of G. We will get that
$z j ( G ) = z j ( T ) ≥ 0$, $2 ≤ j ≤ t$;
$z t + 1 ( G ) = z t ( G ) + [ 1 − d H ] = z t ( T ) + 1 − d H > 0$;
$z t + j ( G ) = z t ( T ) + 1 − d H + z j ( H ) > 0$, $2 ≤ j ≤ s$.
By Theorem 5, $s t r ( G ) ≤ | V ( G ) | + d T$. The lower bound follows from Lemma 2. □
Remark 2.
In Theorem 6, suppose that H and T are connected. Let $x ∈ V ( T t − 1 )$ be chosen to construct $T t$, let $v ∈ V ( T t − 1 )$ be a neighbor of x, and let $u ∈ V ( H )$, which is not chosen to construct $H 2$. We add an edge $v u$ to the graph $H + T$. All of the $y j$ values of this connected graph are the same as those of $H + T$.
Theorem 7.
Keep all notations defined in Theorem 1 and Lemma 4. Let H be a graph with $δ ( H ) ≥ 1$. Suppose that ${ H = H 1 , ⋯ , H s }$ is a d-sequence of H. Suppose
$Z = min { z i ( H ) | 2 ≤ i ≤ s } ≤ 0 .$
Let $G = H + T$, where T is a forest without an isolated vertex of an order of at least 3. If $| P T | − | N T ( P T ) | ≥ d H − Z$; then, $s t r ( G ) = | V ( G ) | + 1$.
Proof.
By Lemma 4, there is a $δ$-sequence ${ T 1 , ⋯ , T t }$ of T such that $z ˜ t ( T ) ≥ | P T | − | N T ( P T ) |$ and all $z ˜ i ( T )$ satisfy (1). Since this $δ$-sequence is a particular d-sequence of T, it satisfies the condition of Theorem 6. Since $d T = 1$ now, by Theorem 6, we have $s t r ( G ) ≤ | V ( G ) | + 1$. Hence, we have the theorem, since $δ ( G ) = 1$. □
Corollary 8.
Let H be a graph with $δ ( H ) ≥ 1$. Suppose that ${ H = H 1 , ⋯ , H s }$ is a d-sequence of H.
Suppose
$Z = min { z i ( H ) | 2 ≤ i ≤ s } ≤ 0 .$
Let $G = H + K 1 , k$. If $k ≥ d H − Z$, then $s t r ( G ) = | V ( G ) | + 1$.
Let $G = K m , n$ for $n ≥ m ≥ 1$. It is proven in [7] (Theorem 3.5) that $s t r ( G ) = | V ( G ) | + m$. So, we have the following.
Corollary 9.
There exists a graph G with $s t r ( G ) = | V ( G ) | + δ ( G )$ for each $δ ( G ) ≥ 1$.
Suppose that H is a graph with $s t r ( H ) > | V ( H ) | + δ ( H )$. So, there is a d-sequence of H with $d H = δ ( H )$ such that Z is non-positive, where Z is defined in (3). Let $T = K m , n$. Since ${ T , ( n − 1 ) K 1 }$ is a d-sequence of T, $z 2 ( T ) = n$. Suppose that $n ≥ d H − Z$ and $m = δ ( H )$. Since $n ≥ d H − Z ≥ δ ( H )$, $δ ( H + T ) = δ ( H )$. By Theorem 6, we have $s t r ( H + T ) = | V ( H + T ) | + δ ( H )$. So, together with Remark 2, we have the following theorem.
Theorem 10.
For every graph H, either $s t r ( H ) = | V ( H ) | + δ ( H )$ or H is a proper subgraph of a graph G such that $s t r ( G ) = | V ( G ) | + δ ( G )$ with $δ ( G ) = δ ( H )$.
Example 5.
It is easy to obtain a δ-sequence of $Q 4$ with
$δ 1 = 4 ; δ 2 = 2 , m 2 = 0 , s o z ˜ 2 = − 1 ; δ 3 = 1 , m 3 = 0 , s o z ˜ 3 = − 1 ; δ 4 = 1 , m 4 = 1 , s o z ˜ 4 = 0 ; δ 5 = 0 , m 5 = 3 , s o z ˜ 5 = 4 .$
Thus, $Z = − 1$. Let $n = 4 + 1 = 5$. By the construction before Theorem 10, we have $s t r ( Q 4 + K 4 , 5 ) = 25 + 4 = 29$. A required labeling can be obtained by similarly following the proof of Theorem 1. Moreover, adding an edge joining a vertex of degree 5 of $K 4 , 5$ and a vertex of $Q 4$ gives a connected graph G that contains $Q 4$ as a proper subgraph with $s t r ( G ) = s t r ( Q 4 + K 4 , 5 )$, as required.
Example 6.
For 2-regular graphs $C k$ with exactly $k ≥ 1$ odd cycles, we have $Z = − k + 1$. Let $n = k + 1$. By the construction before Theorem 10, we have $s t r ( C k + K 2 , k + 1 ) = | V ( C k ) | + ( k + 3 ) + 2$.

