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none of these options are correct
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Minimum Hamming Distance is 1 &

Maximum Hamming Distance is 3
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Option should be (A)
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minimum hamming distance is 1.
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Yes, minimum Hamming distance is 1, but according to question answer should be A.
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how?
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hamming distance between 10101 and 10001 is 1
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ya bro I know.
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@thepeeyoosh can you please explain how A is correct ?

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four valid code words? can someone please explain me, what they really meant by the valid code words.
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here it is not correct the error because min hamming distance is already 1.

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Okk let me tell few points!

"Parity Scheme", it is a error detection technique. It can detect only odd number of errors. I.e. if noise modified even number of errors this parity Scheme will not work.( Drawback of parity Scheme)

Now to detect errors we use the Hamming distance.

Rule:- "To detect 'd' errors the minimum Hamming distance should be 'd+1'" (why? You can read from some standard book)

So in this question you can see

Hamming distance as follows

(I) 10101 & 11011 = 3 [ I assume you know how to calculate Hamming distance].

(II) 11011 & 10111 = 2 ... So on

One more

(III) 10101 & 10001 = 1

But we can see here 10101 is invalid code word. If noise modified the valid code word to invalid code word it easily detect by receiver.

So here you can see minimum Hamming distance is 2 ( as per option of the question) so by applying rule to detect errors of 1 bit

So p= 2 & q= 1
by Active (2k points)
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But we can see here 10101 is invalid code word. If noise modified the valid code word to invalid code word it easily detect by receiver.

How?

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In parity scheme, codeword is generated by XOR gate,

X xor Y = parity bit (we know that XOR gate is also called odd function)

Now check in 10101 contain (1 xor 0 xor 1 xor 0) = 0 but question contain 1 which leads to invalid codeword.
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So they shoulnt be mentioning that there are 4 valid codeword?

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