Plotkin bound

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

Statement of the bound

A code is considered "binary" if the codewords use symbols from the binary alphabet $\{0,1\}$. In particular, if all codewords have a fixed length n, then the binary code has length n. Equivalently, in this case the codewords can be considered 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.

$d=\underset{x,y\in C,x\ne y}{\min }d(x,y)$

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

$A_2(n,d)\le 2\lfloor \frac{d}{2d-n}\rfloor .$

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

$A_2(n,d)\le 2\lfloor \frac{d+1}{2d+1-n}\rfloor .$

iii) If $d$ is even, then

$A_2(2d,d)\le 4d.$

iv) If $d$ is odd, then

$A_2(2d+1,d)\le 4d+4$

where $\lfloor \rfloor$ 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 $\sum _{(x,y)\in C^2,x\ne y}d(x,y)$ 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 $d(x,y)\ge d$ for all $x$ and $y$ ($x\ne y$), it follows that

$\sum _{(x,y)\in C^2,x\ne y}d(x,y)\ge M(M-1)d.$

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 $\sum _{(x,y)\in C,x\ne y}d(x,y)$ and therefore

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

The quantity on the right is maximized if and only if $s_i=M/2$ holds for all $i$ (at this point of the proof we ignore the fact, that the $s_i$ are integers), then

$\sum _{(x,y)\in C,x\ne y}d(x,y)\le \frac{1}{2}n{M}^2.$

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

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

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

$M\le \frac{2d}{2d-n}.$

Since $M$ is even, it follows that

$M\le 2\lfloor \frac{d}{2d-n}\rfloor .$

This completes the proof of the bound.

See also

• Diamond code
• Elias Bassalygo bound
• Gilbert–Varshamov bound
• Griesmer bound
• Hamming bound
• Johnson bound
• Singleton bound

References

• "Binary codes with specified minimum distance". IRE Transactions on Information Theory 6 (4): 445–450. 1960. doi:10.1109/TIT.1960.1057584.