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Task 2 Add comments and code task3

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Krzysztof Rudnicki 2021-11-11 20:12:08 +01:00
parent a7d9a9da18
commit ec05aa206f
16 changed files with 898 additions and 129 deletions
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24
ENUME/projectA/jacobiMethod.m
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@ -9,7 +9,7 @@ function [L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, flag] =
[L, D, U] = decomposeMatrix(Matrix); [L, D, U] = decomposeMatrix(Matrix);
initial_x = ones(Rows, 1); initial_x = ones(Rows, 1);
whichIterationAreWeOn = 0; whichIterationAreWeOn = 0;
demandedTolerance = 1e-10; demandedTolerance = 1e-10; % as per task description
flag = 0; flag = 0;
end end
@ -24,7 +24,7 @@ function [L, D, U] = decomposeMatrix(Matrix)
end end
function [x, whichIterationAreWeOn, demandedTolerance] = jacobiLoop(Matrix, L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, Vector, flag) function [x, whichIterationAreWeOn, demandedTolerance] = jacobiLoop(Matrix, L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, Vector, flag)
while flag ~= 1 while flag ~= 1 % flag denotes whether norm(Matrix*x-Vector) <= demandedTolerance
[x, whichIterationAreWeOn, demandedTolerance, flag, initial_x] = jacobiInsideLoop(Matrix, L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, Vector); [x, whichIterationAreWeOn, demandedTolerance, flag, initial_x] = jacobiInsideLoop(Matrix, L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, Vector);
end end
end end
@ -42,16 +42,16 @@ function x = jacobiEquation(D, L, U, initial_x, Vector)
end end
function [flag, demandedTolerance] = checkError(x, initial_x, demandedTolerance, Matrix, Vector) function [flag, demandedTolerance] = checkError(x, initial_x, demandedTolerance, Matrix, Vector)
flag = 0; flag = 0;
currentError = norm(x - initial_x); currentError = norm(x - initial_x);
if currentError <= demandedTolerance if currentError <= demandedTolerance
currentError = norm(Matrix*x-Vector); currentError = norm(Matrix*x-Vector);
if currentError <= demandedTolerance if currentError <= demandedTolerance % if sequence as per textbook
flag = 1; flag = 1;
else else
demandedTolerance = demandedTolerance * 2; demandedTolerance = demandedTolerance * 2; % arbitrary value
end
end end
end
end end
function [initial_x, whichIterationAreWeOn, flag] = endOfLoop(x, whichIterationAreWeOn) function [initial_x, whichIterationAreWeOn, flag] = endOfLoop(x, whichIterationAreWeOn)
@ -60,8 +60,6 @@ function [initial_x, whichIterationAreWeOn, flag] = endOfLoop(x, whichIterationA
flag = 0; flag = 0;
end end
function dispFinalResults(demandedTolerance, whichIterationAreWeOn, Matrix, Vector) function dispFinalResults(demandedTolerance, whichIterationAreWeOn, Matrix, Vector)
disp("Final demandedTolerance"); disp("Final demandedTolerance");
disp(demandedTolerance); disp(demandedTolerance);

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ENUME/projectA/projectA.out
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@ -39,3 +39,13 @@
\BOOKMARK [2][-]{subsection.5.1.10}{backSubstitutionPhase}{section.5.1}% 39 \BOOKMARK [2][-]{subsection.5.1.10}{backSubstitutionPhase}{section.5.1}% 39
\BOOKMARK [2][-]{subsection.5.1.11}{iterativeResidualCorrection}{section.5.1}% 40 \BOOKMARK [2][-]{subsection.5.1.11}{iterativeResidualCorrection}{section.5.1}% 40
\BOOKMARK [2][-]{subsection.5.1.12}{improveSolution}{section.5.1}% 41 \BOOKMARK [2][-]{subsection.5.1.12}{improveSolution}{section.5.1}% 41
\BOOKMARK [1][-]{section.5.2}{Task 3e code}{chapter.5}% 42
\BOOKMARK [2][-]{subsection.5.2.1}{jacobiMethod}{section.5.2}% 43
\BOOKMARK [2][-]{subsection.5.2.2}{initializeValues}{section.5.2}% 44
\BOOKMARK [2][-]{subsection.5.2.3}{decomposeMatrix}{section.5.2}% 45
\BOOKMARK [2][-]{subsection.5.2.4}{jacobiLoop}{section.5.2}% 46
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\BOOKMARK [2][-]{subsection.5.2.7}{checkError}{section.5.2}% 49
\BOOKMARK [2][-]{subsection.5.2.8}{endOfLoop}{section.5.2}% 50
