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Consider a $3$-bit error detection and $1$-bit error correction hamming code for
$4$-bit data. The extra parity bits required would be ___ and the $3$-bit error
detection is possible because the code has a minimum distance of ____
asked in Computer Networks by Veteran (68.8k points)
edited by | 1.7k views
@habibkhan , @Arjun
check this
see my answer @anirudh

hope it is clear.

3 Answers

+9 votes
Best answer

The Hamming distance between two bit strings is the number of bits that would have
 to flip to make the strings identical.

To  detect d errors requires a minimum Hamming distance of $d + 1$.

Correcting d bit flips requires a minimum Hamming distance of $2\times d + 1,$
where $d$ is number of bit in errors .

 For the first blank , each error detection we need $1$ parity bit

for $3\ bit$ error detection we need $3$ parity bit $\ldots\ldots$ So, $3$ parity bit
requires here. answer is $3$.

Also we can calculate this way, formula is $d+p+1 <= 2^p$
where $\text{d=data bits , p = parity bits , d=4 bit given}$.

according to $1^{st}$ question, d$=4$  so $4+p+1<= 2^p$ 

$p+5 <= 2^p$  now if  p$=2$ it becomes 7 <= 4 , Not satisfy . p$=3$
it becomes $8<= 8,$ satisfy.

  So $p$ must be $3.$[ Minimum value of $p$ is $3$ ]

The second blank the $3$-bit error detection is possible because the code has
a minimum distance of $\underline{\qquad}$answer is $3+1=4,$ where d$=3.$ Formula used $d+1.$

Answer for 2 blanks are  [ 3,4 ]

answered by Veteran (67k points)
edited by
Can u please explain the reason behind the formulas d+1 and 2*d+1

Can someone give any resources to confirm this line? Thank you..!!

each error detection we need 1 parity bit 
for 3 bit error detection we need 3 parity bit  ......  So, 3 parity bit requires here. answer is 3.

+14 votes
let minimum Hamming distance is $t.$
so with this hamming distance $t-1\ bit$ error detection as well as $\dfrac{ (t-1)}{2}$
bit error correction is possible..

for $3\ bit$ error detection minimum Hamming distance $=3+1=4$
for $1\ bit$ error correction  minimum Hamming distance $=2\times1+1= 3$
no. of parity bits $=p$
$p + t + 1 <= 2^p$
$p + 4 +1 <= 2^p$
$p=3$
answered by Veteran (49.1k points)
edited by
It is 3 bit error detection and 1 bit error correction.
check answer now..

Hamming Rule , parity bit "p"

d + p + 1 < = 2 ,  Where d is the number of data bits and p is the number of parity bits 

hamming minimum distance , if d bit is the detection bit = d + 1 bit ,

...................is correct....??? 

http://cs.stackexchange.com/questions/32025/hamming-distance-required-for-error-detection-and-correction http://courses.cs.vt.edu/cs2506/Spring2009/Notes/Lecture7.pdf

http://iete-elan.ac.in/SolQP/soln/soln_ac07.pdf http://en.wikipedia.org/wiki/Hamming_code 

                               

 

yes..
How is detection and correction bits related to data bits Plz explain

Standard formula is 2 power p >=p+m+1
for 1 bit error correction  minimum Hamming distance = 2*1+1 =4 ???

how come 2*1 + 1 became 4 here ?
@Akash No, ans is correct.

and min distance will be 1 bit error correction

http://courses.cs.vt.edu/cs2506/Spring2009/Notes/Lecture7.pdf
0 votes

Now, (binary) Hamming Codes are usually depicted using the following notation: (n,k,d). Here, n is the number of bits in the code-word, k is the number of information/data bits, and d is again the Minimum Hamming Distance. The Hamming Bound for this code is a mathematical relation between n,k and d. The code can exist if and only is the Hamming Bound is satisfied.

Refer --> 

Harsh Aurora's answer on https://www.quora.com/How-many-check-bits-are-required-for-16-bit-data-word-to-detect-2-bit-errors-and-single-bit-error-correction-using-hamming-code

Now n = ? , k = 4, d = 3+1 (because this much minimum distance is required for 1 bit correction and 3 bit error detection)

so putting all values in the formula you will get 

$2^{n-k} \geqslant \binom{n}{0} + \binom{n}{1}$

n = 7. so extra parity bits are n-k = 3.

So final answer is [3,4]

answered by Veteran (11.7k points)


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