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

One way to interpret the classical secretary problem (CSP) is to consider itas a special case of the following problem. We observe $n$ independentindicator variables $I_1,I_2,\dotsc,I_n$ sequentially and we try to stop on thelast variable being equal to 1. If $I_k=1$ it means that the $k$-th observedsecretary has smaller rank than all previous ones (and therefore is a bettersecretary). In the CSP $p_k=E(I_k)=1/k$ and the last $k$ with $I_k=1$ standsfor the best candidate. The more general problem of stopping on a last "1" wasstudied by Bruss(2000). In what we will call Weber's problem the variables$I_k$ can take more than two values and we try to stop on the last occurence of\textit{one} of these values. Notice that we do not know in advance the valuetaken by the variable on which we stop.  We can solve this problem in some cases and provide algorithms to compute theoptimal stopping rule. These cases carry enough generality to be applicable inconcrete situations.

Weber's optimal stopping problem and generalizations