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-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-%Tutorial slides on Python.
-%
-% Author: FOSSEE
-% Copyright (c) 2009-2017, FOSSEE, IIT Bombay
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-
-\documentclass[14pt,compress]{beamer}
-
-\input{macros.tex}
-
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-% Title page
-\title[Control flow]{Python language: Control Flow}
-
-\author[FOSSEE Team] {The FOSSEE Group}
-
-\institute[FOSSEE -- IITB] {Department of Aerospace Engineering\\IIT Bombay}
-\date[] {Mumbai, India}
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-
-
-%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-% DOCUMENT STARTS
-\begin{document}
-
-\begin{frame}
-  \titlepage
-\end{frame}
-
-\begin{frame}
-  \frametitle{Outline}
-  \tableofcontents
-  % You might wish to add the option [pausesections]
-\end{frame}
-
-\section{Control flow}
-
-\begin{frame}
-  \frametitle{Control flow constructs}
-  \begin{itemize}
-  \item \kwrd{if/elif/else}: branching
-  \item \kwrd{while}: looping
-  \item \kwrd{for}: iterating
-  \item \kwrd{break, continue}: modify loop
-  \item \kwrd{pass}: syntactic filler
-  \end{itemize}
-\end{frame}
-
-\subsection{Basic Conditional flow}
-\begin{frame}[fragile]
-  \frametitle{\typ{if...elif...else} example}
-Type the following code in an editor \& save as \alert{ladder.py}
-{  \small
-\begin{lstlisting}
-x = int(input("Enter an integer: "))
-if x < 0:
-    print('Be positive!')
-elif x == 0:
-    print('Zero')
-elif x == 1:
-    print('Single')
-else:
-    print('More')
-\end{lstlisting}
-}
-  %% \inctime{10}
-\pause
-\begin{itemize}
-\item Run in IPython: \typ{\%run ladder.py}
-\item Run on terminal: \typ{python ladder.py}
-\end{itemize}
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{Ternary operator}
-  \begin{itemize}
-  \item \texttt{score\_str} is either \texttt{'AA'} or a string of one
-    of the numbers in the range 0 to 100.
-  \item We wish to convert the string to a number using \texttt{int}
-  \item Convert it to 0, when it is \texttt{'AA'}
-  \item \texttt{if-else} construct or the ternary operator
-  \end{itemize}
-  \begin{lstlisting}
-In []: if score_str != 'AA':
-.....:     score = int(score_str)
-.....: else:
-.....:     score = 0
-  \end{lstlisting}
-\end{frame}
-\begin{frame}[fragile]
-  \frametitle{Ternary operator}
-  With the ternary operator you can do this:
-  \small
-  \begin{lstlisting}
-In []: ss = score_str
-In []: score = int(ss) if ss != 'AA' else 0
-  \end{lstlisting}
-\end{frame}
-
-
-\section{Control flow}
-\subsection{Basic Looping}
-
-\begin{frame}
-  \frametitle{\typ{while}: motivational problem}
-\begin{block}{Example: Fibonacci series}
-  Sum of previous two elements defines the next:
-  \begin{center}
-    $0,\ \ 1,\ \ 1,\ \ 2,\ \ 3,\ \ 5,\ \ 8,\ \ 13,\ \ 21,\ \ ...$
-  \end{center}
-\end{block}
-\end{frame}
-
-\begin{frame}
-  \frametitle{How do you solve this?}
-  \begin{itemize}
-  \item Task: Give computer, instructions to solve this
-    \vspace*{2em}
-  \item How would you solve it?
-  \item How would you tell someone to solve it?
-    \vspace*{1em}
-  \item Assume you are given the starting values, 0 and 1.
-  \end{itemize}
-
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{\typ{while}: Fibonacci}
-\begin{block}{Example: Fibonacci series}
-  Sum of previous two elements defines the next:
-  \begin{center}
-    $0,\ \ 1,\ \ 1,\ \ 2,\ \ 3,\ \ 5,\ \ 8,\ \ 13,\ \ 21,\ \ ...$
-  \end{center}
-\end{block}
-\pause
-\begin{enumerate}
-\item Start with: \typ{a, b = 0, 1}
-  \pause
-\item Next element: \typ{next = a + b}
-  \pause
-\item Shift \typ{a, b} to next values
-  \begin{itemize}
-  \item \typ{a = b}
-  \item \typ{b = next}
-  \end{itemize}
-  \pause
-\item Repeat steps 2, 3 when \typ{b < 30}
-\end{enumerate}
-\end{frame}
-
-
-\begin{frame}[fragile]
-  \frametitle{\typ{while}}
-  \begin{lstlisting}
-In []: a, b = 0, 1
-In []: while b < 30:
-  ...:     print(b, end=' ')
-  ...:     next = a + b
-  ...:     a = b
-  ...:     b = next
-  ...:
-  ...:
-\end{lstlisting}
-  \begin{itemize}
-  \item Do this manually to check logic
-  \item Note: Indentation determines scope
-  \end{itemize}
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{\typ{while}}
-  We can eliminate the temporary \typ{next}:
-  \begin{lstlisting}
-In []: a, b = 0, 1
-In []: while b < 30:
-  ...:     print(b, end=' ')
-  ...:     a, b = b, a + b
-  ...:
-  ...:
-1 1 2 3 5 8 13 21
-\end{lstlisting}
-  Simple!
-\end{frame}
-
-
-\begin{frame}[fragile]
-  \frametitle{\typ{for} \ldots \typ{range()}}
-Example: print squares of first \typ{5} numbers
-  \begin{lstlisting}
-In []: for i in range(5):
- ....:     print(i, i * i)
- ....:
- ....:
-0 0
-1 1
-2 4
-3 9
-4 16
-\end{lstlisting}
-\end{frame}
-
-\begin{frame}[fragile]
-\frametitle{\typ{range()}}
-\kwrd{range([start,] stop[, step])}\\
-\begin{itemize}
-  \item \typ{range()} returns a sequence of integers
-  \item The \typ{start} and the \typ{step} arguments are optional
-  \item \typ{stop} is not included in the sequence
-\end{itemize}
-\vspace*{.5in}
-\begin{block}{Documentation convention}
-  \begin{itemize}
-    \item \alert{Anything within \typ{[]} is optional}
-    \item Nothing to do with Python
-  \end{itemize}
-\end{block}
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{\typ{for} \ldots \typ{range()}}
-Example: print squares of odd numbers from 3 to 9
-  \begin{lstlisting}
-In []: for i in range(3, 10, 2):
- ....:     print(i, i * i)
- ....:
- ....:
-3 9
-5 25
-7 49
-9 81
-\end{lstlisting}
-%% \inctime{5}
-\end{frame}
-
-\begin{frame}
-  \frametitle{Exercise with \typ{for}}
-
-Convert the Fibonnaci sequence example to use a \typ{for} loop with range.
-
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{Solution}
-\begin{lstlisting}
-a, b = 0, 1
-for i in range(10):
-    print(b, end=' ')
-    a, b = b, a + b
-\end{lstlisting}
-  \vspace*{2em}
-Note that the \typ{while} loop is a more natural fit here
-\end{frame}
-
-\begin{frame}
-  \frametitle{\typ{break}, \typ{continue}, and\  \typ{pass}}
-  \begin{itemize}
-  \item Use \typ{break} to break out of loop
-  \item Use \typ{continue} to skip an iteration
-  \item Use \typ{pass} as syntactic filler
-  \end{itemize}
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{\typ{break} example}
-  Find first number in Fibonnaci sequence < 100 divisible by 4:
-\begin{lstlisting}
-a, b = 0, 1
-while b < 500:
-    if b % 4 == 0:
-        print(b)
-        break
-    a, b = b, a + b
-
-\end{lstlisting}
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{\texttt{continue}}
-  \begin{itemize}
-  \item Skips execution of rest of the loop on current iteration
-  \item Jumps to the end of this iteration
-  \item Squares of all odd numbers below 10, not multiples of 3
-  \end{itemize}
-  \begin{lstlisting}
-  In []: for n in range(1, 10, 2):
-  .....:     if n%3 == 0:
-  .....:         continue
-  .....:     print(n*n)
-  \end{lstlisting}
-\end{frame}
-
-
-\begin{frame}[fragile]
-  \frametitle{\typ{pass} example}
-Try this:
-\begin{lstlisting}
-for i in range(5):
-    if i % 2 ==  0:
-        pass
-    else:
-        print(i, 'is Odd')
-\end{lstlisting}
-\begin{itemize}
-\item \typ{pass}: does nothing
-\item Keep Python syntactically happy
-\end{itemize}
-Another example:
-\begin{lstlisting}
-while True:
-    pass
-\end{lstlisting}
-\end{frame}
-
-
-
-\section{Exercises}
-
-\begin{frame}{Problem 1.1: \emph{Armstrong} numbers}
-  Write a program that displays all three digit numbers that are equal to the sum of the cubes of their digits. That is, print numbers $abc$ that have the property $abc = a^3 + b^3 + c^3$\\
-For example, $153 = 1^3 + 5^3 + 3^3$\\
-\vspace*{0.2in}
-
-\begin{block}{Hints}
-  \begin{itemize}
-  \item Break problem into easier pieces
-  \item How would you solve the problem?
-  \item Can you explain to someone else how to solve it?
-  \end{itemize}
-\end{block}
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{Some hints}
-  \begin{itemize}
-  \item What are the possible three digit numbers?
-  \item Can you split 153 into its respective digits, $a=1, b=5, c=3$?
-  \item With $a, b, c$ can you test if it is an Armstrong number?
-  \end{itemize}
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{Solution: part 1}
-\begin{lstlisting}
-x = 153
-a = x//100
-b = (x%100)//10
-c = x%10
-
-(a**3 + b**3 + c**3) == x
-\end{lstlisting}
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{Solution: part 2}
-\begin{lstlisting}
-x = 100
-while x < 1000:
-    print(x)
-    x += 1  # x = x + 1
-\end{lstlisting}
-\end{frame}
-
-\begin{frame}[fragile]
-  \frametitle{Solution}
-\begin{lstlisting}
-x = 100
-while x < 1000:
-    a = x//100
-    b = (x%100)//10
-    c = x%10
-    if (a**3 + b**3 + c**3) == x:
-        print(x)
-    x += 1
-\end{lstlisting}
-\end{frame}
-
-
-\begin{frame}{Problem 1.2: Collatz sequence}
-\begin{enumerate}
-  \item Start with an arbitrary (positive) integer.
-  \item If the number is even, divide by 2; if the number is odd, multiply by 3 and add 1.
-  \item Repeat the procedure with the new number.
-  \item It appears that for all starting values there is a cycle of 4, 2, 1 at which the procedure loops.
-\end{enumerate}
-    Write a program that accepts the starting value and prints out the Collatz sequence.
-%% \inctime{5}
-\end{frame}
-
-
-\begin{frame}[fragile]
-  \frametitle{What did we learn?}
-  \begin{itemize}
-    \item Conditionals: \kwrd{if elif else}
-    \item Looping: \kwrd{while} \& \kwrd{for}
-    \item \typ{range}
-    \item \kwrd{break, continue, pass}
-    \item Solving simple problems
-  \end{itemize}
-\end{frame}
-
-\end{document}