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Reed -Muller Codesite.iugaza.edu.ps/ahdrouss/files/2011/03/Reed-Muller-Code.pdf · ٣ Slide ٥...
Transcript of Reed -Muller Codesite.iugaza.edu.ps/ahdrouss/files/2011/03/Reed-Muller-Code.pdf · ٣ Slide ٥...
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Ammar Abh-Hhdrohss Islamic University -Gaza ١
Reed-Muller Codes
Slide ٢Channel Coding Theory
Reed-Muller Codes
These codes were discovered by Muller and the decoding by Reed in 1954.
Code length: n = 2m,
Dimension:
Minimum Distance
For m =5 and r = 2 then n = 32, k = 16, and dmin = 8.
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Slide ٣Channel Coding Theory
Reed-Muller Codes
For m =5 and r = 2 then n = 32, k = 16, and dmin = 8.
For 1≤ i ≤ m, let vi be a 2m-tuple over GF(2) of the following form:
For m =4, we have the following four vectors v4 = (0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1) v3 = (0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1) v2 = (0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1) v1 = (0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1)
Slide ٤Channel Coding Theory
Reed-Muller Codes
Let a = (a0, a1, …. , an-1) and b = (b0, b1, …., bn-1) be a two binary n-tuples. The Boolean product of a and b is defined as:
For example
The product vector vi1vi2…vil is said to have degree l.
111100 .,,.,.b.a
nn bababa
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Slide ٥Channel Coding Theory
Reed-Muller Codes
Because the weights of v1, v2, …. , vm are even and powers of 2, their products have weights even and a power of 2.
The r th order RM is spanned by the following set of independent vectors
We have the following inclusion chain
Slide ٦Channel Coding Theory
Example Let m =4 and r = 2 , the RM code is generated by the following
vectors 1514131211109876543210
1111111111111111v0
1111111100000000v4
1111000011110000v3
1100110011001100v2
1010101010101010v1
1111000000000000v3v4
1100110000000000v2v4
1010101000000000v1v4
1100000011000000v2v3
1010000010100000v1v3
1000100010001000v1v2
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Slide ٧Channel Coding Theory
Reed Decoding The importance of Muller code stems from the reduced
complexity of the decoding process.
This process pioneered by Reed is better explained by an example
Consider the RM(2, 4) and consider the message to be encoded
The corresponding codeword is
Slide ٨Channel Coding Theory
The sum of the first four components of each generator vector is zero except for v1v2.
The same applied three groups of four consecutive components. Or
Now let r = (r 0, r 1, … , r 15) be the received vector. a 12 can be decoded using
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Slide ٩Channel Coding Theory
a 12 is taken to be the majority of {A1, A2, A3, A4)
If there is only one error a 12 can be detected correctly.
The same can be applied for a 13
The same can be applied for a12, a13,a23, a14, a24 will be decoded correctly.
Slide ١٠Channel Coding Theory
After decoding a12, a13,a23, a14, a24, the following vector is subtracted from r
The result r (1) can be expressed as
a1 can be calculated using the following equations
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Slide ١١Channel Coding Theory
At the receiver side, a1 can be decoded using majority decoding from :
Similarly, we can decode the following information bits a1, a2, a3 and a4.
We modify the received vector
Slide ١٢Channel Coding Theory
In absence of error r (2) is given by
a0 can also be determined from majority logic decoding
This is three steps majority logic decoding
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Slide ١٣Channel Coding Theory
General Muller decdoing
To specify the group of check sum, for 1 ≤ i1 < i2 < ir-l ≤ m with 0 ≤ l ≤ r. the following index set can be formed
Let us define E as
We form the following set of integers
Slide ١٤Channel Coding Theory
General Muller decdoing
We have just completed the lth step of decoding the modified received vector is
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Slide ١٥Channel Coding Theory
Example: Consider RM code of length 16 given in the previous example.
If we want to construct the check sum of a12 (i1 = 1, i2 = 2)
The index of the forming sets are
Slide ١٦Channel Coding Theory
With l =0, the check sum of a12 is given by
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Slide ١٧Channel Coding Theory
For a13, i1 = 1. i2 =3 and
The index of the check sum is given by
Slide ١٨Channel Coding Theory
We obtain the following check sum for a13
Follow the same procedure for a24, a34 then
Now if we want to check a3
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Slide ١٩Channel Coding Theory
The index sets of the check sum are
And the check sum are given by
Similary we can for the check sums of a1, a2, and a4 .
