2023-11-15

CS1029/notes-2023-11-01

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+Edge contractions: merge two vertices, removing an edge between them
+Representations: adjacency list
+				 adjacency matrix
+					What does the determinant of an adjacency matrix mean?
+				incidence matrix: vertices against edges, 1 where edge is connected to vertex
+

CS1029/notes-2023-11-07

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+

CS1029/notes-2023-11-08

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+

CS1029/notes-2023-11-14

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+

CS1029/practical-2023-11-03/relations

@@ -0,0 +1,58 @@
+1. a) {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5) }
+	Reflexive: \forall x. (x, x) \in R
+	Antisymmetric (& not symmetric): all pairs have x <= y. forall x, y. x <= y, !(y <= x) || y == x
+	Transitive
+   b) {(1, 2), (1, 4), (2, 1), (2, 3), (3, 2), (3, 4), (4, 1), (4, 3)}
+	Not reflexive: there are no pairs of (x, x)
+	Symmetric - all pairs have their reverse represented
+	Not antisymmetric: symmetric and anti-symmetric are mutually exclusive
+	Not transitive: (1, 2) and (2, 3) - 1 + 3 is not odd
+
+2. {(item, quantity)}
+   {(Name, {(key, value)})}
+   {(Name, Address, {(Room type, price, {(key, value)})}, ...)}
+
+3. 105 305 306 505 705 707 905 906 909
+
+4. a) ab ac bc cb
+   b) {(a, a), (a, b), (a, c), (b, b), (b, a), (b, c), (c, a), (c, b), (d, d)}
+
+5.a) {(1, 1), (1, 2), (1, 4), (2, 1), (2, 3), (3, 2), (3, 3), (3, 4), (4, 1), (4, 3), (4, 4)}
+	Not reflexive: (2, 2) is not present
+	Symmetric, therefore not anti-symmetric
+	Not transitive: (1, 2) and (2, 3), but not (1, 3)
+
+  b) 12 21 14 41 32 23 43 34
+	Not reflexive
+	Symmetric, therefore not antisymmetric
+	Not transitive: 12 and 23 but not 13
+
+6. 1 1 0 0
+   1 1 0 0
+   1 0 1 1
+   0 0 0 1
+
+Yes, it's reflexive
+
+8. {(a, b) | a divides b OR b divides a}
+9. No, it's not transitive. (a, b) & (b, d), but not (a, d)
+10. a) Not equivalence relation: missing transitivity
+		(1, 3) and (3, 2), but not (1, 2)
+	b) {0}, {1, 2}, {3}
+11. \forall n \in N_0:
+	0 + 3n
+	1 + 3n
+	2 + 3n
+
+12. a) Y
+    b) N: 0 is in both - not disjoint
+	c) Y
+	d) N: 0 is missing
+
+13. a) 00 11 22 33 44 55 12 21 34 43 35 53 45 54 
+    b) 00 11 22 33 44 55 01 10 23 32 45 54
+	c) 00 11 22 33 44 55 01 10 02 20 12 21 34 43 35 53 45 54
+
+14. a) Y, trivially
+    b) N: not antisymmetric ((2, 3) and (3, 2))
+	c) N: not reflexive (no (3, 3))

CS1029/practical-2023-11-10/graphs

@@ -0,0 +1,64 @@
+1. Undirected, unlooped, multi-edged: multigraph
+b) directed, looped, multi-edged: directed pseudo-multigraph
+
+2. ac bd
+b) cd cd dd ee ab bc
+
+3. {{paper}}
+
+4. vertices: 6
+   edges: 6
+   degree: a: 2 b: 4 c: 1 f: 3 e: 2 d: 0
+   isolated: d
+   pendant: c
+b) vertices: 5
+   edges: 14
+   degree: a: 6 b: 6 c: 6 d: 5 e: 3
+   isolated: -
+   pendant: -
+
+5. vertices: 4
+   in-a : 2
+   out-a: 2
+   in-b:  3
+   out-b: 4
+   in-c:  2
+   out-c: 1
+   in-d:  1
+   out-d: 1
+
+6. {ac} {bde}
+b) Not bipartite: 3-loop bcf would require 3 sets
+
+7. {{ paper }}
+
+8. a -> abcd
+   b ->    d
+   c -> ab  
+   d ->  bcd
+
+9.    | a b c d
+    --+--------
+    a | 1 1 1 1
+    b | 0 0 0 1
+    c | 1 1 0 0
+    d | 0 1 1 1
+
+10. {{ paper }}
+
+11. v1 -> u1
+    v2 -> u4
+	v3 -> u2
+	v4 -> u5
+	v5 -> u3
+
+12. v1 -> u4
+	v2 -> u3
+	v3 -> u1
+	v4 -> u2
+
+13. PSCL
+ a) YNN4
+ b) N---
+ c) N---
+ d) YYY5

