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The hamming distance between the octets of $\textsf{0xAA}$ and $\textsf{0x55}$ is

1. $7$
2. $5$
3. $8$
4. $6$

### 1 comment

I think its option C 8

By performing XOR between 2 valid codeword will give another valid codeword,the no of 1's in the valid codeword will indicate hamming distance(d).

0xAA=1010 1010

and 0x55=0101 0101

performing XOR between them we get Hamming distance(d)=no of 1's in (1111 1111)=8

Why did you write 0 as 1010 and AA as 1010?

number starting from $0x/0X$ is the hexadecimal number.$1010 1010$ is the binary representation of $AA.$

U didn't wrote 0 in binary form @Hira Thakur

OxAA = 1010 1010

Ox55  =  0101 0101

As all the 8 pairs are different , so minimum hamming distance is 8.

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Anser is [c] 8

the Hamming distance between two strings of equal length is the number of positions at which the corresponding symbols are different.

"I do not seek. I find."

A is 10 =1010
edited
Thanks for pointing it out.. @shivanisrivarshini  . ANswer edited..

PERFORM XOR OPERATION BETWEEN EACH BIT TO GET HAMMING DISTANCE

OXAA= 10101010

XOR

OX55=  01010101

ANS =   11111111

THEREFORE HAMMING DISTANCE=8

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