Merge pull request #20 from FYS3150-G2-2023/9-solve-problem-9

9 solve problem 9

latex/problems/problem6.tex

@@ -17,6 +17,7 @@ Following Thomas algorithm for gaussian elimination, we first perform a forward
         \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 $\hat{b}_{i} \leftarrow b_{i} - d \cdot c_{i-1}$

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 $6(n-1)+1$.
+