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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 ____
edited | 1.9k views
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@habibkhan , @Arjun
check this
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hope it is clear.
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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 ]

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Can u please explain the reason behind the formulas d+1 and 2*d+1
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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.

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 $=p$
$p + t + 1 <= 2^p$
$p + 4 +1 <= 2^p$
$p=3$
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It is 3 bit error detection and 1 bit error correction.
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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

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

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

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

and min distance will be 1 bit error correction

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

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 -->

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.

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I think there is misinterpretation here.The code requires to satisfy both those properties.3 parity bits won't suffice for this(if only ensures 1 bit correction).You could easily create a counter-case.I think 5 parity bits are enough.Will post the formal proof soon.