Slide ٢٠Channel Coding Theory
performance
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Slide ٢١Channel Coding Theory
Other Construction of Reed Muller code
Definition of kroncker product of two matrices A, B
For example
,,2221
1211
2221
1211
bbbb
Baaaa
A
,2221
1211
BaBaBaBa
BA
Slide ٢٢Channel Coding Theory
Krocker product construction
The generator matrix for RM(2,4) is knocker product
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Slide ٢٣Channel Coding Theory
lu lu+v l construction
Let u = (u0, u1, …, un-1) and v = (v0, v1, …, vn-1) be two vectors over GF(2). From u and v we form the following 2n-tuple vector
Let C1 and C2 are (n, k) code over GF(2) with d2 > d1, we form the following linear code of length 2n
Where C is (2n, 2k) linear code, with generator matrix
Slide ٢٤Channel Coding Theory
And minimum distance
Example: let C1 be the binary (8,4) linear code of minimum distance 4 generated by
And C2 be (8, 1) reptilian code of minimum distance 8 generated by
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Slide ٢٥Channel Coding Theory
The resultant code has generator matrix
Slide ٢٦Channel Coding Theory
lu lu+v l construction
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Slide ٢٧Channel Coding Theory
Squaring construction
Consider a binary (n, k) linear code C with generator matrix G.
For 0 ≤ k1 ≤ k, let C1 be an (n, k1) linear sub-code of C that is spanned by k1 rows of G.
This portion of C with respect to C1 is denoted by C/C1. Each coset of C1 is of the following form:
1 ≤ l ≤ 2k-k1, where for vl ≠ 0, vl is in C but not in C1. vl is called leader (representative) of the coset.
The 0 vector is representative of C1
Slide ٢٨Channel Coding Theory
The set of representatives for the cosets in the partition C/C1is denoted [C/C1] which is called coset representative space for the partition C/C1, such that
The rows of the generator matrix G is divided k1 rows that generate C1 and k – k1 that generate C/C1.
Let C2 be an (n, k2) with 0 ≤ k2 ≤ k1. we can further partition each coset in partition C/C1 based on C2 into 2k1-k2 cosets of C2;
Each coset consists of the following codeword in C
Now we can express C in the following form
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Slide ٢٩Channel Coding Theory
let C1, C2, …., Cm be a sequence of linear subcode of C with dimensions, k1, k2, …., km, respectively, such that:
We can form series of partitions
Ands C can be expressed in the following form
We now can present another method for constructing long codes from a sequence of subcodes.
Slide ٣٠Channel Coding Theory
Let C0 be a binary (n, k0) linear block code with minimum Hamming Distance d0, let C1, C2, … , Cm be a sequence of sub-code of C0 such that:
We form a series of construction like follows
Where the generator matrix for each partition has the following property
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Slide ٣١Channel Coding Theory
One level square construction is based on C1 and the partition C0/C1. let a = (a0, a1, … , an-1) and b = (b0, b1, … , bn-1) be two binary n – tuples, and let (a, b) denote the 2n – tuple (a0, a1, … , an-1,b0, b1, … , bn-1) .
We form the following set of 2n-tuples
Which is a (2n, k0 + k1) linear block code with minimum Hamming distance.
And generator matrix
Slide ٣٢Channel Coding Theory
Let M1 and M2 be two matrices with the same number of columns. The matrix
is called direct sum and denoted as
Then the generator matrix of the squared code can be expressed as
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Slide ٣٣Channel Coding Theory
Let M1 and M2 be two matrices with the same number of columns. The matrix
is called direct sum and denoted as
Then the generator matrix of the squared code can be expressed as
Slide ٣٤Channel Coding Theory
Double square construction
Now if we repeat the one-level construction based on V and U/V. This code is a (4n, k0 + 2k1 + k2) linear block code with minimum
Hamming distance
And generator matrix
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Slide ٣٥Channel Coding Theory
Which can simplifies as:
Or
Slide ٣٦Channel Coding Theory
This method can be generalized and can be applied to form longer Reed-Muller Codes if we notice