Below is a collection of Sbit words ri l 15 received after

Below is a collection of S-bit words {r_i| l = 1.5} received after being encoded with Hamming Code II using the matrix G2 (see Lecture 5). For each word assume there could be only 0.1 or 2 errors. For each of these words, determine how many errors happened during the transmittion. For those words that have been transmitted without an error or with 1 error, correct the word if needed and recover the encoded 4 -bit original (meaningful) sequences w_1 by eliminating the parity bits from the 8-bit For those words >^, that have been received with 2 errors, show that the correct 8-bit words cannot be recovered unambiguously by identifying 2 (or more) different 8-bit words that would produce the same received (incorrect) word after 2 errors. Explain your solution.

Solution

Hamming codes are a class of codes that use a combination of paritybits and data bits to detect and correct the errors of the given codes.

In the given question we have 5 8-bit codes.

r1= 1 0 1 0 1 1 1 0

Now we create the data word for r1 as follows:

1. The word with parity bits will be : _ _ 1 _ 0 1 0 _ 1 1 1 0

2⁰   2¹ 2² 2³

P₁ P₂ D₃ P₄ D₅ D₆ D₇ P₈

This is in binary digit form, we consider it as 2⁰   2¹ 2² 2³  and P₁ P₂ D₃ P₄ D₅ D₆ D₇ P₈ indicate parity and data bits.

2. Each and every parity bit gives the parity for some of the bits in the code and this can be calculated by the position of the parity bit and the sequence of bits it checks and skips alternatively.

For 1st position : It check 1 bit and skip 1 bit alternatively ie

X _ 1 _ 0 1 0 _ 1 1 1 0

Here it checks the letters in bold and skips the normal letters. This is even parity so we set 1st position to 0.

0 _ 1 _ 0 1 0 _ 1 1 1 0

For 2nd position: It checks 2 bits and skips 2 bits alternatively ie

0 X 1 _ 0 1 0 _ 1 1 1 0

Here it checks the letters in bold and skips the normal letters. This is odd parity so we set 2nd position to 1.

0 1 1 _ 0 1 0 _ 1 1 1 0

For 4th position: It checks 4 bits and skips 4 bits alternatively ie

0 1 1 X 0 1 0 _ 1 1 1 0

This is odd parity so we set 4th position to 1.

0 1 1 1 0 1 0 _ 1 1 1 0

For 8th position:

0 1 1 1 0 1 0 X 1 1 1 0

This is again even parity so we set it to 0.

The code word thus obtained is 0 1 1 1 0 1 0 0 1 1 1 0.

In the similar way we can calculate for rest of the words from r2 to r5.