CO 331 Course Notes - Student.math.uwaterloo.ca

CO 331 Course Notes
Ifaz David Gwen
April 4, 2012
Lecture 26 March 9th, Friday
Last Lecture: Let C be an (n, k)-cyclic code over F with gen. poly g (x).
Theorem (Computing Syndromes). Let r (x) be a polynomial with syndrome
s (x) = s0 + s1x + · · · + sn−k−1x n−k−1 . The syndrome of x · r (x) is
�
x · s (x) if sn−k−1 = 0
x · s (x) − sn−k−1g (x) if sn−k−1 �= 0
(H = � In−k| − R T � - the columns of H are x i mod g (x))
0.1 Burst Error Correction
Definition. Let e ∈ Vn (F ). The cyclic burst length of e is the length of the shortest cyclic
block in e which contains all the non-zero coordinates in e.
We say e is a cyclic burst error of length “t” if the cyclic burst length of e is “t”.
−→
−→
� �� � ����
Example. e = 01101 � �� 000 10�
lenght 8
(cyclic block of length 7) is a cyclic burst (error) of length 7.
Definition. A linear code C is a t-cyclic burst error correcting code if all cyclic burst
errors of length ≤ t are in different cosets of C. The largest such t gives the cyclic burst
error correction capability of C.
Example. (15, 9)-cyclic code C over Z2 with gen. poly. g (x) = 1 + x + x 2 + x 3 + x 6 is a
3-cyclic burst error correcting code.
(We’ll see later that t = 3 is the largest such integer)
Example. g (x) = 1 + x 4 + x 6 + x 7 + x 8 generates a binary (15, 7)-cyclic code C with a
cyclic burst error correcting capability of 4. (Verify)
1

Theorem. let C be (n, k, d)-code (linear) over GF (q), with cyclic burst error correcting
capability t. Then � �
d − 1
≤ t ≤ (n − k)
2
Proof. We know that all words of weight ≤ � �
d−1 lie in different cosets of C. So, all cyclic
2
burst errors of length ≤ � �
d−1 are in different cosets
2
=⇒ t ≥ � �
d−1
2
By definition (of “t”) all cyclic burst errors of length≤ t must be in different cosets of C. In
particular, all the words in which all the non zero elements (words) occur in the first t
positions must be in different cosets of C. We have q t such words, and we have q n−k cosets
of C. Hence,
q t ≤ g n−k =⇒ t ≤ (n − k)
Example. Going back to our example of g (x) = 1 + x 4 + x 6 + x 7 + x 8 we get that burst
error correcting capability t ≤ 6.
Theorem. Under the same setting as the previous theorem we have
t ≤
(n − k)
2
Example. Same example t ≤ 3 and then plug into 15, 9
Proof. C cannot have a non-zero codeword with cyclic burst error length ≤ 2t. Otherwise,
one can construct two different words of cyclic burst length ≤ t which are in the same coset
w1 w2
���� ����
of C. (c = 000 · · · / · · · 000 = w1 − w2)
� �� �
2t
So all the words in which all the non-zero components occur in the first 2t positions must
be in different cosets of C.
So � �
d−1
n−k ≤ t ≤ 2
2
0.2 Error Correction
=⇒ q 2t ≤ q n−k =⇒ t ≤
n − k
2
• C is an (n, x, d)-cyclic code with cyclic burst error correction capability t.
• c ∈ C is being sent, and r = c + e where e is a cyclic burst error of length ≤ t ≤ n − k
2

Recall that PCM for C is H = � In−k| − R T � the columns of H are x i mod g (x)
i = 0, . . . , n − 1.
Observation. For some integer i, 0 ≤ i ≤ n − 1, the vector ei corresponding to
ei (x) = x i e (x) will have all its nonzero coordinates in the first (n − k) positions.
Then si = H · e T i = (si, 0)
Lecture 27 March 12th, Monday
0.3 Error Trapping (for cyclic codes)
Last Lecture Let C be a (n, k)-code over GF (q) with generating polynomial g(x).
• C is a t-cyclic burst error correcting code (with t ≤ (n−k)
2 )
• H = � In−k| − R T � is a PCM for C (columns are x i mod g (x), syndrome of r (x) is
r (x) mod g (x))
Senario
• c ∈ C is being sent, r = c + e is received.
• e is a burst error of length ≤ t
Note that we may use the following interchangeably
c ←→ c (x) , e ←→ e (x) , r ←→ r (x) ,
where the left side of each correspondence is a vector and the right side is the associated
polynomial.
Observation. There exists an interger 0 ≤ i ≤ n such that ei = x i e has all its non-zero
entries in the first t positions or in the first (n − k)-positions.
The syndrome of ei is si = He T i . Note that
• si has burst length ≤ t
• ei = (si, 0)
So, the syndrome si determines ei which determines the error vector e. Hence, r can be
corrected to c = r − e.
3

