1. EECS 122: Introduction to Computer Networks Error Detection and Reliable Transmission
Computer Science Division
Department of Electrical Engineering and Computer Sciences
University of California, Berkeley
Berkeley, CA

2. High Level View Goal: transmit correct information
Problem: bits can get corrupted
Electrical interference, thermal noise
Solution
Detect errors
Recover from errors
Correct errors
Retransmission  already done this

3. Overview
Error detection & recovery
Reliable Transmission

4. Error Detection Problem: detect bit errors in packets (frames)
Solution: add extra bits to each packet
Goals:
Reduce overhead, i.e., reduce the number of redundancy bits
Increase the number and the type of bit error patterns that can be detected
Examples:
Two-dimensional parity
Checksum
Cyclic Redundancy Check (CRC)
Hamming Codes

5. Two-dimensional Parity
Add one extra bit to a 7-bit code such that the number of 1’s in the resulting 8 bits is even (for even parity, and odd for odd parity)
Add a parity byte for the packet
Example: five 7-bit character packet, even parity 1 1 1 1

6. How Many Errors Can you Detect?
All 1-bit errors
Example: 1
odd number of 1’s 1
error bit 1 1

7. How Many Errors Can you Detect?
All 2-bit errors
Example: 1 1
error bits 1 1
odd number of 1’s on columns

8. How Many Errors Can you Detect?
All 3-bit errors
Example: 1 1
error bits 1 1
odd number of 1’s on column

9. How Many Errors Can you Detect?
Most 4-bit errors
Example of 4-bit error that is not detected: 1 1
error bits 1 1
How many errors can you correct?

10. Checksum
Sender: add all words of a packet and append the result (checksum) to the packet
Receiver: add all words of a packet and compare the result with the checksum
Can detect all 1-bit errors
Example: Internet checksum
Use 1’s complement addition

11. 1’s Complement Revisited
Negative number –x is x with all bits inverted
When two numbers are added, the carry-on is added to the result
Example: ; assume 8-bit representation
15 =  -15 =
= 1 +
16 = 1 + 1

12. Cyclic Redundancy Check (CRC)
Represent a (n+1)-bit message as an n-degree polynomial M(x)
E.g.,  M(x) = x7 + x5 + x3 + x2 + x0
Choose a divisor k-degree polynomial C(x)
Compute reminder R(x) of M(x)*xk / C(x), i.e., compute A(x) such that
M(x)*xk = A(x)*C(x) + R(x), where degree(R(x)) < k
Let
T(x) = M(x)*xk – R(x) = A(x)*C(x)
Then
T(x) is divisible by C(x)
First n coefficients of T(x) represent M(x)

13. Cyclic Redundancy Check (CRC)
Sender:
Compute and send T(x), i.e., the coefficients of T(x)
Receiver:
Let T’(x) be the (n+k)-degree polynomial generated from the received message
If C(x) divides T’(x)  no errors; otherwise errors
Note: all computations are modulo 2

14. Arithmetic Modulo 2
Like binary arithmetic but without borrowing/carrying from/to adjacent bits
Examples:
Addition and subtraction in binary arithmetic modulo 2 is equivalent to XOR
101 +
010
111
101 +
001
100
1011 +
0111
1100
101 -
010
111
101 -
001
100
1011 -
0111
1100 a b
a b 1

15. Some Polynomial Arithmetic Modulo 2 Properties
If C(x) divides B(x), then degree(B(x)) >= degree(C(x))
Subtracting/adding C(x) from/to B(x) modulo 2 is equivalent to performing an XOR on each pair of matching coefficients of C(x) and B(x)
E.g.:
B(x) = x7 + x5 + x3 + x x0 ( )
C(x) = x x1 + x0 ( )
B(x) - C(x) = x7 + x x2 + x ( )

16. Example (Sender Operation)
Send packet ; choose C(x) = 101
k = 2, M(x)*xK 
Compute the reminder R(x) of M(x)*xk / C(x)
Compute T(x) = M(x)*xk - R(x)  xor 1 =
Send T(x)
101)
101
111
100 1
R(x)

17. Example (Receiver Operation)
Assume T’(x) =
C(x) divides T’(x)  no errors
Assume T’(x) =
Reminder R’(x) = 1  error!
101)
101
110
111 1
R’(x)
Note: an error is not detected iff C(x) divides T’(x) – T(x)

18. CRC Properties
Detect all single-bit errors if coefficients of xk and x0 of C(x) are one
Detect all double-bit errors, if C(x) has a factor with at least three terms
Detect all number of odd errors, if C(x) contains factor (x+1)
Detect all burst of errors smaller than k bits

19. Code words
Combination of the n payload bits and the k check bits as being a n+k bit code word
For any error correcting scheme, not all n+k bit strings will be valid code words
Errors can be detected if and only if the received string is not a valid code word
Example: even parity check only detects an odd number of bit errors