## 3. New Lower Bounds

Theorem 11.
Suppose that G is a graph of order p with an independent number α; then, $s t r ( G ) ≥ 2 p − 2 α + 1$.
Proof.
For any numbering of G, by the pigeonhole principle, at least two integers in $[ p − α , p ] Z$ are assigned to two adjacent vertices. So, the induced edge label is at least $2 p − 2 α + 1$. This completes the proof. □
Corollary 12.
Suppose that G is a graph of order p with minimum degree δ. Suppose that $s t r ( G ) = p + δ$; then, $α ≥ p − δ + 1 2$, where α is the independence number of G.
Proof.
From Theorem 11, we have $α ≥ p − δ + 1 2$. □
Let G be a graph of order p. Let
$x i = min | N G ( S ) ∖ S | : | S | = i ; ξ = ξ ( G ) = max { x i − i + 1 | 1 ≤ i ≤ p − 1 } .$
Theorem 13.
Let G be a graph of order p; then, $s t r ( G ) ≥ p + ξ$.
Proof.
Let $ξ = x i − i + 1$ for some i. Let f be a strength labeling of G. Consider the labels in $[ p − i + 1 , p ] Z$. Let $T = f − 1 ( [ p − i + 1 , p ] )$; then, $| T | = i$. Now, $| f ( N G ( T ) ∖ T ) | = | N G ( T ) ∖ T | ≥ x i$. Let a be the largest label in $f ( N G ( T ) ∖ T )$. There is a vertex $u ∈ N G ( T ) ∖ T$ such that $f ( u ) = a$. Moreover, u is adjacent to $v ∈ T$. Thus,
$s t r f ( G ) ≥ f ( v ) + f ( u ) ≥ p − i + 1 + a ≥ p − i + 1 + x i = p + ξ .$
□
Thus, we provided two good bounds for the strength of a graph, as raised in Problem 2. Note that Lemma 2 is a corollary of Theorem 13 when $ξ = δ = x 1 ≥ x i − i + 1$ for $i ≥ 2$.
Theorem 14.
If $G = ∑ i = 1 h C 2 m i + ∑ j = 1 k C 2 n j + 1$, where $m i ≥ 2$, $n j ≥ 1$, and $h + k ≥ 1$, then $s t r ( G ) = max { p + 2 , p + 1 + k }$.
Proof.
Note that if $h = 0$, then the first summand does not appear, similarly for the second summand. Now, $α ( G ) = ∑ i = 1 h m i + ∑ j = 1 k n j$. By Theorem 11, we have $s t r ( G ) ≥ p + 1 + k$.
Let $H = ∑ i = 1 h C 2 m i$ and $K = ∑ j = 1 k C 2 n j + 1$. Let $M = ∑ i = 1 h m i$ such that $M = 0$ when $h = 0$.
We shall construct a numbering f on G. If $h ≥ 1$, we first label H with integers in $[ 1 , M ] Z ∪ [ p − M + 1 , p ] Z$ as follows.
Label $C 2 m 1$ with $1 , p , 2 , ⋯ , m 1 , p − m 1 + 1$ in the natural order. In general, for $i ≥ 2$, we label the vertices of even cycle $C 2 m i$ with $1 + ∑ l = 1 i − 1 m l$, $p − ∑ l = 1 j − 1 m l$, $2 + ∑ l = 1 j − 1 m l$, ⋯, $∑ l = 1 i m l$, $p − ∑ l = 1 j m l + 1$ in the natural order. Continue this process until $i = h$. Hence, the maximum induced edge label is $p + 2$.
