Bounding the Volume of Hamming Balls

August 13, 2010

In this post I will derive an oft used inequality in Theoretical computer science, using the Chernoff technique. The following function turns up in a few places in combinatorial geometry and other places.

\displaystyle \begin{array}{rcl} \Phi_d(m) = \sum_{i=0}^d {m \choose i}. \end{array}

Here we assume {m \geq d} since otherwise the function is trivially seen to be equal to {2^m}. Here we will show that,

\displaystyle \begin{array}{rcl} \Phi_d(m) \leq \left ( \frac{me}{d} \right )^d \end{array}

Two easy interpretations of {\Phi_d(m)} are as follows. First, it counts the number of points in the hamming cube {{\bf H}^m = \{0,1\}^m} whose hamming distance from the origin is at most {d}. This interpretation also allows us to derive the nice identity,

\displaystyle \begin{array}{rcl} \Phi_d(m) = \sum_{h=0}^d \sum_{l = 0}^h {s \choose l}{m - s \choose h - l}. \end{array}

for any {0 \leq s \leq m}. In the above expression we obey the convention {{p \choose q} = 0} when {p < q}. To see this, interpret the above sum as the distance from a fixed point {p \in {\bf H}^m} whose distance from the origin is {s}. The {h} is the distance from {p} and the inner summation counts the number of points at distance {h}. Then noting that the hamming cube looks the same from every point so the number of points of distance at most {d} is the same from the origin or from {p}. Hence the identity.

This definition has a nice geometric interpretation because {\Phi_d(m)} is also the number of points from {{\bf H}^n} inside any ball of radius {d}. The second interpretation is as follows. Consider the discrete probability space {\Omega = \{ 0, 1 \}} with the measure {p(\{0\}) = p(\{1\}) = 1/2}. Let {X_1, X_2, \dots, X_m} be {m} independent random variables {X_i : \Omega \rightarrow \{0,1\}} defined as {X_i(y) = y} for {y \in \Omega}. Let {X = \sum X_i}. Then we have,

\displaystyle \Pr\{ X_1 = a_1, X_2 = a_2, \dots, X_m = a_m \} = \frac{1}{2^m},

for any {a_i \in \{0,1\}}. As such we have,

\displaystyle \frac{\Phi_d(m)}{2^m} = \Pr\{ X \leq d\}.

We will use this interpretation to derive our inequality.

Lemma: {\Phi_d(m) \leq \left ( \frac{me}{d} \right )^d}.

Proof: Our first observation is that we may take {d \leq m/2}. The function {\left ( \frac{me}{d} \right )^d} is an increasing function of {d}. Denoting this by {f(d)} and extending it to a function on {[0,m]} we have by logarithmic differentiation,

\displaystyle \begin{array}{rcl} \frac{f'(d)}{f(d)} = \ln \left ( \frac{m}{d} \right ) \geq 0 \end{array}

Thus for {d > m/2} we have {f(d) \geq f(m/2) \geq (2e)^{m/2} > 2^m}. On the other hand {\Phi_d(m) \leq 2^m} trivially.

Thus assuming {d} is {0 \leq d \leq m/2}. Now we have the following identity,

\displaystyle \begin{array}{rcl} \sum_{i=0}^d {m \choose i} = \sum_{j=m-d}^m {m \choose j} = 2^m \Pr\{ X \geq m-d \}. \end{array}

so it sufficies to upper bound {\Pr \{ X \geq m - d \}}. We now use the Chernoff technique. For any {\lambda > 0},

\displaystyle \begin{array}{rcl} \Phi_d(m) &=& 2^m \Pr\{ X \geq m - d \} \\\\ &=& 2^m \Pr \{ e^{\lambda X} \geq e^{\lambda(m - d)} \} \\\\ &\leq& 2^m \frac{{\bf E}(e^{\lambda X})}{e^{\lambda(m - d)}} ~~ \text{[By Markov's Inequality]} \\\\ &=& 2^m \frac{\prod_{i=1}^m {\bf E}(e^{\lambda X_i})}{e^{\lambda(m-d)}} ~~ \text{[By Independence of the} ~ X_i] \end{array}

Now for any {i} we have,

\displaystyle {{\bf E}(e^{\lambda X_i}) = \frac{e^{\lambda} + 1}{2}}

and so we have,

\displaystyle \begin{array}{rcl} \Phi_d(m) \leq \frac{(e^{\lambda} + 1)^m}{e^{\lambda(m-d)}} \end{array}

We try to optimize the above expression by finding the {\lambda} which minimizes the expression above. Noting that {m/2 \geq d} we have, by differentiation and observing the sign of the derivative, that the {\lambda} minimizing the expression is given by,

\displaystyle \begin{array}{rcl} e^{\lambda} = \frac{m}{d} - 1 \end{array}

and substituing that in the estimate we have,

\displaystyle \begin{array}{rcl} \Phi_d(m) &\leq& \frac{\left ( \frac{m}{d} \right )^m}{\left ( \frac{m-d}{d}\right )^{m-d}} \\ &=& \left ( 1 + \frac{d}{m-d} \right )^d \left ( \frac{m}{d} \right )^d \leq \left ( \frac{me}{d} \right )^d. \end{array}

The last step follows from the simple fact that,

\displaystyle \begin{array}{rcl} \left ( 1 + \frac{d}{m-d} \right )^{m - d} \leq \left ( e^{\frac{d}{m-d}} \right )^{m-d} = e^d. \end{array}

\Box