Finished exercise 6

Coryab/implement problem 2

.gitignore

@@ -36,3 +36,11 @@
 *.log
 *.out
 *.bib
+*.synctex.gz
+*.bbl 
+
+# C++ specifics
+src/*
+!src/Makefile
+!src/*.cpp
+!src/*.py

latex/assignment_1.tex

@@ -1,7 +1,9 @@
 \documentclass[english,notitlepage]{revtex4-1}  % defines the basic parameters of the document
 %For preview: skriv i terminal: latexmk -pdf -pvc filnavn
 
-
+% Silence warning of revtex4-1
+\usepackage{silence}
+\WarningFilter{revtex4-1}{Repair the float}
 
 % if you want a single-column, remove reprint
 
@@ -13,7 +15,7 @@
 %% I recommend downloading TeXMaker, because it includes a large library of the most common packages.
 
 \usepackage{physics,amssymb}  % mathematical symbols (physics imports amsmath)
-\include{amsmath}
+\usepackage{amsmath}
 \usepackage{graphicx}         % include graphics such as plots
 \usepackage{xcolor}           % set colors
 \usepackage{hyperref}         % automagic cross-referencing (this is GODLIKE)
@@ -72,6 +74,9 @@
 %%
 %% Don't ask me why, I don't know.
 
+% custom stuff
+\graphicspath{{./images/}}
+
 \begin{document}
 
 \title{Project 1}      % self-explanatory
@@ -84,84 +89,22 @@
     
 \textit{https://github.uio.no/FYS3150-G2-2023/Project-1}
     
-\section*{Problem 1}
+\input{problems/problem1}
 
-% Do the double integral
+\input{problems/problem2}
-\begin{align*}
-    u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2\\
-         &= \int \int -100 e^{-10x} dx^2 \\
-         &= \int \frac{-100 e^{-10x}}{-10} + c_1 dx \\
-         &= \int 10 e^{-10x} + c_1 dx \\
-         &= \frac{10 e^{-10x}}{-10} + c_1 x + c_2 \\
-         &= -e^{-10x} + c_1 x + c_2
-\end{align*}
 
-Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
+\input{problems/problem3}
 
-\begin{align*}
+\input{problems/problem4}
-    u(0) &= 0 \\
-    -e^{-10 \cdot 0} + c_1 \cdot 0 + c_2 &= 0 \\
-    -1 + c_2 &= 0 \\
-    c_2 &= 1
-\end{align*}
 
-\begin{align*}
+\input{problems/problem5}
-    u(1) &= 0 \\
-    -e^{-10 \cdot 1} + c_1 \cdot 1 + c_2 &= 0 \\
-    -e^{-10} + c_1 + c_2 &= 0 \\
-    c_1 &= e^{-10} - c_2\\
-    c_1 &= e^{-10} - 1\\
-\end{align*}
 
-Using the values that we found for $c_1$ and $c_2$, we get
+\input{problems/problem6}
 
-\begin{align*}
+\input{problems/problem7}
-    u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
-         &= 1 - (1 - e^{-10}) - e^{-10x}
-\end{align*}
 
-\section*{Problem 2}
+\input{problems/problem8}
 
-% Write which .cpp/.hpp/.py (using a link?) files are relevant for this and show the plot generated.
+\input{problems/problem9}
-
-\section*{Problem 3}
-
-% Show how it's derived and where we found the derivation.
-
-\section*{Problem 4}
-
-% Show that each iteration of the discretized version naturally creates a matrix equation.
-
-\section*{Problem 5}
-
-\subsection*{a)}
-
-\subsection*{b)}
-
-\section*{Problem 6}
-
-\subsection*{a)}
-
-% Use Gaussian elimination, and then use backwards substitution to solve the equation
-
-\subsection*{b)}
-
-% Figure it out
-
-\section*{Problem 7}
-
-% Link to relevant files on gh and possibly add some comments
-
-\section*{Problem 8}
-
-%link to relvant files and show plots
-
-\section*{Problem 9}
-
-% Show the algorithm, then calculate FLOPs, then link to relevant files
-
-\section*{Problem 10}
-
-% Time and show result, and link to relevant files
 
