diff options
Diffstat (limited to 'appendix.tex')
| -rw-r--r-- | appendix.tex | 669 |
1 files changed, 664 insertions, 5 deletions
diff --git a/appendix.tex b/appendix.tex index 2c83e04..dd643d7 100644 --- a/appendix.tex +++ b/appendix.tex @@ -1,6 +1,558 @@ +\subsection{Proof of Proposition~\ref{prop:relaxation}} + +\begin{lemma}\label{lemma:relaxation-ratio} +For all $\lambda\in[0,1]^{n},$ + $ \frac{1}{2} + \,L(\lambda)\leq + F(\lambda)\leq L(\lambda)$. +\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} + +We now prove that $F$ admits the following exchange 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}. + +\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} + +\subsubsection*{End of 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 proposition.\qedhere + +\begin{proof}[Proof of Lemma~\ref{lemma:derivative-bounds}] + 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} + +\subsection{Proof of Proposition~\ref{prop:monotonicity}} + +The $\log\det$ function is concave and self-concordant (see +\cite{boyd2004convex}), in this case, the analysis of the barrier method in +in \cite{boyd2004convex} (Section 11.5.5) can be summarized in the following +lemma: + +\begin{lemma}\label{lemma:barrier} +For any $\varepsilon>0$, the barrier method computes an $\varepsilon$-accurate +approximation of $L^*_c$ in time $O(poly(n,d,\log\log\varepsilon^{-1})$. +\end{lemma} + +We show that the optimal value of \eqref{eq:perturbed-primal} 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 $\tilde{L}^*_c$ be the approximation computed by +Algorithm~\ref{alg:monotone}. +\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} +which proves the $\varepsilon$-accuracy. + +\item for the $\delta$-decreasingness, 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} + +Note that: +\begin{displaymath} + \log\log A^{-1} = O\bigg(\log\log\frac{1}{\varepsilon\delta b} + \log n\bigg) +\end{displaymath} +Using Lemma~\ref{lemma:barrier} concludes the proof of the running time.\qed +\end{enumerate} + +\subsection{Proof of Theorem~\ref{thm:main}}\label{sec:proofofmainthm} + +We now present the proof of Theorem~\ref{thm:main}. $\delta$-truthfulness and +individual rationality follow from $\delta$-monotonicity and threshold +payments. $\delta$-monotonicity and budget feasibility follow the same steps as the +analysis of \citeN{chen}; for the sake of completeness, we restate their proof +here. + \begin{lemma}\label{lemma:monotone} Our mechanism for \EDP{} is $\delta$-monotone and budget feasible. \end{lemma} + \begin{proof} Consider an agent $i$ with cost $c_i$ that is selected by the mechanism, and suppose that she reports @@ -28,11 +580,6 @@ Our mechanism for \EDP{} is $\delta$-monotone and budget feasible. 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. -%\end{proof} -%\begin{lemma}\label{lemma:budget-feasibility} -%The mechanism is budget feasible. -%\end{lemma} -%\begin{proof} To show budget feasibility, suppose that $OPT'_{-i^*} < C V(i^*)$. Then the mechanism selects $i^*$. Since the bid of $i^*$ does not affect the above condition, the threshold payment of $i^*$ is $B$ and the mechanism is budget feasible. Suppose that $OPT'_{-i^*} \geq C V(i^*)$. @@ -57,4 +604,116 @@ Hence, the total payment is bounded by the telescopic sum: \end{displaymath} \end{proof} +The complexity of the mechanism is given by the following lemma. + +\begin{lemma}[Complexity]\label{lemma:complexity} + For any $\varepsilon > 0$ and any $\delta>0$, the complexity of the mechanism is + $O\big(poly(n, d, \log\log\frac{1}{b\varepsilon\delta})\big)$ +\end{lemma} + +\begin{proof} + The value function $V$ in \eqref{modified} can be computed in time + $O(\text{poly}(n, d))$ and the mechanism only involves a linear + number of queries to the function $V$. + + By Proposition~\ref{prop:monotonicity}, line 3 of Algorithm~\ref{mechanism} + can be computed in time + $O(\text{poly}(n, d, \log\log \varepsilon^{-1}))$. Hence