Wrote a rough draft of problem 5

latex/assignment_1.tex

@@ -84,22 +84,90 @@
     
 \textit{https://github.uio.no/FYS3150-G2-2023/Project-1}
     
-\input{problems/problem1}
+\section*{Problem 1}
 
-\input{problems/problem2}
+% 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*}
 
-\input{problems/problem3}
+Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
 
-\input{problems/problem4}
+\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*}
 
-\input{problems/problem5}
+\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*}
 
-\input{problems/problem6}
+Using the values that we found for $c_1$ and $c_2$, we get
 
-\input{problems/problem7}
+\begin{align*}
+    u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
+         &= 1 - (1 - e^{-10}) - e^{-10x}
+\end{align*}
 
-\input{problems/problem8}
+\section*{Problem 2}
 
-\input{problems/problem9}
+% Write which .cpp/.hpp/.py (using a link?) files are relevant for this and show the plot generated.
+
+\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)}
+
+% Phrase it better
+$n = m - 2$, since $\textbf{A}$ is used to solve for all of the points in 
+between the end-points $0$ and $1$. For the complete solution, we need to add
+$u(0)$ and $u(1)$.
+
+\subsection*{b)}
+
+When solving for $\vec{v}$, we find the approximate solutions for $u(x)$
+that are in between the end-points, but not the end-points themselves.
+
+\section*{Problem 6}
+
+\subsection*{a)}
+
+% Use Gaussian elimination, and then use backwards substitution to solve the equation
+
+\subsection*{b)}
+
+% Figure it out
+
+% Linelevant files on gh and possibly add some comments
+
+\section*{Problem 8}
+
+%link to relvant files and show plots
+
+\section*{Problem 9}
+
+% Shon*{Proalgorithm, 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

@@ -1,35 +0,0 @@
-\section*{Problem 1}
-
-% 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 $c_2$ as shown below:
-
-\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

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

latex/problems/problem2.tex

@@ -1,3 +0,0 @@
-\section*{Problem 2}
-
-% Write which .cpp/.hpp/.py (using a link?) files are relevant for this and show the plot generated.

latex/problems/problem3.tex

@@ -1,4 +0,0 @@
-
-\section*{Problem 3}
-
-% Show how it's derived and where we found the derivation.

latex/problems/problem4.tex

@@ -1,3 +0,0 @@
-\section*{Problem 4}
-
-% Show that each iteration of the discretized version naturally creates a matrix equation.

latex/problems/problem5.tex

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

latex/problems/problem6.tex

@@ -1,9 +0,0 @@
-\section*{Problem 6}
-
-\subsection*{a)}
-
-% Use Gaussian elimination, and then use backwards substitution to solve the equation
-
-\subsection*{b)}
-
-% Figure it out

latex/problems/problem7.tex

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

latex/problems/problem8.tex

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

latex/problems/problem9.tex

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

src/analyticPlot.cpp

@@ -1,52 +0,0 @@
-#include <iostream>
-#include <cmath>
-#include <vector>
-#include <string>
-#include <numeric>
-#include <fstream>
-#include <iomanip>
-
-double u(double x);
-void generate_range(std::vector<double> &vec, double start, double stop, int n);
-
-int main() {
-    int n = 1000;
-
-    std::vector<double> x(n), y(n);
-    generate_range(x, 0.0, 1.0, n);
-
-    // Set up output file and strem
-    std::string filename = "datapoints.txt";
-    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();
-
-    return 0;
-};
-
-double u(double x) {
-    double result;
-
-    result = 1 - (1 - exp(-10))*x - exp(-10*x);
-    return result;
-};
-
-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;
-    }
-}

src/analyticPlot.py

@@ -1,17 +0,0 @@
-import numpy as np 
-import matplotlib.pyplot as plt
-
-x = []
-y = []
-v = []
-with open('testdata.txt') as f:
-    for line in f:
-        a, b, c = line.strip().split()
-        x.append(float(a))
-        # y.append(float(b))
-        v.append(float(c))
-
-fig, ax = plt.subplots()
-ax.plot(x, v)
-plt.show()
-# plt.savefig("main.png")

src/main.cpp

@@ -1,24 +0,0 @@
-#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

@@ -1,162 +0,0 @@
-#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();
-}