I have a question regarding exercise 14.17 in An Introduction to Information Retrieval by Manning et al.

The problem is:

"Assuming two classes, show that the percentage of non-separable assignments of the vertices of a hypercube decreases with dimensionality M for M > 1. For example, for M = 1 the proportion of non-separable assignments is 0, for M = 2, it is 2/16. Solve the exercise either analytically or by simulation."

The total number of assignments of vertices of an N-dimensional hypercube is: $2^{(2^N)}$

And as I found in here the number of separable assignments is O($2^{(N^2)}$). So the percentage of non-separable assignments is increasing with N (which is the opposite of what is said in the exercise).

What am I missing here?

  • $\begingroup$ I agree, this seems incorrect. $\endgroup$
    – Ben Reiniger
    Commented Jun 19, 2019 at 21:36

1 Answer 1


You appear to be correct. I believe this is a simple proof of the opposite fact:

Let $s(n)$ denote the number of separable sets in the $n$-cube. Let $Q_0, Q_1$ denote the subcubes of one smaller dimension, with say last coordinates equal to 0, 1 respectively. The key fact is that any separable set of the whole cube intersects with each of $Q_0, Q_1$ in separable sets. (The separating hyperplane intersects the codimension-1 spaces to form separating hyperplanes.) So $$s(n)\leq [s(n-1)]^2.$$

Now the proportion $r(n)$ satisfies $$r(n)= \frac{s(n)}{2^{2^n}}\leq \frac{[s(n-1)]^2}{(2^{2^{n-1}})^2}= [r(n-1)]^2< r(n-1),$$ where the last inequality holds because $r(n-1)<1$ for $n>2$.

  • $\begingroup$ Thank you for your answer! $\endgroup$
    – Mahsa
    Commented Jun 29, 2019 at 21:06

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