491 lines
18 KiB
TeX
491 lines
18 KiB
TeX
\documentclass[unicode,11pt,a4paper,oneside,numbers=endperiod,openany]{scrartcl}
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\usepackage{graphicx}
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\usepackage{subcaption}
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\usepackage{amsmath}
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\input{assignment.sty}
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\usepackage{pgfplots}
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\pgfplotsset{compat=newest}
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\usetikzlibrary{plotmarks}
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\usetikzlibrary{arrows.meta}
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\usepgfplotslibrary{patchplots}
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\usepackage{grffile}
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\usepackage{amsmath}
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\hyphenation{PageRank}
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\hyphenation{PageRanks}
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\begin{document}
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\setassignment
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\setduedate{Wednesday, 14 October 2020, 11:55 PM}
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\serieheader{Numerical Computing}{2020}{Student: Claudio Maggioni}{Discussed with: FULL NAME}{Solution for Project 2}{}
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\newline
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\assignmentpolicy
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The purpose of this assignment\footnote{This document is originally
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based on a blog from Cleve Moler, who wrote a fantastic blog post about the Lake Arrowhead graph, and John
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Gilbert, who initially created the coauthor graph from the 1993 Householder Meeting. You can find more information
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at \url{http://blogs.mathworks.com/cleve/2013/06/10/lake-arrowhead-coauthor-graph/}. Most of this assignment is derived
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from this archived work.} is to learn the importance of sparse linear algebra algorithms to solve fundamental
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questions in social network analyses.
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We will use the coauthor graph from the Householder Meeting and the social network of friendships from Zachary's karate club~\cite{karate}.
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These two graphs are one of the first examples where matrix methods were used in computational social network analyses.
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\section{The Reverse Cuthill McKee Ordering [10 points]}
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The Reverse Cuthill McKee Ordering of matrix \texttt{A\_SymPosDef} is computed with MATLAB's \texttt{sysrcm(\ldots)} and
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the matrix is rearranged accordingly. Here are the spy plot of these matrices:
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\begin{figure}[h]
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\centering
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\begin{subfigure}{0.49\textwidth}
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\centering
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\includegraphics[width = \textwidth]{1_spy_a}
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\caption{Spy plot of \texttt{A\_SymPosDef}}
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\end{subfigure}
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\begin{subfigure}{0.49\textwidth}
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\centering
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\includegraphics[width = \textwidth]{1_spy_rcm}
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\caption{Spy plot of \texttt{sysrcm(\ldots)} rearranged version of \texttt{A\_SymPosDef}}
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\end{subfigure}
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\caption{Spy plots of the two matrices}
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\label{fig:1}
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\end{figure}
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And the spy plots of the corresponding Cholesky factor are listed in figure~\ref{fig:1chol}.
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\begin{figure}[h]
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\centering
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\begin{subfigure}{0.49\textwidth}
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\centering
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\includegraphics[width = \textwidth]{1_spy_chol_a}
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\caption{Spy plot of \texttt{chol(A\_SymPosDef)}}
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\end{subfigure}
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\begin{subfigure}{0.49\textwidth}
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\centering
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\includegraphics[width = \textwidth]{1_spy_chol_rcm}
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\caption{Spy plot of \texttt{chol(A\_SymPosDef(sysrcm(A\_SymPosDef), sysrcm(A\_SymPosDef)))}}
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\end{subfigure}
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\caption{Spy plots of the two Cholesky factors}
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\label{fig:1chol}
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\end{figure}
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The number of nonzero elements in the Cholesky factor of the RCM optimized
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matrix are significantly lower (circa 0.1x) of the ones in the vanilla process.
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The respective nonzero counts can be found in figure~\ref{fig:1chol}.