If $k ≥ 1$, then we label the vertices of odd cycle $C 2 n 1 + 1$ by $M + 1 , p − M , M + 2 , ⋯ , p − M − n 1 + 1 , M + n 1 + 1$ in the natural order. Up to now, the maximum induced edge label is still $p + 2$.
Now, we label the vertices of odd cycle $C 2 n 2 + 1$ with $M + n 1 + 2$, $p − M − n 1$, $M + n 1 + 3$, ⋯, $p − M − n 1 − n 2 + 1$, $M + n 1 + n 2 + 2$ in the natural order. Note that $( M + n 1 + 2 ) + ( M + n 1 + n 2 + 2 ) = 2 M + ( 2 n 1 + 1 ) + n 2 + 2 ≤ p$. So, the current maximum induced edge label is $p + 3$.
In general, for $j ≥ 2$, we label the vertices of odd cycle $C 2 n j + 1$ with $M + 1 + ∑ l = 1 j − 1 ( n l + 1 )$, $p − M − ∑ l = 1 j − 1 n l$, $M + 2 + ∑ l = 1 j − 1 ( n l + 1 )$, ⋯, $p − M − ∑ l = 1 j n l + 1$, $M + ∑ l = 1 j ( n l + 1 )$ in the natural order. Note that $( M + 1 + ∑ l = 1 j − 1 ( n l + 1 ) ) + ( M + ∑ l = 1 j ( n l + 1 ) ) = 2 M + ∑ l = 1 j − 1 ( 2 n l + 1 ) + n j + j ≤ p − 2 + j$. So, the current maximum induced edge label is $p + j + 1$.
Continue this process until $j = k$. Hence, we have $s t r f ( G ) = p + k + 1$. □
Example 7.
Consider $G = C 4 + C 6 + C 5 + C 5 + C 7$. Now, $p = 27$ and $k = 3$.
We label the vertices of
$C 4$ with integers in $[ 1 , 27 , 2 , 26 ]$ (max. induced edge label is 29);
$C 6$ with integers in $[ 3 , 25 , 4 , 24 , 5 , 23 ]$ (max. induced edge label is 29);
$C 5$ with integers in $[ 6 , 22 , 7 , 21 , 8 ]$; (max. induced edge label is 29);
$C 5$ with integers in $[ 9 , 20 , 10 , 19 , 11 ]$ (max. induced edge label is 30);
$C 7$ with integers in $[ 12 , 18 , 13 , 17 , 14 , 16 , 15 ]$ (max. induced edge label is 31).
So, $s t r ( G ) = 31$.
Let $G × H$ be the Cartesian product of graphs G and H.
Lemma 5.
Let G be a bipartite graph with bipartition $( X , Y )$ such that $| X | = | Y | = m$. Suppose that there is a numbering f of G such that $f ( X ) = [ 1 , m ] Z$; then, there is a numbering F of $G × K 2$ such that $F ( [ 1 , 2 m ] ) Z = X ˜$, where $( X ˜ , Y ˜ )$ is a bipartition of $G × K 2$. Moreover, $s t r F ( G × K 2 ) = 5 m + 1$.
Note that the following proof is modified from the proof of Theorem 3.10 in [7].
Proof.
Note that, from the hypotheses, $f ( Y ) = [ m + 1 , 2 m ] Z$. Let u and v be vertices of $K 2$. Then,
$X ˜ = { ( x , u ) | x ∈ X } ∪ { ( y , v ) | y ∈ Y }$ and $Y ˜ = { ( y , u ) | y ∈ Y } ∪ { ( x , v ) | x ∈ X }$.
