Understanding the math of Reed-Solomon code (based on FEC code RS(544,514) used in IEEE 200/400GBASE-R)

1. Understanding Reed-Solomon code
Jeffrey.Wang
2019.12.16

2. Abstraction
• Reed-Solomon code RS(544,514) is applied in 200/400GBASE-R
FEC(Forward Error Correction).
• The big picture of the coding math is like the following:
514-symbol
block
bits stream
514-symbol
message block
30-symbol
parity block
divide by and
keep the remainder
g(x)
544-symbol codeword block 544-symbol received block
Errors
Syndromes
Calculate the syndromes with g(x)
Error location
Error locator polynomial
544-symbol Corrected codeword
Error value
514-symbol message
drop parity block
Encoder Decoder

3. Finite Field
• Reed-Solomon algorithm is defined on finite field. So what is that?
• First of all, the ‘field’ is defined as a set with the following character:
• two operations ‘Addition’ and ‘Multiplication’ defined on it.
• closed on these operations.
• commutative, associative and distributive laws hold.
• additive and multiplicative identity elements.
• additive inverse for every element.
• multiplicative inverse for every non-zero element.
• ‘Real number’ set is a field.
• ‘Integer’ set is not a field. (e.g. Number 2 doesn’t have a
multiplicative inverse which is also an integer)

4. Finite Field
• The ‘Finite field’ is a field with finite elements, as it says in its
name. (Finite field is also called ‘Galois field’ named after
Evariste Galois (1811–1832))
• How can a field be ‘finite’? The key is ‘modulo’. Operations
become ‘addition with modulo ’ and multiplication with
modulo ’. ( is short for prime)
• An example, a finite field has 3 elements: 0, 1, 2.
Operations with modulo 3. The operations meet all the items in
field definition.
• or is a finite field if is a prime number.
p
p p
GF(3)
GF(p) GF(pm
) p
1 + 2 = 3 (mod 3) = 0
2 × 2 = 4 (mod 3) = 1

5. Finite Field
• About the prime number .
• integer set {0, 1, 2} is a finite field,
called . (3 is a prime number)
• integer set {0, 1, 2, 3} is not a finite
field, (4 is not a prime number),
because element 2 doesn’t have a
multiplicative inverse. So there’s
no .
• But we can make the field with 4
elements a finite field if we change the
rules a little
• a set with four elements {00, 01, 10,
11} is a finite field, called
p
GF(3)
GF(4)
GF(22
)
+ 0 1 2
0 0 1 2
1 1 2 0
2 2 0 1
x 0 1 2
0 0 0 0
1 0 1 2
2 0 2 1
+ 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2
x 0 1 2 3
0 0 0 0 0
1 0 1 2 3
2 0 2 0 2
3 0 3 2 1
+ 00 01 10 11
00 00
01 10 11
01 01 00 11 10
10 10 11 00 01
11 11 10 01 00
x 00 01 10 11
00 00 00 00 00
01 00 01 10 11
10 00 10 11 01
11 00 11 01 10

6. Finite Field
• Here we introduce the ‘Polynomial’. We can use polynomials
to present the elements in with coefficients from
.
• Operations will be like the following: (Note that the
coefficients are from )
GF (2m
)
GF (2)
GF (2)
11000001 : x7
+ x6
+ 1 = 1x7
+ 1x6
+ 0x5
+ 0x4
+ 0x3
+ 0x2
+ 0x1
+ 1x0
Addition Multiplication

7. Finite Field
• Then we get back to the add and multiply operation matrix
in .
• Represent the elements with polynomials
• We should use modulo to keep the elements ‘finite’. And the
polynomial is used, which also called the
primitive polynomial.
• Take the red cell for example:
GF(22
)
x2
+ x + 1
+ 00 01 10 11
00 00
01 10 11
01 01 00 11 10
10 10 11 00 01
11 11 10 01 00
x 00 01 10 11
00 00 00 00 00
01 00 01 10 11
10 00 10 11 01
11 00 11 01 10 00 ⇒ 0x + 0
01 ⇒ 0x + 1
10 ⇒ 1x + 0
11 ⇒ 1x + 1
x + 1
× x
= x2
+ x (mod x2
+ x + 1)
= 1

8. Finite Field
• Another character is, let be the primitive element, then we can use power of this
element to generate all the non-zero elements in the finite field . See the following
example in :
α = x
GF (2m
)
GF (24
)
Similar polynomial also used for PRBS
Prime polynomial

9. Encoder
• 400GBASE-R module FEC encoder defined in IEEE 802.3 119.2.4.6.
• RS(544,514). Reed-Solomon code
• k=514 message symbols, code word length 544 symbols
• m=10 means each symbol has 10 bits.
• 2t=30 symbols, can correct t=15 symbols.

