diff --git a/general-template/homework.tex b/general-template/homework.tex
deleted file mode 100644
index 872e6f5..0000000
--- a/general-template/homework.tex
+++ /dev/null
@@ -1,539 +0,0 @@
-\documentclass{article}
-
-\usepackage{fancyhdr}
-\usepackage{extramarks}
-\usepackage{amsmath}
-\usepackage{amsthm}
-\usepackage{amsfonts}
-\usepackage{tikz}
-\usepackage[plain]{algorithm}
-\usepackage{algpseudocode}
-
-\usetikzlibrary{automata,positioning}
-
-%
-% Basic Document Settings
-%
-
-\topmargin=-0.45in
-\evensidemargin=0in
-\oddsidemargin=0in
-\textwidth=6.5in
-\textheight=9.0in
-\headsep=0.25in
-
-\linespread{1.1}
-
-\pagestyle{fancy}
-\lhead{\hmwkAuthorName}
-\chead{\hmwkClass\ (\hmwkClassInstructor\ \hmwkClassTime): \hmwkTitle}
-\rhead{\firstxmark}
-\lfoot{\lastxmark}
-\cfoot{\thepage}
-
-\renewcommand\headrulewidth{0.4pt}
-\renewcommand\footrulewidth{0.4pt}
-
-\setlength\parindent{0pt}
-
-%
-% Create Problem Sections
-%
-
-\newcommand{\enterProblemHeader}[1]{
-    \nobreak\extramarks{}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
-    \nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
-}
-
-\newcommand{\exitProblemHeader}[1]{
-    \nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
-    \stepcounter{#1}
-    \nobreak\extramarks{Problem \arabic{#1}}{}\nobreak{}
-}
-
-\setcounter{secnumdepth}{0}
-\newcounter{partCounter}
-\newcounter{homeworkProblemCounter}
-\setcounter{homeworkProblemCounter}{1}
-\nobreak\extramarks{Problem \arabic{homeworkProblemCounter}}{}\nobreak{}
-
-%
-% Homework Problem Environment
-%
-% This environment takes an optional argument. When given, it will adjust the
-% problem counter. This is useful for when the problems given for your
-% assignment aren't sequential. See the last 3 problems of this template for an
-% example.
-%
-\newenvironment{homeworkProblem}[1][-1]{
-    \ifnum#1>0
-        \setcounter{homeworkProblemCounter}{#1}
-    \fi
-    \section{Problem \arabic{homeworkProblemCounter}}
-    \setcounter{partCounter}{1}
-    \enterProblemHeader{homeworkProblemCounter}
-}{
-    \exitProblemHeader{homeworkProblemCounter}
-}
-
-%
-% Homework Details
-%   - Title
-%   - Due date
-%   - Class
-%   - Section/Time
-%   - Instructor
-%   - Author
-%
-
-\newcommand{\hmwkTitle}{Homework\ \#2}
-\newcommand{\hmwkDueDate}{February 12, 2014}
-\newcommand{\hmwkClass}{Calculus}
-\newcommand{\hmwkClassTime}{Section A}
-\newcommand{\hmwkClassInstructor}{Professor Isaac Newton}
-\newcommand{\hmwkAuthorName}{\textbf{Josh Davis} \and \textbf{Davis Josh}}
-
-%
-% Title Page
-%
-
-\title{
-    \vspace{2in}
-    \textmd{\textbf{\hmwkClass:\ \hmwkTitle}}\\
-    \normalsize\vspace{0.1in}\small{Due\ on\ \hmwkDueDate\ at 3:10pm}\\
-    \vspace{0.1in}\large{\textit{\hmwkClassInstructor\ \hmwkClassTime}}
-    \vspace{3in}
-}
-
-\author{\hmwkAuthorName}
-\date{}
-
-\renewcommand{\part}[1]{\textbf{\large Part \Alph{partCounter}}\stepcounter{partCounter}\\}
-
-%
-% Various Helper Commands
-%
-
-% Useful for algorithms
-\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
-
-% For derivatives
-\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
-
-% For partial derivatives
-\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
-
-% Integral dx
-\newcommand{\dx}{\mathrm{d}x}
-
-% Alias for the Solution section header
-\newcommand{\solution}{\textbf{\large Solution}}
-
-% Probability commands: Expectation, Variance, Covariance, Bias
-\newcommand{\E}{\mathrm{E}}
-\newcommand{\Var}{\mathrm{Var}}
-\newcommand{\Cov}{\mathrm{Cov}}
-\newcommand{\Bias}{\mathrm{Bias}}
-
-\begin{document}
-
-\maketitle
-
-\pagebreak
-
-\begin{homeworkProblem}
-    Give an appropriate positive constant \(c\) such that \(f(n) \leq c \cdot
-    g(n)\) for all \(n > 1\).
-
-    \begin{enumerate}
-        \item \(f(n) = n^2 + n + 1\), \(g(n) = 2n^3\)
-        \item \(f(n) = n\sqrt{n} + n^2\), \(g(n) = n^2\)
-        \item \(f(n) = n^2 - n + 1\), \(g(n) = n^2 / 2\)
-    \end{enumerate}
-
-    \textbf{Solution}
-
-    We solve each solution algebraically to determine a possible constant
-    \(c\).
-    \\
-
-    \textbf{Part One}
-
-    \[
-        \begin{split}
-            n^2 + n + 1 &=
-            \\
-            &\leq n^2 + n^2 + n^2
-            \\
-            &= 3n^2
-            \\
-            &\leq c \cdot 2n^3
-        \end{split}
-    \]
-
-    Thus a valid \(c\) could be when \(c = 2\).
-    \\
-
-    \textbf{Part Two}
-
-    \[
-        \begin{split}
-            n^2 + n\sqrt{n} &=
-            \\
-            &= n^2 + n^{3/2}
-            \\
-            &\leq n^2 + n^{4/2}
-            \\
-            &= n^2 + n^2
-            \\
-            &= 2n^2
-            \\
-            &\leq c \cdot n^2
-        \end{split}
-    \]
-
-    Thus a valid \(c\) is \(c = 2\).
-    \\
-
-    \textbf{Part Three}
-
-    \[
-        \begin{split}
-            n^2 - n + 1 &=
-            \\
-            &\leq n^2
-            \\
-            &\leq c \cdot n^2/2
-        \end{split}
-    \]
-
-    Thus a valid \(c\) is \(c = 2\).
-
-\end{homeworkProblem}
-
-\pagebreak
-
-\begin{homeworkProblem}
-    Let \(\Sigma = \{0, 1\}\). Construct a DFA \(A\) that recognizes the
-    language that consists of all binary numbers that can be divided by 5.
-    \\
-
-    Let the state \(q_k\) indicate the remainder of \(k\) divided by 5. For
-    example, the remainder of 2 would correlate to state \(q_2\) because \(7
-    \mod 5 = 2\).
-
-    \begin{figure}[h]
-        \centering
-        \begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto]
-            \node[state, accepting, initial] (q_0)   {$q_0$};
-            \node[state] (q_1) [right=of q_0] {$q_1$};
-            \node[state] (q_2) [right=of q_1] {$q_2$};
-            \node[state] (q_3) [right=of q_2] {$q_3$};
-            \node[state] (q_4) [right=of q_3] {$q_4$};
-            \path[->]
-                (q_0)
-                    edge [loop above] node {0} (q_0)
-                    edge node {1} (q_1)
-                (q_1)
-                    edge node {0} (q_2)
-                    edge [bend right=-30] node {1} (q_3)
-                (q_2)
-                    edge [bend left] node {1} (q_0)
-                    edge [bend right=-30] node {0} (q_4)
-                (q_3)
-                    edge node {1} (q_2)
-                    edge [bend left] node {0} (q_1)
-                (q_4)
-                    edge node {0} (q_3)
-                    edge [loop below] node {1} (q_4);
-        \end{tikzpicture}
-        \caption{DFA, \(A\), this is really beautiful, ya know?}
-        \label{fig:multiple5}
-    \end{figure}
-
-    \textbf{Justification}
-    \\
-
-    Take a given binary number, \(x\). Since there are only two inputs to our
-    state machine, \(x\) can either become \(x0\) or \(x1\). When a 0 comes
-    into the state machine, it is the same as taking the binary number and
-    multiplying it by two. When a 1 comes into the machine, it is the same as
-    multipying by two and adding one.
-    \\
-
-    Using this knowledge, we can construct a transition table that tell us
-    where to go:
-
-    \begin{table}[ht]
-        \centering
-        \begin{tabular}{c || c | c | c | c | c}
-            & \(x \mod 5 = 0\)
-            & \(x \mod 5 = 1\)
-            & \(x \mod 5 = 2\)
-            & \(x \mod 5 = 3\)
-            & \(x \mod 5 = 4\)
-            \\
-            \hline
-            \(x0\) & 0 & 2 & 4 & 1 & 3 \\
-            \(x1\) & 1 & 3 & 0 & 2 & 4 \\
-        \end{tabular}
-    \end{table}
-
-    Therefore on state \(q_0\) or (\(x \mod 5 = 0\)), a transition line should
-    go to state \(q_0\) for the input 0 and a line should go to state \(q_1\)
-    for input 1. Continuing this gives us the Figure~\ref{fig:multiple5}.
-\end{homeworkProblem}
-
-\begin{homeworkProblem}
-    Write part of \alg{Quick-Sort($list, start, end$)}
-
-    \begin{algorithm}[]
-        \begin{algorithmic}[1]
-            \Function{Quick-Sort}{$list, start, end$}
-                \If{$start \geq end$}
-                    \State{} \Return{}
-                \EndIf{}
-                \State{} $mid \gets \Call{Partition}{list, start, end}$
-                \State{} \Call{Quick-Sort}{$list, start, mid - 1$}
-                \State{} \Call{Quick-Sort}{$list, mid + 1, end$}
-            \EndFunction{}
-        \end{algorithmic}
-        \caption{Start of QuickSort}
-    \end{algorithm}
-\end{homeworkProblem}
-
-\pagebreak
-
-\begin{homeworkProblem}
-    Suppose we would like to fit a straight line through the origin, i.e.,
-    \(Y_i = \beta_1 x_i + e_i\) with \(i = 1, \ldots, n\), \(\E [e_i] = 0\),
-    and \(\Var [e_i] = \sigma^2_e\) and \(\Cov[e_i, e_j] = 0, \forall i \neq
-    j\).
-    \\
-
-    \part
-
-    Find the least squares esimator for \(\hat{\beta_1}\) for the slope
-    \(\beta_1\).
-    \\
-
-    \solution
-
-    To find the least squares estimator, we should minimize our Residual Sum
-    of Squares, RSS:
-
-    \[
-        \begin{split}
-            RSS &= \sum_{i = 1}^{n} {(Y_i - \hat{Y_i})}^2
-            \\
-            &= \sum_{i = 1}^{n} {(Y_i - \hat{\beta_1} x_i)}^2
-        \end{split}
-    \]
-
-    By taking the partial derivative in respect to \(\hat{\beta_1}\), we get:
-
-    \[
-        \pderiv{
-            \hat{\beta_1}
-        }{RSS}
-        = -2 \sum_{i = 1}^{n} {x_i (Y_i - \hat{\beta_1} x_i)}
-        = 0
-    \]
-
-    This gives us:
-
-    \[
-        \begin{split}
-            \sum_{i = 1}^{n} {x_i (Y_i - \hat{\beta_1} x_i)}
-            &= \sum_{i = 1}^{n} {x_i Y_i} - \sum_{i = 1}^{n} \hat{\beta_1} x_i^2
-            \\
-            &= \sum_{i = 1}^{n} {x_i Y_i} - \hat{\beta_1}\sum_{i = 1}^{n} x_i^2
-        \end{split}
-    \]
-
-    Solving for \(\hat{\beta_1}\) gives the final estimator for \(\beta_1\):
-
-    \[
-        \begin{split}
-            \hat{\beta_1}
-            &= \frac{
-                \sum {x_i Y_i}
-            }{
-                \sum x_i^2
-            }
-        \end{split}
-    \]
-
-    \pagebreak
-
-    \part
-
-    Calculate the bias and the variance for the estimated slope
-    \(\hat{\beta_1}\).
-    \\
-
-    \solution
-
-    For the bias, we need to calculate the expected value
-    \(\E[\hat{\beta_1}]\):
-
-    \[
-        \begin{split}
-            \E[\hat{\beta_1}]
-            &= \E \left[ \frac{
-                \sum {x_i Y_i}
-            }{
-                \sum x_i^2
-            }\right]
-            \\
-            &= \frac{
-                \sum {x_i \E[Y_i]}
-            }{
-                \sum x_i^2
-            }
-            \\
-            &= \frac{
-                \sum {x_i (\beta_1 x_i)}
-            }{
-                \sum x_i^2
-            }
-            \\
-            &= \frac{
-                \sum {x_i^2 \beta_1}
-            }{
-                \sum x_i^2
-            }
-            \\
-            &= \beta_1 \frac{
-                \sum {x_i^2 \beta_1}
-            }{
-                \sum x_i^2
-            }
-            \\
-            &= \beta_1
-        \end{split}
-    \]
-
-    Thus since our estimator's expected value is \(\beta_1\), we can conclude
-    that the bias of our estimator is 0.
-    \\
-
-    For the variance:
-
-    \[
-        \begin{split}
-            \Var[\hat{\beta_1}]
-            &= \Var \left[ \frac{
-                \sum {x_i Y_i}
-            }{
-                \sum x_i^2
-            }\right]
-            \\
-            &=
-            \frac{
-                \sum {x_i^2}
-            }{
-                \sum x_i^2 \sum x_i^2
-            } \Var[Y_i]
-            \\
-            &=
-            \frac{
-                \sum {x_i^2}
-            }{
-                \sum x_i^2 \sum x_i^2
-            } \Var[Y_i]
-            \\
-            &=
-            \frac{
-                1
-            }{
-                \sum x_i^2
-            } \Var[Y_i]
-            \\
-            &=
-            \frac{
-                1
-            }{
-                \sum x_i^2
-            } \sigma^2
-            \\
-            &=
-            \frac{
-                \sigma^2
-            }{
-                \sum x_i^2
-            }
-        \end{split}
-    \]
-
-\end{homeworkProblem}
-
-\pagebreak
-
-\begin{homeworkProblem}
-    Prove a polynomial of degree \(k\), \(a_kn^k + a_{k - 1}n^{k - 1} + \hdots
-    + a_1n^1 + a_0n^0\) is a member of \(\Theta(n^k)\) where \(a_k \hdots a_0\)
-    are nonnegative constants.
-
-    \begin{proof}
-        To prove that \(a_kn^k + a_{k - 1}n^{k - 1} + \hdots + a_1n^1 +
-        a_0n^0\), we must show the following:
-
-        \[
-            \exists c_1 \exists c_2 \forall n \geq n_0,\ {c_1 \cdot g(n) \leq
-            f(n) \leq c_2 \cdot g(n)}
-        \]
-
-        For the first inequality, it is easy to see that it holds because no
-        matter what the constants are, \(n^k \leq a_kn^k + a_{k - 1}n^{k - 1} +
-        \hdots + a_1n^1 + a_0n^0\) even if \(c_1 = 1\) and \(n_0 = 1\).  This
-        is because \(n^k \leq c_1 \cdot a_kn^k\) for any nonnegative constant,
-        \(c_1\) and \(a_k\).
-        \\
-
-        Taking the second inequality, we prove it in the following way.
