Polar Coding Notes: Channel Combining and Channel Splitting

Lyons ZhangOctober 19, 20181 comment

Channel Combining  

Channel combining is a step that combines copies of a given B-DMC $W$ in a recursive manner to produce a vector channel $W_N : {\cal X}^N \to {\cal Y}^N$, where $N$ can be any power of two, $N=2^n, n\le0^{[1]}$.  

The notation $u_1^N$ as shorthand for denoting a row vector $(u_1, \dots , u_N)$.  

The vector channel $W_N$ is the virtual channel between the input sequence $u_1^N$ to a linear encoder and the output sequence $y^N_1$ of $N$ copies of the original channel $W, W_1 = W$ is one copy of $W$.  

We use $W^N : {\cal X}^N \to {\cal Y}^N$ to denote the vector channel between the input sequence $x_1^N$ and the output sequence $y_1^N$ of $N$ copies of the original channel $W$.  

$W_N(y_1^N|u_1^N) = W^N(y_1^N|u_1^NG_N) = W^N(y_1^N|x_1^N) = \sum_{i=1}^N W(y_i|x_i)$

for any $N = 2^n, n \le 0$,  

$G_N = B_N F^{\otimes n}$

where $B_N$ is a permutation matrix known as bit-reversal, $F \triangleq \left[\begin{smallmatrix}1 & 0\cr 1 & 1 \end{smallmatrix}\right]$ and $\otimes n$ is Kronecker power, which is a Hadamard transform.  

The bit-reversal is shown as below from$^{[8]}$  

Sort a data sequence in nomal order by successive examination of the MSB as the left in the picture.  

Sort the sequence $x[n]$ as the permutation matrix(sequence A030109 in the OEIS) order, such a separation of odd and even can be carried out by examining the LSB.  

The recursive construction of $W_N$ from two copies of $W_{N/2}$ is useful for software and hardware$^{[9]}$. But sometimes for hardware implementation, the "rewire" structure may be more attractive:  

We denote random variables (RVs) by upper-case letters, the coding scheme for $U_1,U_2 \, \overset{i.i.d.}{\sim} \, Unif\{0,1\}$.  

It is easy to see that $U_1, U_2$ and $X_1, X_2$ have a bijection, and further coupling this with the fact $X_1,X_2 \, \overset{i.i.d.}{\sim} \, Unif\{0,1\}$. We have  

$I(U_1,U_2;Y_1,Y_2) = I(X_1,X_2;Y_1,Y_2) = 2C = 2I(W)$

and for $U_1, \dots, U_N \, \overset{i.i.d.}{\sim} \, Unif\{0,1\}$, then $X_1, \dots, X_N \, \overset{i.i.d.}{\sim} \, Unif\{0,1\}$,   

$\begin{align} I(U^N;Y^N) &= I(X^N;Y^N) \\&= H(Y^N) - H(Y^N|X^N) \\&= H(Y^N) - \sum_{i=1}^N H(Y_i|Y_1, ... , Y_{i-1}, X^N) \tag{Chain rule} \\&= \sum_{i=1}^N H(Y_i) - \sum_{i=1}^N H(Y_i|X_i) \tag{i.i.d.} \\&= \sum_{i=1}^N I(X_i;Y_i) \\&= NI(W) \end{align}$ 

Channel Splitting

Channel Splitting is to split $W_N$ back into a set of $N$ binary-input coordinate channels $W^{(i)}_N : X \to {\cal Y}^N \times {\cal X}^{i−1}, 1 \le i \le N$.  

The transition probability of the $W_N^{(i)}$ is defined as$^{[1]}$ 

$W_N(u_1^N;y_1^N) = \sum_{i=1}^N W_N^{(i)}(u_i;y_1^Nu_1^{i-1})$

For the $u_i$ is $i.i.d. , I(u_i;u_1^{i-1}) = 0$,  

$\begin{align} I(u_1^N;y_1^N) &= \sum_{i=1}^N I(u_i;y_1^N|u_1^{i-1}) \tag{Chain rule} \\ &= \sum_{i=1}^N \{I(u_i;y_1^N,u_1^{i-1}) - I(u_i;u_1^{i-1})\} \tag{Chain rule} \\ &= \sum_{i=1}^N I(u_i;y_1^Nu_1^{i-1})  \end{align}$

