Merge branch 'main' into coryab/implement-problem-5

9 solve problem 9

.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)
@@ -72,10 +74,13 @@
 %%
 %% Don't ask me why, I don't know.
 
+% custom stuff
+\graphicspath{{./images/}}
+
 \begin{document}
 
 \title{Project 1}      % self-explanatory
-\author{Cory Balaton \& Janita Willumsen}          % self-explanatory
+\author{Cory Alexander Balaton \& Janita Ovidie Sandtrøen Willumsen}          % self-explanatory
 \date{\today}                             % self-explanatory
 \noaffiliation                            % ignore this, but keep it.
 
@@ -102,4 +107,6 @@
 
 \input{problems/problem9}
 
+\input{problems/problem10}
+
 \end{document}

latex/problems/problem1.tex

@@ -1,8 +1,17 @@
 \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\\
+    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 \\
@@ -10,7 +19,7 @@
          &= -e^{-10x} + c_1 x + c_2
 \end{align*}
 
-Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
+Using the boundary conditions, we can find $c_1$ and $c_2$
 
 \begin{align*}
     u(0) &= 0 \\
@@ -31,5 +40,5 @@ 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}
+         &= 1 - (1 - e^{-10})x - e^{-10x}
 \end{align*}

latex/problems/problem2.tex

@@ -1,3 +1,11 @@
 \section*{Problem 2}
 
-% Write which .cpp/.hpp/.py (using a link?) files are relevant for this and show the plot generated.
+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

@@ -1,4 +1,37 @@
 
 \section*{Problem 3}
 
-% Show how it's derived and where we found the derivation.
+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

@@ -1,3 +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

@@ -1,6 +1,6 @@
 
 \section*{Problem 5}
 
-\subsection*{a)}
+\subsection*{a \& b)}
 
-\subsection*{b)}
+$n = m - 2$ since when solving for $\vec{v}$, we are finding the solutions for all the points that are in between the boundaries and not the boundaries themselves. $\vec{v}^*$ on the other hand includes the boundary points.

latex/problems/problem6.tex

@@ -1,9 +1,47 @@
 \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}
+        \State $\hat{g}_{1} \leftarrow g_{1}$
+        \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 it out
+% 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/problem9.tex

@@ -1,3 +1,55 @@
 \section*{Problem 9}
 
-% Show the algorithm, then calculate FLOPs, then link to relevant files
+\subsection*{a)}
+% Specialize algorithm
+The special algorithm does not require the values of all $a_{i}$, $b_{i}$, $c_{i}$. 
+We find the values of $\hat{b}_{i}$ from simplifying the general case
+\begin{align*}
+    \hat{b}_{i} &= b_{i} - \frac{a_{i} \cdot c_{i-1}}{\hat{b}_{i-1}} \\
+    \hat{b}_{i} &= 2 - \frac{1}{\hat{b}_{i-1}}
+\end{align*} 
+Calculating the first values to see a pattern 
+\begin{align*}
+    \hat{b}_{1} &= 2 \\
+    \hat{b}_{2} &= 2 - \frac{1}{2} = \frac{3}{2} \\
+    \hat{b}_{3} &= 2 - \frac{1}{\frac{3}{2}} = \frac{4}{3} \\
+    \hat{b}_{4} &= 2 - \frac{1}{\frac{4}{3}} = \frac{5}{4} \\
+    \vdots & \\
+    \hat{b}_{i} &= \frac{i+1}{i} && \text{for $i = 1, 2, ..., n$}
+\end{align*}
+
+
+\begin{algorithm}[H]
+    \caption{Special algorithm}\label{algo:special}
+    \begin{algorithmic}
+        \Procedure{Forward sweep}{$\vec{b}$}
+        \State $n \leftarrow$ length of $\vec{b}$
+        \State $\vec{\hat{b}}$, $\vec{\hat{g}} \leftarrow$ vectors of length $n$.
+        \State $\hat{b}_{1} \leftarrow 2$ \Comment{Handle first element in main diagonal outside loop}
+        \State $\hat{g}_{1} \leftarrow g_{1}$
+        \For{$i = 2, 3, ..., n$}
+        \State $\hat{b}_{i} \leftarrow \frac{i+1}{i}$
+        \State $\hat{g}_{i} \leftarrow g_{i} + \frac{\hat{g}_{i-1}}{\hat{b}_{i-1}}$
+        \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 $v_{i} \leftarrow \frac{\hat{g}_{i} + v_{i+1}}{\hat{b}_{i}}$ 
+        \EndFor
+        \Return $\vec{v}$
+        \EndProcedure
+    \end{algorithmic}
+\end{algorithm}
+
+
+\subsection*{b)}
+% Find FLOPs
+For every iteration of i in forward sweep we have 2 divisions, and 2 additions, resulting in $4(n-1)$ FLOPs. 
+For backward sweep we have 1 division, and for every iteration of i we have 1 addition, and 1 division, resulting in $2(n-1)+1$ FLOPs. 
+Total FLOPs for the special algorithm is $6(n-1)+1$.
+

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

@@ -6,17 +6,36 @@
 #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() {
-    int n = 1000;
+    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::string filename = "datapoints.txt";
     std::ofstream outfile;
     outfile.open(filename);
 
@@ -32,21 +51,5 @@ int main() {
             << 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 +1,19 @@
 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))
+def main():
+    FILENAME = "analytical_solution.pdf"
+    x = []
+    v = []
 
-fig, ax = plt.subplots()
-ax.plot(x, v)
-plt.show()
-# plt.savefig("main.png")
+    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()