Reed–Muller code

A block code can have one or more encoding functions ${\textstyle C:\{0,1\}^{k}\to \{0,1\}^{n}}$ that map messages ${\textstyle x\in \{0,1\}^{k}}$ to codewords ${\textstyle C(x)\in \{0,1\}^{n}}$. The Reed–Muller code RM(r, m) has message length ${\displaystyle \textstyle k=\sum _{i=0}^{r}{\binom {m}{i}}}$ and block length ${\displaystyle \textstyle n=2^{m}}$. One way to define an encoding for this code is based on the evaluation of multilinear polynomials with m variables and total degree r. Every multilinear polynomial over the finite field with two elements can be written as follows:

$${\displaystyle p_{c}(Z_{1},\dots ,Z_{m})=\sum _{\underset {|S|\leq r}{S\subseteq \{1,\dots ,m\}}}c_{S}\cdot \prod _{i\in S}Z_{i}\,.}$$

The ${\textstyle Z_{1},\dots ,Z_{m}}$ are the variables of the polynomial, and the values ${\textstyle c_{S}\in \{0,1\}}$ are the coefficients of the polynomial. Since there are exactly ${\textstyle k}$ coefficients, the message ${\textstyle x\in \{0,1\}^{k}}$ consists of ${\textstyle k}$ values that can be used as these coefficients. In this way, each message ${\textstyle x}$ gives rise to a unique polynomial ${\textstyle p_{x}}$ in m variables. To construct the codeword ${\textstyle C(x)}$, the encoder evaluates ${\textstyle p_{x}}$ at all evaluation points ${\textstyle a\in \{0,1\}^{m}}$, where it interprets the sum as addition modulo two in order to obtain a bit ${\textstyle (p_{x}(a){\bmod {2}})\in \{0,1\}}$. That is, the encoding function is defined via

$${\displaystyle C(x)=\left(p_{x}(a){\bmod {2}}\right)_{a\in \{0,1\}^{m}}\,.}$$

The fact that the codeword ${\displaystyle C(x)}$ suffices to uniquely reconstruct ${\displaystyle x}$ follows from Lagrange interpolation, which states that the coefficients of a polynomial are uniquely determined when sufficiently many evaluation points are given. Since ${\displaystyle C(0)=0}$ and ${\displaystyle C(x+y)=C(x)+C(y){\bmod {2}}}$ holds for all messages ${\displaystyle x,y\in \{0,1\}^{k}}$, the function ${\displaystyle C}$ is a linear map. Thus the Reed–Muller code is a linear code.

For the code RM(2, 4), the parameters are as follows:

${\textstyle {\begin{aligned}r&=2\\m&=4\\k&=\textstyle {\binom {4}{2}}+{\binom {4}{1}}+{\binom {4}{0}}=6+4+1=11\\n&=2^{m}=16\\\end{aligned}}}$

Let ${\textstyle C:\{0,1\}^{11}\to \{0,1\}^{16}}$ be the encoding function just defined. To encode the string x = 1 1010 010101 of length 11, the encoder first constructs the polynomial ${\textstyle p_{x}}$ in 4 variables:

$${\displaystyle {\begin{aligned}p_{x}(Z_{1},Z_{2},Z_{3},Z_{4})&=1+(1\cdot Z_{1}+0\cdot Z_{2}+1\cdot Z_{3}+0\cdot Z_{4})+(0\cdot Z_{1}Z_{2}+1\cdot Z_{1}Z_{3}+0\cdot Z_{1}Z_{4}+1\cdot Z_{2}Z_{3}+0\cdot Z_{2}Z_{4}+1\cdot Z_{3}Z_{4})\\&=1+Z_{1}+Z_{3}+Z_{1}Z_{3}+Z_{2}Z_{3}+Z_{3}Z_{4}\end{aligned}}}$$

