hw2: preparing for initial submission
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Claudio Maggioni (maggicl)
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3 changed files with 26 additions and 249 deletions
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@ -130,9 +130,33 @@ The code to generate the plots below can be found in Section 1.5 of the script \
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\subsection{Finally, explain why the Conjugate Gradient method is a Krylov subspace method.}
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Because theorem 5.3 holds, which itself holds mainly because of this (5.10, page 106 [127]):
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We can say that the Conjugate Gradient method is a Krylov subspace method since we can say that both all residuals $r_i$ and all search directions $p_i$ are contained in the span of $0$ to $k$ repeated applications of matrix transformer $A$ onto the initial search direction $p_0$. Broadly speaking, this property is directly connected with the fact that the CG methods performs up to $n$ steps exactly for a $n$ by $n$ matrix, walking steps that are all $A$-orthogonal with each other (i.e. for all couples of search steps $p_i$, $p_j$ where $i \neq j$, $\langle Ap_i, p_j \rangle = 0$).
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To sum up our claim, we can say CG is indeed a Krylov subspace method because:
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\[r_{k+1} = r_k + a_k * A * p_k\]
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\[\text{span}\{r_0, r_1, \ldots, r_n\} = \text{span}\{r_0, A r_0, \ldots, A^k r_0\}\]
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\[\text{span}\{p_0, p_1, \ldots, p_n\} = \text{span}\{r_0, A r_0, \ldots, A^k r_0\}\]
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These statements have been already proven in class and the proof can be found in Theorem 5.3 of Nocedal's book.
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% These statements can be proven by induction over $k$. The base case holds trivially for $k=0$, while we have by the induction hypothesis for any $k$ that:
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% \[r_k \in \text{span}\{r_0, A r_0, \ldots, A^k r_0\}\;\;\; p_k \in \text{span}\{r_0, A r_0, \ldots, A^k r_0\}\]
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% We'd like to prove that the two properties hold from $k+1$ starting from the hypothesis on $k$. We first multiply the first hypothesis by $A$ from the left:
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% \[A r_k \in \text{span}\{A r_0, A^2 r_0, \ldots, A^{k+1} r_0\}\]
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% By the alternative definition of the residual for the CG method (i.e. $r_{k+1} = r_k + \alpha_k A p_k$), we find that:
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% \[r_{k+1} \in \text{span}\{r_0, A r_0, A^2 r_0, \ldots, A^{k+1} r_0\}\]
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% We need to add $r_0$ in the span again since one of the components that defines $r_1$ is indeed $r_0$. We don't need to add other terms in the span since $r_1$ to $r_k$ are in the span already by the induction hypothesis.
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% Combining this expression with the induction hypothesis we prove induction of the first statement by having:
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% \[\text{span}\{r_0, r_1, \ldots, r_n, r_{n+1}\} \subseteq \text{span}\{r_0, A r_0, \ldots, A^k r_0, A^{k+1} r_0\}\]
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% To prove $\supseteq$ as well to achieve equality, we use the induction hypothesis for the second statement to find that $A^kr_0 \in \text{span}\{A$:
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\section{Exercise 2}
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@ -1,247 +0,0 @@
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\documentclass{scrartcl}
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\usepackage[utf8]{inputenc}
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\usepackage{float}
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\usepackage{graphicx}
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\usepackage[ruled,vlined]{algorithm2e}
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\usepackage{subcaption}
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\usepackage{amsmath}
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\usepackage{pgfplots}
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\pgfplotsset{compat=newest}
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\usetikzlibrary{plotmarks}
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\usetikzlibrary{arrows.meta}
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\usepgfplotslibrary{patchplots}
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\usepackage{grffile}
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\usepackage{amsmath}
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\usepackage{subcaption}
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\usepgfplotslibrary{external}
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\tikzexternalize
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\usepackage[margin=2.5cm]{geometry}
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% To compile:
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% sed -i 's#title style={font=\\bfseries#title style={yshift=1ex, font=\\tiny\\bfseries#' *.tex
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% luatex -enable-write18 -shellescape main.tex
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\pgfplotsset{every x tick label/.append style={font=\tiny, yshift=0.5ex}}
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\pgfplotsset{every title/.append style={font=\tiny, align=center}}
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\pgfplotsset{every y tick label/.append style={font=\tiny, xshift=0.5ex}}
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\pgfplotsset{every z tick label/.append style={font=\tiny, xshift=0.5ex}}
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\setlength{\parindent}{0cm}
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\setlength{\parskip}{0.5\baselineskip}
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\title{Optimization methods -- Homework 2}
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\author{Claudio Maggioni}
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\begin{document}
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\maketitle
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\section{Exercise 1}
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\subsection{Implement the matrix $A$ and the vector $b$, for the moment, without taking into consideration the
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boundary conditions. As you can see, the matrix $A$ is not symmetric. Does an energy function of
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the problem exist? Consider $N = 4$ and show your answer, explaining why it can or cannot exist.}
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The implementation of a function that generates $A$ w.r.t. the parameter $N$ can be found in the MATLAB script
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\texttt{main.m} under Section 1.1.
