This is a generalization of this question. So in $\mathbb{R}^2$, the problem is illustrated like so: enter image description here

Here, $n = 3$ lines divides $\mathbb{R}^2$ into $N_2=7$ regions. For general $n$ in the case of $\mathbb{R}^2$, the number of regions $N_2$ is $\binom{n+1}{2}+1$. But what about if we consider the case of $\mathbb{R}^m$, partitioned using $n$ hyperplanes?

Is the answer $N_m$ still $\binom{n+1}{2}+1$, or will it be a function of $m$?

up vote 7 down vote accepted

Hint: Let's consider how to prove that in $\mathbb{R}^2$, the maximum number of distinct regions is $\frac { n^2 + n + 2 } { 2} $.

First, since there are finitely many lines, we can tilt any configuration such that no line is parallel to the horizontal. Now, let's consider the 'lowest' point of any region.

If the region is bound below, than such a point must exist, is unique, and is a point of intersection of 2 lines. In fact, there is a one-to-one relation between regions that are bounded below, and these points of intersections, which gives us $ { n \choose 2 } = \frac{n^2-n} {2} $ regions.

What happens if the regions are not bounded below? How many are there? Well, let's simply insert a horizontal line way down below, and count the number of unbounded regions, by associating them to these new bounded regions. Now, let's consider the 'lowest, leftmost' point of any region. If the region is bound to the left, then such a point must exist, is unique, and is appoint of intersection of the original $n$ lines with the new horizontal line. In fact, there is a one-to-one relation between regions that are unbounded below, bounded to the left, and these points of intersections, which gives us ${n \choose 1} = n$ regions.

Finally, how many regions are not bounded below, and not bounded to the left? You should convince yourself that there is ${n \choose 0} = 1 $ regions.

Hence, the total number of ways is ${n \choose 2} + {n\choose 1} + {n \choose 0}$.


Now, you can easily see how this shows that the number of regions for the hyperplane version of your problem is $$\sum_{i=0}^m {n \choose i}. $$

  • What if $n < m$? Should the general formula not be $\sum^{\min(m, n)}_{i = 0} \binom{n}{i}$ so that we have $N_m = 2^n$ ($n<m$) when there are not enough hyperplanes to finish cutting all $m$ dimensions in half, only the first $n$ of them hence $2^n = \sum^{n}_{i = 0} \binom{n}{i}$ – mchen Jun 3 '13 at 15:53
  • @MiloChen Note that if $a < b$, then ${ a \choose b } = 0 $. Hence the formula still holds. You might need to do some work to check that the formula is still valid, and indeed it is. Basically, for those low values, the answer is just $2^n$, because each hyperplane can cut each existing region into 2 (e.g. slice along each axis-hyperplane). – Calvin Lin Jun 3 '13 at 15:56
  • Ahh, I didn't know the binomial coefficient had a definition of $0$ under the case $a<b$ - thanks – mchen Jun 3 '13 at 15:59
  • @MiloChen The binomial coefficient can be generalized to where $a$ is any real value. – Calvin Lin Jun 3 '13 at 16:01
  • It must pass through $\sum_{i=0}^{m} {n+1 \choose i} - { n \choose i}$, by just taking the difference of the formula above. – Calvin Lin Jun 3 '13 at 16:49

Denote this number as $A(m, n)$. We will prove $A(m, n) = A(m, n-1) + A(m-1, n-1)$.

Consider removing one of the hyperplanes, the maximum number is $A(m, n-1)$. Then, we add the hyperplane back. The number of regions on the hyperplane is the same as the number of newly-added regions. Since this hyperplane is $m-1$ dimensional, this maximum number is $A(m-1, n-1)$.

So, this number satisfies $A(m, n) = A(m, n-1) + A(m-1, n-1)$, and the formula can be simply derived by induction as $\sum_{i=0}^m \binom{n}{i}$.

I will try to solve the problem in terms of "how many regions add if we insert an extra hyperplane L (Let) in a space of k-1 hyperplanes in n-dimensions".

Let T (n,k) be the maximum number of regions formed in R^n by k hyper-planes.

Now we need to find T (n,k) - T(n,k-1)=S (Let).

Infact the answer will come out to be T (n-1,k-1).

Let us see how.

I have tried to give a purely Geometrical argument.

Some steps:

  1. The number S is actually the maximum number of regions that L can pass through.[Because every region it passes through divides into 2 and hence the count for that region is 1]

  2. Now , After drawing the new hyperplane, in your mind's eye ,consider only the intersections of L with the remaining (k-1) hyperplanes on L.

  3. So, L which is a (n-1) dimensional Space now have a set of (k-1) intersections [ so (k-2) dimensional lines drawn on it.]

The next step is the crucial one. 4. The number of regions formed by the (k-1) lines in on L (n-1 dimensional space) is bijective to the number of regions L passes through.

So only we need to verify why 4 happens.Others are easy to infer.

Now every region formed by the (k-1) hyperplanes in R^n has a boundary of atleast 2 hyperplanes.

Now consider L ,as L passes through such a region.Observe what happens on L, at least 2 (k-2) dimesional lines occur on L.

Now to maximize the number of regions L passes through it will intersect with all the (k-1) hyperplanes.

So consider what happens when L enters one region to another.

New lines appear.

Call 2 regions adjacent in R^n if they have atleast one common hyperplane.

Now,See that L always passes through one region to its adjacent region. As a result the intersections all give rise to adjacent regions.

And as the passes from one region to other the one- one correspondence with regions formed by the lines on L and the regions L passes through happens (due to this adjacent behaviour of the regions,L passes through)

So the maximum number of regions that the (k-1) lines on L form is actually the number T (n-1,k-1),because L is n-1 dimensional and the lines on L are its k-1 hyperplanes.

Now we need to define T (0,k) and T (k,0).

Define T (0,k)=T (k,0)=1.(quite natural)

This seems like a kind of a hand waving argument because we cant actaully visualize n>4.

But this can be easily checked in 3 dimensions and 2 dimensions.

Note: I have mentioned lines and planes to let you visulaize in 3 dimesions.

NOTE:

Now in accordance with the previous answer we know that there exists a similar recursion in the combinatorial domain.

If we solve the recursion,it comes out to be the same as above.

Moreover you can substitute and check that it holds.

Will like to hear how to verify 4 in other different ways in the comments!

Will like to hear any good suggestions to this answer!

I am new to this ,If you want to downvote my answer (please mention the reason,otherwise I will repeat the same mistakes again).

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