GATE1994-9

Question

Following $7$ bit single error correcting hamming coded message is received.

$$\overset{7\qquad 6\qquad 5 \qquad 4\qquad 3 \qquad 2 \qquad 1}{\boxed{1 \qquad 0\qquad 0 \qquad 0 \qquad 1  \qquad 1 \qquad 0}} \qquad \overset{\textbf{bit No.}}{\boxed{X}}$$

Determine if the message is correct (assuming that at most $1$ bit could be corrupted).
If the message contains an error find the bit which is erroneous and gives correct message.

Answer 1

Here, answer is yes. There is error in This message. Error is in bit $6$.

How to calculate it? First of all reverse given input to get it in correct position from $1$ to $7$.

$0110001$

$\text{Bit-1, Bit-2 & Bit-4}$ are partity bits.

Calculating position of error $\Rightarrow$

$\text{$\large c_4 c_2 c_1$}$

$\text{1   1   0}$

Here, $c4 = bit4\oplus bit5 \oplus bit6 \oplus bit7=0\oplus 0\oplus 0\oplus 1=1$
                   $\text{ (Taking Even parity )}$

          $c2  = bit2\oplus bit3\oplus bit6\oplus bit7=1\oplus 1\oplus 0\oplus 1=1$

          $c1= bit1\oplus bit3\oplus bit5\oplus bit7=0\oplus 1\oplus 0\oplus 1=0$

Reference: $\Rightarrow$ https://en.wikipedia.org/wiki/Hamming%287,4%29

When you correct bit $6$.

You get message as $0110011$.

If you calculate $\large c_4,c_2,c_1$ all will be $0$ now!

Comments

http://codegolf.stackexchange.com/questions/45684/correct-errors-using-hamming7-4

See example here..

good read if u want to grasp it completely.

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@Rajesh This link is so "Big" to read :P

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if assume odd parity then bit 1 has error ,it should be zero

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by default which one to choose odd or given ?

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The bits at which correct bits depend, I think is given wrong!!

$C_1: 1\oplus3\oplus5\oplus7 $

$C_2:2\oplus3\oplus6\oplus7$

$C_4:4 \oplus5\oplus6\oplus7$

C1 is like begin from data bit 1, take 1 bit, leave 1 bit, then take 1 bit and so on.

C2 is like from bit position 2 take 2 bits, then leave next two bits, then take next two bits and so on....

Easy and nice to read with a good Example

 

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

 

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if we take the odd parity then bit 1 should be corrected

not the bit 6

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Yes you are correct. But if nothing is mentioned in the question, then we take Even parity by default.

Answer 2

Error bit is 6 ... Msg is 1100110

Answer 3

Yes there is an error in this message at 6-bit position so the correct message is 1100110

Answer 4

6th bit is having error both parity bit 2 and 4 support this so 6 bit need to be 1 and data must be 1100110

Answer 5

Number the bits as given in the question :

Now take only those number in which bit is 1 ( in this case 7, 3, 2). Write them in binary and apply XOR. If we get 0 there is no error. If we get a number that is the bit where error is present.

7 – 111

3 – 011

2 – 010

Result : 110( which is 6).

So error is present in bit 6.

Answer 6

Answer: Bit 5 has the error. It should be 1.

Here, parity bits are 1,2,4 (power of 2).

Hamming code based on data bits received = 110
Bit 1 = Bit 3 $\oplus$ Bit 5 = 1 $\oplus$ 0 = 1. (XOR of all bits having the least significant digit as 1)
Bit 2 = Bit 3 $\oplus$ Bit 6 = 1 $\oplus$ 0 = 1. (XOR of all bits having the second least significant digit as 1)
Bit 7 = Bit 4 $\oplus$ Bit 5 $\oplus$ Bit 6 = 0 $\oplus$ 0 $\oplus$ 0 = 0. (XOR of all bits having the third least significant digit as 1)

But the received message has Bit 1 = 0, Bit 2 = 1, Bit 7 = 1 (Actual hamming code)

Bit error = Hamming code based on data bits received $\oplus$ Actual hamming code = 110 $\oplus$ 011 = 101 (i.e. Bit 5 has the error)

Comments

the answer is wrong.it should be bit 6.if you consider forouzan's book of data networking, the answer should be bit 6.then the correct message will be 1100110.

her the parity bits are bit 1,bit 2, bit 4.

now to calculate bit 1=bit1⊕bit3⊕bit5⊕bit7=0⊕1⊕0⊕1=0

again,to calculate bit 2=bit2⊕bit3⊕bit6⊕bit7=1⊕1⊕0⊕1=1

again,to calculate bit 3=bit4⊕bit5⊕bit6⊕bit7=0⊕0⊕0⊕1=1

therefore ,the error is at the position 110 = 6 

please ,think of it mr @jon_snow.

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Sudipta Paul

Your answer is correct !

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is it correct?

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yes it is

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Superb way to solve! @Sudipta Paul By this we can get the answer in just one go.

Thank u !