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@@ -26,156 +26,7 @@ Sylvester's determinant identity~\cite{sylvester}. Monotonicity therefore follow
 and submodularity also follows, as a function is submodular if and only if the marginal contributions are non-increasing in $S$. \qed
 
 
-\section{Proofs of Statements in Section~\ref{sec:concave}}
-\subsection{Proof of Lemma~\ref{lemma:relaxation-ratio}}\label{proofofrelaxation-ratio}
 %\begin{proof}
-    The bound $F(\lambda)\leq L(\lambda)$ follows by the concavity of the $\log\det$ function and Jensen's inequality.
-    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 we use
-    $\partial_i\, \cdot$ as a shorthand for  $\frac{\partial}{\partial \lambda_i}$, the partial derivative  with respect to the
-    $i$-th variable. 
-
-   Let us start by computing the partial 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}
-Recall from \eqref{eq:marginal_contrib} that the marginal contribution of $i$ to $S$ is given by
-$$V(S\cup \{i\}) - V(S) =\frac{1}{2}\log(1 + \T{x_i}A(S)^{-1}x_i), $$
-where $A(S) = I_d+ \T{X_S}X_S$.
-%  $  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 $\tilde{A}(\lambda) \defeq I_d+\sum_{i\in
-    \mathcal{N}}\lambda_ix_i\T{x_i}$,
-    \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}
-Recall from \eqref{eq:inverse} that the monotonicity of the matrix inverse over positive definite matrices implies 
-\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 since $1\leq \lambda_i\leq 1$, 
-$P_{\mathcal{N}\setminus\{i\}}^\lambda(S) \geq P_\mathcal{N}^\lambda(S)$ and
-$P_{\mathcal{N}\setminus\{i\}}^\lambda(S)\geq P_{\mathcal{N}}^\lambda(S\cup\{i\})$
-for all $S\subseteq\mathcal{N}\setminus\{i\}$. Hence,
-\begin{align*}
-    \partial_i F(\lambda) 
-    & \geq \frac{1}{4}
-    \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)\\
-    &+\frac{1}{4}
-    \sum_{\substack{S\subseteq\mathcal{N}\\ i\in\mathcal{N}\setminus S}}
-    P_{\mathcal{N}}^\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 (see Appendix~\ref{app:properties}):
-\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 three cases.
-
-First, 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 $\lambda$.
-
-Second, if the minimum of the ratio $F(\lambda)/L(\lambda)$ is attained at
-a vertex of the hypercube $[0,1]^n$ different from 0. $F$ and $L$ being
-relaxations of the value function $V$, they are equal to $V$ on the vertices
-which are exactly the binary points. Hence, the minimum is equal to 1 in this
-case; in particular, it is greater than $1/2$.
-
-Finally, if the minimum is attained at a point $\lambda^*$ with at least one
-coordinate belonging to $(0,1)$, let $i$ be one such coordinate and consider
-the function $G_i$:
-\begin{displaymath}
-    G_i: x \mapsto \frac{F}{L}(\lambda_1^*,\ldots,\lambda_{i-1}^*, x,
-    \lambda_{i+1}^*, \ldots, \lambda_n^*).
-\end{displaymath}
-Then this function attains a minimum at $\lambda^*_i\in(0,1)$ and its
-derivative is zero at this point. Hence:
-\begin{displaymath}
-    0 = G_i'(\lambda^*_i) = \partial_i\left(\frac{F}{L}\right)(\lambda^*).
-\end{displaymath}
-But $\partial_i(F/L)(\lambda^*)=0$ implies that
-\begin{displaymath}
-    \frac{F(\lambda^*)}{L(\lambda^*)} = \frac{\partial_i
-    F(\lambda^*)}{\partial_i L(\lambda^*)}\geq \frac{1}{2}
-\end{displaymath}
-using the lower bound on the ratio of the partial derivatives. This concludes
-the proof of the lemma. \qed
 %\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}.
@@ -186,74 +37,8 @@ the proof of the lemma. \qed
 %    fractional %, that is, lies in $(0,1)$ and:
 %     and $F_{\mathcal{N}}(\lambda)\leq F_{\mathcal{N}}(\bar{\lambda})$.