\BOOKMARK [2][-]{subsection.5.2.9}{dispFinalResults}{section.5.2}% 51

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ENUME/projectA/projectA.tex
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@ -748,6 +748,124 @@ end % end function
\end{lstlisting} \end{lstlisting}
\end{simplechar} \end{simplechar}
\section{Task 3e code}
\subsection{jacobiMethod}
\begin{simplechar}
\begin{lstlisting}
function x = jacobiMethod(Matrix, Vector)
[L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, flag] = initializeValues(Matrix);
[x, whichIterationAreWeOn, demandedTolerance] = jacobiLoop(Matrix, L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, Vector, flag);
dispFinalResults(demandedTolerance, whichIterationAreWeOn, Matrix, Vector);
end
\end{lstlisting}
\end{simplechar}
\subsection{initializeValues}
\begin{simplechar}
\begin{lstlisting}
function [L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, flag] = initializeValues(Matrix)
[Rows, ~] = size(Matrix);
[L, D, U] = decomposeMatrix(Matrix);
initial_x = ones(Rows, 1);
whichIterationAreWeOn = 0;
demandedTolerance = 1e-10; % as per task description
flag = 0;
end
\end{lstlisting}
\end{simplechar}
\subsection{decomposeMatrix}
\begin{simplechar}
\begin{lstlisting}
function [L, D, U] = decomposeMatrix(Matrix)
D = diag(diag(Matrix));
U = triu(Matrix, 1); % Generates upper triangular part of matrix
% where the second variable denotes on which diagonal of matrix should we
% start
L = tril(Matrix, -1); % Generates lower triangular part of matrix
% where the second variable denotes on which diagonal of matrix should we
% start
end
\end{lstlisting}
\end{simplechar}
\subsection{jacobiLoop}
\begin{simplechar}
\begin{lstlisting}
function [x, whichIterationAreWeOn, demandedTolerance] = jacobiLoop(Matrix, L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, Vector, flag)
while flag ~= 1 % flag denotes whether norm(Matrix*x-Vector) <= demandedTolerance
[x, whichIterationAreWeOn, demandedTolerance, flag, initial_x] = jacobiInsideLoop(Matrix, L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, Vector);
end
end
\end{lstlisting}
\end{simplechar}
\subsection{jacobiInsideLoop}
\begin{simplechar}
\begin{lstlisting}
function [x, whichIterationAreWeOn, demandedTolerance, flag, initial_x] = jacobiInsideLoop(Matrix, L, D, U, initial_x, whichIterationAreWeOn, demandedTolerance, Vector)
x = jacobiEquation(D, L, U, initial_x, Vector);
[flag, demandedTolerance] = checkError(x, initial_x, demandedTolerance, Matrix, Vector);
[initial_x, whichIterationAreWeOn] = endOfLoop(x, whichIterationAreWeOn);
end
\end{lstlisting}
\end{simplechar}
\subsection{jacobiEquation}
\begin{simplechar}
\begin{lstlisting}
function x = jacobiEquation(D, L, U, initial_x, Vector)
x = - D \ ( L + U ) * initial_x + D \ Vector; % As per formula
% We will be using D \ Vector and D \ ( ) instead of inverseD since
% this is faster according to matlab
end
\end{lstlisting}
\end{simplechar}
\subsection{checkError}
\begin{simplechar}
\begin{lstlisting}
function [flag, demandedTolerance] = checkError(x, initial_x, demandedTolerance, Matrix, Vector)
flag = 0;
currentError = norm(x - initial_x);
if currentError <= demandedTolerance
currentError = norm(Matrix*x-Vector);
if currentError <= demandedTolerance % if sequence as per textbook
flag = 1;
else
demandedTolerance = demandedTolerance * 2; % arbitrary value
end
end
end
\end{lstlisting}
\end{simplechar}
\subsection{endOfLoop}
\begin{simplechar}
\begin{lstlisting}
function [initial_x, whichIterationAreWeOn, flag] = endOfLoop(x, whichIterationAreWeOn)
initial_x = x;
whichIterationAreWeOn = whichIterationAreWeOn + 1;
flag = 0;
end
\end{lstlisting}
\end{simplechar}
\subsection{dispFinalResults}
\begin{simplechar}
\begin{lstlisting}
function dispFinalResults(demandedTolerance, whichIterationAreWeOn, Matrix, Vector)
disp("Final demandedTolerance");
disp(demandedTolerance);
disp("Final Iteration: ");
disp(whichIterationAreWeOn);
disp("A\b matlab:");
disp(Matrix \ Vector);
end
\end{lstlisting}
\end{simplechar}
\begin{thebibliography}{9} \begin{thebibliography}{9}
