Finish algo and FLOPs

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Janita Willumsen 2023-09-09 17:27:04 +02:00
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\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 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