Then it evaluates this polynomial at all 16 evaluation points (0101 means ${\displaystyle Z_{1}=0,Z_{2}=1,Z_{3}=0,Z_{4}=1)}$:

$${\displaystyle p_{x}(0000)=1,\;p_{x}(0001)=1,\;p_{x}(0010)=0,\;p_{x}(0011)=1,\;}$$

$${\displaystyle p_{x}(0100)=1,\;p_{x}(0101)=1,\;p_{x}(0110)=1,\;p_{x}(0111)=0,\;}$$

$${\displaystyle p_{x}(1000)=0,\;p_{x}(1001)=0,\;p_{x}(1010)=0,\;p_{x}(1011)=1,\;}$$

$${\displaystyle p_{x}(1100)=0,\;p_{x}(1101)=0,\;p_{x}(1110)=1,\;p_{x}(1111)=0\,.}$$

As a result, C(1 1010 010101) = 1101 1110 0001 0010 holds.

As was already mentioned, Lagrange interpolation can be used to efficiently retrieve the message from a codeword. However, a decoder needs to work even if the codeword has been corrupted in a few positions, that is, when the received word is different from any codeword. In this case, a local decoding procedure can help.

The algorithm from Reed is based on the following property: you start from the code word, that is a sequence of evaluation points from an unknown polynomial ${\textstyle p_{x}}$ of ${\textstyle {\mathbb {F} }_{2}[X_{1},X_{2},...,X_{m}]}$ of degree at most ${\textstyle r}$ that you want to find. The sequence may contains any number of errors up to ${\textstyle 2^{m-r-1}-1}$ included.

If you consider a monomial ${\textstyle \mu }$ of the highest degree ${\textstyle d}$ in ${\textstyle p_{x}}$ and sum all the evaluation points of the polynomial where all variables in ${\textstyle \mu }$ have the values 0 or 1, and all the other variables have value 0, you get the value of the coefficient (0 or 1) of ${\textstyle \mu }$ in ${\textstyle p_{x}}$ (There are ${\textstyle 2^{d}}$ such points). This is due to the fact that all lower monomial divisors of ${\textstyle \mu }$ appears an even number of time in the sum, and only ${\textstyle \mu }$ appears once.

To take into account the possibility of errors, you can also remark that you can fix the value of other variables to any value. So instead of doing the sum only once for other variables not in ${\textstyle \mu }$ with 0 value, you do it ${\textstyle 2^{m-d}}$ times for each fixed valuations of the other variables. If there is no error, all those sums should be equals to the value of the coefficient searched. The algorithm consists here to take the majority of the answers as the value searched. If the minority is larger than the maximum number of errors possible, the decoding step fails knowing there are too many errors in the input code.

Once a coefficient is computed, if it's 1, update the code to remove the monomial ${\textstyle \mu }$ from the input code and continue to next monomial, in reverse order of their degree.

Let's consider the previous example and start from the code. With ${\textstyle r=2,m=4}$ we can fix at most 1 error in the code. Consider the input code as 1101 1110 0001 0110 (this is the previous code with one error).

We know the degree of the polynomial ${\textstyle p_{x}}$ is at most ${\textstyle r=2}$, we start by searching for monomial of degree 2.

• ${\textstyle \mu =X_{3}X_{4}}$
  • we start by looking for evaluation points with ${\textstyle X_{1}=0,X_{2}=0,X_{3}\in \{0,1\},X_{4}\in \{0,1\}}$. In the code this is: 1101 1110 0001 0110. The first sum is 1 (odd number of 1).
  • we look for evaluation points with ${\textstyle X_{1}=0,X_{2}=1,X_{3}\in \{0,1\},X_{4}\in \{0,1\}}$. In the code this is: 1101 1110 0001 0110. The second sum is 1.
  • we look for evaluation points with ${\textstyle X_{1}=1,X_{2}=0,X_{3}\in \{0,1\},X_{4}\in \{0,1\}}$. In the code this is: 1101 1110 0001 0110. The third sum is 1.
  • we look for evaluation points with ${\textstyle X_{1}=1,X_{2}=1,X_{3}\in \{0,1\},X_{4}\in \{0,1\}}$. In the code this is: 1101 1110 0001 0110. The third sum is 0 (even number of 1).