 \end{document}

latex/problems/problem1.tex

@@ -0,0 +1,44 @@
+\section*{Problem 1}
+
+First, we rearrange the equation.
+
+\begin{align*}
+    - \frac{d^2u}{dx^2} &= 100 e^{-10x} \\
+    \frac{d^2u}{dx^2} &= -100 e^{-10x} \\
+\end{align*}
+
+Now we find $u(x)$.
+
+% Do the double integral
+\begin{align*}
+    u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2 \\
+         &= \int \int -100 e^{-10x} dx^2 \\
+         &= \int \frac{-100 e^{-10x}}{-10} + c_1 dx \\
+         &= \int 10 e^{-10x} + c_1 dx \\
+         &= \frac{10 e^{-10x}}{-10} + c_1 x + c_2 \\
+         &= -e^{-10x} + c_1 x + c_2
+\end{align*}
+
+Using the boundary conditions, we can find $c_1$ and $c_2$
+
+\begin{align*}
+    u(0) &= 0 \\
+    -e^{-10 \cdot 0} + c_1 \cdot 0 + c_2 &= 0 \\
+    -1 + c_2 &= 0 \\
+    c_2 &= 1
+\end{align*}
+
+\begin{align*}
+    u(1) &= 0 \\
+    -e^{-10 \cdot 1} + c_1 \cdot 1 + c_2 &= 0 \\
+    -e^{-10} + c_1 + c_2 &= 0 \\
+    c_1 &= e^{-10} - c_2\\
+    c_1 &= e^{-10} - 1\\
+\end{align*}
+
+Using the values that we found for $c_1$ and $c_2$, we get
+
+\begin{align*}
+    u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
+         &= 1 - (1 - e^{-10}) - e^{-10x}
+\end{align*}

latex/problems/problem10.tex

@@ -0,0 +1,3 @@
+\section*{Problem 10}
+
+% Time and show result, and link to relevant files

latex/problems/problem2.tex

@@ -0,0 +1,11 @@
+\section*{Problem 2}
+
+The code for generating the points and plotting them can be found under.
+
+Point generator code: https://github.uio.no/FYS3150-G2-203/Project-1/blob/main/src/analyticPlot.cpp
+
+Plotting code: https://github.uio.no/FYS3150-G2-2023/Project-1/blob/main/src/analyticPlot.py
+
+Here is the plot of the analytical solution for $u(x)$.
+
+\includegraphics[scale=.5]{analytical_solution.pdf}

latex/problems/problem3.tex

@@ -0,0 +1,37 @@
+
+\section*{Problem 3}
+
+To derive the discretized version of the Poisson equation, we first need
+the Taylor expansion for $u(x)$ around $x$ for $x + h$ and $x - h$.
+
+\begin{align*}
+    u(x+h) &= u(x) + u'(x) h + \frac{1}{2} u''(x) h^2 + \frac{1}{6} u'''(x) h^3 + \mathcal{O}(h^4)
+\end{align*}
+
+\begin{align*}
+    u(x-h) &= u(x) - u'(x) h + \frac{1}{2} u''(x) h^2 - \frac{1}{6} u'''(x) h^3 + \mathcal{O}(h^4)
+\end{align*}
+
+If we add the equations above, we get this new equation:
+
+\begin{align*}
+    u(x+h) + u(x-h) &= 2 u(x) + u''(x) h^2 + \mathcal{O}(h^4) \\
+    u(x+h) - 2 u(x) + u(x-h) + \mathcal{O}(h^4) &= u''(x) h^2 \\
+    u''(x) &= \frac{u(x+h) - 2 u(x) + u(x-h)}{h^2} + \mathcal{O}(h^2) \\
+    u_i''(x) &= \frac{u_{i+1} - 2 u_i + u_{i-1}}{h^2} + \mathcal{O}(h^2) \\
+\end{align*}
+
+We can then replace $\frac{d^2u}{dx^2}$ with the RHS (right-hand side) of the equation:
+
+\begin{align*}
+    - \frac{d^2u}{dx^2} &= f(x) \\
+    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= f_i \\
+\end{align*}
+
+And lastly, we leave out $\mathcal{O}(h^2)$ and change $u_i$ to $v_i$ to 
+differentiate between the exact solution and the approximate solution, 
+and get the discretized version of the equation:
+
+\begin{align*}
+    \frac{ - v_{i+1} + 2 v_i - v_{i-1}}{h^2} &= 100 e^{-10x_i} \\
+\end{align*}

latex/problems/problem4.tex

@@ -0,0 +1,44 @@
+\section*{Problem 4}
+
+% Show that each iteration of the discretized version naturally creates a matrix equation.
+
+The value of $u(x_{0})$ and $u(x_{1})$ is known, using the discretized equation we solve for $\vec{v}$. This will result in a set of equations 
+\begin{align*}
+    - v_{0} + 2 v_{1} - v_{2} &= h^{2} \cdot f_{1} \\
+    - v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
+    \vdots & \\
+    - v_{m-2} + 2 v_{m-1} - v_{m} &= h^{2} \cdot f_{m-1} \\
+\end{align*}
+
+where $v_{i} = v(x_{i})$ and $f_{i} = f(x_{i})$. Rearranging the first and last equation, moving terms of known boundary values to the RHS
+\begin{align*}
+    2 v_{1} - v_{2} &= h^{2} \cdot f_{1} + v_{0} \\
+    - v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
+    \vdots & \\
+    - v_{m-2} + 2 v_{m-1} &= h^{2} \cdot f_{m-1} + v_{m} \\
+\end{align*}
+
+We now have a number of linear eqations, corresponding to the number of unknown values, which can be represented as an augmented matrix 
+\begin{align*}
+    \left[
+      \begin{matrix}
+        2v_{1} & -v_{2} & 0 & \dots & 0 \\
+        -v_{1} & 2v_{2} & -v_{3} & 0 & \\
+        0 & -v_{2} & 2v_{3} & -v_{4} & \\
+        \vdots  & & & \ddots & \vdots \\
+        0 & & & -v_{m-2} & 2v_{m-1} \\
+      \end{matrix}
+      \left|
+        \,
+        \begin{matrix}
+          g_{1}  \\
+          g_{2}  \\
+          g_{2}  \\
+          \vdots  \\
+          g_{m-1}  \\
+        \end{matrix}
+      \right.
+    \right]
+    \end{align*}
+    Since the boundary values are equal to $0$ the RHS can be renamed $g_{i} = h^{2} f_{i}$ for all $i$. An augmented matrix can be represented as $\boldsymbol{A} \vec{x} = \vec{b}$. In this case $\boldsymbol{A}$ is the coefficient matrix with a tridiagonal signature $(-1, 2, -1)$ and dimension $n \cross n$, where $n=m-2$.
+

latex/problems/problem5.tex

@@ -0,0 +1,6 @@
+
+\section*{Problem 5}
+
+\subsection*{a)}
+
+\subsection*{b)}

latex/problems/problem6.tex

@@ -0,0 +1,46 @@
+\section*{Problem 6}
+
+\subsection*{a)}
+% Use Gaussian elimination, and then use backwards substitution to solve the equation
+Renaming the sub-, main-, and supdiagonal of matrix $\boldsymbol{A}$ 
+\begin{align*}
+    \vec{a} &= [a_{2}, a_{3}, ..., a_{n-1}, a_{n}] \\
+    \vec{b} &= [b_{1}, b_{2}, b_{3}, ..., b_{n-1}, b_{n}] \\
+    \vec{c} &= [c_{1}, c_{2}, c_{3}, ..., c_{n-1}] \\
+\end{align*}
+
+Following Thomas algorithm for gaussian elimination, we first perform a forward sweep followed by a backward sweep to obtain $\vec{v}$
+\begin{algorithm}[H]
+    \caption{General algorithm}\label{algo:general}
+    \begin{algorithmic}
+        \Procedure{Forward sweep}{$\vec{a}$, $\vec{b}$, $\vec{c}$}
+        \State $n \leftarrow$ length of $\vec{b}$
+        \State $\vec{\hat{b}}$, $\vec{\hat{g}} \leftarrow$ vectors of length $n$.
+        \State $\hat{b}_{1} \leftarrow b_{1}$ \Comment{Handle first element in main diagonal outside loop}
+        \For{$i = 2, 3, ..., n$}
+        \State $d \leftarrow \frac{a_{i}}{\hat{b}_{i-1}}$ \Comment{Calculating common expression}
+        \State $\hat{b}_{i} \leftarrow b_{i} - d \cdot c_{i-1}$
+        \State $\hat{g}_{i} \leftarrow g_{i} - d \cdot \hat{g}_{i-1}$
+        \EndFor
+        \Return $\vec{\hat{b}}$, $\vec{\hat{g}}$
+        \EndProcedure
+
+        \Procedure{Backward sweep}{$\vec{\hat{b}}$, $\vec{\hat{g}}$}
+        \State $n \leftarrow$ length of $\vec{\hat{b}}$
+        \State $\vec{v} \leftarrow$ vector of length $n$.
+        \State $v_{n} \leftarrow \frac{\hat{g}_{n}}{\hat{b}_{n}}$
+        \For{$i = n-1, n-2, ..., 1$}
+        \State $v_{i} \leftarrow \frac{\hat{g}_{i} - c_{i} \cdot v_{i+1}}{\hat{b}_{i}}$ 
+        \EndFor
+        \Return $\vec{v}$
+        \EndProcedure
+    \end{algorithmic}
+\end{algorithm}
+
+
+\subsection*{b)}
+% Figure out FLOPs
+Counting the number of FLOPs for the general algorithm by looking at one procedure at a time. 
+For every iteration of i in forward sweep we have 1 division, 2 multiplications, and 2 subtractions, resulting in $5(n-1)$ FLOPs. 
+For backward sweep we have 1 division, and for every iteration of i we have 1 subtraction, 1 multiplication, and 1 division, resulting in $3(n-1)+1$ FLOPs. 
+Total FLOPs for the general algorithm is $8(n-1)+1$.

