\documentclass[final]{beamer} \usepackage[utf8]{inputenc} \usepackage[scale=1.6]{beamerposter} % Use the beamerposter package for laying out the poster \usetheme{confposter} % Use the confposter theme supplied with this template \usepackage{color, bbm} \setbeamercolor{block title}{fg=dblue,bg=white} % Colors of the block titles \setbeamercolor{block body}{fg=black,bg=white} % Colors of the body of blocks \setbeamercolor{block alerted title}{fg=white,bg=dblue!70} % Colors of the highlighted block titles \setbeamercolor{block alerted body}{fg=black,bg=dblue!10} % Colors of the body of highlighted blocks % Many more colors are available for use in beamerthemeconfposter.sty %----------------------------------------------------------- % Define the column widths and overall poster size % To set effective sepwid, onecolwid and twocolwid values, first choose how many columns you want and how much separation you want between columns % In this template, the separation width chosen is 0.024 of the paper width and a 4-column layout % onecolwid should therefore be (1-(# of columns+1)*sepwid)/# of columns e.g. (1-(4+1)*0.024)/4 = 0.22 % Set twocolwid to be (2*onecolwid)+sepwid = 0.464 % Set threecolwid to be (3*onecolwid)+2*sepwid = 0.708 \newlength{\sepwid} \newlength{\onecolwid} \newlength{\twocolwid} \newlength{\threecolwid} \setlength{\paperwidth}{48in} % A0 width: 46.8in \setlength{\paperheight}{40in} % A0 height: 33.1in \setlength{\sepwid}{0.024\paperwidth} % Separation width (white space) between columns \setlength{\onecolwid}{0.22\paperwidth} % Width of one column \setlength{\twocolwid}{0.464\paperwidth} % Width of two columns \setlength{\threecolwid}{0.708\paperwidth} % Width of three columns \setlength{\topmargin}{-1in} % Reduce the top margin size %----------------------------------------------------------- \usepackage{graphicx} % Required for including images \usepackage{booktabs} % Top and bottom rules for tables %---------------------------------------------------------------------------------------- % TITLE SECTION %---------------------------------------------------------------------------------------- \title{Sparse Recovery for Graph Reconstruction } % Poster title \author{Eric Balkanski, Jean Pouget-Abadie} % Author(s) \institute{Harvard University} % Institution(s) %---------------------------------------------------------------------------------------- \begin{document} \addtobeamertemplate{block end}{}{\vspace*{2ex}} % White space under blocks \addtobeamertemplate{block alerted end}{}{\vspace*{2ex}} % White space under highlighted (alert) blocks \setlength{\belowcaptionskip}{2ex} % White space under figures \setlength\belowdisplayshortskip{2ex} % White space under equations \begin{frame}[t] % The whole poster is enclosed in one beamer frame \begin{columns}[t] % The whole poster consists of three major columns, the second of which is split into two columns twice - the [t] option aligns each column's content to the top \begin{column}{\sepwid}\end{column} % Empty spacer column \begin{column}{\onecolwid} % The first column %---------------------------------------------------------------------------------------- % INTODUCTION %---------------------------------------------------------------------------------------- %\vspace{- 15.2 cm} %\begin{center} %{\includegraphics[height=7em]{logo.png}} % First university/lab logo on the left %\end{center} %\vspace{4.6 cm} \begin{block}{Problem Statement} \begin{center} \bf{How can we reconstruct a graph on which observed cascades spread?} \end{center} \end{block} %{\bf Graph Reconstruction}: %\begin{itemize} %\item \{${\cal G}, \vec p$\}: directed graph, edge probabilities %\item $F$: cascade generating model %\item ${\cal M} := F\{{\cal G}, \vec p\}$: cascade %\end{itemize} %{\bf Objective}: %\begin{itemize} %\item Find algorithm which computes $F^{-1}({\cal M}) = \{{\cal G}, \vec p\}$ w.h.p., i.e. recovers graph from cascades. %\end{itemize} %{\bf Approach} %\begin{itemize} %\item Frame graph reconstruction as a {\it Sparse Recovery} problem for two cascade generating models. %\end{itemize} %Given a set of observed cascades, the \textbf{graph reconstruction problem} consists of finding the underlying graph on which these cascades spread. We assume that these cascades come from the classical \textbf{Independent Cascade Model} where at each time step, newly infected nodes infect each of their neighbor with some probability. %In previous work, this problem has been formulated in different ways, including a convex optimization and a maximum likelihood problem. However, there is no known algorithm for graph reconstruction with theoretical guarantees and with a reasonable required sample size. %We formulate a novel approach to this problem in which we use \textbf{Sparse Recovery} to find the edges in the unknown underlying