# β-Differential of a Graph

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^{2}

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## Abstract

**:**

## 1. Introduction

## 2. The Function ${\mathit{f}}_{\mathit{G}}(\mathbf{\beta})={\mathbf{\partial}}_{\mathbf{\beta}}(\mathit{G})$

**Proposition**

**1.**

**Proof**

**1.**

**Proposition**

**2.**

**Proof**

**2.**

**Lemma**

**1.**

**Proof**

**3.**

**Proposition**

**3.**

**Proof**

**4.**

**Lemma**

**2.**

**Proof**

**5.**

**Proposition**

**4.**

**Proof**

**6.**

**Lemma**

**3.**

**Proof**

**7.**

**Lemma**

**4.**

**Proof**

**8.**

**Proposition**

**5.**

**Proof**

**9.**

**Theorem**

**1.**

**Proof**

**10.**

## 3. Bounds on the $\mathbf{\beta}$-Differential of a Graph

**Proposition**

**6.**

**Proof**

**11.**

**Proposition**

**7.**

- (a)
- ${\partial}_{\beta}(G)=n-(1+\beta )$ if and only if $\Delta =n-1$.
- (b)
- ${\partial}_{\beta}(G)=n-(2+\beta )$ if and only if $\Delta =n-2$.
- (c)
- If $\beta >1$, then ${\partial}_{\beta}(G)=n-(3+\beta )$ if and only if $\Delta =n-3$.

**Proof**

**12.**

- (a)
- If $\Delta =n-1$, by Proposition 6 we have $n-1-\beta \le {\partial}_{\beta}(G)\le n-1-\beta $, then ${\partial}_{\beta}(G)=n-1-\beta $. If ${\partial}_{\beta}(G)=n-1-\beta $ and D is a $\beta $-differential set, then we have $n-1-\beta =|B(D)|-\beta |D|\le n-1-\beta |D|$. Therefore, $|D|\le 1$, that is, $|D|=1$ and $|B(D)|=n-1$, which means that $\Delta =n-1$.
- (b)
- If $\Delta =n-2$, by Proposition 6 and (a) we have $n-2-\beta \le {\partial}_{\beta}(G)<n-1-\beta $. If D is a $\beta $-differential set such that $|D|\ge 2$, then $n-2-\beta \le |B(D)|-\beta |D|\le |B(D)|-2\beta $, consequently, $n-2-\beta \le |B(D)|$. Since $\beta >0$, we have $n-1\le |B(D)|$, which is a contradiction. If D is a $\beta $-differential set such that $|D|=1$, by $n-2-\beta \le |B(D)|-\beta |D|$ and (a) we obtain $|B(D)|=n-2$, so ${\partial}_{\beta}(G)=n-(2+\beta )$. Now, if ${\partial}_{\beta}(G)=n-2-\beta $, there exists a $\beta $-differential set D such that $n-2-\beta =|B(D)|-\beta |D|\le |B(D)|-\beta $, therefore $|B(D)|\ge n-2$. By (a) we know that $|B(D)|\ne n-1$, then $|B(D)|=n-2$ and, using again that $|B(D)|-\beta |D|=n-2-\beta $, we conclude that $|D|=1$, which means that $\Delta =n-2$.
- (c)
- If ${\partial}_{\beta}(G)=n-(3+\beta )$, there exists a $\beta $-differential set D such that $n-(3+\beta )=|B(D)|-\beta |D|\le |B(D)|-\beta $, then $|B(D)|\ge n-3$. By (a) we know that $|B(D)|\ne n-1$. If $|B(D)|=n-2$, then we have $|D|\le 2$ and $|D|=1+\frac{1}{\beta}$, which is a contradiction with the fact that $\beta >1$. Therefore, $|B(D)|=n-3$ and, consequently, $|D|=1$, which means that $\Delta =n-3$. Finally, if $\Delta =n-3$ and D is a $\beta $-differential set, by Proposition 6 we have $n-3-\beta \le |B(D)|-\beta |D|\le |B(D)|-\beta $, then $|B(D)|=n-2$ or $|B(D)|=n-3$. If $|B(D)|=n-2$, since $\Delta =n-3$, we have $|D|=2$ and $-1-\beta \le -\beta |D|=-2\beta $, a contradiction. If $|B(D)|=n-3$, since $n-3-\beta \le |B(D)|-\beta |D|=n-3-\beta |D|$, we have $|D|=1$ and ${\partial}_{\beta}(G)=n-3-\beta $.

**Proposition**

**8.**

- (a)
- $\Delta (G[V\setminus N[v]])\le \beta $.
- (b)
- ${\delta}_{\overline{N[v]}}(u)\le \beta +1$ for every $u\in N(v)$.
- (c)
- $|{N}_{\overline{N[v]}}(A)|+|{N}_{N(v)\setminus A}(A)|\le \Delta -1+\beta (|A|-1)$ for every $A\subseteq N(v)$.