Computing si Note that
r = c + e
=⇒ x i ���� r = x
ri
i ���� c + x
codeword
i ���� e
ei
=⇒ ri − ei ∈ C,
which implies that ri and ei have the same syndrome si.
Error Trapping Algorithm for Cyclic Burst Error Correcting Codes
• r(x) is a received word
• si(x) denote the syndrome of ri(x) := x i r(x)
1. For i from 0 to n − 1 do :
(a) Compute si(x)
(b) If si has burst length ≤ t, let e(x) = x n−1 si(x) and correct r(x) to
c(x) = r(x) − e(x)
2. Reject r(x)
Example. Let C be a (15, 9) cyclic code over Z2 with generator polynomial
g(x) = 1 + x + x 2 + x 3 + x 6 . Recall that C is a 3-cyclic burst error correcting code. Decode
r = (11101 11011 00000)
r(x) = 1 + x + x 2 + x 4 + x 5 + x 6 + x 8 + x 9
=⇒ s(x) = r(x) mod g(x) = 1 + x + x 4 + x 5 now with burst length ≤ 3
We compute si(x) until we find one with burst length ≤ 3 (Theorem: Computing
Syndromes).
i si = syndrome of ri(x) = x i r(x)
0 110011
1 100101
2 101110
3 010111
4 110111
5 100111
6 101111
7 101011
8 101001
9 101000
4

Note that s9 is a syndrom with burst length ≤ 3. So we have
s9
� �� �
e9 = ( 101000
0
� �� �
000000000)
=⇒ e = x 15−9 e9 = x 6 e9 = (00000 01010 00000)
=⇒ c = r − e = (11101 10001 00000)
Interleaving Let C be a (n, k) cyclic code with t-cyclic burst error correction capability.
Consider
v1 = (v11 v12 . . . v1n) ∈ C
v2 = (v21 v22 . . . v2n) ∈ C
. . . . . . . .
vs = (vs1 vs2 . . . vsn) ∈ C
Each vi can correct burst errors of length ≤ t.
Consider the following word of length ns.
v = (v11v21...vs1v21v22...v2s...v1nv2n...vsn)
Observation. Any cyclic burst error of length ≤ st in V corresponds to individual cyclic
burst errors of length at most t in vi’s (verify!). Therefore, v “can correct” cyclic burst
errors of length ≤ st.
Definition. Reordering of the coordinates as above is called interleaving to a depth s.
Theorem. Let C be an (n, k) code over GF(q) with cyclic burst error correcting capability
t. Let C ∗ be a the code obtained by interleaving C to a depth s. Then C ∗ is a (ns, ks)-code
over GF(q) with cyclic burst error correcting capability is st. Furthermore, if C is a cyclic
code with generating polynomial g(x) then C ∗ is a cyclic code with generating polynomial
g(x s ).
Lecture 28 March 14th, Wednesday
Last lecture: (Interleavin)
Theorem. Let C be an (n, k)-code over F with cyclic burst error correcting capability t.
Let C∗ be the code obtained by interleaving C to a to a depth of S. Then C ∗ is an
(ns, ks)-code over F with cyclic burst error correcting capability st. Furthermore, if C is
cyclic with generator polynomial g (x) then C∗ is cyclic with generator polynomial g (x s ).
Sketch of the Proof:
• Show that C ∗ is a linear code of length ns and dimension ks.
• Show that the cyclic burst error correcting capabilitiy is st (Shown in previous
lecture)
• Show that C ∗ is generated by g(x s )
5

BCH Codes: (Base and / ’60 Ray-Chandhur)
Block codes
Linear codes
Perfect Codes
Minimal Polynomials
Cyclic codes
BCH codes
GF (p n ) -vector space over GF (p) of dim = n
|
GF (p)
Figure: Codes
GF (q m ) -vector space over GF (q) of dim = m
|m
GF (q) = GF (p n )
Definition. Let α ∈ GF(q m ). The minimal polynomial of α over GF(q)[x] is defined
to be the non-zero, monic polynomial, mα(x) ∈ GF(q)[x] that is of smallest degree
satisfying mα(α) = 0.
mα(x) exists for all α ∈ GF(q m ) since
• α = 0 =⇒ mα(x) = x
• α �= 0, ord(α) = t (that is α t = 1), then f(x) = x t − 1 is such that f(α) = 0 and is
monic.
Example. GF (4) =
• m0 (x) = x
• m1 (x) = x + 1
Z2
(x 2 +x+1)
• mx (x) = x 2 + x + 1
• mx+1 (x) = x 2 + x + 1
f (x) = x − α
Theorem. Basic Properties of minimum polynomials:
Let α ∈ GF(q m ).
6

