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@@ -3,4 +3,6 @@ *.aux *.bbl *.blg +*.out +*.brf /*.pdf diff --git a/appendix.tex b/appendix.tex index b1d35d6..53bc328 100644 --- a/appendix.tex +++ b/appendix.tex @@ -22,7 +22,8 @@ Our mechanism for \EDP{} is monotone and budget feasible. \frac{B}{2}\frac{V(S_i\cup\{i\})-V(S_i)}{V(S_i\cup\{i\})} \leq \frac{B}{2}\frac{V(S_i'\cup\{i\})-V(S_i')}{V(S_i'\cup\{i\})} \end{align*} - by the monotonicity and submodularity of $V$. Hence $i\in S_G'$. As $OPT'_{-i^*}$, is the optimal value of \eqref{relax} under relaxation $L$ when $i^*$ is excluded from $\mathcal{N}$, reducing the costs can only increase this value, so under $c'_i\leq c_i$ the greedy set is still allocated and $s_i(c_i',c_{-i}) =1$. + by the monotonicity and submodularity of $V$. Hence $i\in S_G'$. As + $OPT'_{-i^*}$, is the optimal value of \eqref{eq:primal} under relaxation $L$ when $i^*$ is excluded from $\mathcal{N}$, reducing the costs can only increase this value, so under $c'_i\leq c_i$ the greedy set is still allocated and $s_i(c_i',c_{-i}) =1$. Suppose now that when $i$ reports $c_i$, $OPT'_{-i^*} < C V(i^*)$. Then $s_i(c_i,c_{-i})=1$ iff $i = i^*$. Reporting $c_{i^*}'\leq c_{i^*}$ does not change $V(i^*)$ nor $OPT'_{-i^*} \leq C V(i^*)$; thus $s_{i^*}(c_{i^*}',c_{-i^*})=1$, so the mechanism is monotone. diff --git a/approximation.tex b/approximation.tex index 8b13789..ced53ed 100644 --- a/approximation.tex +++ b/approximation.tex @@ -1 +1,600 @@ +Even though \EDP{} is NP-hard, designing a mechanism for this problem will +involve being able to find an approximation $\tilde{L}^*(c)$ of $OPT$ +monotonous with respect to coordinate-wise changes of the cost: if $c$ and $c'$ +are two cost vectors such that $c'=(c_i', c_{-i})$ with $c_i' \leq c_i$, then +we want $\tilde{L}(c')\geq \tilde{L}(c)$. Furthermore, we seek an approximation +that can be computed in polynomial time. +This approximation will be obtained by introducing a concave optimization +problem with a constant approximation ratio to \EDP{} +(Proposition~\ref{prop:relaxation}). Using Newton's method, it is then +possible to solve this concave optimization problem to an arbitrary precision. +However, this approximation breaks the monotonicity of the approximation. +Finding a monotone approximate solution to the concave problem will be the +object of (Section~\ref{sec:monotonicity}). + +\subsection{A concave relaxation of \EDP}\label{sec:concave} + +Let us introduce a new function $L$: +\begin{equation}\label{eq:our-relaxation} +\forall\,\lambda\in[0,1]^n,\quad L(\lambda) \defeq +\log\det\left(I_d + \sum_{i\in\mathcal{N}} \lambda_i x_i\T{x_i}\right), +\end{equation} + +This function is a relaxation of the value function $V$ defined in +\eqref{modified} in the following sense: $L(\id_S) = V(S)$ for all +$S\subseteq\mathcal{N}$, where $\id_S$ denotes the indicator vector of $S$. + +The optimization program \eqref{eq:non-strategic} extends naturally to such +a relaxation. We define: +\begin{equation}\tag{$P_c$}\label{eq:primal} + L^*_c \defeq \max_{\lambda\in[0, 1]^{n}} + \left\{L(\lambda) \Big| \sum_{i=1}^{n} \lambda_i c_i + \leq B\right\} +\end{equation} + +\begin{proposition}\label{prop:relaxation} + $ L^*(c) \leq 2 OPT + + 2\max_{i\in\mathcal{N}}V(i)$. +\end{proposition} + +The proof of this proposition follows the \emph{pipage rounding} framework of +\citeN{pipage}. + +This framework uses the \emph{multi-linear} extension $F$ of the submodular +function $V$. Let $P_\mathcal{N}^\lambda(S)$ be the probability of choosing the +set $S$ if we select each element $i$ in $\mathcal{N}$ independently with +probability $\lambda_i$: +\begin{displaymath} + P_\mathcal{N}^\lambda(S) \defeq \prod_{i\in S} \lambda_i + \prod_{i\in\mathcal{N}\setminus S}( 1 - \lambda_i). +\end{displaymath} +Then, the \emph{multi-linear} extension $F$ is defined by: +\begin{displaymath} + F(\lambda) + \defeq \mathbb{E}_{S\sim P_\mathcal{N}^\lambda}\big[V(S)\big] + = \sum_{S\subseteq\mathcal{N}} P_\mathcal{N}^\lambda(S) V(S) +\end{displaymath} + +For \EDP{} the multi-linear extension can be written: +\begin{equation}\label{eq:multi-linear-logdet} + F(\lambda) = \mathbb{E}_{S\sim + P_\mathcal{N}^\lambda}\bigg[\log\det \big(I_d + \sum_{i\in S} x_i\T{x_i}\big) \Big]. +\end{equation} +Note that the relaxation $L$ that we introduced in \eqref{eq:our-relaxation}, +follows naturally from the \emph{multi-linear} relaxation by swapping the +expectation and the $\log\det$ in \eqref{eq:multi-linear-logdet}: +\begin{displaymath} + L(\lambda) = \log\det\left(\mathbb{E}_{S\sim + P_\mathcal{N}^\lambda}\bigg[I_d + \sum_{i\in S} x_i\T{x_i} \bigg]\right). +\end{displaymath} + +The proof proceeds as follows: +\begin{itemize} +\item First, we prove that $F$ admits the following rounding property: let +$\lambda$ be a feasible element of $[0,1]^n$, it is possible to trade one +fractional component of $\lambda$ for another until one of them becomes +integral, obtaining a new element $\tilde{\lambda}$ which is both feasible and +for which $F(\tilde{\lambda})\geq F(\lambda)$. Here, by feasibility of a point +$\lambda$, we mean that it satisfies the budget constraint $\sum_{i=1}^n +\lambda_i c_i \leq B$. This rounding property is referred to in the literature +as \emph{cross-convexity} (see, \emph{e.g.}, \cite{dughmi}), or +$\varepsilon$-convexity by \citeN{pipage}. This is stated and proven in +Lemma~\ref{lemma:rounding} and allows us to bound $F$ in terms of $OPT$. +\item Next, we prove the central result of bounding $L$ appropriately in terms +of the multi-linear relaxation $F$ (Lemma \ref{lemma:relaxation-ratio}). +\item Finally, we conclude the proof of Proposition~\ref{prop:relaxation} by +combining Lemma~\ref{lemma:rounding} and Lemma~\ref{lemma:relaxation-ratio}. +\end{itemize} + +\begin{lemma}[Rounding]\label{lemma:rounding} + For any feasible $\lambda\in[0,1]^{n}$, there exists a feasible + $\bar{\lambda}\in[0,1]^{n}$ such that at most one of its components is + fractional %, that is, lies in $(0,1)$ and: + and $F_{\mathcal{N}}(\lambda)\leq F_{\mathcal{N}}(\bar{\lambda})$. +\end{lemma} +\begin{proof} + We give a rounding procedure which, given a feasible $\lambda$ with at least + two fractional components, returns some feasible $\lambda'$ with one less fractional + component such that $F(\lambda) \leq F(\lambda')$. + + Applying this procedure recursively yields the lemma's result. + Let us consider such a feasible $\lambda$. Let $i$ and $j$ be two + fractional components of $\lambda$ and let us define the following + function: + \begin{displaymath} + F_\lambda(\varepsilon) = F(\lambda_\varepsilon) + \quad\textrm{where} \quad + \lambda_\varepsilon = \lambda + \varepsilon\left(e_i-\frac{c_i}{c_j}e_j\right) + \end{displaymath} + It is easy to see that if $\lambda$ is feasible, then: + \begin{equation}\label{eq:convex-interval} + \forall\varepsilon\in\Big[\max\Big(-\lambda_i,(\lambda_j-1)\frac{c_j}{c_i}\Big), \min\Big(1-\lambda_i, \lambda_j + \frac{c_j}{c_i}\Big)\Big],\; + \lambda_\varepsilon\;\;\textrm{is feasible} + \end{equation} + Furthermore, the function $F_\lambda$ is convex; indeed: + \begin{align*} + F_\lambda(\varepsilon) + & = \mathbb{E}_{S'\sim P_{\mathcal{N}\setminus\{i,j\}}^\lambda(S')}\Big[ + (\lambda_i+\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i,j\})\\ + & + (\lambda_i+\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i\}) + + (1-\lambda_i-\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{j\})\\ + & + (1-\lambda_i-\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S')\Big] + \end{align*} + Thus, $F_\lambda$ is a degree 2 polynomial whose dominant coefficient is: + \begin{displaymath} + \frac{c_i}{c_j}\mathbb{E}_{S'\sim + P_{\mathcal{N}\setminus\{i,j\}}^\lambda(S')}\Big[ + V(S'\cup\{i\})+V(S'\cup\{i\})\\ + -V(S'\cup\{i,j\})-V(S')\Big] + \end{displaymath} + which is positive by submodularity of $V$. Hence, the maximum of + $F_\lambda$ over the interval given in \eqref{eq:convex-interval} is + attained at one of its limit, at which either the $i$-th or $j$-th component of + $\lambda_\varepsilon$ becomes integral. +\end{proof} + +\begin{lemma}\label{lemma:relaxation-ratio} + % The following inequality holds: +For all $\lambda\in[0,1]^{n},$ + %\begin{displaymath} + $ \frac{1}{2} + \,L(\lambda)\leq + F(\lambda)\leq L(\lambda)$. + %\end{displaymath} +\end{lemma} +\begin{proof} + The bound $F_{\mathcal{N}}(\lambda)\leq L_{\mathcal{N}(\lambda)}$ follows by the concavity of the $\log\det$ function. + To show the lower bound, + we first prove that $\frac{1}{2}$ is a lower bound of the ratio $\partial_i + F(\lambda)/\partial_i L(\lambda)$, where + $\partial_i\, \cdot$ denotes the partial derivative with respect to the + $i$-th variable. + + Let us start by computing the derivatives of $F$ and + $L$ with respect to the $i$-th component. + Observe that + \begin{displaymath} + \partial_i P_\mathcal{N}^\lambda(S) = \left\{ + \begin{aligned} + & P_{\mathcal{N}\setminus\{i\}}^\lambda(S\setminus\{i\})\;\textrm{if}\; + i\in S, \\ + & - P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\;\textrm{if}\; + i\in \mathcal{N}\setminus S. \\ + \end{aligned}\right. + \end{displaymath} + Hence, + \begin{displaymath} + \partial_i F(\lambda) = + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S\setminus\{i\})V(S) + - \sum_{\substack{S\subseteq\mathcal{N}\\ i\in \mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S)V(S). + \end{displaymath} + Now, using that every $S$ such that $i\in S$ can be uniquely written as + $S'\cup\{i\}$, we can write: + \begin{displaymath} + \partial_i F(\lambda) = + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\big(V(S\cup\{i\}) + - V(S)\big). + \end{displaymath} + The marginal contribution of $i$ to + $S$ can be written as +\begin{align*} +V(S\cup \{i\}) - V(S)& = \frac{1}{2}\log\det(I_d + + \T{X_S}X_S + x_i\T{x_i}) + - \frac{1}{2}\log\det(I_d + \T{X_S}X_S)\\ + & = \frac{1}{2}\log\det(I_d + x_i\T{x_i}(I_d + +\T{X_S}X_S)^{-1}) + = \frac{1}{2}\log(1 + \T{x_i}A(S)^{-1}x_i) +\end{align*} +where $A(S) \defeq I_d+ \T{X_S}X_S$, and the last equality follows from the +Sylvester's determinant identity~\cite{sylvester}. +% $ V(S\cup\{i\}) - V(S) = \frac{1}{2}\log\left(1 + \T{x_i} A(S)^{-1}x_i\right)$. +Using this, + \begin{displaymath} + \partial_i F(\lambda) = \frac{1}{2} + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S) + \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big) + \end{displaymath} + The computation of the derivative of $L$ uses standard matrix + calculus: writing $\tilde{A}(\lambda) = I_d+\sum_{i\in + \mathcal{N}}\lambda_ix_i\T{x_i}$, + \begin{displaymath} + \det \tilde{A}(\lambda + h\cdot e_i) = \det\big(\tilde{A}(\lambda) + + hx_i\T{x_i}\big) + =\det \tilde{A}(\lambda)\big(1+ + h\T{x_i}\tilde{A}(\lambda)^{-1}x_i\big). + \end{displaymath} + Hence, + \begin{displaymath} + \log\det\tilde{A}(\lambda + h\cdot e_i)= \log\det\tilde{A}(\lambda) + + h\T{x_i}\tilde{A}(\lambda)^{-1}x_i + o(h), + \end{displaymath} + which implies + \begin{displaymath} + \partial_i L(\lambda) + =\frac{1}{2} \T{x_i}\tilde{A}(\lambda)^{-1}x_i. + \end{displaymath} + +For two symmetric matrices $A$ and $B$, we write $A\succ B$ ($A\succeq B$) if +$A-B$ is positive definite (positive semi-definite). This order allows us to +define the notion of a \emph{decreasing} as well as \emph{convex} matrix +function, similarly to their real counterparts. With this definition, matrix +inversion is decreasing and convex over symmetric positive definite +matrices (see Example 3.48 p. 110 in \cite{boyd2004convex}). +In particular, +\begin{gather*} + \forall S\subseteq\mathcal{N},\quad A(S)^{-1} \succeq A(S\cup\{i\})^{-1} +\end{gather*} +as $A(S)\preceq A(S\cup\{i\})$. Observe that + % \begin{gather*} + % \forall +$P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\geq P_{\mathcal{N}\setminus\{i\}}^\lambda(S\cup\{i\})$ for all $S\subseteq\mathcal{N}\setminus\{i\}$, and + % ,\\ + $P_{\mathcal{N}\setminus\{i\}}^\lambda(S) \geq P_\mathcal{N}^\lambda(S),$ for all $S\subseteq\mathcal{N}$. + %\end{gather*} + Hence, + \begin{align*} + \partial_i F(\lambda) + % & = \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} P_\mathcal{N}^\lambda(S) \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big)\\ + & \geq \frac{1}{4} + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S) + \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big)\\ + &\hspace{-3.5em}+\frac{1}{4} + \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} + P_{\mathcal{N}\setminus\{i\}}^\lambda(S\cup\{i\}) + \log\Big(1 + \T{x_i}A(S\cup\{i\})^{-1}x_i\Big)\\ + &\geq \frac{1}{4} + \sum_{S\subseteq\mathcal{N}} + P_\mathcal{N}^\lambda(S) + \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big). + \end{align*} + Using that $A(S)\succeq I_d$ we get that $\T{x_i}A(S)^{-1}x_i \leq + \norm{x_i}_2^2 \leq 1$. Moreover, $\log(1+x)\geq x$ for all $x\leq 1$. + Hence, + \begin{displaymath} + \partial_i F(\lambda) \geq + \frac{1}{4} + \T{x_i}\bigg(\sum_{S\subseteq\mathcal{N}}P_\mathcal{N}^\lambda(S)A(S)^{-1}\bigg)x_i. + \end{displaymath} + Finally, using that the inverse is a matrix convex function over symmetric + positive definite matrices: + \begin{displaymath} + \partial_i F(\lambda) \geq + \frac{1}{4} + \T{x_i}\bigg(\sum_{S\subseteq\mathcal{N}}P_\mathcal{N}^\lambda(S)A(S)\bigg)^{-1}x_i + = \frac{1}{4}\T{x_i}\tilde{A}(\lambda)^{-1}x_i + = \frac{1}{2} + \partial_i L(\lambda). + \end{displaymath} + +Having bound the ratio between the partial derivatives, we now bound the ratio $F(\lambda)/L(\lambda)$ from below. Consider the following cases. + First, if the minimum of the ratio + $F(\lambda)/L(\lambda)$ is attained at a point interior to the hypercube, then it is + a critical point, \emph{i.e.}, $\partial_i \big(F(\lambda)/L(\lambda)\big)=0$ for all $i\in \mathcal{N}$; hence, at such a critical point: + \begin{equation}\label{eq:lhopital} + \frac{F(\lambda)}{L(\lambda)} + = \frac{\partial_i F(\lambda)}{\partial_i + L(\lambda)} \geq \frac{1}{2}. + \end{equation} + Second, if the minimum is attained as + $\lambda$ converges to zero in, \emph{e.g.}, the $l_2$ norm, by the Taylor approximation, one can write: + \begin{displaymath} + \frac{F(\lambda)}{L(\lambda)} + \sim_{\lambda\rightarrow 0} + \frac{\sum_{i\in \mathcal{N}}\lambda_i\partial_i F(0)} + {\sum_{i\in\mathcal{N}}\lambda_i\partial_i L(0)} + \geq \frac{1}{2}, + \end{displaymath} + \emph{i.e.