\bibitem{texbook} \bibitem{texbook}
Piotr Tatjewski (2014) \emph{Numerical Methods}, Oficyna Wydawnicza Politechniki Warszawskiej Piotr Tatjewski (2014) \emph{Numerical Methods}, Oficyna Wydawnicza Politechniki Warszawskiej

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@ -51,3 +51,13 @@
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\contentsline {subsection}{\numberline {5.1.12}improveSolution}{22}{subsection.5.1.12}% \contentsline {subsection}{\numberline {5.1.12}improveSolution}{22}{subsection.5.1.12}%
\contentsline {section}{\numberline {5.2}Task 3e code}{23}{section.5.2}%
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\contentsline {subsection}{\numberline {5.2.9}dispFinalResults}{25}{subsection.5.2.9}%

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@ -14,16 +14,16 @@
Project number 62} Project number 62}
\date{November 12, 2020} \date{November 12, 2020}
\maketitle \maketitle
\numberwithin{equation}{section} \numberwithin{equation}{section}
\setcounter{section}{-1} \setcounter{section}{-1}
\section{Abstract} \section{Abstract}
This project explores the numerical methods of solving systems of linear This project explores the numerical methods of solving systems of linear
equations and finding eigenvalues of matrices. The following concepts are equations and finding eigenvalues of matrices. The following concepts are
discussed within this document: discussed within this document:
\begin{itemize} \begin{itemize}
\item Computing the machine epsilon of an environment \item Computing the machine epsilon of an environment
\item Solving linear equation systems using Gaussian elimination with \item Solving linear equation systems using Gaussian elimination with
@ -33,22 +33,22 @@
\item Finding the eigenvalues of a symmetric matrix using the QR method \item Finding the eigenvalues of a symmetric matrix using the QR method
with and without shifts with and without shifts
\end{itemize} \end{itemize}
\newpage \newpage
\section{Task 1: Computing the machine epsilon} \section{Task 1: Computing the machine epsilon}
\subsection{Overview} \subsection{Overview}
The goal of this task was to find the machine epsilon of the MATLAB The goal of this task was to find the machine epsilon of the MATLAB
environment, i.e. the maximum relative error of a floating-point environment, i.e. the maximum relative error of a floating-point
representation of a number. representation of a number.
\subsection{Implementation} \subsection{Implementation}
The computation of the machine epsilon is based on the following two The computation of the machine epsilon is based on the following two
observations: observations:
\begin{enumerate} \begin{enumerate}
\item In binary computer environments, the epsilon is a negative power \item In binary computer environments, the epsilon is a negative power
of two. of two.
@ -59,43 +59,43 @@
i.e. the smallest representable number such that, when added to $1$, i.e. the smallest representable number such that, when added to $1$,
has a floating point representation greater than $1$. has a floating point representation greater than $1$.
\end{enumerate} \end{enumerate}
An arbitrary starting point, in this case $2^0$, can thus be chosen, and its An arbitrary starting point, in this case $2^0$, can thus be chosen, and its
value can then be sequentially halved until the criterion $fl(1 + x) > 1$ is value can then be sequentially halved until the criterion $fl(1 + x) > 1$ is
no longer met. no longer met.
\subsection{Program output} \subsection{Program output}
\begin{verbatim} \begin{verbatim}
Found epsilon is 2.2204e-16 Found epsilon is 2.2204e-16
Epsilon is 2.2204e-16 Epsilon is 2.2204e-16
\end{verbatim} \end{verbatim}
\subsection{Observations} \subsection{Observations}
An epsilon of $2.2204 \cdot 10^{-16} = 2^{-52}$ is indicative of a standard An epsilon of $2.2204 \cdot 10^{-16} = 2^{-52}$ is indicative of a standard
double-precision IEEE 754 floating point binary representation. double-precision IEEE 754 floating point binary representation.