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\section{Sparse Matrix Factorization [10 points]}
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\subsection{Show that $A \in R^{n x n}$ has exactly $5n - 6$ nonzero elements.}
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The given description of $A$ says that all the element at the edges of the
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matrix (rows and columns 1 and $n$) plus all the elements on the main diagonal
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are the only nonzero elements of $A$. Therefore, this cells can be counted as
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the 4 vertex cells in the matrix square plus 5 $n-2$-long segments,
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corresponding to all edges and the main diagonal. Therefore:
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\[4 + 5 \dot (n - 2) = 5n - 6\]
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\subsection{Write a short Matlab script to construct this matrix and visualize
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its non-zero structure(you can use, e.g., the command \texttt{spy()}).}
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The MATLAB script can be found in file \texttt{ex3.m}.
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Here is a spy plot of the nonzero values of $A$, for $n = 5$:
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\centering{\input{ex2_2_spy.tex}}
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The matrix $A \in R^{n x n}$ looks like this (zero entries are represented as
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blanks):
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\[
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A := \begin{bmatrix}
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n & 1 & 1 & \hdots & 1 \\
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1 & n + 1 & && 1 \\
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1 & & n + 2 && 1 \\
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\vdots & & & \ddots & \vdots \\
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1 & 1 & 1 & \hdots & 2n - 1 \\
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\end{bmatrix}
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\]
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\subsection{Using again the \texttt{spy()} command, visualize side by side the
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original matrix $A$ and the result of the Cholesky factorization (\texttt{chol()}
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in Matlab). Then explain why for n = $100000$ using Matlab’s \texttt{chol(\ldots)}
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to solve $Ax = b$
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for a given righthand-side vector would be problematic.}
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Here is the plot of \texttt{spy(A)} (on the left) and \texttt{chol(spy(A))} (on
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the right).
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\centering{\input{ex2_3_spy.tex}}
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Solving $Ax = b$ would be a costly operation since the a Cholesky
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decomposition of matrix $A$ (performed using MATLAB's \texttt{chol(\ldots)})
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would drastically reduce the number of zero elements in the matrix in the very
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first iteration. This is due to the fact that the first row, by definition, is
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made of of only nonzero elements (namely 1s) and by subtracting the first row to
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every other row (as what would effectively happen in the first iteration of the
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Cholesky decomposition of A) the zero elements would become (negative) nonzero
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elements, thus making all columns but the first almost empty of 0s.
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\section{Degree Centrality [10 points]}
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Assuming that the degree of the Householder graph is the number of co-authors of
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each author and that an author is not co-author of himself, the degree
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centralities of all authors sorted in descending order are below.
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This output has been obtained by running \texttt{ex3.m}.
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\begin{verbatim}
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Author Centrality: Coauthors...
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Golub 31: Wilkinson TChan Varah Overton Ernst VanLoan Saunders Bojanczyk
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Dubrulle George Nachtigal Kahan Varga Kagstrom Widlund
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OLeary Bjorck Eisenstat Zha VanDooren Tang Reichel Luk Fischer
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Gutknecht Heath Plemmons Berry Sameh Meyer Gill
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Demmel 15: Edelman VanLoan Bai Schreiber Kahan Kagstrom Barlow
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NHigham Arioli Duff Hammarling Bunch Heath Greenbaum Gragg
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Plemmons 13: Golub Nagy Harrod Pan Funderlic Bojanczyk George Barlow Heath
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Berry Sameh Meyer Nichols
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Heath 12: Golub TChan Funderlic George Gilbert Eisenstat Ng Liu Laub Plemmons
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Paige Demmel
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Schreiber 12: TChan VanLoan Moler Gilbert Pothen NTrefethen Bjorstad NHigham
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Eisenstat Tang Elden Demmel
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Hammarling 10: Wilkinson Kaufman Bai Bjorck VanHuffel VanDooren Duff Greenbaum
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Gill Demmel
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VanDooren 10: Golub Boley Bojanczyk Kagstrom VanHuffel Luk Hammarling Laub