The four sums don't agree (so we know there is an error), but the minority report is not larger than the maximum number of error allowed (1), so we take the majority and the coefficient of ${\textstyle \mu }$ is 1.

We remove ${\textstyle \mu }$ from the code before continue : code : 1101 1110 0001 0110, valuation of ${\textstyle \mu }$ is 0001000100010001, the new code is 1100 1111 0000 0111

• ${\textstyle \mu =X_{2}X_{4}}$
  • 1100 1111 0000 0111. Sum is 0
  • 1100 1111 0000 0111. Sum is 0
  • 1100 1111 0000 0111. Sum is 1
  • 1100 1111 0000 0111. Sum is 0

One error detected, coefficient is 0, no change to current code.

• ${\textstyle \mu =X_{1}X_{4}}$
  • 1100 1111 0000 0111. Sum is 0
  • 1100 1111 0000 0111. Sum is 0
  • 1100 1111 0000 0111. Sum is 1
  • 1100 1111 0000 0111. Sum is 0

One error detected, coefficient is 0, no change to current code.

• ${\textstyle \mu =X_{2}X_{3}}$
  • 1100 1111 0000 0111. Sum is 1
  • 100 1111 0000 0111. Sum is 1
  • 1100 1111 0000 0111. Sum is 1
  • 1100 1111 0000 0111. Sum is 0

One error detected, coefficient is 1, valuation of ${\textstyle \mu }$ is 0000 0011 0000 0011, current code is now 1100 1100 0000 0100.

• ${\textstyle \mu =X_{1}X_{3}}$
  • 1100 1100 0000 0100. Sum is 1
  • 1100 1100 0000 0100. Sum is 1
  • 1100 1100 0000 0100. Sum is 1
  • 1100 1100 0000 0100. Sum is 0

One error detected, coefficient is 1, valuation of ${\textstyle \mu }$ is 0000 0000 0011 0011, current code is now 1100 1100 0011 0111.

• ${\textstyle \mu =X_{1}X_{2}}$
  • 1100 1100 0011 0111. Sum is 0
  • 1100 1100 0011 0111. Sum is 1
  • 1100 1100 0011 0111. Sum is 0
  • 1100 1100 0011 0111. Sum is 0

One error detected, coefficient is 0, no change to current code. We know now all coefficient of degree 2 for the polynomial, we can start mononials of degree 1. Notice that for each next degree, there are twice as much sums, and each sums is half smaller.

• ${\textstyle \mu =X_{4}}$
  • 1100 1100 0011 0111. Sum is 0
  • 1100 1100 0011 0111. Sum is 0
  • 1100 1100 0011 0111. Sum is 0
  • 1100 1100 0011 0111. Sum is 0
  • 1100 1100 0011 0111. Sum is 0
  • 1100 1100 0011 0111. Sum is 0
  • 1100 1100 0011 0111. Sum is 1
  • 1100 1100 0011 0111. Sum is 0

One error detected, coefficient is 0, no change to current code.

• ${\textstyle \mu =X_{3}}$
  • 1100 1100 0011 0111. Sum is 1
  • 1100 1100 0011 0111. Sum is 1
  • 1100 1100 0011 0111. Sum is 1
  • 1100 1100 0011 0111. Sum is 1
  • 1100 1100 0011 0111. Sum is 1
  • 1100 1100 0011 0111. Sum is 1
  • 1100 1100 0011 0111. Sum is 1
  • 1100 1100 0011 0111. Sum is 0

One error detected, coefficient is 1, valuation of ${\textstyle \mu }$ is 0011 0011 0011 0011, current code is now 1111 1111 0000 0100.

Then we'll find 0 for ${\textstyle \mu =X_{2}}$, 1 for ${\textstyle \mu =X_{1}}$ and the current code become 1111 1111 1111 1011.