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The matrix $A$ and the vector $b$ appear in the following form:
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\[A = \begin{bmatrix}
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1 & 0 & 0 & 0 & \ldots & 0 \\
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-1 & 2 & -1 & 0 & \ldots & 0 \\
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0 & -1 & 2 & -1 & \ldots & 0 \\
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\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
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0 & 0 & 0 & 0 & \ldots & 1 \\
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\end{bmatrix}\;\;\;b = \begin{bmatrix}0\\h^2\\h^2\\\vdots\\0\end{bmatrix}\]
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For $N = 4$, we can attempt to build a minimizer to the solve the system $Ax = b$. In order to find an $x$ such that $Ax = b$, we could define a minimizer $\phi(x)$ such that:
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\[\phi(x) = \|b - Ax\|^2\]
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Here $\phi(x) = 0$ would mean that $x$ is an exact solution of the system $Ax = b$. We can then attempt to write such minimizer for $N = 4$:
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\[\phi(x) = \|b - Ax\|^2 = \left|\begin{bmatrix}0 - x_1\\\frac19 -x_1 + 2x_2 -x_3\\
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\frac19 -x_2 +2x_3 -x_4\\0 - x_4\end{bmatrix}\right|^2 =\]\[= x_1^2 + \left(\frac19 - x_1 + 2x_2 - x_3\right)^2 + \left(\frac19 - x_2 + 2x_3 - x_4\right)^2 + x_4^2\]
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\[\Delta \phi(x) = \begin{bmatrix}4x_1 - 4x_2 + 2x_3 -\frac29\\
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-4x_1 +10x_2 -8x_3 + 2x_4 +\frac29\\
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2x_1 -8x_2 +10x_3 + -4x_4 +\frac29\\
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2x_2 - 4x_3 + 4x_4 -\frac29\end{bmatrix}\;\;\;\Delta^2 \phi(x) = \begin{bmatrix}4&-4&2&0\\-4&10&-8&2\\2&-8&10&-4\\0&2&-4&4\end{bmatrix}\]
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As it can be seen from the Hessian calculation, the Hessian is positive definite forall $x$s. This means, by the sufficient condition of minimizers, that we can find a minimizer by solving $\Delta \phi(x) = 0$ (i.e. finding stationary points in the hyperspace defined by $\phi(x)$. Solving that, we find:
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\[x = \begin{bmatrix}0\\\frac19\\\frac19\\0\end{bmatrix}\;\;\]
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which is indeed the minimizer and solution of $Ax = b$. Therefore, $\phi(x)$ is a valid energy function and we can say an energy function for the problem exists.
