From 07d48e21fb6fc62b1a85b9d80c25560529a9a0b5 Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Fri, 28 Jun 2013 00:16:44 +0200
Subject: Moving the proofs to the appendix, improving the flow

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 appendix.tex | 669 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++-
 1 file changed, 664 insertions(+), 5 deletions(-)

(limited to 'appendix.tex')

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@@ -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
-- 
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