finished 2.2-2

Foundations/overview/README.md

@@ -131,5 +131,11 @@ appear in A.
 ### 2.2-2
 > Consider sorting `n` numbers stored in array `A` by first finding the smallest element of `A` and exchanging it with the element in `A[1]`. Then find the second smallest element of `A`, and exchange it with `A[2]`􏰃. Continue in this manner for the first `n-1` elements of `A`. Write pseudocode for this algorithm, which is known as **selection sort**. What loop invariant does this algorithm maintain? Why does it need to run for only the first `n-1` elements, rather than for all `n` elements? Give the best-case and worst-case running times of selection sort in O-notation.
 
-[pseudocode](selection-sort.pdf) | [code](insertion_sort.py) |
+**loop invariant**: At the start of each iteration `for` loop, the subarray `A[1..j-1]` consists of the smallest elements of the array, and in sorted order.
+
+Because we always find the smallest element of `A`, when the first `n-1` elements in sorted order. The last element must bigger than before. Thus, we don't need to run for all `n` elements.
+
+[O-notation and pseudocode](selection-sort.pdf) | [code](insertion_sort.py) |
 [unittest](insertion_sort_unittest.py)
+
+Selection sort is always a ![image](https://cloud.githubusercontent.com/assets/1147451/7592456/dd19381e-f905-11e4-8089-8ec444fca706.png) algorithm.

