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+\documentclass[12pt]{article}
+
+\usepackage{karnaugh-map}
+\usepackage[utf8]{inputenc}
+\usepackage[margin=2cm]{geometry}
+
+\title{Howework 5 -- Computer Architecture}
+\author{Claudio Maggioni \and Tommaso Rodolfo Masera}
+
+\newcommand{\bn}[1]{
+	\overline{#1}
+}
+
+\begin{document}
+\maketitle
+\section{Question 1}
+\subsection{\label{sec:11}Sub-question 1}
+\paragraph{Starting with:}
+$f(A, B, C, D) = \bn{A} \times \bn{B} \times \bn{C} \times \bn{D} + 
+\bn{A} \times \bn{B} \times C \times \bn{D} + \bn{A} \times B \times C \times \bn{D} +
+A \times \bn{B} \times \bn{C} \times \bn{D} + A \times \bn{B} \times \bn{C} \times D +
+A \times \bn{B} \times C \times \bn{D} + A \times B \times C \times D + 
+\bn{A} \times \bn{B} \times \bn{C} \times D + \bn{A} \times B \times C \times D =$
+
+\paragraph{apply the distributive law:}
+$\bn{A} \times (\bn{B} \times \bn{C} \times \bn{D} + \bn{B} \times C \times \bn{D} + 
+B \times C \times \bn{D} + \bn{B} \times \bn{C} \times D + B \times C \times D) +
+A \times (\bn{B} \times \bn{C} \times \bn{D} + \bn{B} \times \bn{C} \times D +
+\bn{B} \times C \times \bn{D} + B \times C \times D) =$
+
+\paragraph{apply it again:}
+$\bn{A} \times (\bn{B} \times (\bn{C} \times \bn{D} + C \times \bn{D} + \bn{C} \times D)
++ B \times C \times (\bn{D} + D)) + A \times (\bn{B} \times (\bn{C} \times \bn{D} + \bn{C} \times D + 
+C \times \bn{D}) + B \times C \times D) = $
+
+
+\subsubsection{\label{sec:cd}Solving $\bn{C} \times \bn{D} + C \times \bn{D} + \bn{C} \times D$}   
+
+\paragraph{Apply the distributive law:}
+$\bn{C} \times (\bn{D} + D) + C \times \bn{D} = $i
+
+\paragraph{then apply the inverse law:}
+$\bn{C} \times 1 + C \times \bn{D} =$
+
+\paragraph{then apply the identity law:}
+$\bn{C} + C \times \bn{D} =$
+
+\paragraph{then apply De Morgan's law:}
+$\bn{C \times \bn{C \times \bn{D}}} =$
+
+\paragraph{then apply it again:}
+$\bn{C \times (\bn{C} + D)} =$
+
+\paragraph{then apply the distributive law:}
+$\bn{C \times \bn{C} + C \times D} =$
+
+\paragraph{then apply the inverse law:}
+$\bn{0 + C \times D} =$
+
+\paragraph{then, finally, apply the identity law, obtaining:}
+$\bn{C \times D}$
+    
+\subsubsection{Back to the main function} 
+\paragraph{by \ref{sec:cd} and the inverse law, we continue this way:}
+$\bn{A} \times (\bn{B} \times \bn{C \times D} + B \times C \times 1) +
+A \times (\bn{B} \times \bn{C \times D}) + B \times C \times D) =$
+
+\paragraph{then apply the identity law:}
+$\bn{A} \times (\bn{B} \times \bn{C \times D} + B \times C) +
+A \times (\bn{B} \times \bn{C \times D}) + B \times C \times D) =$
+
+\paragraph{then apply the distributive law:}
+$\bn{A} \times \bn{B} \times \bn{C \times D}) + \bn{A} \times B \times C +
+A \times \bn{B} \times \bn{C \times D} + A \times B \times C \times D =$
+
+\paragraph{then apply it again:}
+$\bn{B} \times \bn{C \times D} \times (\bn{A} + A) + B \times C \times (\bn{A} + A \times D) =$
+
+\paragraph{then apply the inverse law:}
+$\bn{B} \times \bn{C \times D} \times 1 + B \times C \times (\bn{A} + A \times D) =$
+
+\paragraph{then apply the identity law:}
+$\bn{B} \times \bn{C \times D} + B \times C \times (\bn{A} + A \times D) =$
+
+\paragraph{then apply De Morgan's law:}
+$\bn{B} \times \bn{C \times D} + B \times C \times \bn{A \times \bn{A \times D}} =$
+
+\paragraph{then apply it again:}
+$\bn{B} \times \bn{C \times D} + B \times C \times \bn{A \times (\bn{A} + \bn{D})} =$
+
+\paragraph{then apply the distributive law:}
+$\bn{B} \times \bn{C \times D} + B \times C \times \bn{A \times \bn{A} + A \times \bn{D}} =$
+
+\paragraph{then apply the inverse law:}
+$\bn{B} \times \bn{C \times D} + B \times C \times \bn{0 + A \times \bn{D}} =$
+
+\paragraph{then apply the identity law:}
+$\bn{B} \times \bn{C \times D} + B \times C \times \bn{A \times \bn{D}} =$
+
+\paragraph{then, finally, apply De Morgan's law, we find the result:}
+$\bn{B} \times \bn{C \times D} + B \times C \times (\bn{A} + D)$
+
+\subsection{Sub-question 2}
+Using the minterm normal form given in section \ref{sec:11}, we can deduct the truth table of the function. Using that, we can find the following Karnaugh map and the following prime implicants:
+
+\begin{figure}[h]
+\centering{
+	\begin{karnaugh-map}[4][4][1][AB][CD]
+		\maxterms{1,3,5,7,11,12,14}
+		\minterms{0,2,4,6,8,9,10,13,15}
+		\implicant{1}{7}
+		\implicantedge{12}{12}{14}{14}
+		\implicantedge{3}{3}{11}{11}
+	\end{karnaugh-map}
+}
+\end{figure}
+
+Using those prime implicants we find the boolean function $(\bn{B} + C) \times (\bn{A} + \bn{B}
++ D) \times (B + \bn{C} + \bn{D})$.