Define $F : V ( G × K 2 ) → [ 1 , 4 m ] Z$ by
$F ( x , u ) = f ( x ) ∈ [ 1 , m ] Z F ( y , v ) = ( 3 m + 1 ) − f ( y ) ∈ [ m + 1 , 2 m ] Z F ( x , v ) = ( 3 m + 1 ) − f ( x ) ∈ [ 2 m + 1 , 3 m ] Z F ( y , u ) = 2 m + f ( y ) ∈ [ 3 m + 1 , 4 m ] Z .$
Clearly, $F ( X ˜ ) = [ 1 , 2 m ] Z$.
Now, $F ( x , u ) + F ( x , v ) = f ( x ) + ( 3 m + 1 ) − f ( x ) = 3 m + 1$ and $F ( y , u ) + F ( y , v ) = 2 m + f ( y ) + ( 3 m + 1 ) − f ( y ) = 5 m + 1$. Suppose that $( x 1 , u )$ and $( x 2 , u )$ are adjacent in $G × K 2$. By definition, $F ( x 1 , u ) + F ( x 2 , u ) ≤ 2 m$. Similarly, if $( y 1 , v )$ and $( y 2 , v )$ are adjacent in $G × K 2$, then $F ( y 1 , v ) + F ( y 2 , v ) ≤ 4 m$. Thus, $s t r F ( G × K 2 ) = 5 m + 1$. □
Let $Q n$ be a hypercube of dimension n, $n ≥ 2$. Since there is a strength numbering f of $Q 2$ satisfying the hypotheses of Lemma 5, by applying this lemma repeatedly, we get the following.
Theorem 15.
For $n ≥ 2$, $s t r ( Q n ) ≤ 2 n + 2 n − 2 + 1$.
This is a known result in [7] (Theorem 3.10).
We shall improve the lower bound of the strength of $Q n$. The vertices of $Q n$ often used the elements of the vector space $Z 2 n$ over $Z 2$. Two vertices u and v are adjacent if and only if $u + v = e i$, where $e i$ is the standard basis of $Z 2 n$. Note that $v = − v$ for any vector $v ∈ Z 2 n$. In the proofs of the following lemmas, all algebra involving vectors is over $Z 2$.
For any vertex v, we let $N G [ v ] = N G ( v ) ∪ { v }$, the closed neighborhood of v. Hence, for any subset of vertices S, $N G ( S ) ∖ S = ⋃ v ∈ S N G [ v ] ∖ S$. We shall omit the subscript G when there is no ambiguity.
Lemma 6.
If u and v are two distinct vertices of $Q n$, $n ≥ 3$, then $| N [ u ] ∩ N [ v ] |$ is either 0 or 2.
Proof.
Suppose that u and v are adjacent. Clearly, $| N [ u ] ∩ N [ v ] | = 2$. Suppose that u and v are not adjacent. If $z ∈ N [ u ] ∩ N [ v ]$, then the distance between u and v is two. Hence, $u ∉ N ( v )$, $v ∉ N ( u )$ and $u + v = e i + e j$ (equivalently, $u + e i = v + e j$), where $i ≠ j$. Since $z ∈ N ( u ) ∩ N ( v )$, $z = u + e k = v + e l$ for some $k , l$. So, $u + e k + v + e l = 0$ or $u + v = e k + e l$. Thus, ${ i , j } = { k , l }$. Hence, $| N [ u ] ∩ N [ v ] | = | { u + e i , u + e j } | = 2$. □
Lemma 7.
For any three distinct vertices u, v, and w of $Q n$, $n ≥ 3$,
Proof.
