# Graphs with Minimal Strength

^{1}

^{2}

^{3}

^{4}

^{*}

## Abstract

**:**

## 1. Introduction

**Definition**

**1.**

**Lemma**

**1.**

**Lemma**

**2.**

**Lemma**

**3.**

**Proof.**

**Problem**

**1.**

**Problem**

**2.**

**Problem**

**3.**

**Problem**

**4.**

## 2. Sufficient Condition

**Example**

**1.**

1. | ||||||||

→ | → | → | → | |||||

${\mathcal{G}}_{1}={G}_{1}$ | ${\mathcal{G}}_{2}={G}_{2}$ | ${\mathcal{G}}_{3}={G}_{3}$ | ${\mathcal{G}}_{4}={K}_{1}+{G}_{4}$ | ${\mathcal{G}}_{5}={K}_{1}$ | ||||

${\delta}_{1}=2$ | ${\delta}_{2}=1,{m}_{2}=0$ | ${\delta}_{3}=2,{m}_{3}=0$ | ${\delta}_{4}=1,{m}_{4}=1$ | ${\delta}_{5}=0,{m}_{5}=1$ |

2. | ||||||

→ | → | → | ||||

${\mathcal{G}}_{1}={G}_{1}$ | ${\mathcal{G}}_{2}={K}_{1}+{G}_{2}$ | ${\mathcal{G}}_{3}={K}_{1}+{G}_{3}$ | ${\mathcal{G}}_{4}={K}_{2}$ | |||

${\delta}_{1}=2$ | ${\delta}_{2}=2,{m}_{2}=1$ | ${\delta}_{3}=1,{m}_{3}=1$ | ${\delta}_{4}=1,{m}_{4}=0$ |

**Example**

**2.**

→ | → | $\underrightarrow{\mathit{choose}\mathit{any}\mathit{degree}4\mathit{vertex}}$ | K_{1} | |||

${\mathcal{G}}_{1}=G$ | ${\mathcal{G}}_{2}=3{K}_{1}+{G}_{2}$ | ${\mathcal{G}}_{3}={G}_{3}$ | ${\mathcal{G}}_{4}={K}_{1}$ | |||

${\delta}_{1}=2$ | ${\delta}_{2}=2,{m}_{2}=3$ | ${\delta}_{3}=4,{m}_{3}=0$ | ${\delta}_{4}=0,{m}_{4}=1$ |

→ | → | $\underrightarrow{\mathit{choose}\mathit{any}\mathit{degree}4\mathit{vertex}}$ | K_{1} | |||

${\mathcal{G}}_{1}=G$ | ${\mathcal{G}}_{2}={G}_{2}$ | ${\mathcal{G}}_{3}=3{K}_{3}{G}_{3}$ | ${\mathcal{G}}_{4}={K}_{1}$ | |||

${\delta}_{1}=2$ | ${\delta}_{2}=2,{m}_{2}=0$ | ${\delta}_{3}=4,{m}_{3}=3$ | ${\delta}_{4}=0,{m}_{4}=1$ |

**Theorem**

**1.**

**Proof.**

**Example**

**3.**

**Corollary**

**2.**

**Corollary**

**3.**

**Proof.**

**Corollary**

**4.**

**Theorem**

**5.**

**Example**

**4.**

→ | → | → | ||||

${\mathcal{G}}_{1}=G$ | ${\mathcal{G}}_{2}=3{K}_{1}+{G}_{2}$ | ${\mathcal{G}}_{3}={G}_{3}$ | ${\mathcal{G}}_{4}={K}_{2}$ | |||

${d}_{G}={d}_{1}=\delta \left(G\right)=2$ | ${d}_{2}=3,{m}_{2}=3$ | ${d}_{3}=2,{m}_{3}=0$ | ${d}_{4}=1,{m}_{4}=0$ |

