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+14 votes
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What is the distance of the following code $000000$, $010101$, $000111$, $011001$, $111111$?

  1. $2$
  2. $3$
  3. $4$
  4. $1$
asked in Computer Networks by Veteran (59.5k points)
edited by | 1.1k views

6 Answers

+23 votes
Best answer

Distance (also called min-distance) of a block code is the minimum number of positions in which any two distinct codes differ. Here, min-distance occurs for the codes $2$ and $3$ and they differ only in $2$ positions. So, $d = 2$.

https://en.wikipedia.org/wiki/Block_code

answered by Veteran (355k points)
edited by
0
@ Arjun sir,

codes 2 & 4 also has distance 2. right?

 010101

    XOR

 011001

= 001100
0

@Arjun Sir, do we need to check each codeword distance to every other codeword, or just go sequentially?

0
every code with every other code.
+5 votes
(A) Distance = minimum hamming distance = $2$.

$010101  ⊕  011001  =  001100$
answered by Junior (715 points)
edited by
0 votes

Check the solution

answered by (451 points)
0 votes

Hamming distance = minimum distance between keywords.

Here minimum distance occur between 010101
and 011001

and Answer is 2

answered ago by (243 points)
–1 vote
I think it is (b). Do you know the correct answer?
answered by Junior (817 points)
–1 vote
all of them differ by 3 bits if taken in same order ....

then how it is 2
answered by (49 points)
+1
we have to choose the minimum distance by performing an Ex-OR operation between all pair of valid codes. now we have to select which pair has given the minimum number of 1's after Ex-OR operation. count that minimum number of 1's and that will be the distance. in this case,   010101  ⊕  011001  =  001100 number of 1's is 2 so HD is 2,


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