-        By summation, \(\sum\limits_{i=0}^k a_i\) will give us a new constant,
-        \(A\). By taking this value of \(A\), we can then do the following:
-
-        \[
-            \begin{split}
-                a_kn^k + a_{k - 1}n^{k - 1} + \hdots + a_1n^1 + a_0n^0 &=
-                \\
-                &\leq (a_k + a_{k - 1} \hdots a_1 + a_0) \cdot n^k
-                \\
-                &= A \cdot n^k
-                \\
-                &\leq c_2 \cdot n^k
-            \end{split}
-        \]
-
-        where \(n_0 = 1\) and \(c_2 = A\). \(c_2\) is just a constant. Thus the
-        proof is complete.
-    \end{proof}
-\end{homeworkProblem}
-
-\pagebreak
-
-%
-% Non sequential homework problems
-%
-
-% Jump to problem 18
-\begin{homeworkProblem}[18]
-    Evaluate \(\sum_{k=1}^{5} k^2\) and \(\sum_{k=1}^{5} (k - 1)^2\).
-\end{homeworkProblem}
-
-% Continue counting to 19
-\begin{homeworkProblem}
-    Find the derivative of \(f(x) = x^4 + 3x^2 - 2\)
-\end{homeworkProblem}
-
-% Go back to where we left off
-\begin{homeworkProblem}[6]
-    Evaluate the integrals
-    \(\int_0^1 (1 - x^2) \dx\)
-    and
-    \(\int_1^{\infty} \frac{1}{x^2} \dx\).
-\end{homeworkProblem}
-
-\end{document}
diff --git a/general-template/letterfonts.tex b/general-template/letterfonts.tex
new file mode 100644
index 0000000..cf91649
--- /dev/null
+++ b/general-template/letterfonts.tex
@@ -0,0 +1,136 @@
+% number sets
+\newcommand{\RR}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{R}}{\mathbb{R}^{#1}}}}
+\newcommand{\NN}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{N}}{\mathbb{N}^{#1}}}}
+\newcommand{\ZZ}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{Z}}{\mathbb{Z}^{#1}}}}
+\newcommand{\QQ}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{Q}}{\mathbb{Q}^{#1}}}}
+\newcommand{\CC}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{C}}{\mathbb{C}^{#1}}}}
+\newcommand{\PP}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{P}}{\mathbb{P}^{#1}}}}
+\newcommand{\HH}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{H}}{\mathbb{H}^{#1}}}}
+\newcommand{\FF}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{F}}{\mathbb{F}^{#1}}}}
+% expected value
+\newcommand{\EE}{\ensuremath{\mathbb{E}}}
+
+%---------------------------------------
+% BlackBoard Math Fonts :-
+%---------------------------------------
+
+%Captital Letters
+\newcommand{\bbA}{\mathbb{A}}	\newcommand{\bbB}{\mathbb{B}}
+\newcommand{\bbC}{\mathbb{C}}	\newcommand{\bbD}{\mathbb{D}}
+\newcommand{\bbE}{\mathbb{E}}	\newcommand{\bbF}{\mathbb{F}}
+\newcommand{\bbG}{\mathbb{G}}	\newcommand{\bbH}{\mathbb{H}}
+\newcommand{\bbI}{\mathbb{I}}	\newcommand{\bbJ}{\mathbb{J}}
+\newcommand{\bbK}{\mathbb{K}}	\newcommand{\bbL}{\mathbb{L}}
+\newcommand{\bbM}{\mathbb{M}}	\newcommand{\bbN}{\mathbb{N}}
+\newcommand{\bbO}{\mathbb{O}}	\newcommand{\bbP}{\mathbb{P}}
+\newcommand{\bbQ}{\mathbb{Q}}	\newcommand{\bbR}{\mathbb{R}}
+\newcommand{\bbS}{\mathbb{S}}	\newcommand{\bbT}{\mathbb{T}}
+\newcommand{\bbU}{\mathbb{U}}	\newcommand{\bbV}{\mathbb{V}}
+\newcommand{\bbW}{\mathbb{W}}	\newcommand{\bbX}{\mathbb{X}}
+\newcommand{\bbY}{\mathbb{Y}}	\newcommand{\bbZ}{\mathbb{Z}}
+
+%---------------------------------------
+% MathCal Fonts :-
+%---------------------------------------
+
+%Captital Letters
+\newcommand{\mcA}{\mathcal{A}}	\newcommand{\mcB}{\mathcal{B}}
+\newcommand{\mcC}{\mathcal{C}}	\newcommand{\mcD}{\mathcal{D}}
+\newcommand{\mcE}{\mathcal{E}}	\newcommand{\mcF}{\mathcal{F}}
+\newcommand{\mcG}{\mathcal{G}}	\newcommand{\mcH}{\mathcal{H}}
+\newcommand{\mcI}{\mathcal{I}}	\newcommand{\mcJ}{\mathcal{J}}
+\newcommand{\mcK}{\mathcal{K}}	\newcommand{\mcL}{\mathcal{L}}
+\newcommand{\mcM}{\mathcal{M}}	\newcommand{\mcN}{\mathcal{N}}
+\newcommand{\mcO}{\mathcal{O}}	\newcommand{\mcP}{\mathcal{P}}
+\newcommand{\mcQ}{\mathcal{Q}}	\newcommand{\mcR}{\mathcal{R}}
+\newcommand{\mcS}{\mathcal{S}}	\newcommand{\mcT}{\mathcal{T}}
+\newcommand{\mcU}{\mathcal{U}}	\newcommand{\mcV}{\mathcal{V}}
+\newcommand{\mcW}{\mathcal{W}}	\newcommand{\mcX}{\mathcal{X}}
+\newcommand{\mcY}{\mathcal{Y}}	\newcommand{\mcZ}{\mathcal{Z}}
+
+
+
+%---------------------------------------
+% Bold Math Fonts :-
+%---------------------------------------
+
+%Captital Letters
+\newcommand{\bmA}{\boldsymbol{A}}	\newcommand{\bmB}{\boldsymbol{B}}
+\newcommand{\bmC}{\boldsymbol{C}}	\newcommand{\bmD}{\boldsymbol{D}}
+\newcommand{\bmE}{\boldsymbol{E}}	\newcommand{\bmF}{\boldsymbol{F}}
+\newcommand{\bmG}{\boldsymbol{G}}	\newcommand{\bmH}{\boldsymbol{H}}
+\newcommand{\bmI}{\boldsymbol{I}}	\newcommand{\bmJ}{\boldsymbol{J}}
+\newcommand{\bmK}{\boldsymbol{K}}	\newcommand{\bmL}{\boldsymbol{L}}
+\newcommand{\bmM}{\boldsymbol{M}}	\newcommand{\bmN}{\boldsymbol{N}}
+\newcommand{\bmO}{\boldsymbol{O}}	\newcommand{\bmP}{\boldsymbol{P}}
+\newcommand{\bmQ}{\boldsymbol{Q}}	\newcommand{\bmR}{\boldsymbol{R}}
+\newcommand{\bmS}{\boldsymbol{S}}	\newcommand{\bmT}{\boldsymbol{T}}
+\newcommand{\bmU}{\boldsymbol{U}}	\newcommand{\bmV}{\boldsymbol{V}}
+\newcommand{\bmW}{\boldsymbol{W}}	\newcommand{\bmX}{\boldsymbol{X}}
+\newcommand{\bmY}{\boldsymbol{Y}}	\newcommand{\bmZ}{\boldsymbol{Z}}
+%Small Letters
+\newcommand{\bma}{\boldsymbol{a}}	\newcommand{\bmb}{\boldsymbol{b}}
+\newcommand{\bmc}{\boldsymbol{c}}	\newcommand{\bmd}{\boldsymbol{d}}
+\newcommand{\bme}{\boldsymbol{e}}	\newcommand{\bmf}{\boldsymbol{f}}
+\newcommand{\bmg}{\boldsymbol{g}}	\newcommand{\bmh}{\boldsymbol{h}}
+\newcommand{\bmi}{\boldsymbol{i}}	\newcommand{\bmj}{\boldsymbol{j}}
+\newcommand{\bmk}{\boldsymbol{k}}	\newcommand{\bml}{\boldsymbol{l}}
+\newcommand{\bmm}{\boldsymbol{m}}	\newcommand{\bmn}{\boldsymbol{n}}
+\newcommand{\bmo}{\boldsymbol{o}}	\newcommand{\bmp}{\boldsymbol{p}}
+\newcommand{\bmq}{\boldsymbol{q}}	\newcommand{\bmr}{\boldsymbol{r}}
+\newcommand{\bms}{\boldsymbol{s}}	\newcommand{\bmt}{\boldsymbol{t}}
+\newcommand{\bmu}{\boldsymbol{u}}	\newcommand{\bmv}{\boldsymbol{v}}
+\newcommand{\bmw}{\boldsymbol{w}}	\newcommand{\bmx}{\boldsymbol{x}}
+\newcommand{\bmy}{\boldsymbol{y}}	\newcommand{\bmz}{\boldsymbol{z}}
+
+%---------------------------------------
+% Scr Math Fonts :-
+%---------------------------------------
+
+\newcommand{\sA}{{\mathscr{A}}}   \newcommand{\sB}{{\mathscr{B}}}
+\newcommand{\sC}{{\mathscr{C}}}   \newcommand{\sD}{{\mathscr{D}}}
+\newcommand{\sE}{{\mathscr{E}}}   \newcommand{\sF}{{\mathscr{F}}}
+\newcommand{\sG}{{\mathscr{G}}}   \newcommand{\sH}{{\mathscr{H}}}
+\newcommand{\sI}{{\mathscr{I}}}   \newcommand{\sJ}{{\mathscr{J}}}
+\newcommand{\sK}{{\mathscr{K}}}   \newcommand{\sL}{{\mathscr{L}}}
+\newcommand{\sM}{{\mathscr{M}}}   \newcommand{\sN}{{\mathscr{N}}}
+\newcommand{\sO}{{\mathscr{O}}}   \newcommand{\sP}{{\mathscr{P}}}
+\newcommand{\sQ}{{\mathscr{Q}}}   \newcommand{\sR}{{\mathscr{R}}}
+\newcommand{\sS}{{\mathscr{S}}}   \newcommand{\sT}{{\mathscr{T}}}
+\newcommand{\sU}{{\mathscr{U}}}   \newcommand{\sV}{{\mathscr{V}}}
+\newcommand{\sW}{{\mathscr{W}}}   \newcommand{\sX}{{\mathscr{X}}}
+\newcommand{\sY}{{\mathscr{Y}}}   \newcommand{\sZ}{{\mathscr{Z}}}
+
+
+%---------------------------------------
+% Math Fraktur Font
+%---------------------------------------
+
+%Captital Letters
+\newcommand{\mfA}{\mathfrak{A}}	\newcommand{\mfB}{\mathfrak{B}}
+\newcommand{\mfC}{\mathfrak{C}}	\newcommand{\mfD}{\mathfrak{D}}
+\newcommand{\mfE}{\mathfrak{E}}	\newcommand{\mfF}{\mathfrak{F}}
+\newcommand{\mfG}{\mathfrak{G}}	\newcommand{\mfH}{\mathfrak{H}}
+\newcommand{\mfI}{\mathfrak{I}}	\newcommand{\mfJ}{\mathfrak{J}}
+\newcommand{\mfK}{\mathfrak{K}}	\newcommand{\mfL}{\mathfrak{L}}
+\newcommand{\mfM}{\mathfrak{M}}	\newcommand{\mfN}{\mathfrak{N}}
+\newcommand{\mfO}{\mathfrak{O}}	\newcommand{\mfP}{\mathfrak{P}}
+\newcommand{\mfQ}{\mathfrak{Q}}	\newcommand{\mfR}{\mathfrak{R}}
+\newcommand{\mfS}{\mathfrak{S}}	\newcommand{\mfT}{\mathfrak{T}}
+\newcommand{\mfU}{\mathfrak{U}}	\newcommand{\mfV}{\mathfrak{V}}
+\newcommand{\mfW}{\mathfrak{W}}	\newcommand{\mfX}{\mathfrak{X}}
+\newcommand{\mfY}{\mathfrak{Y}}	\newcommand{\mfZ}{\mathfrak{Z}}
+%Small Letters
+\newcommand{\mfa}{\mathfrak{a}}	\newcommand{\mfb}{\mathfrak{b}}
+\newcommand{\mfc}{\mathfrak{c}}	\newcommand{\mfd}{\mathfrak{d}}
+\newcommand{\mfe}{\mathfrak{e}}	\newcommand{\mff}{\mathfrak{f}}
+\newcommand{\mfg}{\mathfrak{g}}	\newcommand{\mfh}{\mathfrak{h}}
+\newcommand{\mfi}{\mathfrak{i}}	\newcommand{\mfj}{\mathfrak{j}}
+\newcommand{\mfk}{\mathfrak{k}}	\newcommand{\mfl}{\mathfrak{l}}
+\newcommand{\mfm}{\mathfrak{m}}	\newcommand{\mfn}{\mathfrak{n}}
+\newcommand{\mfo}{\mathfrak{o}}	\newcommand{\mfp}{\mathfrak{p}}
+\newcommand{\mfq}{\mathfrak{q}}	\newcommand{\mfr}{\mathfrak{r}}
+\newcommand{\mfs}{\mathfrak{s}}	\newcommand{\mft}{\mathfrak{t}}
+\newcommand{\mfu}{\mathfrak{u}}	\newcommand{\mfv}{\mathfrak{v}}
+\newcommand{\mfw}{\mathfrak{w}}	\newcommand{\mfx}{\mathfrak{x}}
+\newcommand{\mfy}{\mathfrak{y}}	\newcommand{\mfz}{\mathfrak{z}}
diff --git a/general-template/macros.tex b/general-template/macros.tex
new file mode 100644
index 0000000..5ebc567
--- /dev/null
+++ b/general-template/macros.tex
@@ -0,0 +1,33 @@
+\newcommand{\eps}{\epsilon}
+\newcommand{\veps}{\varepsilon}
+\newcommand{\Qed}{\begin{flushright}\qed\end{flushright}}
+
+\newcommand{\parinn}{\setlength{\parindent}{1cm}}
+\newcommand{\parinf}{\setlength{\parindent}{0cm}}
+
+% \newcommand{\norm}{\|\cdot\|}
+\newcommand{\inorm}{\norm_{\infty}}
+\newcommand{\opensets}{\{V_{\alpha}\}_{\alpha\in I}}
+\newcommand{\oset}{V_{\alpha}}
+\newcommand{\opset}[1]{V_{\alpha_{#1}}}
+\newcommand{\lub}{\text{lub}}
+\newcommand{\del}[2]{\frac{\partial #1}{\partial #2}}
+\newcommand{\Del}[3]{\frac{\partial^{#1} #2}{\partial^{#1} #3}}
+\newcommand{\deld}[2]{\dfrac{\partial #1}{\partial #2}}
+\newcommand{\Deld}[3]{\dfrac{\partial^{#1} #2}{\partial^{#1} #3}}
+\newcommand{\der}[2]{\frac{\mathrm{d} #1}{\mathrm{d} #2}}
+% \newcommand{\ddd}[3]{\frac{\mathrm{d}^{#3} #1}{\mathrm{d}^{#3} #2}}
+\newcommand{\lm}{\lambda}
+\newcommand{\uin}{\mathbin{\rotatebox[origin=c]{90}{$\in$}}}
+\newcommand{\usubset}{\mathbin{\rotatebox[origin=c]{90}{$\subset$}}}
+\newcommand{\lt}{\left}
+\newcommand{\rt}{\right}
+\newcommand{\bs}[1]{\boldsymbol{#1}}
+\newcommand{\exs}{\exists}
+\newcommand{\st}{\strut}
+\newcommand{\dps}[1]{\displaystyle{#1}}
+\newcommand{\id}{\text{id}}
+
+
+\newcommand{\sol}{\setlength{\parindent}{0cm}\textbf{\textit{Solution:}}\setlength{\parindent}{1cm} }
+\newcommand{\solve}[1]{\setlength{\parindent}{0cm}\textbf{\textit{Solution: }}\setlength{\parindent}{1cm}#1 \Qed}
diff --git a/general-template/preamble.tex b/general-template/preamble.tex
new file mode 100644
index 0000000..bc65870
--- /dev/null
+++ b/general-template/preamble.tex
@@ -0,0 +1,746 @@
+\usepackage[tmargin=2cm,rmargin=1in,lmargin=1in,margin=0.85in,bmargin=2cm,footskip=.2in]{geometry}
+\usepackage{amsmath,amsfonts,amsthm,amssymb,mathtools}
+\usepackage{gensymb}
+
+\usepackage[varbb]{newpxmath}
+\usepackage{xfrac}
+\usepackage[makeroom]{cancel}
+\usepackage{mathtools}
+\usepackage{bookmark}
+\usepackage{enumitem}
+\usepackage{hyperref,theoremref}
+\hypersetup{
+	pdftitle={assignment},
+	colorlinks=true, linkcolor=doc!90,
+	bookmarksnumbered=true,
+	bookmarksopen=true
+}
+\usepackage[most,many,breakable]{tcolorbox}
+\usepackage{xcolor}
+\usepackage{varwidth}
+\usepackage{varwidth}
+\usepackage{etoolbox}
+%\usepackage{authblk}
+\usepackage{nameref}
+\usepackage{multicol,array}
+\usepackage[ruled,vlined,linesnumbered]{algorithm2e}
+\usepackage{comment} % enables the use of multi-line comments (\ifx \fi) 
+\usepackage{import}
+\usepackage{xifthen}
+\usepackage{pdfpages}
+\usepackage{transparent}
+\usepackage{chngcntr}
+\usepackage{tikz}
+\usepackage{titletoc}
+
+\newcommand\mycommfont[1]{\footnotesize\ttfamily\textcolor{blue}{#1}}
