error detection and correction. 10.1 introduction data can be corrupted single-bit error: burst...

ERROR DETECTION AND CORRECTION

10.1 Introduction• Data can be corrupted

• Single-bit error:

• Burst error:

Redundancy• Extra information (redundant bits) can be used to “detect”

and “correct” errors• Coding (to add redundant bits)

Encoding Decoding

10.2 Block Coding• Coding schemes can be divided into two broad

categories: block coding and convolution coding• Background

• Modular arithmetic (modulus N): Use only a limited number of integers [0, 1, 2, …, N-1]• E.g., 7 + 6 in modular-9 arithmetic is 4, or “(7+6) % 9 = 4”

• Exclusive OR (XOR) = modular-2 arithmetic

Block Coding• Datawords: messages divided into k-bits blocks• Codewords: k-bit datawords + r-bits redundancy

• We have 2n codewords, but use only 2k datawords• There are 2n – 2k invalid codewords

(i.e., unused codewords)

Example of Error Detection• Datawords and codewords

• Example: sender encodes dataword 01 as 0111. If receiver receives 011, decodes it to 01

2. If receiver receives 111 (i.e., there is a single bit error), which is not a valid codeword, detects the error

3. If receiver receivers 000 (i.e., there are two bits error), decodes it to 00 undetectable errors

Example of Error Correction• Datawords and codewords

• Example: sender encodes dataword 01 as 01011• Receiver received 01001• Finds the codeword that is closest to 01001

• 01011 has a single-bit difference• 00000 2-bit difference, …, 11110 3-bit difference

• Decodes it to 01

Hamming Distance• Hamming distance between two words is the number of

differences between corresponding bits• Minimum hamming distance (dmin) is the “smallest”

Hamming distance between all possible pairs in a set of words

111

000

100

Hamming Distance for Detection• To guarantee detection of up to “s” errors, we should

have dmin > s

Hamming Distance for Correction• To guarantee correction of up to “t” errors, we should

have dmin > 2t

Example

• Minimum Hamming distance?• How many bit errors can be detected?• How many bit errors can be corrected?

10.3 Linear Block Codes• Precise definition is out of scope• Informally, in a linear block code, XOR of any two valid

codewords creates another valid codeword• Previous two sets of codewords are linear block codes

• Minimum Hamming distance of linear block codes is the minimum number of 1’s in a non-zero codeword

Simple Parity-Check Code• A k-bit dataword is encoded into (k+1)-bit codeword• The extra bit is chosen to make the number of 1’s in the

codeword “even”

• Simple parity-check code is a single-bit error-detecting code, with dmin = 2

The last bit (0 or 1) is added

Implementation of Simple Parity-Check Code

• s0 = b3 + b2 + b1 + b0 + q0 (modulo 2)• If s0 is 1, discard the received codeword

Two-Dimensional Parity Check• Example: four 7-bit datawords become five 8-bit

codewords

• Can detect up to 3 errors

Cannot detect 4 errors

Hamming Codes

• We consider hamming codes with dmin = 3• Detect two errors, or correct one single error

• Choose a number m ≥ 3, then• Codeword length n = 2m – 1• Dataword length k = n – m• Number of check bits r = m

Example: m = 3, n = 7 (=23–1), k = 4 (=7–3)• Dataword “a3a2a1a0” Codeword “a3a2a1a0 + r2r1r0”

• r0 = a2 + a1+ a0 (modulo-2)

• r1 = a3 + a2+ a1 (modulo-2)

• r2 = a1 + a0+ a3 (modulo-2)Each bit (e.g., a0) is

covered by two parity bits (e.g., r0 and r2)

Hamming Codes

Hamming Codes• Calculation of syndrome

• s0 = b2 + b1+ b0 + q0 (modulo-2)

• s1 = b3 + b2+ b1 + q1 (modulo-2)

• s2 = b1 + b0+ b3 + q2 (modulo-2)

• Decision to correct a single-bit error

r0 = a2 + a1+ a0 (modulo-2)r1 = a3 + a2+ a1 (modulo-2)r2 = a1 + a0+ a3 (modulo-2)

Hamming Codes• Example

• Received codeword 0100011 and syndrome is 000, what is the dataword?

• Received codeword 0011001 and syndrome is 011, what is the dataword?

Hamming Codes• Detect a burst error

10.4 Cyclic Codes• Linear block codes with the following property

• If a codeword is cyclically shifted, the result is another codeword• Our discussion is limited to a typical example, called Cyclic

Redundancy Check (CRC)

shift

Implementation• Case of CRC with 4-bit dataword, 7-bit codeword

CRC Encoder

Result of XOR

CRC Decoder

The same divider as encoder

Corrupted bitsNon-corrupted bits

Hardware Design• CRC encoder and decoder can be easily implemented

using shift registers

• Encoder example: divisor = 1011

It takes n-k times to get

the result

CRC Encoder using Shift Registers• Divisor = 1011

Polynomials• A pattern can be considered as coefficients of a

polynomial

• Adding and subtracting are done between the coefficients of the same power• Ex:

Division using Polynomials• Encoder

1011 =

= 1001

Cyclic Code Analysis• Dataword d(x), codeword c(x), divisor g(x), syndrome s(x),

error e(x)• At receiver, received codeword = c(x) + e(x)• After decoding, we have

• An error cannot be detected if e(x) is divisible by g(x)

Design of Divisor• Single-bit error: e(x) = xa for some a

• Divisor has more than one term and the least coefficient (i.e., of x0) is 1

• Two-bit error: e(x) = xa + xb for some a, b • Divisor should not divide xt+1

• Odd numbers of errors• Divisor should have the factor (x+1)

• Burst errors: e(x) = xa (1+ … + xb ) for some a, b• All burst errors of b-a+1 <= r will be detected• All other errors will be detected with probability

1-(1/2)r-1-(1/2)r

10.5 Checksum• Often used in the Internet protocols• Idea

• For data (7, 11, 12, 0, 6), • send (7, 11, 12, 0, 6, -36), where 36 = 7+11+12+0+6• Receiver checks error by adding all elements

• One’s complement• For the case, when the addition results in a large number more

than n bits

Checksum Procedure

One’s complement for

4-bit binary

9 = – 6 in one’s complement

representation

1 1 1 10 0 0 01 0 0 1