From 9ca69da4794f62a9b66c8814fba15d4f194eed7c Mon Sep 17 00:00:00 2001
From: Andreas Tsouchlos <an.tsouchlos@gmail.com>
Date: Thu, 26 Jan 2023 00:13:16 +0100
Subject: [PATCH] Changed pseudocode to not use indexing, double return instead
 of loop break

---
 latex/presentations/midterm/presentation.tex  |  2 +-
 .../midterm/sections/decoding_algorithms.tex  | 15 +++--
 .../midterm/sections/examination_results.tex  | 63 ++++++++-----------
 3 files changed, 35 insertions(+), 45 deletions(-)

diff --git a/latex/presentations/midterm/presentation.tex b/latex/presentations/midterm/presentation.tex
index 8449d46..3566de2 100644
--- a/latex/presentations/midterm/presentation.tex
+++ b/latex/presentations/midterm/presentation.tex
@@ -195,7 +195,7 @@
         basicstyle=\normalfont,
         columns=fullflexible,
         keywordstyle=\color{black}\bfseries,
-        keywords={a, for, end, do, b} % add the keywords you want, or load
+        keywords={a, for, end, do, return, b} % add the keywords you want, or load
                                       % a language as Rubens explains in his comment above.
         numbers=left,
         xleftmargin=.04\textwidth,
diff --git a/latex/presentations/midterm/sections/decoding_algorithms.tex b/latex/presentations/midterm/sections/decoding_algorithms.tex
index c16cce3..6ff7555 100644
--- a/latex/presentations/midterm/sections/decoding_algorithms.tex
+++ b/latex/presentations/midterm/sections/decoding_algorithms.tex
@@ -94,15 +94,14 @@
             
     \vspace{2mm}
     \begin{algorithm}[caption={}, label={}]
-$\boldsymbol{s}^{\left( 0 \right)} = \boldsymbol{0}$
-for $k=0$ to $K-1$ do
-    $\boldsymbol{r}^{\left( k+1 \right)} = \boldsymbol{s}^{(k)} - \omega \nabla L \left( \boldsymbol{y} \mid \boldsymbol{s}^{(k)} \right) $
-    Compute $\nabla h\left( \boldsymbol{r}^{\left( k+1 \right) } \right)$
-    $\boldsymbol{s}^{\left( k+1 \right)} = \boldsymbol{r}^{(k+1)} - \gamma \nabla h\left( \boldsymbol{r}^{\left( k+1 \right) } \right) $
-    $\boldsymbol{\hat{x}} = \text{sign}\left( \boldsymbol{s}^{\left( k+1 \right) } \right) $
-    If $\boldsymbol{\hat{x}}$ passes the parity check condition, break the loop.
+$\boldsymbol{s} \leftarrow \boldsymbol{0}$
+for $K$ iterations do
+    $\boldsymbol{r} \leftarrow \boldsymbol{s} - \omega \nabla L \left( \boldsymbol{y} \mid \boldsymbol{s} \right) $
+    $\boldsymbol{s} \leftarrow \boldsymbol{r} - \gamma \nabla h\left( \boldsymbol{r} \right) $
+    $\boldsymbol{\hat{x}} \leftarrow \text{sign}\left( \boldsymbol{s} \right) $
+    If $\boldsymbol{H}\boldsymbol{\hat{c}} = \boldsymbol{0}$, then return $\boldsymbol{\hat{c}}$
 end for
-Output $\boldsymbol{\hat{x}}$
+return $\boldsymbol{\hat{c}}$
     \end{algorithm}
 \end{frame}
 
diff --git a/latex/presentations/midterm/sections/examination_results.tex b/latex/presentations/midterm/sections/examination_results.tex
index 9890a0c..0e7f7f9 100644
--- a/latex/presentations/midterm/sections/examination_results.tex
+++ b/latex/presentations/midterm/sections/examination_results.tex
@@ -379,18 +379,15 @@
         \begin{figure}[htpb]
             \centering
 
