The Math Behind the Compact Disc Linear Algebra and Error-Correcting Codes william j. martin. mathematical sciences. wpi wednesday december 3. 2008 fairfield

The Math Behind the Compact Disc

Linear Algebra and Error-Correcting Codes

william j. martin. mathematical sciences. wpi

wednesday december 3. 2008

fairfield university

04/21/23 W J Martin Mathematical Sciences WPI

How the device works

The compact disc is a complex system incorporating interesting ideas from engineering, physics, CS and math. We will focus only on the mathematics of the error- correction strategy.

For more info on the CD, see Kelin Kuhn’s book “Laser Engineering”:


Borrowed from K J Kuhn’s book “Laser Engineering”


The Pits

Each pit is 0.5 microns wide…and 0.83 to 3.56 microns long.Tracks are separated by 1.6 microns of “land”Wavelength of green light is about 0.5 micron40 tracks under one strand of human hair


Modelling a CommunicationsChannel

Linear algebra model: r = m+e (vector add.)


Channel with Error Correction


Turn it into an algebra problem!

A number system that the computer can understand:F = { 0, 1 }Ordinary multiplicationAddition: 1+1=0

Now music is turned into binary vectors!


A bit (or a nibble?) of graph theory

The n-cube is a type of Hamming graphVertices are all binary n-tuplesn-tuples are adjacent if they differ in only one coordinateNice ‘eigenvalues’!


Binary Vector Spaces

The vectors are all possible binary n-tuples

0 0 1 0 1 1 1 0 1 0 1 1 0 0 0

+

0 0 1 1 1 1 0 0 0 0 0 0 0 0 1

=

0 0 0 1 0 0 1 0 1 0 1 1 0 0 1


Hamming Distance

The distance between two binary n-tuples x and y is the number of coordinates in which they differ

This is a metric: dist( x, y ) 0 with dist( x, y ) = 0 iff x=y dist( x, y ) = dist( y, x ) Triangle inequality dist( x, z ) dist( x, y ) + dist( y, z )

dist( 001100, 001011 ) = 3


Theorem

Let C (the “code”) be a subset of F with minimum distance between any two codewords equal to d.

Then there exists an algorithm which corrects up to t errors per transmitted codeword if and only if d 2t + 1.

n


Proof

If x and y are distinct codewords, then the balls of radius t around them are disjoint. So if the received vector is within distance t of x, it must be at distance > t from any other codeword. So decoding is unique.


A Useful Extension of the Theorem

The above (computationally infeasible) decoding algorithm also correctly recovers from any t symbol errors and any s symbol erasures provided d > 2t+s.

transmit: 0 1 1 2 2 3 0

receive: 0 1 3 3 ? ? ?

(here, t=2 errors and s=3 erasures)


Small Example

Let C denote the “rowspace” of the matrix

Then C = { 000000, 110100, 011010, 101110, 001101, 111001, 010111, 100011 } and C has minimum distance 3 so C allows correction of

any single-bit error in any transmitted codeword.


The binary Hamming code

Codewords: 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 1 0 0 0 0 0 1 0 1 1 1 0 1 1 0 1 0 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 1 1 0 0 1 0 1 0 0 0 1 1 0 1 1 1 1 0 0 1 0 1 0 0 0 1 1 0 0 1 1 1 0 0 1 0 1 0 0 0 1 1 1 0 1 1 1 0 0 1 0 1 0 0 0 1 0 1 0 1 1 1 0

Quadratic Residues!

In we have

1 = 1 6 = 1

2 = 4 5 = 4

3 = 2 4 = 2

ZZ 7

2

2

2

2

2

2


The Fano projective plane

Vector Space: FF

“Poynts”: 1-dim. subspaces

“Lynes”: 2-dim. subspaces

3

2


All codewords:

0 0 0 0 0 0 0 1 1 1 1 1 1 1

0 0 0 1 1 1 1 1 1 1 0 0 0 0

0 1 1 0 0 1 1 1 0 0 1 1 0 0

0 1 1 1 1 0 0 1 0 0 0 0 1 1

1 0 1 0 1 0 1 0 1 0 1 0 1 0

1 0 1 1 0 1 0 0 1 0 0 1 0 1

1 1 0 0 1 1 0 0 0 1 1 0 0 1

1 1 0 1 0 0 1 0 0 1 0 1 1 0

C = nullsp(H) where


Codes from polynomialsLet’s replace F={0,1} with F={0,1,…,6} (with

modular arithmetic). Now consider the vector space F[z] of all polynomials in z with coefficients in F. For any subset N of F, we have a linear transformation

L: F[z] F via f(z) [ f(0), f(1), f(2), f(3), f(4), f(5) ]

(Here, we use, N={0,1,2,3,4,5}.)This is a Reed-Solomon code.

N


Polynomials to Codewords

Example:

Let the message be [1, 2, 2] (working mod 7)Polynomial is f(z) = z + 2 z + 2Codeword is

[f(0), f(1), f(2), f(3), f(4), f(5)] = [ 2, 5, 3, 3, 5, 2]

2


Reed-Solomon Codes

FACT: Two polynomials of degree less than k having k points of intersection must be equal.

SO: Reed-Solomon code of length n<q and dim k has min. dist. n-k+1


Compact Disc Parameters

SONY/Philips design (1980)Music is sampled 44,100 times per secondEach sample consists of 32 bits, representing

left and right channel signal magnitude 0—65535 (Pulse Code Modulation – PCM)So chip must process 1,411,200 raw data bits per secondBut it gets much worse!


Cross-Interleaved RS Codes

Inner code is a 28-dimensional subspace of a32-dimensional vector space over a finite field of

size 256.Outer code is a 24-dimensional subspace of a 28-dimensional vector space.Six 32-bit samples make up a 192-bit frame which is encoded as a 224-bit codeword. (Eventually, codewords have length 588 bits!)


Encoding – The numbers

The codewords from the first code are interleaved into a virtually infinite array of 28 rows of symbols over GF(256).We pull out 8 binary columns (one symbol) to obtain a 28x8=224-bit frame which is then encoded using another Reed-Solomon code to obtain a codeword of length 256 bits.


Interleaving to disperse errors

Codewords of first code are stacked like bricks 28 rows of vectors over GF(256)Extract columns and re-encode using second Reed-Solomon code


Splitting Odd and Even Bits


Back to the Pits

Each pit is 0.5 microns wide…and 0.83 to 3.56 microns long.Tracks are separated by 1.6 microns of “land”Not all 01-sequences can be recorded


EFM: Eight-to-Fourteen Modulation

This encoding scheme can only store sequences where each consecutive pair of ones is separated by at least 2 and at most 10 zerosThis is achieved by a mapping F F

which is given by a lookup table. 2 2

148


Further Processing

Three more ‘merge bits’ are added to each of these 14So 256+8=264=33x8 bits, carrying six samples, or 192 information bits, gets encoded as 588 channel bits on the diskThis represents 0.000136 seconds of music


What actually goes on the disc?

We must do this 7,350 times per secondSo CD player reads 4,321,800 bits per second of music producedTo get 74 minutes of music, we must store

74x60x4321800 = 19,188,792,000bits of data on the compact disc!


When in doubt, erase

Inner code has minimum distance 5 (over GF(256))Rather than correct two-symbol errors, the CD just erases the entire received vector.


So…how good is it?

The two Reed-Solomon codes team up to correct ‘burst’ errors of up to 4000 consecutive data bits (2.5 mm scratch on disc)If signal at time t cannot be recovered, interpolate

With smart data distribution, this allows for recovery from burst errors of up to 12,000 data bits (7.5 mm track length on disc)If all else fails, mute, giving 0.00028 sec of silence.


Other Applications

Space communications (Mariner,Voyager,etc.)DVD, CD-R, CD-ROMCell phones, internet packetsMemory: chips, hard drives, USB sticksRAID disk arraysQuantum computing


The Last Slide

Thank You All!

Documents

The Math Behind the Compact Disc Linear Algebra and Error-Correcting Codes william j. martin. mathematical sciences. wpi wednesday december 3. 2008 fairfield