CS1032/notes-2023-11-02

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+

CS1032/notes-2023-11-03

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+Pencil & eraser for final exam

CS1032/notes-2023-11-09

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+

CS1032/practical-2023-11-01/head.py

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+import sys, itertools
+
+def open_arg(_arg0, filename = "-"): # Default of - means stdin
+    if filename == "-":
+        return sys.stdin
+
+    return open(filename, 'r')
+
+def head(file):
+    for line in file.readlines()[:10]:
+        print(line, end = "")
+
+if __name__ == '__main__':
+    try:
+        file = open_arg(*sys.argv)
+        head(file)
+    except FileNotFoundError:
+        print("File not found!", file=sys.stderr)
+        sys.exit(1)
+    except:
+        print("Some other error occured", file=sys.stderr)

CS1032/practical-2023-11-01/tail.py

@@ -0,0 +1,21 @@
+import sys, itertools
+
+def open_arg(_arg0, filename = "-"): # Default of - means stdin
+    if filename == "-":
+        return sys.stdin
+
+    return open(filename, 'r')
+
+def tail(file):
+    for line in file.readlines()[-10:]:
+        print(line, end = "")
+
+if __name__ == '__main__':
+    try:
+        file = open_arg(*sys.argv)
+        tail(file)
+    except FileNotFoundError:
+        print("File not found!", file=sys.stderr)
+        sys.exit(1)
+    except:
+        print("Some other error occured", file=sys.stderr)

CS1032/practical-2023-11-08/calc.py

@@ -0,0 +1,70 @@
+# This function adds two numbers
+def add(x, y):
+    return x + y
+
+# This function subtracts two numbers
+def subtract(x, y):
+    return x - y
+
+# This function multiplies two numbers
+def multiply(x, y):
+    return x * y
+
+# This function divides two numbers
+def divide(x, y):
+    return x / y
+
+def avg(x, y):
+    return (x + y) / 2
+
+def sci(x, y):
+    return x * 10 ** y
+
+
+print("Select operation.")
+print("1.Add")
+print("2.Subtract")
+print("3.Multiply")
+print("4.Divide")
+print("5.Average")
+print("6.Scientific Notation")
+
+while True:
+    # take input from the user
+    choice = input("Enter choice(1/2/3/4/5/6): ")
+
+    # check if choice is one of the four options
+    if choice in ('1', '2', '3', '4', '5', '6'):
+        try:
+            num1 = float(input("Enter first number: "))
+            num2 = float(input("Enter second number: "))
+        except ValueError:
+            print("Invalid input. Please enter a number.")
+            continue
+
+        if choice == '1':
+            print(num1, "+", num2, "=", add(num1, num2))
+
+        elif choice == '2':
+            print(num1, "-", num2, "=", subtract(num1, num2))
+
+        elif choice == '3':
+            print(num1, "*", num2, "=", multiply(num1, num2))
+
+        elif choice == '4':
+            print(num1, "/", num2, "=", divide(num1, num2))
+
+        elif choice == '5':
+            print(f"avg({num1}, {num2})", "=", avg(num1, num2))
+
+        elif choice == '6':
+            print(f"{num1}e{num2}", "=", sci(num1, num2))
+        
+        # check if user wants another calculation
+        # break the while loop if answer is no
+        next_calculation = input("Let's do next calculation? (yes/no): ")
+        if next_calculation.lower().startswith('n'):
+          break
+    else:
+        print("Invalid Input")
+

MA1006/decls.tex

@@ -8,8 +8,9 @@
 \date{}
 \author{}
 
-\renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}(#1)}
-\renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}(#1)}
+\newcommand{\paren}[1]{\left(#1\right)}
+\renewcommand{\Re}[1]{\operatorname{\mathbb{R}e}\paren{#1}}
+\renewcommand{\Im}[1]{\operatorname{\mathbb{{I}}m}\paren{#1}}
 \newcommand{\C}{\mathbb{C}}
 \newcommand{\N}{\mathbb{N}}
 \newcommand{\Z}{\mathbb{Z}}
@@ -18,8 +19,8 @@
 \newcommand{\conj}[1]{\overline{#1}}
 \renewcommand{\mod}[1]{\left|#1\right|}
 \newcommand{\abs}[1]{\left|#1\right|}
-\newcommand{\paren}[1]{\left(#1\right)}
 \newcommand{\polar}[2]{#1\paren{\cos{\paren{#2}} + i\sin{\paren{#2}}}}
+\newcommand{\adj}[1]{\operatorname{adj}#1}
 
