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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