1. The minimum polynomial mα(x) of α over GF(q) is unique
2. mα(x) is irreducible over GF(q)
3. deg(mα) ≤ m
4. If f(x) ∈ GF(q)[x], then f(α) = 0 ⇔ mα(x)|f(x)
Proof.
1. Let m1 (x) and m2 (x) be two minimal polynomials of α over GF (q). Then
m(x) := m1 (x) − m2 (x) has degree less than �= deg m1 = deg m2. Moreover,
m (α) = 0.
=⇒ m (x) = 0 (zero as a poly)
=⇒ m1 (x) = m2 (x) (zero as a polynomials)
2. If mα is reducible over GF(q) then mα(x) = s(x)t(x) with deg(s) ≥ 1, deg(t) ≥ 1 and
deg(s), deg(t) < deg(mα) (s and t may be assumed to be monic). Since mα(α) = 0,
we have that s(α) = 0 or t(α) = 0. This is a contradiction with mα being α’s minimal
polynomial. Hence , mα is irreducible.
3. Recall that GF (q m ) is a vector space over GF (q) with dimension = m. Consider
(m + 1)-elements in GF (q m )
1 = α 0 , α 1 , α 2 , α 3 , . . . , α m
which must be linearly dependent over GF (q). That is, ∃a0, a1, . . . , am ∈ GF (q)
(some ai �= 0) such that
a0 + aiα + a2α 2 + · · · + amα m = 0
=⇒ f (x) = a0 + a1x + · · · + amx m ∈ GF (q) [x], f (α) = 0.
deg (f) ≤ m, f is non-zero
=⇒ deg mα ≤ m
4. We write f(x) = q(x)mα(x) + r(x) where deg(r) < deg(mα). If f(α) = 0, we have
that r(α) = 0 so r(x) = 0 (as a polynomial) which implies that mα|f. If we have
mα|f then f(α) = 0 trivially follows. Hence, f(α) = 0 ⇔ mα(x)|f(x).
Corollary. Let α ∈ GF (q m ). If f(x) ∈ GF (q)[x] is a monic irreducible polynomial that has
α as a root then f(x) = mα(x).
7

Proof. From the previous theorem, part 4, we have
mα(x)|f(x)
But f(x) is irreducible and monic, which implies that mα = f(x)
Lecture 29 March 16th, Friday
Last Lecture:
Corollary. Let α ∈ GF (q m ). If f(x) ∈ GF (q)[x] is a monic irreducible polynomial having
α as a root then f(x) is the minimal polynomial of α over GF (q)
Proposition. Let α ∈ GF (q m ). α ∈ GF (q) ⇐⇒ α q = α
Proof. Consider f(x) = x q − x ∈ GF (q)[x]. Recall that, for all α ∈ GF (q) we have α q = α.
So all the elements in GF (q) are roots of f(x) and f(x) cannot have any other root
because deg(f) = q. So α ∈ GF (q) ⇐⇒ α q = α
Definition. Let α ∈ GF (q m ) and t be tha smallest positive interger such that α qt = α.
(note: t ≤ m because α qm = α). Then the conjugates of α with respect to GF (q) is the set
C(α) = {α, α q , α q2 , ... , α qt−1 }
(Note that the elements of the set are pairwise disjoint)
Theorem. Let α ∈ GF (qm ). The minimal polynomial of α over GF (q) is
mα(x) = �
β∈C(α)
(x − β) = (x − α)(x − α q ) ... (x − α qt−1 )
Proof. Clearly mα(x) is monic, mα(α) = 0, and mα is not the zero polynomial.
We have to show that mα ∈ GF (q)[x].
We have
(mα) q ⎛
= ⎝ �
β∈C(α)
⎞
(x − β) ⎠
q
= �
β∈C(α)
(x − β) q = �
β∈C(α)
(x q − β q ) = �
β∈C(α)
(x q − β) = mα(x q ).
Therfore, if mα(x) = � t
i=0 mix i , then mi = m q
i which implies that mi ∈ GF (q) for all
0 ≤ i ≤ t so mα ∈ GF (q)[x].
We now have to show that mα is minimal.
8

