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.gitignore vendored
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@ -36,12 +36,3 @@
*.log *.log
*.out *.out
*.bib *.bib
*.synctex.gz
*.bbl
# C++ specifics
src/*
!src/Makefile
!src/*.cpp
!src/*.hpp
!src/*.py

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@ -3,36 +3,3 @@
## Practical information ## Practical information
- [Project](https://anderkve.github.io/FYS3150/book/projects/project1.html) - [Project](https://anderkve.github.io/FYS3150/book/projects/project1.html)
## How to compile C++ code
Make sure that you are inside the **src** directory before compiling the code.
Now you can execute the command shown under to compile:
```
make
```
This will create object files and link them together into 2 executable files.
These files are called **main** and **analyticPlot**.
To run them, you can simply use the commands below:
```
./main
```
```
./analyticPlot
```
## How to generate plots
For generating the plots, there are 4 Python scripts.
You can run each one of them by using this command:
```
python <PythonFile>
```
The plots will be saved inside **latex/images**.

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@ -1,9 +1,7 @@
\documentclass[english,notitlepage]{revtex4-1} % defines the basic parameters of the document \documentclass[english,notitlepage]{revtex4-1} % defines the basic parameters of the document
%For preview: skriv i terminal: latexmk -pdf -pvc filnavn %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 % if you want a single-column, remove reprint
@ -15,7 +13,7 @@
%% I recommend downloading TeXMaker, because it includes a large library of the most common packages. %% I recommend downloading TeXMaker, because it includes a large library of the most common packages.
\usepackage{physics,amssymb} % mathematical symbols (physics imports amsmath) \usepackage{physics,amssymb} % mathematical symbols (physics imports amsmath)
\usepackage{amsmath} \include{amsmath}
\usepackage{graphicx} % include graphics such as plots \usepackage{graphicx} % include graphics such as plots
\usepackage{xcolor} % set colors \usepackage{xcolor} % set colors
\usepackage{hyperref} % automagic cross-referencing (this is GODLIKE) \usepackage{hyperref} % automagic cross-referencing (this is GODLIKE)
@ -74,13 +72,10 @@
%% %%
%% Don't ask me why, I don't know. %% Don't ask me why, I don't know.
% custom stuff
\graphicspath{{./images/}}
\begin{document} \begin{document}
\title{Project 1} % self-explanatory \title{Project 1} % self-explanatory
\author{Cory Alexander Balaton \& Janita Ovidie Sandtrøen Willumsen} % self-explanatory \author{Cory Balaton \& Janita Willumsen} % self-explanatory
\date{\today} % self-explanatory \date{\today} % self-explanatory
\noaffiliation % ignore this, but keep it. \noaffiliation % ignore this, but keep it.
@ -107,6 +102,4 @@
\input{problems/problem9} \input{problems/problem9}
\input{problems/problem10}
\end{document} \end{document}

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@ -1,17 +1,8 @@
\section*{Problem 1} \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 % Do the double integral
\begin{align*} \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 \int -100 e^{-10x} dx^2 \\
&= \int \frac{-100 e^{-10x}}{-10} + c_1 dx \\ &= \int \frac{-100 e^{-10x}}{-10} + c_1 dx \\
&= \int 10 e^{-10x} + c_1 dx \\ &= \int 10 e^{-10x} + c_1 dx \\
@ -19,7 +10,7 @@ Now we find $u(x)$.
&= -e^{-10x} + c_1 x + c_2 &= -e^{-10x} + c_1 x + c_2
\end{align*} \end{align*}
Using the boundary conditions, we can find $c_1$ and $c_2$ Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
\begin{align*} \begin{align*}
u(0) &= 0 \\ u(0) &= 0 \\
@ -40,5 +31,5 @@ Using the values that we found for $c_1$ and $c_2$, we get
\begin{align*} \begin{align*}
u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\ u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
&= 1 - (1 - e^{-10})x - e^{-10x} &= 1 - (1 - e^{-10}) - e^{-10x}
\end{align*} \end{align*}

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@ -1,8 +1,3 @@
\section*{Problem 10} \section*{Problem 10}
% Time and show result, and link to relevant files % Time and show result, and link to relevant files
\begin{figure}[H]
\centering
\includegraphics[width=0.7\linewidth]{images/problem10.pdf}
\caption{Timing of general algorithm vs. special for step sizes $n_{steps}$}
\end{figure}

