Discrete Borsuk Conjecture 3

For bounded $S\subset\mathbb Z^d$ , is $\beta_{\mathbb Z}(S)=2^d$ if and only if $\operatorname{conv}(S)$ is unimodularly equivalent to $[0,m]^d$ ?

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Open

No public discussion or published progress appears to have been found on this conjecture.

No public discussion or published progress was found for the stated conjecture.

Current status (as of November 2025): the conjecture appears open, with no recorded public activity or progress.

No solutions have been posted yet.