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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 C {\displaystyle C} be a binary code of length n {\displaystyle n} , i.e. a subset of F 2 n {\displaystyle \mathbb {F} _{2}^{n}} . 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 d ( x , y ) {\displaystyle \sum _{x,y\in C}d(x,y)} in two different ways.

On the one hand, there are r {\displaystyle r} choices for x {\displaystyle x} and for each such choice, there are r 1 {\displaystyle r-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} , it follows that

x , y C , x y d ( x , y ) M ( M 1 ) d . {\displaystyle \sum _{x,y\in C,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 d ( x , y ) {\displaystyle \sum _{x,y\in C}d(x,y)} and therefore

x , y C d ( x , y ) = i = 1 n 2 s i ( M s i ) . {\displaystyle \sum _{x,y\in C}d(x,y)=\sum _{i=1}^{n}2s_{i}(M-s_{i}).}

If M {\displaystyle M} is even, then the quantity on the right is maximized if and only if s i = M / 2 {\displaystyle s_{i}=M/2} holds for all i, then

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

Combining the upper and lower bounds for x , y C d ( x , y ) {\displaystyle \sum _{x,y\in C}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\lfloor {\frac {d}{2d-n}}\rfloor .}

On the other hand, if M {\displaystyle M} is odd, then i = 1 n 2 s i ( M s i ) {\displaystyle \sum _{i=1}^{n}2s_{i}(M-s_{i})} is maximized when s i = M ± 1 2 {\displaystyle s_{i}={\frac {M\pm 1}{2}}} which implies that

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

Combining the upper and lower bounds for x , y C d ( x , y ) {\displaystyle \sum _{x,y\in C}d(x,y)} , this means that

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

or, using that 2 d > n {\displaystyle 2d>n} ,

M 2 d 2 d n 1. {\displaystyle M\leq {\frac {2d}{2d-n}}-1.}

Since M is an integer,

M 2 d 2 d n 1 = 2 d 2 d n 1 2 d 2 d n . {\displaystyle M\leq \lfloor {\frac {2d}{2d-n}}-1\rfloor =\lfloor {\frac {2d}{2d-n}}\rfloor -1\leq 2\lfloor {\frac {d}{2d-n}}\rfloor .}

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.
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