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 Plus annual

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

Global: Zendy Plus monthly

Global: Zendy Tools annual

Global: Zendy Tools monthly

Global: Zendy Free Plan

Suppose $G$ is a connected simple graph with the vertex set $V( G ) = \{v_1,v_2,\cdots ,v_n \} $. Let $d_G( v_i,v_j ) $ be the least distance between$v_i$ and $v_j$ in $G$. Then the distance matrix of $G$ is $D( G ) =( d_{ij} )_{n\times n}$, where $d_{ij}=d_G( v_i,v_j ) $. Since $D( G )$ is a non-negativereal symmetric matrix, its eigenvalues can be arranged as $\lambda_1(G)\ge\lambda_2(G)\ge \cdots \ge \lambda_n(G)$, where eigenvalue $\lambda_n(G)$ iscalled the least distance eigenvalue of $G$. In this paper we determine theunique graph whose least distance eigenvalue attains maximum among allcomplements of graphs of diameter greater than three.

The least distance eigenvalue of the complements of graphs of diameter greater than three