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39 votes
39 votes

Consider the binary code that consists of only four valid codewords as given below:

$00000, 01011, 10101, 11110$

Let the minimum Hamming distance of the code $p$ and the maximum number of erroneous bits that can be corrected by the code be $q$. Then the values of $p$ and $q$ are

  1. $p=3$ and $q=1$
  2. $p=3$ and $q=2$
  3. $p=4$ and $q=1$
  4. $p=4$ and $q=2$
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3 Answers

Best answer
52 votes
52 votes

$00000$(code1), $01011$(code2), $10101$(code3), $11110$ (code4) 

Haming distance = min of all hamming distances.

Which is $3$ b/w (code1) and (code2) so, $p = 3$

Now to correct d bit error we need hamming distance $= 2d+1$

So, $2d+1 = 3$ will gives $d= 1$.

A is answer.

edited by
14 votes
14 votes

The Minimum Hamming Distance should be 3.

Lets take the CODE

00000 and

01011

cleary you can see the Minimum hamming distance is 3 and using code with hamming distance 3 we can correct 1 bit error only because

D>= 2*T + 1

where D is minimum Hamming Distance,T is no of bit need to be corrected...

So the answer will be A only.

edited by
3 votes
3 votes

Ans: A

Hamming Distance:- You have to do  XOR  between each Code , then count number of 1’s in output

$\Rightarrow$ Hamming distance between 00000 and 01011

00000 01011 = 01011 no. of 1’s in(01011) = 3  i:e   3

$\Rightarrow$ Hamming distance between 01011 and 10101 

01011 10101 = 11110 no. of 1’s in(11110) = 4  i:e   4

if you do for each code then you will find that minimum hamming distance = 3

Maximum number of erroneous bits that can be corrected by the code = d

(2*d + 1) = Minimum Hamming distance

2*d + 1 = 3

d = 1

Answer:

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