**Lemma**

**4.**

**Proof.**

**Remark**

**1.**

**Theorem**

**6.**

**Proof.**

**Remark**

**2.**

**Theorem**

**7.**

**Proof.**

**Corollary**

**8.**

**Corollary**

**9.**

**Theorem**

**10.**

**Example**

**5.**

**Example**

**6.**

## 3. New Lower Bounds

**Theorem**

**11.**

**Proof.**

**Corollary**

**12.**

**Proof.**

**Theorem**

**13.**

**Proof.**

**Theorem**

**14.**

**Proof.**

**Example**

**7.**

**Lemma**

**5.**

**Proof.**

**Theorem**

**15.**

**Lemma**

**6.**

**Proof.**

**Lemma**

**7.**

**Proof.**

- (1).
- Suppose that only one pair of u, v, and w are adjacent—say, $uv$ 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, $uv$ and $uw$ are edges. Note that v and w cannot be adjacent. Then, $N\left[u\right]\cap N\left[v\right]=\{u,v\}$ and $N\left[u\right]\cap N\left[w\right]=\{u,w\}$. Hence, $u\in N\left[v\right]\cap N\left[w\right]$. This implies that $N\left[u\right]\cap N\left[v\right]\cap N\left[w\right]=\left\{u\right\}$.
- (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 ${i}_{1},{i}_{2},{j}_{1},{j}_{2}$, ${i}_{1}\ne {j}_{1}$, and ${i}_{2}\ne {j}_{2}$. 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, $|\{{i}_{1},{j}_{1}\}\cap \{{i}_{2},{j}_{2}\}|=1$. Without loss of generality, we may assume that ${i}_{1}={i}_{2}$. Now, $N\left[u\right]\cap N\left[v\right]=\{u+{e}_{{i}_{1}},u+{e}_{{j}_{1}}\}$, $N\left[u\right]\cap N\left[w\right]=\{u+{e}_{{j}_{1}},u+{e}_{{j}_{2}}\}$, and $N\left[v\right]\cap N\left[w\right]=\{v+{e}_{{i}_{2}},v+{e}_{{j}_{2}}\}$. Here, $v+{e}_{{i}_{2}}=v+{e}_{{i}_{1}}=u+{e}_{{j}_{1}}$. Hence, $u+{e}_{{j}_{1}}\in N\left[u\right]\cap N\left[v\right]\cap N\left[w\right]$. Since $u+{e}_{{j}_{2}}\notin N\left[u\right]\cap N\left[v\right]$, $N\left[u\right]\cap N\left[v\right]\cap N\left[w\right]=\{u+{e}_{{j}_{1}}\}$.

**Theorem**

**16.**

**Proof.**

- (1).
- If all $\left|N\right[u]\cap N[v\left]\right|$, $\left|N\right[u]\cap N[w\left]\right|$ and $\left|N\right[v]\cap N[w\left]\right|$ are not zero, then by Lemma 7, $\left|N\right[u]\cap N[v]\cap N[w\left]\right|=1$. Thus,$$\begin{array}{cc}\hfill \left|N\right[u]\cup N[v]\cup N[w\left]\right|& =\left|N\right[u\left]\right|+\left|N\right[v\left]\right|+\left|N\right[w\left]\right|\hfill \\ & \phantom{\rule{1.em}{0ex}}-\left|N\right[u]\cap N[v\left]\right|-\left|N\right[u]\cap N[w\left]\right|-\left|N\right[v]\cap N[w\left]\right|+\left|N\right[u]\cap N[v]\cap N[w\left]\right|\hfill \\ & =3(n+1)-3\times 2+1=3n-2.\hfill \end{array}$$Actually, $S=\{\mathbf{0},{e}_{1}+{e}_{2},{e}_{1}+{e}_{3}\}$.
- (2).
- If all $\left|N\right[u]\cap N[v\left]\right|$, $\left|N\right[u]\cap N[w\left]\right|$ and $\left|N\right[v]\cap N[w\left]\right|$ are zero, then $\left|N\right[u]\cup N[v]\cup N[w\left]\right|=3(n+1)$.
- (3).
- If at least one of $\left|N\right[u]\cap N[v\left]\right|$, $\left|N\right[u]\cap N[w\left]\right|$, and $\left|N\right[v]\cap N[w\left]\right|$ is not zero and at least one of them is zero, then$$\begin{array}{cc}\hfill \left|N\right[u]\cup N[v]\cup N[w\left]\right|& =\left|N\right[u\left]\right|+\left|N\right[v\left]\right|+\left|N\right[w\left]\right|\hfill \\ & \phantom{\rule{1.em}{0ex}}-\left|N\right[u]\cap N[v\left]\right|-\left|N\right[u]\cap N[w\left]\right|-\left|N\right[v]\cap N[w\left]\right|+\left|N\right[u]\cap N[v]\cap N[w\left]\right|\hfill \\ & \ge 3(n+1)-2\times 2=3n-1.\hfill \end{array}$$