If one of $| N [ u ] ∩ N [ v ] |$, $| N [ u ] ∩ N [ w ] |$ and $| N [ v ] ∩ N [ w ] |$ is 0, then $| N [ u ] ∩ N [ v ] ∩ N [ w ] | = 0$. Otherwise, Lemma 6 implies that $| N [ u ] ∩ N [ v ] | = | N [ u ] ∩ N [ w ] | = | N [ v ] ∩ N [ w ] | = 2$.
(1).
Suppose that only one pair of u, v, and w are adjacent—say, $u v$ is an edge—then, the distances from w to u and to v are 2. This creates a five-cycle, which is impossible.
(2).
Suppose that two pairs of u, v, and w are adjacent—say, $u v$ and $u w$ are edges. Note that v and w cannot be adjacent. Then, $N [ u ] ∩ N [ v ] = { u , v }$ and $N [ u ] ∩ N [ w ] = { u , w }$. Hence, $u ∈ N [ v ] ∩ N [ w ]$. This implies that $N [ u ] ∩ N [ v ] ∩ N [ w ] = { u }$.
(3).
Suppose that none of u, v, and w are adjacent. By the proof of Lemma 6, we have $u + v = e i 1 + e j 1$ and $v + w = e i 2 + e j 2$ for some $i 1 , i 2 , j 1 , j 2$, $i 1 ≠ j 1$, and $i 2 ≠ j 2$. This implies that $u + w = e i 1 + e j 1 + e i 2 + e j 2$. Since the distance of u and w is 2, $| { i 1 , j 1 } ∩ { i 2 , j 2 } | = 1$. Without loss of generality, we may assume that $i 1 = i 2$. Now, $N [ u ] ∩ N [ v ] = { u + e i 1 , u + e j 1 }$, $N [ u ] ∩ N [ w ] = { u + e j 1 , u + e j 2 }$, and $N [ v ] ∩ N [ w ] = { v + e i 2 , v + e j 2 }$. Here, $v + e i 2 = v + e i 1 = u + e j 1$. Hence, $u + e j 1 ∈ N [ u ] ∩ N [ v ] ∩ N [ w ]$. Since $u + e j 2 ∉ N [ u ] ∩ N [ v ]$, $N [ u ] ∩ N [ v ] ∩ N [ w ] = { u + e j 1 }$.
This completes the proof. □
Theorem 16.
For the hypercube $Q n$, $n ≥ 2$, we have $s t r ( Q 2 ) ≥ 6$; $s t r ( Q 3 ) ≥ 11$; $s t r ( Q 4 ) ≥ 21$; and $s t r ( Q n ) ≥ 2 n + 4 n − 12$ for $n ≥ 5$.
Proof.
Keeping the notations defined in Theorem 13, we want to compute $x i$ and $ξ$. Clearly, $x 1 = δ = n$.
Suppose that $S = { u , v }$ with $u ≠ v$.
$| N [ u ] ∪ N [ v ] | = | N [ u ] | + | N [ v ] | − | N [ u ] ∩ N [ v ] | ≥ 2 ( n + 1 ) − 2 = 2 n .$
So, $| N ( S ) ∖ S | = | N [ u ] ∪ N [ v ] | − | S | ≥ 2 n − 2$. Actually, when $S = { 0 , e 1 + e 2 }$, $| N ( S ) ∖ S | = 2 n − 2$. Thus, $x 2 = 2 n − 2$.
Suppose that $S = { u , v , w }$, where $u , v , w$ are distinct.
(1).
If all $| N [ u ] ∩ N [ v ] |$, $| N [ u ] ∩ N [ w ] |$ and $| N [ v ] ∩ N [ w ] |$ are not zero, then by Lemma 7, $| N [ u ] ∩ N [ v ] ∩ N [ w ] | = 1$. Thus,
$| N [ u ] ∪ N [ v ] ∪ N [ w ] | = | N [ u ] | + | N [ v ] | + | N [ w ] | − | N [ u ] ∩ N [ v ] | − | N [ u ] ∩ N [ w ] | − | N [ v ] ∩ N [ w ] | + | N [ u ] ∩ N [ v ] ∩ N [ w ] | = 3 ( n + 1 ) − 3 × 2 + 1 = 3 n − 2 .$