\newpage \newpage
\section{Task 2: Solving systems of linear equations using Gaussian \section{Task 2: Solving systems of linear equations using Gaussian
elimination} elimination}
\subsection{Overview} \subsection{Overview}
The task consisted of two subtasks, each specifying a different system The task consisted of two subtasks, each specifying a different system
of equations to solve. The systems were to be solved using Gaussian of equations to solve. The systems were to be solved using Gaussian
elimination with partial pivoting. The size of the system was sequentially elimination with partial pivoting. The size of the system was sequentially
doubled until the computation time became prohibitive. The error of each doubled until the computation time became prohibitive. The error of each
solution was to be noted. solution was to be noted.
Additionally, for systems whose number of equations was equal to 10, Additionally, for systems whose number of equations was equal to 10,
residual correction was to be applied and the solution was to be noted. residual correction was to be applied and the solution was to be noted.
The systems were given by: The systems were given by:
\begin{align*} \begin{align*}
&A_{ij}^{(a)} = &A_{ij}^{(a)} =
\begin{cases} \begin{cases}
4 & \text{for } i = j \\ 4 & \text{for } i = j \\
1 & \text{for } i = j \pm 1 \\ 1 & \text{for } i = j \pm 1 \\
0 & \text{otherwise} 0 & \text{otherwise}
@ -106,27 +106,27 @@
&A_{ij}^{(b)} = \frac{6}{7(i + j + 1)} &A_{ij}^{(b)} = \frac{6}{7(i + j + 1)}
&& &&
b_i^{(b)} = b_i^{(b)} =
\begin{cases} \begin{cases}
\frac{1}{3i} & \text{for even } i\\ \frac{1}{3i} & \text{for even } i\\
0 & \text{for odd } i 0 & \text{for odd } i
\end{cases} \end{cases}
\end{align*} \end{align*}
\subsection{Implementation} \subsection{Implementation}
Gaussian elimination is a process that takes a matrix describing a linear Gaussian elimination is a process that takes a matrix describing a linear
equation system and returns its LU (lower-upper) decomposition, enabling the equation system and returns its LU (lower-upper) decomposition, enabling the
system to be cheaply solved for any vector $b$. The triangular matrices $L$ system to be cheaply solved for any vector $b$. The triangular matrices $L$
and $U$ can be supplemented with $b$ to find the solution vector $x$ through and $U$ can be supplemented with $b$ to find the solution vector $x$ through
a process called back-substitution. a process called back-substitution.
\subsubsection{LU decomposition of a matrix} \subsubsection{LU decomposition of a matrix}
The function \texttt{gausseli} takes an equation system and returns the $L$ The function \texttt{gausseli} takes an equation system and returns the $L$
matrix, the $U$ matrix, the permutation matrix $P$ as well as a matrix matrix, the $U$ matrix, the permutation matrix $P$ as well as a matrix
describing the entire system post-elimination, to be immediately handed to describing the entire system post-elimination, to be immediately handed to
back-substitution for the calculation of the result. back-substitution for the calculation of the result.
The elimination itself is done by traveling along the diagonal of The elimination itself is done by traveling along the diagonal of
\texttt{eqsys} and for each encountered cell, eliminating (reducing to $0$) \texttt{eqsys} and for each encountered cell, eliminating (reducing to $0$)
the coefficients directly below it. The program does so by multiplying the the coefficients directly below it. The program does so by multiplying the
@ -134,54 +134,54 @@
This constant, called \texttt{reductor} in the code, is given as the This constant, called \texttt{reductor} in the code, is given as the
ratio of the cell to be reduced and the current diagonal cell. The reductor ratio of the cell to be reduced and the current diagonal cell. The reductor
is then stored in the $L$ matrix. is then stored in the $L$ matrix.
\subsubsection{Partial pivoting} \subsubsection{Partial pivoting}
The above algorithm alone is prone to errors when a value on the diagonal is The above algorithm alone is prone to errors when a value on the diagonal is
equal to $0$ and the reductor cannot be constructed. For this reason, equal to $0$ and the reductor cannot be constructed. For this reason,
partial pivoting is added. partial pivoting is added.
For each cell on the diagonal, the program looks at the cells beneath it and For each cell on the diagonal, the program looks at the cells beneath it and
finds the one with the greatest absolute value. If it exceeds the absolute finds the one with the greatest absolute value. If it exceeds the absolute
value of the current cell, the corresponding two rows are swapped. This value of the current cell, the corresponding two rows are swapped. This
ensures that the divisor making up the reductor is as far from zero as ensures that the divisor making up the reductor is as far from zero as
possible. possible.