10. Encoder
• The big picture of the encoder is divide the message polynomial
(multiply ) by the generating polynomial , then use the remainder
as parity to combine with message to get the final code word.
x2t
g(x)
p(x)
mk−1xn−1
+ ⋯ + m1x2t+1
+ m0x2t
+ 0x2t−1
+ ⋯ + 0x + 0
% g2tx2t
+ ⋯ + g1x + g0
p2t−1x2t−1
+ ⋯ + p1x + p0
(mk−1xk−1
+ ⋯ + m1x + m0) × x2t
=
g(x) =
2t−1
∏
j=0
(x − αj
)
mk−1xn−1
+ ⋯ + m1x2t+1
+ m0x2t
+ p2t−1x2t−1
+ ⋯ + p1x + p0
Message symbols Parity symbols
Code word symbols

11. Encoder - Example
• Let’s take a simple example to look
into the division operation.
• Message is {1,2,3,4,5,6,7,8,9,10,11}
• Parity is {3,3,12,12}
• The final code word is
{1,2,3,4,5,6,7,8,9,10,11,3,3,12,12}
• Generating polynomial
• Prime polynomial is
• RS(15,11) on
g(x) = x4
+ 15x3
+ 3x2
+ x + 12
x4
+ x + 1
GF(24
)

12. Encoder
• IEEE gives the definition of generating polynomial .
• Use the simple one in last page to see how the calculation done.
g(x)
g(x) =
2t−1
∏
j=0
(x − αj
) = g2tx2t
+ ⋯ + g1x + g0
g(x) =
3
∏
j=0
(x − 2j
) = (x − 1) (x − 2) (x − 4) (x − 8)
= (x2
− 3x + 2) (x − 4) (x − 8)
= (x3
− 7x2
+ 14x − 8) (x − 8)
= (x4
− 7x3
+ 14x2
− 8x − 8x3
+ 13x2
− 9x + 12)
= x4
+ 15x3
+ 3x2
+ x + 12
7 × 8 = (x2
+ x + 1)x3
= x5 + x4 + x3
= x(x + 1) + (x + 1) + x3
= x2
+ x + x + 1 + x3
= x3
+ x2
+ 1
= 13
−7 − 8 = 7 + 8
= (x2
+ x + 1) + x3
= x3
+ x2
+ x + 1
= 15
• Keep in mind that all the addition and multiplication are defined in GF(24
)

13. Encoder
• We can calculate the using the same method with the following
parameters:
• Coefficients calculated with python(use modified pyfinite lib).
• Coefficients in IEEE.
gi
α = 2 x10
+ x3
+ 1
Exactly the same
(Primitive polynomial)

14. Encoder
• After we figure out all the detail of the calculation, let’s hold on a moment
to see what’s the mathematical idea behind the calculation.
• The final codeword is combined with message and parity symbols. And
the parity comes from the remainder. So the codeword can be divided by
. That means are roots of the codeword polynomial.g(x) αj
% = 0
2t−1
∏
j=0
(x − αj
)mk−1xn−1
+ ⋯ + m1x2t+1
+ m0x2t
+ p2t−1x2t−1
+ ⋯ + p1x + p0
Message symbols Parity symbols
C(x) = cn−1xn−1
+ ⋯ + c1x + c0
C(αj
) = CHT
= [c0 c1 ⋯ cn−1]
1 1 1 ⋯ 1
1 α1
α2
⋯ α2t−1
1 α2
α4
⋯ α(2t−1)2
⋯ ⋯ ⋯ ⋯ ⋯
1 αn−1
α2(n−1)
⋯ α(2t−1)(n−1)
= O1×2t
• Rewrite the equation in matrix

15. Encoder
• This equation means the codeword subspace is perpendicular
to the ’s row space.HT
[c0 c1 ⋯ cn−1]
1 1 1 ⋯ 1
1 α1
α2
⋯ α2t−1
1 α2
α4
⋯ α(2t−1)2
⋯ ⋯ ⋯ ⋯ ⋯
1 αn−1
α2(n−1)
⋯ α(2t−1)(n−1)
= O1×2t
• The encoder mapping the -dimensional original message into
-dimensional space. The -dimensional codeword subspace is
perpendicular to -dimensional subspace.
• If the transmission introduce any error make to , then
they’re not perpendicular any more. Then we know there are
errors, and maybe we can correct the errors.
k mk
n k
2t (2t = n − k)
C(x) R(x)

16. Decoder
• The result of is no longer Zero-matrix if there are errors.
• The result will be S. (short for Syndromes)
RHT
[r0 r1 ⋯ rn−1]
1 1 1 ⋯ 1
1 α1
α2
⋯ α2t−1
1 α2
α4
⋯ α(2t−1)2
⋯ ⋯ ⋯ ⋯ ⋯
1 αn−1
α2(n−1)
⋯ α(2t−1)(n−1)
= S
• We can correct the message if we know the error locations and
error values.

