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Apparently, there seems to be many online resources that talk of the process of encoding into Reed-Muller code. However, I was not able to find online resources that explain the process of decoding Reed-Muller code clearly.

Can anyone explain the decoding process?

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The section on decoding in the wiki page does not provide enough details, examples, is unclear? –  Sasha Aug 29 '12 at 13:19
    
@Sasha But it does not say it in mathematical way –  Halo Dotcom Aug 29 '12 at 13:27
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@Sasha I too think that the wikipedia page is rather terse on the subject of decoding Reed-Muller codes. For this reason, I posted a long answer to the question. –  Dilip Sarwate Aug 29 '12 at 21:54
    
+1 to Dilip. Sasha, from Dilip's answer you hopefully infer that this is usually a chapter's worth of material in a coding theory textbook. Admittedly the question was not necessarily a good fit to a Q&A site. –  Jyrki Lahtonen Aug 30 '12 at 14:12
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1 Answer

up vote 2 down vote accepted

I assume that the encoding process for a $r$-th order Reed-Muller code is the following.

  • Denote the $\sum_{i=0}^r\binom{m}{i}$ data bits as $$\begin{align*} &d_0,\\ &d_1, \quad d_2, \quad \ldots, \quad d_m,\\ &d_{1,2}, d_{1,3}, \ldots, d_{1,m},\quad d_{2,3}, d_{2,4},\ldots, d_{2,m}, \quad\ldots,\quad d_{m-1, m}\\ &\vdots \qquad \ddots\\ &d_{1,2,\ldots, r}, \quad\ldots, \quad d_{m-r+1, m-r+2, \ldots, m}, \end{align*}$$ where the $i$-th row above lists $\binom{m}{i}$ data bits.

  • Form the polynomial $$\begin{align*} d(x_1, x_2, \ldots, x_m) &= \qquad d_0 \\ &\quad \oplus\quad d_1 x_1 \oplus d_2x_2 \oplus \cdots \oplus d_mx_m\\ &\quad \oplus \quad d_{1,2}x_1x_2 \oplus d_{1.3}x_1x_3 \oplus \cdots d_{m-1,m}x_{m-1}x_m\\ &\quad \oplus \quad \cdots \\ &\quad \oplus \quad d_{1,2,\ldots,r} x_1x_2\cdots x_r \oplus \cdots d_{m-r+1,m-r+2,\ldots,m}x_{m-r+1}x_{m-r+2}\cdots x_m \end{align*}$$ of degree $r$ in $m$ binary variables $x_1, x_2, \ldots, x_m$. Note that the $i$-th row has all the $\binom{m}{i}$ terms of degree $i$.

  • Transmit (or record) the $2^m$ values of this polynomial (Boolean function) as the codeword. That is, $$\mathbf{d} = \left(d(0,0,\ldots,0), d(0,0,\ldots, 0,1), d(0, 0, \ldots, 1, 0),\ldots, d(1,1, \ldots, 1)\right) $$ is transmitted. Note that this just the truth table for this Boolean function.

The Hamming weight of $(d(x_1, x_2, \ldots, x_m)$ is the number of entries in $\mathbf d$ that have value $1$. In what follows, we will need the following property: The Hamming weight of any polynomial is odd if and only if the polynomial is of degree $m$, that is, $d_{1,2,\ldots, m} = 1$ and thus the term $x_1x_2\cdots x_m$ is included in the polynomial


Given the codeword (or truth table) $\mathbf d$, how can we determine the value of the data bit $d_{1,2,\ldots,r}$ which is the coefficient of the term $x_1x_2\cdots x_r$ in $d(x_1,x_2,\ldots, x_m)$? Well, let us consider the polynomial $$g(x_1,x_2,\ldots, x_m)=x_{r+1}x_{r+2}\cdots x_md(x_1,x_2,\ldots, x_m)$$ where we assume that for all $i, 1 \leq i \leq m$, all occurrences of $x_i^2$ on the right side of the above equation have been replaced by $x_i$ since $x_i$ takes on values $0$ and $1$ only. Since $d(x_1, \ldots, x_m)$ is of degree $r$ of the form shown above, we conclude that $x_{r+1}x_{r+2}\cdots x_md(x_1,x_2,\ldots, x_m)$ includes the term $x_1x_2\cdots x_m$ if and only if $d_{1,2,\ldots,r} = 1$, that is, $d(x_1, \ldots, x_m)$ contains the term $x_1x_2\cdots x_r$. It follows that the truth table of $x_{r+1}x_{r+2}\cdots x_md(x_1,x_2,\ldots, x_m)$ contains an odd number of $1$ entries if and only if $d_{1,2,\ldots,r} = 1$. Now, the truth table of $x_{r+1}x_{r+2}\cdots x_md(x_1,x_2,\ldots, x_m)$ has $0$s wherever at least one of $x_{r+1}, x_{r+2}, \ldots, x_m$ has value $0$, and the nonzero entries, if any, must be in the $2^{r}$ positions where $x_{r+1}=x_{r+2} = \cdots = x_m = 1$ are fixed, and only $x_1, x_2, \ldots, x_r$ are varying. Furthermore, the entries in these $2^r$ positions must be the same as the entries in $\mathbf d$ -- which we know. Are there an odd number of $1$ entries in $\mathbf d$ in these $2^r$ positions? We can find out by computing the parity (also known as the Exclusive-OR sum) of these $2^r$ bits in $\mathbf d$. Formally, we can write $$\begin{align*} d_{1,2,\ldots, r} = \quad &d(0,0,\ldots, 0, 0, 1, 1, \ldots, 1)\\ \quad \oplus &d(0,0,\ldots, 0, 1, 1, 1, \ldots, 1)\\ \quad \oplus &d(0,0,\ldots, 1, 0, 1, 1, \ldots, 1)\\ \quad \oplus &d(0,0,\ldots, 1, 1, 1, 1, \ldots, 1)\\ \quad \oplus &\cdots\\ \quad \oplus &d(1,1,\ldots, 1, 1, 1, 1, \ldots, 1).\\ \end{align*}$$ Any reader who has not been totally bored out of his gourd by the prolix presentation above will have realized that similar parity sums can be set up for the coefficients of all the other terms of degree $r$. All we have to do is create a $g(x_1, x_2, \ldots, x_m)$ by multiplying $d(x_1, x_2, \ldots, x_m)$ by all the missing variables and then compute the parity of the appropriate $2^r$ bits in $\mathbf d$. (To paint refined gold and gild the lily, the missing variables were $x_{r+1}, x_{r+2},\ldots,x_m$ when we were trying to determine the coefficient of $x_1x_2\cdots x_r$).