network. Sparse Recovery is the problem of finding the sparsest vector $x$ such that $\mathbf{M x =b}$. In our case, for each node $i$, we wish to recover the vector $x = p_i$ where $p_{i_j}$ is the probability that node $j$ infects node $i$ if $j$ is active. To recover this vector, we are given $M$, where row $M_{t,k}$ indicates which nodes are infected at time $t$ in observed cascade $k$, and $b$, where $b_{t+1,k}$ indicates if node $i$ is infected at time $t+1$ in cascade $k$. Since most nodes have a small number of neighbors in large networks, we can assume that these vectors are sparse. Sparse Recovery is a well studied problem which can be solved efficiently and with small error if $M$ satisfies certain properties. In this project, we empirically study to what extent $M$ satisfies the Restricted Isometry Property. %--------------------------------------------------------------------------------- %--------------------------------------------------------------------------------- \begin{block}{Voter Model} \begin{figure} \centering \includegraphics[width=0.6\textwidth]{images/voter.png} \end{figure} \vspace{0.5 cm} {\bf Description} \vspace{0.5 cm} \begin{itemize} \item $\mathbb{P}$({\color{blue} blue} at $t=0) = p_{\text{init}}$ \item $\mathbb{P}$({\color{blue} blue} at $t+1) = \frac{\text{Number of {\color{blue}blue} neighbors}}{\text{Total number of neighbors}}$ \end{itemize} \vspace{0.5 cm} {\bf Sparse Recovery Formulation} \vspace{0.5 cm} To recover the neighbors of $v_1$, observe which nodes are {\color{red} red} (1) or {\color{blue} blue} (0) at time step $t$: \begin{align*} &v_2 \hspace{0.2 cm} v_3 \hspace{0.2 cm} v_4 \hspace{0.2 cm} v_5 &\\ \vspace{1 cm} M = & \left( \begin{array}{cccc} 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 \\ \end{array} \right) & \begin{array}{l} \hspace{ - 5cm} \text{time step 0} \\ \hspace{ - 5cm} \text{time step 1} \\ \end{array} \end{align*} and which color $v_1$ is at time step $t+1$ due to $M$: \begin{align*} b_1 = & \left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right) & \begin{array}{l} \hspace{ - 5cm} \text{time step 1} \\ \hspace{ - 5cm} \text{time step 2} \\ \end{array} \end{align*} Then , \begin{equation} \boxed{M \vec x_1 = \vec b_1 + \epsilon \nonumber} \end{equation} where $(\vec x_{1})_j := \frac{\text{1}}{\text{deg}(i)} \cdot \left[\text{j parent of 1 in }{\cal G}\right] $ \end{block} %--------------------------------------------------------------------------------- %--------------------------------------------------------------------------------- %--------------------------------------------------------------------------------- %--------------------------------------------------------------------------------- %--------------------------------------------------------------------------------- %--------------------------------------------------------------------------------- \end{column} % End of the first column \begin{column}{\sepwid}\end{column} % Empty spacer column \begin{column}{\onecolwid} % The first column %---------------------------------------------------------------------------------------- % CONSTRAINT SATISFACTION - BACKTRACKING %---------------------------------------------------------------------------------------- \begin{block}{Independent Cascades Model} \begin{figure} \centering \includegraphics[width=0.6\textwidth]{images/icc.png} \end{figure} \vspace{0.5 cm} {\bf Description} \vspace{0.5 cm} \begin{itemize} \item Three possible states: {\color{blue} susceptible}, {\color{red} infected}, {\color{yellow} inactive } \item $\mathbb{P}$(infected at t=0)$=p_{\text{init}}$ \item Infected node $j$ infects its susceptible neighbors $i$ with probability $p_{j,i}$ independently \end{itemize} \vspace{0.5 cm} {\bf Sparse Recovery Formulation} \vspace{0.5 cm} To recover the neighbors of $v_5$,observe which nodes are {\color{red} red} (1), {\color{blue} blue} (0), or {\color{yellow} yellow} (0) at time step $t$: \begin{align*} &v_1 \hspace{0.2 cm} v_2 \hspace{0.2 cm} v_3 \hspace{0.2 cm} v_4 \\ \vspace{1 cm} M = & \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ \end{array} \right) \begin{array}{l} \hspace{ 1cm} \text{time step 0} \\ \hspace{ 1cm} \text{time step 1} \\ \end{array} \end{align*} and if $M$ caused $v_5$ to be infected at time step $t+1$: \begin{align*} b_5 = & \left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right) \begin{array}{l} \hspace{ 1cm} \text{time step 1} \\ \hspace{ 1cm} \text{time step 2} \\ \end{array} \end{align*} Then, \begin{equation} \boxed{e^{M \vec \theta_5} = (1 - \vec b_5) + \epsilon} \nonumber \end{equation} where $(\vec \theta_5)_j := \log ( 1 - p_{j,5}) $ \vspace{1 cm} This problem is a {\bf Noisy Sparse Recovery} problem, which has been studied extensively. Here, the vectors $\vec x_i$ are