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Given a $d$-dimensional box, using only orthogonal cuts and not weird cuts to try to be 'creative', it takes $\sum\limits_{i=1}^{d}\lceil \log_2(s_i)\rceil$ cuts given that the sidelengths of the box are $(s_1,s_2,...,s_d)$. For example, the minimum number of cuts to turn a cube with side lengths of $4$ into $64$ unit cubes would be $2+2+2$ (from the formula above), giving $6$ cuts. I'll put a copy of the formula below for aesthetic purposes.

$$\sum\limits_{i=1}^{d}\lceil \log_2(s_i)\rceil$$

I don't need to understand the abstract explanation for dimensions higher than $3$, but can somebody please explain how a formula like this is derived from a problem like this?

Lucien Jaccon
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  • I think that turning a side-length four cube into $64$ unit cubes takes nine cuts. Consider one unit cube not on any face. It takes six cuts just to free that one unit cube, and those six cuts won't free any of the other unit cubes that don't reach a face of the big cube. Are you allowed to move pieces after you've made a cut, or is that too "weird" and "creative"? If you're allowed to rearrange, see https://math.stackexchange.com/questions/2950625/mathcounts-cutting-a-larger-cube-into-smaller-cubes – Gerry Myerson Mar 27 '24 at 02:42
  • You are allowed to rearrange. – Lucien Jaccon Mar 27 '24 at 11:59

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