+\SetCommentSty{mycommfont}
+\newcommand{\incfig}[1]{%
+    \def\svgwidth{\columnwidth}
+    \import{./figures/}{#1.pdf_tex}
+}
+
+\usepackage{tikzsymbols}
+\tikzset{
+	symbol/.style={
+			draw=none,
+			every to/.append style={
+					edge node={node [sloped, allow upside down, auto=false]{$#1$}}}
+		}
+}
+\tikzstyle{c} = [circle,fill=black,scale=0.5]
+\tikzstyle{b} = [draw, thick, black, -]
+\tikzset{
+    vertex/.style={
+        circle,
+        draw,
+        minimum size=6mm,
+        inner sep=0pt
+    }
+}
+
+\renewcommand\qedsymbol{$\Laughey$}
+
+% \usepackage{xparse}
+% \usepackage{pgffor}
+% \usepackage{emoji}
+%
+% \NewDocumentCommand{\memoji}{m o}{%
+%   \IfNoValueTF{#2}
+%     {\ifmmode \text{\emoji{#1}} \else \emoji{#1}\fi} % Single emoji case
+%     {\foreach \index in {1,...,#2}{\ifmmode \text{\emoji{#1}} \else \emoji{#1}\fi}} % Repeated emoji case
+% }
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% SELF MADE COLORS
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\definecolor{doc}{RGB}{0,60,110}
+\definecolor{myg}{RGB}{56, 140, 70}
+\definecolor{myb}{RGB}{45, 111, 177}
+\definecolor{myr}{RGB}{199, 68, 64}
+\definecolor{mytheorembg}{HTML}{F2F2F9}
+\definecolor{mytheoremfr}{HTML}{00007B}
+\definecolor{mylemmabg}{HTML}{FFFAF8}
+\definecolor{mylemmafr}{HTML}{983b0f}
+\definecolor{mypropbg}{HTML}{f2fbfc}
+\definecolor{mypropfr}{HTML}{191971}
+\definecolor{myexamplebg}{HTML}{F2FBF8}
+\definecolor{myexamplefr}{HTML}{88D6D1}
+\definecolor{myexampleti}{HTML}{2A7F7F}
+\definecolor{mydefinitbg}{HTML}{E5E5FF}
+\definecolor{mydefinitfr}{HTML}{3F3FA3}
+\definecolor{notesgreen}{RGB}{0,162,0}
+\definecolor{myp}{RGB}{197, 92, 212}
+\definecolor{mygr}{HTML}{2C3338}
+\definecolor{myred}{RGB}{127,0,0}
+\definecolor{myyellow}{RGB}{169,121,69}
+\definecolor{myexercisebg}{HTML}{F2FBF8}
+\definecolor{myexercisefg}{HTML}{88D6D1}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% TCOLORBOX SETUPS
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\setlength{\parindent}{1cm}
+%================================
+% THEOREM BOX
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Theorem}{Theorem}
+{%
+	enhanced,
+	breakable,
+	colback = mytheorembg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mytheoremfr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mytheoremfr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mytheoremfr},
+}
+{th}
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{theorem}{Theorem}
+{%
+	enhanced,
+	breakable,
+	colback = mytheorembg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mytheoremfr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mytheoremfr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mytheoremfr},
+}
+{th}
+
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcolorbox{Theoremcon}
+{%
+	enhanced
+	,breakable
+	,colback = mytheorembg
+	,frame hidden
+	,boxrule = 0sp
+	,borderline west = {2pt}{0pt}{mytheoremfr}
+	,sharp corners
+	,description font = \mdseries
+	,separator sign none
+}
+
+%================================
+% Corollery
+%================================
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Corollary}{Corollary}
+{%
+	enhanced
+	,breakable
+	,colback = myp!10
+	,frame hidden
+	,boxrule = 0sp
+	,borderline west = {2pt}{0pt}{myp!85!black}
+	,sharp corners
+	,detach title
+	,before upper = \tcbtitle\par\smallskip
+	,coltitle = myp!85!black
+	,fonttitle = \bfseries\sffamily
+	,description font = \mdseries
+	,separator sign none
+	,segmentation style={solid, myp!85!black}
+}
+{th}
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{corollary}{Corollary}
+{%
+	enhanced
+	,breakable
+	,colback = myp!10
+	,frame hidden
+	,boxrule = 0sp
+	,borderline west = {2pt}{0pt}{myp!85!black}
+	,sharp corners
+	,detach title
+	,before upper = \tcbtitle\par\smallskip
+	,coltitle = myp!85!black
+	,fonttitle = \bfseries\sffamily
+	,description font = \mdseries
+	,separator sign none
+	,segmentation style={solid, myp!85!black}
+}
+{th}
+
+
+%================================
+% LEMMA
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Lemma}{Lemma}
+{%
+	enhanced,
+	breakable,
+	colback = mylemmabg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mylemmafr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mylemmafr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mylemmafr},
+}
+{th}
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{lemma}{lemma}
+{%
+	enhanced,
+	breakable,
+	colback = mylemmabg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mylemmafr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mylemmafr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mylemmafr},
+}
+{th}
+
+%================================
+% Exercise
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Exercise}{Exercise}
+{%
+	enhanced,
+	breakable,
+	colback = myexercisebg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{myexercisefg},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = myexercisefg,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, myexercisefg},
+}
+{th}
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{exercise}{Exercise}
+{%
+	enhanced,
+	breakable,
+	colback = myexercisebg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{myexercisefg},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = myexercisefg,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, myexercisefg},
+}
+{th}
+
+
+%================================
+% PROPOSITION
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Prop}{Proposition}
+{%
+	enhanced,
+	breakable,
+	colback = mypropbg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mypropfr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mypropfr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mypropfr},
+}
+{th}
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{prop}{Proposition}
+{%
+	enhanced,
+	breakable,
+	colback = mypropbg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mypropfr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mypropfr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mypropfr},
+}
+{th}
+
+
+%================================
+% CLAIM
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{claim}{Claim}
+{%
+	enhanced
+	,breakable
+	,colback = myg!10
+	,frame hidden
+	,boxrule = 0sp
+	,borderline west = {2pt}{0pt}{myg}
+	,sharp corners
+	,detach title
+	,before upper = \tcbtitle\par\smallskip
+	,coltitle = myg!85!black
+	,fonttitle = \bfseries\sffamily
+	,description font = \mdseries
+	,separator sign none
+	,segmentation style={solid, myg!85!black}
+}
+{th}
+
+
+
+%================================
+% EXAMPLE BOX
+%================================
+
+\newtcbtheorem[number within=section]{Example}{Example}
+{%
+	colback = myexamplebg
+	,breakable
+	,colframe = myexamplefr
+	,coltitle = myexampleti
+	,boxrule = 1pt
+	,sharp corners
+	,detach title
+	,before upper=\tcbtitle\par\smallskip
+	,fonttitle = \bfseries
+	,description font = \mdseries
+	,separator sign none
+	,description delimiters parenthesis
+}
+{ex}
+
+\newtcbtheorem[number within=chapter]{example}{Example}
+{%
+	colback = myexamplebg
+	,breakable
+	,colframe = myexamplefr
+	,coltitle = myexampleti
+	,boxrule = 1pt
+	,sharp corners
+	,detach title
+	,before upper=\tcbtitle\par\smallskip
+	,fonttitle = \bfseries
+	,description font = \mdseries
+	,separator sign none
+	,description delimiters parenthesis
+}
+{ex}
+
+%================================
+% DEFINITION BOX
+%================================
+
+\newtcbtheorem[number within=section]{Definition}{Definition}{enhanced,
+	before skip=2mm,after skip=2mm, colback=red!5,colframe=red!80!black,boxrule=0.5mm,
+	attach boxed title to top left={xshift=1cm,yshift*=1mm-\tcboxedtitleheight}, varwidth boxed title*=-3cm,
+	boxed title style={frame code={
+					\path[fill=tcbcolback]
+					([yshift=-1mm,xshift=-1mm]frame.north west)
+					arc[start angle=0,end angle=180,radius=1mm]
+					([yshift=-1mm,xshift=1mm]frame.north east)
+					arc[start angle=180,end angle=0,radius=1mm];
+					\path[left color=tcbcolback!60!black,right color=tcbcolback!60!black,
+						middle color=tcbcolback!80!black]
+					([xshift=-2mm]frame.north west) -- ([xshift=2mm]frame.north east)
+					[rounded corners=1mm]-- ([xshift=1mm,yshift=-1mm]frame.north east)
+					-- (frame.south east) -- (frame.south west)
+					-- ([xshift=-1mm,yshift=-1mm]frame.north west)
+					[sharp corners]-- cycle;
+				},interior engine=empty,
+		},
+	fonttitle=\bfseries,
+	title={#2},#1}{def}
+\newtcbtheorem[number within=chapter]{definition}{Definition}{enhanced,
+	before skip=2mm,after skip=2mm, colback=red!5,colframe=red!80!black,boxrule=0.5mm,
+	attach boxed title to top left={xshift=1cm,yshift*=1mm-\tcboxedtitleheight}, varwidth boxed title*=-3cm,
+	boxed title style={frame code={
+					\path[fill=tcbcolback]
+					([yshift=-1mm,xshift=-1mm]frame.north west)
+					arc[start angle=0,end angle=180,radius=1mm]
+					([yshift=-1mm,xshift=1mm]frame.north east)
+					arc[start angle=180,end angle=0,radius=1mm];
+					\path[left color=tcbcolback!60!black,right color=tcbcolback!60!black,
+						middle color=tcbcolback!80!black]
+					([xshift=-2mm]frame.north west) -- ([xshift=2mm]frame.north east)
+					[rounded corners=1mm]-- ([xshift=1mm,yshift=-1mm]frame.north east)
+					-- (frame.south east) -- (frame.south west)
+					-- ([xshift=-1mm,yshift=-1mm]frame.north west)
+					[sharp corners]-- cycle;
+				},interior engine=empty,
+		},
+	fonttitle=\bfseries,
+	title={#2},#1}{def}
+
+
+%================================
+% EXERCISE BOX
+%================================
+
+\newcounter{questioncounter}
+\counterwithin{questioncounter}{chapter}
+% \counterwithin{questioncounter}{section}
+
+\makeatletter
+\newtcbtheorem[use counter=questioncounter]{question}{Question}{enhanced,
+	breakable,
+	colback=white,
+	colframe=myb!80!black,
+	attach boxed title to top left={yshift*=-\tcboxedtitleheight},
+	fonttitle=\bfseries,
+	title={#2},
+	boxed title size=title,
+	boxed title style={%
+			sharp corners,
+			rounded corners=northwest,
+			colback=tcbcolframe,
+			boxrule=0pt,
+		},
+	underlay boxed title={%
+			\path[fill=tcbcolframe] (title.south west)--(title.south east)
+			to[out=0, in=180] ([xshift=5mm]title.east)--
+			(title.center-|frame.east)
+			[rounded corners=\kvtcb@arc] |-
+			(frame.north) -| cycle;
+		},
+	#1
+}{def}
+\makeatother
+
+%================================
+% SOLUTION BOX
+%================================
+
+\makeatletter
+\newtcolorbox{solution}{enhanced,
+	breakable,
+	colback=white,
+	colframe=myg!80!black,
+	attach boxed title to top left={yshift*=-\tcboxedtitleheight},
+	title=Solution,
+	boxed title size=title,
+	boxed title style={%
+			sharp corners,
+			rounded corners=northwest,
+			colback=tcbcolframe,
+			boxrule=0pt,
+		},
+	underlay boxed title={%
+			\path[fill=tcbcolframe] (title.south west)--(title.south east)
+			to[out=0, in=180] ([xshift=5mm]title.east)--
+			(title.center-|frame.east)
+			[rounded corners=\kvtcb@arc] |-
+			(frame.north) -| cycle;
+		},
+}
+\makeatother
+
+%================================
+% Question BOX
+%================================
+
+\makeatletter
+\newtcbtheorem{qstion}{Question}{enhanced,
+	breakable,
+	colback=white,
+	colframe=mygr,
+	attach boxed title to top left={yshift*=-\tcboxedtitleheight},
+	fonttitle=\bfseries,
+	title={#2},
+	boxed title size=title,
+	boxed title style={%
+			sharp corners,
+			rounded corners=northwest,
+			colback=tcbcolframe,
+			boxrule=0pt,
+		},
+	underlay boxed title={%
+			\path[fill=tcbcolframe] (title.south west)--(title.south east)
+			to[out=0, in=180] ([xshift=5mm]title.east)--
+			(title.center-|frame.east)
+			[rounded corners=\kvtcb@arc] |-
+			(frame.north) -| cycle;
