From bcf05d26afbf0ae1bdb4ee5f005192c2ca288461 Mon Sep 17 00:00:00 2001 From: Andreas Tsouchlos Date: Sat, 15 Apr 2023 18:56:55 +0200 Subject: [PATCH] Fixed mistake in implementation details; |N_v(i)| -> d_i --- latex/thesis/chapters/lp_dec_using_admm.tex | 23 ++++++++++----------- 1 file changed, 11 insertions(+), 12 deletions(-) diff --git a/latex/thesis/chapters/lp_dec_using_admm.tex b/latex/thesis/chapters/lp_dec_using_admm.tex index 1ad0d63..9547d6b 100644 --- a/latex/thesis/chapters/lp_dec_using_admm.tex +++ b/latex/thesis/chapters/lp_dec_using_admm.tex @@ -134,8 +134,8 @@ examplary code, which is described by the generator and parity-check matrices% and has only two possible codewords: % \begin{align*} -\mathcal{C} = \left\{ \begin{bmatrix} 0 & 0 & 0 \end{bmatrix}, - \begin{bmatrix} 0 & 1 & 1 \end{bmatrix} \right\} +\mathcal{C} = \left\{ \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \right\} .\end{align*} % Figure \ref{fig:lp:poly:exact_ilp} shows the domain of exact \ac{ML} decoding. @@ -631,7 +631,7 @@ The same is true for $\left( \boldsymbol{\lambda}_j \right)_i$.} \cite[Sec. III. B.]{original_admm}:% % \begin{alignat*}{3} - \tilde{c}_i &\leftarrow \frac{1}{\left| N_v\left( i \right) \right|} \left( + \tilde{c}_i &\leftarrow \frac{1}{d_i} \left( \sum_{j\in N_v\left( i \right) } \Big( \left( \boldsymbol{z}_j \right)_i - \frac{1}{\mu} \left( \boldsymbol{\lambda}_j \right)_i \Big) - \frac{\gamma_i}{\mu} \right) @@ -652,7 +652,7 @@ This representation can be slightly simplified by substituting $\boldsymbol{\lambda}_j = \mu \cdot \boldsymbol{u}_j \,\forall\,j\in\mathcal{J}$:% % \begin{alignat*}{3} - \tilde{c}_i &\leftarrow \frac{1}{\left| N_v\left( i \right) \right|} \left( + \tilde{c}_i &\leftarrow \frac{1}{d_i} \left( \sum_{j\in N_v\left( i \right) } \Big( \left( \boldsymbol{z}_j \right)_i - \left( \boldsymbol{u}_j \right)_i \Big) - \frac{\gamma_i}{\mu} \right) @@ -692,7 +692,7 @@ while $\sum_{j\in\mathcal{J}} \lVert \boldsymbol{T}_j\tilde{\boldsymbol{c}} - \b - \boldsymbol{z}_j$ end for for $i$ in $\mathcal{I}$ do - $\tilde{c}_i \leftarrow \frac{1}{\left| N_v\left( i \right) \right|} \left( + $\tilde{c}_i \leftarrow \frac{1}{d_i} \left( \sum_{j\in N_v\left( i \right) } \Big( \left( \boldsymbol{z}_j \right)_i - \left( \boldsymbol{u}_j \right)_i @@ -724,7 +724,7 @@ The method chosen here is the one presented in \cite{lautern}. \section{Implementation Details}% \label{sec:lp:Implementation Details} -The development process used to implement this decoding algorithm is the same +The development process used to implement this decoding algorithm was the same as outlined in section \ref{sec:prox:Implementation Details} for proximal decoding. At first, an initial version was implemented in Python, before repeating the @@ -752,11 +752,10 @@ Using this observation, the sum can be written as% - \boldsymbol{u}_j \right) \right)_i .\end{align*} Further noticing that the vectors -$\boldsymbol{T}_j^\text{T}\left( \boldsymbol{z}_j - \boldsymbol{u}_j \right), -\hspace{1mm} j\in\mathcal{J} $ +$\boldsymbol{T}_j^\text{T}\left( \boldsymbol{z}_j - \boldsymbol{u}_j \right)$ unrelated to \ac{VN} $i$ have $0$ as the $i$th component, the set of indices the summation takes place over can be extended to $\mathcal{J}$, allowing the -expression to be rewritten to% +expression to be rewritten as% % \begin{align*} \sum_{j\in \mathcal{J}}\left( \boldsymbol{T}_j^\text{T} \left( \boldsymbol{z}_j @@ -771,12 +770,12 @@ Defining% \boldsymbol{D} := \begin{bmatrix} d_1 \\ \vdots \\ - d_m + d_n \end{bmatrix}% \hspace{5mm}% \text{and}% \hspace{5mm}% - \boldsymbol{M} := \sum_{j\in\mathcal{J}} \boldsymbol{T}_j^\text{T} + \boldsymbol{s} := \sum_{j\in\mathcal{J}} \boldsymbol{T}_j^\text{T} \left( \boldsymbol{z}_j - \boldsymbol{u}_j \right) \end{align*}% % @@ -784,7 +783,7 @@ the $\tilde{\boldsymbol{c}}$ update can then be rewritten as% % \begin{align*} \tilde{\boldsymbol{c}} \leftarrow \boldsymbol{D}^{\circ -1} \circ - \left( \boldsymbol{M} - \frac{1}{\mu}\boldsymbol{\gamma} \right) + \left( \boldsymbol{s} - \frac{1}{\mu}\boldsymbol{\gamma} \right) .\end{align*} %