midterm: done 1
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@ -48,6 +48,13 @@ activated uses a fixed $\alpha=1$ despite the indications on the assignment
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sheet. This was done in order to comply with the forum post on iCorsi found
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sheet. This was done in order to comply with the forum post on iCorsi found
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here: \url{https://www.icorsi.ch/mod/forum/discuss.php?d=81144}.
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here: \url{https://www.icorsi.ch/mod/forum/discuss.php?d=81144}.
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Here is a plot of the Rosenbrock function in 3d, with our starting point in red ($(0,0)$),
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and the true minimizer in black ($(1,1)$):
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\begin{center}
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\resizebox{0.6\textwidth}{0.6\textwidth}{\includegraphics{rosenb.jpg}}
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\end{center}
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\subsection{Exercise 1.2}
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\subsection{Exercise 1.2}
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Please consult the MATLAB implementation in the file \texttt{main.m} in section
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Please consult the MATLAB implementation in the file \texttt{main.m} in section
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@ -131,6 +138,12 @@ zig-zagging pattern with iterates in the vicinity of the Netwon + backtracking
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iterates. Finally, GD without backtracking quickly degenerates as it can be
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iterates. Finally, GD without backtracking quickly degenerates as it can be
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seen by the enlarged plot.
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seen by the enlarged plot.
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From the initial plot in the Rosenbrock's function, we can see why algorithms using
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backtracking follow roughly the $(0,0) - (1,1)$ diagonal: since this region is effectively a
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``valley'' in the 3D energy landscape, due to the nature of the backtracking methods and their
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strict adherence of the Wolfe conditions these methods avoid wild ``climbs'' (unlike the Netwon method without backtracking)
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and instead finding iterates with either sufficient decrease or a not to high increase.
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When looking at gradient norms and objective function
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When looking at gradient norms and objective function
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values (figure \ref{fig:gsppn}) over time, the degeneration of GD without
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values (figure \ref{fig:gsppn}) over time, the degeneration of GD without
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backtracking and the inefficiency of GD with backtracking can clearly be seen.
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backtracking and the inefficiency of GD with backtracking can clearly be seen.
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@ -141,6 +154,16 @@ at gradient $\nabla f(x_1) \approx 450$ and objective value $f(x_1) \approx
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100$, but quickly has both values decrease to 0 for its second iteration
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100$, but quickly has both values decrease to 0 for its second iteration
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achieving convergence.
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achieving convergence.
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What has been observed in this assignment matches with the theory behind the methods.
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Since the Newton method has quadratic convergence and uses quadratic information (i.e.
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using the hessian in the step direction calculation) the number of iterations required
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to find the minimizer when already close to it (as we are with $x_0 = [0;0]$) is significantly
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less than the ones required for linear methods, like gradient descent. However, it must be
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said that for an objective with an high number of dimensions a single iteration of a quadratic
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method is significantly more costly than a single iteration of a linear method due to the
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quadratically growing number of cells in the hessian matrix, which makes it harder and harder
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to compute as the number of dimensions increase.
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\section{Exercise 2}
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\section{Exercise 2}
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\subsection{Exercise 2.1}
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\subsection{Exercise 2.1}
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@ -212,4 +235,10 @@ the opposite direction of the minimizer. This behaviour can also be observed in
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the plots in figure \ref{fig:4}, where several bumps and spikes are present in
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the plots in figure \ref{fig:4}, where several bumps and spikes are present in
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the gradient norm plot and small plateaus can be found in the objective
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the gradient norm plot and small plateaus can be found in the objective
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function value plot.
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function value plot.
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Comparing these results with the theory behind BFGS, we can say the results that have been
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obtained fall within what we expect from the theory. Since BFGS is a superlinear but not quadratic
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method of convergence, its ``speed'' in terms of number of iterations falls within linear methods (like GD) and
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quadratic methods (like Newton).
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\end{document}
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\end{document}
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@ -1,176 +0,0 @@
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%% Homework 3 - Optimization Methods
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% Author: Claudio Maggioni
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%
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% Note: exercises are not in the right order due to matlab constraints of
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% functions inside of scripts.
