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