20. Hamming Distance
Given code words A and B, the Hamming distance between them is the number of bits in A that need to be flipped to turn it into B
E.g., H(011101,000000) = 4
If all code words are at least d Hamming distance apart, then up to d-1 bit errors can be detected

21. Error Correction
If all the code words are at least a hamming distance of 2d+1 apart then up to d bit errors can be corrected
Just pick the codeword closest to the one received!
How many bits are required to correct d errors when there are n bits in the payload?
Example: d=1: Suppose n=3. Then any payload can be transformed into 3 other payload strings (e.g., 000 into 001, 010 or 100).
Need at least two extra bits to differentiate between 4 possibilities
In general need at least k ≥ log2(n+1) bits
A scheme that is optimal is called a perfect parity code

22. Perfect Parity Codes Consider a codeword of n+k bits
b1 b2 b3 b4 b5 b6 b7 b8 b9 b10 b11…
Parity bits are in positions 20, 21, 22 ,23 ,24…
A parity bit in position 2h, checks all data bits bp such that if you write out p in binary, the hth place in p’s binary representation is a one

23. Example: (7,4)-Parity Code
n=4, k=3
Corrects one error
log2(1+n) = 2.32  k = 3, perfect parity code
data payload = 1010
For each error there is a unique combination of checks that fail
E.g., 3rd bit is in error,:1000  both b2 and b4 fail (single case in which only b2 and b4 fail) b1 b2 b3 b4 b5 b6 b7
Position 1 10 11
100
101
110
111
Check:b1 x
Check:b2
Check:b4 b1 b2 1 b4
Position 10 11
100
101
110
111
Check:b1
Check:b2
Check:b4

24. Overview
Error detection & recovery
Reliable transmission

25. Reliable Transmission
Problem: obtain correct information once errors are detected
Solutions:
Use error correction codes
E.g. perfect parity codes, erasure codes (see next)
Use retransmission (we have studied this already)
Algorithmic challenges:
Achieve high link utilization, and low overhead

26. Error correction or Retransmission?
Error Correction requires a lot of redundancy
Wasteful if errors are unlikely
Retransmission strategies are more popular
As links get reliable this is just done at the transport layer
Error correction is useful when retransmission is costly (satellite links, multicast)

27. Erasure Codes Content Encoding Transmission Received Decoding Content
n blocks
Encoding
n+k blocks
Received
Transmission
>= n blocks
Content
Decoding
original n blocks
(*Michael Luby slide)

28. Example: Digital Fountain
Use erasure codes
for reliable data
distribution
Intuition:
- Goal is to fill the cup
- Full cup = content recovered
(*Michael Luby slide)

29. Erasure Coding Approaches
Reed-Solomon codes
Complex encoding/decoding algorithm and analysis
Tornado codes
Simple encoding/decoding algorithm
Complexity and theory in design and analysis
LT codes
Simpler design and analysis
(*Michael Luby slide)

30. LT encoding Content Degree Dist. (*Michael Luby slide) Choose 2 random
content symbols
Choose degree
XOR content
symbols 2
Insert header, and send
Degree
Prob 1
0.055
0.0004
0.3 2
0.1 3
0.08 4
100000
Degree Dist.
(*Michael Luby slide)

31. LT Encoding Content Degree Dist. (*Michael Luby slide) Choose 1 random
content symbol
Choose degree
Copy content
symbol 1
Insert header, and send
Degree
Prob 1
0.055
0.0004
0.3 2
0.1 3
0.08 4
100000
Degree Dist.
(*Michael Luby slide)

32. LT Encoding Content Degree Dist. (*Michael Luby slide) Choose 4 random
content symbols
Choose degree
XOR content
symbols 4
Insert header, and send
Degree
Prob 1
0.055
0.0004
0.3 2
0.1 3
0.08 4
100000
Degree Dist.
(*Michael Luby slide)

33. LT Decoding Content (unknown)
Collect enough encoding symbols and set up graph between encoding symbols and content symbols to be recovered
Identify encoding symbol of degree 1. STOP if none exists
3. Copy value of encoding symbol into unique neighbor, XOR value of newly recovered content symbol into encoding symbol neighbors and delete edges emanating from content symbol
4. Go to Step 2.
(*Michael Luby slide)

34. LT Encoding Properties
Encoding symbols generated independently of each other
Any number of encoding symbols can be generated on the fly
Reception overhead independent of loss patterns
The success of the decoding process depends only on the degree distribution of received encoding symbols.
The degree distribution on received encoding symbols is the same as the degree distribution on generated encoding symbols.
(*Michael Luby slide)

35. What Do You Need To Know? Understand
2-dimensional parity
CRC
Hamming codes
Tradeoff between achieving reliability via retransmission vs. error correction