-            \begin{algorithm}[caption={}, label={},
-            basicstyle=\fontsize{7.5}{9.5}\selectfont
-            ]
-$\boldsymbol{s}^{\left( 0 \right)} = \boldsymbol{0}$
-for $k=0$ to $K-1$ do
-    $\boldsymbol{r}^{\left( k+1 \right)} = \boldsymbol{s}^{(k)} - \omega \nabla L \left( \boldsymbol{y} \mid \boldsymbol{s}^{(k)}\right) $
-    Compute $\nabla h\left( \boldsymbol{r}^{\left( k+1 \right) } \right)$
-    $\boldsymbol{s}^{\left( k+1 \right)} = \boldsymbol{r}^{(k+1)} - \gamma \nabla h\left( \boldsymbol{r}^{\left( k+1 \right) } \right) $
-    $\boldsymbol{\hat{x}} = \text{sign}\left( \boldsymbol{s}^{\left( k+1 \right) } \right) $
-    If $\boldsymbol{\hat{x}}$ passes the parity check condition, break the loop.
+            \begin{algorithm}[caption={}, label={}]
+$\boldsymbol{s} \leftarrow \boldsymbol{0}$
+for $K$ iterations do
+    $\boldsymbol{r} \leftarrow \boldsymbol{s} - \omega \nabla L \left( \boldsymbol{y} \mid \boldsymbol{s} \right) $
+    $\boldsymbol{s} \leftarrow \boldsymbol{r} - \gamma \nabla h\left( \boldsymbol{r} \right) $
+    $\boldsymbol{\hat{x}} \leftarrow \text{sign}\left( \boldsymbol{s} \right) $
+    If $\boldsymbol{H}\boldsymbol{\hat{c}} = \boldsymbol{0}$, then return $\boldsymbol{\hat{c}}$
 end for
-Output $\boldsymbol{\hat{x}}$
+return $\boldsymbol{\hat{c}}$
             \end{algorithm}
 
             \caption{Proximal decoding algorithm \cite{proximal_paper}}
@@ -862,18 +859,15 @@ Output $\boldsymbol{\hat{x}}$
         \begin{figure}
            \centering
 
-            \begin{algorithm}[caption={}, label={},
-            basicstyle=\fontsize{7.5}{9.5}\selectfont
-            ]
-$\boldsymbol{s}^{\left( 0 \right)} = \boldsymbol{0}$
-for $k=0$ to $K-1$ do
-    $\boldsymbol{r}^{\left( k+1 \right)} = \boldsymbol{s}^{(k)} - \omega \nabla L \left( \boldsymbol{y} \mid \boldsymbol{s}^{(k)} \right) $
-    Compute $\nabla h\left( \boldsymbol{r}^{\left( k+1 \right) } \right)$
-    $\boldsymbol{s}^{\left( k+1 \right)} = \boldsymbol{r}^{(k+1)} - \gamma \nabla h\left( \boldsymbol{r}^{\left( k+1 \right) } \right) $
-    $\boldsymbol{\hat{x}} = \text{sign}\left( \boldsymbol{s}^{\left( k+1 \right) } \right) $
-    If $\boldsymbol{\hat{x}}$ passes the parity check condition, break the loop.
+           \begin{algorithm}[caption={}, label={}]
+$\boldsymbol{s} \leftarrow \boldsymbol{0}$
+for $K$ iterations do
+    $\boldsymbol{r} \leftarrow \boldsymbol{s} - \omega \nabla L \left( \boldsymbol{y} \mid \boldsymbol{s} \right) $
+    $\boldsymbol{s} \leftarrow \boldsymbol{r} - \gamma \nabla h\left( \boldsymbol{r} \right) $
+    $\boldsymbol{\hat{x}} \leftarrow \text{sign}\left( \boldsymbol{s} \right) $
+    If $\boldsymbol{H}\boldsymbol{\hat{c}} = \boldsymbol{0}$, then return $\boldsymbol{\hat{c}}$
 end for
-Output $\boldsymbol{\hat{x}}$
+return $\boldsymbol{\hat{c}}$
             \end{algorithm}
 
