# What’s the best way to cut an apple?

Take the apple in one hand, and the knife in the other. In the first cut, the apple is divided in two pieces: a small one that drops into the plate and a big one that is still hold with the hand. This big piece is again divided in two pieces, a small one that falls into the plate and a big one that is held in the hand. We continue doing cuts until we hold a small piece with the hand that we just drop into the plate. We don’t want to take a dropped piece from the plate to recut it. We generate a definitive piece with each cut, except for the last cut that we generate two pieces.

The objective is to generate $N$ pieces from the apple so that they are most close to a sphere of volume $\dfrac{V}{N}$ where $V$ is the volume of the whole apple.

Mathematically, we can approximate an apple with a sphere, and I would like to minimize the error function compared with the perfect solution.

In other words, for every small (differential) piece of volume that is outside of the sphere of volume $\dfrac{V}{N}$ and radius $a$ and for every piece of volume inside this sphere that it’s not filled, we calculate:

$$e= (r-a)^2$$ where $r$ is the distance of the small volume to the centre of the sphere. We then add (integrate) all errors of all the small volumes of all the pieces that we generated. This result is the value that we want to minimize.

The question is, what’s the best approach to cut the sphere (apple) in order to minimize the error function (in order to be more palatable)?

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Safer to use a cutting board. – Will Jagy Jun 18 '13 at 1:43
May be the first step is to try to solve the problem in 2D. Instead of a sphere, take a circle, and instead of cutting planes, take cutting rects, and instead of differential volumes, take differential surfaces... – jbaylina Jun 18 '13 at 10:09
I recommend that you solve this for the circle for, say, $N=2,3,4,$ explicitly and post the best solutions as jpegs within your question. Right now I cannot tell what you want, I don't know why you want it, and I suspect that there will be multiple competing algorithms, no one of which gives the best solution for all possible $N.$ – Will Jagy Jun 18 '13 at 17:07