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diff --git a/basic_python/control_flow.tex b/basic_python/control_flow.tex new file mode 100644 index 0000000..f53e028 --- /dev/null +++ b/basic_python/control_flow.tex @@ -0,0 +1,373 @@ +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%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} |