Actually, $S = { 0 , e 1 + e 2 , e 1 + e 3 }$.
(2).
If all $| N [ u ] ∩ N [ v ] |$, $| N [ u ] ∩ N [ w ] |$ and $| N [ v ] ∩ N [ w ] |$ are zero, then $| N [ u ] ∪ N [ v ] ∪ N [ w ] | = 3 ( n + 1 )$.
(3).
If at least one of $| N [ u ] ∩ N [ v ] |$, $| N [ u ] ∩ N [ w ] |$, and $| N [ v ] ∩ N [ w ] |$ is not zero and at least one of them is zero, then
$| N [ u ] ∪ N [ v ] ∪ N [ w ] | = | N [ u ] | + | N [ v ] | + | N [ w ] | − | N [ u ] ∩ N [ v ] | − | N [ u ] ∩ N [ w ] | − | N [ v ] ∩ N [ w ] | + | N [ u ] ∩ N [ v ] ∩ N [ w ] | ≥ 3 ( n + 1 ) − 2 × 2 = 3 n − 1 .$
Thus, $x 3 = 3 n − 5$.
Let us consider $S = { u 1 , u 2 , u 3 , u 4 }$, where $u 1 , u 2 , u 3 , u 4$ are distinct. Then,
$⋃ l = 1 4 N [ u l ] = ∑ l = 1 4 | N [ u l ] ) | − ∑ 1 ≤ j < l ≤ 4 | N [ u j ] ∩ N [ u l ] | + ∑ 1 ≤ h < j < l ≤ 4 | N [ u h ] ∩ N [ u j ] ∩ N [ u l ] | − | N [ u 1 ] ∩ N [ u 2 ] ∩ N [ u 3 ] ∩ N [ u 4 ] | .$
(1).
If only one of $N [ u j ] ∩ N [ u l ] = ⌀$, then by Lemma 7, the third summand is 1 and the fourth summand is 0. Then, $⋃ l = 1 4 N [ u l ] ≥ 4 n + 4 − 5 × 2 + 1 = 4 n − 5$.
Actually, $S = { 0 , e 1 + e 2 , e 1 + e 3 , e 1 + e 2 + e 3 + e 4 }$.
(2).
If more than one of $N [ u j ] ∩ N [ u l ] = ⌀$, then the third and fourth summands are 0. Thus, $⋃ l = 1 4 N [ u l ] ≥ 4 n + 4 − 4 × 2 = 4 n − 4$.
(3).
If all of $N [ u j ] ∩ N [ u l ] ≠ ⌀$, then
$⋃ l = 1 4 N [ u l ] = 4 n + 4 − 6 × 2 + 4 − | N [ u 1 ] ∩ N [ u 2 ] ∩ N [ u 3 ] ∩ N [ u 4 ] | ≥ 4 n − 4 − 1 = 4 n − 5 .$
Therefore, $x 4 = 4 n − 9$.
Hence, by Theorem 13, we have $ξ ≥ 2$ when $n = 2$; $ξ ≥ 3$ when $n = 3$; $ξ ≥ 5$ when $n = 4$; and $ξ ≥ 4 n − 12$ when $n ≥ 5$. Thus, we have $s t r ( Q 2 ) ≥ 6$; $s t r ( Q 3 ) ≥ 11$; $s t r ( Q 4 ) ≥ 21$; and $s t r ( Q n ) ≥ 2 n + 4 n − 12$ when $n ≥ 5$. □
From the proof of Theorem 16, we have $x 1 = n$, $x 2 = 2 n − 2$, $x 3 = 3 n − 5$, and $x 4 = 4 n − 9$ for $Q n$.
Suppose that $x i + 1 = | N ( S ) ∖ S |$ for some subset of vertices S with $| S | = i + 1$. Let $S = { u 1 , ⋯ , u i + 1 }$ with $| S | = i + 1$.
So, $x i + 1 ≥ n + x i − 2 i$. Since $x 4 = 4 n − 9$, by induction, we will get $x i ≥ i n + 3 − ( i − 1 ) i$, where $i ≥ 4$.
Let $η i = x i − i + 1$; then, $η i + 1 ≥ η i + n − 2 i − 1$. So, $η i$ is increasing when $i ≤ ( n − 1 ) / 2$.