}, the ratio $\frac{F(\lambda)}{L(\lambda)}$ is necessarily bounded from below by 1/2 for small enough $\lambda$. + Finally, if the minimum is attained on a face of the hypercube $[0,1]^n$ (a face is + defined as a subset of the hypercube where one of the variable is fixed to + 0 or 1), without loss of generality, we can assume that the minimum is + attained on the face where the $n$-th variable has been fixed + to 0 or 1. Then, either the minimum is attained at a point interior to the + face or on a boundary of the face. In the first sub-case, relation + \eqref{eq:lhopital} still characterizes the minimum for $i< n$. + In the second sub-case, by repeating the argument again by induction, we see + that all is left to do is to show that the bound holds for the vertices of + the cube (the faces of dimension 1). The vertices are exactly the binary + points, for which we know that both relaxations are equal to the value + function $V$. Hence, the ratio is equal to 1 on the vertices. +\end{proof} + +To conclude the proof of Proposition~\ref{prop:relaxation}, let us consider +a feasible point $\lambda^*\in[0,1]^{n}$ such that $L(\lambda^*) = L^*_c$. By +applying Lemma~\ref{lemma:relaxation-ratio} and Lemma~\ref{lemma:rounding} we +get a feasible point $\bar{\lambda}$ with at most one fractional component such +that +\begin{equation}\label{eq:e1} + L(\lambda^*) \leq 2 F(\bar{\lambda}). +\end{equation} + Let $\lambda_i$ denote the fractional component of $\bar{\lambda}$ and $S$ + denote the set whose indicator vector is $\bar{\lambda} - \lambda_i e_i$. + By definition of the multi-linear extension $F$: + \begin{displaymath} + F(\bar{\lambda}) = (1-\lambda_i)V(S) +\lambda_i V(S\cup\{i\}). + \end{displaymath} + By submodularity of $V$, $V(S\cup\{i\})\leq V(S) + V(\{i\})$. Hence, + \begin{displaymath} + F(\bar{\lambda}) \leq V(S) + V(i). + \end{displaymath} + Note that since $\bar{\lambda}$ is feasible, $S$ is also feasible and + $V(S)\leq OPT$. Hence, + \begin{equation}\label{eq:e2} + F(\bar{\lambda}) \leq OPT + \max_{i\in\mathcal{N}} V(i). + \end{equation} + Together, \eqref{eq:e1} and \eqref{eq:e2} imply the lemma. \hspace*{\stretch{1}}\qed + +\subsection{A monotonous Newton's estimator}\label{sec:monotonicity} + +\textbf{TODO} Explain that we only get approximate monotonicity, but even +that is not immediate since the variation induced by a change of cost on +coordinate $i$ depends on the allocation at this coordinate which can be +arbitrarily small. + +For the ease of presentation, we normalize the costs by dividing them by the +budget $B$ so that the budget constraint in \eqref{eq:primal} now reads +$\T{c}\lambda\leq 1$. + +\begin{proposition}\label{prop:monotonicity} + Let $\delta\in(0,1]$. For any $\varepsilon\in(0,1]$, there exists + an algorithm which computes an approximate solution $\tilde{L}^*_c$ to + \eqref{eq:primal} such that: + \begin{enumerate} + \item $|\tilde{L}^*_c - L^*_c| \leq \varepsilon$ + \item for all $c' = (c_i', c_{-i})$ with $c_i'\leq c_i-\delta$, $\tilde{L}^*_c \leq \tilde{L}^*_{c'}$ + \item the routine's running time is $O\big(poly(n, d, \log\log\frac{1}{b\varepsilon\delta})\big)$ + \end{enumerate} +\end{proposition} + +We consider a perturbation of \eqref{eq:primal} by introducing: +\begin{equation}\tag{$P_{c, \alpha}$}\label{eq:perturbed-primal} + L^*_c(\alpha) \defeq \max_{\lambda\in[\alpha, 1]^{n}} + \left\{L(\lambda) \Big| \sum_{i=1}^{n} \lambda_i c_i + \leq B\right\} +\end{equation} +Note that we have $L^*_c = L^*_c(0)$. We will also assume that +$\alpha<\frac{1}{n}$ so that \eqref{eq:perturbed-primal} has at least one +feasible point: $(\frac{1}{n},\ldots,\frac{1}{n})$. + +Having introduced this perturbed problem, we show that its optimal value is +close to the optimal value of \eqref{eq:primal} (Lemma~\ref{lemma:proximity}) +while being well-behaved with respect to changes of the cost +(Lemma~\ref{lemma:monotonicity}). These lemmas together imply +Proposition~\ref{prop:monotonicity}. + + +\begin{lemma}\label{lemma:derivative-bounds} + Let $\partial_i L(\lambda)$ denote the $i$-th derivative of $L$, for $i\in\{1,\ldots, n\}$, then: + \begin{displaymath} + \forall\lambda\in[0, 1]^n,\;\frac{b}{2^n} \leq \partial_i L(\lambda) \leq 1 + \end{displaymath} +\end{lemma} + +\begin{proof} + Let us define: + \begin{displaymath} + S(\lambda)\defeq I_d + \sum_{i=1}^n \lambda_i x_i\T{x_i} + \quad\mathrm{and}\quad + S_k \defeq I_d + \sum_{i=1}^k x_i\T{x_i} + \end{displaymath} + + We have $\partial_i L(\lambda) = \T{x_i}S(\lambda)^{-1}x_i$. Since + $S(\lambda)\geq I_d$, $\partial_i L(\lambda)\leq \T{x_i}x_i \leq 1$, which + is the right-hand side of the lemma. + + For the left-hand side, note that $S(\lambda) \leq S_n$. Hence + $\partial_iL(\lambda)\geq \T{x_i}S_n^{-1}x_i$. + + Using the Sherman-Morrison formula, for all $k\geq 1$: + \begin{displaymath} + \T{x_i}S_k^{-1} x_i = \T{x_i}S_{k-1}^{-1}x_i + - \frac{(\T{x_i}S_{k-1}^{-1}x_k)^2}{1+\T{x_k}S_{k-1}^{-1}x_k} + \end{displaymath} + + By the Cauchy-Schwarz inequality: + \begin{displaymath} + (\T{x_i}S_{k-1}^{-1}x_k)^2 \leq \T{x_i}S_{k-1}^{-1}x_i\;\T{x_k}S_{k-1}^{-1}x_k + \end{displaymath} + + Hence: + \begin{displaymath} + \T{x_i}S_k^{-1} x_i \geq \T{x_i}S_{k-1}^{-1}x_i + - \T{x_i}S_{k-1}^{-1}x_i\frac{\T{x_k}S_{k-1}^{-1}x_k}{1+\T{x_k}S_{k-1}^{-1}x_k} + \end{displaymath} + + But $\T{x_k}S_{k-1}^{-1}x_k\leq 1$ and $\frac{a}{1+a}\leq \frac{1}{2}$ if + $0\leq a\leq 1$, so: + \begin{displaymath} + \T{x_i}S_{k}^{-1}x_i \geq \T{x_i}S_{k-1}^{-1}x_i + - \frac{1}{2}\T{x_i}S_{k-1}^{-1}x_i\geq \frac{\T{x_i}S_{k-1}^{-1}x_i}{2} + \end{displaymath} + + By induction: + \begin{displaymath} + \T{x_i}S_n^{-1} x_i \geq \frac{\T{x_i}x_i}{2^n} + \end{displaymath} + + Using that $\T{x_i}{x_i}\geq b$ concludes the proof of the left-hand side + of the lemma's inequality. +\end{proof} + +Let us introduce the lagrangian of problem, \eqref{eq:perturbed-primal}: + +\begin{displaymath} + \mathcal{L}_{c, \alpha}(\lambda, \mu, \nu, \xi) \defeq L(\lambda) + + \T{\mu}(\lambda-\alpha\mathbf{1}) + \T{\nu}(\mathbf{1}-\lambda) + \xi(1-\T{c}\lambda) +\end{displaymath} +so that: +\begin{displaymath} + L^*_c(\alpha) = \min_{\mu, \nu, \xi\geq 0}\max_\lambda \mathcal{L}_{c, \alpha}(\lambda, \mu, \nu, \xi) +\end{displaymath} +Similarly, we define $\mathcal{L}_{c}\defeq\mathcal{L}_{c, 0}$ the lagrangian of \eqref{eq:primal}. + +Let $\lambda^*$ be primal optimal for \eqref{eq:perturbed-primal}, and $(\mu^*, +\nu^*, \xi^*)$ be dual optimal for the same problem. In addition to primal and +dual feasibility, the KKT conditions give $\forall i\in\{1, \ldots, n\}$: +\begin{gather*} + \partial_i L(\lambda^*) + \mu_i^* - \nu_i^* - \xi^* c_i = 0\\ + \mu_i^*(\lambda_i^* - \alpha) = 0\\ + \nu_i^*(1 - \lambda_i^*) = 0 +\end{gather*} + +\begin{lemma}\label{lemma:proximity} +We have: +\begin{displaymath} + L^*_c - \alpha n^2\leq L^*_c(\alpha) \leq L^*_c +\end{displaymath} +In particular, $|L^*_c - L^*_c(\alpha)| \leq \alpha n^2$. +\end{lemma} + +\begin{proof} + $\alpha\mapsto L^*_c(\alpha)$ is a decreasing function as it is the + maximum value of the $L$ function over a set-decreasing domain, which gives + the rightmost inequality. + + Let $\mu^*, \nu^*, \xi^*$ be dual optimal for $(P_{c, \alpha})$, that is: + \begin{displaymath} + L^*_{c}(\alpha) = \max_\lambda \mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*) + \end{displaymath} + + Note that $\mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*) + = \mathcal{L}_{c}(\lambda, \mu^*, \nu^*, \xi^*) + - \alpha\T{\mathbf{1}}\mu^*$, and that for any $\lambda$ feasible for + problem \eqref{eq:primal}, $\mathcal{L}_{c}(\lambda, \mu^*, \nu^*, \xi^*) + \geq L(\lambda)$. Hence, + \begin{displaymath} + L^*_{c}(\alpha) \geq L(\lambda) - \alpha\T{\mathbf{1}}\mu^* + \end{displaymath} + for any $\lambda$ feasible for \eqref{eq:primal}. In particular, for $\lambda$ primal optimal for $\eqref{eq:primal}$: + \begin{equation}\label{eq:local-1} + L^*_{c}(\alpha) \geq L^*_c - \alpha\T{\mathbf{1}}\mu^* + \end{equation} + + Let