# On (2-d)-Kernels in the Tensor Product of Graphs

## Abstract

**:**

## 1. Introduction and Preliminary Results

- We partially solve the problem of the existence of $(2$-$d)$-kernels in the tensor product, in particular giving the complete solutions for cases where factors are paths, cycles or complete graphs.
- We obtain new interpretations of the Padovan and Perrin numbers related to $(2$-$d)$-kernels in the tensor product of basic classes of graphs, i.e., paths, cycles and complete graphs.

## 2. $\mathbf{(}\mathbf{2}$-$\mathit{d}\mathbf{)}$-Kernels in the Tensor Product of Graphs

**Theorem**

**1**

**.**Let G and H be connected non-trivial graphs. If at least one of them has an odd cycle, then $G\times H$ is connected. If both G and H are bipartite, then $G\times H$ has exactly two connected components.

**Theorem**

**3**

**.**The graph ${P}_{n}$ has a $(2$-$d)$-kernel J if and only if $n=2p+1$, $p\ge 1$, and this kernel is unique. If it holds, then $\left|J\right|=p+1$.

**Theorem**

**4**

**.**Let $G=G({V}_{1},{V}_{2})$ be a bipartite graph. If for each two vertices $x,y\in L\left(G\right)$, ${d}_{G}(x,y)\equiv 0\left(\mathrm{mod}\phantom{\rule{4pt}{0ex}}2\right)$, then G has a $(2$-$d)$-kernel.

**Corollary**

**1.**

**Theorem**

**5.**

**Proof.**

**Corollary**

**2.**

- (i)
- $\sigma ({P}_{n}\times {P}_{m})=\left\{\begin{array}{cc}1\hfill & ifmisevenandnisodd,\hfill \\ 2\hfill & ifbothnandmareodd,\hfill \end{array}\right.$
- (ii)
- $j({P}_{n}\times {P}_{m})=\left\{\begin{array}{cc}m\xb7\u2308\frac{n}{2}\u2309\hfill & ifmisevenandnisodd,\hfill \\ min\left\{m\xb7\u2308\frac{n}{2}\u2309,n\xb7\u2308\frac{m}{2}\u2309\right\}\hfill & ifbothnandmareodd,\hfill \end{array}\right.$
- (iii)
- $J({P}_{n}\times {P}_{m})=\left\{\begin{array}{cc}m\xb7\u2308\frac{n}{2}\u2309\hfill & ifmisevenandnisodd,\hfill \\ max\left\{m\xb7\u2308\frac{n}{2}\u2309,n\xb7\u2308\frac{m}{2}\u2309\right\}\hfill & ifbothnandmareodd.\hfill \end{array}\right.$

**Proof.**

- $({x}_{1},{y}_{2})\notin {J}_{2}$.Then, $({x}_{2},{y}_{2k-1})\in {J}_{2}$, $k=1,2,\dots ,\u2308\frac{m}{2}\u2309$. Otherwise, $({x}_{1},{y}_{2i})$, $i=1,2,\dots ,\u230a\frac{m}{2}\u230b$ are not two-dominated by ${J}_{2}$. This means that $({x}_{3},{y}_{2i})\notin {J}_{2}$, $i=1,2,\dots ,\u230a\frac{m}{2}\u230b$. Then, by analogous reasoning, we obtain ${J}_{2}=\left\{({x}_{2a},{y}_{2b-1});a=1,2,\dots ,\u230a\frac{n}{2}\u230b,b=1,2,\dots ,\u2308\frac{m}{2}\u2309\right\}$ is the unique $(2$-$d)$-kernel of the graph ${G}_{2}$ not containing the vertex $({x}_{1},{y}_{2})$.
- $({x}_{1},{y}_{2})\in {J}_{2}$.Proving analogously as in case 1, we obtain that ${J}_{2}=\left\{({x}_{2a-1},{y}_{2b});a=1,2,\dots ,\u2308\frac{n}{2}\u2309,b=1,2,\dots ,\u230a\frac{m}{2}\u230b\right\}$ is the unique $(2$-$d)$-kernel of the graph ${G}_{2}$ containing the vertex $({x}_{1},{y}_{2})$.