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clc
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clear
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close all
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% Set to non-zero to generate LaTeX for graphs
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enable_m2tikz = 0;
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if enable_m2tikz
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addpath /home/claudio/git/matlab2tikz/src
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else
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matlab2tikz = @(a,b,c) 0;
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end
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syms x y;
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f = (1 - x)^2 + 100 * (y - x^2)^2;
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global fl
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fl = matlabFunction(f);
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%% 1.3 - Newton and GD solutions and energy plot
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[x1, xs1, gs1] = Newton(f, [0;0], 50000, 1e-6, true);
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plot(xs1(1, :), xs1(2, :), 'Marker', '.');
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fprintf("Newton backtracking: it=%d\n", size(xs1, 2)-1);
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xlim([-0.01 1.01])
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ylim([-0.01 1.01])
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hold on;
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[x2, xs2, gs2] = Newton(f, [0;0], 50000, 1e-6, false);
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plot(xs2(1, :), xs2(2, :), 'Marker', '.');
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fprintf("Newton: it=%d\n", size(xs2, 2)-1);
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[x3, xs3, gs3] = GD(f, [0;0], 50000, 1e-6, true);
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plot(xs3(1, :), xs3(2, :), 'Marker', '.');
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fprintf("GD backtracking: it=%d\n", size(xs3, 2)-1);
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[x4, xs4, gs4] = GD(f, [0;0], 50000, 1e-6, false);
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plot(xs4(1, :), xs4(2, :), 'Marker', '.');
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fprintf("GD: it=%d\n", size(xs4, 2)-1);
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hold off;
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legend('Newton + backtracking', 'Newton', 'GD + backtracking', 'GD (alpha=1)')
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sgtitle("Iterations of Newton and Gradient descent methods over 2D energy landscape");
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matlab2tikz('showInfo', false, './ex1-3.tex');
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figure;
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plot(xs4(1, :), xs4(2, :), 'Marker', '.');
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legend('GD (alpha=1)')
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sgtitle("Iterations of Newton and Gradient descent methods over 2D energy landscape");
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matlab2tikz('showInfo', false, './ex1-3-large.tex');
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%% 2.3 - BGFS solution and energy plot
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figure;
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[x5, xs5, gs5] = BGFS(f, [0;0], eye(2), 50000, 1e-6);
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xlim([-0.01 1.01])
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ylim([-0.01 1.01])
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plot(xs5(1, :), xs5(2, :), 'Marker', '.');
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fprintf("BGFS backtracking: it=%d\n", size(xs5, 2)-1);
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sgtitle("Iterations of BGFS method over 2D energy landscape");
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matlab2tikz('showInfo', false, './ex2-3.tex');
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%% 1.4 - Newton and GD gradient norm log
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figure;
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semilogy(0:size(xs1, 2)-1, gs1, 'Marker', '.');
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ylim([5e-10, 1e12]);
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xlim([-1, 30]);
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hold on
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semilogy(0:size(xs2, 2)-1, gs2, 'Marker', '.');
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semilogy(0:size(xs3, 2)-1, gs3, 'Marker', '.');
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semilogy(0:size(xs4, 2)-1, gs4, 'Marker', '.');
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hold off
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legend('Newton + backtracking', 'Newton', 'GD + backtracking', 'GD (alpha=1)')
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sgtitle("Gradient norm w.r.t. iteration number for Newton and GD methods");
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matlab2tikz('showInfo', false, './ex1-4-grad.tex');
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figure;
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plot(0:size(xs1, 2)-1, gs1, 'Marker', '.');
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ylim([5e-10, 25]);
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xlim([-1, 30]);
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hold on
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plot(0:size(xs2, 2)-1, gs2, 'Marker', '.');
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plot(0:size(xs3, 2)-1, gs3, 'Marker', '.');
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plot(0:size(xs4, 2)-1, gs4, 'Marker', '.');
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hold off
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legend('Newton + backtracking', 'Newton', 'GD + backtracking', 'GD (alpha=1)')
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sgtitle("Gradient norm w.r.t. iteration number for Newton and GD methods");