+
+\section{Question 2}
+\subsection{Sub-question 1}
+The truth table of the function is:
+
+\begin{figure}[h]
+\centering{
+	\begin{tabular}{cccc|c}
+	\textbf{$X_3$} & \textbf{$X_2$} & \textbf{$X_1$} & \textbf{$X_0$} & \textbf{$Y$} \\ \hline
+	0 & 0 & 0 & 0 & 0 \\
+	0 & 0 & 0 & 1 & 1 \\
+	0 & 0 & 1 & 0 & 0 \\
+	0 & 0 & 1 & 1 & 0 \\
+	0 & 1 & 0 & 0 & 0 \\
+	0 & 1 & 0 & 1 & 0 \\
+	0 & 1 & 1 & 0 & 1 \\
+	0 & 1 & 1 & 1 & 1 \\
+	1 & 0 & 0 & 0 & 1 \\
+	1 & 0 & 0 & 1 & 1 \\
+	1 & 0 & 1 & 0 & 1 \\
+	1 & 0 & 1 & 1 & 1 \\
+	1 & 1 & 0 & 0 & 1 \\
+	1 & 1 & 0 & 1 & 1 \\
+	1 & 1 & 1 & 0 & 1 \\
+	1 & 1 & 1 & 1 & 1 \\
+	\end{tabular}	
+}
+\end{figure}
+
+\subsection{Sub-question 2}
+The Conjunctive Normal Form, or the maxterm expansion of the function, is:
+
+\begin{figure}[h]
+$(X_3 + X_2 + X_1 + X_0) \times
+(X_3 + X_2 + \bn{X_1} + X_0) \times (X_3 + X_2 + \bn{X_1} + \bn{X_0}) \times
+(X_3 + \bn{X_2} + X_1 + X_0) \times (X_3 + \bn{X_2} + X_1 + \bn{X_0})$ 
+\end{figure}
+
+The Disjunctive Normal Form, or the minterm expansion of the function, is:
+
+\begin{figure}[h]
+$(\bn{X_3} \times \bn{X_2} \times \bn{X_1} \times X_0) + (\bn{X_3} \times X_2 \times X_1 \times \bn{X_0}) +
+(\bn{X_3} \times X_2 \times X_1 \times X_0) + (X_3 \times \bn{X_2} \times \bn{X_1} \times \bn{X_0}) +
+(X_3 \times \bn{X_2} \times \bn{X_1} \times X_0) + (X_3 \times \bn{X_2} \times X_1 \times \bn{X_0}) +
+(X_3 \times \bn{X_2} \times X_1 \times X_0) + (X_3 \times X_2 \times \bn{X_1} \times \bn{X_0}) +
+(X_3 \times X_2 \times \bn{X_1} \times X_0) + (X_3 \times X_2 \times X_1 \times \bn{X_0}) +
+(X_3 \times X_2 \times X_1 \times X_0)$
+\end{figure}
+
+\subsection{Sub-question 3}
+The maxterm expansion of the function above is clearly the best approach between the two,
+ since it contains fewer terms.
+ 
+\subsection{Sub-question 4}
+TODO
+
+\subsection{Sub-question 5}
+TODO 
+ 
+\section{Question 3}
+The bomb will not explode if either the first, the second or the fourth cable from the left
+are cut. For the first and the second cable this happens because the NOR inside the circuit will
+get at least a 1 as input, and therefore it will produce a 0 as output, pulling the final AND
+output to 0. When the fourth cable is cut, the NOT will give always 0 as output and therefore
+the final AND output will be always 0.
+\end{document}
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