**Theorem**

**6.**

- (i)
- ${P}_{n}\times {C}_{m}$, for $n\ge 1$, $m\ge 3$,
- (ii)
- ${C}_{n}\times {C}_{m}$, for $n,m\ge 3$,
- (iii)
- ${P}_{n}\times {K}_{m}$, for $n\ge 1$, $m\ge 3$,
- (iv)
- ${C}_{n}\times {K}_{m}$, for $n,m\ge 3$,
- (v)
- ${K}_{n}\times {K}_{m}$, for $n\ge 1$, $m\ge 3$,

**Proof.**

- (i)
- Let $n\ge 1$, $m\ge 3$ be integers. We will show that the graph ${P}_{n}\times {C}_{m}$ has a $(2$-$d)$-kernel. Suppose that $V\left({P}_{n}\right)=\{{x}_{1},{x}_{2},\dots ,{x}_{n}\}$, $V\left({C}_{m}\right)=\{{y}_{1},{y}_{2},\dots ,{y}_{m}\}$ with the numbering of vertices in the natural order. If $n=1$, then the graph ${P}_{1}\times {C}_{m}$ is an edgeless graph, so $V({P}_{1}\times {C}_{m})$ is a $(2$-$d)$-kernel. Let $n=2$. If m is odd, then ${P}_{2}\times {C}_{m}\cong {C}_{2m}$, so it has a $(2$-$d)$-kernel. If m is even, then ${P}_{2}\times {C}_{m}={C}_{m}\cup {C}_{m}$. Thus, ${P}_{2}\times {C}_{m}$, $m\ge 3$ has a $(2$-$d)$-kernel. Let $n\ge 3$. Suppose that ${V}_{i}=\left\{{x}_{i}\right\}\times V\left({C}_{m}\right)$, $i=1,2,\dots ,n$. We will show that the set $J={\bigcup}_{j=1}^{\u2308\frac{n}{2}\u2309}{V}_{2j-1}$ is a $(2$-$d)$-kernel of the graph ${P}_{n}\times {C}_{m}$. By the definition of the tensor product, it follows that ${V}_{i}$ is an independent set, for $i=1,2,\dots ,n$. Moreover, the factor ${P}_{n}$ ensures that J is independent. We will show that J is two-dominating. Let $({x}_{a},{y}_{b})\in V({P}_{n}\times {C}_{m})\backslash J$, $a=1,2,\dots ,2j$, $j=1,2,\dots ,\u230a\frac{n}{2}\u230b$, $b=1,2,\dots ,m$. Then, $({x}_{a},{y}_{b})\in {\bigcup}_{j=1}^{\u230a\frac{n}{2}\u230b}{V}_{2j}$. Since for all $b=1,2,\dots ,m$, it holds that ${deg}_{{C}_{m}}\left({y}_{b}\right)=2$, and each vertex from ${V}_{2j}$, $j=1,2,\dots ,\u230a\frac{n}{2}\u230b$ is two-dominated by ${V}_{2j-1}\subseteq J$. This means that J is a $(2$-$d)$-kernel of the graph ${P}_{n}\times {C}_{m}$.
- (ii)
- Let $n,m\ge 3$. Proving analogously as in Case (i), we can show that $J={\bigcup}_{i=1}^{\u230a\frac{n}{2}\u230b}{V}_{2i-1}$ is a $(2$-$d)$-kernel of the graph ${C}_{n}\times {C}_{m}$.
- (iii)
- Let $n\ge 1,m\ge 3$. Proving analogously as in Case (i), we can show that $J={\bigcup}_{i=1}^{\u2308\frac{n}{2}\u2309}{V}_{2i-1}$ is a $(2$-$d)$-kernel of the graph ${P}_{n}\times {K}_{m}$.
- (iv)
- Let $n,m\ge 3$. Proving analogously as in Case (i), we can show that $J=V\left({C}_{n}\right)\times \left\{{y}_{1}\right\}$ is a $(2$-$d)$-kernel of the graph ${C}_{n}\times {K}_{m}$.
- (v)
- Let $n\ge 1,m\ge 3$. Proving analogously as in Case (i), we can show that $J=V\left({K}_{n}\right)\times \left\{{y}_{1}\right\}$ is a $(2$-$d)$-kernel of the graph ${K}_{n}\times {K}_{m}$.

**Theorem**

**7.**

- (i)
- $\sigma ({P}_{n}\times {C}_{m})=\left\{\begin{array}{cc}{S}_{1}(n+1)\hfill & ifmisodd,\hfill \\ {\left({S}_{1}(n+1)\right)}^{2}\hfill & ifmiseven,\hfill \end{array}\right.$
- (ii)
- $j({P}_{n}\times {C}_{m})=m\xb7\u2308\frac{n}{3}\u2309$,
- (iii)
- $J({P}_{n}\times {C}_{m})=m\xb7\u2308\frac{n}{2}\u2309$.

