Hamming Code

Question

Comments

11..?

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FOr odd parity I also got error at 11th bit

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yes me too was gtting 11th but ans is given as 4th

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can anyone explain how to solve this question..?

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@Ashutosh ,here ,For odd parity ,Check bits are :-

C₁ = (8,9,10,11)  = (0,1,1,0) = 1

C₂ = (4,5,6,7) = (0,0,1,0) = 0

C₃ = (2,3,6,7,10,11) = (1,1,1,0,1,0) = 1

C₄ = (1,3,5,7,9,11) =(0,1,0,0,1,0) = 1

So, bit error position will be C₁C₂C₃C₄ = 1011 = 11 

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@ ankitgupta.1729

C for carry bit?

Can u tell me the calculation part?

It is a single string message, that is why getting problem

chk here http://web.mit.edu/6.02/www/f2006/handouts/bits_ecc.pdf

but they are more than single string

plz elaborate

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@srestha,  sorry , I m not getting ur.question...C is error correction bit here..

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@ ankitgupta.1729
why for C1 u r calculating  8,9,10,11?
plz explain in detail
not getting details in wikipedia too :(
https://en.wikipedia.org/wiki/Parity_bit

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@srestha ,  check it here :- https://www.youtube.com/watch?v=wbH2VxzmoZk

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@ srestha  you can refer to this https://en.wikipedia.org/wiki/Hamming_code

and http://logos.cs.uic.edu/366/notes/ErrorCorrectionAndDetectionSupplement.pdf

This hamming codes problems are explained in different ways and i have often seen them conflicting with each other. But the method that i now follow has so far helped me to get the answers correct. You can do like this ->

Bits positions:-     1  2  3  4  5  6  7  8  9  10  11

Bits:-                         0  1  1  0  0  1  0  0  1  1    0

c1,c2,c4,c8 are check bits.

c1: bits at positions (1,3,5,7,9,11) = (010010) => even no. of ones but odd parity is required so c1=1.

c2: bits at positions (2,3,6,7,10,11)= (111010) => even no. of ones but odd parity is required so c2=1.

c4:bits at position (4,5,6,7)= (0010) => odd no. of ones so c4=0.

c8: bits at positions(8,9,10,11) => (0110) =>even no. of ones but odd parity is required so c8=1.

Hence c8c4c2c1=1011.

(1011)₂= (11)₁₀

The bit error position is at 11.

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I think ,all things are same ..when we do  numbering on data from right to left then in the end (explained in video) , we have to do c₄c₃c₂c₁ ...And if we do numbering from left to right then in the end , we have to do c₁c₂c₃c₄...

As mini panda did numbering for  c₁, c₂, c₃ ,c₄ in reverse way ..so he has taken c₄c₃c₂c₁ for error correction..

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Yeah the numbering creates confusion :P

Answer 1

c1 c2 c3 c4 = (1 0 1 1) = 11 so the error are at the position of 11 bit .as more than 8 bit are there so we need atleast 4 parity from the formula   2^p >= m+p+1 where the m= no of message bit , p= #parity bit and 1 i have included in this for handling the case of no error.

as we got the 4 parity bit we get 2^4 = 16 (0-15)  combination but we need only 12 combination (0-11)  . Let me draw the table so that u will get better understanding.f

p4	p3	p2	p1
check bit 1	check bit 2	check bit 3	check bit 4
0	0	0	0
0	0	0	1
0	0	1	0
0	0	1	1
0	1	0	0
0	1	0	1
0	1	1	0
0	1	1	1
1	0	0	0
1	0	0	1
1	0	1	0
1	0	1	1

 

Here P4 bit is firstly arrived at the side of the receiver that why we have taken the parity p4 bit as check bit 1 .

And in every column the bit  1 is the indication that check bit will take care of that bit position on the actual message and here it is said that it for odd parity .

now here c1 will take care of 8 9 10 11 so what u all need to do is just try to see the how many number of  ones is available on that all position it should counted for odd number of times if the number of 1 present is even then we have to put 1 for check bit 1(c1) similarly try to find the for all .

U will endup with this c1 c2 c3 c4 = 1 0 1 1 = 11 so the error is happened at the bit position 11.