17. Decoder
• If there are errors.
• We can see that the ‘Syndromes’ are only affected by errors, no
matter what the codeword is.
• If there are errors and , then we can find the errors and
correct them.
• What we are going to do is to solve equations to get errors location
and errors value , with the known ‘Syndromes’
v v ≤ t
Xj Yj Si
R(x) = C(x) + E(x)
R(αi
) = C(αi
) + E(αi
) = 0 + E(αi
) = Si
Si = E(αi
)
= Y1αie1 + Y2αie2 + ⋯ + Yvαiev
= Y1Xi
1 + Y2Xi
2 + ⋯ + YvXi
v =
v
∑
j=1
YjXi
j

18. Decoder - Error locator
• Errors locator polynomial
• Since are roots of this equation
• With , we got equation group with equations.
And solve the equation group to get the coefficients
X−1
j
i = (0,1,⋯, v − 1) v
(Λ1, Λ2, ⋯, Λv)
Λ(x) = (1 + X1x)(1 + X2x)⋯(1 + Xvx)
= 1 + Λ1x + Λ2x2
+ ⋯ + Λvxv
0 = 1 + Λ1X−1
j + Λ2X−2
j + ⋯ + ΛvX−v
j
0 = YjXi+v
j + Λ1YjXi+v−1
j + ⋯ + ΛvYjXi
j
0 =
v
∑
j=1
YjXi+v
j +
v
∑
j=1
Λ1YjXi+v−1
j + ⋯ +
v
∑
j=1
ΛvYjXi
j
0 = Si+v + Λ1Si+v−1 + ⋯ + ΛvSi
Multiply YjXi+v
j
Add up for all j

19. Decoder - Error evaluation
• Get the roots of the locator equation, we got the error locations
.
• After substitute the error locations with solved values, here’s coming to the
final calculation: find the error values.
• Recover the codeword with error polynomial and drop the parity symbols. We
get the transmitted message.
• It’s better to go through all the calculation with our simple example. (P.S. the
algorithms for RS(544,514) are much more complex than in the simple
example)
(X1, X2, ⋯, Xv)
Xj
(1 + X1x)(1 + X2x)⋯(1 + Xvx) = 1 + Λ1x + Λ2x2
+ ⋯ + Λvxv
Si = Y1Xi
1 + Y2Xi
2 + ⋯ + YvXi
v
C(x) = R(x) − E(x)

20. Decoder - Example
• First, use received symbols to get the ‘Syndromes’.
[12 12 7 3 11 10 9 8 7 0 5 4 3 2 1]
1 1 1 1
1 2 4 8
1 4 3 12
1 8 12 10
1 3 5 15
1 6 7 1
1 12 15 8
1 11 9 12
1 5 2 10
1 10 8 15
1 7 6 1
1 14 11 8
1 15 10 12
1 13 14 10
1 9 13 15
= [2 10 9 1]
errors
Syndromesreceived codeword
parity check matrix
Here we use the parity check matrix. Divide the codeword
with with do the same thing.g(x)

21. Decoder - Example
• Error locator polynomial
0 = Si+v + Λ1Si+v−1 + ⋯ + ΛvSi
v = 2
0 = S2 + Λ1S1 + Λ2S0
0 = S3 + Λ1S2 + Λ2S1
S0 = 2
S1 = 10
S2 = 9
S3 = 1
9 = 10Λ1 + 2Λ2
1 = 9Λ1 + 10Λ2
Λ1 = 14
Λ2 = 14
Λ(x) = 1 + 14x + 14x2
= (1 + 4x)(1 + 10x)
Si = Y14i
+ Y210i
2 = Y1 + Y2
10 = 4Y1 + 10Y2
Y1 = 4
Y2 = 6
E(x) = Y1xe1 + Y2xe2 = 4x2
+ 6x9
X1 = 4
X2 = 10
e1 = 2
e2 = 9

22. Decoder - Example
• Decode finish
E(x) = 6x9
+ 4x2
R(x) = 1x14
+ 2x13
+ 3x12
+ 4x11
+ 5x10
+ 0x9
+ 7x8
+ 8x7
+ 9x6
+ 10x5
+ 11x4
+ 3x3
+ 7x2
+ 12x + 12
C(x) = 1x14
+ 2x13
+ 3x12
+ 4x11
+ 5x10
+ 6x9
+ 7x8
+ 8x7
+ 9x6
+ 10x5
+ 11x4
+ 3x3
+ 3x2
+ 12x + 12
[1 2 3 4 5 6 7 8 9 10 11]

23. Reference
• IEEE 802.3 119.2.4.6
• https://content.sakai.rutgers.edu/access/content/user/ak892/
Reed-SolomonProjectReport.pdf
• SC390 Introduction to Forward Error Correction, Frank
Kschischang, Univ. of Toronto, Canada.
• https://users.math.msu.edu/users/jhall/classes/codenotes/
GRS.pdf
• http://downloads.bbc.co.uk/rd/pubs/whp/whp-pdf-files/
WHP031.pdf