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The hamming distance between the octets of 0xAA and 0x55 is

1. 7
2. 5
3. 8
4. 6
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0
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
by Active (3k points)
selected by

OxAA = 1010 1010

Ox55  =  0101 0101

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

by Veteran (50.9k points)

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."

by Loyal (8.1k points)
edited
0
A is 10 =1010
+1
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

by Boss (11.1k points)