Keep remote branch up to date

ENUME/projectA/debilMode

@@ -0,0 +1,71 @@
+% ------- Solving a system of n linear equations Ax = b. ------- %
+% -------------------- The Indicated Method -------------------- %
+function x = Indicated_Method(A, b)
+    
+[M,N] = size(A);
+if M ~= N
+      
+error ('A is not square matrix!');
+end
+for j = 1 : N - 1  
+%--- Find the greatest value within column ---%
+        
+m = max(A(j:N,j));
+%--- Find row whitin greatest value occure ---%
+for k = j : N
+if A(k,j) == m
+%--- SwapRow in matrix A ---%
+                
+tempRow =  A(j , :);
+                
+A(j , :) = A(k, :);
+                
+A(k, :) = tempRow;
+%--- SwapValue in vector b ---%
+                
+tempVal = b(j);
+                
+b(j) = b(k);
+                
+b(k) = tempVal;
+break;
+end
+end
+for i = j + 1 : N
+            
+l = A(i,j) / A(j,j);
+            
+b(i,1) = b(i,1) - l * b(j, 1);
+for t = 1 : N
+                
+A(i,t) = A(i,t) - l * A(j, t);
+end
+end
+end
+    
+x = zeros(N,1);
+% -------------------- The back-substitution phase -------------------- %
+for k = N : -1 : 1
+ 
+E = 0;
+for iter = k+1 : N
+    
+ 
+ 
+E = E + A(k,iter) * x(iter,1);
+end
+ 
+x(k, 1) = (b(k,1) - E) / A(k,k);
+end
+
+euclideanNormOfR = norm(r);
+
+   r = A*x - b;
+   euclideanNormOfR = norm(r);
+   new_euclideanNormOfR = euclideanNormOfR;    
+while new_euclideanNormOfR <= euclideanNormOfR
+       euclideanNormOfR = new_euclideanNormOfR;
+       r = A*x - b;
+       x = x - r;
+       new_euclideanNormOfR = norm(r);
+end

ENUME/projectA/debilMode.asv

@@ -0,0 +1,71 @@
+% ------- Solving a system of n linear equations Ax = b. ------- %
+% -------------------- The Indicated Method -------------------- %
+function x = debilMode(A, b)
+        
+    [M,N] = size(A);
+    if M ~= N
+        error ('A is not square matrix!');
+    end % end if
+
+    for j = 1 : N - 1  
+        %--- Find the greatest value within column ---%
+                
+        m = max(A(j:N,j));
+        %--- Find row whitin greatest value occure ---%
+        for k = j : N
+            if A(k,j) == m
+            %--- SwapRow in matrix A ---%
+                            
+            tempRow =  A(j , :);
+                            
+            A(j , :) = A(k, :);
+                            
+            A(k, :) = tempRow;
+            %--- SwapValue in vector b ---%
+                            
+            tempVal = b(j);
+                            
+            b(j) = b(k);
+                            
+            b(k) = tempVal;
+            break;
+            end % end if
+        end % end for
+        for i = j + 1 : N
+                        
+            l = A(i,j) / A(j,j);
+                        
+            b(i,1) = b(i,1) - l * b(j, 1);
+            for t = 1 : N
+                            
+            A(i,t) = A(i,t) - l * A(j, t);
+        end % end for
+    end % end for
+    end
+        
+    x = zeros(N,1);
+    % -------------------- The back-substitution phase -------------------- %
+    for k = N : -1 : 1
+     
+    E = 0;
+    for iter = k+1 : N
+        
+     
+     
+    E = E + A(k,iter) * x(iter,1);
+    end
+     
+    x(k, 1) = (b(k,1) - E) / A(k,k);
+    end
+    
+       r = A*x - b;
+       euclideanNormOfR = norm(r);
+       new_euclideanNormOfR = euclideanNormOfR;    
+    while new_euclideanNormOfR <= euclideanNormOfR
+           euclideanNormOfR = new_euclideanNormOfR;
+           r = A*x - b;
+           x = x - r;
+           new_euclideanNormOfR = norm(r);
+    end
+    
+end % end function