us denote by the $M$ the support of $\mu^*$, that is $M\defeq + \{i|\mu_i^* > 0\}$, and by $\lambda^*$ a primal optimal point for + $\eqref{eq:perturbed-primal}$. From the KKT conditions we see that: + \begin{displaymath} + M \subseteq \{i|\lambda_i^* = \alpha\} + \end{displaymath} + + + Let us first assume that $|M| = 0$, then $\T{\mathbf{1}}\mu^*=0$ and the lemma follows. + + We will now assume that $|M|\geq 1$. In this case $\T{c}\lambda^* + = 1$, otherwise we could increase the coordinates of $\lambda^*$ in $M$, + which would increase the value of the objective function and contradict the + optimality of $\lambda^*$. Note also, that $|M|\leq n-1$, otherwise, since + $\alpha< \frac{1}{n}$, we would have $\T{c}\lambda^*\ < 1$, which again + contradicts the optimality of $\lambda^*$. Let us write: + \begin{displaymath} + 1 = \T{c}\lambda^* = \alpha\sum_{i\in M}c_i + \sum_{i\in \bar{M}}\lambda_i^*c_i + \leq \alpha |M| + (n-|M|)\max_{i\in \bar{M}} c_i + \end{displaymath} + That is: + \begin{equation}\label{local-2} + \max_{i\in\bar{M}} c_i \geq \frac{1 - |M|\alpha}{n-|M|}> \frac{1}{n} + \end{equation} + where the last inequality uses again that $\alpha<\frac{1}{n}$. From the + KKT conditions, we see that for $i\in M$, $\nu_i^* = 0$ and: + \begin{equation}\label{local-3} + \mu_i^* = \xi^*c_i - \partial_i L(\lambda^*)\leq \xi^*c_i\leq \xi^* + \end{equation} + since $\partial_i L(\lambda^*)\geq 0$ and $c_i\leq 1$. + + Furthermore, using the KKT conditions again, we have that: + \begin{equation}\label{local-4} + \xi^* \leq \inf_{i\in \bar{M}}\frac{\partial_i L(\lambda^*)}{c_i}\leq \inf_{i\in \bar{M}} \frac{1}{c_i} + = \frac{1}{\max_{i\in\bar{M}} c_i} + \end{equation} + where the last inequality uses Lemma~\ref{lemma:derivative-bounds}. + + Combining \eqref{local-2}, \eqref{local-3} and \eqref{local-4}, we get that: + \begin{displaymath} + \sum_{i\in M}\mu_i^* \leq |M|\xi^* \leq n\xi^*\leq \frac{n}{\max_{i\in\bar{M}} c_i} \leq n^2 + \end{displaymath} + + This implies that: + \begin{displaymath} + \T{\mathbf{1}}\mu^* = \sum_{i=1}^n \mu^*_i = \sum_{i\in M}\mu_i^*\leq n^2 + \end{displaymath} + which in addition to \eqref{eq:local-1} proves the lemma. +\end{proof} + +\begin{lemma}\label{lemma:monotonicity} + If $c'$ = $(c_i', c_{-i})$, with $c_i'\leq c_i - \delta$, we have: + \begin{displaymath} + L^*_{c'}(\alpha) \geq L^*_c(\alpha) + \frac{\alpha\delta b}{2^n} + \end{displaymath} +\end{lemma} + +\begin{proof} + Let $\mu^*, \nu^*, \xi^*$ be dual optimal for $(P_{c', \alpha})$. Noting that: + \begin{displaymath} + \mathcal{L}_{c', \alpha}(\lambda, \mu^*, \nu^*, \xi^*) \geq + \mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*) + \lambda_i\xi^*\delta, + \end{displaymath} + we get similarly to Lemma~\ref{lemma:proximity}: + \begin{displaymath} + L^*_{c'}(\alpha) \geq L(\lambda) + \lambda_i\xi^*\delta + \end{displaymath} + for any $\lambda$ feasible for \eqref{eq:perturbed-primal}. In particular, for $\lambda^*$ primal optimal for \eqref{eq:perturbed-primal}: + \begin{displaymath} + L^*_{c'}(\alpha) \geq L^*_{c}(\alpha) + \alpha\xi^*\delta + \end{displaymath} + since $\lambda_i^*\geq \alpha$. + + Using the KKT conditions for $(P_{c', \alpha})$, we can write: + \begin{displaymath} + \xi^* = \inf_{i:\lambda^{'*}_i>\alpha} \frac{\T{x_i}S(\lambda^{'*})^{-1}x_i}{c_i'} + \end{displaymath} + with $\lambda^{'*}$ optimal for $(P_{c', \alpha})$. Since $c_i'\leq 1$, using Lemma~\ref{lemma:derivative-bounds}, we get that $\xi^*\geq \frac{b}{2^n}$, which concludes the proof. +\end{proof} + + +\subsubsection*{End of the proof of Proposition~\ref{prop:monotonicity}} + +Let $\varepsilon$ in $(0, 1]$. The routine works as follows: set $\alpha\defeq +\varepsilon(\delta + n^2)^{-1}$ and return an approximation $\tilde{L}^*_c$ of +$L^*_c(\alpha)$ with an accuracy $\frac{1}{2^{n+1}}\alpha\delta b$ computed by +a standard convex optimization algorithm. Note that this choice of $\alpha$ +implies $\alpha<\frac{1}{n}$ as required. + +\begin{enumerate} + \item using Lemma~\ref{lemma:proximity}: +\begin{displaymath} + |\tilde{L}^*_c - L^*_c| \leq |\tilde{L}^*_c - L^*_c(\alpha)| + |L^*_c(\alpha) - L^*_c| + \leq \alpha\delta + \alpha n^2 = \varepsilon +\end{displaymath} + +\item let $c' = (c_i', c_{-i})$ with $c_i'\leq c_i-\delta$, then: +\begin{displaymath} + \tilde{L}^*_{c'} \geq L^*_{c'} - \frac{\alpha\delta b}{2^{n+1}} + \geq L^*_c + \frac{\alpha\delta b}{2^{n+1}} + \geq \tilde{L}^*_c +\end{displaymath} +where the first and inequality come from the accuracy of the approximation, and +the inner inequality follows from Lemma~\ref{lemma:monotonicity}. + +\item the accuracy of the approximation $\tilde{L}^*_c$ is: +\begin{displaymath} + A\defeq\frac{\varepsilon\delta b}{2^{n+1}(\delta + n^2)} +\end{displaymath} + +The function $L$ is well-known to be concave and even self-concordant (see +\emph{e.g.}, \cite{boyd2004convex}). In this case, the analysis of Newton's +method for self-concordant functions in \cite{boyd2004convex}, shows that +finding the maximum of $L$ to any precision $A$ can be done in +$O(\log\log A^{-1})$ iterations. Note that: +\begin{displaymath} + \log\log A^{-1} = O\bigg(\log\log\frac{1}{\varepsilon\delta b} + \log n\bigg) +\end{displaymath} +Furthermore, each iteration of Newton's method can be done in time $O\big(poly(n, +d)\big)$.\qed +\end{enumerate} diff --git a/definitions.tex b/definitions.tex index 616012e..281b290 100644 --- a/definitions.tex +++ b/definitions.tex @@ -1,6 +1,7 @@ \newcommand{\mutual}{\ensuremath{{I}}} \newcommand{\entropy}{\ensuremath{{H}}} \newtheorem{lemma}{Lemma} +\newtheorem{proposition}{Proposition} \newtheorem{fact}{Fact} \newtheorem{example}{Example} \newtheorem{prop}{Proposition} diff --git a/dual.tex b/dual.tex deleted file mode 100644 index a2d28c6..0000000 --- a/dual.tex +++ /dev/null @@ -1,263 +0,0 @@ -\documentclass{article} -\usepackage[T1]{fontenc} -\usepackage[utf8]{inputenc} -\usepackage{amsmath, amsfonts, amsthm} -\newtheorem{proposition}{Proposition} -\input{definitions} - -\begin{document} -Let $c$ be a cost vector in $[0,1]^n$, and $x_1,\ldots,x_n$, $n$ vectors in -$\mathbf{R}^d$ such that for all $i\in\{1,\ldots,n\}$, $b\leq \T{x_i}{x_i}\leq -1$ for some $b\in(0,1]$. Let us consider the following convex optimization -problem: -\begin{equation}\tag{$P_c$}\label{eq:primal} - \begin{split} - \mathrm{maximize}\quad & L(\lambda) \defeq\log\det\left(I_d + \sum_{i=1}^n - \lambda_i x_i x_i^T\right)\\ - \textrm{subject to}\quad & c^T\lambda \leq 1 \;;\; 0\leq\lambda\leq \mathbf{1} -\end{split} -\end{equation} -We denote by $L^*_c$ its optimal value. - -Let $\alpha\in\mathbf{R}^+$, consider the perturbed optimization problem: -\begin{equation}\tag{$P_{c, \alpha}$}\label{eq:perturbed-primal} - \begin{split} - \mathrm{maximize}\quad & L(\lambda) \defeq\log\det\left(I_d + \sum_{i=1}^n - \lambda_i x_i x_i^T\right)\\ - \textrm{subject to}\quad & c^T\lambda \leq 1 \;;\; \alpha\leq\lambda\leq \mathbf{1} -\end{split} -\end{equation} -and denote by $L^*_c(\alpha)$ its optimal value. Note that we have $L^*_c = L^*_c(0)$. - -We will assume that $\alpha<\frac{1}{n}$ so that \eqref{eq:perturbed-primal} has at -least one feasible point: $(\frac{1}{n},\ldots,\frac{1}{n})$. - -Let us introduce the lagrangian of problem, \eqref{eq:perturbed-primal}: -\begin{displaymath} - \mathcal{L}_{c, \alpha}(\lambda, \mu, \nu, \xi) \defeq L(\lambda) - + \T{\mu}(\lambda-\alpha\mathbf{1}) + \T{\nu}(\mathbf{1}-\lambda) + \xi(1-\T{c}\lambda) -\end{displaymath} -so