 %\end{lemma}
-\subsection{Proof of Lemma~\ref{lemma:rounding}}\label{proofoflemmarounding}
 %\begin{proof}
-    We give a rounding procedure which, given a feasible $\lambda$ with at least
-    two fractional components, returns some feasible $\lambda'$ with one fewer 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 limits, at which either the $i$-th or $j$-th component of
-    $\lambda_\varepsilon$ becomes integral. \qed
 %\end{proof}
-\subsection{Proof of Proposition~\ref{prop:relaxation}}\label{proofofproprelaxation}
-The lower bound on $L^*_c$ follows immediately from the fact that $L$ extends $V$ to $[0,1]^n$. For the upper bound, let us consider a feasible point $\lambda^*\in \dom_c$ 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.\qed
-
-\section{Proof of Proposition~\ref{prop:monotonicity}}\label{proofofpropmonotonicity}
 
 %The $\log\det$ function is concave and self-concordant (see
 %\cite{boyd2004convex}), in this case, the analysis of the barrier method in
@@ -264,7 +49,6 @@ Together, \eqref{eq:e1} and \eqref{eq:e2} imply the proposition.\qed
 %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 begin by a description of Algorithm~\ref{alg:monotone} which computes an approximation of $L^*_c$, which is arbitrarily accurate \emph{and} $\delta$-decreasing. 
 %Note, that the feasible set in Problem~\eqref{eq:primal} increases (for the
 %inclusion) when the cost decreases. 
 %non-increasing.
@@ -311,243 +95,6 @@ We begin by a description of Algorithm~\ref{alg:monotone} which computes an appr
 %an approximation of $L^*_{c,\alpha}$ as in Algorithm~\ref{alg:monotone}, we
 %obtain a $\delta$-decreasing approximation of $L^*_c$.
 
-\begin{algorithm}[t]
-    \caption{}\label{alg:monotone}
-    \begin{algorithmic}[1]
-	\Require{ $B\in \reals_+$, $c\in[0,B]^n$, $\delta\in (0,1]$, $\epsilon\in (0,1]$ }
-        \State $\alpha \gets \varepsilon (\delta/B+n^2)^{-1}$ 
-        \State Use the barrier method to solve \eqref{eq:perturbed-primal} with
-        accuracy $\varepsilon'=\frac{1}{2^{n+1}B}\alpha\delta b$; denote the output by $\hat{L}^*_{c,\alpha}$
-	\State \textbf{return} $\hat{L}^*_{c,\alpha}$
-    \end{algorithmic}
-\end{algorithm}
-
-Our construction of a $\delta$-decreasing, $\varepsilon$-accurate approximator of $L_c^*$ proceeds as follows: first, it computes an appropriately selected lower bound $\alpha$; using this bound, it solves the perturbed problem \eqref{eq:perturbed-primal} using the barrier method, also at an appropriately selected accuracy $\varepsilon'$, obtaining thus a $\varepsilon'$-accurate approximation of $L^*_{c,\alpha}\defeq \max_{\lambda\in \dom_{c,\alpha}} L(\lambda)$ . The corresponding output is returned as an approximation of $L^*_c$. A summary of the algorithm and the specific choices of $\alpha$ and $\varepsilon'$ are given in Algorithm~\ref{alg:monotone}. We proceed by showing 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}.
-
-We note that the execution of the barrier method on the restricted set $\dom_{c,\alpha}$ is necessary. The algorithm's output when executed over the entire domain may not necessarily be $\delta$-decreasing, even when the approximation accuracy is small. This is because costs become saturated when the optimal $\lambda\in \dom_c$ lies at the boundary: increasing them has no effect on the objective. Forcing the optimization to happen ``off'' the boundary ensures that this does not occur, while taking $\alpha$ to be small ensures that this perturbation does not cost much in terms of approximation accuracy.
-
-Note that the choice of $\alpha$ given in Algorithm~\ref{alg:monotone} implies
-that $\alpha<\frac{1}{n}$. This in turn implies that the feasible set
-$\mathcal{D}_{c, \alpha}$ of \eqref{eq:perturbed-primal} is non-empty: it
-contains the strictly feasible point $\lambda=(\frac{1}{n},\ldots,\frac{1}{n})$.
-
-\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}
-    Recall that we had defined:
-    \begin{displaymath}
-        \tilde{A}(\lambda)\defeq I_d + \sum_{i=1}^n \lambda_i x_i\T{x_i}
-        \quad\mathrm{and}\quad
-        A(S) \defeq I_d + \sum_{i\in S} x_i\T{x_i}
-    \end{displaymath}
-    Let us also define $A_k\defeq A(\{x_1,\ldots,x_k\})$.
-    We have $\partial_i L(\lambda) = \T{x_i}\tilde{A}(\lambda)^{-1}x_i$. Since
-    $\tilde{A}(\lambda)\succeq I_d$, $\partial_i L(\lambda)\leq \T{x_i}x_i \leq 1$, which
-    is the right-hand side  of the lemma.
-    For the left-hand side, note that $\tilde{A}(\lambda) \preceq A_n$. Hence
-    $\partial_iL(\lambda)\geq \T{x_i}A_n^{-1}x_i$.
-    Using the Sherman-Morrison formula \cite{sm}, for all $k\geq 1$:
-    \begin{displaymath}
-        \T{x_i}A_k^{-1} x_i = \T{x_i}A_{k-1}^{-1}x_i 
-        - \frac{(\T{x_i}A_{k-1}^{-1}x_k)^2}{1+\T{x_k}A_{k-1}^{-1}x_k}
-    \end{displaymath}
-    By the Cauchy-Schwarz inequality:
-    \begin{displaymath}
-        (\T{x_i}A_{k-1}^{-1}x_k)^2 \leq \T{x_i}A_{k-1}^{-1}x_i\;\T{x_k}A_{k-1}^{-1}x_k
-    \end{displaymath}
-    Hence:
-    \begin{displaymath}
-        \T{x_i}A_k^{-1} x_i \geq \T{x_i}A_{k-1}^{-1}x_i 
-        - \T{x_i}A_{k-1}^{-1}x_i\frac{\T{x_k}A_{k-1}^{-1}x_k}{1+\T{x_k}A_{k-1}^{-1}x_k}
-    \end{displaymath}
-    But $\T{x_k}A_{k-1}^{-1}x_k\leq 1$ and $\frac{a}{1+a}\leq \frac{1}{2}$ if
-    $0\leq a\leq 1$, so:
-    \begin{displaymath}
-        \T{x_i}A_{k}^{-1}x_i \geq \T{x_i}A_{k-1}^{-1}x_i
-        - \frac{1}{2}\T{x_i}A_{k-1}^{-1}x_i\geq \frac{\T{x_i}A_{k-1}^{-1}x_i}{2}
-    \end{displaymath}
-    By induction:
-    \begin{displaymath}
-        \T{x_i}A_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(B-\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 Karush-Kuhn-Tucker (KKT) conditions \cite{boyd2004convex} 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 $\T{c}\lambda^*
-    = B$, 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 $\T{c}\lambda^*\ < B$, which again
-    contradicts the optimality of $\lambda^*$. Let us write:
-    \begin{displaymath}
-        B = \T{c}\lambda^* = \alpha\sum_{i\in M}c_i + \sum_{i\in \bar{M}}\lambda_i^*c_i
-        \leq \alpha |M|B + (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{B - B|M|\alpha}{n-|M|}> \frac{B}{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^*B
-    \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^*B \leq n\xi^*B\leq \frac{nB}{\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 along with \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^nB}
-    \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 $c_i'\leq B$,
-    using Lemma~\ref{lemma:derivative-bounds}, we get that $\xi^*\geq
-    \frac{b}{2^nB}$, which concludes the proof.
-\end{proof}
-
-%\subsection*{End of the proof of Proposition~\ref{prop:monotonicity}}
-We are now ready to conclude the  proof of Proposition~\ref{prop:monotonicity}.
-Let $\hat{L}^*_{c,\alpha}$ be the approximation computed by
-Algorithm~\ref{alg:monotone}.
