Wednesday, April 30, 2014

Into How Many Regions Can Six Planes Divide Space?

I recently misread an easy recreational puzzle as 'Into how many regions can six planes divide space?'.

It took me about half an hour to get an answer. (Which was the correct answer to the wrong question, but sadly the wrong answer to the right question). But it turned out to be great fun to think about.

1,2,4,8,..... how does this sequence continue?

So far Tim and Sips and Paul Cook have had a go. Paul guessed the answer but didn't know why it was true and I haven't heard anything from the other two.

Have fun! It's not actually that hard.


  1. 1 ,2 ,4 ,8 , then 15 given the first part of your post. And then I ran out of time to generalise.

  2. After 15 the next number is 26.

  3. ( I considered the problem of cutting the plane with lines, which is fairly simple to solve: if a(n) is the number of areas generated by n lines, then a(0) = 1, a(1) = 2, a(n) = a(n-1) + n, which can be rewritten as a(n) = nC2+nC1+nC0. Then I made a leap of faith that for space rather than plane it would be nC3+nC2+nC1+nC0.)

  4. Keir: That is the correct answer. I think there ought to be a really elegant argument that goes directly to the sum-of-binomial-coefficients answer, for any number of dimensions (it's not hard to do less directly by induction). Something along these lines: Pick a direction that isn't parallel to any of the (hyper)planes; given any region, head in that direction, doing "the obvious thing" when you hit any of the (hyper)planes; before hitting a corner and being unable to proceed you'll encounter some number (at least 0, at most d) of the hyperplanes; this establishes a bijection between the regions and the sets of up to d hyperplanes. But so far (after, admittedly, not a huge amount of thought) I haven't found a really clear short way to turn that handwaving into a rigorous proof.