Because the input matrix changes while being decomposed, a permutation Because the input matrix changes while being decomposed, a permutation
matrix $P$ is introduced to enable its recomposition from $L$ and $U$, matrix $P$ is introduced to enable its recomposition from $L$ and $U$,
satisfying the equation: satisfying the equation:
\begin{equation} \begin{equation}
LU = PA LU = PA
\end{equation} \end{equation}
Finally, the matrix $L$ is also permuted to reflect changes in the layout Finally, the matrix $L$ is also permuted to reflect changes in the layout
of the system. of the system.
\subsubsection{Back-substitution} \subsubsection{Back-substitution}
A triangular matrix supplied with a constant vector $b$ can be easily solved A triangular matrix supplied with a constant vector $b$ can be easily solved
by traveling along its diagonal and eliminating the factors directly above by traveling along its diagonal and eliminating the factors directly above
the visited cell. In the end, the matrix is transformed into an identity the visited cell. In the end, the matrix is transformed into an identity
matrix and the vector alongside it is transformed into $x$. matrix and the vector alongside it is transformed into $x$.
\subsubsection{Residual correction} \subsubsection{Residual correction}
Residual correction is a method of improving the accuracy of the results Residual correction is a method of improving the accuracy of the results
obtained from Gaussian elimination. By reusing the already obrained matrices obtained from Gaussian elimination. By reusing the already obrained matrices
$L$, $U$ and $P$, the system is solved once again, this time with $b$ $L$, $U$ and $P$, the system is solved once again, this time with $b$
replaced with the vector of errors of the solution. The resulting vector replaced with the vector of errors of the solution. The resulting vector
$\delta x$ is then subtracted from the previous solution. This procedure may $\delta x$ is then subtracted from the previous solution. This procedure may
be repeated as needed. be repeated as needed.
\subsection{Program output} \subsection{Program output}
\subsubsection{Error plots} \subsubsection{Error plots}
\includegraphics[width=\textwidth]{aerror} \includegraphics[width=\textwidth]{aerror}
\includegraphics[width=\textwidth]{berror} \includegraphics[width=\textwidth]{berror}
\subsubsection{Solutions for $n = 10$} \subsubsection{Solutions for $n = 10$}
\begin{multicols}{2} \begin{multicols}{2}
\begin{verbatim} \begin{verbatim}
Solution to subtask a: Solution to subtask a:
@ -250,37 +250,37 @@
Error: 0.00020161 Error: 0.00020161
\end{verbatim} \end{verbatim}
\end{multicols} \end{multicols}
\newpage \newpage
\subsection{Observations} \subsection{Observations}
Matrices from both subtasks a and b lent themselves nicely to being solved Matrices from both subtasks a and b lent themselves nicely to being solved
using Gaussian elimination. Their absolute errors, when divided by the using Gaussian elimination. Their absolute errors, when divided by the
solutions themselves, resulted in very small relative errors, nearing the solutions themselves, resulted in very small relative errors, nearing the
machine epsilon. machine epsilon.
The computation time became prohibitive beyond $n = 320$, showing that The computation time became prohibitive beyond $n = 320$, showing that
Gaussian elimination is an effective method of solving large systems Gaussian elimination is an effective method of solving large systems
of equations. of equations.
Residual correction was very effective for the matrix from subtask a, Residual correction was very effective for the matrix from subtask a,
reducing the error to a perfect $0$. For subtask b, on the other hand, no reducing the error to a perfect $0$. For subtask b, on the other hand, no
matter the amount of iterations, the error could not be reduced any further. matter the amount of iterations, the error could not be reduced any further.
This might be related to a high condition number of the matrix, This might be related to a high condition number of the matrix,
$2.4225 \cdot 10^{14}$, prohibiting it from producing accurate enough $2.4225 \cdot 10^{14}$, prohibiting it from producing accurate enough
results. results.
\newpage \newpage
\section{Task 3: Solving systems of linear equations using the Jacobi and \section{Task 3: Solving systems of linear equations using the Jacobi and
Gauss-Seidel algorithms} Gauss-Seidel algorithms}
\subsection{Overview} \subsection{Overview}
In this task, a predefined, fixed-size system of equations was to be solved In this task, a predefined, fixed-size system of equations was to be solved
using the Jacobi and Gauss-Seidel iterative algorithms. The algorithms were using the Jacobi and Gauss-Seidel iterative algorithms. The algorithms were
to be iterated until the accuracy $\lVert Ax_k - b \rVert < 10^{-10}$ was to be iterated until the accuracy $\lVert Ax_k - b \rVert < 10^{-10}$ was
achieved. The error of each iteration was to be noted. achieved. The error of each iteration was to be noted.