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@ -1,13 +1,3 @@
\section*{Problem 2} \section*{Problem 2}
The code for generating the points and plotting them can be found under. % Write which .cpp/.hpp/.py (using a link?) files are relevant for this and show the plot generated.
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
\begin{figure}[H]
\centering
\includegraphics[width=0.8\linewidth]{images/analytical_solution.pdf}
\caption{Plot of the analytical solution $u(x)$.}
\end{figure}

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@ -1,37 +1,4 @@
\section*{Problem 3} \section*{Problem 3}
To derive the discretized version of the Poisson equation, we first need % Show how it's derived and where we found the derivation.
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*}

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\section*{Problem 4} \section*{Problem 4}
% Show that each iteration of the discretized version naturally creates a matrix equation. % 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 $n=m-2$.

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\section*{Problem 5} \section*{Problem 5}
\subsection*{a \& b)} \subsection*{a)}
$n = m - 2$ 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. \subsection*{b)}

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@ -1,47 +1,9 @@
\section*{Problem 6} \section*{Problem 6}
\subsection*{a)} \subsection*{a)}
% Use Gaussian elimination, and then use backwards substitution to solve the equation % 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)} \subsection*{b)}
% Figure out FLOPs
Counting the number of FLOPs for the general algorithm by looking at one procedure at a time. % Figure it out
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 $8(n-1)+1$.

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@ -1,14 +1,3 @@
\section*{Problem 7} \section*{Problem 7}
\subsection*{a)}
% Link to relevant files on gh and possibly add some comments % Link to relevant files on gh and possibly add some comments
The code can be found at https://github.uio.no/FYS3150-G2-2023/Project-1/blob/main/src/generalAlgorithm.cpp
\subsection*{b)}
Increasing the number of steps results in an approximation close to the exact solution.
\begin{figure}[H]
\centering
\includegraphics[width=0.8\linewidth]{images/problem7.pdf}
\caption{Plot showing the numeric solution of $u_{approx}$ for $n_{steps}$ and the exact solution $u_{exact}$.}
\end{figure}

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@ -1,27 +1,3 @@
\section*{Problem 8} \section*{Problem 8}
%link to relvant files and show plots %link to relvant files and show plots
\subsection*{a)}
Increasing number of steps result in a decrease of absolute error.
\begin{figure}[H]
\centering
\includegraphics[width=0.8\linewidth]{images/problem8_a.pdf}
\caption{Absolute error for different step sizes $n_{steps}$.}
\end{figure}
\subsection*{b)}
Increasing number of steps also result in a decrease of absolute error.
\begin{figure}[H]
\centering
\includegraphics[width=0.8\linewidth]{images/problem8_b.pdf}
\caption{Relative error for different step sizes $n_{steps}$.}
\end{figure}
\subsection*{c)}
Increasing number of steps result in a decrease of maximum relative error, up to a certain number of steps. At $n_{steps} \approx 10^{5}$ the maximumrelative error increase.
This can be related to loss of numerical precicion when step size is small.
\begin{figure}[H]
\centering
\includegraphics[width=0.7\linewidth]{images/problem8_c.pdf}
\caption{Maximum relative error for each step sizes $n_{steps}$.}
\end{figure}

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@ -1,55 +1,3 @@
\section*{Problem 9} \section*{Problem 9}
\subsection*{a)} % Show the algorithm, then calculate FLOPs, then link to relevant files
% 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 $6(n-1)+1$.

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CC=g++
CCFLAGS= -std=c++11
OBJS=generalAlgorithm.o specialAlgorithm.o funcs.o
EXEC=main analyticPlot
.PHONY: clean create_dirs
all: create_dirs $(EXEC)
main: main.o $(OBJS)
$(CC) $(CCFLAGS) -o $@ $^
analyticPlot: analyticPlot.o
$(CC) $(CCFLAGS) -o $@ $^
%.o: %.cpp
$(CC) $(CCFLAGS) -c -o $@ $^
clean:
rm *.o
rm $(EXEC)
rm -r output
create_dirs:
mkdir -p output/general
mkdir -p output/special
mkdir -p output/error

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@ -1,55 +0,0 @@
#include <iostream>
#include <cmath>
#include <vector>
#include <string>
#include <numeric>
#include <fstream>
#include <iomanip>
#define RANGE 1000
#define FILENAME "output/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();
}

