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In the ] of ], the '''Plotkin bound''' is a bound on the size of ] ]s of length <math>n</math> and minimum distance <math>d</math>. | In the ] of ], the '''Plotkin bound''' is a bound on the size of ] ]s of length <math>n</math> and minimum distance <math>d</math>. | ||
== Statement of the |
== Statement of the bound == | ||
Let <math>C</math> be a binary code of length <math>n</math>, i.e. a subset of <math>\mathbb{F}_2^n</math>. Let <math>d</math> be the minimum distance of <math>C</math>, i.e. | Let <math>C</math> be a binary code of length <math>n</math>, i.e. a subset of <math>\mathbb{F}_2^n</math>. Let <math>d</math> be the minimum distance of <math>C</math>, i.e. | ||
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where <math> \left\lfloor ~ \right\rfloor</math> denotes the ]. | where <math> \left\lfloor ~ \right\rfloor</math> denotes the ]. | ||
== Proof of case i== | == Proof of case ''i''== | ||
Let <math>d(x,y)</math> be the ] of <math>x</math> and <math>y</math>, and <math>M</math> be the number of elements in <math>C</math> (thus, <math>M</math> is equal to <math>A_{2}(n,d)</math>). The bound is proved by bounding the quantity <math>\sum_{x,y \in C} d(x,y)</math> in two different ways. | Let <math>d(x,y)</math> be the ] of <math>x</math> and <math>y</math>, and <math>M</math> be the number of elements in <math>C</math> (thus, <math>M</math> is equal to <math>A_{2}(n,d)</math>). The bound is proved by bounding the quantity <math>\sum_{x,y \in C} d(x,y)</math> in two different ways. |
Revision as of 05:32, 8 March 2009
In the mathematics of coding theory, the Plotkin bound is a bound on the size of binary codes of length and minimum distance .
Statement of the bound
Let be a binary code of length , i.e. a subset of . Let be the minimum distance of , i.e.
where is the Hamming distance between and . The expression represents the maximum number of possible codewords in a binary code of length and minimum distance . The Plotkin bound places a limit on this expression.
Theorem (Plotkin bound):
i) If is even and , then
ii) If is odd and , then
iii) If is even, then
iv) If is odd, then
where denotes the floor function.
Proof of case i
Let be the Hamming distance of and , and be the number of elements in (thus, is equal to ). The bound is proved by bounding the quantity in two different ways.
On the one hand, there are choices for and for each such choice, there are choices for . Since by definition for all and , it follows that
On the other hand, let be an matrix whose rows are the elements of . Let be the number of zeros contained in the 'th column of . This means that the 'th column contains ones. Each choice of a zero and a one in the same column contributes exactly (because ) to the sum and therefore
If is even, then the quantity on the right is maximized if and only if holds for all i, then
Combining the upper and lower bounds for that we have just derived,
which given that is equivalent to
Since is even, it follows that
On the other hand, if is odd, then is maximized when which implies that
Combining the upper and lower bounds for , this means that
or, using that ,
Since M is an integer,
This completes the proof of the bound.
References
- Binary codes with specified minimum distance, M. Plotkin, IRE Transactions on Information Theory, 6:445-450, 1960.