+		},
+	#1
+}{def}
+\makeatother
+
+\newtcbtheorem[number within=chapter]{wconc}{Wrong Concept}{
+	breakable,
+	enhanced,
+	colback=white,
+	colframe=myr,
+	arc=0pt,
+	outer arc=0pt,
+	fonttitle=\bfseries\sffamily\large,
+	colbacktitle=myr,
+	attach boxed title to top left={},
+	boxed title style={
+			enhanced,
+			skin=enhancedfirst jigsaw,
+			arc=3pt,
+			bottom=0pt,
+			interior style={fill=myr}
+		},
+	#1
+}{def}
+
+
+%================================
+% NOTE BOX
+%================================
+
+\usetikzlibrary{arrows,calc,shadows.blur}
+\tcbuselibrary{skins}
+\newtcolorbox{note}[1][]{%
+	enhanced jigsaw,
+	colback=gray!20!white,%
+	colframe=gray!80!black,
+	size=small,
+	boxrule=1pt,
+	title=\textbf{Note:-},
+	halign title=flush center,
+	coltitle=black,
+	breakable,
+	drop shadow=black!50!white,
+	attach boxed title to top left={xshift=1cm,yshift=-\tcboxedtitleheight/2,yshifttext=-\tcboxedtitleheight/2},
+	minipage boxed title=1.5cm,
+	boxed title style={%
+			colback=white,
+			size=fbox,
+			boxrule=1pt,
+			boxsep=2pt,
+			underlay={%
+					\coordinate (dotA) at ($(interior.west) + (-0.5pt,0)$);
+					\coordinate (dotB) at ($(interior.east) + (0.5pt,0)$);
+					\begin{scope}
+						\clip (interior.north west) rectangle ([xshift=3ex]interior.east);
+						\filldraw [white, blur shadow={shadow opacity=60, shadow yshift=-.75ex}, rounded corners=2pt] (interior.north west) rectangle (interior.south east);
+					\end{scope}
+					\begin{scope}[gray!80!black]
+						\fill (dotA) circle (2pt);
+						\fill (dotB) circle (2pt);
+					\end{scope}
+				},
+		},
+	#1,
+}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% SELF MADE COMMANDS
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\newcommand{\thm}[2]{\begin{Theorem}{#1}{}#2\end{Theorem}}
+\newcommand{\cor}[2]{\begin{Corollary}{#1}{}#2\end{Corollary}}
+\newcommand{\mlemma}[2]{\begin{Lemma}{#1}{}#2\end{Lemma}}
+\newcommand{\mer}[2]{\begin{Exercise}{#1}{}#2\end{Exercise}}
+\newcommand{\mprop}[2]{\begin{Prop}{#1}{}#2\end{Prop}}
+\newcommand{\clm}[3]{\begin{claim}{#1}{#2}#3\end{claim}}
+\newcommand{\wc}[2]{\begin{wconc}{#1}{}\setlength{\parindent}{1cm}#2\end{wconc}}
+\newcommand{\thmcon}[1]{\begin{Theoremcon}{#1}\end{Theoremcon}}
+\newcommand{\ex}[2]{\begin{Example}{#1}{}#2\end{Example}}
+\newcommand{\dfn}[2]{\begin{Definition}[colbacktitle=red!75!black]{#1}{}#2\end{Definition}}
+\newcommand{\dfnc}[2]{\begin{definition}[colbacktitle=red!75!black]{#1}{}#2\end{definition}}
+\newcommand{\qs}[2]{\begin{question}{#1}{}#2\end{question}}
+\newcommand{\pf}[2]{\begin{myproof}[#1]#2\end{myproof}}
+\newcommand{\nt}[1]{\begin{note}#1\end{note}}
+
+\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
+		\node[shape=circle,draw,inner sep=1pt] (char) {#1};}}
+\newcommand\getcurrentref[1]{%
+	\ifnumequal{\value{#1}}{0}
+	{??}
+	{\the\value{#1}}%
+}
+\newcommand{\getCurrentSectionNumber}{\getcurrentref{section}}
+\newenvironment{myproof}[1][\proofname]{%
+	\proof[\bfseries #1: ]%
+}{\endproof}
+
+\newcommand{\mclm}[2]{\begin{myclaim}[#1]#2\end{myclaim}}
+\newenvironment{myclaim}[1][\claimname]{\proof[\bfseries #1: ]}{}
+\newenvironment{iclaim}[1][\claimname]{\bfseries #1\mdseries:}{}
+\newcommand{\iclm}[2]{\begin{iclaim}[#1]#2\end{iclaim}}
+
+\newcounter{mylabelcounter}
+
+\makeatletter
+\newcommand{\setword}[2]{%
+	\phantomsection
+	#1\def\@currentlabel{\unexpanded{#1}}\label{#2}%
+}
+\makeatother
+
+% deliminators
+\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
+\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
+
+\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}
+\DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor}
+\DeclarePairedDelimiter{\round}{\lfloor}{\rceil}
+
+\newsavebox\diffdbox
+\newcommand{\slantedromand}{{\mathpalette\makesl{d}}}
+\newcommand{\makesl}[2]{%
+\begingroup
+\sbox{\diffdbox}{$\mathsurround=0pt#1\mathrm{#2}$}%
+\pdfsave
+\pdfsetmatrix{1 0 0.2 1}%
+\rlap{\usebox{\diffdbox}}%
+\pdfrestore
+\hskip\wd\diffdbox
+\endgroup
+}
+\newcommand{\dd}[1][]{\ensuremath{\mathop{}\!\ifstrempty{#1}{%
+\slantedromand\@ifnextchar^{\hspace{0.2ex}}{\hspace{0.1ex}}}%
+{\slantedromand\hspace{0.2ex}^{#1}}}}
+\ProvideDocumentCommand\dv{o m g}{%
+  \ensuremath{%
+    \IfValueTF{#3}{%
+      \IfNoValueTF{#1}{%
+        \frac{\dd #2}{\dd #3}%
+      }{%
+        \frac{\dd^{#1} #2}{\dd #3^{#1}}%
+      }%
+    }{%
+      \IfNoValueTF{#1}{%
+        \frac{\dd}{\dd #2}%
+      }{%
+        \frac{\dd^{#1}}{\dd #2^{#1}}%
+      }%
+    }%
+  }%
+}
+\providecommand*{\pdv}[3][]{\frac{\partial^{#1}#2}{\partial#3^{#1}}}
+%  - others
+\DeclareMathOperator{\Lap}{\mathcal{L}}
+\DeclareMathOperator{\Var}{Var} % varience
+\DeclareMathOperator{\Cov}{Cov} % covarience
+\DeclareMathOperator{\E}{E} % expected
+
+% Since the amsthm package isn't loaded
+
+% I prefer the slanted \leq
+\let\oldleq\leq % save them in case they're every wanted
+\let\oldgeq\geq
+\renewcommand{\leq}{\leqslant}
+\renewcommand{\geq}{\geqslant}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% TABLE OF CONTENTS
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\contentsmargin{0cm}
+\titlecontents{chapter}[3.7pc]
+{\addvspace{30pt}%
+	\begin{tikzpicture}[remember picture, overlay]%
+		\draw[fill=doc!60,draw=doc!60] (-7,-.1) rectangle (-0.9,.5);%
+		\pgftext[left,x=-3.7cm,y=0.2cm]{\color{white}\Large\sc\bfseries Chapter\ \thecontentslabel};%
+	\end{tikzpicture}\color{doc!60}\large\sc\bfseries}%
+{}
+{}
+{\;\titlerule\;\large\sc\bfseries Page \thecontentspage
+	\begin{tikzpicture}[remember picture, overlay]
+		\draw[fill=doc!60,draw=doc!60] (2pt,0) rectangle (4,0.1pt);
+	\end{tikzpicture}}%
+\titlecontents{section}[3.7pc]
+{\addvspace{2pt}}
+{\contentslabel[\thecontentslabel]{2pc}}
+{}
+{\hfill\small \thecontentspage}
+[]
+\titlecontents*{subsection}[3.7pc]
+{\addvspace{-1pt}\small}
+{}
+{}
+{\ --- \small\thecontentspage}
+[ \textbullet\ ][]
+
+\makeatletter
+\renewcommand{\tableofcontents}{%
+	\chapter*{%
+	  \vspace*{-20\p@}%
+	  \begin{tikzpicture}[remember picture, overlay]%
+		  \pgftext[right,x=15cm,y=0.2cm]{\color{doc!60}\Huge\sc\bfseries \contentsname};%
+		  \draw[fill=doc!60,draw=doc!60] (13,-.75) rectangle (20,1);%
+		  \clip (13,-.75) rectangle (20,1);
+		  \pgftext[right,x=15cm,y=0.2cm]{\color{white}\Huge\sc\bfseries \contentsname};%
+	  \end{tikzpicture}}%
+	\@starttoc{toc}}
+\makeatother
diff --git a/general-template/template.tex b/general-template/template.tex
new file mode 100644
index 0000000..970913a
--- /dev/null
+++ b/general-template/template.tex
@@ -0,0 +1,128 @@
+\documentclass{report}
+
+\input{preamble}
+\input{macros}
+\input{letterfonts}
+
+\title{\Huge{Some Class}\\Random Examples}
+\author{\huge{Your Name}}
+\date{}
+
+\begin{document}
+
+\maketitle
+\newpage% or \cleardoublepage
+% \pdfbookmark[<level>]{<title>}{<dest>}
+\pdfbookmark[section]{\contentsname}{toc}
+\tableofcontents
+\pagebreak
+
+\chapter{}
+\section{Random Examples}
+\dfn{Limit of Sequence in $\bs{\bbR}$}{Let $\{s_n\}$ be a sequence in $\bbR$. We say $$\lim_{n\to\infty}s_n=s$$ where $s\in\bbR$ if $\forall$ real numbers $\eps>0$ $\exists$ natural number $N$ such that for $n>N$ $$s-\eps<s_n<s+\eps\text{ i.e. }|s-s_n|<\eps$$}
+\qs{}{Is the set ${x-}$axis${\setminus\{\text{Origin}\}}$ a closed set}
+\sol We have to take its complement and check whether that set is a open set i.e. if it is a union of open balls
+\nt{We will do topology in Normed Linear Space  (Mainly $\bbR^n$ and occasionally $\bbC^n$)using the language of Metric Space}
+\clm{Topology}{}{Topology is cool}
+\ex{Open Set and Close Set}{
+	\begin{tabular}{rl}
+		Open Set:   & $\bullet$ $\phi$                                              \\
+		            & $\bullet$ $\bigcup\limits_{x\in X}B_r(x)$ (Any $r>0$ will do) \\[3mm]
+		            & $\bullet$ $B_r(x)$ is open                                    \\
+		Closed Set: & $\bullet$ $X,\ \phi$                                          \\
+		            & $\bullet$ $\overline{B_r(x)}$                                 \\
+		            & $x-$axis $\cup$ $y-$axis
+	\end{tabular}}
+\thm{}{If $x\in$ open set $V$ then $\exists$ $\delta>0$ such that $B_{\delta}(x)\subset V$}
+\begin{myproof}By openness of $V$, $x\in B_r(u)\subset V$
+	\begin{center}
+		\begin{tikzpicture}
+			\draw[red] (0,0) circle [x radius=3.5cm, y radius=2cm] ;
+			\draw (3,1.6) node[red]{$V$};
+			\draw [blue] (1,0) circle (1.45cm) ;
+			\filldraw[blue] (1,0) circle (1pt) node[anchor=north]{$u$};
+			\draw (2.9,0.4) node[blue]{$B_r(u)$};
+			\draw [green!40!black] (1.7,0) circle (0.5cm) node [yshift=0.7cm]{$B_{\delta}(x)$} ;
+			\filldraw[green!40!black] (1.7,0) circle (1pt) node[anchor=west]{$x$};
+		\end{tikzpicture}
+	\end{center}
+
+	Given $x\in B_r(u)\subset V$, we want $\delta>0$ such that $x\in B_{\delta} (x)\subset B_r(u)\subset V$. Let $d=d(u,x)$. Choose $\delta $ such that $d+\delta<r$ (e.g. $\delta<\frac{r-d}{2}$)
+
+	If $y\in B_{\delta}(x)$ we will be done by showing that $d(u,y)<r$ but $$d(u,y)\leq d(u,x)+d(x,y)<d+\delta<r$$
+\end{myproof}
+
+\cor{}{By the result of the proof, we can then show...}
+\mlenma{}{Suppose $\vec{v_1}, \dots, \vec{v_n} \in \RR[n]$ is subspace of $\RR^n$.}
+\mprop{}{$1 + 1 = 2$.}
+
+\section{Random}
+\dfn{Normed Linear Space and Norm $\boldsymbol{\|\cdot\|}$}{Let $V$ be a vector space over $\bbR$ (or $\bbC$). A norm on $V$ is function $\|\cdot\|\ V\to \bbR_{\geq 0}$ satisfying \begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+		\item \label{n:1}$\|x\|=0 \iff x=0$ $\forall$ $x\in V$
+		\item \label{n:2}	$\|\lambda x\|=|\lambda|\|x\|$ $\forall$ $\lambda\in\bbR$(or $\bbC$), $x\in V$
+		\item \label{n:3} $\|x+y\| \leq \|x\|+\|y\|$ $\forall$ $x,y\in V$ (Triangle Inequality/Subadditivity)
+	\end{enumerate}And $V$ is called a normed linear space.
+
+	$\bullet $ Same definition works with $V$ a vector space over $\bbC$ (again $\|\cdot\|\to\bbR_{\geq 0}$) where \ref{n:2} becomes $\|\lambda x\|=|\lambda|\|x\|$ $\forall$ $\lambda\in\bbC$, $x\in V$, where for $\lambda=a+ib$, $|\lambda|=\sqrt{a^2+b^2}$ }
+
+
+\ex{$\bs{p-}$Norm}{\label{pnorm}$V={\bbR}^m$, $p\in\bbR_{\geq 0}$. Define for $x=(x_1,x_2,\cdots,x_m)\in\bbR^m$ $$\|x\|_p=\Big(|x_1|^p+|x_2|^p+\cdots+|x_m|^p\Big)^{\frac1p}$$(In school $p=2$)}
+\textbf{Special Case $\bs{p=1}$}: $\|x\|_1=|x_1|+|x_2|+\cdots+|x_m|$ is clearly a norm by usual triangle inequality. \par
+\textbf{Special Case $\bs{p\to\infty\ (\bbR^m$ with $\|\cdot\|_{\infty})}$}: $\|x\|_{\infty}=\max\{|x_1|,|x_2|,\cdots,|x_m|\}$\\
+For $m=1$ these $p-$norms are nothing but $|x|$.
+Now exercise
+\qs{}{\label{exs1}Prove that triangle inequality is true if $p\geq 1$ for $p-$norms. (What goes wrong for $p<1$ ?)}
+\sol{\textbf{For Property \ref{n:3} for norm-2}	\subsubsection*{\textbf{When field is $\bbR:$}} We have to show\begin{align*}
+		         & \sum_i(x_i+y_i)^2\leq \left(\sqrt{\sum_ix_i^2} +\sqrt{\sum_iy_i^2}\right)^2                                       \\
+		\implies & \sum_i (x_i^2+2x_iy_i+y_i^2)\leq \sum_ix_i^2+2\sqrt{\left[\sum_ix_i^2\right]\left[\sum_iy_i^2\right]}+\sum_iy_i^2 \\
+		\implies & \left[\sum_ix_iy_i\right]^2\leq \left[\sum_ix_i^2\right]\left[\sum_iy_i^2\right]
+	\end{align*}So in other words prove $\langle x,y\rangle^2 \leq \langle x,x\rangle\langle y,y\rangle$ where
+	$$\langle x,y\rangle =\sum\limits_i x_iy_i$$
+
+	\begin{note}
+		\begin{itemize}
+			\item $\|x\|^2=\langle x,x\rangle$
+			\item $\langle x,y\rangle=\langle y,x\rangle$
+			\item $\langle \cdot,\cdot\rangle$ is $\bbR-$linear in each slot i.e. \begin{align*}
+				      \langle rx+x',y\rangle=r\langle x,y\rangle+\langle x',y\rangle	\text{ and similarly for second slot}
+			      \end{align*}Here in $\langle x,y\rangle$ $x$ is in first slot and $y$ is in second slot.
