User:ECCclass/Plotkin

In the mathematics of coding theory, the Plotkin bound, named after Morris Plotkin, is a bound on the maximum possible number of codewords in binary codes of given length n and given minimum distance d.

Statement of the bound
We say a code is binary, if the codewords use symbols from the binary alphabet $$\{0,1\}$$. In particular, if all codewords have a fixed length n, we speak of a binary code of length n. Equivalently, we may consider in this case the codewords as elements of vector space $$\mathbb{F}_2^n$$ over the finite field $$\mathbb{F}_2$$. Let $$d$$ be the minimum distance of $$C$$, i.e.
 * $$d = \min_{x,y \in C, x \neq y} d(x,y)$$

where $$d(x,y)$$ is the Hamming distance between $$x$$ and $$y$$. The expression $$A_{2}(n,d)$$ represents the maximum number of possible codewords in a binary code of length $$n$$ and minimum distance $$d$$. The Plotkin bound places a limit on this expression.

Theorem (Plotkin bound):

i) If $$d$$ is even and $$ 2d > n $$, then


 * $$ A_{2}(n,d) \leq 2 \left\lfloor\frac{d}{2d-n}\right\rfloor. $$

ii) If $$d$$ is odd and $$ 2d+1 > n $$, then


 * $$ A_{2}(n,d) \leq 2 \left\lfloor\frac{d+1}{2d+1-n}\right\rfloor. $$

iii) If $$d$$ is even, then


 * $$ A_{2}(2d,d) \leq 4d. $$

iv) If $$d$$ is odd, then


 * $$ A_{2}(2d+1,d) \leq 4d+4 $$

where $$ \left\lfloor ~ \right\rfloor$$ denotes the floor function.

Proof of case i
Let $$d(x,y)$$ be the Hamming distance of $$x$$ and $$y$$, and $$M$$ be the number of elements in $$C$$ (thus, $$M$$ is equal to $$A_{2}(n,d)$$). The bound is proved by bounding the quantity $$\sum_{x,y \in C} d(x,y)$$ in two different ways.

On the one hand, there are $$M$$ choices for $$x$$ and for each such choice, there are $$M-1$$ choices for $$y$$. Since by definition $$d(x,y) \geq d$$ for all $$x$$ and $$y$$, it follows that


 * $$ \sum_{x,y \in C, x\neq y} d(x,y) \geq M(M-1) d. $$

On the other hand, let $$A$$ be an $$M \times n$$ matrix whose rows are the elements of $$C$$. Let $$s_i$$ be the number of zeros contained in the $$i$$'th column of $$A$$. This means that the $$i$$'th column contains $$M-s_i$$ ones. Each choice of a zero and a one in the same column contributes exactly $$2$$ (because $$d(x,y)=d(y,x)$$) to the sum $$\sum_{x,y \in C} d(x,y)$$ and therefore


 * $$ \sum_{x,y \in C} d(x,y) = \sum_{i=1}^n 2s_i (M-s_i).$$

If $$M$$ is even, then the quantity on the right is maximized if and only if $$s_i = M/2$$ holds for all $$i$$, then


 * $$ \sum_{x,y \in C} d(x,y) \leq \frac{1}{2} n M^2.$$

Combining the upper and lower bounds for $$ \sum_{x,y \in C} d(x,y) $$ that we have just derived,


 * $$ M(M-1) d \leq \frac{1}{2} n M^2$$

which given that $$2d>n$$ is equivalent to


 * $$ M \leq \frac{2d}{2d-n}.$$

Since $$M$$ is even, it follows that


 * $$ M \leq 2 \left\lfloor \frac{d}{2d-n} \right\rfloor. $$

On the other hand, if $$M$$ is odd, then $$\sum_{i=1}^n 2s_i (M-s_i)$$ is maximized when $$s_i = \frac{M \pm 1}{2}$$ which implies that


 * $$ \sum_{x,y \in C} d(x,y) \leq \frac{1}{2} n (M^2-1).$$

Combining the upper and lower bounds for $$ \sum_{x,y \in C} d(x,y)$$, this means that


 * $$ M(M-1) d \leq \frac{1}{2} n (M^2-1)$$

or, using that $$2d > n$$,


 * $$ M \leq \frac{2d}{2d-n} - 1.$$

Since $$M$$ is an integer,


 * $$ M \leq \left\lfloor \frac{2d}{2d-n} - 1 \right\rfloor = \left\lfloor \frac{2d}{2d-n} \right\rfloor -1 \leq 2 \left\lfloor \frac{d}{2d-n} \right\rfloor. $$

This completes the proof of the bound.