From 01b0bc9e7642730dbe8820ea557835febb7d51d1 Mon Sep 17 00:00:00 2001 From: Jeffery Russell Date: Sat, 9 May 2020 11:16:23 -0400 Subject: [PATCH] Initial Overleaf Import --- main.tex | 400 +++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 400 insertions(+) create mode 100644 main.tex diff --git a/main.tex b/main.tex new file mode 100644 index 0000000..11b3540 --- /dev/null +++ b/main.tex @@ -0,0 +1,400 @@ +\documentclass[12pt]{article}% +\usepackage{amsfonts} +\usepackage{fancyhdr} +\usepackage{comment} +\usepackage[a4paper, top=2.5cm, bottom=2.5cm, left=2.2cm, right=2.2cm]% +{geometry} +\usepackage{times} +\usepackage{amsmath} +\usepackage{changepage} +\usepackage{amssymb} +\usepackage{enumitem} + + +\usepackage{algorithm} +\usepackage[noend]{algpseudocode} + + +\usepackage{scrextend} + +\usepackage{graphicx}% +\setcounter{MaxMatrixCols}{30} +\newtheorem{theorem}{Theorem} +\newtheorem{acknowledgement}[theorem]{Acknowledgement} +\newtheorem{algorithm}[theorem]{Algorithm} +\newtheorem{axiom}{Axiom} +\newtheorem{case}[theorem]{Case} +\newtheorem{claim}[theorem]{Claim} +\newtheorem{conclusion}[theorem]{Conclusion} +\newtheorem{condition}[theorem]{Condition} +\newtheorem{conjecture}[theorem]{Conjecture} +\newtheorem{corollary}[theorem]{Corollary} +\newtheorem{criterion}[theorem]{Criterion} +\newtheorem{definition}[theorem]{Definition} +\newtheorem{example}[theorem]{Example} +\newtheorem{exercise}[theorem]{Exercise} +\newtheorem{lemma}[theorem]{Lemma} +\newtheorem{notation}[theorem]{Notation} +\newtheorem{problem}[theorem]{Problem} +\newtheorem{proposition}[theorem]{Proposition} +\newtheorem{remark}[theorem]{Remark} +\newtheorem{solution}[theorem]{Solution} +\newtheorem{summary}[theorem]{Summary} +\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} + +\newcommand{\Q}{\mathbb{Q}} +\newcommand{\R}{\mathbb{R}} +\newcommand{\C}{\mathbb{C}} +\newcommand{\Z}{\mathbb{Z}} + + +\documentclass{minimal} +\usepackage{mathtools} +\DeclarePairedDelimiter\ceil{\lceil}{\rceil} +\DeclarePairedDelimiter\floor{\lfloor}{\rfloor} +\begin{document} + +\title{Homework 3} +\author{Jeffery Russell} +\date{\today} +\maketitle + +\section{Strassen's algorithm} + + + +\begin{enumerate}[label=(\alph*)] + +\item + +$$ +\begin{bmatrix} +1 & 3\\ +7 & 5 +\end{bmatrix} \odot +\begin{bmatrix} +6 & 8\\ +4 & 2 +\end{bmatrix}\\* +$$ + +$S_1 = 6, S_2 = 4$\\ +$S_3 = 12, S_4 = -2$\\ +$S_5 = 5, S_6 = 8$\\ +$S_9 = -6, S_{10} = 14$\\ +\\ +$P_1 = 1*6 = 6, P_2 = 4*2 = 8$\\ +$P_3 = 6*12 = 72, P_4 = -2*5 = -10$\\ +$P_5 = 6*8 = 48, P_6 = -2*6 = -12$\\ +$P_7 = -6*14 = -84$\\ + +$$ += +\begin{bmatrix} +c_{11} & c_{12}\\ +c_{21} & c_{22} +\end{bmatrix} = +\begin{bmatrix} +(P_5 + P_4 - P_2 + P_6) & (P_1+P_2)\\ +(P_3+P_4) & (P-5+P_1 - P_3 + P_7) +\end{bmatrix} +$$ + +$$ += +\begin{bmatrix} +(48 + (-10) - 8 + (-12)) & (6 + 8)\\ +(72 + (-10)) & (48 + 6 - 72 - (-84)) +\end{bmatrix} = +\begin{bmatrix} +18 & 14\\ +62 & 66 +\end{bmatrix} +$$ + + +\item + +$$ + A \odot B = + \begin{cases} + A*B & \text{if } A_{rows} = 1 \\ + \begin{bmatrix} +c_{11} & c_{12}\\ +c_{21} & c_{22} +\end{bmatrix} & \text{otherwise} + \end{cases} +$$ + +Where: +$A = +\begin{bmatrix} +A_{11} & A_{12}\\ +A_{21} & A_{22} +\end{bmatrix}$ + +$B = +\begin{bmatrix} +B_{11} & B_{12}\\ +B_{21} & B_{22} +\end{bmatrix}$ + +$c_{11} = A_{11} \odot B_{11} + A_{12} \odot B_{21}$\\ +$c_{12} = A_{11} \odot B_{12} + A_{12} \odot B_{22}$\\ +$c_{21} = A_{21} \odot B_{11} + A_{22} \odot B_{21}$\\ +$c_{22} = A_{21} \odot B_{12} + A_{22} \odot B_{22}$ + +\newpage + +\begin{algorithm} +\caption{Strassen's Algorithm}\label{euclid} +\begin{algorithmic}[1] +\Procedure{Strassen(A,B):}{} +\State $\textit{n} \gets \text{number of rows in}\textit{ A}$ +\State $C \gets \textit{new n by n matrix}$ +\If {$\textit{n} = 1$} +\State $c \gets A[1][1] * B[1][1]$ +\Else{} +\State Sub partition A into 4 equal matrix quadrants A11, A12, A21, A22 +\State Sub partition B into 4 equal matrix quadrants B11, B12, B21, B22 + +\State $s1 \gets B12 - B22$ +\State $s2 \gets A11 + A12$ +\State $s3 \gets A21 + A22$ +\State $s4 \gets B21 - B11$ +\State $s5 \gets A11 + A22$ +\State $s6 \gets B11 + B22$ +\State $s7 \gets A12 - A22$ +\State $s8 \gets B21 + B22$ +\State $s9 \gets A11 - A21$ +\State $s10 \gets B11 + B12$ +\State $p1 \gets Strassen(A11,S1)$ +\State $p2 \gets Strassen(S2,B22)$ +\State $p3 \gets Strassen(S3,B11)$ +\State $p4 \gets Strassen(A22,S4)$ +\State $p5 \gets Strassen(S5,S6)$ +\State $p6 \gets Strassen(S7,S8)$ +\State $p7 \gets Strassen(S9,S10)$ +\State $C[1][1] \gets p5 + p4 - p2 + p6$ +\State $C[1][2] \gets p1 + p2$ +\State $C[2][1] \gets p3 + p4$ +\State $C[2][2] \gets p5 + p1 - p3 + p7$ +\EndIf +\State \textbf{return} C +\EndProcedure +\end{algorithmic} +\end{algorithm} + +\item + +$T(1) = 1$\\ +$T(n) = 7T(\frac{n}{2}) + \frac{9}{2}n^2$\\ +$T(2^0) = 1$\\ +$T(2^m) = 7T(2^{m-1}) + \frac{9}{2}(2^m)^2$\\ +$= 7(7T(n^{m-2}) + \frac{9}{2}(2^{m-1})^2) + \frac{9}{2}(2^m)^2$\\ +$= 7^2T(n^{m-2}) + 7*\frac{9}{2}(2^{m-1})^2 + 7^0*\frac{9}{2}(2^m)^2$\\ +$= 7^2(7T(2^{m-3}) + \frac{9}{2}(2^{m-2})^2) + 7*\frac{9}{2}(2^{m-1})^2 + 7^0*\frac{9}{2}(2^m)^2$\\ +$= 7^3T(2^{m-3}) + 7^2\frac{9}{2}(2^{m-2})^2 + 7*\frac{9}{2}(2^{m-1})^2 + 7^0*\frac{9}{2}(2^m)^2$\\ +$= 7^kT(2^{m-k}) + 7^{(k-1)}\frac{9}{2}(2^{m-(k-1)})^2 + ... + 7^{k-k}*\frac{9}{2}(2^{m-(k-k)})^2$\\ + +Let $m = k$\\ +$T(2^m)= 7^mT(2^{m-m}) + 7^{(m-1)}\frac{9}{2}(2^{m-(m-1)})^2 + ... + 7^{m-m}*\frac{9}{2}(2^{m-(m-m)})^2$\\ +$T(2^m)= 7^m + 7^{(m-1)}\frac{9}{2}(2^{1})^2 + ... + 7^{0}*\frac{9}{2}(2^{m})^2$\\ +$= 7^m + \frac{9}{2}\sum_{k=0}^{m-1}7^k(2^{m-k})^2$\\ +$= \frac{9}{2}\sum_{k=0}^{m}7^k(2^{2(m-k)}) - \frac{9}{2}7^m$\\ +$= \frac{9}{2}\sum_{k=0}^{m}7^k(2^{2m})(2^{-2k}) - \frac{9}{2}7^m$\\ +$= (2^{2m})\frac{9}{2}\sum_{k=0}^{m}7^k(2^{-2k}) - \frac{9}{2}7^m + 7^m$\\ +$= (2^{2m})\frac{9}{2}\sum_{k=0}^{m}\frac{7^k}{2^{2k}}) - \frac{7}{2}7^m$\\ +$= (2^{2m})\frac{9}{2}\sum_{k=0}^{m}(\frac{7}{4})^k - \frac{7}{2}7^m$\\ +$= (2^{2m})\frac{9}{2}(\frac{\frac{7}{4}^{m+1} -1)}{1.75-1}) - \frac{7}{2}7^m$\\ +$= (2^m)^2\frac{9}{2}\frac{4}{3}((\frac{7}{4})^{m+1}-1) - (\frac{7}{2})7^m$\\ +$= 6(2^m)^2((\frac{7}{4})^{m+1}-1) - (\frac{7}{2})7^m$\\ +$= 6(2^m)^2\frac{7}{4} - 6(2^m)^2 - (\frac{7}{2})7^m$\\ +$= \frac{21*7^m}{2} - 6(2^m)^2 - (\frac{7}{2})7^m$\\ +$= 7*7^m - 6 * 4^m$\\ +$T(n)= 7*n^{lg(7)} - 6 * n^2$\\ +$T(n) \approx 7*n^{2.81} - 6 * n^2$\\ + +\item Modifying Strassen's Algorithm + +To make Strassen's algorithm to work with any n x n matrix we would pad the two matrices being multiplied with zeros to become two m x m matrix where m is a power of 2. The answer would be the result but only taking out the n x n section out of the m x m product.\\ + +To prove that this is still asymptotically equal we will demonstrate that at most the dimensions of the matrix being multiply double. + +let m = length of new padded matrices being multiplied \\ +let n = length of original matrices being multiplied \\ +Since: +$2^{k-1} < n < 2^k = m$\\ +Note: +$2n > 2^{k+1} > m$\\ +Hence: +$T(n) \in \Theta((2n)^{lg(7)}) = \Theta(n^{lg7})$ + + +\item + +$(a+bi)(c+di)$\\ +$= ac + adi + bci + bdi^2$\\ +$= ac + adi + bci - bd$\\ +$Real Part= ac - bd$\\ +$Imaginary Part= (a + b)(c+d) -ac - bd$\\ + +Consider:\\ +$A_1 = ac$\\ +$A_2 = bd$\\ +$A_3 = (a+b)(c+d)$\\ + +Then:\\ +Real Part = $A_1 - A_2$\\ +Imaginary Part = $A_3 - A_1 - A_2$\\ + +\end{enumerate} + + +\section{Recurrence} + +\begin{enumerate}[label=(\alph*)] + +\item +\textbf{Lemma 1}: For any n $\in \mathbb{N}, \left\lfloor\dfrac{n+1}{2}\right\rfloor = \left\lceil\dfrac{n}{2}\right\rceil$\\ + +\textbf{Proof By Cases:}\\ +\textbf{Even Case:}\\ +n can be represented as 2m for some value of m\\ +$LHS = \left\lfloor\dfrac{n+1}{2}\right\rfloor = \left\lfloor\dfrac{2m+1}{2}\right\rfloor$\\ +$=\left\lfloor m + \dfrac{1}{2}\right\rfloor = m$\\ +$RHS = \left\lceil\dfrac{n}{2}\right\rceil = \left\lceil\dfrac{2m}{2}\right\rceil$ +$=\left\lceil m \right\rceil = m = LHS$ + +\textbf{Odd Case:}\\ +n can be represented as (2m + 1) for some value of m\\ +$LHS = \left\lfloor\dfrac{n+1}{2}\right\rfloor = \left\lfloor\dfrac{2m +1+1}{2}\right\rfloor$\\ +$=\left\lfloor m + 1\right\rfloor = m +1$\\ +$RHS = \left\lceil\dfrac{n}{2}\right\rceil = \left\lceil\dfrac{2m +1}{2}\right\rceil$ +$=\left\lceil m + \dfrac{1}{2} \right\rceil = m +1 = LHS$ + +$\Box$\\ + +\item +\textbf{Lemma 2}: For any n $\in \mathbb{N}, \left\lfloor\dfrac{n}{2}\right\rfloor + 1 = \left\lceil\dfrac{n+1}{2}\right\rceil$ + +\textbf{Proof By Cases:}\\ +\textbf{Even Case:}\\ +n can be represented as 2m for some value of m\\ +$LHS = \left\lfloor\dfrac{n}{2}\right\rfloor +1= \left\lfloor\dfrac{2m}{2}\right\rfloor + 1$\\ +$=\left\lfloor m \right\rfloor = m + 1$\\ +$RHS = \left\lceil\dfrac{n + 1}{2}\right\rceil = \left\lceil\dfrac{2m + 1}{2}\right\rceil$ +$=\left\lceil m + \dfrac{1}{2} \right\rceil = m + 1 = LHS$ + +\textbf{Odd Case:}\\ +n can be represented as (2m + 1) for some value of m\\ +$LHS = \left\lfloor\dfrac{n}{2}\right\rfloor + 1 = \left\lfloor\dfrac{2m + 1}{2}\right\rfloor + 1 = \left\lfloor m +\dfrac{1}{2}\right\rfloor + 1$\\ +$=\left\lfloor m \right\rfloor + 1 = m +1$\\ +$RHS = \left\lceil\dfrac{n + 1}{2}\right\rceil = \left\lceil\dfrac{2m + 1 +1}{2}\right\rceil$ +$=\left\lceil m + 1 \right\rceil = m +1 = LHS$\\ + +$\Box$\\ + +\item +\textbf{Let:} $T(1) = 0, T(n) = T(\left\lfloor\dfrac{n}{2}\right\rfloor) + T(\left\lceil\dfrac{n}{2}\right\rceil) + n$\\ +\textbf{Let:} $D(n) = T(n+1) - T(n)$\\ + +\textbf{Lemma 3:} $D(1) = 2$\\ +\textbf{Direct Proof:}\\ +$D(1) = T(2) - T(1)$\\ +$= T(\left\lfloor\dfrac{2}{2}\right\rfloor) + T(\left\lceil\dfrac{2}{2}\right\rceil) + 2 - T(0)$\\ +$= T(1) + T(1) + 2 - T(0)$\\ +$= 2$\\ +$\Box$\\ + + +\textbf{Lemma 4:} $D(n) = D(\left\lfloor\dfrac{n}{2}\right\rfloor) + 1$\\ +\textbf{Direct Proof:}\\ +$D(n) = T(n+1) - T(n)$\\ +$=T(\left\lfloor\dfrac{n + 1}{2}\right\rfloor) + T(\left\lceil\dfrac{n + 1}{2}\right\rceil) + (n+1) - T(n)$\\ +$=T(\left\lceil\dfrac{n}{2}\right\rceil) + T(\left\lfloor\dfrac{n}{2}\right\rfloor + 1) + (n+1) - T(n)$ \textit{by lemma 1, 2}\\ +$=T(\left\lceil\dfrac{n}{2}\right\rceil) + T(\left\lfloor\dfrac{n}{2}\right\rfloor + 1) + (n+1) - T(\left\lfloor\dfrac{n}{2}\right\rfloor) - T(\left\lceil\dfrac{n}{2}\right\rceil) - n$\\ +$=T(\left\lfloor\dfrac{n}{2}\right\rfloor + 1) - T(\left\lfloor\dfrac{n}{2}\right\rfloor) +1$\\ +$= D(\left\lfloor\dfrac{n}{2}\right\rfloor) + 1$\\ +$\Box$\\ + +\item +\textbf{Lemma 5:} For any $n \in \mathbb{N}$, if $n > 1$ and n is even then :\\ +$\left\lfloor lg(\left\lfloor\dfrac{n}{2}\right\rfloor)\right\rfloor = \left\lfloor\ lg(n) -1\right\rfloor$\\ +\textbf{Direct Proof:}\\ +Suppose that n is even:\\ +$\left\lfloor lg(\left\lfloor\dfrac{n}{2}\right\rfloor)\right\rfloor = \left\lfloor lg(\left\dfrac{n}{2}\right)\right\rfloor$ \textit{since n is even}\\ +$= \left\lfloor lg(n) - lg(2)\right\rfloor$\\ +$= \left\lfloor lg(n) - 1\right\rfloor$\\ +$= \left\lfloor lg(n)\right\rfloor -1$\\ +$\Box$\\ + + + +\textbf{Lemma 6:} For any $m \in \mathbb{N}$, if $m > 0$ then:\\ +$\left\lfloor lg(\left\lfloor 2m+1 \right\rfloor)\right\rfloor = \left\lfloor\ lg(2m)\right\rfloor$\\ +\textbf{Direct