+		\end{itemize}
+	\end{note}Now the statement is just the Cauchy-Schwartz Inequality. For proof $$\langle x,y\rangle^2\leq \langle x,x\rangle\langle y,y\rangle $$ expand everything of $\langle x-\lambda y,x-\lambda y\rangle$ which is going to give a quadratic equation in variable $\lambda $ \begin{align*}
+		\langle x-\lambda y,x-\lambda y\rangle & =\langle x,x-\lambda y\rangle-\lambda\langle y,x-\lambda y\rangle                                       \\
+		                                       & =\langle x ,x\rangle -\lambda\langle x,y\rangle -\lambda\langle y,x\rangle +\lambda^2\langle y,y\rangle \\
+		                                       & =\langle x,x\rangle -2\lambda\langle x,y\rangle+\lambda^2\langle y,y\rangle
+	\end{align*}Now unless $x=\lambda y$ we have $\langle x-\lambda y,x-\lambda y\rangle>0$ Hence the quadratic equation has no root therefore the discriminant is greater than zero.
+
+	\subsubsection*{\textbf{When field is $\bbC:$}}Modify the definition by $$\langle x,y\rangle=\sum_i\overline{x_i}y_i$$Then we still have $\langle x,x\rangle\geq 0$}
+
+\section{Algorithms}
+\begin{algorithm}[H]
+\KwIn{This is some input}
+\KwOut{This is some output}
+\SetAlgoLined
+\SetNoFillComment
+\tcc{This is a comment}
+\vspace{3mm}
+some code here\;
+$x \leftarrow 0$\;
+$y \leftarrow 0$\;
+\uIf{$ x > 5$} {
+    x is greater than 5 \tcp*{This is also a comment}
+}
+\Else {
+    x is less than or equal to 5\;
+}
+\ForEach{y in 0..5} {
+    $y \leftarrow y + 1$\;
+}
+\For{$y$ in $0..5$} {
+    $y \leftarrow y - 1$\;
+}
+\While{$x > 5$} {
+    $x \leftarrow x - 1$\;
+}
+\Return Return something here\;
+\caption{what}
+\end{algorithm}
+
+\end{document}
diff --git a/labs-set-1/homework.pdf b/labs-set-1/homework.pdf
deleted file mode 100644
index c8b27ff..0000000
Binary files a/labs-set-1/homework.pdf and /dev/null differ
diff --git a/labs-set-1/homework.tex b/labs-set-1/homework.tex
deleted file mode 100644
index c8823d2..0000000
--- a/labs-set-1/homework.tex
+++ /dev/null
@@ -1,541 +0,0 @@
-\documentclass{article}
-
-\usepackage{fancyhdr}
-\usepackage{extramarks}
-\usepackage{amsmath}
-\usepackage{amsthm}
-\usepackage{amsfonts}
-\usepackage{tikz}
-\usepackage[plain]{algorithm}
-\usepackage{algpseudocode}
-\usepackage{comment}
-\usetikzlibrary{automata,positioning}
-
-%
-% Basic Document Settings
-%
-
-\topmargin=-0.45in
-\evensidemargin=0in
-\oddsidemargin=0in
-\textwidth=6.5in
-\textheight=9.0in
-\headsep=0.25in
-
-\linespread{1.1}
-
-\pagestyle{fancy}
-\lhead{\hmwkAuthorName}
-\chead{\hmwkClass\ (\hmwkClassInstructor\ \hmwkClassTime): \hmwkTitle}
-\rhead{\firstxmark}
-\lfoot{\lastxmark}
-\cfoot{\thepage}
-
-\renewcommand\headrulewidth{0.4pt}
-\renewcommand\footrulewidth{0.4pt}
-
-\setlength\parindent{0pt}
-
-%
-% Create Problem Sections
-%
-
-\newcommand{\enterProblemHeader}[1]{
-    \nobreak\extramarks{}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
-    \nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
-}
-
-\newcommand{\exitProblemHeader}[1]{
-    \nobreak\extramarks{Problem \arabic{#1} (continued)}{Problem \arabic{#1} continued on next page\ldots}\nobreak{}
-    \stepcounter{#1}
-    \nobreak\extramarks{Problem \arabic{#1}}{}\nobreak{}
-}
-
-\setcounter{secnumdepth}{0}
-\newcounter{partCounter}
-\newcounter{homeworkProblemCounter}
-\setcounter{homeworkProblemCounter}{1}
-\nobreak\extramarks{Problem \arabic{homeworkProblemCounter}}{}\nobreak{}
-
-%
-% Homework Problem Environment
-%
-% This environment takes an optional argument. When given, it will adjust the
-% problem counter. This is useful for when the problems given for your
-% assignment aren't sequential. See the last 3 problems of this template for an
-% example.
-%
-\newenvironment{homeworkProblem}[1][-1]{
-    \ifnum#1>0
-        \setcounter{homeworkProblemCounter}{#1}
-    \fi
-    \section{Problem \arabic{homeworkProblemCounter}}
-    \setcounter{partCounter}{1}
-    \enterProblemHeader{homeworkProblemCounter}
-}{
-    \exitProblemHeader{homeworkProblemCounter}
-}
-
-%
-% Homework Details
-%   - Title
-%   - Due date
-%   - Class
-%   - Section/Time
-%   - Instructor
-%   - Author
-%
-
-\newcommand{\hmwkTitle}{Homework\ \#1}
-\newcommand{\hmwkDueDate}{September 21, 2025}
-\newcommand{\hmwkClass}{Calc III}
-\newcommand{\hmwkClassTime}{$6^{\text{th}}$ Hour}}
-\newcommand{\hmwkClassInstructor}{Professor Foresee}
-\newcommand{\hmwkAuthorName}{\textbf{Krishna A.}}
-
-%
-% Title Page
-%
-
-\title{
-    \vspace{2in}
-    \textmd{\textbf{\hmwkClass:\ \hmwkTitle}}\\
-    \normalsize\vspace{0.1in}\small{Due\ on\ \hmwkDueDate\ at 3:10pm}\\
-    \vspace{0.1in}\large{\textit{\hmwkClassInstructor\ \hmwkClassTime}}
-    \vspace{3in}
-}
-
-\author{\hmwkAuthorName}
-\date{}
-
-\renewcommand{\part}[1]{\textbf{\large Part \Alph{partCounter}}\stepcounter{partCounter}\\}
-
-%
-% Various Helper Commands
-%
-
-% Useful for algorithms
-\newcommand{\alg}[1]{\textsc{\bfseries \footnotesize #1}}
-
-% For derivatives
-\newcommand{\deriv}[1]{\frac{\mathrm{d}}{\mathrm{d}x} (#1)}
-
-% For partial derivatives
-\newcommand{\pderiv}[2]{\frac{\partial}{\partial #1} (#2)}
-
-% Integral dx
-\newcommand{\dx}{\mathrm{d}x}
-
-% Alias for the Solution section header
-\newcommand{\solution}{\textbf{\large Solution}}
-
-% Probability commands: Expectation, Variance, Covariance, Bias
-\newcommand{\E}{\mathrm{E}}
-\newcommand{\Var}{\mathrm{Var}}
-\newcommand{\Cov}{\mathrm{Cov}}
-\newcommand{\Bias}{\mathrm{Bias}}
-
-\begin{document}
-
-\maketitle
-
-\pagebreak
-
-\begin{comment}
-
-\begin{homeworkProblem}
-    Give an appropriate positive constant \(c\) such that \(f(n) \leq c \cdot
-    g(n)\) for all \(n > 1\).
-
-    \begin{enumerate}
-        \item \(f(n) = n^2 + n + 1\), \(g(n) = 2n^3\)
-        \item \(f(n) = n\sqrt{n} + n^2\), \(g(n) = n^2\)
-        \item \(f(n) = n^2 - n + 1\), \(g(n) = n^2 / 2\)
-    \end{enumerate}
-
-    \textbf{Solution}
-
-    We solve each solution algebraically to determine a possible constant
-    \(c\).
-    \\
-
-    \textbf{Part One}
-
-    \[
-        \begin{split}
-            n^2 + n + 1 &=
-            \\
-            &\leq n^2 + n^2 + n^2
-            \\
-            &= 3n^2
-            \\
-            &\leq c \cdot 2n^3
-        \end{split}
-    \]
-
-    Thus a valid \(c\) could be when \(c = 2\).
-    \\
-
-    \textbf{Part Two}
-
-    \[
-        \begin{split}
-            n^2 + n\sqrt{n} &=
-            \\
-            &= n^2 + n^{3/2}
-            \\
-            &\leq n^2 + n^{4/2}
-            \\
-            &= n^2 + n^2
-            \\
-            &= 2n^2
-            \\
-            &\leq c \cdot n^2
-        \end{split}
-    \]
-
-    Thus a valid \(c\) is \(c = 2\).
-    \\
-
-    \textbf{Part Three}
-
-    \[
-        \begin{split}
-            n^2 - n + 1 &=
-            \\
-            &\leq n^2
-            \\
-            &\leq c \cdot n^2/2
-        \end{split}
-    \]
-
-    Thus a valid \(c\) is \(c = 2\).
-
-\end{homeworkProblem}
-
-\pagebreak
-
-\begin{homeworkProblem}
-    Let \(\Sigma = \{0, 1\}\). Construct a DFA \(A\) that recognizes the
-    language that consists of all binary numbers that can be divided by 5.
-    \\
-
-    Let the state \(q_k\) indicate the remainder of \(k\) divided by 5. For
-    example, the remainder of 2 would correlate to state \(q_2\) because \(7
-    \mod 5 = 2\).
-
-    \begin{figure}[h]
-        \centering
-        \begin{tikzpicture}[shorten >=1pt,node distance=2cm,on grid,auto]
-            \node[state, accepting, initial] (q_0)   {$q_0$};
-            \node[state] (q_1) [right=of q_0] {$q_1$};
-            \node[state] (q_2) [right=of q_1] {$q_2$};
-            \node[state] (q_3) [right=of q_2] {$q_3$};
-            \node[state] (q_4) [right=of q_3] {$q_4$};
-            \path[->]
-                (q_0)
-                    edge [loop above] node {0} (q_0)
-                    edge node {1} (q_1)
-                (q_1)
-                    edge node {0} (q_2)
-                    edge [bend right=-30] node {1} (q_3)
-                (q_2)
-                    edge [bend left] node {1} (q_0)
-                    edge [bend right=-30] node {0} (q_4)
-                (q_3)
-                    edge node {1} (q_2)
-                    edge [bend left] node {0} (q_1)
-                (q_4)
-                    edge node {0} (q_3)
-                    edge [loop below] node {1} (q_4);
-        \end{tikzpicture}
-        \caption{DFA, \(A\), this is really beautiful, ya know?}
-        \label{fig:multiple5}
-    \end{figure}
-
-    \textbf{Justification}
-    \\
-
-    Take a given binary number, \(x\). Since there are only two inputs to our
-    state machine, \(x\) can either become \(x0\) or \(x1\). When a 0 comes
-    into the state machine, it is the same as taking the binary number and
-    multiplying it by two. When a 1 comes into the machine, it is the same as
-    multipying by two and adding one.
-    \\
-
-    Using this knowledge, we can construct a transition table that tell us
-    where to go:
-
-    \begin{table}[ht]
-        \centering
-        \begin{tabular}{c || c | c | c | c | c}
-            & \(x \mod 5 = 0\)
-            & \(x \mod 5 = 1\)
-            & \(x \mod 5 = 2\)
-            & \(x \mod 5 = 3\)
-            & \(x \mod 5 = 4\)
-            \\
-            \hline
-            \(x0\) & 0 & 2 & 4 & 1 & 3 \\
-            \(x1\) & 1 & 3 & 0 & 2 & 4 \\
-        \end{tabular}
-    \end{table}
-
-    Therefore on state \(q_0\) or (\(x \mod 5 = 0\)), a transition line should
-    go to state \(q_0\) for the input 0 and a line should go to state \(q_1\)
-    for input 1. Continuing this gives us the Figure~\ref{fig:multiple5}.
-\end{homeworkProblem}
-
-\begin{homeworkProblem}
-    Write part of \alg{Quick-Sort($list, start, end$)}
-
-    \begin{algorithm}[]
-        \begin{algorithmic}[1]
-            \Function{Quick-Sort}{$list, start, end$}
-                \If{$start \geq end$}
-                    \State{} \Return{}
-                \EndIf{}
-                \State{} $mid \gets \Call{Partition}{list, start, end}$
-                \State{} \Call{Quick-Sort}{$list, start, mid - 1$}
-                \State{} \Call{Quick-Sort}{$list, mid + 1, end$}
-            \EndFunction{}
-        \end{algorithmic}
-        \caption{Start of QuickSort}
-    \end{algorithm}
-\end{homeworkProblem}
-
-\pagebreak
-
-\begin{homeworkProblem}
-    Suppose we would like to fit a straight line through the origin, i.e.,
-    \(Y_i = \beta_1 x_i + e_i\) with \(i = 1, \ldots, n\), \(\E [e_i] = 0\),
-    and \(\Var [e_i] = \sigma^2_e\) and \(\Cov[e_i, e_j] = 0, \forall i \neq
-    j\).
-    \\
-
-    \part
-
-    Find the least squares esimator for \(\hat{\beta_1}\) for the slope
-    \(\beta_1\).
-    \\
-
-    \solution
-
-    To find the least squares estimator, we should minimize our Residual Sum
-    of Squares, RSS:
-
-    \[
-        \begin{split}
-            RSS &= \sum_{i = 1}^{n} {(Y_i - \hat{Y_i})}^2
-            \\
-            &= \sum_{i = 1}^{n} {(Y_i - \hat{\beta_1} x_i)}^2
-        \end{split}
-    \]
-
-    By taking the partial derivative in respect to \(\hat{\beta_1}\), we get:
-
-    \[
-        \pderiv{
-            \hat{\beta_1}
-        }{RSS}
-        = -2 \sum_{i = 1}^{n} {x_i (Y_i - \hat{\beta_1} x_i)}
-        = 0
-    \]
-
-    This gives us:
-
-    \[
-        \begin{split}
-            \sum_{i = 1}^{n} {x_i (Y_i - \hat{\beta_1} x_i)}
-            &= \sum_{i = 1}^{n} {x_i Y_i} - \sum_{i = 1}^{n} \hat{\beta_1} x_i^2
-            \\
-            &= \sum_{i = 1}^{n} {x_i Y_i} - \hat{\beta_1}\sum_{i = 1}^{n} x_i^2
-        \end{split}
-    \]
-
-    Solving for \(\hat{\beta_1}\) gives the final estimator for \(\beta_1\):
-
-    \[
-        \begin{split}
-            \hat{\beta_1}
-            &= \frac{
-                \sum {x_i Y_i}
-            }{
-                \sum x_i^2
-            }
-        \end{split}
-    \]
-
-    \pagebreak
-
-    \part
-
-    Calculate the bias and the variance for the estimated slope
-    \(\hat{\beta_1}\).