Suppose that $n = 2 m$, $m ≥ 2$; then, $η m ≥ η m − 1 + 2 m − 2 ( m − 1 ) − 1 = η m − 1 + 1$. So, $ξ ≥ η m = x m − m + 1 ≥ m ( 2 m ) + 3 − ( m − 1 ) m − m + 1 = m 2 + 4$. We have $s t r ( Q 2 m ) ≥ 2 2 m + m 2 + 4$.
Suppose that $n = 2 m − 1$, $m ≥ 2$; then, $η m ≥ η m − 1 + ( 2 m − 1 ) − 2 ( m − 1 ) − 1 = η m − 1$. So, $ξ ≥ η m = x m − m + 1 ≥ m 2 − m − 4$. We have $s t r ( Q 2 m − 1 ) ≥ 2 2 m − 1 + m 2 − m + 4$.
Combining with Theorem 16, we have the following.
Theorem 17.
For $n ≥ 2$,
1.
$s t r ( Q 2 ) ≥ 6$; $s t r ( Q 3 ) ≥ 11$; $s t r ( Q 4 ) ≥ 21$;
2.
$s t r ( Q n ) ≥ 2 n + 4 n − 12$ for $5 ≤ n ≤ 9$;
3.
$s t r ( Q 2 m ) ≥ 2 2 m + m 2 + 4$ for $m ≥ 5$.
4.
$s t r ( Q 2 m − 1 ) ≥ 2 2 m − 1 + m 2 − m + 4$ for $m ≥ 6$.
Corollary 18.
$s t r ( Q 2 ) = 6$, $s t r ( Q 3 ) = 11$, $s t r ( Q 4 ) = 21$, and $s t r ( Q 5 ) = 40$.
Proof.
Combining with Theorems 15 and 16, we have $s t r ( Q 2 ) = 6$, $s t r ( Q 3 ) = 11$, and $s t r ( Q 4 ) = 21$. By considering the following labeling of $Q 5$, we have $s t r ( Q 5 ) = 40$. □
 $Q 3$ 000 100 110 010 001 101 111 011 $Q 2$ 00 00000 10000 11000 01000 00100 10100 11100 01100 1 32 3 31 26 2 30 6 37 10 00010 10010 11010 01010 00110 10110 11110 01110 21 4 29 7 12 27 9 24 39 11 10011 10011 11011 01011 00111 10111 11111 01111 16 19 11 20 17 13 18 15 36 01 00001 10001 11001 01001 00101 10101 11101 01101 22 5 28 8 14 25 10 23 39 38 37 40 39 40 40 40 39 max. induced edge label
Note that $Q 5 ≅ Q 3 × Q 2$. The first row and the first column are vertices of $Q 3$ and $Q 2$, respectively. The following is the corresponding figure.
The following is a numbering for $Q 6 ≅ Q 3 × Q 3$.
 $Q 3$ 000 100 110 010 001 101 111 011 000 000000 100000 110000 010000 001000 101000 111000 011000 1 64 3 63 62 2 52 7 70 100 000100 100100 110100 010100 001100 101100 111100 011100 61 4 45 8 11 57 17 58 75 110 000110 100110 110110 010110 001110 101110 111110 011110 15 49 28 38 56 21 33 19 75 010 000010 100010 110010 010010 001010 101010 111010 011010 60 6 50 10 13 54 25 46 79 001 000001 100001 110001 010001 001001 101001 111001 011001 59 5 37 9 12 53 23 48 76 101 000101 100101 110101 010101 001101 101101 111101 011101 14 44 30 43 55 20 42 18 75 111 000111 100111 110111 010111 001111 101111 111111 011111 39 26 35 31 22 40 32 34 75 011 000011 100011 110011 010011 001011 101011 111011 011011 16 51 27 41 47 24 36 29 78 76 77 78 73 78 78 77 77 max. induced edge label
Thus, $76 ≤ s t r ( Q 6 ) ≤ 79$.

