diff --git a/latex/assignment_1.pdf b/latex/assignment_1.pdf
index 80e2ee3..c29ca0d 100644
Binary files a/latex/assignment_1.pdf and b/latex/assignment_1.pdf differ
diff --git a/latex/assignment_1.tex b/latex/assignment_1.tex
index 211f82a..87de2c0 100644
--- a/latex/assignment_1.tex
+++ b/latex/assignment_1.tex
@@ -84,84 +84,22 @@
     
 \textit{https://github.uio.no/FYS3150-G2-2023/Project-1}
     
-\section*{Problem 1}
+\input{problems/problem1}
 
-% 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/problem2}
 
-Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
+\input{problems/problem3}
 
-\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/problem4}
 
-\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/problem5}
 
-Using the values that we found for $c_1$ and $c_2$, we get
+\input{problems/problem6}
 
-\begin{align*}
-    u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
-         &= 1 - (1 - e^{-10}) - e^{-10x}
-\end{align*}
+\input{problems/problem7}
 
-\section*{Problem 2}
+\input{problems/problem8}
 
-% 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)}
-
-\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
+\input{problems/problem9}
 
 \end{document}
diff --git a/latex/problems/problem1.tex b/latex/problems/problem1.tex
new file mode 100644
index 0000000..779a58f
--- /dev/null
+++ b/latex/problems/problem1.tex
@@ -0,0 +1,35 @@
+\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*}
diff --git a/latex/problems/problem10.tex b/latex/problems/problem10.tex
new file mode 100644
index 0000000..72de7f2
--- /dev/null
+++ b/latex/problems/problem10.tex
@@ -0,0 +1,3 @@
+\section*{Problem 10}
+
+% Time and show result, and link to relevant files
diff --git a/latex/problems/problem2.tex b/latex/problems/problem2.tex
new file mode 100644
index 0000000..b90aa58
--- /dev/null
+++ b/latex/problems/problem2.tex
@@ -0,0 +1,3 @@
+\section*{Problem 2}
+
+% Write which .cpp/.hpp/.py (using a link?) files are relevant for this and show the plot generated.
diff --git a/latex/problems/problem3.tex b/latex/problems/problem3.tex
new file mode 100644
index 0000000..03fc02f
--- /dev/null
+++ b/latex/problems/problem3.tex
@@ -0,0 +1,4 @@
+
+\section*{Problem 3}
+
+% Show how it's derived and where we found the derivation.
diff --git a/latex/problems/problem4.tex b/latex/problems/problem4.tex
new file mode 100644
index 0000000..2f690f3
--- /dev/null
+++ b/latex/problems/problem4.tex
@@ -0,0 +1,3 @@
+\section*{Problem 4}
+
+% Show that each iteration of the discretized version naturally creates a matrix equation.
diff --git a/latex/problems/problem5.tex b/latex/problems/problem5.tex
new file mode 100644
index 0000000..5d32e6f
--- /dev/null
+++ b/latex/problems/problem5.tex
@@ -0,0 +1,6 @@
+
+\section*{Problem 5}
+
+\subsection*{a)}
+
+\subsection*{b)}
diff --git a/latex/problems/problem6.tex b/latex/problems/problem6.tex
new file mode 100644
index 0000000..d3798e5
--- /dev/null
+++ b/latex/problems/problem6.tex
@@ -0,0 +1,9 @@
+\section*{Problem 6}
+
+\subsection*{a)}
+
+% Use Gaussian elimination, and then use backwards substitution to solve the equation
+
+\subsection*{b)}
+
+% Figure it out
diff --git a/latex/problems/problem7.tex b/latex/problems/problem7.tex
new file mode 100644
index 0000000..a39867c
--- /dev/null
+++ b/latex/problems/problem7.tex
@@ -0,0 +1,3 @@
+\section*{Problem 7}
+
+% Link to relevant files on gh and possibly add some comments
diff --git a/latex/problems/problem8.tex b/latex/problems/problem8.tex
new file mode 100644
index 0000000..4d20743
--- /dev/null
+++ b/latex/problems/problem8.tex
@@ -0,0 +1,3 @@
+\section*{Problem 8}
+
+%link to relvant files and show plots
diff --git a/latex/problems/problem9.tex b/latex/problems/problem9.tex
new file mode 100644
index 0000000..e5bd862
--- /dev/null
+++ b/latex/problems/problem9.tex
@@ -0,0 +1,3 @@
+\section*{Problem 9}
+
+% Show the algorithm, then calculate FLOPs, then link to relevant files
diff --git a/src/main.cpp b/src/main.cpp
new file mode 100644
index 0000000..65bbf4a
--- /dev/null
+++ b/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;
+
+}
diff --git a/src/simpleFile.cpp b/src/simpleFile.cpp
new file mode 100644
index 0000000..7ad5b38
--- /dev/null
+++ b/src/simpleFile.cpp
@@ -0,0 +1,147 @@
+#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();
+}