2
$\begingroup$

Suppose I have been given two codes $x$ and $y$ such that $x = x_1 x_2 . . . .x_m$ and $y = y_1 y_2 . . . .y_m$ where ${x_i} 's$ and ${y_i} 's$ are binary digits. Then we define the concept of distance between $x$ and $y$ as $$\delta (x,y) = |\{ j : x_j \neq y_j \}|$$

i.e. we compare $x$ and $y$ term by term and count how many digits are different at corresponding positions in $x$ and $y$. That count gives the value of distance between $x$ and $y$ .

I am unable to understand the concept of this distance. From above theory I can only understand the way in which distance between two codes is calculated but not what actually a distance between two codes mean ?

$\endgroup$

5 Answers 5

3
$\begingroup$

If it helps in terms of traditional metrics, this is equivalent to 1-norm (which is sometimes called the taxicab norm), which for $\vec{x},\vec{y} \in \mathbb{R}^n$ is defined as $$\left\|\vec{x} - \vec{y} \right\|_1 = \sum_{j=1}^n \left| x_j - y_j \right|.$$

Note that since you are dealing with binary sequences, $\left| x_j - y_j \right|$ is nonzero when $x_j \ne y_j$, in which case $\left| x_j - y_j \right| = 1....$

$\endgroup$
3
$\begingroup$

The calculation of the "distance" between two codes is pretty literal in that you are simply adding up the number of ways in which the two codes differ. However, I can see how the usage of the word distance can be fairly misleading. You can also think of it as the number of "flips" (0 -> 1 , or 1 -> 0), are required to transform one code into the other.

$\endgroup$
2
$\begingroup$

Normally, the distance measure is used to calculate the minimum of distance of any two code words of one and the same code. It is a measure of how well errors in transmitted code words can be detected/corrected.

$\endgroup$
2
$\begingroup$

Interpretation

The distance between two codewords is the number of positions in which the codewords differ.

Example

The codewords 11000011 and 11010010 differ in two positions (No. 4 and 8), so their distance is 2.

$\endgroup$
1
  • $\begingroup$ Thanks. The interpretation(how we are calculating the distance) is clear to me. I am just asking what does it mean? $\endgroup$ Apr 7, 2016 at 14:30
1
$\begingroup$

You interpret it to mean how "far apart" the codewords are. In this measurement of distance, they are far apart if you have to change a lot of the symbols in order to change one codeword into the other.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .