From 809715c9c344092fc5fb1709f29be3ee27f7e7d6 Mon Sep 17 00:00:00 2001
From: Prabhu Ramachandran
Date: Sun, 22 Jan 2017 16:44:02 +0530
Subject: Add initial slides for basic python.

---
 basic_python/control_flow.tex | 373 ++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 373 insertions(+)
 create mode 100644 basic_python/control_flow.tex

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diff --git a/basic_python/control_flow.tex b/basic_python/control_flow.tex
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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}
+
+\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{b = next}
+  \item \typ{a = b}
+  \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{\typ{continue} example}
+Try this:
+\begin{lstlisting}
+a = 0
+while a < 10:
+    if a % 2 == 0:
+        continue
+    print(a, end=' ')
+    a += 1
+\end{lstlisting}
+Carefully notice what it does
+\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, while}
+    \item Solving simple problems
+  \end{itemize}
+\end{frame}
+
+\end{document}
-- 
cgit 

