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In the ] of ], the '''Plotkin bound''' is a bound on the size of ] ]s of length ''n'' and minimum distance ''d''.
{{One source|date=October 2024}}
{{Use dmy dates|date=May 2020|cs1-dates=y}}
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 == == 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'',
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. 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&nbsp;<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 <math>d</math>. The Plotkin bound places a limit on this expression.

<strong>Theorem (Plotkin bound):</strong>


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.


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>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>
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> 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 (M^2-1)</math>

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.


==See also== ==See also==
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*] * ]
*] * ]
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==References== ==References==
*<em>Binary codes with specified minimum distance,</em> M. Plotkin, IRE Transactions on Information Theory, 6:445-450, 1960. * {{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 { 0 , 1 } {\displaystyle \{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 F 2 n {\displaystyle \mathbb {F} _{2}^{n}} over the finite field F 2 {\displaystyle \mathbb {F} _{2}} . Let d {\displaystyle d} be the minimum distance of C {\displaystyle C} , i.e.

d = min x , y C , x y d ( x , y ) {\displaystyle d=\min _{x,y\in C,x\neq y}d(x,y)}

where d ( x , y ) {\displaystyle d(x,y)} is the Hamming distance between x {\displaystyle x} and y {\displaystyle y} . The expression A 2 ( n , d ) {\displaystyle A_{2}(n,d)} represents the maximum number of possible codewords in a binary code of length n {\displaystyle n} and minimum distance  d {\displaystyle d} . The Plotkin bound places a limit on this expression.

Theorem (Plotkin bound):

i) If d {\displaystyle d} is even and 2 d > n {\displaystyle 2d>n} , then

A 2 ( n , d ) 2 d 2 d n . {\displaystyle A_{2}(n,d)\leq 2\left\lfloor {\frac {d}{2d-n}}\right\rfloor .}

ii) If d {\displaystyle d} is odd and 2 d + 1 > n {\displaystyle 2d+1>n} , then

A 2 ( n , d ) 2 d + 1 2 d + 1 n . {\displaystyle A_{2}(n,d)\leq 2\left\lfloor {\frac {d+1}{2d+1-n}}\right\rfloor .}

iii) If d {\displaystyle d} is even, then

A 2 ( 2 d , d ) 4 d . {\displaystyle A_{2}(2d,d)\leq 4d.}

iv) If d {\displaystyle d} is odd, then

A 2 ( 2 d + 1 , d ) 4 d + 4 {\displaystyle A_{2}(2d+1,d)\leq 4d+4}

where   {\displaystyle \left\lfloor ~\right\rfloor } denotes the floor function.

Proof of case i

Let d ( x , y ) {\displaystyle d(x,y)} be the Hamming distance of x {\displaystyle x} and y {\displaystyle y} , and M {\displaystyle M} be the number of elements in C {\displaystyle C} (thus, M {\displaystyle M} is equal to A 2 ( n , d ) {\displaystyle A_{2}(n,d)} ). The bound is proved by bounding the quantity ( x , y ) C 2 , x y d ( x , y ) {\displaystyle \sum _{(x,y)\in C^{2},x\neq y}d(x,y)} in two different ways.

On the one hand, there are M {\displaystyle M} choices for x {\displaystyle x} and for each such choice, there are M 1 {\displaystyle M-1} choices for y {\displaystyle y} . Since by definition d ( x , y ) d {\displaystyle d(x,y)\geq d} for all x {\displaystyle x} and y {\displaystyle y} ( x y {\displaystyle x\neq y} ), it follows that

( x , y ) C 2 , x y d ( x , y ) M ( M 1 ) d . {\displaystyle \sum _{(x,y)\in C^{2},x\neq y}d(x,y)\geq M(M-1)d.}

On the other hand, let A {\displaystyle A} be an M × n {\displaystyle M\times n} matrix whose rows are the elements of C {\displaystyle C} . Let s i {\displaystyle s_{i}} be the number of zeros contained in the i {\displaystyle i} 'th column of A {\displaystyle A} . This means that the i {\displaystyle i} 'th column contains M s i {\displaystyle M-s_{i}} ones. Each choice of a zero and a one in the same column contributes exactly 2 {\displaystyle 2} (because d ( x , y ) = d ( y , x ) {\displaystyle d(x,y)=d(y,x)} ) to the sum ( x , y ) C , x y d ( x , y ) {\displaystyle \sum _{(x,y)\in C,x\neq y}d(x,y)} and therefore

( x , y ) C , x y d ( x , y ) = i = 1 n 2 s i ( M s i ) . {\displaystyle \sum _{(x,y)\in C,x\neq 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 {\displaystyle s_{i}=M/2} holds for all i {\displaystyle i} (at this point of the proof we ignore the fact, that the s i {\displaystyle s_{i}} are integers), then

( x , y ) C , x y d ( x , y ) 1 2 n M 2 . {\displaystyle \sum _{(x,y)\in C,x\neq y}d(x,y)\leq {\frac {1}{2}}nM^{2}.}

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

M ( M 1 ) d 1 2 n M 2 {\displaystyle M(M-1)d\leq {\frac {1}{2}}nM^{2}}

which given that 2 d > n {\displaystyle 2d>n} is equivalent to

M 2 d 2 d n . {\displaystyle M\leq {\frac {2d}{2d-n}}.}

Since M {\displaystyle M} is even, it follows that

M 2 d 2 d n . {\displaystyle M\leq 2\left\lfloor {\frac {d}{2d-n}}\right\rfloor .}

This completes the proof of the bound.

See also

References

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