**Proof.**

- m is odd.Then, the graph ${P}_{n}\times {C}_{m}$ is connected for each $n\ge 2$. If $n=1,2$, then $\sigma ({P}_{1}\times {C}_{m})=1={S}_{1}\left(2\right)$ and $\sigma ({P}_{2}\times {C}_{m})=2={S}_{1}\left(3\right)$, respectively. If $n=3$, then the family $\mathcal{J}({P}_{3}\times {C}_{m})$ of all $(2$-$d)$-kernels has the form $\mathcal{J}({P}_{3}\times {C}_{m})=\{{V}_{1}\cup {V}_{3},{V}_{2}\}$. This means that $\sigma ({P}_{3}\times {C}_{m})=2={S}_{1}\left(4\right)$. Let $n\ge 4$. We will show that $\sigma ({P}_{n}\times {C}_{m})={S}_{1}(n+1)$. Let $J\subset V({P}_{n}\times {C}_{m})$ be a $(2$-$d)$-kernel of the graph ${P}_{n}\times {C}_{m}$. We have two possibilities.
- 1.1
- $({x}_{n},{y}_{1})\in J$.Then, $({x}_{n-1},{y}_{2})\notin J$. Since ${deg}_{{P}_{n}\times {C}_{m}}\left(({x}_{n},{y}_{3})\right)=2$ and vertices $({x}_{n-1},{y}_{2})$, $({x}_{n},{y}_{3})$ are adjacent, the vertex $({x}_{n},{y}_{3})$ has to belong to a $(2$-$d)$-kernel J. Otherwise, $({x}_{n},{y}_{3})$ is not two-dominated by J. Since m is odd, by analogous reasoning, we obtain $({x}_{n},{y}_{k})\in J$, $k=1,2,\dots ,m$. Thus, ${V}_{n}\subset J$. Then, ${V}_{n-1}\cap J=\varnothing $, and moreover, every vertex from ${V}_{n-1}$ is two-dominated by J. Hence, $J={J}^{*}\cup {V}_{n}$, where ${J}^{*}$ is any $(2$-$d)$-kernel of the graph ${P}_{n-2}\times {C}_{m}$. Thus, ${\sigma}_{({x}_{n},{y}_{1})}({P}_{n}\times {C}_{m})=\sigma ({P}_{n-2}\times {C}_{m})$.
- 1.2
- $({x}_{n},{y}_{1})\notin J$.Then, $({x}_{n-1},{y}_{2}),({x}_{n-1},{y}_{m})\in J$. Proving analogously as in Case 1.1, we obtain that ${V}_{n-1}\subset J$ and ${V}_{n}\cap J=\varnothing ={V}_{n-2}\cap J$. Then, every vertex from ${V}_{n-2}\cup {V}_{n}$ is two-dominated by J. Hence, $J={J}^{**}\cup {V}_{n-1}$, where ${J}^{**}$ is any $(2$-$d)$-kernel of the graph ${P}_{n-3}\times {C}_{m}$. Thus, ${\sigma}_{-({x}_{n},{y}_{1})}({P}_{n}\times {C}_{m})=\sigma ({P}_{n-3}\times {C}_{m})$.

Consequently, $\sigma ({P}_{n}\times {C}_{m})=\sigma ({P}_{n-2}\times {C}_{m})+\sigma ({P}_{n-3}\times {C}_{m})$, $n\ge 4$. By the initial conditions, it follows that $\sigma ({P}_{n}\times {C}_{m})={S}_{1}(n+1)$, for $n\ge 1$. - m is even.Then, by Theorem 1, it follows that the graph ${P}_{n}\times {C}_{m}$, $n\ge 2$ is disconnected and contains two isomorphic connected components ${G}_{1}$, ${G}_{2}$. Thus, $\sigma ({P}_{n}\times {C}_{m})=\sigma \left({G}_{1}\right)\sigma \left({G}_{2}\right)={\left(\sigma \left({G}_{1}\right)\right)}^{2}$. Let ${V}_{i}^{1}=\left\{{x}_{i}\right\}\times \{{y}_{2t-1};t=1,2,\dots ,\frac{m}{2}\}$, ${V}_{i}^{2}=\left\{{x}_{i}\right\}\times \{{y}_{2t};t=1,2,\dots ,\frac{m}{2}\}$, for $i=1,2,\dots ,n$. Without loss of generality, assume that $({x}_{1},{y}_{1})\in V\left({G}_{1}\right)$. If $n=1,2$, then $\sigma ({P}_{1}\times {C}_{m})=1={\left({S}_{1}\left(2\right)\right)}^{2}$ and $\sigma ({P}_{2}\times {C}_{m})=4={\left({S}_{1}\left(3\right)\right)}^{2}$, respectively. If $n=3$, then the family $J({P}_{3}\times {C}_{m})$ of all $(2$-$d)$-kernels has the form $J({P}_{3}\times {C}_{m})=\{{V}_{1}^{1}\cup {V}_{3}^{1},{V}_{2}^{2}\}$. This means that $\sigma ({P}_{3}\times {C}_{m})={2}^{2}={\left({S}_{1}\left(4\right)\right)}^{2}$. Let $n\ge 4$. Proving analogously as in Case 1, we obtain that $\sigma \left({G}_{1}\right)={S}_{1}(n+1)$. Hence, $\sigma ({P}_{n}\times {C}_{m})={\left({S}_{1}(n+1)\right)}^{2}$, for $n\ge 1$.

