English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Tools annual

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Tools monthly

Global: Zendy Plus monthly

Presents the access of premium content as premium feature

Premium Content

Global: Zendy Plus annual

Global: Zendy Free Plan

It is known that for any graph $G,$ $\gamma (G\square P_2)\geq \gamma (G)$where $\gamma$ stands for the domination number, $\square$ for the cartesianproduct and $P_2$ is the path graph on two vertices. In an attempt to proveVizing's conjecture, Clark and Suen proved in $2000$ that $\gamma (X\squareY)\geq \frac{1}{2}\gamma (X)\gamma (Y)$ for any pair of graphs $X$ and $Y.$Combining these two inequalities, we have $\gamma (X\square Y\square P_2)\geq\frac{1}{2}\gamma (X)\gamma (Y).$ In this paper, we use space projections toimprove this lower bound and show that $\gamma (X\square Y\square P_2)\geq\frac{2}{3}\gamma (X)\gamma (Y)$ for any pair of graphs $X$ and $Y.$ Inaddition, we prove that $\gamma (X\square Y\square P_{n})\geq c_n\gamma(X)\gamma (Y)\gamma (P_{n}),$ where $c_n$ is almost $\frac{3}{4}$ when $n$ isbig enough.

On the domination number of the cartesian product of the path graph and any pair of graphs