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matlab2tikz('showInfo', false, './ex1-4-grad.tex');
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figure;
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semilogy(0:size(xs3, 2)-1, gs3, 'Marker', '.');
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ylim([1e-7, 1e10]);
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hold on
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semilogy(0:size(xs4, 2)-1, gs4, 'Marker', '.');
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hold off
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legend('GD + backtracking', 'GD (alpha=1)')
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sgtitle("Gradient norm w.r.t. iteration number for Newton and GD methods");
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matlab2tikz('showInfo', false, './ex1-4-grad-large.tex');
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figure;
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ys1 = funvalues(xs1);
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ys2 = funvalues(xs2);
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ys3 = funvalues(xs3);
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ys4 = funvalues(xs4);
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ys5 = funvalues(xs5);
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semilogy(0:size(xs1, 2)-1, ys1, 'Marker', '.');
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ylim([5e-19, 1e12]);
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xlim([-1, 30]);
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hold on
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semilogy(0:size(xs2, 2)-1, ys2, 'Marker', '.');
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semilogy(0:size(xs3, 2)-1, ys3, 'Marker', '.');
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semilogy(0:size(xs4, 2)-1, ys4, 'Marker', '.');
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hold off
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legend('Newton + backtracking', 'Newton', 'GD + backtracking', 'GD (alpha=1)')
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sgtitle("Objective function value w.r.t. iteration number for Newton and GD methods");
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matlab2tikz('showInfo', false, './ex1-4-ys.tex');
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ys1 = funvalues(xs1);
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ys2 = funvalues(xs2);
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ys3 = funvalues(xs3);
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ys4 = funvalues(xs4);
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ys5 = funvalues(xs5);
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semilogy(0:size(xs1, 2)-1, ys1, 'Marker', '.');
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ylim([5e-19, 20]);
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xlim([-1, 30]);
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hold on
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semilogy(0:size(xs2, 2)-1, ys2, 'Marker', '.');
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semilogy(0:size(xs3, 2)-1, ys3, 'Marker', '.');
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semilogy(0:size(xs4, 2)-1, ys4, 'Marker', '.');
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hold off
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legend('Newton + backtracking', 'Newton', 'GD + backtracking', 'GD (alpha=1)')
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sgtitle("Objective function value w.r.t. iteration number for Newton and GD methods");
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matlab2tikz('showInfo', false, './ex1-4-ys.tex');
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figure;
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semilogy(0:size(xs3, 2)-1, ys3, 'Marker', '.');
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semilogy(0:size(xs4, 2)-1, ys4, 'Marker', '.');
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legend('GD + backtracking', 'GD (alpha=1)')
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sgtitle("Objective function value w.r.t. iteration number for Newton and GD methods");
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matlab2tikz('showInfo', false, './ex1-4-ys-large.tex');
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%% 2.4 - BGFS gradient norms plot
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figure;
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semilogy(0:size(xs5, 2)-1, gs5, 'Marker', '.');
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sgtitle("Gradient norm w.r.t. iteration number for BGFS method");
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matlab2tikz('showInfo', false, './ex2-4-grad.tex');
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%% 2.4 - BGFS objective values plot
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figure;
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semilogy(0:size(xs5, 2)-1, ys5, 'Marker', '.');
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sgtitle("Objective function value w.r.t. iteration number for BGFS methods");
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matlab2tikz('showInfo', false, './ex2-4-ys.tex');
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function ys = funvalues(xs)
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ys = zeros(1, size(xs, 2));
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global fl
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for i = 1:size(xs, 2)
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ys(i) = fl(xs(1,i), xs(2,i));
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end
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end
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clc
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clc
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clear
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clear
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close all
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% Set to non-zero to generate LaTeX for graphs
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% Set to non-zero to generate LaTeX for graphs
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enable_m2tikz = 0;
