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⚫ | In the ] of ], the '''Plotkin bound''' is a bound on the |
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{{One source|date=October 2024}} | |||
{{Use dmy dates|date=May 2020|cs1-dates=y}} | |||
⚫ | == Statement of the |
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⚫ | In the ] of ], the '''Plotkin bound''', named after Morris Plotkin, is a limit (or bound) on the maximum possible number of codewords in ] ]s 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 ] <math>\{0,1\}</math>. 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 ] <math>\mathbb{F}_2^n</math> over the ] <math>\mathbb{F}_2</math>. Let <math>d</math> be the minimum distance of <math>C</math>, i.e. | ||
:<math>d = \min_{x,y \in C, x \neq y} d(x,y)</math> | :<math>d = \min_{x,y \in C, x \neq y} d(x,y)</math> | ||
⚫ | where <math>d(x,y)</math> is the ] between <math>x</math> and <math>y</math>. The expression <math>A_{2}(n,d)</math> represents the maximum number of possible codewords in a binary code of length <math>n</math> and minimum distance <math>d</math>. The Plotkin bound places a limit on this expression. | ||
⚫ | '''Theorem (Plotkin bound):''' | ||
⚫ | where <math>d(x,y)</math> is the ] between <math>x</math> and <math>y</math>. The expression <math>A_{2}(n,d)</math> represents the maximum number of possible codewords in a binary code of length <math>n</math> and minimum distance |
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⚫ | |||
i) If <math>d</math> is even and <math> 2d > n </math>, then | i) If <math>d</math> is even and <math> 2d > n </math>, then | ||
:<math> A_{2}(n,d) \leq 2 \left\lfloor\frac{d}{2d-n}\right\rfloor </math> | :<math> A_{2}(n,d) \leq 2 \left\lfloor\frac{d}{2d-n}\right\rfloor. </math> | ||
ii) If <math>d</math> is odd and <math> 2d+1 > n </math>, then | ii) If <math>d</math> is odd and <math> 2d+1 > n </math>, then | ||
:<math> A_{2}(n,d) \leq 2 \left\lfloor\frac{d+1}{2d+1-n}\right\rfloor </math> | :<math> A_{2}(n,d) \leq 2 \left\lfloor\frac{d+1}{2d+1-n}\right\rfloor. </math> | ||
iii) If <math>d</math> is even, then | iii) If <math>d</math> is even, then | ||
:<math> A_{2}(2d,d) \leq 4d </math> | :<math> A_{2}(2d,d) \leq 4d. </math> | ||
iv) If <math>d</math> is odd, then | iv) If <math>d</math> is odd, then | ||
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:<math> A_{2}(2d+1,d) \leq 4d+4 </math> | :<math> A_{2}(2d+1,d) \leq 4d+4 </math> | ||
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^2, x\neq y} d(x,y)</math> in two different ways. | |||
On the one hand, there are <math>M</math> choices for <math>x</math> and for each such choice, there are <math>M-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> (<math> x\neq y </math>), it follows that | |||
⚫ | :<math> \sum_{(x,y) \in C^2, x\neq y} d(x,y) \geq M(M-1) d. </math> | ||
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 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, x \neq y} d(x,y)</math> and therefore | ||
⚫ | :<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) = \sum_{i=1}^n 2s_i (M-s_i).</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 | ||
⚫ | The quantity on the right is maximized if and only if <math>s_i = M/2</math> holds for all <math>i</math> (at this point of the proof we ignore the fact, that the <math>s_i</math> are integers), then | ||
⚫ | :<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, x \neq y} d(x,y) \leq \frac{1}{2} n M^2.</math> | ||
⚫ | |||
Combining the upper and lower bounds for <math> \sum_{(x,y) \in C, x \neq y} 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, | |||
:<math> M(M-1) d \leq \frac{1}{2} n M^2</math> | :<math> M(M-1) d \leq \frac{1}{2} n M^2</math> | ||
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 \left\lfloor \frac{d}{2d-n} \right\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 | |||
⚫ | :<math> \sum_{x,y \in C} d(x,y) \leq \frac{1}{2} n |
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Combining the upper and lower bounds for <math> \sum_{x,y \in C} d(x,y)</math>, this means that | |||
:<math> M(M-1) d \leq \frac{1}{2} n (M^2-1)</math> | |||
or, using that <math>2d > n</math>, | |||
:<math> M \leq \frac{2d}{2d-n} - 1</math> | |||
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> | |||
This completes the proof of the bound. | This completes the proof of the bound. | ||
⚫ | ==References== | ||
*<em>Binary codes with specified minimum distance,</em> M. Plotkin, IRE Transactions on Information Theory, 6:445-450, 1960. | |||
==See also== | ==See also== | ||
* ] | |||
⚫ | * ] | ||
* ] | |||
⚫ | * ] | ||
⚫ | * ] | ||
* ] | |||
* ] | |||
⚫ | ==References== | ||
⚫ | *] | ||
* {{cite journal |title=Binary codes with specified minimum distance |author-first=Morris |author-last=Plotkin |journal=] |volume=6 |pages=445–450 |date=1960 |issue=4 |doi=10.1109/TIT.1960.1057584}} | |||
⚫ | *] | ||
*] | |||
⚫ | *] | ||
] | ] | ||
] | |||
] | |||
] | |||
] | |||
] |
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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 over the finite field . 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
The quantity on the right is maximized if and only if holds for all (at this point of the proof we ignore the fact, that the are integers), 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
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
- Plotkin, Morris (1960). "Binary codes with specified minimum distance". IRE Transactions on Information Theory. 6 (4): 445–450. doi:10.1109/TIT.1960.1057584.