Revision as of 17:05, 20 April 2009 editGiftlite (talk | contribs)Autopatrolled, Extended confirmed users, Pending changes reviewers39,632 edits +Morris Plotkin← Previous edit | Revision as of 17:06, 20 April 2009 edit undoGiftlite (talk | contribs)Autopatrolled, Extended confirmed users, Pending changes reviewers39,632 editsm →Proof of case i: +8.Next edit → | ||
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== 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. | ||
On the one hand, there are <math>r</math> choices for <math>x</math> and for each such choice, there are <math>r-1</math> choices for <math>y</math>. Since by definition <math>d(x,y) \geq d</math> for all <math>x</math> and <math>y</math>, it follows that | On the one hand, there are <math>r</math> choices for <math>x</math> and for each such choice, there are <math>r-1</math> choices for <math>y</math>. Since by definition <math>d(x,y) \geq d</math> for all <math>x</math> and <math>y</math>, it follows that | ||
:<math> \sum_{x,y \in C, x\neq y} d(x,y) \geq M(M-1) d </math> | :<math> \sum_{x,y \in C, x\neq y} d(x,y) \geq M(M-1) d. </math> | ||
On the other hand, let <math>A</math> be an <math>M \times n</math> matrix whose rows are the elements of <math>C</math>. Let <math>s_i</math> be the number of zeros contained in the <math>i</math>'th column of <math>A</math>. This means that the <math>i</math>'th column contains <math>M-s_i</math> ones. Each choice of a zero and a one in the same column contributes exactly <math>2</math> (because <math>d(x,y)=d(y,x)</math>) to the sum <math>\sum_{x,y \in C} d(x,y)</math> and therefore | On the other hand, let <math>A</math> be an <math>M \times n</math> matrix whose rows are the elements of <math>C</math>. Let <math>s_i</math> be the number of zeros contained in the <math>i</math>'th column of <math>A</math>. This means that the <math>i</math>'th column contains <math>M-s_i</math> ones. Each choice of a zero and a one in the same column contributes exactly <math>2</math> (because <math>d(x,y)=d(y,x)</math>) to the sum <math>\sum_{x,y \in C} d(x,y)</math> and therefore | ||
:<math> \sum_{x,y \in C} d(x,y) = \sum_{i=1}^n 2s_i (M-s_i)</math> | :<math> \sum_{x,y \in C} d(x,y) = \sum_{i=1}^n 2s_i (M-s_i).</math> | ||
If <math>M</math> is even, then the quantity on the right is maximized if and only if <math>s_i = M/2</math> holds for all i, then | If <math>M</math> is even, then the quantity on the right is maximized if and only if <math>s_i = M/2</math> holds for all i, then | ||
:<math> \sum_{x,y \in C} d(x,y) \leq \frac{1}{2} n M^2</math> | :<math> \sum_{x,y \in C} d(x,y) \leq \frac{1}{2} n M^2.</math> | ||
Combining the upper and lower bounds for <math> \sum_{x,y \in C} d(x,y) </math> that we have just derived, | Combining the upper and lower bounds for <math> \sum_{x,y \in C} d(x,y) </math> that we have just derived, | ||
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which given that <math>2d>n</math> is equivalent to | which given that <math>2d>n</math> is equivalent to | ||
:<math> M \leq \frac{2d}{2d-n}</math> | :<math> M \leq \frac{2d}{2d-n}.</math> | ||
Since <math>M</math> is even, it follows that | Since <math>M</math> is even, it follows that | ||
:<math> M \leq 2 \lfloor \frac{d}{2d-n} \rfloor </math> | :<math> M \leq 2 \lfloor \frac{d}{2d-n} \rfloor. </math> | ||
On the other hand, if <math>M</math> is odd, then <math>\sum_{i=1}^n 2s_i (M-s_i)</math> is maximized when <math>s_i = \frac{M \pm 1}{2}</math> which implies that | On the other hand, if <math>M</math> is odd, then <math>\sum_{i=1}^n 2s_i (M-s_i)</math> is maximized when <math>s_i = \frac{M \pm 1}{2}</math> which implies that | ||
:<math> \sum_{x,y \in C} d(x,y) \leq \frac{1}{2} n (M^2-1)</math> | :<math> \sum_{x,y \in C} d(x,y) \leq \frac{1}{2} n (M^2-1).</math> | ||
Combining the upper and lower bounds for <math> \sum_{x,y \in C} d(x,y)</math>, this means that | Combining the upper and lower bounds for <math> \sum_{x,y \in C} d(x,y)</math>, this means that | ||
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or, using that <math>2d > n</math>, | or, using that <math>2d > n</math>, | ||
:<math> M \leq \frac{2d}{2d-n} - 1</math> | :<math> M \leq \frac{2d}{2d-n} - 1.</math> | ||
Since M is an integer, | Since M is an integer, | ||
:<math> M \leq \lfloor \frac{2d}{2d-n} - 1 \rfloor = \lfloor \frac{2d}{2d-n} \rfloor -1 \leq 2 \lfloor \frac{d}{2d-n} \rfloor </math> | :<math> M \leq \lfloor \frac{2d}{2d-n} - 1 \rfloor = \lfloor \frac{2d}{2d-n} \rfloor -1 \leq 2 \lfloor \frac{d}{2d-n} \rfloor. </math> | ||
This completes the proof of the bound. | This completes the proof of the bound. |
Revision as of 17:06, 20 April 2009
In the mathematics of coding theory, the Plotkin bound, named after Morris Plotkin, is a bound on the size of binary codes of length n and minimum distance d.
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.
See also
References
- Binary codes with specified minimum distance, M. Plotkin, IRE Transactions on Information Theory, 6:445-450, 1960.