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Plotkin bound : ウィキペディア英語版
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 \. 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_2^n over the finite field \mathbb_2. Let d be the minimum
distance of C, i.e.
:d = \min_ d(x,y)
where d(x,y) is the Hamming distance between x and y. The expression A_(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_(n,d) \leq 2 \left\lfloor\frac\right\rfloor.
ii) If d is odd and 2d+1 > n , then
: A_(n,d) \leq 2 \left\lfloor\frac\right\rfloor.
iii) If d is even, then
: A_(2d,d) \leq 4d.
iv) If d is odd, then
: A_(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_(n,d)). The bound is proved by bounding the quantity \sum_ 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 ( x\neq y ), it follows that
: \sum_ 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_ d(x,y) and therefore
: \sum_ d(x,y) = \sum_^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_ d(x,y) \leq \frac n M^2.
Combining the upper and lower bounds for \sum_ d(x,y) that we have just derived,
: M(M-1) d \leq \frac n M^2
which given that 2d>n is equivalent to
: M \leq \frac.
Since M is even, it follows that
: M \leq 2 \left\lfloor \frac \right\rfloor.
On the other hand, if M is odd, then \sum_^n 2s_i (M-s_i) is maximized when s_i = \frac which implies that
: \sum_ d(x,y) \leq \frac n (M^2-1).
Combining the upper and lower bounds for \sum_ d(x,y), this means that
: M(M-1) d \leq \frac n (M^2-1)
or, using that 2d > n,
: M \leq \frac - 1.
Since M is an integer,
: M \leq \left\lfloor \frac - 1 \right\rfloor = \left\lfloor \frac \right\rfloor -1 \leq 2 \left\lfloor \frac \right\rfloor.
This completes the proof of the bound.

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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