**Proof**

**13.**

**Proposition**

**9.**

**Proof**

**14.**

**Lemma**

**5.**

**Proof**

**15.**

**Lemma**

**6**

**Theorem**

**2.**

**Proof**

**16.**

**Proposition**

**10.**

- (a)
- if $\beta \in (0,\delta -1)$, then ${\partial}_{\beta}(G)\ge {\rho}^{o}(G)(\delta -\beta -1)$,
- (b)
- if $\beta \in (0,\delta )$, then ${\partial}_{\beta}(G)\ge \rho (G)(\delta -\beta )$.

**Proof**

**17.**

**Proposition**

**11.**

**Proof**

**18.**

**Proposition**

**12.**

- ${\partial}_{\beta}({K}_{n})={\partial}_{\beta}({W}_{n})=n-1-\beta .$
- ${\partial}_{\beta}({P}_{n})={\partial}_{\beta}({C}_{n})=\left\{\begin{array}{cc}\u230a\frac{n}{3}\u230b(2-\beta )+1-\beta & \phantom{\rule{1.em}{0ex}}if\phantom{\rule{1.em}{0ex}}\beta \in (0,1)\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}and\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}n\equiv 2\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}(\mathrm{mod}\phantom{\rule{0.166667em}{0ex}}3)\hfill \\ \u230a\frac{n}{3}\u230b(2-\beta )& \phantom{\rule{1.em}{0ex}}otherwise.\hfill \end{array}\right.$

- ${\partial}_{\beta}({K}_{n,m})=\left\{\begin{array}{cc}m+n-2(1+\beta )& \phantom{\rule{1.em}{0ex}}if\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}0<\beta <n-2\hfill \\ m-\beta & \phantom{\rule{1.em}{0ex}}if\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\beta \ge n-2.\hfill \end{array}\right.$
- ${\partial}_{\beta}({S}_{n,m})=\left\{\begin{array}{cc}m+n-2\beta & \phantom{\rule{1.em}{0ex}}if\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}0<\beta <n-1\hfill \\ m+1-\beta & \phantom{\rule{1.em}{0ex}}if\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\phantom{\rule{0.166667em}{0ex}}\beta \ge n-1.\hfill \end{array}\right.$

**Proof**

**19.**

**Lemma**

**7.**

**Proof**

**20.**

**Proposition**

**13.**

**Proof**

**21.**

**Theorem**

**3.**

- (i)
- If $\beta \in (0,1]$, then $\frac{(\Delta -\beta )\partial (G)}{\Delta -1}\le {\partial}_{\beta}(G).$
- (ii)
- If $\beta \in (1,\Delta )$, then $\frac{\left(\lfloor \beta \rfloor -\beta +1\right)\partial (G)}{\lfloor \beta \rfloor}\le {\partial}_{\beta}(G).$

**Proof**

**22.**

**Theorem**

**4.**

**Proof**

**23.**

**Theorem**

**5.**

**Proof**

**24.**

- (1)
- $|D|\le {\displaystyle \frac{{\partial}_{\beta}(G)}{\lfloor \beta \rfloor -\beta +1}}$.
- (2)
- If $v\in B(D)$, then ${\delta}_{C(D)}(v)\le \lfloor \beta \rfloor +1$.
- (3)
- If $v\in C(D)$, then ${\delta}_{C(D)}(v)\le \lfloor \beta \rfloor $.

**Lemma**

**8.**

**Proof**

**25.**

**Theorem**

**6.**

**Proof**

**26.**

## Acknowledgments

## Author Contributions

## Conflicts of Interest

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**Figure 4.**On the left $|{D}_{1}^{m}|=1$ and $|{D}_{1}^{M}|=2$, and on the right $|{D}_{\frac{1}{2}}^{m}|=1$ and $|{D}_{\frac{1}{2}}^{M}|=3$.

**Figure 5.**$\beta $-differential set when $\beta \in \left(0,\frac{1}{2}\right]$ (on the left) and $\beta $-differential set when $\beta \in \left(\frac{1}{2},3\right]$ (on the right).

**Figure 6.**An example of a graph ${G}_{r}$ such that ${\partial}_{\beta}({G}_{r})=1+2r-\beta r$ if $\beta \le \frac{1}{r-1}$, and ${\partial}_{\beta}({G}_{r})=2r-\beta $ if $\beta >\frac{1}{r-1}$.

**Figure 8.**This graph show that the bound (ii) in Theorem 3 is attained when $\beta \in \mathbb{N}$ and $r=s=1+\beta $.

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**MDPI and ACS Style**

Basilio, L.A.; Bermudo, S.; Leaños, J.; Sigarreta, J.M.
*β*-Differential of a Graph. *Symmetry* **2017**, *9*, 205.
https://doi.org/10.3390/sym9100205

**AMA Style**

Basilio LA, Bermudo S, Leaños J, Sigarreta JM.
*β*-Differential of a Graph. *Symmetry*. 2017; 9(10):205.
https://doi.org/10.3390/sym9100205

**Chicago/Turabian Style**

Basilio, Ludwin A., Sergio Bermudo, Jesús Leaños, and José M. Sigarreta.
2017. "*β*-Differential of a Graph" *Symmetry* 9, no. 10: 205.
https://doi.org/10.3390/sym9100205