ENUME/projectA/debilMode.m

@@ -0,0 +1,67 @@
+% ------- Solving a system of n linear equations Ax = b. ------- %
+% -------------------- The Indicated Method -------------------- %
+function x = debilMode(A, b)
+        
+    [M,N] = size(A);
+    if M ~= N
+        error ('A is not square matrix!');
+    end % end if
+
+    for j = 1 : N - 1  
+        %--- Find the greatest value within column ---%
+                
+        m = max(A(j:N,j));
+        %--- Find row whitin greatest value occure ---%
+        for k = j : N
+            if A(k,j) == m
+            %--- SwapRow in matrix A ---%
+                            
+            tempRow =  A(j , :);
+                            
+            A(j , :) = A(k, :);
+                            
+            A(k, :) = tempRow;
+            %--- SwapValue in vector b ---%
+                            
+            tempVal = b(j);
+                            
+            b(j) = b(k);
+                            
+            b(k) = tempVal;
+            break;
+            end % end if
+        end % end for
+        for i = j + 1 : N
+                        
+            l = A(i,j) / A(j,j);
+                        
+            b(i,1) = b(i,1) - l * b(j, 1);
+            for t = 1 : N
+                A(i,t) = A(i,t) - l * A(j, t);
+            end % end for
+        end % end for
+    end % end for
+        
+    x = zeros(N,1);
+    % -------------------- The back-substitution phase -------------------- %
+    for k = N : -1 : 1
+         
+        E = 0;
+        for iter = k+1 : N
+            E = E + A(k,iter) * x(iter,1);
+        end % end for
+      
+        x(k, 1) = (b(k,1) - E) / A(k,k);
+    end % end for
+        
+    r = A*x - b;
+    euclideanNormOfR = norm(r);
+    new_euclideanNormOfR = euclideanNormOfR;    
+    while new_euclideanNormOfR <= euclideanNormOfR
+        euclideanNormOfR = new_euclideanNormOfR;
+        r = A*x - b;
+        x = x - r;
+        new_euclideanNormOfR = norm(r);
+    end % end while
+    
+end % end function

ENUME/projectA/findmacheps.m

@@ -1,9 +1,9 @@
 macheps = 1;
-while 1.0 + macheps / 2 > 1.0;
+while 1.0 + macheps / 2 > 1.0
     macheps = macheps/2;
-end;
+end
 
-format long;
+format long; 
 disp("Display calculated macheps:")
 disp(macheps)
 disp("Display actual eps:")

ENUME/projectA/gaussianElimination.asv

@@ -0,0 +1,61 @@
+function x = gaussianElimination(A, b) % A - matrix, b - vector
+    [~, Height] = size(A);
+    % Find column with greatest value element 
+    for i = 1 : Height - 1
+        m = max(A(i:Height, i)); 
+        % Find row with greatest value element 
+        for k = i : Height
+            if A(k, i) == m % If the element has the biggest value we now 
+                % have the value of two rows that we want to swap
+               % Swap row in matrix A %
+                swapRow(A, i, k);
+                swapValue(b, i, k);
+              break;
+            end % end if
+        end % end for
+           % Swap value in vector B %
+       
+    end % end for
+
+    for j = i + 1 : Height
+        l = A(j, i) / A(i, i);
+        b(j, 1) = b(j, 1) - l * b(i, 1);
+
+        for t = 1 : Height
+            A(j, t) = A(j, t) - l * A(i, t);
+        end % end for 
+    end % end for
+
+
+
+x = zeros(Height, 1);
+
+
+
+for k = Height: -1:1
+    E = 0;
+    for iter = k + 1  : Height
+        E = E + A(k, iter) * x(iter, 1);
+    end
+
+    x(k, 1) = (b(k, 1) - E) / A(k, k);
+end
+
+
+end % end function
+
+
+
+
+function swapRow(A, j, k)
+    temp = A(j, :);
+    A(j, :) = A(k, :);
+    A(k, :) = temp;
+end % end function
+
+function swapValue(b, j, k)
+    temp = b(j);
+    b(j) = b(k);
+    b(k) = temp;
+end % end function
+

ENUME/projectA/gaussianElimination.m

@@ -0,0 +1,61 @@
+function x = gaussianElimination(A, b) % A - matrix, b - vector
+    [~, Height] = size(A);
+    % Find column with greatest value element 
+    for i = 1 : Height - 1
+        m = max(A(i:Height, i)); 
+        % Find row with greatest value element 
+        for k = i : Height
+            if A(k, i) == m % If the element has the biggest value we now 
+                % have the value of two rows that we want to swap
+               % Swap row in matrix A %
+                swapRow(A, i, k);
+                swapValue(b, i, k);
+              break;
+            end % end if
+        end % end for
+           % Swap value in vector B %
+       
+    end % end for
+
+    for j = i + 1 : Height
+        l = A(j, i) / A(i, i);
+        b(j, 1) = b(j, 1) - l * b(i, 1);
+
+        for t = 1 : Height
+            A(j, t) = A(j, t) - l * A(i, t);
+        end % end for 
+    end % end for
+
+
+
+x = zeros(Height, 1);
+
+
+
+for k = Height: -1:1
+    E = 0;
+    for iter = k + 1  : Height
+        E = E + A(k, iter) * x(iter, 1);
+    end
+
+    x(k, 1) = (b(k, 1) - E) / A(k, k);
+end
+
+
+end % end function
+
+
+
+
+function swapRow(A, j, k)
+    temp = A(j, :);
+    A(j, :) = A(k, :);
+    A(k, :) = temp;
+end % end function
+
+function swapValue(b, j, k)
+    temp = b(j);
+    b(j) = b(k);
+    b(k) = temp;
+end % end function
+

