1. Computer Architecture and Assembly Language
Practical Session 10

2. Error detection and correction
data transmission via communication channels channel noise  errors may happen
we need techniques that enable reliable delivery of digital data over unreliable communication channels
general idea: add some redundancy (extra data) to a message
this extra data is used
to check consistency of the delivered message
to recover data determined to be corrupted 

3. Hamming Distance
Hamming distance (d) between two code words of equal length is an amount of 1-bit changes required to reach from one word to the other (number of 1’s in the XOR between the words)
d(000, 011) = 2
d(10101, 11110) = 3
Minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words.
Example:
Given the following coding scheme, let’s find the minimum Hamming distance.
00000
01011
10101
11110
d(00000,01011)=3
d(00000,10101)=3
d(00000,11110)=4
d(01011,10101)=4
d(01011,11110)=3
d(10101,11110)=3
d=3

4. Minimum Hamming Distance d can detect d-1 errors
Example:
00000
01011
10101
11110
3 errors can take us from one legal code word to another
can detect at least 3-1=2 errors
d=3
01011
00000
01000
01010
01011
Minimum Hamming Distance d can fix d-1 erasures
3 erasures can take us from one legal code word to another
can detect at least 3-1=2 erasures
01011
00000
0_000
0_0_0
0_0__
Minimum Hamming Distance d can fix errors
2 errors can take us from two different legal code words to the same illegal word
00000
01000
01010
can fix at least floor((3-1)/2)=1 error
11110
01110

5. Answer: yes. We need a hamming distance d=3 to fix 1 error.
Given four data words, can we use 5-bit code words for fixing 1 error?
Answer: yes. We need a hamming distance d=3 to fix 1 error.
d(00000,11100)=3 d(11100, 10111)=3
d(00000,10111)=4 d(11100, 01011)=4
d(00000,01011)=3 d(10111, 01011)=3
We can create code words with independent graphs of a single error code words.
00000
00001
00010
00100
01000
10000
11100
11000
10100
01100
11110
11101
10111
10110
10101
10011
11111
00111
01011
01010
01001
01111
00011
11011

6. Parity Check
Given a data word, a code word is created by adding a parity bit to the end of the data word
d=2 (why?)
Can detect 1 error and fix 1 erasure
Code word
Data word
00000
0000
00011
0001
00101
0010
00110
0011
0001 1
send:
0011 1
00_1 1
receive:
1 error occurs
erased bit is a parity of other bits: parity (0001) = 1

7. Hamming Code (4,3) We assume that there may be at most one error.
if p0 is wrong, we get p2 p1 p0 =√√x = 001b = 1 p0 should be at index 1
if p1 is wrong, we get p1 p1 p0 =√x√ = 010b = 2 p1 should be at index 2
if p2 is wrong, we get p2 p1 p0 =x√√ = 100b = 4 p2 should be at index 4
if x0 is wrong, we get p2 p1 p0 =√xx = 011b = 3 x0 should be at index 3 …
if x3 is wrong, we get p2 p1 p0 =xxx = 111b = 7 x3 should be at index 7
We assume that there may be at most one error.
indexes: 7 6 5 4 3 2 1
code word:
4 data bits: x0 x1 x2 x3
3 parity bits: p0 p1 p2
We calculate:
why we interleave bits of data and parity in this way ?
Example:
We got
Sender sent
p0 = 1 = x3 ⊕x1 ⊕x0 = 0 ⊕ 0 ⊕ 1 = 1  √
p1 = 1 = x3 ⊕x2 ⊕x0 = 0 ⊕ 1 ⊕ 1 = 0  x
p2 = 0 = x3 ⊕x2 ⊕x1 = 0 ⊕ 1 ⊕ 0 = 1  x
p2 p1 p0 =xx√ = 110b = 6
 The error bit is at index 6 (from right to left) in the code word  error bit is x2

8. Cyclic Redundancy Check
CRC is an error-detecting code
Given a data, its non trivial (cryptographic) challenge to find another data with the same CRC code.
 CRC can protect data (even from deliberate
damage)
Explanation is needed

9. Binary Pattern as Polynomial

10. CRC Polynomials (common)
CRC-8-ATM
x8 + x2 + x + 1
CRC-16-IBM
x16 + x15 + x2 + 1
CRC-16-CCITT
x16 + x12 + x5 + 1
CRC-32-IEEE 802.3
x32 + x26 + x23 + x22 + x16 + x12 + x11 + x10 + x8 + x7 + x5 + x4 + x2 + x + 1
which one is used when?

11. Sending – Calculation Steps
Generator is chosen
sequence of bits, of which the first and last are 1
CRC check sequence is computed by long-division of message by generator
check sequence has 1 fewer bits than generator
Check sequence is appended to the original message

12. Sending – Calculation Steps
Compute 8-bit CRC a message ‘W’ (0x57).
Select Generator: CRC-8-ATM
x8 + x2 + x + 1 =
‘W’ is = x6 + x4 + x2 + x + 1
Extend ‘W’ with 8 bits:
Perform XOR of the word get in step (3) by generator
CRC code is the remainder
Append CRC code to ‘W’

13. Generating CRC Code long-division of message by generator
Each time apply XOR on the extended message, when place a generator from the left-most ‘1’ bit of the extended message
xor
xor
xor
xor
Stop when CRC code has 1 fewer bits than generator
CRC Code:

14. Receiving – Calculation Steps
Append CRC code to the message:
Perform long division by the generator
If the reminder is not 0: an error occurred
xor
only a single error occurred? why?
There is no errors in received message.
result: