Updated measured performance; Fixed H matrix indexing; Fixed graph ticks
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@ -15,6 +15,7 @@
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\usepackage{listings}
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\usepackage{graphicx}
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\usepackage{xcolor}
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\usepackage[binary-units]{siunitx}
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%\geometry{textheight=17.07cm,textwidth=6.9cm}
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%\usepackage{pgfpages}
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%\pgfpagesuselayout{resize to}[physical paper height=17.07cm,
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@ -19,7 +19,7 @@
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_{\text{Parity constraint}},
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\hspace{5mm}\mathcal{A}\left( i \right) \equiv \left\{
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j | j\in \mathcal{J},
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\boldsymbol{H}_{i,j} = 1
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\boldsymbol{H}_{j,i} = 1
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\right\},
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i \in \mathcal{I}
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\end{align*}
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@ -82,7 +82,10 @@
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\begin{frame}[t, fragile]
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\frametitle{Proximal Decoding: Algorithm}
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\begin{algorithm}[caption={}, label={}]
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\begin{itemize}
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\item Resulting terative decoding algorithm:
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\vspace{2mm}
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\begin{algorithm}[caption={}, label={}]
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$\boldsymbol{s}^{\left( 0 \right)} = \boldsymbol{0}$
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for $k=0$ to $K-1$ do
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$\boldsymbol{r}^{\left( k+1 \right)} = \boldsymbol{s}^{(k)} - \omega \nabla L \left( \boldsymbol{s}^{(k)}; \boldsymbol{y} \right) $
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@ -92,7 +95,8 @@ for $k=0$ to $K-1$ do
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If $\boldsymbol{\hat{x}}$ passes the parity check condition, break the loop.
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end for
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Output $\boldsymbol{\hat{x}}$
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\end{algorithm}
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\end{algorithm}
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\end{itemize}
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\end{frame}
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@ -25,7 +25,7 @@
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legend style={at={(0.05,0.05)},anchor=south west},
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width=11.5cm,
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height=8cm,
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ytick={0, 10e-1, 10e-2, 10e-3, 10e-4},
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ytick={0, 1e-1, 1e-2, 1e-3, 1e-4},
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xtick={1, 2, 3, 4, 5},
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ymax=1.2, ymin=0.8e-4,
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xmin=0.9, xmax=5.6,
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@ -59,8 +59,13 @@
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\end{figure}
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\item Performance: $2800 \text{ transm.} / s$ - Intel Core i7-7700HQ @ 2.80GHz\\
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($\sim 10s$ for the shown plot)
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\item $\mathcal{O}\left(n \right) $ time complexity - same as BP;
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Only multiplication and addition necessary \cite{proximal_paper}
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\item Measured Performance: Between $\SI{0.5}{\mega\bit / \second}$ and
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$\SI{2.5}{\mega\bit / \second}$ - Intel Core i7-7700HQ @ 2.80GHz\\
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($\sim \SI{10}{\second}$ for the shown plot)
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\todo{Use the shown bitrate, or half?
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($n_{iterations} \cdot n$ or $n_{iterations} \cdot k$?)}
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\end{itemize}
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\end{frame}
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@ -14,7 +14,9 @@
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\item The standard message-passing algorithms used for decoding [LDPC and turbo codes]
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are often difficult to analyze. \cite{feldman_thesis}
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\item The iterative message–passing algorithms preffered in practice do not guarantee
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optimality and may fail to decode correctly when the graph contains cycles \cite{ldpc_conv}
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optimality and may fail to decode correctly when the graph contains cycles
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\cite{ldpc_conv}
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\end{itemize}
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\end{frame}
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@ -62,7 +64,7 @@
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\left(0,\frac{1}{2}\left(\frac{k}{n}\frac{E_b}{N_0}\right)^{-1}\right),
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\hspace{2mm} \boldsymbol{y}, \boldsymbol{n} \in \mathbb{R}^n
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\end{align*}
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\item All zeros assumption:
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\item All-zeros assumption:
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\begin{align*}
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\boldsymbol{c} = 0
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\end{align*}
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@ -172,12 +174,10 @@
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\begin{align*}
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N\left( j \right) \equiv \left\{
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i | i\in \mathcal{I},
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\boldsymbol{H}_{i,j} = 1
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\boldsymbol{H}_{j,i} = 1
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\right\},
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j \in \mathcal{J}
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\end{align*}
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\todo{Is this correct? Shouldn't i and j be switched around?}
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\item ``Illegal configurations''
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\begin{align*}
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S \subseteq N\left( j \right), \left| S \right| \text{odd}
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\end{align*}
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