the allocation + function's complexity is as stated. + %Payments can be easily computed in time $O(\text{poly}(n, d))$ as in prior work. +\junk{ + Using Singer's characterization of the threshold payments + \cite{singer-mechanisms}, one can verify that they can be computed in time + $O(\text{poly}(n, d))$. + } +\end{proof} + +Finally, we prove the approximation ratio of the mechanism. + +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. + +\begin{lemma}[\cite{chen}]\label{lemma:greedy-bound} +Let $S_G$ be the set computed in Algorithm \ref{mechanism} and let +$i^*=\argmax_{i\in\mathcal{N}} V(\{i\})$. We have: +\begin{displaymath} +OPT \leq \frac{e}{e-1}\big( 3 V(S_G) + 2 V(i^*)\big). +\end{displaymath} +\end{lemma} + +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 +$S^*$ allocated by the mechanism is such that: +\begin{equation} \label{approxbound} +OPT +\leq \frac{10e\!-\!3 + \sqrt{64e^2\!-\!24e\!+\!9}}{2(e\!-\!1)} V(S^*)\! ++ \! \varepsilon . +\end{equation} +To see this, let $L^*_{-i^*}$ be the true maximum value of $L$ subject to +$\lambda_{i^*}=0$, $\sum_{i\in \mathcal{N}\setminus{i^*}}c_i\leq B$. From line +3 of Algorithm~\ref{mechanism}, we have +$L^*_{-i^*}-\varepsilon\leq OPT_{-i^*}' \leq L^*_{-i^*}+\varepsilon$. + +If the condition on line 4 of the algorithm holds, then +\begin{displaymath} + V(i^*) \geq \frac{1}{C}L^*_{-i^*}-\frac{\varepsilon}{C} \geq + \frac{1}{C}OPT_{-i^*} -\frac{\varepsilon}{C}. +\end{displaymath} +Indeed, $L^*_{-i^*}\geq OPT_{-i^*}$ as $L$ is a fractional relaxation of $V$. Also, $OPT \leq OPT_{-i^*} + V(i^*)$, +hence, +\begin{equation}\label{eq:bound1} + OPT\leq (1+C)V(i^*) + \varepsilon. +\end{equation} + +If the condition does not hold, by observing that $L^*_{-i^*}\leq L^*_c$ and +applying Proposition~\ref{prop:relaxation}, we get +\begin{displaymath} + V(i^*)\leq \frac{1}{C}L^*_{-i^*} + \frac{\varepsilon}{C} + \leq \frac{1}{C} \big(2 OPT + 2 V(i^*)\big) + \frac{\varepsilon}{C}. +\end{displaymath} +Applying Lemma~\ref{lemma:greedy-bound}, +\begin{displaymath} + V(i^*) \leq \frac{1}{C}\left(\frac{2e}{e-1}\big(3 V(S_G) + + 2 V(i^*)\big) + 2 V(i^*)\right) + \frac{\varepsilon}{C}. +\end{displaymath} +Thus, if $C$ is such that $C(e-1) -6e +2 > 0$, +\begin{align*} + V(i^*) \leq \frac{6e}{C(e-1)- 6e + 2} V(S_G) + + \frac{(e-1)\varepsilon}{C(e-1)- 6e + 2}. +\end{align*} +Finally, using Lemma~\ref{lemma:greedy-bound} again, we get +\begin{equation}\label{eq:bound2} + OPT(V, \mathcal{N}, B) \leq + \frac{3e}{e-1}\left( 1 + \frac{4e}{C (e-1) -6e +2}\right) V(S_G) + + \frac{2e\varepsilon}{C(e-1)- 6e + 2}. +\end{equation} +To minimize the coefficients of $V_{i^*}$ and $V(S_G)$ in \eqref{eq:bound1} +and \eqref{eq:bound2} respectively, we wish to chose $C$ that minimizes +\begin{displaymath} + \max\left(1+C,\frac{3e}{e-1}\left( 1 + \frac{4e}{C (e-1) -6e +2} + \right)\right). +\end{displaymath} +This function has two minima, only one of those is such that $C(e-1) -6e ++2 \geq 0$. This minimum is +\begin{equation}\label{eq:constant} + C = \frac{8e-1 + \sqrt{64e^2-24e + 9}}{2(e-1)}. +\end{equation} +For this minimum, $\frac{2e\varepsilon}{C^*(e-1)- 6e + 2}\leq \varepsilon.$ +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 Theorem \ref{thm:lowerbound}} +Suppose, for contradiction, that such a mechanism exists. Consider two +experiments with dimension $d=2$, such that $x_1 = e_1=[1 ,0]$, $x_2=e_2=[0,1]$ +and $c_1=c_2=B/2+\epsilon$. Then, one of the two experiments, say, $x_1$, must +be in the set selected by the mechanism, otherwise the ratio is unbounded, +a contradiction. If $x_1$ lowers its value to $B/2-\epsilon$, by monotonicity +it remains in the solution; by threshold payment, it is paid at least +$B/2+\epsilon$. So $x_2$ is not included in the solution by budget feasibility +and individual rationality: hence, the selected set attains a value $\log2$, +while the optimal value is $2\log 2$.\hspace*{\stretch{1}}\qed |