ENUME/projectA/matrixA.m

@@ -0,0 +1,14 @@
+function a = matrixA(n) % We want n rows and columns in the matrix
+a = zeros(n, n); % in order to save speed we preallocate zeros to the vector
+    for i = 1 : n % we iterate through rows
+        for j = 1 : n % we iterate through columns
+            if i == j % as per problem description on how to fill the matrix
+                a(i, j) = 9;
+            elseif i == j - 1 || i == j + 1
+                a(i, j) = -3;
+            else
+                a(i, j) = 0;
+            end % end if
+        end % end column for 
+    end % end row for
+end  % end function

ENUME/projectA/matrixB.m

@@ -0,0 +1,8 @@
+function a = matrixB(n) % We want n rows and columns in the matrix
+a = zeros(n, n); % in order to save speed we preallocate zeros to the vector
+    for i = 1 : n % we iterate through rows
+        for j = 1 : n % we iterate through columns
+            a(i, j) = 5 / (8*(i + j + 1));
+        end % end column for 
+    end % end row for
+end  % end function

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@@ -1,5 +1,5 @@
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ENUME/projectA/projectA.fls

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ENUME/projectA/projectA.log

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 []\T1/cmr/bx/n/12 Transform sys-tem of equa-tion into an upper-triangular
  []
 
+
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+ []
+
 [5
 
-] [6]
+]
+Overfull \hbox (2.71124pt too wide) detected at line 161
+ [] 
+ []
+
+[6] [7]
 Chapter 3.
-[7
+[8
 