- (1).
- If only one of $N\left[{u}_{j}\right]\cap N\left[{u}_{l}\right]=\u2300$, then by Lemma 7, the third summand is 1 and the fourth summand is 0. Then, $\left|{\bigcup}_{l=1}^{4}N\left[{u}_{l}\right]\right|\ge 4n+4-5\times 2+1=4n-5$.Actually, $S=\{\mathbf{0},{e}_{1}+{e}_{2},{e}_{1}+{e}_{3},{e}_{1}+{e}_{2}+{e}_{3}+{e}_{4}\}$.
- (2).
- If more than one of $N\left[{u}_{j}\right]\cap N\left[{u}_{l}\right]=\u2300$, then the third and fourth summands are 0. Thus, $\left|{\bigcup}_{l=1}^{4}N\left[{u}_{l}\right]\right|\ge 4n+4-4\times 2=4n-4$.
- (3).
- If all of $N\left[{u}_{j}\right]\cap N\left[{u}_{l}\right]\ne \u2300$, then$$\begin{array}{cc}\hfill \left|{\displaystyle \bigcup _{l=1}^{4}}N\left[{u}_{l}\right]\right|& =4n+4-6\times 2+4-|N\left[{u}_{1}\right]\cap N\left[{u}_{2}\right]\cap N\left[{u}_{3}\right]\cap N\left[{u}_{4}\right]|\hfill \\ & \ge 4n-4-1=4n-5.\hfill \end{array}$$

**Theorem**

**17.**

- 1.
- $str\left({Q}_{2}\right)\ge 6$; $str\left({Q}_{3}\right)\ge 11$; $str\left({Q}_{4}\right)\ge 21$;
- 2.
- $str\left({Q}_{n}\right)\ge {2}^{n}+4n-12$ for $5\le n\le 9$;
- 3.
- $str\left({Q}_{2m}\right)\ge {2}^{2m}+{m}^{2}+4$ for $m\ge 5$.
- 4.
- $str\left({Q}_{2m-1}\right)\ge {2}^{2m-1}+{m}^{2}-m+4$ for $m\ge 6$.

**Corollary**

**18.**

**Proof.**

${\mathit{Q}}_{\mathbf{3}}$ | 000 | 100 | 110 | 010 | 001 | 101 | 111 | 011 | ||

${\mathit{Q}}_{\mathbf{2}}$ | ||||||||||

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 |

${\mathit{Q}}_{\mathbf{3}}$ | 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 |

## 4. Conclusions and Open Problems

**Problem**

**5.**

**Problem**

**6.**

**Problem**

**7.**

**Problem**

**8.**

**Problem**

**9.**

## Author Contributions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

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**MDPI and ACS Style**

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

**AMA Style**

Gao Z-B, Lau G-C, Shiu W-C. Graphs with Minimal Strength. *Symmetry*. 2021; 13(3):513.
https://doi.org/10.3390/sym13030513

**Chicago/Turabian Style**

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