The system to be solved: The system to be solved:
\begin{align*} \begin{align*}
18x_1 + 2x_2 - 3x_3 + x_4 & = 7\\ 18x_1 + 2x_2 - 3x_3 + x_4 & = 7\\
@ -288,12 +288,12 @@
x_1 + 3x_2 + 13x_3 - 8x_4 & = 24\\ x_1 + 3x_2 + 13x_3 - 8x_4 & = 24\\
x_1 + x_2 - 2x_3 - 10x_4 & = 20 x_1 + x_2 - 2x_3 - 10x_4 & = 20
\end{align*} \end{align*}
Afterwards, one of the algorithms was to be applied to matrices from task 2 Afterwards, one of the algorithms was to be applied to matrices from task 2
a and b, for $n = 10$. a and b, for $n = 10$.
\subsection{Implementation} \subsection{Implementation}
The general formula describing iterative methods of solving linear equation The general formula describing iterative methods of solving linear equation
systems is: systems is:
\begin{equation}\label{eq:1} \begin{equation}\label{eq:1}
@ -301,41 +301,41 @@
\end{equation} \end{equation}
Where the matrix $M$ and the vector $w$ are specific to the chosen Where the matrix $M$ and the vector $w$ are specific to the chosen
algorithm. algorithm.
\subsubsection{The Jacobi algorithm} \subsubsection{The Jacobi algorithm}
The first step of the Jacobi algorithm is to split the matrix $A$ into three The first step of the Jacobi algorithm is to split the matrix $A$ into three
matrices: the lower triangular matrix, the upper triangular matrix and the matrices: the lower triangular matrix, the upper triangular matrix and the
diagonal matrix. diagonal matrix.
Because the diagonal matrix is only ever used in its inverse from, the Because the diagonal matrix is only ever used in its inverse from, the
function doing the splitting only returns the inverse diagonal matrix. It function doing the splitting only returns the inverse diagonal matrix. It
utilizes the fact that the inverse of a diagonal matrix is a matrix of the utilizes the fact that the inverse of a diagonal matrix is a matrix of the
inverse of its diagonal elements. inverse of its diagonal elements.
Once the matrices $L$, $U$ and $D^{-1}$ have been obtained, the parameters Once the matrices $L$, $U$ and $D^{-1}$ have been obtained, the parameters
$M$ and $w$ are given by: $M$ and $w$ are given by:
\begin{align*} \begin{align*}
M &= -D^{-1}(L + U)\\ M &= -D^{-1}(L + U)\\
w &= D^{-1}b w &= D^{-1}b
\end{align*} \end{align*}
The step described in \eqref{eq:1} is then performed until the desired The step described in \eqref{eq:1} is then performed until the desired
accuracy is achieved. accuracy is achieved.
\subsubsection{The Gauss-Seidel algorithm} \subsubsection{The Gauss-Seidel algorithm}
The Gauss-Seidel algorithm is similar to the Jacobi algorithm in that it The Gauss-Seidel algorithm is similar to the Jacobi algorithm in that it
utilizes the lower, upper and inverse diagonal matrices to calculate utilizes the lower, upper and inverse diagonal matrices to calculate
subsequent iterations. An important difference is that $M$ and $w$ are not subsequent iterations. An important difference is that $M$ and $w$ are not
constant and they cannot be trivially calculated. constant and they cannot be trivially calculated.
An iteration of the Gauss-Seidel algorithm can be written as: An iteration of the Gauss-Seidel algorithm can be written as:
\begin{align*} \begin{align*}
Dx^{(i + 1)} &= -Lx^{(i + 1)} - (Ux^{(i)} - b)\\ Dx^{(i + 1)} &= -Lx^{(i + 1)} - (Ux^{(i)} - b)\\
&= -Lx^{(i + 1)} - w^{(i)} &= -Lx^{(i + 1)} - w^{(i)}
\end{align*} \end{align*}
The presence of $x^{(i + 1)}$ on both sides of the equation might appear The presence of $x^{(i + 1)}$ on both sides of the equation might appear
problematic, however, it's crucial to observe that individual elements of problematic, however, it's crucial to observe that individual elements of
the vector $x_n^{(i + 1)}$ can be calculated based solely on the knowledge the vector $x_n^{(i + 1)}$ can be calculated based solely on the knowledge
@ -350,19 +350,19 @@
- w_3^{(i)}) \cdot D^{-1}_{33}\\ - w_3^{(i)}) \cdot D^{-1}_{33}\\
&\ldots &\ldots
\end{align*} \end{align*}
And so, on each step, the vector $w^{(i)}$ is calculated first, after which And so, on each step, the vector $w^{(i)}$ is calculated first, after which
the elements of $x_n^{(i + 1)}$ are computed in sequence. the elements of $x_n^{(i + 1)}$ are computed in sequence.