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@ -1,38 +0,0 @@
#include "funcs.hpp"
double f(double x) {
return 100*std::exp(-10*x);
}
double u(double x) {
return 1. - (1. - std::exp(-10.))*x - std::exp(-10.*x);
}
void build_g_vec(int n_steps, arma::vec& g_vec) {
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) = step_size*step_size*f((i+1)*step_size);
}
}
void build_arrays(
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);
build_g_vec(n_steps, g_vec);
}

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#ifndef __FUNCS__
#define __FUNCS__
#include <armadillo>
#include <cmath>
#define PRECISION 8
#define N_STEPS_EXP 7
double f(double x);
double u(double x);
void build_g_vec(int n_steps, arma::vec& g_vec);
void build_arrays(
int n_steps,
arma::vec& sub_diag,
arma::vec& main_diag,
arma::vec& sup_diag,
arma::vec& g_vec
);
#endif

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@ -1,84 +0,0 @@
#include "funcs.hpp"
#include "generalAlgorithm.hpp"
#include <cmath>
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;
}
void general_algorithm_main()
{
arma::vec sub_diag, main_diag, sup_diag, g_vec, v_vec;
std::ofstream ofile;
int steps;
double step_size;
for (int i = 0; i < N_STEPS_EXP; i++) {
steps = std::pow(10, i+1);
step_size = 1./(double) steps;
build_arrays(steps, sub_diag, main_diag, sup_diag, g_vec);
v_vec = general_algorithm(sub_diag, main_diag, sup_diag, g_vec);
ofile.open("output/general/out_" + std::to_string(steps) + ".txt");
for (int j=0; j < v_vec.n_elem; j++) {
ofile << std::setprecision(PRECISION) << std::scientific << step_size*(j+1) << ","
<< std::setprecision(PRECISION) << std::scientific << v_vec(j) << std::endl;
}
ofile.close();
}
}
void general_algorithm_error()
{
arma::vec sub_diag, main_diag, sup_diag, g_vec, v_vec;
std::ofstream ofile;
int steps;
double step_size, abs_err, rel_err, u_i, v_i;
for (int i=0; i < N_STEPS_EXP; i++) {
steps = std::pow(10, i+1);
step_size = 1./(double) steps;
build_arrays(steps, sub_diag, main_diag, sup_diag, g_vec);
v_vec = general_algorithm(sub_diag, main_diag, sup_diag, g_vec);
ofile.open("output/error/out_" + std::to_string(steps) + ".txt");
for (int j=0; j < v_vec.n_elem; j++) {
u_i = u(step_size*(j+1));
v_i = v_vec(j);
abs_err = u_i - v_i;
ofile << std::setprecision(PRECISION) << std::scientific
<< step_size*(j+1) << ","
<< std::setprecision(PRECISION) << std::scientific
<< std::log10(std::abs(abs_err)) << ","
<< std::setprecision(PRECISION) << std::scientific
<< std::log10(std::abs(abs_err/u_i)) << std::endl;
}
ofile.close();
}
}

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#ifndef __GENERAL_ALG__
#define __GENERAL_ALG__
#include <armadillo>
#include <iomanip>
arma::vec& general_algorithm(
arma::vec& sub_diag,
arma::vec& main_diag,
arma::vec& sup_diag,
arma::vec& g_vec
);
void general_algorithm_main();
void general_algorithm_error();
#endif

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#include "GeneralAlgorithm.hpp"
#include <armadillo> #include <armadillo>
#include <cmath> #include <iostream>
#include <ctime>
#include <fstream>
#include <iomanip>
#include <ios>
#include <string>
#include "funcs.hpp" double f(double x) {
#include "generalAlgorithm.hpp" return 100. * std::exp(-10.*x);
#include "specialAlgorithm.hpp"
#define TIMING_ITERATIONS 5
void timing() {
arma::vec sub_diag, main_diag, sup_diag, g_vec;
int n_steps;
std::ofstream ofile;
ofile.open("output/timing.txt");
// Timing
for (int i=1; i < N_STEPS_EXP; i++) {
n_steps = std::pow(10, i);
clock_t g_1, g_2, s_1, s_2;
double g_res = 0, s_res = 0;
// Repeat a number of times to take an average
for (int j=0; j < TIMING_ITERATIONS; j++) {
build_arrays(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;
// Rebuild g_vec for the special alg
build_g_vec(n_steps, 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;
}
// Write the average time to file
ofile
<< n_steps << ","
<< g_res / (double) TIMING_ITERATIONS << ","
<< s_res / (double) TIMING_ITERATIONS << std::endl;
}
ofile.close();
} }
int main() double a_sol(double x) {
{ return 1. - (1. - std::exp(-10)) * x - std::exp(-10*x);
timing(); }
general_algorithm_main();
general_algorithm_error(); int main() {
special_algorithm_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;
} }

