diff --git a/latex/thesis/chapters/decoding_techniques.tex b/latex/thesis/chapters/decoding_techniques.tex
index 429af0e..575b455 100644
--- a/latex/thesis/chapters/decoding_techniques.tex
+++ b/latex/thesis/chapters/decoding_techniques.tex
@@ -147,8 +147,7 @@ which minimizes the objective function $f$ (as shown in figure \ref{fig:dec:spac
 \textit{integer linear program} and subsequently presented a relaxation into
 a \textit{linear program}, lifting the integer requirement.
 The optimization method used to solve this problem that is examined in this
-work is the \ac{ADMM}.
-\todo{With or without 'the'?}
+work is \ac{ADMM}.
 \todo{Why chose ADMM?}
 
 Feldman at al. begin by looking at the \ac{ML} decoding problem%
@@ -164,11 +163,11 @@ They suggest that maximizing the likelihood
 $f_{\boldsymbol{Y} \mid \boldsymbol{C}}\left( \boldsymbol{y} \mid \boldsymbol{c} \right)$
 is equivalent to minimizing the negative log-likelihood.
 
-\ldots (Explaing arriving at cost function from ML decoding problem)
+\ldots (Explain arriving at the cost function from the ML decoding problem)
 
 Based on this, they propose their cost function%
 \footnote{In this context, \textit{cost function} and \textit{objective function}
-mean the same thing.}
+have the same meaning.}
 for the \ac{LP} decoding problem:%
 %
 \begin{align*}
@@ -177,11 +176,11 @@ for the \ac{LP} decoding problem:%
             \frac{f_{\boldsymbol{Y} | \boldsymbol{C}}
                 \left( Y_i = y_i  \mid C_i = 0 \right) }
             {f_{\boldsymbol{Y} | \boldsymbol{C}}
-                \left( Y_i = y_i | C_i = 1 \right) } \right) \\
+                \left( Y_i = y_i | C_i = 1 \right) } \right)
 .\end{align*}
 %
 %
-The exact integer linear program \todo{ILP acronym?} formulation of \ac{ML}
+With this cost function, the exact integer linear program formulation of \ac{ML}
 decoding is the following:%
 %
 \begin{align*}
@@ -190,19 +189,75 @@ decoding is the following:%
 .\end{align*}%
 %
 
-\ldots (LP Relaxation)
+As solving integer linear programs is generally NP-hard, the decoding problem \todo{New \S?}
+has to be approximated by one with looser constraints.
+A technique called \textit{\ac{LP} Relaxation} is applied,
+essentially removing the requirement for the components of $\boldsymbol{c}$
+to be integer values.
+In order to provide a formal definition of the relaxed constraints, the
+authors go on to define the concept of the \textit{codeword polytope}
+(figure \ref{fig:dec:poly}) as
+being the convex hull of all possible codewords:
+%
+\begin{align*}
+    \text{poly}\left( \mathcal{C} \right) = \left\{
+        \sum_{c \in \mathcal{C}} \lambda_{\boldsymbol{c}} \boldsymbol{c}
+            \text{ : } \lambda_{\boldsymbol{c}} \ge 0,
+        \sum_{\boldsymbol{c} \in \mathcal{C}} \lambda_{\boldsymbol{c}} = 1 \right\} 
+.\end{align*}
 
-%They go on to define the constraints under which this minimization is to be
-%accomplished.
-%They define the concept of the \textit{codeword polytope} as a linear
-%combination of all possible codewords, forming their convex hull:%
-%%
-%\begin{align*}
-%    \text{poly}\left( \mathcal{C} \right) = \left\{
-%        \sum_{c \in \mathcal{C}} \lambda_{\boldsymbol{c}} \boldsymbol{c}
-%            \text{ : } \lambda_{\boldsymbol{c}} \ge 0,
-%        \sum_{\boldsymbol{c} \in \mathcal{C}} \lambda_{\boldsymbol{c}} = 1 \right\} 
-%.\end{align*}
+\begin{figure}[H]
+    \centering
+
+    \tikzstyle{codeword} = [color=KITblue, fill=KITblue,
+                        draw, circle, inner sep=0pt, minimum size=4pt]
+
+    \tdplotsetmaincoords{60}{245}
+    \begin{tikzpicture}[scale=1, transform shape, tdplot_main_coords]
+        % Cube
+
+        \draw[dashed] (0, 0, 0) -- (2, 0, 0);
+        \draw[dashed] (2, 0, 0) -- (2, 0, 2);
+        \draw[] (2, 0, 2) -- (0, 0, 2);
+        \draw[] (0, 0, 2) -- (0, 0, 0);
+
+        \draw[] (0, 2, 0) -- (2, 2, 0);
+        \draw[] (2, 2, 0) -- (2, 2, 2);
+        \draw[] (2, 2, 2) -- (0, 2, 2);
+        \draw[] (0, 2, 2) -- (0, 2, 0);
+
+        \draw[] (0, 0, 0) -- (0, 2, 0);
+        \draw[dashed] (2, 0, 0) -- (2, 2, 0);
+        \draw[] (2, 0, 2) -- (2, 2, 2);
+        \draw[] (0, 0, 2) -- (0, 2, 2);
+
+        % Codeword Polytope
+
+        \draw[line width=1pt, color=KITblue] (0, 0, 0) -- (2, 0, 2);
+        \draw[line width=1pt, color=KITblue] (0, 0, 0) -- (2, 2, 0);
+        \draw[line width=1pt, color=KITblue] (0, 0, 0) -- (0, 2, 2);
+
+        \draw[line width=1pt, color=KITblue] (2, 0, 2) -- (2, 2, 0);
+        \draw[line width=1pt, color=KITblue] (2, 0, 2) -- (0, 2, 2);
+
+        \draw[line width=1pt, color=KITblue] (0, 2, 2) -- (2, 2, 0);
+
+        % Polytope Annotations
+
+        \node[codeword] (c000) at (0, 0, 0) {};% {$\left( 0, 0, 0 \right) $};
+        \node[codeword] (c101) at (2, 0, 2) {};% {$\left( 1, 0, 1 \right) $};
+        \node[codeword] (c110) at (2, 2, 0) {};% {$\left( 1, 1, 0 \right) $};
+        \node[codeword] (c011) at (0, 2, 2) {};% {$\left( 0, 1, 1 \right) $};
+
+        \node[color=KITblue, right=0cm of c000] {$\left( 0, 0, 0 \right) $};
+        \node[color=KITblue, above=0cm of c101] {$\left( 1, 0, 1 \right) $};
+        \node[color=KITblue, left=0cm of c110] {$\left( 1, 1, 0 \right) $};
+        \node[color=KITblue, left=-0.1cm of c011] {$\left( 0, 1, 1 \right) $};
+    \end{tikzpicture}
+
+    \caption{Codeword polytope of a single parity-check code}
+    \label{fig:dec:poly}
+\end{figure}
 