\subsection{Program output} \subsection{Program output}
\subsubsection{Error plots} \subsubsection{Error plots}
\includegraphics[width=\textwidth]{jacobierror} \includegraphics[width=\textwidth]{jacobierror}
\includegraphics[width=\textwidth]{gaussseidelerror} \includegraphics[width=\textwidth]{gaussseidelerror}
\subsubsection{Solutions} \subsubsection{Solutions}
\begin{multicols}{2} \begin{multicols}{2}
\begin{verbatim} \begin{verbatim}
Solution using Jacobi: Solution using Jacobi:
@ -387,10 +387,10 @@
Error: 4.6134e-11 Error: 4.6134e-11
\end{verbatim} \end{verbatim}
\end{multicols} \end{multicols}
\begin{verbatim} \begin{verbatim}
Solution to task 2a using Gauss-Seidel: Solution to task 2a using Gauss-Seidel:
i x_i i x_i
1 0.8953 1 0.8953
2 0.71881 2 0.71881
@ -405,33 +405,33 @@
Error: 3.01e-11 Error: 3.01e-11
\end{verbatim} \end{verbatim}
\subsection{Observations} \subsection{Observations}
In both algorithms, the error of the solution decreased exponentially in In both algorithms, the error of the solution decreased exponentially in
relation to the number of iterations. relation to the number of iterations.
The Gauss-Seidel method proved to be faster than the Jacobi method, with the The Gauss-Seidel method proved to be faster than the Jacobi method, with the
desired accuracy achieved in just 32 iterations compared to 43. desired accuracy achieved in just 32 iterations compared to 43.
The matrix from task 2a was solved correctly. The 2b variant, however, could The matrix from task 2a was solved correctly. The 2b variant, however, could
not achieve the desired accuracy and the algorithm became stuck in a not achieve the desired accuracy and the algorithm became stuck in a
loop, iterating infinitely in an attempt to reduce the error. This again loop, iterating infinitely in an attempt to reduce the error. This again
may be related to the high condition number of the matrix, or to the fact may be related to the high condition number of the matrix, or to the fact
that the measured error is absolute rather than relative. that the measured error is absolute rather than relative.
\section{Task 4: Finding the eigenvalues of a symmetric matrix using the QR \section{Task 4: Finding the eigenvalues of a symmetric matrix using the QR
method} method}
\subsection{Overview} \subsection{Overview}
The problem presented in this task was to find the eigenvalues of a The problem presented in this task was to find the eigenvalues of a
symmetric matrix by utilizing the properties of its QR decomposition. Two symmetric matrix by utilizing the properties of its QR decomposition. Two
variants of the algorithm were to be used, one with and one without shifting variants of the algorithm were to be used, one with and one without shifting
the diagonal to accelerate the convergence. The algorithms were to be the diagonal to accelerate the convergence. The algorithms were to be
iterated until the off-diagonal elements decreased below the threshold of iterated until the off-diagonal elements decreased below the threshold of
$10^{-6}$. The resultant final matrices were to be printed. $10^{-6}$. The resultant final matrices were to be printed.
The matrix of choice: The matrix of choice:
\begin{equation*} \begin{equation*}
\begin{bmatrix} \begin{bmatrix}
@ -442,55 +442,55 @@
2 & 4 & 8 & 8 & 1 2 & 4 & 8 & 8 & 1
\end{bmatrix} \end{bmatrix}
\end{equation*} \end{equation*}
\subsection{Implementation} \subsection{Implementation}
The QR method is based on the observation that by repeatedly decomposing the The QR method is based on the observation that by repeatedly decomposing the
matrix into its orthogonal matrix $Q$ and triangular matrix $R$ and matrix into its orthogonal matrix $Q$ and triangular matrix $R$ and
recomposing it as $R * Q$, the matrix will converge into a diagonal matrix recomposing it as $R * Q$, the matrix will converge into a diagonal matrix
of its eigenvalues by means of similarity. of its eigenvalues by means of similarity.