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\subsection{Once the new matrix has been derived, write the energy function related to the new problem
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and the corresponding gradient and Hessian.}
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As by the definitions of $A$ and $b$ given in the assignment, we already enforce $x_1 = x_n = 0$, since the first and the last term of the matrix-vector product $Ax$ are $x_1$ and $x_n$ respectively and the system of equations represented by $Ax = b$ would indeed include the equalities $x_1 = b_1 = 0$ and $x_n = b_n = 0$. Therefore, we can simply alter the matrix $A$ without any need to perform any other transformation in the system.
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We therefore define $\overline{A}$ as a copy of $A$ where $\overline{A}_{2,1} = \overline{A}_{n-1, n} = 0$.
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The objective function then becomes $\phi(x) = \frac12 x^T \overline{A} x - b^T x$.
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And since the objective is a standard quadratic form, the gradient is $\overline{A}x - b$ while the Hessian is $\overline{A}$.
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\subsection{Write the Conjugate Gradient algorithm in the pdf and implement it Matlab code in a function
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called \texttt{CGSolve}.}
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The Conjugate Gradient algorithm is the following:
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\begin{algorithm}[H]
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\SetAlgoLined
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\KwResult{Conjugate Gradient Algorithm}
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Set $r_0 \gets Ax_0 - b$, $p_0 \gets r_0$, $k \gets 0$\;
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\While{$r_k \neq 0$}{%
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$\alpha_k \gets \frac{r^T_kr_k}{p^T_kAp_k}$\;
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$x_{k+1} \gets x_k + \alpha_k p_k$\;
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$r_{k+1} \gets r_k + \alpha_k A p_k$\;
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$\beta_{k+1} \gets \frac{r^T_{k+1}r_{k+1}}{r^T_kr_k}$\;
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$p_{k+1} \gets -r_{k+1} + \beta_{k+1}p_k$\;
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$k \gets k + 1$\;
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}
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\end{algorithm}
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The MATLAB solution of this task can be found in Section 1.3 of the script \texttt{main.m} under the function \texttt{CGSolve}.
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\subsection{Solve the Poisson problem.}
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The solution of this task can be found in Section 1.4 of the script \texttt{main.m}.
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Due to space constraints, the $R^{1000}$ column vector for the solution of $x$ is not shown here but can be easily derived by running the script and checking the variable \texttt{x} after the script execution.
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\subsection{Plot the value of energy function and the norm of the gradient (here,
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use semilogy) as functions of the iterations.}
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The code to generate the plots below can be found in Section 1.5 of the script \texttt{main.m}.
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\begin{figure}[H]
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\begin{subfigure}{0.5\textwidth}
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\resizebox{\textwidth}{\textwidth}{\input{obvalues}}
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\caption{Objective function values w.r.t. iteration number}
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\end{subfigure}
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\begin{subfigure}{0.5\textwidth}
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\resizebox{\textwidth}{\textwidth}{\input{gnorms}}
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\caption{Norm of the gradient w.r.t. iteration number \\ (y-axis is log scaled)}
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\end{subfigure}
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\caption{Plots for Exercise 1.4.}
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\end{figure}
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\subsection{Finally, explain why the Conjugate Gradient method is a Krylov subspace method.}
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Because theorem 5.3 holds, which itself holds mainly because of this (5.10, page 106 [127]):
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\[r_{k+1} = r_k + a_k * A * p_k\]
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\section{Exercise 2}
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Consider the linear system $Ax = b$, where the matrix $A$ is constructed in three different ways:
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\begin{itemize}
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\item $A_1 =$ diag([1:10])
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\item $A_2 =$ diag(ones(1,10))
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\item $A_3 =$ diag([1, 1, 1, 3, 4, 5, 5, 5, 10, 10])
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\item $A_4 =$ diag([1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0])
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\end{itemize}
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\subsection{How many distinct eigenvalues has each matrix?}
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Each matrix has a distinct number of eigenvalues equal to the number of distinct
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elements on its diagonal. So, in order, each A has respectively 10, 1, 5, and 10 distinct eigenvalues.
