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

1. $2$
2. $3$
3. $4$
4. $1$
edited | 1k views

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

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@ Arjun sir,

codes 2 & 4 also has distance 2. right?

010101

XOR

011001

= 001100
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@Arjun Sir, do we need to check each codeword distance to every other codeword, or just go sequentially?

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every code with every other code.
(A) Distance = minimum hamming distance = $2$.

$010101 ⊕ 011001 = 001100$
edited

Check the solution

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

then how it is 2
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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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