But, wait, there is more. Once we have determined the coefficients of the degree $r$ terms, it is necessary to modify $\mathbf d$ to be the truth table not of the $d(x_1, x_2, \ldots, x_m)$ that we began with but rather of a polynomial $\hat{d}(x_1, x_2, \ldots, x_m)$ of degree $r-1$. This new polynomial is just the terms of degree $r-1$ or less in $d(x_1, x_2, \ldots, x_m)$, and the modification is just subtracting off the truth table of the polynomial $$d_{1,2,\ldots,r} x_1x_2\cdots x_r \oplus \cdots d_{m-r+1,m-r+2,\ldots,m}x_{m-r+1}x_{m-r+2}\cdots x_m $$ from $\mathbf d$ (same as Exclusive-ORing in the truth table into $\mathbf d$). If this is not done, then multiplying by the missing variables $x_r, x_{r+1}, \ldots x_m$ in attempting to calculate $d_{1,2,\ldots, r-1}$ will not work because the term $d_{1,2,\ldots,r} x_1x_2\cdots x_r$ will create additional $x_1x_2\cdots x_m$ terms that interfere in the calculation of $d_{1,2,\ldots, r-1}$.

Thus, the decoding works in stages: first determine the coefficients of the degree-$r$ terms and subtract them off, then determine the coefficients of the degree-$(r-1)$ terms and subtract them off, and continue until all the terms have been calculated.


But if you order today, I will throw in error-correction for free! All of the above works if the transmitted (or recorded) codeword $\mathbf d$ is received (or read out) without any errors. But when some of the bits in $\mathbf d$ are incorrect, we cannot be sure that the parity calculation $$\begin{align*} d_{1,2,\ldots, r} = \quad &d(0,0,\ldots, 0, 0, 1, 1, \ldots, 1)\\ \quad \oplus &d(0,0,\ldots, 0, 1, 1, 1, \ldots, 1)\\ \quad \oplus &d(0,0,\ldots, 1, 0, 1, 1, \ldots, 1)\\ \quad \oplus &d(0,0,\ldots, 1, 1, 1, 1, \ldots, 1)\\ \quad \oplus &\cdots\\ \quad \oplus &d(1,1,\ldots, 1, 1, 1, 1, \ldots, 1).\\ \end{align*}$$ will give the correct value of $d_{1,2,\ldots, r}$. After all, an odd number of bits participating in the parity calculation might be wrong. Thus, the above parity calculation should be treated as an estimate of the value of $d_{1,2,\ldots, r}$. Fortunately, we can obtain $2^{m-r} - 1$ other estimates from of $d_{1,2,\ldots, r}$ from other bits in $\mathbf d$. For example, the polynomial $$\begin{align*} g(x_1,x_2,\ldots, x_m) &= \bar{x}_{r+1}x_{r+2}\cdots x_md(x_1,x_2,\ldots, x_m)\\ &= (x_{r+1} \oplus 1)x_{r+2}\cdots x_md(x_1,x_2,\ldots, x_m)\\ &= x_{r+1}x_{r+2}\cdots x_md(x_1,x_2,\ldots, x_m) \oplus x_{r+2}\cdots x_md(x_1,x_2,\ldots, x_m) \end{align*}$$ has degree $m$ if and only if $d_{1,2,\ldots, r} = 1$ but its truth table has nonzero entries in $2^r$ other locations, viz. where $x_{r+1}=0, x_{r+2}= x_{r+3} = \cdots = x_m = 1$. In this fashion, we do $2^{m-r}$ different parity calculations for $d_{1,2,\ldots, r}$ using all the $2^m$ bits in $\mathbf d$, and then take a majority vote of the different estimates to estimate the value of $d_{1,2,\ldots, r}$. As long as no more that $2^{m-r-1}-1$ bits in the received $\mathbf d$ are incorrect, the majority vote results in a correct estimate, that is, up to $2^{m-r-1}-1$ bit errors can be corrected: a tie vote indicates that a detectable but uncorrectable error pattern has occurred.

The rest of the decoding proceeds as described before with the obvious changes of doing multiple parity calculations in estimating each data bit, and what is subtracted off between stages is the estimated terms of that degree. Note that more parity calculations can be made in later stages, which can make assurance doubly sure if the previous stages resulted in correct decoding, but provide no additional error protection if incorrect decoding occurred at an earlier stage.

In summary, Reed-Muller decoding is a lot like peeling an onion. It has to be done one layer at a time, and inexperienced decoders (and cooks) tend to weep a lot.

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Wow, thanks! I really appreciate!!! –  Halo Dotcom Aug 30 '12 at 2:19
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