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Nichols Paige
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TChan 10: Golub Saied Ong Kuo Tong Schreiber Arioli Duff Heath Hansen
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Gragg 9: Borges Kaufman Harrod Reichel Stewart BunseGerstner Ammar Warner Demmel
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Moler 8: Wilkinson VanLoan Gilbert Schreiber Henrici Stewart Bunch Laub
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VanLoan 8: Golub Moler Schreiber Kagstrom Luk Bunch Paige Demmel
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Paige 7: Anjos VanLoan Saunders Bjorck VanDooren Laub Heath
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Gutknecht 7: Golub Ashby Boley NTrefethen Nachtigal Varga Hochbruck
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Luk 7: Golub Overton Boley VanLoan Bojanczyk Park VanDooren
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Eisenstat 7: Golub Gu George Schreiber Liu Heath Ipsen
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George 7: Golub Eisenstat Ng Liu Tang Heath Plemmons
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Meyer 6: Golub Benzi Funderlic Stewart Ipsen Plemmons
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Bunch 6: LeBorne Fierro VanLoan Moler Stewart Demmel
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Stewart 6: Moler Bunch Gragg Meyer Gill Mathias
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Reichel 6: Golub NTrefethen Nachtigal Fischer Gragg Ammar
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Bjorck 6: Golub Park Duff Hammarling Elden Paige
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NTrefethen 6: Schreiber Nachtigal Reichel Gutknecht Greenbaum ATrefethen
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Nichols 5: Byers Barlow VanDooren Plemmons BunseGerstner
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Greenbaum 5: Cullum Strakos NTrefethen Hammarling Demmel
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Ipsen 5: Chandrasekaran Barlow Eisenstat Meyer Jessup
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Laub 5: Kenney Moler VanDooren Heath Paige
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Duff 5: TChan Bjorck Arioli Hammarling Demmel
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Liu 5: George Gilbert Eisenstat Ng Heath
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Park 5: Boley Bjorck VanHuffel Luk Elden
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Zha 5: Golub Bai Barlow VanHuffel Hansen
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Widlund 5: Golub Bjorstad OLeary Smith Szyld
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Barlow 5: Zha Ipsen Plemmons Nichols Demmel
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Kagstrom 5: Golub VanLoan VanDooren Ruhe Demmel
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Varga 5: Golub Marek Young Gutknecht Starke
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Gilbert 5: Moler Schreiber Ng Liu Heath
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Gill 4: Golub Saunders Hammarling Stewart
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Sameh 4: Golub Harrod Plemmons Berry
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Berry 4: Golub Harrod Plemmons Sameh
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BunseGerstner 4: He Byers Gragg Nichols
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Hansen 4: TChan Fierro OLeary Zha
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Ng 4: George Gilbert Liu Heath
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Arioli 4: TChan MuntheKaas Duff Demmel
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VanHuffel 4: Zha Park VanDooren Hammarling
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Nachtigal 4: Golub NTrefethen Reichel Gutknecht
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Bojanczyk 4: Golub VanDooren Luk Plemmons
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Harrod 4: Plemmons Gragg Berry Sameh
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Boley 4: Park VanDooren Luk Gutknecht
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Wilkinson 4: Golub Dubrulle Moler Hammarling
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Ammar 3: He Reichel Gragg
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Elden 3: Schreiber Bjorck Park
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Fischer 3: Golub Modersitzki Reichel
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Tang 3: Golub George Schreiber
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NHigham 3: Schreiber Pothen Demmel
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OLeary 3: Golub Widlund Hansen
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Bjorstad 3: Schreiber Widlund Boman
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Kahan 3: Golub Davis Demmel
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Bai 3: Zha Hammarling Demmel
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Saunders 3: Golub Paige Gill
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Funderlic 3: Heath Plemmons Meyer
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Kaufman 3: Hammarling Gragg Warner
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Starke 2: Varga Hochbruck
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Hochbruck 2: Gutknecht Starke
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Jessup 2: Crevelli Ipsen
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Warner 2: Kaufman Gragg
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Ruhe 2: Wold Kagstrom
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Szyld 2: Marek Widlund
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Young 2: Kincaid Varga
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Pothen 2: Schreiber NHigham
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Tong 2: TChan Kuo
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Kuo 2: TChan Tong
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Marek 2: Varga Szyld
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Dubrulle 2: Golub Wilkinson
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Fierro 2: Bunch Hansen
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Byers 2: BunseGerstner Nichols
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Overton 2: Golub Luk
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He 2: BunseGerstner Ammar