 \makeatletter
 \renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%

MA1006/linear.tex

@@ -219,7 +219,7 @@ $\det{A} = \prod^{N}_{i=0}~a_{ij} \forall~\text{ row-echelon }A$
 \end{description}
 Note: $\det{A} = \det{A^T}~\forall~A$
 \section*{Matrix Multiplication}
-LHS has columns $=$ rows of RHS
+LHS has columns $=$ rows of RHS \\
 It's the cartesian product
 \[A\times B = (a_{i1}b_{j1} + a_{i2}b_{2j} + \ldots + a_{im}b_{mj})_{ij}\]
 \begin{align*}
@@ -240,4 +240,119 @@ It's the cartesian product
 		                                                                                    9 & 10 & 11 & 12 \\
 	                                                                                    \end{pmatrix}
 \end{align*}
+
+\[A\vec{x} = \vec{b}\]
+where $A$ is the coefficient matrix, $\vec{x}$ is the variables, and $\vec{b}$ is the values of the equations of a linear equation system.
+\subsection*{Inverse Matrices}
+The identity matrix exists as $I_n$ for size $n$.
+\[AA^{-1} = I_n = A^{-1}A \quad \forall~\text{matrices }A \text{ of size } n\]
+Assume that $A$ has two distinct inverses, $B$ and $C$.
+\begin{align*}
+	            & \text{matrix multiplication is associative} \\
+	\therefore~ & C(AB) = (CA)B                               \\
+	\therefore~ & C I_n = I_n B                               \\
+	\therefore~ & C = B                                       \\
+	            & \text{
+		As $B = C$, while $B$ and $C$ are assumed to be distinct, matrices have no more than one unique inverse by contradiction
+	}
+\end{align*}
+Matrices are invertible $\iff \det{A} \neq 0$
+\[\det{AB} = \det{A}\det{B}\]
+\[\therefore~ \det{A}\det{A^{-1}} = \det{I_n} = 1\]
+\[\therefore~ \det{A} \neq 0 \]
+\begin{align*}
+	\begin{pmatrix}	a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}
+\end{align*}
+\subsubsection*{Computation thereof}
+\[\det{A} = \sum_{k = 1}^{n}~a_{ik}(-1)^{i+j}\det{A_{ij}} \quad \text{ for any $i$}\]
+\begin{description}
+	\item[Matrix of Cofactors: $C$] determinants of minors \& signs of laplace expansion \\
+		ie. $\sum A \odot C = \det{A}$
+	\item[$\adj{A}$ Adjucate of $A$ =] $C^T$
+\end{description}
+\begin{align*}
+	A    & = \begin{pmatrix}
+		         1  & 0 & 1 \\
+		         -1 & 1 & 2 \\
+		         2  & 0 & 1
+	         \end{pmatrix} \\
+	C(A) & = \begin{pmatrix}
+		         1  & 5  & -2 \\
+		         0  & -1 & 0  \\
+		         -1 & -3 & 1  \\
+	         \end{pmatrix}
+\end{align*}
+$$ A^{-1} = \frac{\adj{A}}{\det{A}} $$
+Gaussian elimination can also be used: augmented matrix with $I_n$ on the right,
+reduce to reduced row-echelon. If the left is of the form $I_n$, the right is
+the inverse. If there is a zero row, $\det{A} = 0$, and the $A$ has no inverse.
+\section*{Linear Transformations}
+\begin{align*}
+	f: & ~ \R^n \to \R^m                                                                                    \\
+	f  & (x_1, \cdots, x_n) = (f_1(x_1, \cdots, x_n), f_2(x_1, \cdots, x_n), \cdots, f_m(x_1, \cdots, x_n))
+\end{align*}
+$f$ is a linear transformation if \(\forall i.~f_i(x_1, \cdots, x_n)\) is a
+linear polynomial in $x_1, \cdots, x_n$ with a zero constant term
+\begin{align*}
+	f(x_1,~ x_2)       & = (x_1 + x_2,~ 3x_1 - x_2,~ 10x_2) \tag{is a linear transformation} \\
+	g(x_1,~ x_2,~ x_3) & = (x_1 x_2,~ x_3^2) \tag{not a linear transformation}               \\
+	h(x_1,~ x_2)       & = (3x_1 + 4,~ 2x_2 - 4) \tag{not a linear transformation}           \\
+\end{align*}
+\[f: \R^n \to \R^m = \vec{x} \to A\vec{x} \]
+\[\exists \text{ a matrix $A$ of dimension $n$x$m$ } \forall\text{ linear transforms } f \]