-    \\
-
-    \solution
-
-    For the bias, we need to calculate the expected value
-    \(\E[\hat{\beta_1}]\):
-
-    \[
-        \begin{split}
-            \E[\hat{\beta_1}]
-            &= \E \left[ \frac{
-                \sum {x_i Y_i}
-            }{
-                \sum x_i^2
-            }\right]
-            \\
-            &= \frac{
-                \sum {x_i \E[Y_i]}
-            }{
-                \sum x_i^2
-            }
-            \\
-            &= \frac{
-                \sum {x_i (\beta_1 x_i)}
-            }{
-                \sum x_i^2
-            }
-            \\
-            &= \frac{
-                \sum {x_i^2 \beta_1}
-            }{
-                \sum x_i^2
-            }
-            \\
-            &= \beta_1 \frac{
-                \sum {x_i^2 \beta_1}
-            }{
-                \sum x_i^2
-            }
-            \\
-            &= \beta_1
-        \end{split}
-    \]
-
-    Thus since our estimator's expected value is \(\beta_1\), we can conclude
-    that the bias of our estimator is 0.
-    \\
-
-    For the variance:
-
-    \[
-        \begin{split}
-            \Var[\hat{\beta_1}]
-            &= \Var \left[ \frac{
-                \sum {x_i Y_i}
-            }{
-                \sum x_i^2
-            }\right]
-            \\
-            &=
-            \frac{
-                \sum {x_i^2}
-            }{
-                \sum x_i^2 \sum x_i^2
-            } \Var[Y_i]
-            \\
-            &=
-            \frac{
-                \sum {x_i^2}
-            }{
-                \sum x_i^2 \sum x_i^2
-            } \Var[Y_i]
-            \\
-            &=
-            \frac{
-                1
-            }{
-                \sum x_i^2
-            } \Var[Y_i]
-            \\
-            &=
-            \frac{
-                1
-            }{
-                \sum x_i^2
-            } \sigma^2
-            \\
-            &=
-            \frac{
-                \sigma^2
-            }{
-                \sum x_i^2
-            }
-        \end{split}
-    \]
-
-\end{homeworkProblem}
-
-\pagebreak
-
-\begin{homeworkProblem}
-    Prove a polynomial of degree \(k\), \(a_kn^k + a_{k - 1}n^{k - 1} + \hdots
-    + a_1n^1 + a_0n^0\) is a member of \(\Theta(n^k)\) where \(a_k \hdots a_0\)
-    are nonnegative constants.
-
-    \begin{proof}
-        To prove that \(a_kn^k + a_{k - 1}n^{k - 1} + \hdots + a_1n^1 +
-        a_0n^0\), we must show the following:
-
-        \[
-            \exists c_1 \exists c_2 \forall n \geq n_0,\ {c_1 \cdot g(n) \leq
-            f(n) \leq c_2 \cdot g(n)}
-        \]
-
-        For the first inequality, it is easy to see that it holds because no
-        matter what the constants are, \(n^k \leq a_kn^k + a_{k - 1}n^{k - 1} +
-        \hdots + a_1n^1 + a_0n^0\) even if \(c_1 = 1\) and \(n_0 = 1\).  This
-        is because \(n^k \leq c_1 \cdot a_kn^k\) for any nonnegative constant,
-        \(c_1\) and \(a_k\).
-        \\
-
-        Taking the second inequality, we prove it in the following way.
-        By summation, \(\sum\limits_{i=0}^k a_i\) will give us a new constant,
-        \(A\). By taking this value of \(A\), we can then do the following:
-
-        \[
-            \begin{split}
-                a_kn^k + a_{k - 1}n^{k - 1} + \hdots + a_1n^1 + a_0n^0 &=
-                \\
-                &\leq (a_k + a_{k - 1} \hdots a_1 + a_0) \cdot n^k
-                \\
-                &= A \cdot n^k
-                \\
-                &\leq c_2 \cdot n^k
-            \end{split}
-        \]
-
-        where \(n_0 = 1\) and \(c_2 = A\). \(c_2\) is just a constant. Thus the
-        proof is complete.
-    \end{proof}
-\end{homeworkProblem}
-
-\pagebreak
-
-%
-% Non sequential homework problems
-%
-
-% Jump to problem 18
-\begin{homeworkProblem}[18]
-    Evaluate \(\sum_{k=1}^{5} k^2\) and \(\sum_{k=1}^{5} (k - 1)^2\).
-\end{homeworkProblem}
-
-% Continue counting to 19
-\begin{homeworkProblem}
-    Find the derivative of \(f(x) = x^4 + 3x^2 - 2\)
-\end{homeworkProblem}
-
-% Go back to where we left off
-\begin{homeworkProblem}[6]
-    Evaluate the integrals
-    \(\int_0^1 (1 - x^2) \dx\)
-    and
-    \(\int_1^{\infty} \frac{1}{x^2} \dx\).
-\end{homeworkProblem}
-\end{comment}
-\end{document}
diff --git a/labs-set-1/letterfonts.tex b/labs-set-1/letterfonts.tex
new file mode 100644
index 0000000..cf91649
--- /dev/null
+++ b/labs-set-1/letterfonts.tex
@@ -0,0 +1,136 @@
+% number sets
+\newcommand{\RR}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{R}}{\mathbb{R}^{#1}}}}
+\newcommand{\NN}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{N}}{\mathbb{N}^{#1}}}}
+\newcommand{\ZZ}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{Z}}{\mathbb{Z}^{#1}}}}
+\newcommand{\QQ}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{Q}}{\mathbb{Q}^{#1}}}}
+\newcommand{\CC}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{C}}{\mathbb{C}^{#1}}}}
+\newcommand{\PP}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{P}}{\mathbb{P}^{#1}}}}
+\newcommand{\HH}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{H}}{\mathbb{H}^{#1}}}}
+\newcommand{\FF}[1][]{\ensuremath{\ifstrempty{#1}{\mathbb{F}}{\mathbb{F}^{#1}}}}
+% expected value
+\newcommand{\EE}{\ensuremath{\mathbb{E}}}
+
+%---------------------------------------
+% BlackBoard Math Fonts :-
+%---------------------------------------
+
+%Captital Letters
+\newcommand{\bbA}{\mathbb{A}}	\newcommand{\bbB}{\mathbb{B}}
+\newcommand{\bbC}{\mathbb{C}}	\newcommand{\bbD}{\mathbb{D}}
+\newcommand{\bbE}{\mathbb{E}}	\newcommand{\bbF}{\mathbb{F}}
+\newcommand{\bbG}{\mathbb{G}}	\newcommand{\bbH}{\mathbb{H}}
+\newcommand{\bbI}{\mathbb{I}}	\newcommand{\bbJ}{\mathbb{J}}
+\newcommand{\bbK}{\mathbb{K}}	\newcommand{\bbL}{\mathbb{L}}
+\newcommand{\bbM}{\mathbb{M}}	\newcommand{\bbN}{\mathbb{N}}
+\newcommand{\bbO}{\mathbb{O}}	\newcommand{\bbP}{\mathbb{P}}
+\newcommand{\bbQ}{\mathbb{Q}}	\newcommand{\bbR}{\mathbb{R}}
+\newcommand{\bbS}{\mathbb{S}}	\newcommand{\bbT}{\mathbb{T}}
+\newcommand{\bbU}{\mathbb{U}}	\newcommand{\bbV}{\mathbb{V}}
+\newcommand{\bbW}{\mathbb{W}}	\newcommand{\bbX}{\mathbb{X}}
+\newcommand{\bbY}{\mathbb{Y}}	\newcommand{\bbZ}{\mathbb{Z}}
+
+%---------------------------------------
+% MathCal Fonts :-
+%---------------------------------------
+
+%Captital Letters
+\newcommand{\mcA}{\mathcal{A}}	\newcommand{\mcB}{\mathcal{B}}
+\newcommand{\mcC}{\mathcal{C}}	\newcommand{\mcD}{\mathcal{D}}
+\newcommand{\mcE}{\mathcal{E}}	\newcommand{\mcF}{\mathcal{F}}
+\newcommand{\mcG}{\mathcal{G}}	\newcommand{\mcH}{\mathcal{H}}
+\newcommand{\mcI}{\mathcal{I}}	\newcommand{\mcJ}{\mathcal{J}}
+\newcommand{\mcK}{\mathcal{K}}	\newcommand{\mcL}{\mathcal{L}}
+\newcommand{\mcM}{\mathcal{M}}	\newcommand{\mcN}{\mathcal{N}}
+\newcommand{\mcO}{\mathcal{O}}	\newcommand{\mcP}{\mathcal{P}}
+\newcommand{\mcQ}{\mathcal{Q}}	\newcommand{\mcR}{\mathcal{R}}
+\newcommand{\mcS}{\mathcal{S}}	\newcommand{\mcT}{\mathcal{T}}
+\newcommand{\mcU}{\mathcal{U}}	\newcommand{\mcV}{\mathcal{V}}
+\newcommand{\mcW}{\mathcal{W}}	\newcommand{\mcX}{\mathcal{X}}
+\newcommand{\mcY}{\mathcal{Y}}	\newcommand{\mcZ}{\mathcal{Z}}
+
+
+
+%---------------------------------------
+% Bold Math Fonts :-
+%---------------------------------------
+
+%Captital Letters
+\newcommand{\bmA}{\boldsymbol{A}}	\newcommand{\bmB}{\boldsymbol{B}}
+\newcommand{\bmC}{\boldsymbol{C}}	\newcommand{\bmD}{\boldsymbol{D}}
+\newcommand{\bmE}{\boldsymbol{E}}	\newcommand{\bmF}{\boldsymbol{F}}
+\newcommand{\bmG}{\boldsymbol{G}}	\newcommand{\bmH}{\boldsymbol{H}}
+\newcommand{\bmI}{\boldsymbol{I}}	\newcommand{\bmJ}{\boldsymbol{J}}
+\newcommand{\bmK}{\boldsymbol{K}}	\newcommand{\bmL}{\boldsymbol{L}}
+\newcommand{\bmM}{\boldsymbol{M}}	\newcommand{\bmN}{\boldsymbol{N}}
+\newcommand{\bmO}{\boldsymbol{O}}	\newcommand{\bmP}{\boldsymbol{P}}
+\newcommand{\bmQ}{\boldsymbol{Q}}	\newcommand{\bmR}{\boldsymbol{R}}
+\newcommand{\bmS}{\boldsymbol{S}}	\newcommand{\bmT}{\boldsymbol{T}}
+\newcommand{\bmU}{\boldsymbol{U}}	\newcommand{\bmV}{\boldsymbol{V}}
+\newcommand{\bmW}{\boldsymbol{W}}	\newcommand{\bmX}{\boldsymbol{X}}
+\newcommand{\bmY}{\boldsymbol{Y}}	\newcommand{\bmZ}{\boldsymbol{Z}}
+%Small Letters
+\newcommand{\bma}{\boldsymbol{a}}	\newcommand{\bmb}{\boldsymbol{b}}
+\newcommand{\bmc}{\boldsymbol{c}}	\newcommand{\bmd}{\boldsymbol{d}}
+\newcommand{\bme}{\boldsymbol{e}}	\newcommand{\bmf}{\boldsymbol{f}}
+\newcommand{\bmg}{\boldsymbol{g}}	\newcommand{\bmh}{\boldsymbol{h}}
+\newcommand{\bmi}{\boldsymbol{i}}	\newcommand{\bmj}{\boldsymbol{j}}
+\newcommand{\bmk}{\boldsymbol{k}}	\newcommand{\bml}{\boldsymbol{l}}
+\newcommand{\bmm}{\boldsymbol{m}}	\newcommand{\bmn}{\boldsymbol{n}}
+\newcommand{\bmo}{\boldsymbol{o}}	\newcommand{\bmp}{\boldsymbol{p}}
+\newcommand{\bmq}{\boldsymbol{q}}	\newcommand{\bmr}{\boldsymbol{r}}
+\newcommand{\bms}{\boldsymbol{s}}	\newcommand{\bmt}{\boldsymbol{t}}
+\newcommand{\bmu}{\boldsymbol{u}}	\newcommand{\bmv}{\boldsymbol{v}}
+\newcommand{\bmw}{\boldsymbol{w}}	\newcommand{\bmx}{\boldsymbol{x}}
+\newcommand{\bmy}{\boldsymbol{y}}	\newcommand{\bmz}{\boldsymbol{z}}
+
+%---------------------------------------
+% Scr Math Fonts :-
+%---------------------------------------
+
+\newcommand{\sA}{{\mathscr{A}}}   \newcommand{\sB}{{\mathscr{B}}}
+\newcommand{\sC}{{\mathscr{C}}}   \newcommand{\sD}{{\mathscr{D}}}
+\newcommand{\sE}{{\mathscr{E}}}   \newcommand{\sF}{{\mathscr{F}}}
+\newcommand{\sG}{{\mathscr{G}}}   \newcommand{\sH}{{\mathscr{H}}}
+\newcommand{\sI}{{\mathscr{I}}}   \newcommand{\sJ}{{\mathscr{J}}}
+\newcommand{\sK}{{\mathscr{K}}}   \newcommand{\sL}{{\mathscr{L}}}
+\newcommand{\sM}{{\mathscr{M}}}   \newcommand{\sN}{{\mathscr{N}}}
+\newcommand{\sO}{{\mathscr{O}}}   \newcommand{\sP}{{\mathscr{P}}}
+\newcommand{\sQ}{{\mathscr{Q}}}   \newcommand{\sR}{{\mathscr{R}}}
+\newcommand{\sS}{{\mathscr{S}}}   \newcommand{\sT}{{\mathscr{T}}}
+\newcommand{\sU}{{\mathscr{U}}}   \newcommand{\sV}{{\mathscr{V}}}
+\newcommand{\sW}{{\mathscr{W}}}   \newcommand{\sX}{{\mathscr{X}}}
+\newcommand{\sY}{{\mathscr{Y}}}   \newcommand{\sZ}{{\mathscr{Z}}}
+
+
+%---------------------------------------
+% Math Fraktur Font
+%---------------------------------------
+
+%Captital Letters
+\newcommand{\mfA}{\mathfrak{A}}	\newcommand{\mfB}{\mathfrak{B}}
+\newcommand{\mfC}{\mathfrak{C}}	\newcommand{\mfD}{\mathfrak{D}}
+\newcommand{\mfE}{\mathfrak{E}}	\newcommand{\mfF}{\mathfrak{F}}
+\newcommand{\mfG}{\mathfrak{G}}	\newcommand{\mfH}{\mathfrak{H}}
+\newcommand{\mfI}{\mathfrak{I}}	\newcommand{\mfJ}{\mathfrak{J}}
+\newcommand{\mfK}{\mathfrak{K}}	\newcommand{\mfL}{\mathfrak{L}}
+\newcommand{\mfM}{\mathfrak{M}}	\newcommand{\mfN}{\mathfrak{N}}
+\newcommand{\mfO}{\mathfrak{O}}	\newcommand{\mfP}{\mathfrak{P}}
+\newcommand{\mfQ}{\mathfrak{Q}}	\newcommand{\mfR}{\mathfrak{R}}
+\newcommand{\mfS}{\mathfrak{S}}	\newcommand{\mfT}{\mathfrak{T}}
+\newcommand{\mfU}{\mathfrak{U}}	\newcommand{\mfV}{\mathfrak{V}}
+\newcommand{\mfW}{\mathfrak{W}}	\newcommand{\mfX}{\mathfrak{X}}
+\newcommand{\mfY}{\mathfrak{Y}}	\newcommand{\mfZ}{\mathfrak{Z}}
+%Small Letters
+\newcommand{\mfa}{\mathfrak{a}}	\newcommand{\mfb}{\mathfrak{b}}
+\newcommand{\mfc}{\mathfrak{c}}	\newcommand{\mfd}{\mathfrak{d}}
+\newcommand{\mfe}{\mathfrak{e}}	\newcommand{\mff}{\mathfrak{f}}
+\newcommand{\mfg}{\mathfrak{g}}	\newcommand{\mfh}{\mathfrak{h}}
+\newcommand{\mfi}{\mathfrak{i}}	\newcommand{\mfj}{\mathfrak{j}}
+\newcommand{\mfk}{\mathfrak{k}}	\newcommand{\mfl}{\mathfrak{l}}
+\newcommand{\mfm}{\mathfrak{m}}	\newcommand{\mfn}{\mathfrak{n}}
+\newcommand{\mfo}{\mathfrak{o}}	\newcommand{\mfp}{\mathfrak{p}}
+\newcommand{\mfq}{\mathfrak{q}}	\newcommand{\mfr}{\mathfrak{r}}
+\newcommand{\mfs}{\mathfrak{s}}	\newcommand{\mft}{\mathfrak{t}}