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Mathias 1: Stewart
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Davis 1: Kahan
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ATrefethen 1: NTrefethen
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Henrici 1: Moler
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Smith 1: Widlund
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MuntheKaas 1: Arioli
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Boman 1: Bjorstad
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Chandrasekaran 1: Ipsen
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Wold 1: Ruhe
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Ong 1: TChan
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Saied 1: TChan
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Strakos 1: Greenbaum
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Cullum 1: Greenbaum
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Edelman 1: Demmel
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Pan 1: Plemmons
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Nagy 1: Plemmons
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Gu 1: Eisenstat
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Benzi 1: Meyer
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Anjos 1: Paige
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Crevelli 1: Jessup
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Kincaid 1: Young
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Borges 1: Gragg
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Ernst 1: Golub
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Modersitzki 1: Fischer
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LeBorne 1: Bunch
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Ashby 1: Gutknecht
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Kenney 1: Laub
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Varah 1: Golub
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\end{verbatim}
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\section{The Connectivity of the Coauthors [10 points]}
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The author indexes of the common authors between the author at index $i$ and the
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author at index $j$ can be computed by listing the indexes of the nonzero
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elements in the Schur product (or element-wise product) between $A_{:,i}$ and
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$A_{:,j}$ (respectively the i-th and j-th column vector of $A$). Therefore the set $C$ of common coauthor's indexes can be defined
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as:
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\[C = \{i \in N_0 \;|\; (A_{:,i} \odot A_{:,j})_i = 1\}\]
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The results below were computing by using the script \texttt{ex4.m}.
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The common Co-authors between Golub and Moler are Wilkinson and Van Loan.
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The common Co-authors between Golub and Saunders are Golub, Saunders and Gill.
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The common Co-authors between TChan and Demmel are Schreiber, Arioli, Duff and
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Heath.
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\section{PageRank of the Coauthor Graph [10 points]}
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The PageRank values for all authors were computing by using the scripts
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\texttt{ex5.m} and \texttt{pagerank.m}, a basically identical version of
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\texttt{pagerank.m} from Mini Project 1. The output is shown below.
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\begin{verbatim}
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page-rank in out author
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1 0.0511 32 32 Golub
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104 0.0261 16 16 Demmel
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86 0.0229 14 14 Plemmons
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44 0.0212 13 13 Schreiber
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3 0.0201 11 11 TChan
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81 0.0198 13 13 Heath
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90 0.0181 10 10 Gragg
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74 0.0177 11 11 Hammarling
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66 0.0171 11 11 VanDooren
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42 0.0152 9 9 Moler
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79 0.0151 8 8 Gutknecht
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32 0.0142 9 9 VanLoan
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59 0.0135 8 8 Eisenstat
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98 0.0133 8 8 Paige
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46 0.0130 7 7 NTrefethen
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49 0.0129 6 6 Varga
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96 0.0128 7 7 Meyer
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77 0.0128 7 7 Stewart
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73 0.0127 8 8 Luk
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78 0.0127 7 7 Bunch
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53 0.0127 6 6 Widlund
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72 0.0125 7 7 Reichel
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41 0.0125 8 8 George
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82 0.0124 6 6 Ipsen
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83 0.0122 6 6 Greenbaum
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58 0.0113 7 7 Bjorck
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97 0.0107 6 6 Nichols
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51 0.0106 6 6 Kagstrom
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80 0.0106 6 6 Laub
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52 0.0104 6 6 Barlow
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60 0.0103 6 6 Zha
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69 0.0102 6 6 Duff
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62 0.0100 6 6 Park
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89 0.0099 5 5 BunseGerstner
