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/

Zendy Plus

Presents the access of premium content as premium feature

Premium Content

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

Zendy Tools

Zendy Open

The Hamilton-Waterloo problem asks for a decomposition of the complete graphinto $r$ copies of a 2-factor $F_{1}$ and $s$ copies of a 2-factor $F_{2}$ suchthat $r+s=\left\lfloor\frac{v-1}{2}\right\rfloor$. If $F_{1}$ consists of$m$-cycles and $F_{2}$ consists of $n$ cycles, then we call such adecomposition a $(m,n)-$HWP$(v;r,s)$. The goal is to find a decomposition forevery possible pair $(r,s)$. In this paper, we show that for odd $x$ and $y$,there is a $(2^kx,y)-$HWP$(vm;r,s)$ if $\gcd(x,y)\geq 3$, $m\geq 3$, and both$x$ and $y$ divide $v$, except possibly when $1\in\{r,s\}$.

On the Hamilton-Waterloo problem: the case of two cycles sizes of different parity