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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 _______.

@habibkhan , @Arjun
check this

hope it is clear.

To calculate the numbers of redundant bits (r) required to correct d data bits, let us find out the relationship between the two. So we have (d+r) as the total number of bits, which are to be transmitted; then r must be able to indicate at least d+r+1 different values. Of these, one value means no error, and remaining d+r values indicate error location of error in each of d+r locations. So, d+r+1 states must be distinguishable by r bits, and r bits can indicates 2 r states. Hence, 2 r must be greater than d+r+1.

$2^{r}$ $\geqslant$ d+r+1

Put d=4 then what is the value of r ?

r=3 in order to satisy the condition.

edited

### Minimum Hamming distance for error detection

To design a code that can detect d single bit errors, the minimum Hamming distance for the set of codewords must be d + 1 (or more). That way, no set of d errors in a single bit could turn one valid codeword into some other valid codeword.

### Minimum Hamming distance for error correction

To design a code that can correct d single bit errors, a minimum distance of 2d + 1 is required. That puts the valid codewords so far apart that even after bit errors in d of the bits, it is still less than half the distance to another valid codeword, so the receiver will be able to determine what the correct starting codeword was.

Redundant bits in hamming code:

2^r >= m + r + 1 , where m = redundant bits, r = data bits

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

To detect $d$ errors we require 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 bits. So, $3$ parity bits requires here.

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

According to $1^{\text{st}}$ question, $d=4$  so $4+p+1\leq 2^p$

$p+5 \leq 2^p$  now if  p$=2$ it becomes $7 \leq 4,$ Not possible.

If $p=3$ it becomes $8\leq 8,$ which is possible.

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 is $d+1.$

The answer for $2$ blanks is $[ 3,4 ].$

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.

@Hemant Parihar  No there is no such rule

@Bikram Sir the "d" bits you used to represent the errors and the "d" used for data bits..they aren't same, right? It might create confusion.

can you please check if the minimum hamming distance of two coded words is 4 ,since i can think of a counter example like M1:0000 encoded to E1:0000000 and M2:0010 encoded to E2:0101010(consider left to right message coding) ,here the hamming distance is only 3 which is not sufficient for 3 bit error detection

@Bikram  "It can be shown that a single parity check code can detect only odd number of errors in a code word."

Doesn't it mean that a single parity bit can be used to detect odd no. of errors and hence 3 bit error?

Which would make your statement "For 3 bit bit error detection we need 3 parity bits" false?

For 3 bit error detection hamming distance should be 4 but if we use parity bit 3 then hamming distance is 3 not 4
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 $=p$
$p + t + 1 <= 2^p$
$p + 4 +1 <= 2^p$
$p=3$

It is 3 bit error detection and 1 bit error correction.
reshown by

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 ?
edited by
@Akash No, ans is correct.

and min distance will be 1 bit error correction

http://courses.cs.vt.edu/cs2506/Spring2009/Notes/Lecture7.pdf
Shouldn't it be,  p+d+1<= 2^p

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.

by

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.
I think this formula is right as it considers minimum hamming distance also.

I think the question is wrong because Hamming codes cannot detect arbitrary 3-bit errors.

Intuition : Suppose that we flip the first three bits (first two bits are redundancy bits and the third bit is a message bit), then the error will go undetected in any version of the Hamming code.

Further explanation : Suppose that we are using Hamming code (7, 4) and the codeword is [1R, 2R, 3M, 4R, 5M, 6M, 7M]. Flipping 1R, 2R and 3M will never be detected. This is true for any arbitrary Hamming code as the first message bit affects only the first two redundancy bits. So maximum distance of any Hamming code can be 2. (Note that maximum distance of any two Hamming codewords can be greater than 2. But distance of a code is defined as the minimum distance of any two valid codewords in it.)

... Hamming codes cannot detect or recover from an arbitrary three-bit error