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63 0.0098 5 5 Arioli
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43 0.0097 6 6 Gilbert
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67 0.0096 6 6 Liu
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87 0.0096 5 5 Hansen
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47 0.0090 5 5 Nachtigal
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54 0.0090 4 4 Bjorstad
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2 0.0088 5 5 Wilkinson
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23 0.0088 5 5 Harrod
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99 0.0087 5 5 Gill
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92 0.0086 5 5 Sameh
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91 0.0086 5 5 Berry
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15 0.0086 5 5 Boley
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76 0.0085 4 4 Fischer
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50 0.0085 3 3 Young
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61 0.0084 5 5 VanHuffel
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100 0.0084 3 3 Jessup
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48 0.0083 4 4 Kahan
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35 0.0083 5 5 Bojanczyk
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65 0.0082 5 5 Ng
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93 0.0082 4 4 Ammar
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55 0.0079 4 4 OLeary
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84 0.0079 3 3 Ruhe
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19 0.0078 4 4 Kaufman
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56 0.0076 4 4 NHigham
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37 0.0075 3 3 Marek
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75 0.0075 3 3 Szyld
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103 0.0074 3 3 Starke
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34 0.0072 4 4 Saunders
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25 0.0072 4 4 Funderlic
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39 0.0072 4 4 Bai
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102 0.0072 3 3 Hochbruck
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88 0.0071 4 4 Elden
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71 0.0070 4 4 Tang
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38 0.0069 3 3 Kuo
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40 0.0069 3 3 Tong
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4 0.0068 3 3 He
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13 0.0067 2 2 Kincaid
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14 0.0067 2 2 Crevelli
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94 0.0065 3 3 Warner
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17 0.0065 3 3 Byers
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21 0.0064 3 3 Fierro
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31 0.0064 2 2 Wold
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45 0.0062 3 3 Pothen
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36 0.0060 3 3 Dubrulle
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57 0.0058 2 2 Boman
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10 0.0058 3 3 Overton
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9 0.0057 2 2 Modersitzki
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68 0.0056 2 2 Smith
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95 0.0056 2 2 Davis
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33 0.0056 2 2 Chandrasekaran
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27 0.0055 2 2 Cullum
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28 0.0055 2 2 Strakos
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64 0.0054 2 2 MuntheKaas
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7 0.0053 2 2 Ashby
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85 0.0053 2 2 ATrefethen
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29 0.0052 2 2 Saied
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30 0.0052 2 2 Ong
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18 0.0052 2 2 Benzi
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101 0.0052 2 2 Mathias
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8 0.0052 2 2 LeBorne
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12 0.0052 2 2 Borges
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6 0.0051 2 2 Kenney
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70 0.0050 2 2 Henrici
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\end{verbatim}
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\section{Zachary's karate club: social network of friendships between 34 members [50 points]}
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\subsection{Write a Matlab code that ranks the five nodes with the largest
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degree centrality? What are their degrees?}
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Results found here can be computed using the file \texttt{ex6.m}.
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Please find the top 5 nodes by degree centrality, with their degree and their
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neighbours listed below:
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\begin{verbatim}
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Node Degree: Neighbours...
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34 16: 9, 10, 14, 15, 16, 19, 20, 21, 23, 24, 27, 28, 29, 30, 31, 32, 33,
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1 15: 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 18, 20, 22, 32,
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33 11: 3, 9, 15, 16, 19, 21, 23, 24, 30, 31, 32, 34,
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3 9: 1, 2, 4, 8, 9, 10, 14, 28, 29, 33,
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2 8: 1, 3, 4, 8, 14, 18, 20, 22, 31,
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\end{verbatim}
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\subsection{Rank the five nodes with the largest eigenvector centrality. What are
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their (properly normalized) eigenvector centralities?}
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Results found here can be computed using the file \texttt{ex6.m}.