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enable_m2tikz = 0;
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else
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else
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matlab2tikz = @(a,b,c) 0;
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matlab2tikz = @(a,b,c) 0;
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end
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end
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%% 1.1 - Rosenbrock function definition and surf plot
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close all
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syms x y;
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syms x y;
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f = (1 - x)^2 + 100 * (y - x^2)^2;
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f = (1 - x)^2 + 100 * (y - x^2)^2;
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global fl
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global fl
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fl = matlabFunction(f);
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fl = matlabFunction(f);
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xs = -0.2:0.01:1.2;
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Z = zeros(size(xs,2), size(xs,2));
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for x = 1:size(xs, 2)
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for y = 1:size(xs, 2)
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Z(x,y) = fl(xs(x), xs(y));
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end
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end
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surf(xs, xs, Z, 'EdgeColor', 'none', 'FaceAlpha', 0.4);
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hold on
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plot3([0,1],[0,1],[fl(0,0),fl(1,1)],'-r.');
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plot3([1],[1],[fl(1,1)],'-k.');
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hold off
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figure;
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%% 1.2 - Minimizing the Rosenbrock function
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%% 1.2 - Minimizing the Rosenbrock function
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[x1, xs1, gs1] = Newton(f, [0;0], 50000, 1e-6, true);
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[x1, xs1, gs1] = Newton(f, [0;0], 50000, 1e-6, true);
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BIN
Claudio_Maggioni_3/rosenb.jpg
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Claudio_Maggioni_3/rosenb.jpg
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Binary file not shown.
After Width: | Height: | Size: 56 KiB |
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@ -18,12 +18,12 @@ header-includes:
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### Question (a)
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### Question (a)
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As already covered in the course, the gradient of a standard quadratic form at a
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As already covered in the course, the gradient of a standard quadratic form at a
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point $ x_0$ is equal to:
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point $x_0$ is equal to:
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$$ \nabla f(x_0) = A x_0 - b $$
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$$ \nabla f(x_0) = A x_0 - b $$
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Plugging in the definition of $x_0$ and knowing that $\nabla f(x_m) = A x_m - b = 0$
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Plugging in the definition of $x_0$ and knowing that $\nabla f(x_m) = A x_m - b
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(according to the first necessary condition for a minimizer), we obtain:
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= 0$ (according to the first necessary condition for a minimizer), we obtain:
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$$ \nabla f(x_0) = A (x_m + v) - b = A x_m + A v - b = b + \lambda v - b =
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$$ \nabla f(x_0) = A (x_m + v) - b = A x_m + A v - b = b + \lambda v - b =
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\lambda v $$
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\lambda v $$
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@ -52,7 +52,90 @@ $\alpha \neq \frac{1}{\lambda}$.
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Therefore, since $x_1 = x_m$, we have proven SD
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Therefore, since $x_1 = x_m$, we have proven SD
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converges to the minimizer in one iteration.
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converges to the minimizer in one iteration.
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### Point 2
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## Point 2
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The right answer is choice (a), since the energy norm of the error indeed always
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The right answer is choice (a), since the energy norm of the error indeed always
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decreases monotonically.
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decreases monotonically.
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To prove that this is true, we first consider a way to express any iterate $x_k$
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in function of the minimizer $x_s$ and of the missing iterations:
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$$x_k = x_s + \sum_{i=k}^{N} \alpha_i A^i p_0$$
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This formula makes use of the fact that step directions in CG are all
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A-orthogonal with each other, so the k-th search direction $p_k$ is equal to
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$A^k p_0$, where $p_0 = -r_0$ and $r_0$ is the first residual.
