1. Practical Session 10 Computer Architecture and Assembly Language

2. Error detection and correction channel noise data transmission via communication channels  channel noise  errors may happen reliable delivery of digital data over unreliable communication channels we need techniques that enable reliable delivery of digital data over unreliable communication channels redundancyextra data 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 – to recover data determined to be corrupted

3. Hamming Distance Hamming distance (d) amount of 1-bit changes 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) 000 011 Minimum Hamming distance smallest Hamming distance all possible pairs 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 d(000, 011) = 2 d(10101, 11110) = 3 10101 11110

4. 000000100001010 Example: 00000 01011 10101 11110 3 errors can take us from one legal code word to another 01011 Minimum Hamming Distance d can detect d-1 errors d=3 can detect at least 3-1=2 errors Minimum Hamming Distance d can fix d-1 erasures 000000_0000_0_0 3 erasures can take us from one legal code word to another 0_0__ 01011 can detect at least 3-1=2 erasures 2 errors can take us from two different legal code words to the same illegal word Minimum Hamming Distance d can fix errors 0000001000 1111001110 01010 can fix at least floor((3-1)/2)=1 error

5. 00000000010001000100 01000 10000 11100 11000 10100 0110011110 11101 10111 10110 10101 1001111111 00111 01011 01010 01001 0111100011 11011 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 independent graphs of a single error code words We can create code words with independent graphs of a single error code words. 00000 11100 10111 01011

6. Parity Check 00011 00111 parity bit Given a data word, a code word is created by adding a parity bit to the end of the data word d=2 d=2 (why?) Can detect 1 error and fix 1 Can detect 1 error and fix 1 erasure send: receive: 1 error occurs Code word Data word 0 000000000 1 000110001 1 001010010 0 001100011 00_11 erased bit is a parity of other bits: parity (0011) = 1

7. Hamming Code (4,3) We calculate: 4 data bits: x0 x1 x2 x33 parity bits: p0 p1 p2 We assume that there may be at most one error. Example: We got 0100111 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  The error bit is at index 6 (from right to left) in the code word  error bit is x2 p2 p1 p0 =xx√ = 110 b = 6 12 code word: indexes: 34567 why we interleave bits of data and parity in this way ? if p0 is wrong, we get p2 p1 p0 =√√x = 001 b = 1  p0 should be at index 1 if p1 is wrong, we get p1 p1 p0 =√x√ = 010 b = 2  p1 should be at index 2 if p2 is wrong, we get p2 p1 p0 =x√√ = 100 b = 4  p2 should be at index 4 if x0 is wrong, we get p2 p1 p0 =√xx = 011 b = 3  x0 should be at index 3 … if x3 is wrong, we get p2 p1 p0 =xxx = 111 b = 7  x3 should be at index 7 Sender sent 0000111

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)

9. Binary Pattern as Polynomial

10. CRC Polynomials (common) CRC-8-ATM x 8 + x 2 + x + 1 CRC-16-IBM x 16 + x 15 + x 2 + 1 CRC-16-CCITT x 16 + x 12 + x 5 + 1 CRC-32-IEEE 802.3 x 32 + x 26 + x 23 + x 22 + x 16 + x 12 + x 11 + x 10 + x 8 + x 7 + x 5 + x 4 + x 2 + x + 1

11. Sending – Calculation Steps 1.Generator 1.Generator is chosen – sequence of bits, of which the first and last are 1 2.CRC check sequence is computed 2.CRC check sequence is computed by long- division of message by generator – check sequence has 1 fewer bits than generator appended 3.Check sequence is appended to the original message

12. Sending – Calculation Steps Compute 8-bit CRC a message ‘W’ (0x57). 1.Select Generator: CRC-8-ATM 1.x 8 + x 2 + x + 1 = 100000111 2.‘W’ is 01010111= x 6 + x 4 + x 2 + x + 1 3.Extend ‘W’ with 8 bits: 0101011100000000 4.Perform XOR of the word get in step (3) by generator – CRC code is the remainder 5.Append CRC code to ‘W’

13. Generating CRC Code Generating CRC Code long-division of message by generator 0101011100000000100000111 10100010 CRC Code: 10100010 100000111 001011011000000 100000111 0011010110000 100000111 01010101100 100000111 0010100010 xor Each time apply XOR on the extended message, when place a generator from the left-most ‘1’ bit of the extended message Stop when CRC code has 1 fewer bits than generator

14. Receiving – Calculation Steps Append CRC code to the message: 0101011110100010 Perform long division by the generator reminder is not 0 If the reminder is not 0: an error occurred 0101011110100010100000111 result: 00000000 100000111 xor There is no errors in received message.