that: -\begin{displaymath} - L^*_c(\alpha) = \min_{\mu, \nu, \xi\geq 0}\max_\lambda \mathcal{L}_{c, \alpha}(\lambda, \mu, \nu, \xi) -\end{displaymath} -Similarly, we define $\mathcal{L}_{c}\defeq\mathcal{L}_{c, 0}$ the lagrangian of \eqref{eq:primal}. - -Let $\lambda^*$ be primal optimal for \eqref{eq:perturbed-primal}, and $(\mu^*, -\nu^*, \xi^*)$ be dual optimal for the same problem. In addition to primal and -dual feasibility, the KKT conditions give $\forall i\in\{1, \ldots, n\}$: -\begin{gather*} - \partial_i L(\lambda^*) + \mu_i^* - \nu_i^* - \xi^* c_i = 0\\ - \mu_i^*(\lambda_i^* - \alpha) = 0\\ - \nu_i^*(1 - \lambda_i^*) = 0 -\end{gather*} - -\begin{lemma}\label{lemma:derivative-bounds} - Let $\partial_i L(\lambda)$ denote the $i$-th derivative of $L$, for $i\in\{1,\ldots, n\}$, then: - \begin{displaymath} - \forall\lambda\in[0, 1]^n,\;\frac{b}{2^n} \leq \partial_i L(\lambda) \leq 1 - \end{displaymath} -\end{lemma} - -\begin{proof} - Let us define: - \begin{displaymath} - S(\lambda)\defeq I_d + \sum_{i=1}^n \lambda_i x_i\T{x_i} - \quad\mathrm{and}\quad - S_k \defeq I_d + \sum_{i=1}^k x_i\T{x_i} - \end{displaymath} - - We have $\partial_i L(\lambda) = \T{x_i}S(\lambda)^{-1}x_i$. Since - $S(\lambda)\geq I_d$, $\partial_i L(\lambda)\leq \T{x_i}x_i \leq 1$, which - is the right-hand side of the lemma. - - For the left-hand side, note that $S(\lambda) \leq S_n$. Hence - $\partial_iL(\lambda)\geq \T{x_i}S_n^{-1}x_i$. - - Using the Sherman-Morrison formula, for all $k\geq 1$: - \begin{displaymath} - \T{x_i}S_k^{-1} x_i = \T{x_i}S_{k-1}^{-1}x_i - - \frac{(\T{x_i}S_{k-1}^{-1}x_k)^2}{1+\T{x_k}S_{k-1}^{-1}x_k} - \end{displaymath} - - By the Cauchy-Schwarz inequality: - \begin{displaymath} - (\T{x_i}S_{k-1}^{-1}x_k)^2 \leq \T{x_i}S_{k-1}^{-1}x_i\;\T{x_k}S_{k-1}^{-1}x_k - \end{displaymath} - - Hence: - \begin{displaymath} - \T{x_i}S_k^{-1} x_i \geq \T{x_i}S_{k-1}^{-1}x_i - - \T{x_i}S_{k-1}^{-1}x_i\frac{\T{x_k}S_{k-1}^{-1}x_k}{1+\T{x_k}S_{k-1}^{-1}x_k} - \end{displaymath} - - But $\T{x_k}S_{k-1}^{-1}x_k\leq 1$ and $\frac{a}{1+a}\leq \frac{1}{2}$ if - $0\leq a\leq 1$, so: - \begin{displaymath} - \T{x_i}S_{k}^{-1}x_i \geq \T{x_i}S_{k-1}^{-1}x_i - - \frac{1}{2}\T{x_i}S_{k-1}^{-1}x_i\geq \frac{\T{x_i}S_{k-1}^{-1}x_i}{2} - \end{displaymath} - - By induction: - \begin{displaymath} - \T{x_i}S_n^{-1} x_i \geq \frac{\T{x_i}x_i}{2^n} - \end{displaymath} - - Using that $\T{x_i}{x_i}\geq b$ concludes the proof of the left-hand side - of the lemma's inequality. -\end{proof} - -\begin{lemma}\label{lemma:proximity} -We have: -\begin{displaymath} - L^*_c - \alpha n^2\leq L^*_c(\alpha) \leq L^*_c -\end{displaymath} -In particular, $|L^*_c - L^*_c(\alpha)| \leq \alpha n^2$. -\end{lemma} - -\begin{proof} - $\alpha\mapsto L^*_c(\alpha)$ is a decreasing function as it is the - maximum value of the $L$ function over a set-decreasing domain, which gives - the rightmost inequality. - - Let $\mu^*, \nu^*, \xi^*$ be dual optimal for $(P_{c, \alpha})$, that is: - \begin{displaymath} - L^*_{c}(\alpha) = \max_\lambda \mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*) - \end{displaymath} - - Note that $\mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*) - = \mathcal{L}_{c}(\lambda, \mu^*, \nu^*, \xi^*) - - \alpha\T{\mathbf{1}}\mu^*$, and that for any $\lambda$ feasible for - problem \eqref{eq:primal}, $\mathcal{L}_{c}(\lambda, \mu^*, \nu^*, \xi^*) - \geq L(\lambda)$. Hence, - \begin{displaymath} - L^*_{c}(\alpha) \geq L(\lambda) - \alpha\T{\mathbf{1}}\mu^* - \end{displaymath} - for any $\lambda$ feasible for \eqref{eq:primal}. In particular, for $\lambda$ primal optimal for $\eqref{eq:primal}$: - \begin{equation}\label{eq:local-1} - L^*_{c}(\alpha) \geq L^*_c - \alpha\T{\mathbf{1}}\mu^* - \end{equation} - - Let us denote by the $M$ the support of $\mu^*$, that is $M\defeq - \{i|\mu_i^* > 0\}$, and by $\lambda^*$ a primal optimal point for - $\eqref{eq:perturbed-primal}$. From the KKT conditions we see that: - \begin{displaymath} - M \subseteq \{i|\lambda_i^* = \alpha\} - \end{displaymath} - - - Let us first assume that $|M| = 0$, then $\T{\mathbf{1}}\mu^*=0$ and the lemma follows. - - We will now assume that $|M|\geq 1$. In this case $\T{c}\lambda^* - = 1$, otherwise we could increase the coordinates of $\lambda^*$ in $M$, - which would increase the value of the objective function and contradict the - optimality of $\lambda^*$. Note also, that $|M|\leq n-1$, otherwise, since - $\alpha< \frac{1}{n}$, we would have $\T{c}\lambda^*\ < 1$, which again - contradicts the optimality of $\lambda^*$. Let us write: - \begin{displaymath} - 1 = \T{c}\lambda^* = \alpha\sum_{i\in M}c_i + \sum_{i\in \bar{M}}\lambda_i^*c_i - \leq \alpha |M| + (n-|M|)\max_{i\in \bar{M}} c_i - \end{displaymath} - That is: - \begin{equation}\label{local-2} - \max_{i\in\bar{M}} c_i \geq \frac{1 - |M|\alpha}{n-|M|}> \frac{1}{n} - \end{equation} - where the last inequality uses again that $\alpha<\frac{1}{n}$. From the - KKT conditions, we see that for $i\in M$, $\nu_i^* = 0$ and: - \begin{equation}\label{local-3} - \mu_i^* = \xi^*c_i - \partial_i L(\lambda^*)\leq \xi^*c_i\leq \xi^* - \end{equation} - since $\partial_i L(\lambda^*)\geq 0$ and $c_i\leq 1$. - - Furthermore, using the KKT conditions again, we have that: - \begin{equation}\label{local-4} - \xi^* \leq \inf_{i\in \bar{M}}\frac{\partial_i L(\lambda^*)}{c_i}\leq \inf_{i\in \bar{M}} \frac{1}{c_i} - = \frac{1}{\max_{i\in\bar{M}} c_i} - \end{equation} - where the last inequality uses Lemma~\ref{lemma:derivative-bounds}. - - Combining \eqref{local-2}, \eqref{local-3} and \eqref{local-4}, we get that: - \begin{displaymath} - \sum_{i\in M}\mu_i^* \leq |M|\xi^* \leq n\xi^*\leq \frac{n}{\max_{i\in\bar{M}} c_i} \leq n^2 - \end{displaymath} - - This implies that: - \begin{displaymath} - \T{\mathbf{1}}\mu^* = \sum_{i=1}^n \mu^*_i = \sum_{i\in M}\mu_i^*\leq n^2 - \end{displaymath} - which in addition to \eqref{eq:local-1} proves the lemma. -\end{proof} - -\begin{lemma}\label{lemma:monotonicity} - If $c'$ = $(c_i', c_{-i})$, with $c_i'\leq c_i - \delta$, we have: - \begin{displaymath} - L^*_{c'}(\alpha) \geq L^*_c(\alpha) + \frac{\alpha\delta b}{2^n} - \end{displaymath} -\end{lemma} - -\begin{proof} - Let $\mu^*, \nu^*, \xi^*$ be dual optimal for $(P_{c', \alpha})$. Noting that, $\mathcal{L}_{c', \alpha}(\lambda, \mu^*, \nu^*, \xi^*) \geq - \mathcal{L}_{c, \alpha}(\lambda, \mu^*, \nu^*, \xi^*) + \lambda_i\xi^*\delta$, we get similarly to Lemma~\ref{lemma:proximity}: - \begin{displaymath} - L^*_{c'}(\alpha) \geq L(\lambda) + \lambda_i\xi^*\delta - \end{displaymath} - for any $\lambda$ feasible for \eqref{eq:perturbed-primal}. In particular, for $\lambda^*$ primal optimal for \eqref{eq:perturbed-primal}: - \begin{displaymath} - L^*_{c'}(\alpha) \geq L^*_{c}(\alpha) + \alpha\xi^*\delta - \end{displaymath} - since $\lambda_i^*\geq \alpha$. - - Using the KKT conditions for $(P_{c', \alpha})$, we can write: - \begin{displaymath} - \xi^* = \inf_{i:\lambda^{'*}_i>\alpha} \frac{\T{x_i}S(\lambda^{'*})^{-1}x_i}{c_i'} - \end{displaymath} - with $\lambda^{'*}$ optimal for $(P_{c', \alpha})$. Since $c_i'\leq 1$, using Lemma~\ref{lemma:derivative-bounds}, we get that $\xi^*\geq \frac{b}{2^n}$, which concludes the proof. -\end{proof} - -\begin{proposition} - Let $\delta\in(0,1]$. For any $\varepsilon\in(0,1]$, there exists - a routine which computes an approximate solution $\tilde{L}^*_c$ to - \eqref{eq:primal} such that: - \begin{enumerate} - \item $|\tilde{L}^*_c - L^*_c| \leq \varepsilon$ - \item for all $c' = (c_i', c_{-i})$ with $c_i'\leq c_i-\delta$, $\tilde{L}^*_c \leq \tilde{L}^*_{c'}$ - \item the routine's running time is $O\big(poly(n, d, \log\log\frac{1}{b\varepsilon\delta})\big)$ - \end{enumerate} -\end{proposition} - -\begin{proof} -Let $\varepsilon$ in $(0, 1]$. The routine works as follows: set $\alpha\defeq -\varepsilon(\delta + n^2)^{-1}$ and return an approximation $\tilde{L}^*_c$ of -$L^*_c(\alpha)$ with an accuracy $\frac{1}{2^{n+1}}\alpha\delta b$ computed by -a standard convex optimization algorithm. Note that this choice of $\alpha$ -implies $\alpha<\frac{1}{n}$ as required. - -\begin{enumerate} - \item using Lemma~\ref{lemma:proximity}: -\begin{displaymath} - |\tilde{L}^*_c - L^*_c| \leq |\tilde{L}^*_c - L^*_c(\alpha)| + |L^*_c(\alpha) - L^*_c| - \leq \alpha\delta + \alpha n^2 = \varepsilon -\end{displaymath} - -\item let $c' = (c_i', c_{-i})$ with $c_i'\leq c_i-\delta$, then: -\begin{displaymath} - \tilde{L}^*_{c'} \geq L^*_{c'} - \frac{\alpha\delta b}{2^{n+1}} - \geq L^*_c + \frac{\alpha\delta b}{2^{n+1}} - \geq \tilde{L}^*_c -\end{displaymath} -where the first and inequality come from the accuracy of the approximation, and -the inner inequality follows from Lemma~\ref{lemma:monotonicity}. - -\item the accuracy of the approximation $\tilde{L}^*_c$ is: -\begin{displaymath} - A\defeq\frac{\varepsilon\delta b}{2^{n+1}(\delta + n^2)} -\end{displaymath} -\sloppy -hence, the standard convex optimization algorithm runs in time $O(poly(n, d,\log\log A^{-1}))$. Note that: -\begin{displaymath} - \log\log A^{-1} = O\bigg(\log\log\frac{1}{\varepsilon\delta b} + \log n\bigg) -\end{displaymath} -which yields the wanted running time for the routine.\qedhere -\end{enumerate} -\end{proof} -\end{document} @@ -61,16 +61,6 @@ following properties: \item $OPT'_{-i^*}$ must be computable in polynomial time. \end{itemize} -Such a quantity can be found by considering relaxations of the -optimization problem \eqref{eq:opt-reduced}. A function $L:[0, 1]^{n}\to\reals_+$ defined on the hypercube $[0, 1]^{n}$ is a \emph{fractional relaxation} of $V$ over the set -$\mathcal{N}$ if $L(\id_S) = V(S)$ for all $S\subseteq\mathcal{N}$, where $\id_S$ -denotes the indicator vector of $S$. The optimization program -\eqref{eq:non-strategic} extends naturally to such relaxations: -\begin{align} - OPT' \defeq \max_{\lambda\in[0, 1]^{n}} - \left\{L(\lambda) \Big| \sum_{i=1}^{n} \lambda_i c_i - \leq B\right\}\label{relax} -\end{align} One of the main technical contributions of \citeN{chen} and \citeN{singer-influence} is to come up with appropriate such relaxations for @@ -78,24 +68,13 @@ One of the main technical contributions of \citeN{chen} and \subsection{Our Approach} -\sloppy -We introduce a relaxation $L$ specifically tailored to the value function of -\SEDP: -\begin{equation}\label{eq:our-relaxation} -\forall\,\lambda\in[0,1]^n,\quad L(\lambda) \defeq -\log\det\left(I_d + \sum_{i\in\mathcal{N}} \lambda_i x_i\T{x_i}\right), -\end{equation} -The function $L$ is well-known to be concave and even self-concordant (see -\emph{e.g.}, \cite{boyd2004convex}). In this case, the analysis of Newton's -method for self-concordant functions in \cite{boyd2004convex}, shows that -finding the maximum of $L$ to any precision $\varepsilon$ can be done in -$O(\log\log\varepsilon^{-1})$ iterations. Being the solution to a maximization -problem, $OPT'_{-i^*}$ satisfies the required monotonicity property. +\begin{comment} The main challenge will be to prove that $OPT'_{-i^*}$, for our relaxation $L$, is close to $OPT_{-i^*}$. We show this by establishing that $L$ is within a constant factor from the so-called multi-linear relaxation of \eqref{modified}, which in turn can be related to \eqref{modified} through pipage rounding. We establish the constant factor to the multi-linear relaxation by bounding the partial derivatives of these two functions; we obtain the latter bound by exploiting convexity properties of matrix functions over the convex cone of positive semidefinite matrices. +\end{comment} \begin{algorithm}[t] \caption{Mechanism for \SEDP{}}\label{mechanism} @@ -1,38 +1,36 @@ -\documentclass[prodmode,acmec]{ec-acmsmall} -%\documentclass{article} -\acmVolume{X} -\acmNumber{X} -\acmArticle{X} -\acmYear{2013} -\acmMonth{2} +\documentclass[11pt]{article} +\usepackage[vmargin=1.5in]{geometry} \usepackage[numbers]{natbib} \usepackage[utf8]{inputenc} \usepackage{amsmath,amsfonts} -%\usepackage{amsthm} -\usepackage{algorithm} -\usepackage{algpseudocode,bbm,color,verbatim} +\usepackage{algorithm, algpseudocode} +\usepackage{bbm,color,verbatim} +\usepackage[pagebackref=true,breaklinks=true,colorlinks=true]{hyperref} +\usepackage{amsthm} \input{definitions} \title{Budget Feasible Mechanisms for Experimental Design} \author{ - Thibaut Horel \affil{École Normale Supérieure} -Stratis Ioannidis \affil{Technicolor} -S. Muthukrishnan \affil{Rutgers University, Microsoft Research}} -%Remove permission block empty space -\makeatletter -\let\@copyrightspace\relax -\makeatother -\markboth{Thibaut Horel, Stratis Ioannidis and S. Muthukrishnan}{Budget Feasible Mechanisms for Experimental Design} -\begin{document} + Thibaut Horel\\École normale supérieure\\\texttt{thibaut.horel@ens.fr} + \and + Stratis Ioannidis\\Technicolor\\\texttt{stratis.ioannidis@technicolor.com} + \and + S. Muthukrishnan\\Rutgers University\\\texttt{muthu@cs.rutgers.edu} +} +\begin{document} +\maketitle \begin{abstract} \input{abstract} \end{abstract} -\maketitle +\newpage + \section{Introduction} \input{intro} \section{Preliminaries}\label{sec:peel} \input{problem} +\section{Approximation results} +\input{approximation} \section{Mechanism for \SEDP{}} \input{main} \section{Proofs}\label{sec:proofs} @@ -41,7 +39,7 @@ S. Muthukrishnan \affil{Rutgers University, Microsoft Research}} \input{general} %\section{Conclusion} %\input{conclusion} -\bibliographystyle{acmsmall} +\bibliographystyle{abbrvnat} \bibliography{notes} \newpage \section*{Appendix} @@ -31,18 +31,9 @@ The complexity of the mechanism is given by the following lemma. } \end{proof} -Finally, we prove the approximation ratio of the mechanism. We use the -following lemma which establishes that $OPT'$, the optimal value \eqref{relax} -of the fractional relaxation $L$ under the budget constraints is not too far -from $OPT$. -\begin{lemma}[Approximation]\label{lemma:relaxation} - $ OPT' \leq 2 OPT - + 2\max_{i\in\mathcal{N}}V(i)$. -\end{lemma} -The proof of Lemma~\ref{lemma:relaxation} is our main technical contribution, -and can be found in Section \ref{sec:relaxation}. +Finally, we prove the approximation ratio of the mechanism. -We also use the following lemma from \cite{chen} which bounds $OPT$ in terms of +We use the following lemma from \cite{chen} which bounds $OPT$ in terms of the value of $S_G$, as computed in Algorithm \ref{mechanism}, and $i^*$, the element of maximum value. @@ -54,7 +45,7 @@ OPT \leq \frac{e}{e-1}\big( 3 V(S_G) + 2 V(i^*)\big). \end{displaymath} \end{lemma} -Using Lemmas~\ref{lemma:relaxation} and~\ref{lemma:greedy-bound} we can complete the proof of +Using Proposition~\ref{prop:relaxation} and~\ref{lemma:greedy-bound} we can complete the proof of Theorem~\ref{thm:main} by showing that, for any $\varepsilon > 0$, if $OPT_{-i}'$, the optimal value of $L$ when $i^*$ is excluded from $\mathcal{N}$, has been computed to a precision $\varepsilon$, then the set @@ -83,7 +74,7 @@ hence, \end{equation} If the condition does not hold, by observing that $OPT'_{-i^*}\leq OPT'$ and -applying Lemma~\ref{lemma:relaxation}, we get +applying Proposition~\ref{prop:relaxation}, we get \begin{displaymath} V(i^*) \stackrel{}\leq \frac{1}{C}OPT_{-i^*}' + \frac{\varepsilon}{C} \leq \frac{1}{C} \big(2 OPT + 2 V(i^*)\big) + \frac{\varepsilon}{C}. @@ -120,316 +111,6 @@ Placing the expression of $C$ in \eqref{eq:bound1} and \eqref{eq:bound2} gives the approximation ratio in \eqref{approxbound}, and concludes the proof of Theorem~\ref{thm:main}.\hspace*{\stretch{1}}\qed -\subsection{Proof of Lemma~\ref{lemma:relaxation}}\label{sec:relaxation} - -We need to prove that for our relaxation $L$ given by -\eqref{eq:our-relaxation}, $OPT'$ is close to $OPT$ as stated in -Lemma~\ref{lemma:relaxation}. Our analysis follows the \emph{pipage rounding} -framework of \citeN{pipage}. - -This framework uses the \emph{multi-linear} extension $F$ of the submodular -function $V$. Let $P_\mathcal{N}^\lambda(S)$ be the probability of choosing the set $S$ if we select each element $i$ in $\mathcal{N}$ independently with probability $\lambda_i$: -\begin{displaymath} - P_\mathcal{N}^\lambda(S) \defeq \prod_{i\in S} \lambda_i - \prod_{i\in\mathcal{N}\setminus S}( 1 - \lambda_i). -\end{displaymath} -Then, the \emph{multi-linear} extension $F$ is defined by: -\begin{displaymath} - F(\lambda) - \defeq \mathbb{E}_{S\sim P_\mathcal{N}^\lambda}\big[V(S)\big] - = \sum_{S\subseteq\mathcal{N}} P_\mathcal{N}^\lambda(S) V(S) -\end{displaymath} - -For \EDP{} the multi-linear extension can be written: -\begin{equation}\label{eq:multi-linear-logdet} - F(\lambda) = \mathbb{E}_{S\sim - P_\mathcal{N}^\lambda}\bigg[\log\det \big(I_d + \sum_{i\in S} x_i\T{x_i}\big) \Big]. -\end{equation} -Note that the relaxation $L$ that we introduced in \eqref{eq:our-relaxation}, -follows naturally from the \emph{multi-linear} relaxation by swapping the -expectation and the $\log\det$ in \eqref{eq:multi-linear-logdet}: -\begin{displaymath} - L(\lambda) = \log\det\left(\mathbb{E}_{S\sim - P_\mathcal{N}^\lambda}\bigg[I_d + \sum_{i\in S} x_i\T{x_i} \bigg]\right). -\end{displaymath} - -The proof proceeds as follows: -\begin{itemize} -\item First, we prove that $F$ admits the following rounding property: let -$\lambda$ be a feasible element of $[0,1]^n$, it is possible to trade one -fractional component of $\lambda$ for another until one of them becomes -integral, obtaining a new element $\tilde{\lambda}$ which is both feasible and -for which $F(\tilde{\lambda})\geq F(\lambda)$. Here, by feasibility of a point -$\lambda$, we mean that it satisfies the budget constraint $\sum_{i=1}^n -\lambda_i c_i \leq B$. This rounding property is referred to in the literature -as \emph{cross-convexity} (see, \emph{e.g.}, \cite{dughmi}), or -$\varepsilon$-convexity by \citeN{pipage}. This is stated and proven in -Lemma~\ref{lemma:rounding} and allows us to bound $F$ in terms of $OPT$. -\item Next, we prove the central result of bounding $L$ appropriately in terms -of the multi-linear relaxation $F$ (Lemma \ref{lemma:relaxation-ratio}). -\item Finally, we conclude the proof of Lemma~\ref{lemma:relaxation} by -combining Lemma~\ref{lemma:rounding} and Lemma~\ref{lemma:relaxation-ratio}. -\end{itemize} - -\begin{comment} -Formally, if we define: -\begin{displaymath} - \tilde{F}_\lambda(\varepsilon) \defeq F\big(\lambda + \varepsilon(e_i - - e_j)\big) -\end{displaymath} -where $e_i$ and $e_j$ are two vectors of the standard basis of -$\reals^{n}$, then $\tilde{F}_\lambda$ is convex. Hence its maximum over the interval: -\begin{displaymath} - I_\lambda = \Big[\max(-\lambda_i,\lambda_j-1), \min(1-\lambda_i, \lambda_j)\Big] -\end{displaymath} -is attained at one of the boundaries of $I_\lambda$ for which one of the $i$-th -or the $j$-th component of $\lambda$ becomes integral. -\end{comment} - -\begin{lemma}[Rounding]\label{lemma:rounding} - For any feasible $\lambda\in[0,1]^{n}$, there exists a feasible - $\bar{\lambda}\in[0,1]^{n}$ such that at most one of its components is - fractional %, that is, lies in $(0,1)$ and: - and $F_{\mathcal{N}}(\lambda)\leq F_{\mathcal{N}}(\bar{\lambda})$. -\end{lemma} -\begin{proof} - We give a rounding procedure which, given a feasible $\lambda$ with at least - two fractional components, returns some feasible $\lambda'$ with one less fractional - component such that $F(\lambda) \leq F(\lambda')$. - - Applying this procedure recursively yields the lemma's result. - Let us consider such a feasible $\lambda$. Let $i$ and $j$ be two - fractional components of $\lambda$ and let us define the following - function: - \begin{displaymath} - F_\lambda(\varepsilon) = F(\lambda_\varepsilon) - \quad\textrm{where} \quad - \lambda_\varepsilon = \lambda + \varepsilon\left(e_i-\frac{c_i}{c_j}e_j\right) - \end{displaymath} - It is easy to see that if $\lambda$ is feasible, then: - \begin{equation}\label{eq:convex-interval} - \forall\varepsilon\in\Big[\max\Big(-\lambda_i,(\lambda_j-1)\frac{c_j}{c_i}\Big), \min\Big(1-\lambda_i, \lambda_j - \frac{c_j}{c_i}\Big)\Big],\; - \lambda_\varepsilon\;\;\textrm{is feasible} - \end{equation} - Furthermore, the function $F_\lambda$ is convex; indeed: - \begin{align*} - F_\lambda(\varepsilon) - & = \mathbb{E}_{S'\sim P_{\mathcal{N}\setminus\{i,j\}}^\lambda(S')}\Big[ - (\lambda_i+\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i,j\})\\ - & + (\lambda_i+\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{i\}) - + (1-\lambda_i-\varepsilon)\Big(\lambda_j-\varepsilon\frac{c_i}{c_j}\Big)V(S'\cup\{j\})\\ - & + (1-\lambda_i-\varepsilon)\Big(1-\lambda_j+\varepsilon\frac{c_i}{c_j}\Big)V(S')\Big] - \end{align*} - Thus, $F_\lambda$ is a degree 2 polynomial whose dominant coefficient is: - \begin{displaymath} - \frac{c_i}{c_j}\mathbb{E}_{S'\sim - P_{\mathcal{N}\setminus\{i,j\}}^\lambda(S')}\Big[ - V(S'\cup\{i\})+V(S'\cup\{i\})\\ - -V(S'\cup\{i,j\})-V(S')\Big] - \end{displaymath} - which is positive by submodularity of $V$. Hence, the maximum of - $F_\lambda$ over the interval given in \eqref{eq:convex-interval} is - attained at one of its limit, at which either the $i$-th or $j$-th component of - $\lambda_\varepsilon$ becomes integral. -\end{proof} - - -\begin{lemma}\label{lemma:relaxation-ratio} - % The following inequality holds: -For all $\lambda\in[0,1]^{n},$ - %\begin{displaymath} - $ \frac{1}{2} - \,L(\lambda)\leq - F(\lambda)\leq L(\lambda)$. - %\end{displaymath} -\end{lemma} -\begin{proof} - The bound $F_{\mathcal{N}}(\lambda)\leq L_{\mathcal{N}(\lambda)}$ follows by the concavity of the $\log\det$ function. - To show the lower bound, - we first prove that $\frac{1}{2}$ is a lower bound of the ratio $\partial_i - F(\lambda)/\partial_i L(\lambda)$, where - $\partial_i\, \cdot$ denotes the partial derivative with respect to the - $i$-th variable. - - Let us start by computing the derivatives of $F$ and - $L$ with respect to the $i$-th component. - Observe that - \begin{displaymath} - \partial_i P_\mathcal{N}^\lambda(S) = \left\{ - \begin{aligned} - & P_{\mathcal{N}\setminus\{i\}}^\lambda(S\setminus\{i\})\;\textrm{if}\; - i\in S, \\ - & - P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\;\textrm{if}\; - i\in \mathcal{N}\setminus S. \\ - \end{aligned}\right. - \end{displaymath} - Hence, - \begin{displaymath} - \partial_i F(\lambda) = - \sum_{\substack{S\subseteq\mathcal{N}\\ i\in