Let f(x) ∈ GF (x)[q] with f(α) = 0 and f �= 0 (as a polynomial). We have to show that
deg(f) ≥ deg(mα) = t. Let f(x) = � d
i=0 fix i . We have
f(α) = 0 =⇒ f0 + f1α + f2α 2 + ... + fdα d = 0
=⇒ (f0 + f1α + f2α 2 + ... + fdα d ) q = 0
=⇒ f0 + f1α q + f2α 2q + ... + fdα dq = 0 (since fi = f q
i )
=⇒ α q is a root of f.
Similarly, we can show that α, α q , ... , α qt−1 are all roots of f. Therefore, deg(f) ≥ t.
This complete the proof.
Example. Let GF (16) = Z2[x]/(x 4 + x + 1). Find the minimal polynomial of
α = x 2 + x 3 ∈ GF (16) over GF (2).
Check that x is a generator for GF (16) and write all powers of x.
x 0 = 1 x 1 = x x 2 = x 2 x 3 = x 3 x 4 = x + 1
x 5 = x + x 2 x 6 = x 2 + x 3 x 7 = 1 + x + x 3 x 8 = 1 + x 2 x 9 = x + x 3
x 10 = 1 + x + x 2 x 11 = x + x 2 + x 3 x 12 = 1 + x + x 2 + x 3 x 13 = 1 + x 2 + x 3 x 14 = 1 + x 3
So we have that C(α) = {α = x 6 , x 12 , x 24 = x 9 , x 48 = x 3 }
=⇒ mα(X) = (X − x 6 )(X − x 12 )(X − x 9 )(X − x 3 ) = X 4 + X 3 + X 2 + X + 1.
Lecture 30 March 19th, Monday
0.4 Factoring x n − 1 over GF (q)
Let the characteristic of GF (q) be p, where q = pl . We will assume that gcd(n, q) = 1. If
not, then n = npl =⇒ xn − 1 = xnpl − 1 = (xn − 1) pl over GF (q) =⇒ gcd(n, q) = 1.
Factorization of x n − 1 over GF (q): Let m be the smallest positive interger such that
q m = 1 (mod n). Since gcd(n, q) = 1, such an interger exists (why?).
Construct GF (q m ) and assume that α is a generator of GF (q m ) ∗ . We set β = α qm −1
n and
order of β is n. Note that β 0 , β, β 2 , β 3 , . . . , β n−1 are pairwise distinct and (β i ) n = 1.
This gives us
� i
β �n − 1
� �� �
Roots of xn−1 are βi =
, 0≤i≤n−1
⎛
⎝ β n
����
1
⎞
⎠
i
− 1 = 1 − 1
Since βi is a root of xn − 1, we have that mβi|xn − 1. The roots of mβi are
C(bi ) = {βi , βiq , βiq2, ... , βiqt−1}, where t is the smallest interger for which βiqt = βi . Since
ord(β) = n, we have that βiqt = β i if and only if iq t ≡ i (mod n) .
Therefore, the value of t can be calculated from i,q, and n, without using β.
9

Definition. The cyclotomic coset of q mod n containing i is defined to be the set
Ci = {i, iq, . . . , iq t−1 } ⊂ Zn where t is the smallest integer such that iq t = i mod n. The set
C = {Ci : 0 ≤ i ≤ n − 1} is called cyclotomic cosets of q mod n
Example. Let q = 2, n = 15. We have
C0 = {0} C1 = {1, 2, 4, 8} C3 = {3, 6, 12, 9} C5 = {5, 10} C7 = {7, 14, 13, 11}
Proposition. If j ∈ Ci, Ci = Cj
Proof. Left as an exercise.
mβi (x) = (x − βi ) (x − βiq �
) x − βiq2 � �
. . . x − βiqt−1 �
= �
j∈Ci (x − βj ) and this
polynomial is an irreducible factor of xn − 1.
Note that the elements of C from a partition of the set {0, 1, . . . , n − 1}. The degree of
m β i(x) is |Ci|, and the number of irreducible factors of x n − 1 over GF (q) is |C|. This
proves the following
Theorem. Suppose that gcd(n, q) = 1. Let p ∈ GF (q m ) have order n, where m is the
smallest positive interger with q m ≡ 1 (mod n). Then the irreducible factors of x n − 1 over
GF (q) are {m β i(x) : 0 ≤ i ≤ n − 1} where m β i(x) = �
j∈Ci (x − βj ).
The number of irreducible factors of x n − 1 is equal to the number of (distinct) cyclotomic
cosets of q (mod n), and the number of irreducible factors of degree d is equal to the
number of (distinct) cyclotomic cosets of q (mod n) of size d.
Example. Factor x 1 5 − 1 over GF (2).
Find m such that 2 m ≡ 1 (mod 15). m = 4 works. Assume α is a generator of GF (2 4 ) ∗ .
� where GF (2 4 ) = Z2[x]/〈x 4 + x + 1〉 �
β = α 24 −1
15 = α, so α 0 , α 1 , ... , α 14 are roots of x 15 − 1 and m α i(x)|(x 15 − 1)
m α 0(x) = x − 1 = x + 1 in GF (2) that divides x 15 − 1.
mα(x) = (x − α)(x − α 2 )(x − α 4 )(x − α 8 ) = (x 4 + x + 1)|x 15 − 1
m α 3(x) = (x − α 3 )(x − α 6 )(x − α 12 )(x − α 9 ) = (x 4 + x 3 + x 2 + x + 1)|x 15 − 1
m α 5(x) = (x − α 5 )(x − α 10 ) = (x 2 + x + 1)|x 15 − 1
m α 7(x) = (x − α 7 )(x − α 14 )(x − α 13 )(x − α 11 ) = (x 4 + x 3 + 1)|x 15 − 1
Lecture 31 March 21th, Wednesday
BCH Codes (Bose-Chaudhuri ’60 & Hocquenghem ’59) A special class of cyclic
codes with a lower bound on the distance (⌊ d−1⌋
≤ t ≤ n − k where t is the cyclic burst
2
error correction capability.)
We’ll proove the “BCH bound”.
10

