Perhaps a math degree and cooking don't go hand in hand, but hopefully they do.
I have been thinking about this problem for some time when in the kitchen without making any real progress: How does one cut onions in a mathematically efficient way?
For simplicity, let's assume that an onion is perfectly spherical with radius $r$, consisting of $n$ layers of thickness $d = \frac{r}{n}$. Each cut consists of a plane $p_i$ orthogonal to the $xy$-plane (the cutting board) intersecting some (or all) parts of the onion, arranged in some way, dividing them.
The goal is to, with a sequence of cuts $\{p_i\}_{i=1}^{N}$, get the onion to consist of a set of pieces (most likely irregular in shape) so that they all have diameter at most $M$. By diameter of a piece, we mean the longest possible distance between two points on the piece.
How should the cuts $p_1, p_2, p_3, \dots, p_N$ of the onion be made so that this criterion is fulfilled (all the pieces are small enough) and $N$ is as small as possible?
In other words, we don't consider time to be an issue, just the smallest number of cuts. You may rearrange the pieces in any which way between cuts as long as you specify which part goes where.
I think it's fair to assume $d < M$, because otherwise the pieces need to be very small and one would just use a blender. Of course, the model isn't perfect, since we have an infinitely long knife on an infinitely large cutting board, but since I never seem to run out of room on it, I think that's OK.
Yes, this question is just as "serious" as it sounds, but I would very much appreciate your thoughts on this. I welcome everything from ideas on how to approach the problem and sketches of an optimal strategy (I'm a Master student in computational science) to a full-blown perfect algorithm.
This could become a great conversation starter.
Best Answer
If we can rearrange the pieces any which way between cuts, as the question states, then we might as well arrange them in a line before each cut, so that the cut intersects every piece of the onion. By choosing each piece's translation perpendicular to the cut, we can guarantee that it is divided into two pieces of equal area. By choosing the piece's orientation, we can also ensure that none of the pieces get too long and skinny, so the diameter of a piece of area $a$ is bounded above by $c\sqrt a$, where $c$ is a universal constant.
Then it is sufficient to get the area of each piece below $M^2/c^2$. The original area of the largest piece is $4\pi r^2$, and each cut reduces it by half, so we can succeed in $\log_2(r^2/M^2)$ cuts plus a constant.
The converse argument shows that $\log_2(r^2/M^2)$ cuts plus a constant are indeed necessary.