Hamming Distance Question

Question

Comments

@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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answer should be 1, 0.

here it is not correct the error because min hamming distance is already 1.

Answer 1

code words :                                -Let minimum hamming distance be p 

10101                              -so we have to apply XOR operation (same value =0 and different value =1) and then count   

11011                                             the number of '1' that will be your min.hamming distance.                    

10111                                            -for example :   10101

10001                                                                     10111  

                                                                     ________________

                                                                   (XOR)  00010

                                                                     ________________

                                                     - now count the number of '1' that is x =1. That means the minimum hamming distance is 1

 

                                                     - For maximum errors that can be detected : - 

                                            - maximum hamming distance  for example : -      10101

                                                                                                                                          11011

                                                                                                                                       __________

                                                                                                                       (XOR)     01110 

                                                                                                                                        __________

                                                   - now count the number of '1' that is q =3.That means the maximum hamming distance is 3.

                                                   - no.of errors detected = [d-1]  = 3-1 = 2.

                    So the final answer will be (min.hamming distance , max no.of errors detected) =  (1,2)     

Answer 2

Answer should be (A)

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

Comments

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?