Minimum proper-subbox counts for exact double covers of ternary cubes
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Let f(d) be the fewest proper subboxes giving an exact double cover of the ternary cube {0,1,2}ᵈ, so every cube point belongs to exactly two boxes. Must f(d)≥2ᵈ?
Why it matters and what counts as progress
Why it matters: Beyond-volume lower bounds measure a combinatorial obstruction invisible to total box sizes alone.
What counts as progress: New rigorous lower bounds, a proof of the 2ᵈ threshold in further dimensions, or a double-cover construction below it; a bound of 60 at d=6 does not yet reach 64.
How the problem developed
- 29 Jan 2017 · Question [1]
Imre Leader, Luka Milićević and Ta Sheng Tan Asked, in a more general form, whether an exact double cover of a ternary box needs at least 2^d proper sub-boxes.
- 10 Jul 2026 · Partial result [2]
Patrick White Proved f(4) ≥ 19 and f(5) ≥ 33, resolving the 2^d question in dimensions 4 and 5; also proved f(6) ≥ 60 and f(6) ≤ 81, leaving the conjecture open for d ≥ 6.
AI contribution
New bounds for double covers of the discrete box {0,1,2}^d
For exact double covers of {0,1,2}^d by proper subboxes, Patrick White’s model-origin modular and line-rigidity argument gives f(4)≥19, f(5)≥33 and the unconditional bound f(6)≥60.
Sources
Website timeline
Publication dates and editorial changes.
AI-assisted argument raises the six-dimensional double-cover bound