Lecture 19:
Midterm on Thursday.
Office hours this week: Tuesday, Wednesday 1:30- 3:30
Practice midterm:
We have been using the following useful interpretations of matrix
multiplication:
Ax = y
— means that y is a linear combination of columns of A where
the coefficients are the entries of x
— and it also means that the entries of y are the dot products of
x with the rows of A.
Recall: For a linear code C, generator matrix G:
C = {uG : u ∈ V (k, q)}
parity check matrix H:
C = {x ∈ V (n, q) : HxT = 0}
Fact (see Lecture 18 notes): if for a linear code C, H = [B | In−k ]
is a parity check matrix in standard form, then a generator matrix
for C in standard form is:
G = [Ik | − B T ]
Recall: The Hamming code is
C = {x : x1+x2+x3+x5 = 0, x1+x2+x4+x6 = 0, x1+x3+x4+x7 = 0}
Cluster picture: Clusters are
{x1, x2, x3, x5}, {x1, x2, x4, x6}, {x1, x3, x4, x7}
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Each cluster has even parity.
Equivalently, the Hamming code is defined by the parity check
matrix
1 1 1 0 1 0 0
H= 1 1 0 1 0 1 0
1 0 1 1 0 0 1
So n = 7 and n − k = 3 (the number of rows of H) and so k = 4.
And d(C) = 3 because there is no zero column, no two columns
are identical and K 1 + K 4 + K 5 = 0.
So, the Hamming code is a [7, 4, 3]-binary code and |C| = 24 = 16.
Note that H = [B | I3] is a parity check matrix in standard form.
So, a generator matrix in standard form is:
1
0
G = [I4 | − B T ] =
0
0
0
1
0
0
0
0
1
0
0
0
0
1
1
1
1
0
1
1
0
1
1
0
1
1
equivalently,
C = {x1x2x3x4x5x6x7 : x5 = x1+x2+x3, x6 = x1+x2+x4, x7 = x1+x3+x4}
Note that the variables x1, x2, x3, x4 are free (i.e., we can choose any
values for them). For each such choice, the other variables x5, x6, x7
are determined by the equations above.
Decoding of linear codes:
Recall: cosets of a subgroup: G is a group, K is a subgroup, a+K
is a coset (a ∈ G); here,
a + K = {a + k : k ∈ K}
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Also, given x, y ∈ G, x and y belong to the same coset of K iff
x − y ∈ K.
Let C be a linear code in V (n, q). Then C is an (additive) subgroup of V (n, q), and thus V (n, q) is the disjoint union of the cosets,
a + C, of C, each of which has the same size as |C| = q k .
How many cosets are there?: q n/q k = q n−k .
Note: the cosets have to do with the (additive) group structure of
V (n, q) but nothing to do with scalar multiplication.
Defn: A coset leader of a coset is an element of the coset with
minimum weight.
Note: C is a coset of itself and its coset leader is 0.
Standard Array of a code: a list of all elements of V (n, q) with
the rows being the cosets, the first row being the code and the first
column being the coset leaders.
Example: C3 handout
C3 : {00000, 10110, 01101, 11011}
Standard Array for C3:
coset leader
C3 :
00000
10000
01000
00100
00010
00001
11000
10001
01101
11101
00101
01001
01111
01100
10101
11100
3
10110
00110
11110
10010
10100
10111
01110
00111
11011
01011
10011
11111
11001
11010
00011
01010
Note: Coset leaders need not be unique (see the last two rows of
the standard array above).
Defn: Let C be a linear code in V (n, q) with parity check matrix
H. The syndrome of x ∈ V (n, q) is defined
S(x) = SH (x) = HxT
Note: computation of the syndromes depends on the field structure
(both addition and multiplication).
Note: a syndrome is a column vector in V (n − k, q) but often we
write it as a row vector.
Proposition: Let C be a linear code in V (n, q). Let x, y ∈ V (n, q).
Then
1. S(x + y) = S(x) + S(y)
2. S(x) = 0 iff x ∈ C.
3. Let x, y ∈ V (n, q). Then x, y belong to the same coset of C iff
S(x) = S(y)
Proof:
1. H(x + y)T = H(xT + y T ) = HxT + Hy T .
2. S(x) = 0 iff HxT = 0 iff x ∈ C.
3. S(x) = S(y) iff H(x − y)T = 0 iff x − y ∈ C. Note: meaning of “syndrome” (symptom of disease).
By Part 2, a nonzero syndrome is the “symptom” that that the
received word x is not a codeword.
Part 3 says that the syndromes are in 1-1 correspondence with the
cosets. Since the syndromes are in V (n − k, q) and there are q n−k
cosets, we have
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Proposition: The syndromes exhaust all of V (n − k, q)
Alternative Proof: The set S of all syndromes is the set of all linear
combinations, over GF (q), of columns of H. So, S is the column
space of the matrix H.
dim(S) = dim( column space of H) = dim( row space of H)
= dim(C ⊥) = n − k.
But S ⊆ V (n − k, q). Thus, S = V (n − k, q). Return to Example C3:
[5, 2, 3]-code over GF (2) (a 1-error-correcting code)
Generator matrix (in standard form):
1 0 1 1 0
G=
0 1 1 0 1
Parity Check matrix (in standard form):


1 1 1 0 0
H=1 0 0 1 0
0 1 0 0 1
Syndrome Table for C3:
coset leader syndrome
C3 :
00000
000
10000
110
01000
101
00100
100
00010
010
00001
001
11000
011
10001
111
Proposition: Let x ∈ V (n, q).
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1. There is a coset leader y such that S(y) = S(x).
2. Let c := x − y. Then c ∈ C and c is a nearest nbr. of x, i.e.,
d(x, c) = minc0∈C d(x, c0).
Proof: 1. y is the coset leader of the coset with syndrome S(x).
2. S(c) = S(x − y) = S(x) − S(y) = 0. So, c ∈ C.
d(x, c) = wt(x − c) = wt(y).
for all c0 ∈ C, d(x, c0) = wt(x − c0).
Both y and x − c0 belong to x + C.
Since y is a coset leader, wt(y) ≤ wt(x − c0).
Thus, d(x, c) ≤ d(x, c0). Complete Syndrome Decoding Algorithm:
1. Given received vector x, compute S(x).
2. Find coset leader, y, corresponding to y in the syndrome table.
3. Decode x to c := x − y.
Corollary: Complete syndrome decoding is a specific way to implement complete nearest neighbour decoding.
Proof: This follows from previous proposition. Idea: y is the best guess for the “error vector”.
Example for C3:
Transmit c = 11011;
error vector e = 01000;
received vector x = c + e = 10011.
Syndrome: 101
Corresponding coset leader: y = 01000.
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Decode to c = x − y = 11011. Correct!
Another Example for C3:
Transmit c = 11011;
error vector e = 01010;
received vector x = c + e = 10001.
Syndrome: 111
Corresponding coset leader: y = 10001.
Decode to c = x − y = 00000. Incorrect!
Note: in first example, one error was made and code was able to
correct; in second example, two errors were made.
Advantage 4 of linear codes: “easy” to decode.
Lecture 20: Midterm 2
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