Merge pull request #19 from FYS3150-G2-2023/6-solve-problem-6

Finished exercise 6

latex/problems/problem6.tex

@@ -1,7 +1,6 @@
 \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*}
@@ -10,40 +9,38 @@ Renaming the sub-, main-, and supdiagonal of matrix $\boldsymbol{A}$
     \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
+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{Foreward sweep}\label{algo:foreward}
+    \caption{General algorithm}\label{algo:general}
     \begin{algorithmic}
-        \State Create new vector \vec{\hat{b}} of length n.
+        \Procedure{Forward sweep}{$\vec{a}$, $\vec{b}$, $\vec{c}$}
-        \State \hat{b}[0] &= b[0] \Comment{Handle first element in main diagonal outside loop}
+        \State $n \leftarrow$ length of $\vec{b}$
-        \For{$i = 1, ..., n-1$}
+        \State $\vec{\hat{b}}$, $\vec{\hat{g}} \leftarrow$ vectors of length $n$.
-        \State d = \frac{a[i-1]}{b[i-1]}
+        \State $\hat{b}_{1} \leftarrow b_{1}$ \Comment{Handle first element in main diagonal outside loop}
-        \State b -= d*(*sup_diag)(i-1);
+        \For{$i = 2, 3, ..., n$}
-        (*g_vec)(i) -= d*(*g_vec)(i-1);
+        \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
-        \While{Some condition}
+        \Return $\vec{\hat{b}}$, $\vec{\hat{g}}$
-        \State Do something more here 
+        \EndProcedure
-        \EndWhile
+
-        \State Maybe even some more math here, e.g $\int_0^1 f(x) \dd x$
+        \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}
-Here the index i does not refer to the element in the vector, ie. $b_{i}$, but to the index in the vector where the element is found ie. $b[i] = b_{i+1}$.  
 
-\begin{align*}
-    \hat{b}_{1} &= b_{1} \\
-    \hat{b}_{2} &= b_{2} - \frac{a_{2}}{\hat{b}_{1}} \cdot c_{1} \\
-    \vdots & \\
-    \hat{b}_{i} &= b_{i} - \frac{a_{i}}{\hat{b}_{i-1}} \cdot c_{i-1} \\
-    \vdots & \\
-\end{align*}
-
-
-% as g_hat also make use of the ratio (ai/bi-1_hat) it only needs to be calculated once
-
-% for n steps we have n+1 points, that is values of x to evaluate u(x) at, when boundary values are known 
-% we have to perform (n+1)-2 = n-1 calculations (equal number of col in matrix). Dealing
-% with a quadratic matrix number of col = number of rows 
 
 \subsection*{b)}
-
+% Figure out FLOPs
-% Figure it out
+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 $8(n-1)+1$.