Merge branch 'main' into coryab/implement-problem-5

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 +9,39 @@ 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 $b -= d*(*sup_diag)(i-1);$
+        \State $\hat{g}_{1} \leftarrow g_{1}$
-        $(*g_vec)(i) -= d*(*g_vec)(i-1);$
+        \For{$i = 2, 3, ..., n$}
+        \State $d \leftarrow \frac{a_{i}}{\hat{b}_{i-1}}$ \Comment{Calculating common expression}
+        \State $\hat{b}_{i} \leftarrow b_{i} - d \cdot c_{i-1}$
+        \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 $8(n-1)+1$.

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