For the degree 0, we have 16 sums of only 1 bit. The minority is still of size 1, and we found ${\textstyle p_{x}=1+X_{1}+X_{3}+X_{1}X_{3}+X_{2}X_{3}+X_{3}X_{4}}$ and the corresponding initial word 1 1010 010101

Using low-degree polynomials over a finite field ${\displaystyle \mathbb {F} }$ of size ${\displaystyle q}$, it is possible to extend the definition of Reed–Muller codes to alphabets of size ${\displaystyle q}$. Let ${\displaystyle m}$ and ${\displaystyle d}$ be positive integers, where ${\displaystyle m}$ should be thought of as larger than ${\displaystyle d}$. To encode a message ${\textstyle x\in \mathbb {F} ^{k}}$ of width ${\displaystyle k=\textstyle {\binom {m+d}{m}}}$, the message is again interpreted as an ${\displaystyle m}$-variate polynomial ${\displaystyle p_{x}}$ of total degree at most ${\displaystyle d}$ and with coefficient from ${\displaystyle \mathbb {F} }$. Such a polynomial indeed has ${\displaystyle \textstyle {\binom {m+d}{m}}}$ coefficients. The Reed–Muller encoding of ${\displaystyle x}$ is the list of all evaluations of ${\displaystyle p_{x}(a)}$ over all ${\displaystyle a\in \mathbb {F} ^{m}}$. Thus the block length is ${\displaystyle n=q^{m}}$.

The Reed–Muller RM(r, m) code of order r and length N = 2^m is the code generated by v₀ and the wedge products of up to r of the v_i, 1 ≤ i ≤ m (where by convention a wedge product of fewer than one vector is the identity for the operation). In other words, we can build a generator matrix for the RM(r, m) code, using vectors and their wedge product permutations up to r at a time ${\displaystyle {v_{0},v_{1},\ldots ,v_{n},\ldots ,(v_{i_{1}}\wedge v_{i_{2}}),\ldots (v_{i_{1}}\wedge v_{i_{2}}\ldots \wedge v_{i_{r}})}}$, as the rows of the generator matrix, where 1 ≤ i_k ≤ m.

Let m = 3. Then N = 8, and

${\displaystyle X=\mathbb {F} _{2}^{3}=\{(0,0,0),(0,0,1),(0,1,0)\ldots ,(1,1,1)\},}$

and

${\displaystyle {\begin{aligned}v_{0}&=(1,1,1,1,1,1,1,1)\\[2pt]v_{1}&=(1,0,1,0,1,0,1,0)\\[2pt]v_{2}&=(1,1,0,0,1,1,0,0)\\[2pt]v_{3}&=(1,1,1,1,0,0,0,0).\end{aligned}}}$

The RM(1,3) code is generated by the set

${\displaystyle \{v_{0},v_{1},v_{2},v_{3}\},\,}$

or more explicitly by the rows of the matrix:

${\displaystyle {\begin{pmatrix}1&1&1&1&1&1&1&1\\1&0&1&0&1&0&1&0\\1&1&0&0&1&1&0&0\\1&1&1&1&0&0&0&0\end{pmatrix}}}$

The RM(2,3) code is generated by the set:

${\displaystyle \{v_{0},v_{1},v_{2},v_{3},v_{1}\wedge v_{2},v_{1}\wedge v_{3},v_{2}\wedge v_{3}\}}$

or more explicitly by the rows of the matrix:

${\displaystyle {\begin{pmatrix}1&1&1&1&1&1&1&1\\1&0&1&0&1&0&1&0\\1&1&0&0&1&1&0&0\\1&1&1&1&0&0&0&0\\1&0&0&0&1&0&0&0\\1&0&1&0&0&0&0&0\\1&1&0&0&0&0&0&0\\\end{pmatrix}}}$

The following properties hold:

1. The set of all possible wedge products of up to m of the v_i form a basis for ${\displaystyle \mathbb {F} _{2}^{N}}$.
2. The RM (r, m) code has rank

   ${\displaystyle \sum _{s=0}^{r}{m \choose s}.}$

3. RM (r, m) = RM (r, m − 1) | RM (r − 1, m − 1) where '|' denotes the bar product of two codes.
4. RM (r, m) has minimum Hamming weight 2^{m − r}.