+\[\forall \text{ matrices $A$ of dimension $n$x$m$ } \exists \text{ a linear transform $f$ of dimension $n$x$m$ such that } f(\vec{x}) = A\vec{x} \]
+Function composition of linear translations is is just matrix multiplication:
+\begin{align*}
+	f(\vec{x})          & = A\vec{x}                  \\
+	g(\vec{y})          & = B\vec{y}                  \\
+	(f\cdot g)(\vec{x}) & = g(f(\vec{x})) = BA\vec{x}
+\end{align*}
+A function \(f: \R^n \to \R^m\) is a linear transformation iff:
+\begin{enumerate}
+	\item $f(\vec{x} + \vec{y}) = f(\vec{x}) + f(\vec{y}) \quad \forall~\vec{x},~\vec{y} \in \R^n $
+	\item $f(r\vec{x}) = r\cdot f(\vec{x}) \quad \forall~\vec{x} \in \R^n, r \in \R $
+\end{enumerate}
+\subsection*{Building the matrix of a linear transform}
+\[ f(\vec{x}) = f(x_1\vec{e}_1 + x_2\vec{e}_2) = f(x_1\vec{e}_1) + f(x_2\vec{e}_2) = x_1f(\vec{e}_1) + x_2f(\vec{e}_2) \]
+\[ A = \begin{pmatrix} f(\vec{e}_1) & f(\vec{e}_2) \end{pmatrix} \]
+\begin{align*}
+	 & \vec{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
+	\\ & \vec{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}
+	\\ & \vdots
+	\\ & \forall \vec{x}.~ \vec{x} = \sum_{i}^{n}~\vec{e}_i x_i
+\end{align*}
+\subsection*{Composition}
+\[ \paren{f \cdot g}\paren{\vec{x}} = f(g(\vec{x})) = AB\vec{x} \]
+where: $f(\vec{x}) = A\vec{x}$, $g(\vec{x}) = B\vec{x}$
+\subsection*{Geometry}
+\begin{description}
+	\item[rotation of $x$ by $\theta$ anticlockwise] \( = R_\theta = \begin{pmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{pmatrix} \)
+	\item[reflection about a line at angle $\alpha$ from the $x$-axis] \( = T_\alpha = R_{\alpha}T_0R_{-\alpha}\) where \( T_0 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \)
+	\item[scaling by $\lambda \in \R$] \( = S_\lambda = \lambda I_n\)
+	\item[Skew by $\alpha$ in $x$ and $\gamma$ in $y$] \( \begin{pmatrix} \alpha & 0 \\ 0 & \gamma \end{pmatrix}\)
+\end{description}
+The image of the unit square under the linear transform $A$ is a parallelogram of $(0, 0)$, $(a_{11}, a_{21})$, $(a_{12}, a_{22})$, $(a_{11} + a_{12}, a_{21} + a_{22})$, with area $ \abs{\det{A}} $
+\subsection*{Inversion}
+Inversion of a linear transformation is equivalent to inversion of its representative matrix
+\subsection*{Eigen\{values, vectors\}}
+\[ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} a \\ 0 \end{pmatrix} = a\vec{e}_1\]
+\[ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ b \end{pmatrix} = b\vec{e}_2\]
+\[ T_\alpha \vec{x} = \vec{x} \text{ for $\vec{x}$ along the line of transformation }\]
+\begin{description}
+	\item[Eigenvector (of some transformation $f$)] A non-zero vector $\vec{x}$ such that $f(\vec{x}) = \lambda\vec{x}$ for some value $\lambda$
+	\item[Eigenvalue] $\lambda$ as above
+\end{description}
+\[ \forall \text{ eigenvectors of $A$ } \vec{x}, c \in R, \neq 0 .~ c\vec{x} \text{ is an eigenvector with eigenvalue } \lambda\]
+
+\[ \forall A: \text{$n$x$n$ matrix}.\quad P_A\paren{\lambda} = \det{\paren{A - \lambda I_n}} \tag{characteristic polynomial in $\lambda$}\]
+Eigenvalues of $A$ are the solutions of $P_A\paren{\lambda} = 0$
+\begin{align*}
+	& A\vec{x} = \lambda\vec{x} & x \neq 0\\
+	\iff & A\vec{x} - \lambda\vec{x} = 0 \\
+	\iff & (A - \lambda I_n)\vec{x} = 0 \\
+	\iff & \det{\paren{A - \lambda I_n}} = 0 \\
+	& \quad \text{ or $\paren{A - \lambda I_n}$ is invertible and $x = 0$ }
+\end{align*}
+\[ P_{R\theta}(\lambda) = \frac{2\cos{\theta} \pm \sqrt{-4\lambda^2\sin^2{\theta}}}{2}\]
+\[ R_\theta \text{ has eigenvalues }\iff \sin{\theta} = 0 \]
 \end{document}