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Given that definition of iterates, we're able to express the error after
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iteration $k$ $e_k$ in a similar fashion:
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$$e_k = x_k - x_s = \sum_{i=k}^{N} \alpha_i A^i p_0$$
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We then recall the definition of energy norm $\|e_k\|_A$:
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$$\|e_k\|_A = \sqrt{\langle Ae_k, e_k \rangle}$$
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We then want to show that $\|e_k\|_A = \|x_k - x_s\|_A > \|e_{k+1}\|_A$, which
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in turn is equivalent to claim that:
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|
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$$\langle Ae_k, e_k \rangle > \langle Ae_{k+1}, e_{k+1} \rangle$$
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Knowing that the dot product is linear w.r.t. either of its arguments, we pull
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out the sum term related to the k-th step (i.e. the first term in the sum that
|
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makes up $e_k$) from both sides of $\langle Ae_k, e_k \rangle$,
|
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obtaining the following:
|
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|
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|
$$\langle Ae_{k+1}, e_{k+1} \rangle + \langle \alpha_k A^{k+1} p_0, e_k \rangle
|
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|
+ \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > \langle Ae_{k+1}, e_{k+1}
|
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|
\rangle$$
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|
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|
which in turn is equivalent to claim that:
|
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|
|
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|
$$\langle \alpha_k A^{k+1} p_0, e_k \rangle
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|
+ \langle Ae_{k+1},\alpha_k A^k p_0 \rangle > 0$$
|
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|
|
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|
From this expression we can collect term $\alpha_k$ thanks to linearity of the
|
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|
dot-product:
|
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|
|
||||||
|
$$\alpha_k (\langle A^{k+1} p_0, e_k \rangle
|
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|
+ \langle Ae_{k+1}, A^k p_0 \rangle) > 0$$
|
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|
|
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|
and we can further "ignore" the $\alpha_k$ term since we know that all
|
||||||
|
$\alpha_i$s are positive by definition:
|
||||||
|
|
||||||
|
$$\langle A^{k+1} p_0, e_k \rangle
|
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|
+ \langle Ae_{k+1}, A^k p_0 \rangle > 0$$
|
||||||
|
|
||||||
|
Then, we convert the dot-products in their equivalent vector to vector product
|
||||||
|
form, and we plug in the definitions of $e_k$ and $e_{k+1}$:
|
||||||
|
|
||||||
|
$$p_0^T (A^{k+1})^T (\sum_{i=k}^{N} \alpha_i A^i p_0) +
|
||||||
|
p_0^T (A^{k})^T (\sum_{i=k+1}^{N} \alpha_i A^i p_0) > 0$$
|
||||||
|
|
||||||
|
We then pull out the sum to cover all terms thanks to associativity of vector
|
||||||
|
products:
|
||||||
|
|
||||||
|
$$\sum_{i=k}^N (p_0^T (A^{k+1})^T A^i p_0) \alpha_i+ \sum_{i=k+1}^N
|
||||||
|
(p_0^T (A^{k})^T A^i p_0) \alpha_i > 0$$
|
||||||
|
|
||||||
|
We then, as before, can "ignore" all $\alpha_i$ terms since we know by
|
||||||
|
definition that
|
||||||
|
they are all strictly positive. We then recalled that we assumed that A is
|
||||||
|
symmetric, so $A^T = A$. In the end we have to show that these two
|
||||||
|
inequalities are true:
|
||||||
|
|
||||||
|
$$p_0^T A^{k+1+i} p_0 > 0 \; \forall i \in [k,N]$$
|
||||||
|
$$p_0^T A^{k+i} p_0 > 0 \; \forall i \in [k+1,N]$$
|
||||||
|
|
||||||
|
To show these inequalities are indeed true, we recall that A is symmetric and
|
||||||
|
positive definite. We then consider that if a matrix A is SPD, then $A^i$ for
|
||||||
|
any positive $i$ is also SPD[^1]. Therefore, both inequalities are trivially
|
||||||
|
true due to the definition of positive definite matrices.
|
||||||
|
|
||||||
|
[^1]: source: [Wikipedia - Definite Matrix $\to$ Properties $\to$
|
||||||
|
Multiplication](
|
||||||
|
https://en.wikipedia.org/wiki/Definite_matrix#Multiplication)
|
||||||
|
|
||||||
|
Thanks to this we have indeed proven that the delta $\|e_k\|_A - \|e_{k+1}\|_A$
|
||||||
|
is indeed positive and thus as $i$ increases the energy norm of the error
|
||||||
|
monotonically decreases.
|
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|
|
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Reference in a new issue