 ]
 Chapter 4.
-[8
+[9
 
-] [9
+] [10
 
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ENUME/projectA/projectA.tex

@@ -116,17 +116,111 @@ Write a program solving a system of \textit{n} linear equations Ax = b using the
 \section{Theoretical Introduction}
 Gaussian elimination with partial pivoting consists of 3 main steps:
 \subsection{Transform system of equation into an upper-triangular matrix}
+
+\subsubsection{Starting conditions}
 We start with the system of linear equations looking like this:
-\begin{align*}
-a_{11}^{(1)} x_1 + a_{12}^{(1)}  x_2 + \cdots + a_{1n}^{(1)}  x_n &= b_1^{(1)}, \\
-a_{21}^{(1)} x_1 + a_{22}^{(1)}  x_2 + \cdots + a_{2n}^{(1)}  x_n &= b_2^{(1)}, \\
-&\vdotswithin{=} \\
-a_{n1}^{(1)} x_1 + a_{n2}^{(1)}  x_2 + \cdots + a_{nn}^{(1)}  x_n &= b_n^{(1)}, \\
-\end{align*}
+
+\[
+\begin{matrix}
+
+&a_{11}x_1 &{}+&a_{12}x_2&+&\dots&+&a_{1n}x_n &=&b_1,\\
+
+&a_{21}x_1 &{}+&a_{22}x_2&+&\dots&+&a_{2n}x_n &=&b_2,\\
+
+&\vdots    &&\vdots      & &     & &  \vdots  & &\vdots\\
+
+&a_{n1}x_1&{}+&a_{n2}x_2&+&\dots &+&a_{nn}x_n&=&b_n.
+
+\end{matrix}
+\]
+
+In order for this method to work all the elements of "diagonal" line - $ a_{11}, a_{22}, \dots, a_{nn} $ must be different from zero since we will be dividing by them.
+
+We will denote rows as '$w_i$' where 'i' is number of the row.
+
+\subsubsection{Zeroing first column}
+We start transforming the system by "zeroing" elements in first column excluding first row element. We do it by multiplying first row by $l_{i1}$, where:
+\[ l_{i1} = \frac{ a_{i1}^{(1)} }
+{ a_{11}^{(1)} }  \]
+And then substracting what we got ($ l_{i1}w_1 $), from \textit{i} row.
+
+Doing so we obtain a system of linear equations:
+
+\[
+\begin{matrix}
+
+&a_{11}x_1 &{}+&a_{12}x_2&+&\dots&+&a_{1n}x_n &=&b_1,\\
+
+&0 &{}+&(a_{22} - a_{12}l_{21})x_2&{}+&\dots&{}+&(a_{2n} - a_{1n}l_{21})x_n &=&b_2 - b_{1}l_{21},\\
+
+&\vdots    &&\vdots      & &     & &  \vdots  & &\vdots\\
+
+&0&{}+&(a_{n2} - a_{12}l_{n1})x_2&+&\dots &+&(a_{nn} - a_{1n}l_{n1})x_n&=&b_n - b_{1}l_{n1}.
+
+\end{matrix}
+\]
+
+\subsubsection{Zeroing second column}
+We continue onto the second column, this time we will zero all elements except first and second rows.
+Row multiplier becomes:
+\[ l_{i2} = \frac{ a_{i2}^{(2)} }{ a_{22}^{(2)} } \]
+
+Where:
+\[ a_{22}^{(2)} = (a_{22} - a_{12}l_{21}) \]
+And:
+\[ a_{i2}^{(2)} = (a_{i2} - a_{12}l_{i1}) \]
+They are modified values obtained from previous step.
+We continue as in the first step and we end up with:
+
+\[
+\begin{matrix}
+
+&a_{11}x_1 &{}+&a_{12}x_2&+&\dots&+&a_{1n}x_n &=&b_1,\\
+
+&0 &{}+& a_{22}^{(2)}x_2&{}+&\dots&{}+& a_{2n}^{(2)}x_n &=&b_2^{(2)},\\
+
+&\vdots    &&\vdots      & &     & &  \vdots  & &\vdots\\
+
+&0 &{}+& 0 &{}+&\dots&{}+& a_{nn}^{(3)}x_n &=&b_2^{(3)},
+
+\end{matrix}
+\]
+
+\subsubsection{Zeroing next columns}
+We repeat this process $n-1$ times and we end up with upper triangular matrix:
+
+\[
+\begin{matrix}
+
+&a_{11}x_1 &{}+&a_{12}x_2&+&\dots&+&a_{1n}x_n &=&b_1,\\
+
+&0 &{}+& a_{22}^{(2)}x_2&{}+&\dots&{}+& a_{i2}^{(2)}x_n &=&b_2^{(2)},\\
+
+&\vdots    &&\vdots      & &     & &  \vdots  & &\vdots\\
+
+&0 &{}+& 0 &{}+&\dots&{}+& a_{nn}^{(n)}x_n &=&b_2^{(n)},
+
+\end{matrix}
+\]
 
 \subsection{Backward substitution}
+After transforming the system we solve the system from last to first. \\
+First we calculate value of last element:
+\[ x_n = \frac{b_n}{a_{nn}} \]
+Then one "above":
+\[ x_{n-1} = \frac{ b_{n-1} - a_{n-1, n}x_n}{a_{n-1, n-1}} \]
+And so on, for $x_k$:
+\[ x_{k} = \frac{b_k - \sum_{j = k + 1}^n a_{kj}x_j}{a_{kk}} \]
 
 \subsection{Partial Pivoting}
+Gaussian elimination method has one flaw, where it can come into halt if:
+\[ a_{kk}^{(k)} = 0 \]
+To avoid it we use method of pivoting, in our case we will use partial pivoting method.
+Before each Gaussian elimination step, we do it before each Gaussian elimination step since this will lead to smaller error.
+\\We first find a row $i$ such that:
+\[ |{a_{ik}^{k}}| = \underset{j}{max} \{ |{a_{kk}^{k}}|, |{a_{k+1, k}^{k}}|, \cdots, |{a_{nk}^{k}}|\} \]
+
+Then we swap this row with k-th row. Since the matrix we use is assumed to be nonsingular then $|{a_{ik}^{k}}| \neq 0$ will be always true. After that we continue with the Gaussian elimination method.
 