RM(r, m) codes can be decoded using majority logic decoding. The basic idea of majority logic decoding is to build several checksums for each received code word element. Since each of the different checksums must all have the same value (i.e. the value of the message word element weight), we can use a majority logic decoding to decipher the value of the message word element. Once each order of the polynomial is decoded, the received word is modified accordingly by removing the corresponding codewords weighted by the decoded message contributions, up to the present stage. So for a rth order RM code, we have to decode iteratively r+1, times before we arrive at the final received code-word. Also, the values of the message bits are calculated through this scheme; finally we can calculate the codeword by multiplying the message word (just decoded) with the generator matrix.

One clue if the decoding succeeded, is to have an all-zero modified received word, at the end of (r + 1)-stage decoding through the majority logic decoding. This technique was proposed by Irving S. Reed, and is more general when applied to other finite geometry codes.

All RM(r, m) codes with ${\displaystyle 0\leq m\leq 5}$ and alphabet size 2 are displayed here, annotated with the standard [n,k,d] coding theory notation for block codes. The code RM(r, m) is a ${\displaystyle \textstyle [2^{m},k,2^{m-r}]_{2}}$-code, that is, it is a linear code over a binary alphabet, has block length ${\displaystyle \textstyle 2^{m}}$, message length (or dimension) k, and minimum distance ${\displaystyle \textstyle 2^{m-r}}$.

0	1	2	3	4	5	m
						RM(m,m) (2m, 2m, 1)	universe codes
					RM(5,5) (32,32,1)
				RM(4,4) (16,16,1)		RM(m − 1, m) (2m, 2m−1, 2)	SPC codes
			RM(3,3) (8,8,1)		RM(4,5) (32,31,2)
		RM(2,2) (4,4,1)		RM(3,4) (16,15,2)		RM(m − 2, m) (2m, 2m−m−1, 4)	extended Hamming codes
	RM(1,1) (2,2,1)		RM(2,3) (8,7,2)		RM(3,5) (32,26,4)
RM(0,0) (1,1,1)		RM(1,2) (4,3,2)		RM(2,4) (16,11,4)
	RM(0,1) (2,1,2)		RM(1,3) (8,4,4)		RM(2,5) (32,16,8)	RM(r, m=2r+1) (22r+1, 22r, 2r+1)	self-dual codes
RM(−1,0) (1,0,${\displaystyle \infty }$)		RM(0,2) (4,1,4)		RM(1,4) (16,5,8)
	RM(−1,1) (2,0,${\displaystyle \infty }$)		RM(0,3) (8,1,8)		RM(1,5) (32,6,16)
		RM(−1,2) (4,0,${\displaystyle \infty }$)		RM(0,4) (16,1,16)		RM(1,m) (2m, m+1, 2m−1)	punctured Hadamard codes
			RM(−1,3) (8,0,${\displaystyle \infty }$)		RM(0,5) (32,1,32)
				RM(−1,4) (16,0,${\displaystyle \infty }$)		RM(0,m) (2m, 1, 2m)	repetition codes
					RM(−1,5) (32,0,${\displaystyle \infty }$)
						RM(−1,m) (2m, 0, ∞)	trivial codes

• RM(0, m) codes are repetition codes of length N = 2^m, rate ${\displaystyle {R={\tfrac {1}{N}}}}$ and minimum distance ${\displaystyle d_{\min }=N}$.
• RM(1, m) codes are parity check codes of length N = 2^m, rate ${\displaystyle R={\tfrac {m+1}{N}}}$ and minimum distance ${\displaystyle d_{\min }={\tfrac {N}{2}}}$.
• RM(m − 1, m) codes are single parity check codes of length N = 2^m, rate ${\displaystyle R={\tfrac {N-1}{N}}}$ and minimum distance ${\displaystyle d_{\min }=2}$.
• RM(m − 2, m) codes are the family of extended Hamming codes of length N = 2^m with minimum distance ${\displaystyle d_{\min }=4}$.[7]