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

We give two combinatorial proofs of an elegant product formula for the numberof spanning trees of the $n$-dimensional hypercube. The first proof is based onthe assertion that if one chooses a uniformly random rooted spanning tree ofthe hypercube and orient each edge from parent to child, then the paralleledges of the hypercube get orientations which are independent of one another.This independence property actually holds in a more general context and hasintriguing consequences. The second proof uses some "killing involutions" inorder to identify the factors in the product formula. It leads to anenumerative formula for the spanning trees of the $n$-dimensional hypercubeaugmented with diagonals edges, counted according to the number of edges ofeach type. We also discuss more general formulas, obtained using a matrix-treeapproach, for the number of spanning trees of the Cartesian product of completegraphs.

On the spanning trees of the hypercube and other products of graphs