 \section{Solution}
 

ENUME/projectA/projectA.toc

@@ -10,17 +10,21 @@
 \contentsline {section}{\numberline {2.1}Problem}{5}{section.2.1}%
 \contentsline {section}{\numberline {2.2}Theoretical Introduction}{5}{section.2.2}%
 \contentsline {subsection}{\numberline {2.2.1}Transform system of equation into an upper-triangular matrix}{5}{subsection.2.2.1}%
-\contentsline {subsection}{\numberline {2.2.2}Backward substitution}{6}{subsection.2.2.2}%
-\contentsline {subsection}{\numberline {2.2.3}Partial Pivoting}{6}{subsection.2.2.3}%
-\contentsline {section}{\numberline {2.3}Solution}{6}{section.2.3}%
-\contentsline {section}{\numberline {2.4}Discussion of the result}{6}{section.2.4}%
-\contentsline {chapter}{\numberline {3}Problem 3 - Solving a system of n linear equations - iterative algorithm}{7}{chapter.3}%
-\contentsline {section}{\numberline {3.1}Problem}{7}{section.3.1}%
-\contentsline {section}{\numberline {3.2}Theoretical introduction}{7}{section.3.2}%
-\contentsline {section}{\numberline {3.3}Solution}{7}{section.3.3}%
-\contentsline {section}{\numberline {3.4}Discussion of the result}{7}{section.3.4}%
-\contentsline {chapter}{\numberline {4}Problem 4 - QR method of finding eigenvalues}{8}{chapter.4}%
-\contentsline {section}{\numberline {4.1}Problem}{8}{section.4.1}%
-\contentsline {section}{\numberline {4.2}Theoretical introduction}{8}{section.4.2}%
-\contentsline {section}{\numberline {4.3}Solution}{8}{section.4.3}%
-\contentsline {section}{\numberline {4.4}Discussion of the result}{8}{section.4.4}%
+\contentsline {subsubsection}{Starting conditions}{5}{section*.2}%
+\contentsline {subsubsection}{Zeroing first column}{6}{section*.3}%
+\contentsline {subsubsection}{Zeroing second column}{6}{section*.4}%
+\contentsline {subsubsection}{Zeroing next columns}{6}{section*.5}%
+\contentsline {subsection}{\numberline {2.2.2}Backward substitution}{7}{subsection.2.2.2}%
+\contentsline {subsection}{\numberline {2.2.3}Partial Pivoting}{7}{subsection.2.2.3}%
+\contentsline {section}{\numberline {2.3}Solution}{7}{section.2.3}%
+\contentsline {section}{\numberline {2.4}Discussion of the result}{7}{section.2.4}%
+\contentsline {chapter}{\numberline {3}Problem 3 - Solving a system of n linear equations - iterative algorithm}{8}{chapter.3}%
+\contentsline {section}{\numberline {3.1}Problem}{8}{section.3.1}%
+\contentsline {section}{\numberline {3.2}Theoretical introduction}{8}{section.3.2}%
+\contentsline {section}{\numberline {3.3}Solution}{8}{section.3.3}%
+\contentsline {section}{\numberline {3.4}Discussion of the result}{8}{section.3.4}%
+\contentsline {chapter}{\numberline {4}Problem 4 - QR method of finding eigenvalues}{9}{chapter.4}%
+\contentsline {section}{\numberline {4.1}Problem}{9}{section.4.1}%
+\contentsline {section}{\numberline {4.2}Theoretical introduction}{9}{section.4.2}%
+\contentsline {section}{\numberline {4.3}Solution}{9}{section.4.3}%
+\contentsline {section}{\numberline {4.4}Discussion of the result}{9}{section.4.4}%

ENUME/projectA/usefullinks.txt

@@ -29,3 +29,7 @@ https://tex.stackexchange.com/questions/32217/3-dots-in-matrix/32221 - 3 dots la
 https://stackoverflow.com/questions/48240982/how-do-you-align-a-system-of-equations - System of equations latex
 
 https://tex.stackexchange.com/questions/526025/problem-with-vdotswithin - vdots alligned to equal sign
+
+https://www.overleaf.com/learn/latex/Brackets_and_Parentheses - Curly brackets latex
+
+https://tex.stackexchange.com/questions/130516/how-to-write-something-vertically-below-another-math-symbol - Symbol vertically below another 

ENUME/projectA/vectorA.m

@@ -0,0 +1,8 @@
+function b = vectorA(n) % We want n elements in the vector 
+b = zeros(n, 1); % in order to save speed we preallocate zeros to the vector
+    for i = 1 : n % We loop from first element to the last element of the vector
+        b(i, 1) = -5 + 0.3 * i; % Formula as in the task description
+        % We need to put '1' in b(i, 1) so that matlab acknowledges that
+        % this is a column and not a row
+    end % end for
+end % end function

ENUME/projectA/vectorB.m

@@ -0,0 +1,10 @@
+function b = vectorB(n) % We want n elements in the vector 
+b = zeros(n, 1); % in order to save speed we preallocate zeros to the vector
+    for i = 1 : n % We loop from first element to the last element of the vector
+        if mod(i, 2) == 0 % as per problem description
+            b(i, 1) = 9 / (2*i);
+        else
+            b(i, 1) = 0;
+        end % end if
+    end % end for
+end % end function