+\newcommand{\mfu}{\mathfrak{u}}	\newcommand{\mfv}{\mathfrak{v}}
+\newcommand{\mfw}{\mathfrak{w}}	\newcommand{\mfx}{\mathfrak{x}}
+\newcommand{\mfy}{\mathfrak{y}}	\newcommand{\mfz}{\mathfrak{z}}
diff --git a/labs-set-1/macros.tex b/labs-set-1/macros.tex
new file mode 100644
index 0000000..5ebc567
--- /dev/null
+++ b/labs-set-1/macros.tex
@@ -0,0 +1,33 @@
+\newcommand{\eps}{\epsilon}
+\newcommand{\veps}{\varepsilon}
+\newcommand{\Qed}{\begin{flushright}\qed\end{flushright}}
+
+\newcommand{\parinn}{\setlength{\parindent}{1cm}}
+\newcommand{\parinf}{\setlength{\parindent}{0cm}}
+
+% \newcommand{\norm}{\|\cdot\|}
+\newcommand{\inorm}{\norm_{\infty}}
+\newcommand{\opensets}{\{V_{\alpha}\}_{\alpha\in I}}
+\newcommand{\oset}{V_{\alpha}}
+\newcommand{\opset}[1]{V_{\alpha_{#1}}}
+\newcommand{\lub}{\text{lub}}
+\newcommand{\del}[2]{\frac{\partial #1}{\partial #2}}
+\newcommand{\Del}[3]{\frac{\partial^{#1} #2}{\partial^{#1} #3}}
+\newcommand{\deld}[2]{\dfrac{\partial #1}{\partial #2}}
+\newcommand{\Deld}[3]{\dfrac{\partial^{#1} #2}{\partial^{#1} #3}}
+\newcommand{\der}[2]{\frac{\mathrm{d} #1}{\mathrm{d} #2}}
+% \newcommand{\ddd}[3]{\frac{\mathrm{d}^{#3} #1}{\mathrm{d}^{#3} #2}}
+\newcommand{\lm}{\lambda}
+\newcommand{\uin}{\mathbin{\rotatebox[origin=c]{90}{$\in$}}}
+\newcommand{\usubset}{\mathbin{\rotatebox[origin=c]{90}{$\subset$}}}
+\newcommand{\lt}{\left}
+\newcommand{\rt}{\right}
+\newcommand{\bs}[1]{\boldsymbol{#1}}
+\newcommand{\exs}{\exists}
+\newcommand{\st}{\strut}
+\newcommand{\dps}[1]{\displaystyle{#1}}
+\newcommand{\id}{\text{id}}
+
+
+\newcommand{\sol}{\setlength{\parindent}{0cm}\textbf{\textit{Solution:}}\setlength{\parindent}{1cm} }
+\newcommand{\solve}[1]{\setlength{\parindent}{0cm}\textbf{\textit{Solution: }}\setlength{\parindent}{1cm}#1 \Qed}
diff --git a/labs-set-1/preamble.tex b/labs-set-1/preamble.tex
new file mode 100644
index 0000000..5ba904f
--- /dev/null
+++ b/labs-set-1/preamble.tex
@@ -0,0 +1,745 @@
+\usepackage[tmargin=2cm,rmargin=1in,lmargin=1in,margin=0.85in,bmargin=2cm,footskip=.2in]{geometry}
+\usepackage{amsmath,amsfonts,amsthm,amssymb,mathtools}
+\usepackage{gensymb}
+\usepackage[varbb]{newpxmath}
+\usepackage{xfrac}
+\usepackage[makeroom]{cancel}
+\usepackage{mathtools}
+\usepackage{bookmark}
+\usepackage{enumitem}
+\usepackage{hyperref,theoremref}
+\hypersetup{
+	pdftitle={assignment},
+	colorlinks=true, linkcolor=doc!90,
+	bookmarksnumbered=true,
+	bookmarksopen=true
+}
+\usepackage[most,many,breakable]{tcolorbox}
+\usepackage{xcolor}
+\usepackage{varwidth}
+\usepackage{varwidth}
+\usepackage{etoolbox}
+%\usepackage{authblk}
+\usepackage{nameref}
+\usepackage{multicol,array}
+\usepackage[ruled,vlined,linesnumbered]{algorithm2e}
+\usepackage{comment} % enables the use of multi-line comments (\ifx \fi) 
+\usepackage{import}
+\usepackage{xifthen}
+\usepackage{pdfpages}
+\usepackage{transparent}
+\usepackage{chngcntr}
+\usepackage{tikz}
+\usepackage{titletoc}
+
+\newcommand\mycommfont[1]{\footnotesize\ttfamily\textcolor{blue}{#1}}
+\SetCommentSty{mycommfont}
+\newcommand{\incfig}[1]{%
+    \def\svgwidth{\columnwidth}
+    \import{./figures/}{#1.pdf_tex}
+}
+
+\usepackage{tikzsymbols}
+\tikzset{
+	symbol/.style={
+			draw=none,
+			every to/.append style={
+					edge node={node [sloped, allow upside down, auto=false]{$#1$}}}
+		}
+}
+\tikzstyle{c} = [circle,fill=black,scale=0.5]
+\tikzstyle{b} = [draw, thick, black, -]
+\tikzset{
+    vertex/.style={
+        circle,
+        draw,
+        minimum size=6mm,
+        inner sep=0pt
+    }
+}
+
+\renewcommand\qedsymbol{$\Laughey$}
+
+% \usepackage{xparse}
+% \usepackage{pgffor}
+% \usepackage{emoji}
+%
+% \NewDocumentCommand{\memoji}{m o}{%
+%   \IfNoValueTF{#2}
+%     {\ifmmode \text{\emoji{#1}} \else \emoji{#1}\fi} % Single emoji case
+%     {\foreach \index in {1,...,#2}{\ifmmode \text{\emoji{#1}} \else \emoji{#1}\fi}} % Repeated emoji case
+% }
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% SELF MADE COLORS
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\definecolor{doc}{RGB}{0,60,110}
+\definecolor{myg}{RGB}{56, 140, 70}
+\definecolor{myb}{RGB}{45, 111, 177}
+\definecolor{myr}{RGB}{199, 68, 64}
+\definecolor{mytheorembg}{HTML}{F2F2F9}
+\definecolor{mytheoremfr}{HTML}{00007B}
+\definecolor{mylemmabg}{HTML}{FFFAF8}
+\definecolor{mylemmafr}{HTML}{983b0f}
+\definecolor{mypropbg}{HTML}{f2fbfc}
+\definecolor{mypropfr}{HTML}{191971}
+\definecolor{myexamplebg}{HTML}{F2FBF8}
+\definecolor{myexamplefr}{HTML}{88D6D1}
+\definecolor{myexampleti}{HTML}{2A7F7F}
+\definecolor{mydefinitbg}{HTML}{E5E5FF}
+\definecolor{mydefinitfr}{HTML}{3F3FA3}
+\definecolor{notesgreen}{RGB}{0,162,0}
+\definecolor{myp}{RGB}{197, 92, 212}
+\definecolor{mygr}{HTML}{2C3338}
+\definecolor{myred}{RGB}{127,0,0}
+\definecolor{myyellow}{RGB}{169,121,69}
+\definecolor{myexercisebg}{HTML}{F2FBF8}
+\definecolor{myexercisefg}{HTML}{88D6D1}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% TCOLORBOX SETUPS
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\setlength{\parindent}{1cm}
+%================================
+% THEOREM BOX
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Theorem}{Theorem}
+{%
+	enhanced,
+	breakable,
+	colback = mytheorembg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mytheoremfr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mytheoremfr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mytheoremfr},
+}
+{th}
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{theorem}{Theorem}
+{%
+	enhanced,
+	breakable,
+	colback = mytheorembg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mytheoremfr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mytheoremfr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mytheoremfr},
+}
+{th}
+
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcolorbox{Theoremcon}
+{%
+	enhanced
+	,breakable
+	,colback = mytheorembg
+	,frame hidden
+	,boxrule = 0sp
+	,borderline west = {2pt}{0pt}{mytheoremfr}
+	,sharp corners
+	,description font = \mdseries
+	,separator sign none
+}
+
+%================================
+% Corollery
+%================================
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Corollary}{Corollary}
+{%
+	enhanced
+	,breakable
+	,colback = myp!10
+	,frame hidden
+	,boxrule = 0sp
+	,borderline west = {2pt}{0pt}{myp!85!black}
+	,sharp corners
+	,detach title
+	,before upper = \tcbtitle\par\smallskip
+	,coltitle = myp!85!black
+	,fonttitle = \bfseries\sffamily
+	,description font = \mdseries
+	,separator sign none
+	,segmentation style={solid, myp!85!black}
+}
+{th}
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{corollary}{Corollary}
+{%
+	enhanced
+	,breakable
+	,colback = myp!10
+	,frame hidden
+	,boxrule = 0sp
+	,borderline west = {2pt}{0pt}{myp!85!black}
+	,sharp corners
+	,detach title
+	,before upper = \tcbtitle\par\smallskip
+	,coltitle = myp!85!black
+	,fonttitle = \bfseries\sffamily
+	,description font = \mdseries
+	,separator sign none
+	,segmentation style={solid, myp!85!black}
+}
+{th}
+
+
+%================================
+% LEMMA
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Lemma}{Lemma}
+{%
+	enhanced,
+	breakable,
+	colback = mylemmabg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mylemmafr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mylemmafr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mylemmafr},
+}
+{th}
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{lemma}{lemma}
+{%
+	enhanced,
+	breakable,
+	colback = mylemmabg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mylemmafr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mylemmafr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mylemmafr},
+}
+{th}
+
+%================================
+% Exercise
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Exercise}{Exercise}
+{%
+	enhanced,
+	breakable,
+	colback = myexercisebg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{myexercisefg},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = myexercisefg,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, myexercisefg},
+}
+{th}
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{exercise}{Exercise}
+{%
+	enhanced,
+	breakable,
+	colback = myexercisebg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{myexercisefg},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = myexercisefg,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, myexercisefg},
+}
+{th}
+
+
+%================================
+% PROPOSITION
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{Prop}{Proposition}
+{%
+	enhanced,
+	breakable,
+	colback = mypropbg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mypropfr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mypropfr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mypropfr},
+}
+{th}
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=chapter]{prop}{Proposition}
+{%
+	enhanced,
+	breakable,
+	colback = mypropbg,
+	frame hidden,
+	boxrule = 0sp,
+	borderline west = {2pt}{0pt}{mypropfr},
+	sharp corners,
+	detach title,
+	before upper = \tcbtitle\par\smallskip,
+	coltitle = mypropfr,
+	fonttitle = \bfseries\sffamily,
+	description font = \mdseries,
+	separator sign none,
+	segmentation style={solid, mypropfr},
+}
+{th}
+
+
+%================================
+% CLAIM
+%================================
+
+\tcbuselibrary{theorems,skins,hooks}
+\newtcbtheorem[number within=section]{claim}{Claim}
+{%
+	enhanced
+	,breakable
+	,colback = myg!10
+	,frame hidden
+	,boxrule = 0sp
+	,borderline west = {2pt}{0pt}{myg}
+	,sharp corners
+	,detach title
+	,before upper = \tcbtitle\par\smallskip
+	,coltitle = myg!85!black
+	,fonttitle = \bfseries\sffamily
+	,description font = \mdseries
+	,separator sign none
+	,segmentation style={solid, myg!85!black}
+}
+{th}
+
+
+
+%================================
+% EXAMPLE BOX
+%================================
+
+\newtcbtheorem[number within=section]{Example}{Example}
+{%
+	colback = myexamplebg
+	,breakable
+	,colframe = myexamplefr
+	,coltitle = myexampleti
+	,boxrule = 1pt
+	,sharp corners
+	,detach title
+	,before upper=\tcbtitle\par\smallskip
+	,fonttitle = \bfseries
+	,description font = \mdseries
+	,separator sign none
+	,description delimiters parenthesis
+}
+{ex}
+
+\newtcbtheorem[number within=chapter]{example}{Example}
+{%
+	colback = myexamplebg
+	,breakable
+	,colframe = myexamplefr
+	,coltitle = myexampleti
+	,boxrule = 1pt
+	,sharp corners
+	,detach title
+	,before upper=\tcbtitle\par\smallskip
+	,fonttitle = \bfseries
+	,description font = \mdseries
+	,separator sign none
+	,description delimiters parenthesis
+}
+{ex}
+
+%================================
+% DEFINITION BOX
+%================================
+
+\newtcbtheorem[number within=section]{Definition}{Definition}{enhanced,
+	before skip=2mm,after skip=2mm, colback=red!5,colframe=red!80!black,boxrule=0.5mm,
+	attach boxed title to top left={xshift=1cm,yshift*=1mm-\tcboxedtitleheight}, varwidth boxed title*=-3cm,
+	boxed title style={frame code={
+					\path[fill=tcbcolback]
+					([yshift=-1mm,xshift=-1mm]frame.north west)
+					arc[start angle=0,end angle=180,radius=1mm]
+					([yshift=-1mm,xshift=1mm]frame.north east)
+					arc[start angle=180,end angle=0,radius=1mm];
+					\path[left color=tcbcolback!60!black,right color=tcbcolback!60!black,
+						middle color=tcbcolback!80!black]
+					([xshift=-2mm]frame.north west) -- ([xshift=2mm]frame.north east)
+					[rounded corners=1mm]-- ([xshift=1mm,yshift=-1mm]frame.north east)
+					-- (frame.south east) -- (frame.south west)
+					-- ([xshift=-1mm,yshift=-1mm]frame.north west)
+					[sharp corners]-- cycle;
+				},interior engine=empty,
+		},