Foundations/overview/selection-sort.tex

@@ -1,9 +1,22 @@
-\documentclass{article}
+\documentclass[twoside]{article}
+\setlength{\oddsidemargin}{0.25 in}
+\setlength{\evensidemargin}{-0.25 in}
+\setlength{\topmargin}{-0.6 in}
+\setlength{\textwidth}{6.5 in}
+\setlength{\textheight}{8.5 in}
+\setlength{\headsep}{0.75 in}
+\setlength{\parindent}{0 in}
+\setlength{\parskip}{0.1 in}
+
 \usepackage{clrscode3e}
+\usepackage{multicol}
+\usepackage{amsmath}
+
 \begin{document}
 	\title{2.2-2}
 	\author{pezy}
 	\maketitle
+	\begin{multicols}{3}
 		\begin{codebox}
 			\Procname{$\proc{Selection-Sort}(A)$}
 			\li \For $j \gets 1$ \To $\attrib{A}{length} - 1$
@@ -15,6 +28,104 @@
 			\End
 			\li			\func{swap}($A[min], A[j]$)
 			\End
-	\li \Return $A$
 		\end{codebox}
+		
+		\columnbreak
+		
+		\begin{center}	% column 2 holds cost of each statement
+			\textbf{\large Cost}
+			\\*$c_{1}$
+			\\*$c_{2}$
+			\\*$c_{3}$
+			\\*$c_{4}$
+			\\*$c_{5}$
+			\\*$c_{6}$
+		\end{center}
+		
+		\columnbreak
+		
+		\begin{flushleft} % displays run time of each statement
+			\textbf{\large Time}
+			\\* $n$
+			\\* $n-1$
+			\\* $\sum _{ j=1 }^{ n-1 } t_j $
+			\\* $\sum _{ j=1 }^{ n-1 } (t_j - 1) $
+			\\* $0 \To \sum _{ j=1 }^{ n-1 } (t_j - 1)$
+			\\* $n-1$
+		\end{flushleft}
+	\end{multicols}
+	
+	We will now calculate the running time, $T(n)$,  of \proc{Selection-Sort}:
+	
+	\begin{align}
+	\notag
+	T(n) &= c_1 n +  c_2 (n-1) + c_3 \sum _{j=1}^{n-1}  t_j + c_4 \sum _{j=1}^{n-1} ( t_j -1 ) + k c_5 \sum _{j=1}^{n-1}  (t_j -1)  + c_6 (n-1),\\
+	\notag
+	&= c_1 n+ (c_2 + c_6)(n-1) + c_3 \sum _{j=1}^{n-1}  t_j + (c_4 + k c_5) \sum _{j=1}^{n-1}  (t_j -1),\\
+	\notag
+	&= c_1 n+ (c_2 + c_6)(n-1) + (c_3 + c_4 + k c_5 ) \sum _{j=1}^{n-1}  t_j - (c_4 + k c_5) (n-1),\\
+	\label{eq:Tn}
+	&= c_1 n +(c_2 + c_6 - c_4 - k c_5) (n-1) + (c_3 + c_4 + k c_5 ) \sum _{j=1}^{n-1}  t_j.
+	\end{align}
+	
+	\section{Best case running time}
+	In the  {\bf BEST CASE} running time, the list of input will already be sorted. Thus, the body of \If is never step in, and $k = 0$. we obtain that $t_j = j+1$, for every choice of j. Thus,
+	
+	\[
+	\sum_{j=1}^{n-1} t_j = \frac{1}{2} n(n+1) - 1 = (\frac{n}{2}+1)(n-1)
+	\]
+	
+	Substituting this into the last term of Eqn.~(\ref{eq:Tn}) yields,
+	\begin{align}
+	T(n) &= c_1 n +(c_2 + c_6 - c_4)(n-1) + (c_3 + c_4)(\frac{n}{2}+1)(n-1) \\
+	&= \frac{c_3+c_4}{2}n^2 + (c_1+c_2+\frac{c_3}{2}-\frac{c_4}{2}+c_6)n - (c_2+c_3+c_6)
+	\end{align}
+	
+	which can be simplified to the linear equation $T(n) = An^2+Bn+C$ where
+	\begin{align*}
+	A &= \frac{c_3+c_4}{2} > 0,\\
+	B &= c_1+c_2+\frac{c_3}{2}-\frac{c_4}{2}+c_6, \quad\text{and,}\\
+	C &= -c_2-c_3-c_6 < 0.
+	\end{align*}
+	
+	Therefore, the {\bf BEST CASE} running time of the \proc{Selection-Sort} Algorithm equals:
+	\begin{center}
+		$\boldsymbol{T(n) = An^2+Bn+C}$
+	\end{center}
+	
+	%
+	%Begin section for Worst Case Running Time
+	%
+	%
+	\section{Worst case running time}
+	We will now look at the {\bf WORST CASE} for \proc{Selection-Sort}:
+	\begin{itemize}
+		\item In the worst case, the \If statement is invoked on every occasion.
+		\item This means $ k = 1$
+	\end{itemize}
+	
+	Substituting $t_{ j }$ with $j$ into the last summation in  Eqn.~(\ref{eq:Tn}) yields,
+	
+	\[
+	\sum_{j=1}^{n-1} t_j = \frac{1}{2} n(n+1) - 1 = (\frac{n}{2}+1)(n-1)
+	\]
+	
+	Thus, Eqn.~(\ref{eq:Tn}) becomes,
+	\begin{align*}
+	T(n)   &=   c_1 n +(c_2+c_6-c_4-c_5)(n-1) + (c_3+c_4+c_5)(\frac{n}{2}+1)(n-1),\\
+	&= \frac{c_3+c_4+c_5}{2} n^2 + (c_1+c_2+\frac{c_3}{2}-\frac{c_4}{2}-\frac{c_5}{2}+c_6) n - (c_2+c_3+c_6)
+	\end{align*}
+	a \emph{quadratic function} of $n$, the input sequence length, where,
+	\begin{align*}
+	A' &= \frac{c_3+c_4+c_5}{2} > 0,\\
+	B' &= c_1+c_2+\frac{c_3}{2}-\frac{c_4}{2}-\frac{c_5}{2}+c_6, \quad\text{and,}\\
+	C' &= -c_{2}-c_{3}-c_{6} < 0.
+	\end{align*}
+	
+	
+	Therefore, the {\bf WORST CASE} running time of the \proc{Selection-Sort} Algorithm also equals:
+	\begin{center}
+		$\boldsymbol{T(n) = An^2+Bn+C}$
+	\end{center}
+	
 \end{document}