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Please find the top 5 nodes by eigenvector centrality (page-rank column)
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listed below:
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\begin{verbatim}
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page-rank in out author
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34 0.1009 17 17 34
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1 0.0970 16 16 1
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33 0.0717 12 12 33
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3 0.0571 10 10 3
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2 0.0529 9 9 2
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\end{verbatim}
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\subsection{Are the rankings in (a) and (b) identical? Give a brief verbal
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explanation of the similarities and differences.}
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The rankings found are identical, even though if we normalize the degree
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centrality to the greatest eigenvector centrality we find slighly different
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values ($[0.1009, 0.0946, 0.0694, 0.0568, 0.0505]$) w.r.t the actual eigenvector
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centrality.
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The identical rankings may be explained by the fact that by computing the
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eigenvector centrality we are effectively applying PageRank to a symmetrical
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matrix, i.e. to a graph with bidirectional links. Since the links are
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bidirectional, we effectively make all the nodes in the graph of the same
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``importance'' to the eyes of PageRank, thus avoiding a case where a node has
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high PageRank thank to connections with few, but very ``important'' nodes.
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Therefore PageRank is simply reduced to a priotarization of nodes with many
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edges, i.e. the degree centrality ranking.
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\subsection{Use spectral graph partitioning to find a near-optimal split of the
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||
network into two groups of 16 and 18 nodes, respectively. List the nodes in the
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||
two groups. How does spectral bisection compare to the real split observed by
|
||
Zachary?}
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||
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||
The spectral bisection of the matrix a in two groups of 16 and 18 members
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||
respectively is identical to the real split observed by Zachary. To compute the
|
||
split, the script \texttt{ex6.m} was used.
|
||
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||
Here are the (sorted) two groups found:
|
||
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||
\begin{gather*}
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||
G_1 = [1, 2, 3, 4, 5, 6, 7, 8, 11, 12, 13, 14, 17, 18, 20, 22] \\
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||
G_2 = [9, 10, 15, 16, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34]
|
||
\end{gather*}
|
||
|
||
Here are the spy plots of the original matrix A (to the left) and the spectral
|
||
bisected permutated matrix (to the right):
|
||
|
||
\centering{\input{ex6_6_spy.tex}}
|
||
|
||
Here is a plot of the sorted elements of the second eigenvector $\lambda_2$:
|
||
|
||
\centering{\input{ex6_6_ev.tex}}
|
||
|
||
and here are the actual (sorted) values of $\lambda_2$:
|
||
|
||
\begin{align*}
|
||
sort(\lambda_2) = [-0.4228, -0.3237, -0.3237, -0.2846,
|
||
-0.2846, -0.2110, -0.1121, -0.1095, -0.1002, \\ -0.1002, -0.0555,
|
||
-0.0526, -0.0413, -0.0147, -0.0136, 0.0232, 0.0516, 0.0735, \\
|
||
0.0928, 0.0952, 0.0988, 0.1189, 0.1277, 0.1303, 0.1530,
|
||
0.1557, 0.1610, \\ 0.1628, 0.1628, 0.1628, 0.1628, 0.1628,
|
||
0.1677, 0.1871]^T
|
||
\end{align*}
|
||
|
||
As it can be seen above, there are only 15 negative values out the 16 we would
|
||
need to obtain a perfect 16/18 partition. We therefore add the index corresponding to
|
||
the smallest positive value in $\lambda_2$ in the set of indexes of group 1.
|
||
This seems to be a good approximation since indeed we get the same partitioning
|
||
as the original Zachary's one.
|
||
|
||
\begin{thebibliography}{99}
|
||
\bibitem{karate} The social network of a karate club at a US university, M.~E.~J. Newman and M. Girvan, Phys. Rev. E 69,026113 (2004)
|
||
pp. 219-229.
|
||
\end{thebibliography}
|
||
|
||
|
||
\end{document}
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