 \begin{itemize}
     \item Equivalent \ac{ML} optimization problem
@@ -254,7 +309,7 @@ continuous optimization problem, and one term representing the parity
 constraint, accomodating the role of the parity-check matrix $\boldsymbol{H}$.
 %
 The equal probability assumption is made on $\mathcal{C}\left( \boldsymbol{H} \right) $.
-The prior \ac{PDF} is then approximated using the code-constraint polynomial\todo{Italic?}:%
+The prior \ac{PDF} is then approximated using the code-constraint polynomial:%
 %
 \begin{align}
     f_{\boldsymbol{X}}\left( \boldsymbol{x} \right) =
@@ -267,12 +322,13 @@ The prior \ac{PDF} is then approximated using the code-constraint polynomial\tod
 %
 The authors justify this approximation by arguing that for
 $\gamma \rightarrow \infty$, the right-hand side aproaches the left-hand
-side. In \ref{eq:prox:vanilla_MAP} the prior \ac{PDF}
+side. In equation \ref{eq:prox:vanilla_MAP}, the prior \ac{PDF}
 $f_{\boldsymbol{X}}\left( \boldsymbol{x} \right) $ can then be subsituted
-for \ref{eq:prox:prior_pdf_approx} and the likelihood can be rewritten using
+for equation \ref{eq:prox:prior_pdf_approx} and the likelihood can be rewritten using
 the negative log-likelihood
-$f_{\boldsymbol{X} \mid \boldsymbol{Y}}\left( \boldsymbol{x} \mid \boldsymbol{y} \right)
-    = e^{- L\left( \boldsymbol{y} \mid \boldsymbol{x} \right) }$:%
+$L \left( \boldsymbol{y} \mid \boldsymbol{x} \right) = -\ln\left(
+        f_{\boldsymbol{X} \mid \boldsymbol{Y}}\left(
+            \boldsymbol{x} \mid \boldsymbol{y} \right) \right) $:%
 %
 \begin{align}
     \hat{\boldsymbol{x}} &= \argmax_{\boldsymbol{x} \in \mathbb{R}^{n}}
@@ -283,7 +339,7 @@ $f_{\boldsymbol{X} \mid \boldsymbol{Y}}\left( \boldsymbol{x} \mid \boldsymbol{y}
         + \gamma h\left( \boldsymbol{x} \right) 
         \right)%
     \label{eq:prox:approx_map_problem}
-\end{align}%
+.\end{align}%
 %
 Thus, with proximal decoding, the objective function
 $f\left( \boldsymbol{x} \right)$ to be minimized is%
@@ -292,11 +348,11 @@ $f\left( \boldsymbol{x} \right)$ to be minimized is%
     f\left( \boldsymbol{x} \right) = L\left( \boldsymbol{x} \mid \boldsymbol{y} \right)
         + \gamma h\left( \boldsymbol{x} \right)%
     \label{eq:prox:objective_function}
-.\end{align}\todo{Dot after equations?}
+.\end{align}
 
 For the solution of the approximalte \ac{MAP} decoding problem, the two parts
-of \ref{eq:prox:approx_map_problem} are considered separately from one
-another: the minimization of the objective function occurs in an alternating
+of equation \ref{eq:prox:approx_map_problem} are considered separately:
+the minimization of the objective function occurs in an alternating
 manner, switching between the minimization of the negative log-likelihood
 $L\left( \boldsymbol{y} \mid \boldsymbol{x} \right) $ and the scaled
 code-constaint polynomial $\gamma h\left( \boldsymbol{x} \right) $.
@@ -332,7 +388,7 @@ The second step thus becomes \todo{Write the formulation optimization problem pr
     \hspace{5mm}\gamma > 0,\text{ small}
 .\end{align*}
 %
-While the approximatin of the prior \ac{PDF} made in \ref{eq:prox:prior_pdf_approx}
+While the approximation of the prior \ac{PDF} made in \ref{eq:prox:prior_pdf_approx}
 theoretically becomes better
 with larger $\gamma$, the constraint that $\gamma$ be small is important,
 as it keeps the effect of $h\left( \boldsymbol{x} \right) $ on the landscape