\section*{Problem 9}

\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 division, 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 $6(n-1)+1$.


\subsection*{)}
% Code