Setting GF (q), n ∈ Z ; m is the smallest integer such that gcd (n, q) = 1, q m ≡ 1 (mod n)
• α ∈ GF (q m ) ∗ and α is a generator.
• β = α qm −1
n ∈ GF (q m ) ∗ , β is an element of order “n”.
• m β i (x) denotes the minimal polynomial of β i over GF (q)
Definition. A BCH code, of block length n, over GF (q) with a designed destance δ, is a
cyclic code generated by
for some a ∈ Z
Remark.
g(x) = lcm{m β i(x) : a ≤ i ≤ a + δ − 2}
1. g (x) is indeed a generator poly. for a cyclic code of block length n, defined over
GF (q) because m β i (x) |x n − 1 for all a ≤ i ≤ a + δ − 2 and m β i (x) is monic,
m β i (x) ∈ GF (q) [x]
2. (δ − 1)-consecutive powers of β are roots of g(x) (# of consecutive powers of β which
are roots of g(x) will determine the lower bound (BCH-bound) for the distance)
3. The cyclic code C (as in the previous definition) has distance at least δ (proof is on
next lecture)
Example. Let q = 3, n = 13 =⇒ m = 3 (q m ≡ 1 mod n)
GF (q m ) = GF (3 3 ) = Z3[x]/(x 3 +2x 2 +1)
• α = x is agenerator in GF (3 3 ) ∗
• β = α qm −1
n = x 2 is an element of order n = 13
0.4.1 CYCLOTOMIC COSETS OF q mod n
minimum polynomial of βi over GF (q) : mβi(X) = (X − βi )(X − βiq )...(X − βiqt−1) β i → β iq → ... → β iqt−1
→ β i ⇔ i → iq → iq 2 → ... → iq t−1 → i
so t is the smallest interger such that i ≡ iq t (mod n)
11

The Cyclotomic cosets of q mod n (q = 3, n = 13)
The minimal polynomials
C0 = {0}
C1 = {1, 3, 9} = C3
C2 {2, 6, 5} = C5 = C6
C4 = {4, 12, 10}
C7 = {7, 8, 11}
m β 0(X) = X − 1 = X + 2
m β 1(X) = (X − β)(X − β 3 )(X − β 9 ) = 2 + 2X + 2X 2 + X 3
m β 2(X) = (X − β 2 )(X − β 6 )(X − β 5 ) = 2 + 2X + X 3
m β 4(X) = (X − β 4 )(X − β 12 )(X − β 10 ) = 2 + X + X 2 + X 3
m β 7(X) = (X − β 7 )(X − β 8 )(X − β 11 ) = 1 + 2X + X 3
Recall. GF (33 ) = Z3[x]/(x 3 +2x2 +1), α = x is a generator, β = α2 = x2 . It helps to write
α = x α 8 = 2 + x + x 2
α 14 = · · ·
some powers of α
α 2 = x 2
α 9 = 2 + 2x + 2x 2 .
α 3 = x 3 = x 2 + 2 α 10 = 1 + 2x + x 2 .
α 4 = 2 + 2x + x 2 α 11 = 2 + x α 26 = 1
α 5 = 2 + 2x α 12 = 2x + x 2
α 6 = 2x + 2x 2
α 7 = 1 + x 2 .
α 13
m β 2(X) = (X − α 4 )(X − α 12 )(X − α 10 ) = (X 2 − (α 4 + α 12 )X + α 16 )(X − α 10 ) = X 3 + 2X + 2
We may set (a = 0, δ = 4, a = 0 ≤ i ≤ a + δ − 2 = 2)
g (x) = lcm (m β 0 (x) , m β 1 (x) , m β 2 (x))
=⇒ g (x) = m β 0 (x) m β 1 (x) m β 2 (x)
= 2 + 2x + x 4 + 2x 7 + x 6 + x 7 and note that
x n − 1 = m β 0 (x) m β 1 (x) m β 2 (x) m β 4 (x) m β 7 (x)
Remark. 1. β 0 , β 1 , β 2 , β 3 , β 5 , β 6 , β 9 are roots of g(X) (In fact the designed distance is
δ = 5 as we have 4 consecutive powers of β which are roots of g(X))
2. g(X) generates a BCH-code (13,6) cyclic code over GF (3) with “distance at least 5”.
Lecture 33 March 23th, Friday
12

