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Matrix Codes
1
G
1
H
Decoding algorithm:
- Compute syn( r ) = H• r
t .
- If syn( r ) = 0, assume no errors in transmission and peel off the first k digits of r as the
correct information digits.
- If syn( r ) ≠ 0 and syn( r ) = column i of H, then assume a single error in transmission,
and it must be in position i. Correct that position, and then peel off the correct
information digits.
- If syn( r ) ≠ 0 and syn( r ) ≠ any column of H, then assume at least 2 errors in
transmission - so you cannot decode correctly.
Theorem 3: An ( n-k )× k parity check matrix H will correctly decode all single errors if
and only if the columns of H are all non-zero and distinct.
Proof: <== Assume the columns of H are non-zero and distinct. Let e (^) i = the row vector
with a 1 in position i and 0's everywhere else.
Let r = c + ei (a received word with a single error)
Then H• r
t = H•[c+ e (^) i ]
t = H•ct + H• e (^) i
t
= 0 + i
th column of H so we will correctly decode the single error.
Proof: ==> (by contradiction):
Assume either the j
th column of H is zero. (1)
Then H• r
t = H•[c + e (^) j ]
t = H•c
t
t
which means we will not correct the single error.
or assume column i = column j [in H] (2)
Then H•]c+ e (^) j ]
t = H•c
t
t
= 0 + j
th column of H
= 0 + i
th column of H = H•c
t
t
= H•[c + e (^) i ]
t
So we cannot tell if the single error is in position i or j.
Dual Codes :
Let C be an (n,k) -code with generator matrix G = (I k | A) and parity check matrix H = (A
t
| I n-k ). Then H = G┴ = G perp is a generator matrix for the dual code C┴ = C perp.
Maple and Matrix Codes:
restart: with(linalg): with(LinearAlgebra):
We define a generating matrix G1 for generating a (6,3) linear binary code C 1 :
G1:=matrix(3,6,[1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,1,1,1]):
Now we want to use this matrix to encode every combination of 3-information digits.
So - we list all the possible information sets:
Info3:=matrix(8,3,[0,0,0,1,0,0,0,1,0,0,0,1,0,1,1,1,1,0,1,0,
1,1,1,1]);
Info3 :=
and then multiply each information triple by the generating matrix ‐ mod 2 to obtain our
list of codewords:
C1:=map(modp,multiply(Info3,G1),2);
C1 :=
Now – to decode with the parity check matrix, we define H 1 (the parity check matrix for
C 1 ) by:
H1:=matrix(3,6,[1,0,1,1,0,0,1,1,1,0,1,0,0,1,1,0,0,1]);
H1 :=
Now for each received word r, define its syndrome, syn(r) to be H 1 *c^t.
Let's compute the syndrome for a received vector r:
r:=matrix(1,6,[0,1,1,0,1,1]);
r :=[ 0 1 1 0 1 1 ]
C2perp :=
Now we check that G 2 is a parity check matrix for C 2 ┴ by computing the syndromes for
all codewords in C 2 ┴:
C2perpsyndrones:=map(modp,multiply(G2,transpose(C2perp)),2)
C2perpsyndrones :=
Homework Exercises #5:
due 3/24/09 ~ 5:00 pm
- Find the parity check matrix (H 1 – H 4 ) for codes (C 1 – C 4 ) generated by the
following generator matrices:
a) 1 b)
1 0 0 1 1 1
0 1 0 0 1 1
0 0 1 1 1 0
G
⎡ ⎤
⎢ ⎥
⎢
⎢ ⎥ ⎣ ⎦
2
1 0 0 0 1 1
0 1 0 1 1 1
0 0 1 1 1 0
G
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥ ⎣ ⎦
⎥
c) d) 3
G
4
1 0 0 0 1 1 0
0 1 0 0 0 1 1
0 0 1 0 1 0 1
0 0 0 1 1 1 1
G
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥
⎢ ⎥ ⎢⎣ ⎥⎦
- Determine the number of errors that can be
a) detected b) corrected by each of the matrix codes C 1 – C 4 defined in #1.
- Use the parity check matrices H 1 – H 3 you found in #1 to determine the location
of any errors in the following received vectors. (If there are double-errors, do not
try to correct). Report the correct information digits for each received vector.
a) C 1 : r 1 = [1 0 0 1 0 1]; r 2 = [1 0 0 1 0 0]; r 3 = [1 1 1 0 1 0 ]
b) C 2 : r 1 = [0 1 0 0 1 1]; r 2 = [1 1 1 0 0 1]; r 3 = [1 0 0 1 0 1 ]
c) C 3 : r 1 = [0 1 1 0 1 1 0]; r 2 = [1 1 1 0 0 0 0]; r 3 = [1 0 0 1 1 1 0 ]
- Let C┴ 5 – C┴ 8 be the codes generated by the parity check matrices H 5 – H 8
shown below.
i) List all the codewords in each code.
ii) Find a parity check matrix for each code.
iii) Use the parity check matrix to determine whether or not the code can correct a
single error. Explain.
a) b) 5
1 1 0 1 0 0
1 0 0 0 1 0
1 1 1 0 0 1
H
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥ ⎣ ⎦
6
1 1 0 1 0 0
1 1 1 0 1 0
1 0 0 0 0 1
H
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥ ⎣ ⎦
c) d) 7
1 1 0 1 0 0
1 1 1 0 1 0
1 0 1 0 0 1
H
⎡ ⎤
⎢ ⎥
⎢ ⎥
⎢ ⎥ ⎣ ⎦
8
1 1 1 1 1 0 0 0
1 1 0 1 0 1 0 0
1 0 1 1 0 0 1 0
0 0 0 1 0 0 0 1
H
⎡ ⎤
⎢ ⎥
= ⎢^ ⎥ ⎢ ⎥
⎢ ⎥ ⎢⎣ ⎥⎦
