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Diffstat (limited to 'homeworks')
| -rw-r--r-- | homeworks/hw-5.org | 2 | ||||
| -rw-r--r-- | homeworks/hw-5.pdf | bin | 81313 -> 81812 bytes | |||
| -rw-r--r-- | homeworks/hw-5.tex | 22 |
3 files changed, 12 insertions, 12 deletions
diff --git a/homeworks/hw-5.org b/homeworks/hw-5.org index d650375..a2339f9 100644 --- a/homeworks/hw-5.org +++ b/homeworks/hw-5.org @@ -19,7 +19,7 @@ Unless the following are met, the resulting solution will be garbage. 1. The matrix $U$ must be not be singular. 2. $U$ must be square (or it will fail the ~assert~). 3. The system created by $Ux = b$ must be consistent. -4. $U$ is in (obviously) upper-triangular form. +4. $U$ is (quite obviously) in upper-triangular form. Thus, the actual calculation performing the $LU$ decomposition (in ~lu_decomp~) does a sanity diff --git a/homeworks/hw-5.pdf b/homeworks/hw-5.pdf Binary files differindex ce43787..a7773bc 100644 --- a/homeworks/hw-5.pdf +++ b/homeworks/hw-5.pdf diff --git a/homeworks/hw-5.tex b/homeworks/hw-5.tex index 8b9d24b..98cca2e 100644 --- a/homeworks/hw-5.tex +++ b/homeworks/hw-5.tex @@ -1,4 +1,4 @@ -% Created 2023-10-30 Mon 19:05 +% Created 2023-11-01 Wed 20:49 % Intended LaTeX compiler: pdflatex \documentclass[11pt]{article} \usepackage[utf8]{inputenc} @@ -29,7 +29,7 @@ \setlength\parindent{0pt} \section{Question One} -\label{sec:org88abf18} +\label{sec:org4e80298} See LIZFCM \(\rightarrow\) Matrix Routines \(\rightarrow\) \texttt{lu decomp} \& \texttt{bsubst}. The test \texttt{UTEST(matrix, lu\_decomp)} is a unit test for the \texttt{lu\_decomp} routine, @@ -39,14 +39,14 @@ and \texttt{UTEST(matrix, bsubst)} verifies back substitution on an upper triang Both can be found in \texttt{tests/matrix.t.c}. \section{Question Two} -\label{sec:org098a7f1} +\label{sec:orga73d05c} Unless the following are met, the resulting solution will be garbage. \begin{enumerate} \item The matrix \(U\) must be not be singular. \item \(U\) must be square (or it will fail the \texttt{assert}). \item The system created by \(Ux = b\) must be consistent. -\item \(U\) is in (obviously) upper-triangular form. +\item \(U\) is (quite obviously) in upper-triangular form. \end{enumerate} Thus, the actual calculation performing the \(LU\) decomposition @@ -55,41 +55,41 @@ check for 1-3 will fail an assert, should a point along the diagonal (pivot) be zero, or the matrix be non-factorable. \section{Question Three} -\label{sec:org40d5983} +\label{sec:org35163c5} See LIZFCM \(\rightarrow\) Matrix Routines \(\rightarrow\) \texttt{fsubst}. \texttt{UTEST(matrix, fsubst)} verifies forward substitution on a lower triangular 3 \texttimes{} 3 matrix with a known solution that can be verified manually. \section{Question Four} -\label{sec:orgf7d23bb} +\label{sec:org79d9061} See LIZFCM \(\rightarrow\) Matrix Routines \(\rightarrow\) \texttt{gaussian\_elimination} and \texttt{solve\_gaussian\_elimination}. \section{Question Five} -\label{sec:org54e966c} +\label{sec:orgc6ac464} See LIZFCM \(\rightarrow\) Matrix Routines \(\rightarrow\) \texttt{m\_dot\_v}, and the \texttt{UTEST(matrix, m\_dot\_v)} in \texttt{tests/matrix.t.c}. \section{Question Six} -\label{sec:org413b527} +\label{sec:org66fedab} See \texttt{UTEST(matrix, solve\_gaussian\_elimination)} in \texttt{tests/matrix.t.c}, which generates a diagonally dominant 10 \texttimes{} 10 matrix and shows that the solution is consistent with the initial matrix, according to the steps given. Then, we do a dot product between each row of the diagonally dominant matrix and the solution vector to ensure it is near equivalent to the input vector. \section{Question Seven} -\label{sec:orgd3d7443} +\label{sec:org6897ff2} See \texttt{UTEST(matrix, solve\_matrix\_lu\_bsubst)} which does the same test in Question Six with the solution according to \texttt{solve\_matrix\_lu\_bsubst} as shown in the Software Manual. \section{Question Eight} -\label{sec:orgf8ac9bf} +\label{sec:org5d529dd} No, since the time complexity for Gaussian Elimination is always less than that of the LU factorization solution by \(O(n^2)\) operations (in LU factorization we perform both backwards and forwards substitutions proceeding the LU decomp, in Gaussian Elimination we only need back substitution). \section{Question Nine, Ten} -\label{sec:orgb270171} +\label{sec:org0fb8e09} See LIZFCM Software manual and shared library in \texttt{dist} after compiling. \end{document}
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