+	fonttitle=\bfseries,
+	title={#2},#1}{def}
+\newtcbtheorem[number within=chapter]{definition}{Definition}{enhanced,
+	before skip=2mm,after skip=2mm, colback=red!5,colframe=red!80!black,boxrule=0.5mm,
+	attach boxed title to top left={xshift=1cm,yshift*=1mm-\tcboxedtitleheight}, varwidth boxed title*=-3cm,
+	boxed title style={frame code={
+					\path[fill=tcbcolback]
+					([yshift=-1mm,xshift=-1mm]frame.north west)
+					arc[start angle=0,end angle=180,radius=1mm]
+					([yshift=-1mm,xshift=1mm]frame.north east)
+					arc[start angle=180,end angle=0,radius=1mm];
+					\path[left color=tcbcolback!60!black,right color=tcbcolback!60!black,
+						middle color=tcbcolback!80!black]
+					([xshift=-2mm]frame.north west) -- ([xshift=2mm]frame.north east)
+					[rounded corners=1mm]-- ([xshift=1mm,yshift=-1mm]frame.north east)
+					-- (frame.south east) -- (frame.south west)
+					-- ([xshift=-1mm,yshift=-1mm]frame.north west)
+					[sharp corners]-- cycle;
+				},interior engine=empty,
+		},
+	fonttitle=\bfseries,
+	title={#2},#1}{def}
+
+
+%================================
+% EXERCISE BOX
+%================================
+
+\newcounter{questioncounter}
+\counterwithin{questioncounter}{chapter}
+% \counterwithin{questioncounter}{section}
+
+\makeatletter
+\newtcbtheorem[use counter=questioncounter]{question}{Question}{enhanced,
+	breakable,
+	colback=white,
+	colframe=myb!80!black,
+	attach boxed title to top left={yshift*=-\tcboxedtitleheight},
+	fonttitle=\bfseries,
+	title={#2},
+	boxed title size=title,
+	boxed title style={%
+			sharp corners,
+			rounded corners=northwest,
+			colback=tcbcolframe,
+			boxrule=0pt,
+		},
+	underlay boxed title={%
+			\path[fill=tcbcolframe] (title.south west)--(title.south east)
+			to[out=0, in=180] ([xshift=5mm]title.east)--
+			(title.center-|frame.east)
+			[rounded corners=\kvtcb@arc] |-
+			(frame.north) -| cycle;
+		},
+	#1
+}{def}
+\makeatother
+
+%================================
+% SOLUTION BOX
+%================================
+
+\makeatletter
+\newtcolorbox{solution}{enhanced,
+	breakable,
+	colback=white,
+	colframe=myg!80!black,
+	attach boxed title to top left={yshift*=-\tcboxedtitleheight},
+	title=Solution,
+	boxed title size=title,
+	boxed title style={%
+			sharp corners,
+			rounded corners=northwest,
+			colback=tcbcolframe,
+			boxrule=0pt,
+		},
+	underlay boxed title={%
+			\path[fill=tcbcolframe] (title.south west)--(title.south east)
+			to[out=0, in=180] ([xshift=5mm]title.east)--
+			(title.center-|frame.east)
+			[rounded corners=\kvtcb@arc] |-
+			(frame.north) -| cycle;
+		},
+}
+\makeatother
+
+%================================
+% Question BOX
+%================================
+
+\makeatletter
+\newtcbtheorem{qstion}{Question}{enhanced,
+	breakable,
+	colback=white,
+	colframe=mygr,
+	attach boxed title to top left={yshift*=-\tcboxedtitleheight},
+	fonttitle=\bfseries,
+	title={#2},
+	boxed title size=title,
+	boxed title style={%
+			sharp corners,
+			rounded corners=northwest,
+			colback=tcbcolframe,
+			boxrule=0pt,
+		},
+	underlay boxed title={%
+			\path[fill=tcbcolframe] (title.south west)--(title.south east)
+			to[out=0, in=180] ([xshift=5mm]title.east)--
+			(title.center-|frame.east)
+			[rounded corners=\kvtcb@arc] |-
+			(frame.north) -| cycle;
+		},
+	#1
+}{def}
+\makeatother
+
+\newtcbtheorem[number within=chapter]{wconc}{Wrong Concept}{
+	breakable,
+	enhanced,
+	colback=white,
+	colframe=myr,
+	arc=0pt,
+	outer arc=0pt,
+	fonttitle=\bfseries\sffamily\large,
+	colbacktitle=myr,
+	attach boxed title to top left={},
+	boxed title style={
+			enhanced,
+			skin=enhancedfirst jigsaw,
+			arc=3pt,
+			bottom=0pt,
+			interior style={fill=myr}
+		},
+	#1
+}{def}
+
+
+%================================
+% NOTE BOX
+%================================
+
+\usetikzlibrary{arrows,calc,shadows.blur}
+\tcbuselibrary{skins}
+\newtcolorbox{note}[1][]{%
+	enhanced jigsaw,
+	colback=gray!20!white,%
+	colframe=gray!80!black,
+	size=small,
+	boxrule=1pt,
+	title=\textbf{Note:-},
+	halign title=flush center,
+	coltitle=black,
+	breakable,
+	drop shadow=black!50!white,
+	attach boxed title to top left={xshift=1cm,yshift=-\tcboxedtitleheight/2,yshifttext=-\tcboxedtitleheight/2},
+	minipage boxed title=1.5cm,
+	boxed title style={%
+			colback=white,
+			size=fbox,
+			boxrule=1pt,
+			boxsep=2pt,
+			underlay={%
+					\coordinate (dotA) at ($(interior.west) + (-0.5pt,0)$);
+					\coordinate (dotB) at ($(interior.east) + (0.5pt,0)$);
+					\begin{scope}
+						\clip (interior.north west) rectangle ([xshift=3ex]interior.east);
+						\filldraw [white, blur shadow={shadow opacity=60, shadow yshift=-.75ex}, rounded corners=2pt] (interior.north west) rectangle (interior.south east);
+					\end{scope}
+					\begin{scope}[gray!80!black]
+						\fill (dotA) circle (2pt);
+						\fill (dotB) circle (2pt);
+					\end{scope}
+				},
+		},
+	#1,
+}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% SELF MADE COMMANDS
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\newcommand{\thm}[2]{\begin{Theorem}{#1}{}#2\end{Theorem}}
+\newcommand{\cor}[2]{\begin{Corollary}{#1}{}#2\end{Corollary}}
+\newcommand{\mlemma}[2]{\begin{Lemma}{#1}{}#2\end{Lemma}}
+\newcommand{\mer}[2]{\begin{Exercise}{#1}{}#2\end{Exercise}}
+\newcommand{\mprop}[2]{\begin{Prop}{#1}{}#2\end{Prop}}
+\newcommand{\clm}[3]{\begin{claim}{#1}{#2}#3\end{claim}}
+\newcommand{\wc}[2]{\begin{wconc}{#1}{}\setlength{\parindent}{1cm}#2\end{wconc}}
+\newcommand{\thmcon}[1]{\begin{Theoremcon}{#1}\end{Theoremcon}}
+\newcommand{\ex}[2]{\begin{Example}{#1}{}#2\end{Example}}
+\newcommand{\dfn}[2]{\begin{Definition}[colbacktitle=red!75!black]{#1}{}#2\end{Definition}}
+\newcommand{\dfnc}[2]{\begin{definition}[colbacktitle=red!75!black]{#1}{}#2\end{definition}}
+\newcommand{\qs}[2]{\begin{question}{#1}{}#2\end{question}}
+\newcommand{\pf}[2]{\begin{myproof}[#1]#2\end{myproof}}
+\newcommand{\nt}[1]{\begin{note}#1\end{note}}
+
+\newcommand*\circled[1]{\tikz[baseline=(char.base)]{
+		\node[shape=circle,draw,inner sep=1pt] (char) {#1};}}
+\newcommand\getcurrentref[1]{%
+	\ifnumequal{\value{#1}}{0}
+	{??}
+	{\the\value{#1}}%
+}
+\newcommand{\getCurrentSectionNumber}{\getcurrentref{section}}
+\newenvironment{myproof}[1][\proofname]{%
+	\proof[\bfseries #1: ]%
+}{\endproof}
+
+\newcommand{\mclm}[2]{\begin{myclaim}[#1]#2\end{myclaim}}
+\newenvironment{myclaim}[1][\claimname]{\proof[\bfseries #1: ]}{}
+\newenvironment{iclaim}[1][\claimname]{\bfseries #1\mdseries:}{}
+\newcommand{\iclm}[2]{\begin{iclaim}[#1]#2\end{iclaim}}
+
+\newcounter{mylabelcounter}
+
+\makeatletter
+\newcommand{\setword}[2]{%
+	\phantomsection
+	#1\def\@currentlabel{\unexpanded{#1}}\label{#2}%
+}
+\makeatother
+
+% deliminators
+\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
+\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
+
+\DeclarePairedDelimiter{\ceil}{\lceil}{\rceil}
+\DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor}
+\DeclarePairedDelimiter{\round}{\lfloor}{\rceil}
+
+\newsavebox\diffdbox
+\newcommand{\slantedromand}{{\mathpalette\makesl{d}}}
+\newcommand{\makesl}[2]{%
+\begingroup
+\sbox{\diffdbox}{$\mathsurround=0pt#1\mathrm{#2}$}%
+\pdfsave
+\pdfsetmatrix{1 0 0.2 1}%
+\rlap{\usebox{\diffdbox}}%
+\pdfrestore
+\hskip\wd\diffdbox
+\endgroup
+}
+\newcommand{\dd}[1][]{\ensuremath{\mathop{}\!\ifstrempty{#1}{%
+\slantedromand\@ifnextchar^{\hspace{0.2ex}}{\hspace{0.1ex}}}%
+{\slantedromand\hspace{0.2ex}^{#1}}}}
+\ProvideDocumentCommand\dv{o m g}{%
+  \ensuremath{%
+    \IfValueTF{#3}{%
+      \IfNoValueTF{#1}{%
+        \frac{\dd #2}{\dd #3}%
+      }{%
+        \frac{\dd^{#1} #2}{\dd #3^{#1}}%
+      }%
+    }{%
+      \IfNoValueTF{#1}{%
+        \frac{\dd}{\dd #2}%
+      }{%
+        \frac{\dd^{#1}}{\dd #2^{#1}}%
+      }%
+    }%
+  }%
+}
+\providecommand*{\pdv}[3][]{\frac{\partial^{#1}#2}{\partial#3^{#1}}}
+%  - others
+\DeclareMathOperator{\Lap}{\mathcal{L}}
+\DeclareMathOperator{\Var}{Var} % varience
+\DeclareMathOperator{\Cov}{Cov} % covarience
+\DeclareMathOperator{\E}{E} % expected
+
+% Since the amsthm package isn't loaded
+
+% I prefer the slanted \leq
+\let\oldleq\leq % save them in case they're every wanted
+\let\oldgeq\geq
+\renewcommand{\leq}{\leqslant}
+\renewcommand{\geq}{\geqslant}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+% TABLE OF CONTENTS
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\contentsmargin{0cm}
+\titlecontents{chapter}[3.7pc]
+{\addvspace{30pt}%
+	\begin{tikzpicture}[remember picture, overlay]%
+		\draw[fill=doc!60,draw=doc!60] (-7,-.1) rectangle (-0.9,.5);%
+		\pgftext[left,x=-3.7cm,y=0.2cm]{\color{white}\Large\sc\bfseries Chapter\ \thecontentslabel};%
+	\end{tikzpicture}\color{doc!60}\large\sc\bfseries}%
+{}
+{}
+{\;\titlerule\;\large\sc\bfseries Page \thecontentspage
+	\begin{tikzpicture}[remember picture, overlay]
+		\draw[fill=doc!60,draw=doc!60] (2pt,0) rectangle (4,0.1pt);
+	\end{tikzpicture}}%
+\titlecontents{section}[3.7pc]
+{\addvspace{2pt}}
+{\contentslabel[\thecontentslabel]{2pc}}
+{}
+{\hfill\small \thecontentspage}
+[]
+\titlecontents*{subsection}[3.7pc]
+{\addvspace{-1pt}\small}
+{}
+{}
+{\ --- \small\thecontentspage}
+[ \textbullet\ ][]
+
+\makeatletter
+\renewcommand{\tableofcontents}{%
+	\chapter*{%
+	  \vspace*{-20\p@}%
+	  \begin{tikzpicture}[remember picture, overlay]%
+		  \pgftext[right,x=15cm,y=0.2cm]{\color{doc!60}\Huge\sc\bfseries \contentsname};%
+		  \draw[fill=doc!60,draw=doc!60] (13,-.75) rectangle (20,1);%
+		  \clip (13,-.75) rectangle (20,1);
+		  \pgftext[right,x=15cm,y=0.2cm]{\color{white}\Huge\sc\bfseries \contentsname};%
+	  \end{tikzpicture}}%
+	\@starttoc{toc}}
+\makeatother
diff --git a/labs-set-1/template.pdf b/labs-set-1/template.pdf
new file mode 100644
index 0000000..1b7e841
Binary files /dev/null and b/labs-set-1/template.pdf differ
diff --git a/labs-set-1/template.tex b/labs-set-1/template.tex
new file mode 100644
index 0000000..596f88b
--- /dev/null
+++ b/labs-set-1/template.tex
@@ -0,0 +1,44 @@
+\documentclass{report}
+
+\input{preamble}
+\input{macros}
+\input{letterfonts}
+
+\title{\Huge{Calculus III}\\Homework \# 1}
+\author{\huge{Krishna Ayyalasomayajula}}
+\date{}
+
+\begin{document}
+
+\maketitle
+\newpage% or \cleardoublepage
+% \pdfbookmark[<level>]{<title>}{<dest>}
+\pdfbookmark[section]{\contentsname}{toc}
+\tableofcontents
+\pagebreak
+
+\chapter{Lab 2 - 13.1 Apps}
+\section{Work}
+
+\qs{}{
+  Let $\vec{T_1}$ represent the tension of the leftmost cable, while $\vec{T_2}$ encodes the tension force experienced by the rightmost cable. Our coordinate system will originate at the intersection of the two cables.
+
+  \begin{align*}
+    \vec{T_1} \coloneqq \langle \|\vec{T_1}\| ; 135\degree \rangle \\
+    \vec{T_2} \coloneqq \langle \|\vec{T_2}|\| ; 15\degree \rangle \\
+    500=\|\vec{T_1}\|\sin{135\degree} + \|\vec{T_2}\|\sin{15} \\
+    0 = \|\vec{T_1}\|\cos{135\degree} + \|\vec{T_2}\|\cos{15} \\
+    \implies 500=\|\vec{T_1}\|\tfrac{\sqrt{2}}{2} + \|\vec{T_2}\|\cdot0.258819045103 \\
+    \implies0 = \|\vec{T_1}\|\tfrac{-\sqrt{2}}{2} + \|\vec{T_2}\|\cdot0.965925826289 \\
+    \text{Solving the system numerically yields: }\\
+    \|\vec{T_1}\| \Rightarrow 557.677\, \mathrm{lb}\\
+    \|\vec{T_2}\| \Rightarrow 408.248\,\mathrm{lb} \\
+    \text{This corresponds to answer choice A}
+  \end{align*}
+  
+}
+
+\qs{}{
+
+}
+\end{document}
diff --git a/labs-set-1/template.toc b/labs-set-1/template.toc
new file mode 100644
index 0000000..853a91d
--- /dev/null
+++ b/labs-set-1/template.toc
@@ -0,0 +1,3 @@
+\contentsline {chapter}{\numberline {1}Lab 2 - 13.1 Apps}{2}{chapter.1}%
+\contentsline {section}{\numberline {1.1}Work}{2}{section.1.1}%
+\contentsfinish