## 4. Conclusions and Open Problems

We obtained a sufficient condition for $s t r ( G ) = | V ( G ) | + δ ( G )$. Many open problems were solved and new results were obtained immediately. A new lower bound of $s t r ( G )$ in terms of $α ( G )$ was obtained. Consequently, the strengths of all 2-regular graphs were determined. An approach for obtaining the lower bound of $s t r ( G )$ in terms of the neighborhood size of all possible subsets of $V ( G )$ was also obtained. This gave us a sharp lower bound of $s t r ( Q n )$ and partially answered Problem 4. The following problems naturally arise.
Problem 5.
Find sufficient and/or necessary conditions such that $s t r ( G ) = 2 p − 2 α ( G ) + 1$ or $s t r ( G ) = p + ξ ( G )$.
Problem 6.
Determine the exact strength of all r-regular graphs for $r ≥ 3$.
Note that for $G = C 2 n + 1 , n ≥ 1$, $s t r ( G ) = 2 n + 3 = | V ( G ) | + δ ( G ) = 2 | V ( G ) | − 2 α ( G ) + 1$.
Problem 7.
Characterize all graphs G of order p with (i) $s t r ( G ) = p + δ ( G ) = 2 p − 2 α ( G ) + 1$, (ii) $s t r ( G ) = p + ξ ( G ) = 2 p − 2 α ( G ) + 1$, or (iii) $s t r ( G ) = p + δ ( G ) = p + ξ ( G )$.
Note that every 2-regular graph $C k$ that has $k ≥ 2$ odd cycles has $s t r ( C k ) = 2 | V ( C k ) | − 2 α ( C k ) + 1 > | V ( C k ) | + δ ( C k )$. Observe that if $C k$ contains an even cycle C, then it contains a component C with $s t r ( C ) = | V ( C ) | + δ ( C )$.
Problem 8.
Prove that if $s t r ( G ) = 2 | V ( G ) | − 2 α ( G ) + 1 > | V ( G ) | + δ ( G )$, then G contains a proper subgraph H with $s t r ( H ) = | V ( H ) | + δ ( H )$.
Problem 9.
Prove that for each graph G, if $s t r ( G ) > | V ( G ) | + δ ( G )$, then $s t r ( G ) = 2 | V ( G ) | − 2 α ( G ) + 1$. Otherwise, either G is a proper subgraph of a graph H with $s t r ( H ) = 2 | V ( H ) | − 2 α ( H ) + 1$ with $α ( H ) ≥ α ( G )$, or else G contains a proper subgraph H with $s t r ( H ) = | V ( H ) | + δ ( H )$.

## Author Contributions

Algorithm development, Z.-B.G.; examples searching and theorems proving, G.-C.L. and W.-C.S.; writing—original draft preparation, G.-C.L. and W.-C.S.; writing—review and editing, G.-C.L. and W.-C.S. All authors have read and agreed to the published version of the manuscript.

## Funding

This research received no external funding.

Not applicable.

Not applicable.

## Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

## Conflicts of Interest

The authors declare no conflict of interest.

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Gao, Z.-B.; Lau, G.-C.; Shiu, W.-C. Graphs with Minimal Strength. Symmetry 2021, 13, 513. https://doi.org/10.3390/sym13030513

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Gao, Zhen-Bin, Gee-Choon Lau, and Wai-Chee Shiu. 2021. "Graphs with Minimal Strength" Symmetry 13, no. 3: 513. https://doi.org/10.3390/sym13030513

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