             \caption{Proximal decoding algorithm \cite{proximal_paper}}
@@ -884,22 +878,19 @@ Output $\boldsymbol{\hat{x}}$
         \centering
         \begin{figure}
            \centering 
-
-            \begin{algorithm}[caption={}, label={},
-            basicstyle=\fontsize{7.5}{9.5}\selectfont
-            ]
-$\boldsymbol{s}^{\left( 0 \right)} = \boldsymbol{0}$
-for $k=0$ to $K-1$ do
-    $\boldsymbol{r}^{\left( k+1 \right)} = \boldsymbol{s}^{(k)} - \omega \nabla L \left( \boldsymbol{y} \mid \boldsymbol{s}^{(k)}\right) $
-    Compute $\nabla h\left( \boldsymbol{r}^{\left( k+1 \right) } \right)$
-    $\boldsymbol{s}^{\left( k+1 \right)} = \boldsymbol{r}^{(k+1)} - \gamma \nabla h\left( \boldsymbol{r}^{\left( k+1 \right) } \right) $
-    $\boldsymbol{\hat{x}} = \text{sign}\left( \boldsymbol{s}^{\left( k+1 \right) } \right) $
-    $\textcolor{KITblue}{\text{If }\boldsymbol{\hat{x}}\text{ passes the parity check condition, output }\boldsymbol{\hat{x}}}$
+            
+           \begin{algorithm}[caption={}, label={}]
+$\boldsymbol{s} \leftarrow \boldsymbol{0}$
+for $K$ iterations do
+    $\boldsymbol{r} \leftarrow \boldsymbol{s} - \omega \nabla L \left( \boldsymbol{y} \mid \boldsymbol{s} \right) $
+    $\boldsymbol{s} \leftarrow \boldsymbol{r} - \gamma \nabla h\left( \boldsymbol{r} \right) $
+    $\boldsymbol{\hat{x}} \leftarrow \text{sign}\left( \boldsymbol{s} \right) $
+    If $\boldsymbol{H}\boldsymbol{\hat{c}} = \boldsymbol{0}$, then return $\boldsymbol{\hat{c}}$
 end for
 $\textcolor{KITblue}{\text{Find }N\text{ most probably wrong bits.}}$
-$\textcolor{KITblue}{\text{Generate variations } \boldsymbol{\tilde{x}}_n\text{ of }\boldsymbol{\hat{x}}\text{ with the }N\text{ bits modified.}}$
-$\textcolor{KITblue}{\text{Compute }d_H\left( \boldsymbol{ \tilde{x}}_n, \boldsymbol{\hat{x}} \right) \forall n \in \left[ 1 : N-1 \right]}$
-$\textcolor{KITblue}{\text{Output }\boldsymbol{\tilde{x}}_n\text{ with lowest }d_H\left( \boldsymbol{ \tilde{x}}_n, \boldsymbol{\hat{x}} \right)}$
+$\textcolor{KITblue}{\text{Generate variations } \boldsymbol{\tilde{c}}_n\text{ of }\boldsymbol{\hat{c}}\text{ with the }N\text{ bits modified.}}$
+$\textcolor{KITblue}{\text{Compute }d_H\left( \boldsymbol{ \tilde{c}}_n, \boldsymbol{\hat{c}} \right) \forall n \in \left[ 1 : N-1 \right]}$
+$\textcolor{KITblue}{\text{Output }\boldsymbol{\tilde{c}}_n\text{ with lowest }d_H\left( \boldsymbol{ \tilde{c}}_n, \boldsymbol{\hat{c}} \right)}$
             \end{algorithm}
 
             \caption{Hybrid proximal \& ML decoding algorithm}