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import numpy as np
import matplotlib.pyplot as plt
def main():
FILENAME = "../latex/images/analytical_solution.pdf"
x = []
v = []
with open('output/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()

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import matplotlib.pyplot as plt
import numpy as np
analytical_func = lambda x: 1 - (1 - np.exp(-10))*x - np.exp(-10*x)
def main():
for i in range(7):
x = []
y = []
x.append(0.)
y.append(0.)
with open(f"output/general/out_{10**(i+1)}.txt", "r") as f:
lines = f.readlines()
for line in lines:
x_i, y_i = line.strip().split(",")
x.append(float(x_i))
y.append(float(y_i))
x.append(1.)
y.append(0.)
plt.plot(x, y, label=f"n$_{{steps}} = 10^{i+1}$")
x = np.linspace(0, 1, 1001)
plt.plot(x, analytical_func(x), label=f"u$_{{exact}}$")
plt.legend()
plt.savefig("../latex/images/problem7.pdf")
if __name__ == "__main__":
main()

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import matplotlib.pyplot as plt
# plt.rc('text', usetex=True)
# plt.rc('font', family='serif')
def main():
for i in range(7):
x = []
abs_err = []
rel_err = []
with open(f"output/error/out_{10**(i+1)}.txt", "r") as f:
lines = f.readlines()
for line in lines:
x_i, abs_err_i, rel_err_i = line.strip().split(",")
x.append(float(x_i))
abs_err.append(float(abs_err_i))
rel_err.append(float(rel_err_i))
plt.figure(1)
plt.plot(x, abs_err, label=f"n$_{{steps}} = 10^{i+1}$")
plt.figure(2)
plt.plot(x, rel_err, label=f"n$_{{steps}} = 10^{i+1}$")
plt.figure(3)
plt.plot(i+1, max(rel_err), marker="o", markersize=10)
plt.figure(1)
plt.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.)
plt.figure(2)
plt.legend(bbox_to_anchor=(1.05, 1), loc=2, borderaxespad=0.)
plt.figure(1)
plt.savefig("../latex/images/problem8_a.pdf", bbox_inches="tight")
plt.figure(2)
plt.savefig("../latex/images/problem8_b.pdf", bbox_inches="tight")
plt.figure(3)
plt.savefig("../latex/images/problem8_c.pdf", bbox_inches="tight")
if __name__ == "__main__":
main()

0
src/problem2.cpp Normal file
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147
src/simpleFile.cpp Normal file
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#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 = main_diag->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
)
{
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();
}

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#include "funcs.hpp"
#include "specialAlgorithm.hpp"
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 special_algorithm_main()
{
arma::vec g_vec, v_vec;
std::ofstream ofile;
int steps;
double sub_sig, main_sig, sup_sig, step_size;
for (int i = 0; i < N_STEPS_EXP; i++) {
steps = std::pow(10, i+1);
step_size = 1./(double) steps;
build_g_vec(steps, g_vec);
v_vec = special_algorithm(sub_sig, main_sig, sup_sig, g_vec);
ofile.open("output/special/out_" + std::to_string(steps) + ".txt");
for (int j=0; j < v_vec.n_elem; j++) {
ofile << std::setprecision(PRECISION) << std::scientific << step_size*(j+1) << ","
<< std::setprecision(PRECISION) << std::scientific << v_vec(j) << std::endl;
}
ofile.close();
}
}

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#ifndef __SPECIAL_ALG__
#define __SPECIAL_ALG__
#include <armadillo>
#include <iomanip>
arma::vec& special_algorithm(
double sub_sig,
double main_sig,
double sup_sig,
arma::vec& g_vec
);
void special_algorithm_main();
#endif

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import matplotlib.pyplot as plt
def main():
x = []
gen_alg = []
spec_alg = []
with open(f"output/timing.txt", "r") as f:
lines = f.readlines()
for line in lines:
x_i, gen_i, spec_i = line.strip().split(",")
x.append(float(x_i))
gen_alg.append(float(gen_i))
spec_alg.append(float(spec_i))
plt.plot(x, gen_alg, label=f"General algorithm")
plt.plot(x, spec_alg, label=f"Special algorithm")
plt.legend()
plt.savefig("../latex/images/problem10.pdf")
if __name__ == "__main__":
main()