The transition probabilities are given by  

$\begin{align} W_N^{(i)} (y_1^N,u_1^{i-1}|u_i) &= {P(y_1^N,u_1^{i-1},u_i) \over P(u_i)} \\&= {P(y_1^N,u_1^i) \over P(u_i)} \\&= \sum_{u_{i+1}^N \in {\cal X}^{N-i}} {P(y_1^N,u_1^i,u_{i+1}^N) \over P(u_i)} \tag{$p_X(x) = \sum_y P_{X,Y}(x,y)$} \\&= \sum_{u_{i+1}^N \in {\cal X}^{N-i}} {P(y_1^N|u_1^N)P(u_1^N) \over P(u_i)} \\&= \sum_{u_{i+1}^N \in {\cal X}^{N-i}} P(y_1^N|u_1^N) {2^{-N} \over 2^{-1}} \\&= \sum_{u_{i+1}^N \in {\cal X}^{N-i}} {1 \over 2^{N-1}} W_N(y_1^N|u_1^N) \tag{$\ast$} \\&= \sum_{u_{i+1}^N \in {\cal X}^{N-i}} {1 \over 2^{N-1}} W^N(y_1^N|x_1^N) \\&= \sum_{u_{i+1}^N \in {\cal X}^{N-i}} {1 \over 2^{N-1}} W(y_1|x_1)W(y_2|x_2) \dots W(y_N|x_N) \end{align}$

For $N=2$,  

$\begin{align} W_2^{(1)}(y_1^2|u_1) &= \sum_{u_2^2 \in {\cal X}^1} {1 \over {2^{2-1}}}W(y_1|x_1)W(y_2|x_2) \tag{$i=1$, get rid of $u_1^0$} \\&= \sum_{u_2} {1\over 2}W(y_1|u_1\oplus u_2)W(y_2|u_2) \end{align}$

$\begin{align} W_2^{(2)}(y_1^2,u_1|u_2) &=  {1 \over 2}W(y_1|x_1)W(y_2|x_2) \tag{$i=2$, get rid of ${\cal X}^0$} \\&= {1\over 2}W(y_1|u_1\oplus u_2)W(y_2|u_2) \end{align}$

Below is from$^{[1]}$  

$\begin{align} W_{2N}^{(2i-1)}(y_1^{2N},u_1^{2i-2}|u_{2i-1}) &=  \sum_{u_{2i}^{2N}} {1\over 2^{2N-1}}W_{2N}(y_1^{2N}|u_1^{2N}) \\&= \sum_{u_{2i,o}^{2N},u_{2i,e}^{2N}} {1\over 2^{2N-1}}W_N(y_1^N|u_{1,o}^{2N}\oplus u_{1,e}^{2N})W_N(y_{N+1}^{2N}|u_{1,e}^{2N}) \\&= \sum_{u_{2i}} {1\over 2} \sum_{u_{2i+1,e}^{2N}} {1\over 2^{N-1}} W_N(y_{N+1}^{2N}|u_{1,e}^{2N})\sum_{u_{2i+1,o}^{2N}} {1\over 2^{N-1}} W_N(y_1^N|u_{1,o}^{2N} \oplus u_{1,e}^{2N}) \end{align}$

Because, as both $u_{2i+1,o}^{2N}$ and $u_{2i+1,o}^{2N} \oplus u_{2i+1,e}^{2N}$ range over ${\cal X}^{N−i}$. And from ($\ast$), the sum over $u_{2i+1,o}^{2N}$ for any fixed $u_{1,e}^{2N}$ is  

$\sum_{u_{2i+1,o}^{2N}} {1\over 2^{N-1}} W_N(y_1^N|u_{1,o}^{2N} \oplus u_{1,e}^{2N}) = W_N^{(i)}(y_1^N,u_{1,o}^{2i-2} \oplus u_{1,e}^{2i-2}|u_{2i-1}\oplus u_{2i})$

So we abtain  

$\begin{align} W_{2N}^{(2i-1)}(y_1^{2N},u_1^{2i-2}|u_{2i-1}) &=  \sum_{u_{2i}} {1\over 2}W_N^{(i)}(y_1^N,u_{1,o}^{2i-2} \oplus u_{1,e}^{2i-2}|u_{2i-1} \oplus u_{2i}) W_N^{(i)}(y_{N+1}^{2N},u_{1,e}^{2i-2}|u_{2i}) \end{align}$

$\begin{align} W_{2N}^{(2i)}(y_1^{2N},u_1^{2i-1}|u_{2i}) &=  {1\over 2}W_N^{(i)}(y_1^N,u_{1,o}^{2i-2} \oplus u_{1,e}^{2i-2}|u_{2i-1} \oplus u_{2i}) W_N^{(i)}(y_{N+1}^{2N},u_{1,e}^{2i-2}|u_{2i}) \end{align}$


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   Polar Coding Notes: A Simple Proof

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