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\subsection{Construct a right-hand side $b=$rand(10,1) and apply the
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Conjugate Gradient method to solve the system for each $A$.}
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The solution of this task can be found in section 2.2 of the \texttt{main.m} MATLAB script.
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Below are the chosen $b$ vector (which is unchanged between executions due to fixing the random generator seed in the script) and the solutions of $x$ for each matrix respectively.
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\[b = \begin{bmatrix}0.814723686393179\\
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0.905791937075619\\
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0.126986816293506\\
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0.913375856139019\\
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0.632359246225410\\
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0.0975404049994095\\
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0.278498218867048\\
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0.546881519204984\\
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0.957506835434298\\
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0.964888535199277\end{bmatrix}\;
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x_1 = \begin{bmatrix}0.814723686393179\\
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0.452895968537810\\
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0.0423289387645020\\
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0.228343964034755\\
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0.126471849245082\\
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0.0162567341665680\\
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0.0397854598381500\\
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0.0683601899006230\\
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0.106389648381589\\
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0.0964888535199280\end{bmatrix}\;x_2 = \begin{bmatrix}0.814723686393179\\
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0.905791937075619\\
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0.126986816293506\\
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0.913375856139019\\
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0.632359246225410\\
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0.0975404049994095\\
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0.278498218867048\\
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0.546881519204984\\
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0.957506835434298\\
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0.964888535199277\end{bmatrix}\]\[x_3 = \begin{bmatrix}0.814723686393179\\
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0.905791937075619\\
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0.126986816293506\\
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0.304458618713007\\
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0.158089811556352\\
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0.0195080809998819\\
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0.0556996437734097\\
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0.109376303840997\\
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0.0957506835434296\\
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0.0964888535199280\end{bmatrix}\;x_4 = \begin{bmatrix}0.740657896716011\\
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0.754826614267239\\
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0.0976821653904972\\
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0.652411326111541\\
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0.421572830214428\\
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0.0609627567866090\\
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0.163822480881755\\
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0.303823066390868\\
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0.503950965995613\\
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0.482444267601989\end{bmatrix}\]
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\subsection{Compute the logarithm energy norm of the error for each matrix
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and plot it with respect to the number of iteration.}
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The code to generate the plots below and to compute the logarithm energy norm of the error can be found in section 2.3 of the \texttt{main.m} MATLAB script.
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\begin{figure}[H]
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\begin{subfigure}{0.5\textwidth}
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\resizebox{\textwidth}{!}{\input{a1}}
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\caption{First matrix}
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\end{subfigure}
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\begin{subfigure}{0.5\textwidth}
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\resizebox{\textwidth}{!}{\input{a2}}
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\caption{Second matrix}
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\end{subfigure}
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\begin{subfigure}{0.5\textwidth}
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\resizebox{\textwidth}{!}{\input{a3}}
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\caption{Third matrix}
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\end{subfigure}
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\begin{subfigure}{0.5\textwidth}
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\resizebox{\textwidth}{!}{\input{a4}}
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\caption{Fourth matrix}
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\end{subfigure}
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\caption{Plots of logarithm energy norm of the error per iteration. Minus infinity logarithms not shown in the plot.}
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\end{figure}
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\subsection{Comment on the convergence of the method for the different matrices. What can you say observing
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the number of iterations obtained and the number of clusters of the eigenvalues of the related
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matrix?}
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The method converges quickly for each matrix. The fastest convergence surely happens for $A2$, which is
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the identity matrix and therefore makes the $Ax = b$ problem trivial.
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For all the other matrices, we observe the energy norm of the error decreasing exponentially as the iterations
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increase, eventually reaching $0$ for the cases where the method converges exactly (namely on matrices $A1$ and $A3$).
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Other than for the fourth matrix, the number of iterations is exactly equal
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to the number of distinct eigenvalues for the matrix. That exception on the fourth matrix is simply due to the
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tolerance termination condition holding true for an earlier iteration, i.e. we terminate early since we find an
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approximation of $x$ with residual norm below $10^{-8}$.
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\end{document}
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