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Task 2 Add comments and code task3

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PolishPigeon 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);
initial_x = ones(Rows, 1);
whichIterationAreWeOn = 0;
demandedTolerance = 1e-10;
demandedTolerance = 1e-10; % as per task description
flag = 0;
end
@ -24,7 +24,7 @@ function [L, D, U] = decomposeMatrix(Matrix)
end
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);
end
end
@ -42,16 +42,16 @@ function x = jacobiEquation(D, L, U, initial_x, Vector)
end
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
flag = 1;
else
demandedTolerance = demandedTolerance * 2;
end
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
function [initial_x, whichIterationAreWeOn, flag] = endOfLoop(x, whichIterationAreWeOn)
@ -60,8 +60,6 @@ function [initial_x, whichIterationAreWeOn, flag] = endOfLoop(x, whichIterationA
flag = 0;
end
function dispFinalResults(demandedTolerance, whichIterationAreWeOn, Matrix, Vector)
disp("Final 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.5.1.11}{iterativeResidualCorrection}{section.5.1}% 40
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\BOOKMARK [2][-]{subsection.5.2.1}{jacobiMethod}{section.5.2}% 43
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ENUME/projectA/projectA.tex
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@ -748,6 +748,124 @@ end % end function
\end{lstlisting}
\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}
\bibitem{texbook}
Piotr Tatjewski (2014) \emph{Numerical Methods}, Oficyna Wydawnicza Politechniki Warszawskiej

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@ -51,3 +51,13 @@
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@ -14,16 +14,16 @@
Project number 62}
\date{November 12, 2020}
\maketitle
\numberwithin{equation}{section}
\setcounter{section}{-1}
\section{Abstract}
This project explores the numerical methods of solving systems of linear
equations and finding eigenvalues of matrices. The following concepts are
discussed within this document:
\begin{itemize}
\item Computing the machine epsilon of an environment
\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
with and without shifts
\end{itemize}
\newpage
\section{Task 1: Computing the machine epsilon}
\subsection{Overview}
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
representation of a number.
\subsection{Implementation}
The computation of the machine epsilon is based on the following two
observations:
\begin{enumerate}
\item In binary computer environments, the epsilon is a negative power
of two.
@ -59,43 +59,43 @@
i.e. the smallest representable number such that, when added to $1$,
has a floating point representation greater than $1$.
\end{enumerate}
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
no longer met.
\subsection{Program output}
\begin{verbatim}
Found epsilon is 2.2204e-16
Epsilon is 2.2204e-16
\end{verbatim}
\subsection{Observations}
An epsilon of $2.2204 \cdot 10^{-16} = 2^{-52}$ is indicative of a standard
double-precision IEEE 754 floating point binary representation.
\newpage
\section{Task 2: Solving systems of linear equations using Gaussian
elimination}
\subsection{Overview}
The task consisted of two subtasks, each specifying a different system
of equations to solve. The systems were to be solved using Gaussian
elimination with partial pivoting. The size of the system was sequentially
doubled until the computation time became prohibitive. The error of each
solution was to be noted.
Additionally, for systems whose number of equations was equal to 10,
residual correction was to be applied and the solution was to be noted.
The systems were given by:
\begin{align*}
&A_{ij}^{(a)} =
\begin{cases}
\begin{cases}
4 & \text{for } i = j \\
1 & \text{for } i = j \pm 1 \\
0 & \text{otherwise}
@ -106,27 +106,27 @@
&A_{ij}^{(b)} = \frac{6}{7(i + j + 1)}
&&
b_i^{(b)} =
\begin{cases}
\begin{cases}
\frac{1}{3i} & \text{for even } i\\
0 & \text{for odd } i
\end{cases}
\end{align*}
\subsection{Implementation}
Gaussian elimination is a process that takes a matrix describing a linear
equation system and returns its LU (lower-upper) decomposition, enabling the
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
a process called back-substitution.
\subsubsection{LU decomposition of a matrix}
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
describing the entire system post-elimination, to be immediately handed to
back-substitution for the calculation of the result.
The elimination itself is done by traveling along the diagonal of
\texttt{eqsys} and for each encountered cell, eliminating (reducing to $0$)
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
ratio of the cell to be reduced and the current diagonal cell. The reductor
is then stored in the $L$ matrix.
\subsubsection{Partial pivoting}
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,
partial pivoting is added.
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
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
possible.
Because the input matrix changes while being decomposed, a permutation
matrix $P$ is introduced to enable its recomposition from $L$ and $U$,
satisfying the equation:
\begin{equation}
LU = PA
\end{equation}
Finally, the matrix $L$ is also permuted to reflect changes in the layout
of the system.
\subsubsection{Back-substitution}
A triangular matrix supplied with a constant vector $b$ can be easily solved
by traveling along its diagonal and eliminating the factors directly above
the visited cell. In the end, the matrix is transformed into an identity
matrix and the vector alongside it is transformed into $x$.
\subsubsection{Residual correction}
Residual correction is a method of improving the accuracy of the results
obtained from Gaussian elimination. By reusing the already obrained matrices
$L$, $U$ and $P$, the system is solved once again, this time with $b$
replaced with the vector of errors of the solution. The resulting vector
$\delta x$ is then subtracted from the previous solution. This procedure may
be repeated as needed.
\subsection{Program output}
\subsubsection{Error plots}
\includegraphics[width=\textwidth]{aerror}
\includegraphics[width=\textwidth]{berror}
\subsubsection{Solutions for $n = 10$}
\begin{multicols}{2}
\begin{verbatim}
Solution to subtask a:
@ -250,37 +250,37 @@
Error: 0.00020161
\end{verbatim}
\end{multicols}
\newpage
\subsection{Observations}
Matrices from both subtasks a and b lent themselves nicely to being solved
using Gaussian elimination. Their absolute errors, when divided by the
solutions themselves, resulted in very small relative errors, nearing the
machine epsilon.