latex/problems/problem7.tex

@@ -0,0 +1,3 @@
+\section*{Problem 7}
+
+% Link to relevant files on gh and possibly add some comments

latex/problems/problem8.tex

@@ -0,0 +1,3 @@
+\section*{Problem 8}
+
+%link to relvant files and show plots

latex/problems/problem9.tex

@@ -0,0 +1,3 @@
+\section*{Problem 9}
+
+% Show the algorithm, then calculate FLOPs, then link to relevant files

src/Makefile

@@ -0,0 +1,17 @@
+CC=g++
+
+.PHONY: clean
+
+all: simpleFile analyticPlot
+
+simpleFile: simpleFile.o
+	$(CC) -o $@ $^
+
+analyticPlot: analyticPlot.o
+	$(CC) -o $@ $^
+
+%.o: %.cpp
+	$(CC) -c $< -o $@
+
+clean:
+	rm *.o

src/analyticPlot.cpp

@@ -0,0 +1,55 @@
+#include <iostream>
+#include <cmath>
+#include <vector>
+#include <string>
+#include <numeric>
+#include <fstream>
+#include <iomanip>
+
+#define RANGE 1000
+#define FILENAME "analytical_solution.txt"
+
+double u(double x);
+void generate_range(std::vector<double> &vec, double start, double stop, int n);
+void write_analytical_solution(std::string filename, int n);
+
+int main() {
+    write_analytical_solution(FILENAME, RANGE);
+
+    return 0;
+};
+
+double u(double x) {
+    return 1 - (1 - exp(-10))*x - exp(-10*x);
+};
+
+void generate_range(std::vector<double> &vec, double start, double stop, int n) {
+    double step = (stop - start) / n;
+
+    for (int i = 0; i <= vec.size(); i++) {
+        vec[i] = i * step;
+    }
+}
+
+void write_analytical_solution(std::string filename, int n) {
+    std::vector<double> x(n), y(n);
+    generate_range(x, 0.0, 1.0, n);
+
+    // Set up output file and strem
+    std::ofstream outfile;
+    outfile.open(filename);
+
+    // Parameters for formatting 
+    int width = 12;
+    int prec = 4;
+
+    // Calculate u(x) and write to file
+    for (int i = 0; i <= x.size(); i++) {
+        y[i] = u(x[i]);
+        outfile << std::setw(width) << std::setprecision(prec) << std::scientific << x[i] 
+            << std::setw(width) << std::setprecision(prec) << std::scientific << y[i] 
+            << std::endl;
+    }
+    outfile.close();
+}
+

src/analyticPlot.py

@@ -0,0 +1,19 @@
+import numpy as np 
+import matplotlib.pyplot as plt
+
+def main():
+    FILENAME = "analytical_solution.pdf"
+    x = []
+    v = []
+
+    with open('analytical_solution.txt') as f:
+        for line in f:
+            a, b = line.strip().split()
+            x.append(float(a))
+            v.append(float(b))
+
+    plt.plot(x, v)
+    plt.savefig(FILENAME)
+
+if __name__ == "__main__":
+    main()

src/main.cpp

@@ -0,0 +1,24 @@
+#include "GeneralAlgorithm.hpp"
+#include <armadillo>
+#include <iostream>
+
+double f(double x) {
+    return 100. * std::exp(-10.*x);
+}
+
+double a_sol(double x) {
+    return 1. - (1. - std::exp(-10)) * x - std::exp(-10*x);
+}
+
+int main() {
+    arma::mat A = arma::eye(3,3);
+
+    GeneralAlgorithm ga(3, &A, f, a_sol, 0., 1.);
+
+    ga.solve();
+    std::cout << "Time: " << ga.time(5) << std::endl;
+    ga.error();
+
+    return 0;
+
+}