S}} - P_{\mathcal{N}\setminus\{i\}}^\lambda(S\setminus\{i\})V(S) - - \sum_{\substack{S\subseteq\mathcal{N}\\ i\in \mathcal{N}\setminus S}} - P_{\mathcal{N}\setminus\{i\}}^\lambda(S)V(S). - \end{displaymath} - Now, using that every $S$ such that $i\in S$ can be uniquely written as - $S'\cup\{i\}$, we can write: - \begin{displaymath} - \partial_i F(\lambda) = - \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} - P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\big(V(S\cup\{i\}) - - V(S)\big). - \end{displaymath} - The marginal contribution of $i$ to - $S$ can be written as -\begin{align*} -V(S\cup \{i\}) - V(S)& = \frac{1}{2}\log\det(I_d - + \T{X_S}X_S + x_i\T{x_i}) - - \frac{1}{2}\log\det(I_d + \T{X_S}X_S)\\ - & = \frac{1}{2}\log\det(I_d + x_i\T{x_i}(I_d + -\T{X_S}X_S)^{-1}) - = \frac{1}{2}\log(1 + \T{x_i}A(S)^{-1}x_i) -\end{align*} -where $A(S) \defeq I_d+ \T{X_S}X_S$, and the last equality follows from the -Sylvester's determinant identity~\cite{sylvester}. -% $ V(S\cup\{i\}) - V(S) = \frac{1}{2}\log\left(1 + \T{x_i} A(S)^{-1}x_i\right)$. -Using this, - \begin{displaymath} - \partial_i F(\lambda) = \frac{1}{2} - \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} - P_{\mathcal{N}\setminus\{i\}}^\lambda(S) - \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big) - \end{displaymath} - The computation of the derivative of $L$ uses standard matrix - calculus: writing $\tilde{A}(\lambda) = I_d+\sum_{i\in - \mathcal{N}}\lambda_ix_i\T{x_i}$, - \begin{displaymath} - \det \tilde{A}(\lambda + h\cdot e_i) = \det\big(\tilde{A}(\lambda) - + hx_i\T{x_i}\big) - =\det \tilde{A}(\lambda)\big(1+ - h\T{x_i}\tilde{A}(\lambda)^{-1}x_i\big). - \end{displaymath} - Hence, - \begin{displaymath} - \log\det\tilde{A}(\lambda + h\cdot e_i)= \log\det\tilde{A}(\lambda) - + h\T{x_i}\tilde{A}(\lambda)^{-1}x_i + o(h), - \end{displaymath} - which implies - \begin{displaymath} - \partial_i L(\lambda) - =\frac{1}{2} \T{x_i}\tilde{A}(\lambda)^{-1}x_i. - \end{displaymath} - -For two symmetric matrices $A$ and $B$, we write $A\succ B$ ($A\succeq B$) if -$A-B$ is positive definite (positive semi-definite). This order allows us to -define the notion of a \emph{decreasing} as well as \emph{convex} matrix -function, similarly to their real counterparts. With this definition, matrix -inversion is decreasing and convex over symmetric positive definite -matrices (see Example 3.48 p. 110 in \cite{boyd2004convex}). -In particular, -\begin{gather*} - \forall S\subseteq\mathcal{N},\quad A(S)^{-1} \succeq A(S\cup\{i\})^{-1} -\end{gather*} -as $A(S)\preceq A(S\cup\{i\})$. Observe that - % \begin{gather*} - % \forall -$P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\geq P_{\mathcal{N}\setminus\{i\}}^\lambda(S\cup\{i\})$ for all $S\subseteq\mathcal{N}\setminus\{i\}$, and - % ,\\ - $P_{\mathcal{N}\setminus\{i\}}^\lambda(S) \geq P_\mathcal{N}^\lambda(S),$ for all $S\subseteq\mathcal{N}$. - %\end{gather*} - Hence, - \begin{align*} - \partial_i F(\lambda) - % & = \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} P_\mathcal{N}^\lambda(S) \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big)\\ - & \geq \frac{1}{4} - \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} - P_{\mathcal{N}\setminus\{i\}}^\lambda(S) - \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big)\\ - &\hspace{-3.5em}+\frac{1}{4} - \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}} - P_{\mathcal{N}\setminus\{i\}}^\lambda(S\cup\{i\}) - \log\Big(1 + \T{x_i}A(S\cup\{i\})^{-1}x_i\Big)\\ - &\geq \frac{1}{4} - \sum_{S\subseteq\mathcal{N}} - P_\mathcal{N}^\lambda(S) - \log\Big(1 + \T{x_i}A(S)^{-1}x_i\Big). - \end{align*} - Using that $A(S)\succeq I_d$ we get that $\T{x_i}A(S)^{-1}x_i \leq - \norm{x_i}_2^2 \leq 1$. Moreover, $\log(1+x)\geq x$ for all $x\leq 1$. - Hence, - \begin{displaymath} - \partial_i F(\lambda) \geq - \frac{1}{4} - \T{x_i}\bigg(\sum_{S\subseteq\mathcal{N}}P_\mathcal{N}^\lambda(S)A(S)^{-1}\bigg)x_i. - \end{displaymath} - Finally, using that the inverse is a matrix convex function over symmetric - positive definite matrices: - \begin{displaymath} - \partial_i F(\lambda) \geq - \frac{1}{4} - \T{x_i}\bigg(\sum_{S\subseteq\mathcal{N}}P_\mathcal{N}^\lambda(S)A(S)\bigg)^{-1}x_i - = \frac{1}{4}\T{x_i}\tilde{A}(\lambda)^{-1}x_i - = \frac{1}{2} - \partial_i L(\lambda). - \end{displaymath} - -Having bound the ratio between the partial derivatives, we now bound the ratio $F(\lambda)/L(\lambda)$ from below. Consider the following cases. - First, if the minimum of the ratio - $F(\lambda)/L(\lambda)$ is attained at a point interior to the hypercube, then it is - a critical point, \emph{i.e.}, $\partial_i \big(F(\lambda)/L(\lambda)\big)=0$ for all $i\in \mathcal{N}$; hence, at such a critical point: - \begin{equation}\label{eq:lhopital} - \frac{F(\lambda)}{L(\lambda)} - = \frac{\partial_i F(\lambda)}{\partial_i - L(\lambda)} \geq \frac{1}{2}. - \end{equation} - Second, if the minimum is attained as - $\lambda$ converges to zero in, \emph{e.g.}, the $l_2$ norm, by the Taylor approximation, one can write: - \begin{displaymath} - \frac{F(\lambda)}{L(\lambda)} - \sim_{\lambda\rightarrow 0} - \frac{\sum_{i\in \mathcal{N}}\lambda_i\partial_i F(0)} - {\sum_{i\in\mathcal{N}}\lambda_i\partial_i L(0)} - \geq \frac{1}{2}, - \end{displaymath} - \emph{i.e.}, the ratio $\frac{F(\lambda)}{L(\lambda)}$ is necessarily bounded from below by 1/2 for small enough $\lambda$. - Finally, if the minimum is attained on a face of the hypercube $[0,1]^n$ (a face is - defined as a subset of the hypercube where one of the variable is fixed to - 0 or 1), without loss of generality, we can assume that the minimum is - attained on the face where the $n$-th variable has been fixed - to 0 or 1. Then, either the minimum is attained at a point interior to the - face or on a boundary of the face. In the first sub-case, relation - \eqref{eq:lhopital} still characterizes the minimum for $i< n$. - In the second sub-case, by repeating the argument again by induction, we see - that all is left to do is to show that the bound holds for the vertices of - the cube (the faces of dimension 1). The vertices are exactly the binary - points, for which we know that both relaxations are equal to the value - function $V$. Hence, the ratio is equal to 1 on the vertices. -\end{proof} - -To conclude the proof of Lemma~\ref{lemma:relaxation}, let us consider -a feasible point $\lambda^*\in[0,1]^{n}$ such that $L(\lambda^*) = OPT'$. By -applying Lemma~\ref{lemma:relaxation-ratio} and Lemma~\ref{lemma:rounding} we -get a feasible point $\bar{\lambda}$ with at most one fractional component such -that -\begin{equation}\label{eq:e1} - L(\lambda^*) \leq 2 F(\bar{\lambda}). -\end{equation} - Let $\lambda_i$ denote the fractional component of $\bar{\lambda}$ and $S$ - denote the set whose indicator vector is $\bar{\lambda} - \lambda_i e_i$. - By definition of the multi-linear extension $F$: - \begin{displaymath} - F(\bar{\lambda}) = (1-\lambda_i)V(S) +\lambda_i V(S\cup\{i\}). - \end{displaymath} - By submodularity of $V$, $V(S\cup\{i\})\leq V(S) + V(\{i\})$. Hence, - \begin{displaymath} - F(\bar{\lambda}) \leq V(S) + V(i). - \end{displaymath} - Note that since $\bar{\lambda}$ is feasible, $S$ is also feasible and - $V(S)\leq OPT$. Hence, - \begin{equation}\label{eq:e2} - F(\bar{\lambda}) \leq OPT + \max_{i\in\mathcal{N}} V(i). - \end{equation} - Together, \eqref{eq:e1} and \eqref{eq:e2} imply the lemma. \hspace*{\stretch{1}}\qed \subsection{Proof of Theorem \ref{thm:lowerbound}} |