The computation time became prohibitive beyond $n = 320$, showing that
Gaussian elimination is an effective method of solving large systems
of equations.
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
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,
$2.4225 \cdot 10^{14}$, prohibiting it from producing accurate enough
results.
\newpage
\section{Task 3: Solving systems of linear equations using the Jacobi and
Gauss-Seidel algorithms}
\subsection{Overview}
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
to be iterated until the accuracy $\lVert Ax_k - b \rVert < 10^{-10}$ was
achieved. The error of each iteration was to be noted.
The system to be solved:
\begin{align*}
18x_1 + 2x_2 - 3x_3 + x_4 & = 7\\
@ -288,12 +288,12 @@
x_1 + 3x_2 + 13x_3 - 8x_4 & = 24\\
x_1 + x_2 - 2x_3 - 10x_4 & = 20
\end{align*}
Afterwards, one of the algorithms was to be applied to matrices from task 2
a and b, for $n = 10$.
\subsection{Implementation}
The general formula describing iterative methods of solving linear equation
systems is:
\begin{equation}\label{eq:1}
@ -301,41 +301,41 @@
\end{equation}
Where the matrix $M$ and the vector $w$ are specific to the chosen
algorithm.
\subsubsection{The Jacobi algorithm}
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
diagonal matrix.
Because the diagonal matrix is only ever used in its inverse from, the
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
inverse of its diagonal elements.
Once the matrices $L$, $U$ and $D^{-1}$ have been obtained, the parameters
$M$ and $w$ are given by:
\begin{align*}
M &= -D^{-1}(L + U)\\
w &= D^{-1}b
\end{align*}
The step described in \eqref{eq:1} is then performed until the desired
accuracy is achieved.
\subsubsection{The Gauss-Seidel algorithm}
The Gauss-Seidel algorithm is similar to the Jacobi algorithm in that it
utilizes the lower, upper and inverse diagonal matrices to calculate
subsequent iterations. An important difference is that $M$ and $w$ are not
constant and they cannot be trivially calculated.
An iteration of the Gauss-Seidel algorithm can be written as:
\begin{align*}
Dx^{(i + 1)} &= -Lx^{(i + 1)} - (Ux^{(i)} - b)\\
&= -Lx^{(i + 1)} - w^{(i)}
\end{align*}
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
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}\\
&\ldots
\end{align*}
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.
\subsection{Program output}
\subsubsection{Error plots}
\includegraphics[width=\textwidth]{jacobierror}
\includegraphics[width=\textwidth]{gaussseidelerror}
\subsubsection{Solutions}
\begin{multicols}{2}
\begin{verbatim}
Solution using Jacobi:
@ -387,10 +387,10 @@
Error: 4.6134e-11
\end{verbatim}
\end{multicols}
\begin{verbatim}
Solution to task 2a using Gauss-Seidel:
i x_i
1 0.8953
2 0.71881
@ -405,33 +405,33 @@
Error: 3.01e-11
\end{verbatim}
\subsection{Observations}
In both algorithms, the error of the solution decreased exponentially in
relation to the number of iterations.
The Gauss-Seidel method proved to be faster than the Jacobi method, with the
desired accuracy achieved in just 32 iterations compared to 43.
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
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
that the measured error is absolute rather than relative.
\section{Task 4: Finding the eigenvalues of a symmetric matrix using the QR
method}
\subsection{Overview}
The problem presented in this task was to find the eigenvalues of a
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
the diagonal to accelerate the convergence. The algorithms were to be
iterated until the off-diagonal elements decreased below the threshold of
$10^{-6}$. The resultant final matrices were to be printed.
The matrix of choice:
\begin{equation*}
\begin{bmatrix}
@ -442,55 +442,55 @@
2 & 4 & 8 & 8 & 1
\end{bmatrix}
\end{equation*}
\subsection{Implementation}
The QR method is based on the observation that by repeatedly decomposing the
matrix into its orthogonal matrix $Q$ and triangular matrix $R$ and
recomposing it as $R * Q$, the matrix will converge into a diagonal matrix
of its eigenvalues by means of similarity.
\subsubsection{QR decomposition using the Gram-Schmidt algorithm}
The columns of the input matrix $A$ are orthogonalized with the modified
Gram-Schmidt algorithm, producing matrices $Q$ and $R$, which can later be
multiplied together to reverse the process.
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
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
process, as well as $R$ as a byproduct.
\subsubsection{QR method without shifts}
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
section. The matrix is repeatedly decomposed and recomposed until its
off-diagonal elements converge below the prescribed threshold.
\subsubsection{QR method with shifts}
The QR method may be accelerated by shifting the diagonal of the matrix
towards an approximation of the expected eigenvalues. The required shift can
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.
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
a quadratic equation whose solutions are guaranteed to be real for a
symmetric matrix.
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
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
further.
\newpage
\subsection{Program output}
\begin{verbatim}
Solution using QR without shifts:
23.3571 -9.3695 -7.8773 5.0307 -1.1410
@ -503,7 +503,7 @@
0.0000 0.0000 0.0000 0.0000 -1.1410
Iteration count: 75
Solution using QR with shifts:
5.0307 -1.1410 -7.8773 -9.3695 23.3571
@ -516,10 +516,10 @@
Iteration count: 9
\end{verbatim}
\subsection{Observations}
The addition of shifting greatly improved the convergence ratio and the same
values could be obtained in just 9 iterations instead of 75.
\end{document}