deg(i)-sparse. \end{block} %---------------------------------------------------------------------------------------- %---------------------------------------------------------------------------------------- % MIP %---------------------------------------------------------------------------------------- % \begin{block}{RIP property} % %The Restricted Isometry Property (RIP) characterizes a quasi-orthonormality of the measurement matrix M on sparse vectors. % For all k, we define $\delta_k$ as the best possible constant such that for all unit-normed ($l$2) and k-sparse vectors $x$ % \begin{equation} % 1-\delta_k \leq \|Mx\|^2_2 \leq 1 + \delta_k % \end{equation} % In general, the smaller $\delta_k$ is, the better we can recover $k$-sparse vectors $x$! % \end{block} %---------------------------------------------------------------------------------------- \end{column} \begin{column}{\sepwid}\end{column} % Empty spacer column \begin{column}{\onecolwid} % The first column within column 2 (column 2.1) %---------------------------------------------------------------------------------------- \begin{block}{Algorithms} {\bf Voter Model} \begin{itemize} \item Solve for each node i: \begin{equation} \min_{\vec x_i} \|\vec x_i\|_1 + \lambda \|M \vec x_i - \vec b_i \|_2 \nonumber \end{equation} \end{itemize} {\bf Independent Cascade Model} \begin{itemize} \item Solve for each node i: \begin{equation} \min_{\vec \theta_i} \|\vec \theta_i\|_1 + \lambda \|e^{M \vec \theta_i} - (1 - \vec b_i) \|_2 \nonumber \end{equation} \end{itemize} \end{block} \begin{block}{Restricted Isometry Property (RIP)} {\bf Definition} \begin{itemize} \item Characterizes a quasi-orthonormality of M on sparse vectors. \item The RIP constant $\delta_k$ is the best possible constant such that for all unit-normed ($l$2) and k-sparse vectors $x$: \begin{equation} 1-\delta_k \leq \|Mx\|^2_2 \leq 1 + \delta_k \nonumber \end{equation} \item The smaller $\delta_k$ is, the better we can recover $k$-sparse vectors $x$. \end{itemize} \end{block} \begin{block}{Theoretical Guarantees} With small RIP constants $(\delta \leq 0.25)$ for $M$ and some assumption on the noise $\epsilon$: {\bf Theorem \cite{candes}} If node $i$ has degree $\Delta$ and $n_{\text{rows}}(M) \geq C_1 \mu \Delta \log n$, then, w.h.p., $$\| \hat x - x^* \|_2 \leq C (1 + \log^{3/2}(n))\sqrt{\frac{\Delta \log n}{n_{\text{rows}}(M) }}$$ \end{block} %---------------------------------------------------------------------------------------- % RESULTS %---------------------------------------------------------------------------------------- \begin{block}{RIP Experiments} \begin{center} \begin{table} \begin{tabular}{c | c | c | c | c } & $c$ = 100, &$c$ = 1000,& $c$ = 100, &$c$ = 1000,\\ & $i$ = 0.1& $i$ = 0.1& $i$ = 0.05& $i$ = 0.05\\ \hline $\delta_4$ & 0.54 & 0.37 &0.43&0.23 \\ \end{tabular} \caption{RIP constant for a small graph. Here, $c$ is the number of cascades and $i$ is $p_{\text{init}}$.} \end{table} \end{center} \end{block} %---------------------------------------------------------------------------------------- \end{column} % End of the second column \begin{column}{\sepwid}\end{column} % Empty spacer column \begin{column}{\onecolwid} % The third column %---------------------------------------------------------------------------------------- % IOVERALL COMPARISON %---------------------------------------------------------------------------------------- %\vspace{- 14.2 cm} %\begin{center} %{\includegraphics[height=7em]{cmu_logo.png}} % First university/lab logo on the left %\end{center} %\vspace{4 cm} \begin{alertblock}{Experimental Results} \end{alertblock} %---------------------------------------------------------------------------------------- %---------------------------------------------------------------------------------------- % CONCLUSION %---------------------------------------------------------------------------------------- \begin{block}{Conclusion} \begin{center} {\bf Graph reconstruction can naturally be expressed as Sparse Recovery. Understanding properties of $M$, for example RIP, leads to theoretical guarantees on the reconstruction.} \end{center} \end{block} %---------------------------------------------------------------------------------------- % REFERENCES %---------------------------------------------------------------------------------------- \begin{block}{References} \begin{thebibliography}{42} \bibitem{candes} Candès, E., and Plan, Y. \newblock {\it A Probabilistic and RIPless Theory of Compressed Sensing} \newblock Information Theory, IEEE Transactions on, 57(11): 7235--7254, \newblock 2011. \end{thebibliography} \end{block} %---------------------------------------------------------------------------------------- \end{column} % End of the third column \end{columns} % End of all the columns in the poster \end{frame} % End of the enclosing frame \end{document}