**Theorem**

**8.**

- (i)
- $\sigma ({P}_{n}\times {K}_{m})={S}_{1}(n+1)$,
- (ii)
- $j({P}_{n}\times {K}_{m})=m\xb7\u2308\frac{n}{3}\u2309$,
- (iii)
- $J({P}_{n}\times {K}_{m})=m\xb7\u2308\frac{n}{2}\u2309$.

**Lemma**

**1.**

**Proof**(induction on n)

**.**

**Theorem**

**9.**

- (i)
- $\sigma ({C}_{n}\times {C}_{3})=3+{S}_{2}\left(n\right),$
- (ii)
- $j({C}_{n}\times {C}_{3})=n,$
- (iii)
- $J({C}_{n}\times {C}_{3})=3\xb7\u230a\frac{n}{2}\u230b$.

**Proof.**

- $({x}_{n},{y}_{1})\in J$.Then, $({x}_{n-1},{y}_{2}),({x}_{n-1},{y}_{3}),({x}_{1},{y}_{2}),({x}_{1},{y}_{3})\notin J$. Consider the vertex $({x}_{n},{y}_{2})$. If $({x}_{n},{y}_{2})\in J$, then $({x}_{n},{y}_{3})\in J$. Otherwise, $({x}_{n},{y}_{3})$ cannot be two-dominated by J. This means that ${X}_{n}\subset J$ and ${X}_{n-1}\cap J=\varnothing ={X}_{1}\cap J$. Moreover, every vertex from ${X}_{1}\cup {X}_{n-1}$ is two-dominated by J. Hence, $J={J}^{*}\cup {X}_{n}$, where ${J}^{*}$ is any $(2$-$d)$-kernel of the graph ${P}_{n-3}\times {C}_{3}$. Thus, $\sigma ({P}_{n-3}\times {C}_{3})$ is the number of $(2$-$d)$-kernels J containing vertices $({x}_{n},{y}_{1})$ and $({x}_{n},{y}_{2})$. If $({x}_{n},{y}_{2})\notin J$, then $({x}_{n-1},{y}_{1}),({x}_{1},{y}_{1})\in J$. Otherwise, $({x}_{n},{y}_{2})$ is not two-dominated by J. Then, $({x}_{n},{y}_{3})\notin J$, and moreover, $\left({x}_{n-2}{y}_{k}\right),({x}_{2},{y}_{k})\notin J$, $k=2,3$. By analogous reasoning, we obtain ${Y}_{1}\subset J$ and ${Y}_{2}\cap J=\varnothing ={Y}_{3}\cap J$. This means that ${Y}_{1}$ is the unique $(2$-$d)$-kernel of the graph ${C}_{n}\times {C}_{3}$ containing the vertex $({x}_{n},{y}_{1})$ and not containing the vertex $({x}_{n},{y}_{2})$. Hence, ${\sigma}_{({x}_{n},{y}_{1})}({C}_{n}\times {C}_{3})=\sigma ({P}_{n-3}\times {C}_{3})+1$.
- $({x}_{n},{y}_{1})\notin J$.Consider four cases.
- 2.1
- $({x}_{n-1},{y}_{1}),({x}_{1},{y}_{1})\notin J$.It is obvious that ${Y}_{2}$, ${Y}_{3}$ are the only $(2$-$d)$-kernels of the graph ${C}_{n}\times {C}_{3}$.
- 2.2
- $({x}_{n-1},{y}_{1})\in J$ and $({x}_{1},{y}_{1})\notin J$.We will show that $({x}_{n},{y}_{1})$ is two-dominated by $({x}_{n-1},{y}_{2}),({x}_{n-1},{y}_{3})\in J$. Otherwise, at least one of the vertices $({x}_{1},{y}_{2})$, $({x}_{1},{y}_{3})$ belongs to a $(2$-$d)$-kernel J. If $({x}_{1},{y}_{2})\in J$, then $({x}_{1},{y}_{3})$ cannot be two-dominated by J. If $({x}_{1},{y}_{2})\in J$ or $({x}_{1},{y}_{3})\in J$, then $({x}_{1},{y}_{1})$ is not dominated by J. This means that ${X}_{n}\cap J=\varnothing ={X}_{1}\cap J$ and ${X}_{n-1}\subset J$. Then, ${X}_{2}\subset J$, otherwise vertices from ${X}_{1}$ are not two-dominated by J. Moreover, ${X}_{n-2}\cap J=\varnothing ={X}_{3}\cap J$, and the vertices from ${X}_{1}\cup {X}_{3}\cup {X}_{n-2}\cup {X}_{n}$ are two-dominated by J. Hence, $J={J}^{**}\cup {X}_{2}\cup {X}_{n-1}$, where ${J}^{**}$ is any $(2$-$d)$-kernel of the graph ${P}_{n-6}\times {C}_{3}$. This means that $\sigma ({P}_{n-6}\times {C}_{3})$ is the number of $(2$-$d)$-kernels J containing $({x}_{n-1},{y}_{1})$ and not containing $({x}_{1},{y}_{1})$, $({x}_{n},{y}_{1})$.
- 2.3
- $({x}_{n-1},{y}_{1})\notin J$ and $({x}_{1},{y}_{1})\in J$.Proving analogously as in Subcase 2.2, we obtain that $\sigma ({P}_{n-6}\times {C}_{3})$ is the number of $(2$-$d)$-kernels J containing $({x}_{1},{y}_{1})$ and not containing $({x}_{n-1},{y}_{1})$, $({x}_{n},{y}_{1})$.
- 2.4
- $({x}_{n-1},{y}_{1}),({x}_{1},{y}_{1})\in J$.Then, ${X}_{n}\cap J=\varnothing $, ${X}_{1}\subset J$ and ${X}_{n-1}\subset J$. Otherwise, J is not two-dominating. This means that ${X}_{2}\cap J=\varnothing ={X}_{n-2}\cap J$. Moreover, every vertex from ${X}_{2}\cup {X}_{n-2}\cup {X}_{n}$ is two-dominated by J. Thus, $J=\tilde{J}\cup {X}_{1}\cup {X}_{n-1}$, where $\tilde{J}$ is any $(2$-$d)$-kernel of the graph ${P}_{n-5}\times {C}_{3}$. This means that $\sigma ({P}_{n-5}\times {C}_{3})$ is the number of $(2$-$d)$-kernels J containing $({x}_{n-1},{y}_{1})$, $({x}_{1},{y}_{1})$ and not containing $({x}_{n},{y}_{1})$.

Hence, ${\sigma}_{-({x}_{n},{y}_{1})}({C}_{n}\times {C}_{3})=2+2\sigma ({P}_{n-6}\times {C}_{3})+\sigma ({P}_{n-5}\times {C}_{3})$.

**Theorem**

**10.**

- (i)
- $\sigma ({C}_{n}\times {K}_{m})={S}_{2}\left(n\right)+m,$
- (ii)
- $j({C}_{n}\times {K}_{m})=n$,
- (iii)
- $J({C}_{n}\times {K}_{m})=m\xb7\u230a\frac{n}{2}\u230b$.

**Theorem**

**11.**

- (i)
- $\sigma ({K}_{n}\times {K}_{m})=n+m,$
- (ii)
- $j({K}_{n}\times {K}_{m})=min\{n,m\},$
- (iii)
- $J({K}_{n}\times {K}_{m})=max\{n,m\}.$

**Proof.**

## 3. Concluding Remarks

## Funding

## Conflicts of Interest

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Bednarz, P.
On (2-*d*)-Kernels in the Tensor Product of Graphs. *Symmetry* **2021**, *13*, 230.
https://doi.org/10.3390/sym13020230

**AMA Style**

Bednarz P.
On (2-*d*)-Kernels in the Tensor Product of Graphs. *Symmetry*. 2021; 13(2):230.
https://doi.org/10.3390/sym13020230

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Bednarz, Paweł.
2021. "On (2-*d*)-Kernels in the Tensor Product of Graphs" *Symmetry* 13, no. 2: 230.
https://doi.org/10.3390/sym13020230