src/simpleFile.cpp

@@ -0,0 +1,162 @@
+#include <armadillo>
+#include <cmath>
+#include <ctime>
+#include <fstream>
+#include <iomanip>
+#include <ios>
+#include <string>
+
+#define TIMING_ITERATIONS 5
+
+arma::vec* general_algorithm(
+    arma::vec* sub_diag,
+    arma::vec* main_diag,
+    arma::vec* sup_diag,
+    arma::vec* g_vec
+)
+{
+    int n = g_vec->n_elem;
+    double d;
+
+    for (int i = 1; i < n; i++) {
+        d = (*sub_diag)(i-1) / (*main_diag)(i-1);
+        (*main_diag)(i) -= d*(*sup_diag)(i-1);
+        (*g_vec)(i) -= d*(*g_vec)(i-1);
+    }
+
+    (*g_vec)(n-1) /= (*main_diag)(n-1);
+
+    for (int i = n-2; i >= 0; i--) {
+        (*g_vec)(i) = ((*g_vec)(i) - (*sup_diag)(i) * (*g_vec)(i+1)) / (*main_diag)(i);
+    }
+    return g_vec;
+}
+
+
+arma::vec* special_algorithm(
+    double sub_sig,
+    double main_sig,
+    double sup_sig,
+    arma::vec* g_vec
+)
+{
+    int n = g_vec->n_elem;
+    arma::vec diag = arma::vec(n);
+
+    for (int i = 1; i < n; i++) {
+        // Calculate values for main diagonal based on indices
+        diag(i-1) = (double)(i+1) / i;
+        (*g_vec)(i) += (*g_vec)(i-1) / diag(i-1);
+    }
+    // The last element in main diagonal has value (i+1)/i = (n+1)/n
+    (*g_vec)(n-1) /= (double)(n+1) / (n);
+
+    for (int i = n-2; i >= 0; i--) {
+        (*g_vec)(i) = ((*g_vec)(i) + (*g_vec)(i+1))/ diag(i);
+    }
+
+    return g_vec;
+}
+
+void error(
+    std::string filename,
+    arma::vec* x_vec,
+    arma::vec* v_vec,
+    arma::vec* a_vec
+)
+{
+    std::ofstream ofile;
+    ofile.open(filename);
+
+    if (!ofile.is_open()) {
+        exit(1);
+    }
+
+    for (int i=0; i < a_vec->n_elem; i++) {
+        double sub = (*a_vec)(i) - (*v_vec)(i);
+        ofile << std::setprecision(8) << std::scientific << (*x_vec)(i)
+            << std::setprecision(8) << std::scientific << std::log10(std::abs(sub))
+            << std::setprecision(8) << std::scientific << std::log10(std::abs(sub/(*a_vec)(i)))
+            << std::endl;
+    }
+    
+    ofile.close();
+
+}
+
+double f(double x) {
+    return 100*std::exp(-10*x);
+}
+
+void build_array(
+    int n_steps, 
+    arma::vec* sub_diag, 
+    arma::vec* main_diag, 
+    arma::vec* sup_diag,
+    arma::vec* g_vec
+) 
+{
+    sub_diag->resize(n_steps-2);
+    main_diag->resize(n_steps-1);
+    sup_diag->resize(n_steps-2);
+
+    sub_diag->fill(-1);
+    main_diag->fill(2);
+    sup_diag->fill(-1);
+
+    g_vec->resize(n_steps-1);
+
+    double step_size = 1./ (double) n_steps;
+    for (int i=0; i < n_steps-1; i++) {
+        (*g_vec)(i) = f((i+1)*step_size);
+    }
+
+}
+
+void timing() {
+    arma::vec sub_diag, main_diag, sup_diag, g_vec;
+    int n_steps;
+
+    std::ofstream ofile;
+    ofile.open("timing.txt");
+
+    // Timing
+    for (int i=1; i <= 8; i++) {
+        n_steps = std::pow(10, i);
+        clock_t g_1, g_2, s_1, s_2;
+        double g_res = 0, s_res = 0;
+      
+        for (int j=0; j < TIMING_ITERATIONS; j++) {
+            build_array(n_steps, &sub_diag, &main_diag, &sup_diag, &g_vec);
+
+            g_1 = clock();
+
+            general_algorithm(&sub_diag, &main_diag, &sup_diag, &g_vec);
+
+            g_2 = clock();
+
+            g_res += (double) (g_2 - g_1) / CLOCKS_PER_SEC;
+            build_array(n_steps, &sub_diag, &main_diag, &sup_diag, &g_vec);
+
+            s_1 = clock();
+
+            special_algorithm(-1., 2., -1., &g_vec);
+
+            s_2 = clock();
+
+            s_res += (double) (s_2 - s_1) / CLOCKS_PER_SEC;
+
+        }
+        ofile 
+            << n_steps << ","
+            << g_res / (double) TIMING_ITERATIONS << ","
+            << s_res / (double) TIMING_ITERATIONS << std::endl;
+    }
+
+    ofile.close();
+}
+
+int main() 
+{
+    timing();
+}