Last Lecture: BCH Codes - Definition
• q - a prime power; 1 ≤ n ∈ Z, gcd(q, n) = 1
• m - the smallest positve integer such that q m ≡ 1 (mod n)
• α ∈ GF (q m ) ∗ a generator
• β = α (qm −1)
n ∈ GF (q m ) ∗ an elementof order n.
• m β i(X) - minimal polynomial of βi over GF (q)
A BCH cod C over GF (q) with block length n, and with designed distance δ is a cyclic
code generated by
g (x) = lcm � m β i (x) : 0 ≤ i ≤ a + δ − 2 �
for some a ∈ Z.
1. g (x) |x n − 1
2. β a , β a+1 , β a+2 , . . . , β a+δ−2
roots of g
� �� �
(δ−1)−consecutive powers of β
3. d (C) ≤ δ
Definition. A Vandermonde matrix over a field F is an (n × n) matrix of the following
form
⎡
⎤
where xi ∈ F .
⎢
A = A(x1, x2, ..., xn) = ⎢
⎣.
Theorem. det (A (x1, x2, . . . , xn)) = �
i≤j≤n (xj − xi)
Proof. Exercise
1 x1 x2 1 ... x n−1
1
1 x2 x2 2 ... x n−1
2
.
.
. ..
1 xn x 2 n ... x n−1
n
Corollary. A (x1, x2, . . . , xn) in non-singular ⇐⇒ x1, x2, . . . , xn are pairwise distinct.
Theorem. Let C be a BCH-code over GF (q) with block length n, and with designed
distance δ. Then the distance of C is at least δ. (d(C) ≥ δ)
Proof. Let a,n, and m be as before. Let β ∈ GF (q m ) ∗ wiht order n, and
g(x) = lcm{m β i(X) : a ≤ i ≤ a + δ − 2} be the generator polynomial for C. Let
r ∈ Vn(GF (a))
13
.
⎥
⎦

Note.
Consider the matrix
⎡
⎢
H = ⎢
⎣.
r ∈ C ⇐⇒ g (x) |r (x)
⇐⇒ m β i (x) |r (x) ∀a ≤ i ≤ a + δ − 2
⇐⇒ r (β i ) = 0 ∀a ≤ i ≤ a + δ − 2 (†)
1 βa (βa ) 2 (βa ) 3 ... (βa ) n−1
1 βa+1 (βa+1 ) 2 (βa+1 ) 3 ... (βa+1 ) n−1
1 βa+2 (βa+2 ) 2 (βa+2 ) 3 ... (βa+2 ) n−1
.
.
.
. .. · · ·
1 β a+δ−2 (β a+δ−2 ) 2 (β a+δ−2 ) 3 ... (β a+δ−2 ) n−1
H is a (δ − 1) × n matrix over GF (q m ).
Observe that H · r T = 0 ⇐⇒ r ∈ C (follows from †)
⎤
⎥
⎦
(δ−1)×n
Now we prove that any t = (δ − 1) columns of H are linearly independent over GF (q m );
and hence they are linearly independent over GF (q) ∈ GF (q m ).
We construct the matrix H ′ (based an any t columns of H) and prove that H ′ is
non-singular
⎡
H ′ ⎢
= ⎢
⎣
(βa ) l1 (βa ) l2 ... (βa ) lt
(βa+1 ) l1 (βa+1 ) l2 ... (βa+1 ) lt
.
.
. ..
(β a+δ−2 ) l1 (β a+δ−2 ) l2 ... (β a+δ−2 ) lt
where 0 ≤ li ≤ (n − 1) and pairwise distinct.
det (H ′ ) = (β a ) l1 (β a ) l2 · · · (β a ) lt
=
�
t�
(β a ) li
�
i=1
� �� �
�=0 as β∈GF (qm ) ∗
(because β has order n and 0 ≤ li ≤ n − 1)
.
⎤
⎥
⎦
t×t=(δ−1)×(δ−1)
�
�
�
1 1 · · · 1
�
�
�
� �
l1
2 � �
l2
2 �
lt
� β β · · · β
�
�
�
. . . .
�
� �
l2
t−1 � �
l2
t−1 �
lt
β β · · · β
βl1 βl2 · · · βlt �2 · det � A � β l1 , β l2 , . . . , β lt ��
� �� �
�=0 ⇐⇒ β l i’s are pairwise distinct
This implies that H ′ is non-singular which implies that any t = (δ − 1) columns of H are
linearly independent over GF (q m ) (over GF (q))
14
� t−1
�
�
�
�
�
�
�
�
�
�
�

Claim: ∀c ∈ C with c �= 0, we have w(c) ≥ δ. ( =⇒ d(C) ≥ δ)
Proof. If c ∈ C, c �= 0, and w(c) < δ, then Hc T = 0 and so there exist (δ − 1) or fewer
columns of H which are linearly dependent over GF (q). Contradiction! Hence ∀c ∈ C with
c �= 0, we have w(c) ≥ δ.
Remark. H is NOT a P.C.M. for C because the entries are in GF (q m ) not in GF (q).
However, H behaves very much like a P.C.M.
Lecture 34 March 26th, Monday
0.5 Decoding BCH-codes: A decoding algorithm for C15
Setting. q = 2, n = 15, m = 4 (m is the smallest positive integer qm = q mod r).
Represent GF (24 ) as Z2[x]/(x4 + x + 1). α = x ∈ GF (24 ) ∗ is a generator.
β = α qn−1 ” 15
n = α 15 = α = x is anelement of order n = 15 in GF (24 ) ∗ .
Definition. C15 is defined to be the (15, 7)-BCH code over GF (2) with the generator
polynomial
g (x) = mβ (x) · m β 3 (x) = � x 4 + x + 1 � · � x 4 + x 3 + x 2 + x + 1 �
= x 8 + x 7 + x 6 + x 4 + 1
Note. • mβ (x) , m β 3 (x) |x 15 − 1, g (x) | (x 15 − 1)
• mβ(X) = (X − β)(X − β 2 )(X − β 4 )(X − β 8 )
• m β 2(X) = (X − β 3 )(X − β 6 )(X − β 12 )(X − β 9 )
• 4 consecutive powers of β: β, β 2 , β 3 , β 4 are roots of g(x)
=⇒ C15 has designed distance δ = 5 and d(C15) ≥ δ = 5. In fact, d(C15) = 5 (because
w(g(x)) = 5 =⇒ d(C15) ≤ 5) C15 has error correcting capability ⌊ d−1⌋
= 2.
2
r ∈ V15(GF (2)), r ∈ C15 ⇐⇒ r(βi ) = 0, ∀i = 1, 2, 3, 4. ( ⇐⇒ : proved in the last lecture)
⇐⇒ r(β) = 0 and r(β3 ) = 0.
Therefore, if we set
⎡
1 β β
H = ⎣
2 β3 · · · β14 1 β3 β6 β9 · · · β 42
⎤
⎦
15
����
β 12
2×15

then, r ∈ C15 ⇐⇒ Hr T − 0 ( ⇐⇒ r(β) = r(β 3 ) = 0. In particular, if c ∈ C15 is being
sent, and r = c + e is received, then
(H behaves like a PCM.)
Hr T = H(c + e) T = Hc T + He T =⇒ He T = Hr T
Remark. Computing HrT is easy because
Hr t �
r(β)
=
r(β3 �
)
0.6 A decoding algorithm for C15
(No need to store H)
Scenario c ∈ C15 is being sent, r = c + e is received where e is an error vector. (The
algorithm will always make the correct decision whenever w(w) ≤ 2.)
1. Compute
Hr T
�
r(β)
r(β3 �
=
)
2. If s1 = s3 = 0, then accept r, STOP (no errors occured or more than 2 errors have
occured)
3. Justification/Observation If w(e)f1, say exactly one error has occured, say in
position i 0 ≤ i ≤ 14 then the error polynomial is e(x) = xi , and
HrT � � �
s1 r(β
= =
r(β3 �
= He
)
T �
e(β)
=
e(β3 � �
i β
=
) β3i �
s3
=⇒ s3 = s 3 1 If s3 = s 3 1 and s1 = β i for some 0 ≤ i ≤ 14 then correc r in position i; STOP
(Also verify that if w(e) = 2 then we never have s3 = s 3 1)
4. If s1 = 0 (and s3 �= 0), reject r; STOP
Justification s3 �= 0 =⇒ r(β 3 ) �= 0 =⇒ e(β 3 ) �= 0 =⇒ e(X) is not the zero polynomial.
s1 = 0 =⇒ e(β) = r(β) = s1 = 0 =⇒ β is a root of e(X). mβ(X) | e)(X) =⇒ e(X) is a
codeword in the codespace generated by mβ(X) which is a BCH code with designed
distance 3 =⇒ w(e) ≥ d(C) ≥ δ = 3.
Lecture 35 March 28th, Wednesday
16
� s1
s3
�

0.7 A BCH CODE: C15
• GF (2 4 ) = Z2[x]/(x 4 + x + 1)
• α = x is a generator in GF (2 4 ) ∗
• β = α = x is an element of order n = 15
C15 is a (15,7)-BCH code generated by g(x) = mβ(X)m β 3(X) = X 8 + X 7 + X 6 + X 4 + 1
Powers of β in GF (2 4 )
β 0 = 1 β 4 = x + 1 β 8 = x 2 + 1 β 12 = x 3 + x 2 + x + 1
β 1 = x β 5 = x 2 + x β 9 = x 3 + x β 13 = x 3 + x 2 + 1
β 2 = x 2 β 6 = x 3 + x 2 β 10 = x 2 + x + 1 β 14 = x 3 + 1
β 3 = x 3 β 7 = x 3 + x + 1 β 11 = x 3 + x 2 + x
A Decoding algorithm for C15 (c ∈ C15 is being sent; r = c + e is received)
1. Compute s1 = r(β) and s3 = r(β 3 )
2. If s1 = s3 = 0; then accpet r, STOP (no error)
3. If s 3 1 − s3, then correct r in position “i” where si = β i ; STOP (single error)
4. If s1 = 0 and s3 �= 0, then reject r; STOP (more than 2 errors)
5. Observation If there are two errors, then the error polynomial e(x) = x i + x j (two
errors have occurred in positions i and j, 0 ≤ i ≤ j ≤ 14)
Then, s1 = r(β) = e(β) = β i + β j and s3 = r(β 3 ) = β 3i + β 3j .
s3 = β 3i + β 3j = (β i + β j )(β 2i + β i+j + β 2j ) = s1((β i + β j ) 2 + β i+j ) = s1(s 2 1 + β i+j )
=⇒ β i+j = s3
s1 + s2 1
Remark. s1 is never zero! - see step 2 and 4.
Therefore, β i and β j are the roots of the following “error locator polynomial”.
σ(z) = (z − β i )(z − β j ) = z 2 + (β i + β j )z + β i+j =⇒ σ(z) = z 2 + s1z + ( s3
s1 + s2 1)
• From the error locator polynomial, solve for B i and B j (find the “two distinct
non-zero roots” B i and B j of σ(z)) and correct r in positions i and j; STOP.
• σ(z) will indeed have two distinct roots. Because otherwise, we would have
si = β i + β j = β i + β i = 0, which is a contradiction.
17

• 0 is not a root of σ because otherwise s3
s1 + s2 1 = 0 =⇒ s 3 1 = s3, another contradiction
6. Reject r; STOP
Example. The received word is
Compute
s1 = r(β) = 1 + β 4 + β 7 + β 8
r = (10001 00110 00000)
= 1 + x 4 + x 7 + x 8
= ✁1 + ✘✘ ✘✘
(x + 1) + (✁1 +✚x + x 3 ) + (x 2 + ✁1) = x 2 + x 3 = β 6
s3 = r(β 3 ) = 1 + β 12 + β 21 + β 24 = 1 + β 12 + β 6 + β 9
= 1 + (x 3 + x 2 + x + 1) + (x 3 + x 2 ) + (x 2 + x) = x 3 + β 3
=⇒ s1 = β 6 and s3 = β 3 . s1 �= 0, s3 �= 0, s 3 1 = β 18 = β 3 = s3.
=⇒ correct r in position i = 6 (because s1 = β i = β 6 )
=⇒ e = (00000 01000 00000) and c = (1001 01110 00000)
Example. The received word is r = (00111 01110 00000) = x 2 + x 3 + x 4 + x 6 + x 7 + x 8
s1 = r(β) = β 13
s3 = r(β 3 ) = β 10
Construct the error locator polynomial, σ(z)
Note.
s3
s1
s1 �= 0, s 3 1 = β 9 �= β3
+ s 2 1 = β10
β 13 + β26 = β −3 + β 26 = β 12 + β 11 = 1
=⇒ σ(z) = z 2 + s1z + ( s3
s1 + s2 1) = z 2 + β 13 z + 1
Find the two distinct non-zero roots, say β i and beta j , 0 ≤ i ≤ j ≤ 14.
i + j ≡ mod15 =⇒ All possible pairs (i, j), 0 ≤ i ≤ j ≤ 14 with (i + j) = 0 mod 15 are
(1, 14), (2, 13), (3, 12), (4, 11), (5, 10), (6, 9), (7, 8)
We find the correct pair (i, j) by checking if β i + β j = β 13 .
In fact β 4 + β 11 = (x + 1) + (x 3 + x 2 + x) = 1 + x 2 + x 3 = β 13 =⇒ β 4 , β 11 are the roots of
σ(z) and e = (000010000001000).
18