Proof:}\\ +Let $k = \left\lfloor lg(\left\lfloor 2m+1 \right\rfloor)\right\rfloor$\\ +$\left\lfloor lg(\left\lfloor 2m+1 \right\rfloor)\right\rfloor = k \xrightarrow[]{} k \leq lg(2m + 1) < k + 1$\\ +$\xrightarrow[]{} 2^k \leq 2m+1 < 2^{k+1}$\\ +$\xrightarrow{} 2^{k} -1 \leq 2m < 2^{k+1} -1 \textit{by transitivity}$\\ +$\xrightarrow{} 2^k -1 \leq 2m < 2^{k+1}$\\ +$\xrightarrow[]{} 2^k \leq 2m < 2^{k+1}$\textit{Since 2m is even}\\ +$\xrightarrow{} k \leq lg(2m) < k + 1$\\ +$\xrightarrow{}\left\lfloor\ lg(2m)\right\rfloor = k$\\ +$\Box$\\ + + +\textbf{Lemma 7:} For any $n \in \mathbb{N}$, if $n > 1$ and n is odd then :\\ +$\left\lfloor lg(\left\lfloor\dfrac{n}{2}\right\rfloor)\right\rfloor = \left\lfloor\ lg(n) -1\right\rfloor$\\ +\textbf{Direct Proof:}\\ +$\left\lfloor lg(\left\lfloor\dfrac{n}{2}\right\rfloor)\right\rfloor = \left\lfloor\ lg(\dfrac{n-1}{2})\right\rfloor$\textit{Gets an even number since n is odd}\\ +$= \left\lfloor\ lg(n-1) - lg(2)\right\rfloor$\\ +$= \left\lfloor\ lg(n-1) - 1\right\rfloor$\\ +$= \left\lfloor\ lg(n-1)\right\rfloor -1$\\ +$= \left\lfloor\ lg(n)\right\rfloor -1$\textit{By lemma 6}\\ +$\Box$\\ + +\textbf{Corollary 1:} +By lemmas 5 and 7, for any $n \in \mathbb{N}$, if $n > 1$ then :\\ +$\left\lfloor lg(\left\lfloor\dfrac{n}{2}\right\rfloor)\right\rfloor = \left\lfloor\ lg(n) -1\right\rfloor$\\ + + +\textbf{Lemma 8:} For any for any $n \in \mathbb{N}$, if $n > 1$ then $D(n) = \left\lfloor\ lg(n)\right\rfloor +2$\\ +\textbf{Proof via Strong Induction:}\\ +\textbf{base case: n =1}\\ +$D(1) = 2$ \textit{lemma 3}\\ +$D(1) = \left\lfloor\ lg(1)\right\rfloor +2$\\ +$= 0 + 2 = 2$\\ +\textbf{Inductive Step:} Assume proposition holds up to but not including n.\\ Show that n follows.\\ +$D(n) = D(\left\lfloor\dfrac{n}{2} \right\rfloor) +1$\\ +$= \left\lfloor lg(\left\lfloor \dfrac{n}{2}\right\rfloor)\right\rfloor + 2 + 1$\textit{ By I.H}\\ +$=\left\lfloor\ lg(n)\right\rfloor -1 + 2 + 1$ \textit{ Corollary 1}\\ +$=\left\lfloor\ lg(n)\right\rfloor +2$\\ +$\Box$\\ + +\item + +\textbf{Lemma 9:} $T(n) - T(1) = \sum^{n-1}_{k = 1}D(k)$\\ +\textbf{Direct Proof:}\\ +$\sum^{n-1}_{k = 1}D(k) = D(1) + D(2) + D(3) + ... + D(n-3) + D(n-2) + D(n-1)$\\ +$= T(2) - T(1) + T(3) - T(2) + T(4) - T(3) +\\ ... + T(n-2) - T(n-3) + T(n-1) - T(n-2) + T(n) - T(n-1)$\\ +$=T(n)-T(1)$\textit{Due to cancellations}\\ +$\Box$\\ +\textbf{Corollary 2:}\\ +By lemmas 9 and 8, $T(n) = \sum^{n-1}_{k = 1}\left\lfloor\ (lg(n) + 2)\right\rfloor$\\ + +\item + +\textbf{Lemma 10:} $T(n) \in O(nlog(n))$\\ +$T(n) = \sum^{n-1}_{k = 1}\left\lfloor\ (lg(n) + 2)\right\rfloor$\\ +$= \sum^{n-1}_{k = 1}\left\lfloor\ (lg(n))\right\rfloor + \sum^{n-1}_{k = 1}2$\\ +$\leq nlg(n) + 2n$\\ +$\xrightarrow[]{} T(n) \in O(nlog(n))$\\ +$\Box$\\ + + +\end{enumerate} + + +\end{document}