\subsubsection{QR decomposition using the Gram-Schmidt algorithm} \subsubsection{QR decomposition using the Gram-Schmidt algorithm}
The columns of the input matrix $A$ are orthogonalized with the modified The columns of the input matrix $A$ are orthogonalized with the modified
Gram-Schmidt algorithm, producing matrices $Q$ and $R$, which can later be Gram-Schmidt algorithm, producing matrices $Q$ and $R$, which can later be
multiplied together to reverse the process. multiplied together to reverse the process.
The algorithm iterates over the columns of A and orthogonalizes every column The algorithm iterates over the columns of A and orthogonalizes every column
in front of it with relation to the currently visited column. This is done in front of it with relation to the currently visited column. This is done
by calculating the dot product of the two columns and using it to subtract by calculating the dot product of the two columns and using it to subtract
the ``common part" from the second column. The matrix $Q$ is produced in the the ``common part" from the second column. The matrix $Q$ is produced in the
process, as well as $R$ as a byproduct. process, as well as $R$ as a byproduct.
\subsubsection{QR method without shifts} \subsubsection{QR method without shifts}
The QR method for finding the eigenvalues of a matrix without shifts is a The QR method for finding the eigenvalues of a matrix without shifts is a
naive implementation of the observation described at the beginning of this naive implementation of the observation described at the beginning of this
section. The matrix is repeatedly decomposed and recomposed until its section. The matrix is repeatedly decomposed and recomposed until its
off-diagonal elements converge below the prescribed threshold. off-diagonal elements converge below the prescribed threshold.
\subsubsection{QR method with shifts} \subsubsection{QR method with shifts}
The QR method may be accelerated by shifting the diagonal of the matrix The QR method may be accelerated by shifting the diagonal of the matrix
towards an approximation of the expected eigenvalues. The required shift can towards an approximation of the expected eigenvalues. The required shift can
be calculated by finding the eigenvalues of the lower right $2 \times 2$ be calculated by finding the eigenvalues of the lower right $2 \times 2$
sub-matrix and choosing the one closer to the lower right corner of said sub-matrix and choosing the one closer to the lower right corner of said
sub-matrix. sub-matrix.
The eigenvalues of the sub-matrix can be found by solving the characteristic The eigenvalues of the sub-matrix can be found by solving the characteristic
equation of the sub-matrix, $\det{(A - \lambda I)} = 0$. This will result in equation of the sub-matrix, $\det{(A - \lambda I)} = 0$. This will result in
a quadratic equation whose solutions are guaranteed to be real for a a quadratic equation whose solutions are guaranteed to be real for a
symmetric matrix. symmetric matrix.
This shift is applied on every iteration of the algorithm. Once all This shift is applied on every iteration of the algorithm. Once all
non-diagonal values of the lowermost row of the matrix have decreased below non-diagonal values of the lowermost row of the matrix have decreased below
the threshold, the matrix is deflated, i.e. its size is reduced by one in the threshold, the matrix is deflated, i.e. its size is reduced by one in
both dimensions. The algorithm stops once the matrix can't be deflated any both dimensions. The algorithm stops once the matrix can't be deflated any
further. further.
\newpage \newpage
\subsection{Program output} \subsection{Program output}
\begin{verbatim} \begin{verbatim}
Solution using QR without shifts: Solution using QR without shifts:
23.3571 -9.3695 -7.8773 5.0307 -1.1410 23.3571 -9.3695 -7.8773 5.0307 -1.1410
@ -503,7 +503,7 @@
0.0000 0.0000 0.0000 0.0000 -1.1410 0.0000 0.0000 0.0000 0.0000 -1.1410
Iteration count: 75 Iteration count: 75
Solution using QR with shifts: Solution using QR with shifts:
5.0307 -1.1410 -7.8773 -9.3695 23.3571 5.0307 -1.1410 -7.8773 -9.3695 23.3571
@ -516,10 +516,10 @@
Iteration count: 9 Iteration count: 9
\end{verbatim} \end{verbatim}
\subsection{Observations} \subsection{Observations}
The addition of shifting greatly improved the convergence ratio and the same The addition of shifting greatly improved the convergence ratio and the same
values could be obtained in just 9 iterations instead of 75. values could be obtained in just 9 iterations instead of 75.
\end{document} \end{document}