-\begin{enumerate}
-    \item using Lemma~\ref{lemma:proximity}:
-\begin{displaymath}
-        |\hat{L}^*_{c,\alpha} - L^*_c| \leq |\hat{L}^*_{c,\alpha} - L^*_{c,\alpha}| + |L^*_{c,\alpha} - L^*_c|
-        \leq \frac{\alpha\delta}{B} + \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}
-    \hat{L}^*_{c',\alpha} \geq L^*_{c',\alpha} - \frac{\alpha\delta b}{2^{n+1}B} 
-                     \geq L^*_{c,\alpha} + \frac{\alpha\delta b}{2^{n+1}B}
-    \geq \hat{L}^*_{c,\alpha}
-\end{displaymath}
-where the first and last inequalities follow from the accuracy of the approximation, and
-the inner inequality follows from Lemma~\ref{lemma:monotonicity}.
-
-\item the accuracy of the approximation $\hat{L}^*_{c,\alpha}$ is:
-\begin{displaymath}
-    \varepsilon' =\frac{\varepsilon\delta b}{2^{n+1}(\delta + n^2B)}
-\end{displaymath}
-
-Note that:
-\begin{displaymath}
-    \log\log (\varepsilon')^{-1} = O\bigg(\log\log\frac{B}{\varepsilon\delta b} + \log n\bigg)
-\end{displaymath}
-Using Lemma~\ref{lemma:barrier} concludes the proof of the running time.\qed
-\end{enumerate}
-
 \section{Budget Feasible Reverse Auction Mechanisms}\label{app:budgetfeasible}
 We review in this appendix the formal definition of a budget feasible reverse auction mechanisms, as introduced by \citeN{singer-mechanisms}. We depart from the definitions in \cite{singer-mechanisms} only in considering $\delta$-truthful, rather than truthful, mechanisms.
 
@@ -582,10 +129,6 @@ Formally, there must exist some $\alpha\geq 1$ and $\beta>0$
     should be computable in time polynomial in various parameters. 
     %time in the number of agents $n$. %\thibaut{Should we say something about the black-box model for $V$?  Needed to say something in general, but not in our case where the value function can be computed in polynomial time}.
 \end{itemize}
-\section{Proof of Lemma~\ref{thm:myerson-variant}}\label{sec:myerson}
-\input{myerson}
-
-\section{Description of our mechanism for \EDP{} and proof of Theorem~\ref{thm:main}}\label{sec:proofofmainthm}
 
 %
 %Instead, \citeN{singer-mechanisms} and \citeN{chen}
@@ -634,216 +177,3 @@ Formally, there must exist some $\alpha\geq 1$ and $\beta>0$
 %variant of Myerson's theorem.
 
 
-\begin{algorithm}[!t]
-    \caption{Mechanism for \SEDP{}}\label{mechanism}
-    \begin{algorithmic}[1]
-    %\State $\mathcal{N} \gets \mathcal{N}\setminus\{i\in\mathcal{N} : c_i > B\}$
-	\Require{ $B\in \reals_+$,$c\in[0,B]^n$, $\delta\in (0,1]$, $\epsilon\in (0,1]$ }
-    \State $i^* \gets \argmax_{j\in\mathcal{N}}V(j)$
-    \State \label{relaxexec}$OPT'_{-i^*} \gets$ using Proposition~\ref{prop:monotonicity},
-    compute a $\varepsilon$-accurate, $\delta$-decreasing
-    approximation of $$\textstyle L^*_{c_{-i^*}}\defeq\max_{\lambda\in[0,1]^{n}} \{L(\lambda)
-                                    \mid \lambda_{i^*}=0,\sum_{i \in \mathcal{N}\setminus\{i^*\}}c_i\lambda_i\leq B\}$$
-        %\Statex
-     \State $C \gets \frac{8e-1 + \sqrt{64e^2-24e + 9}}{2(e-1)}$
-
-        \If{$OPT'_{-i^*} < C \cdot V(i^*)$} \label{c} 
-       \State \textbf{return} $\{i^*\}$ 
-        \Else
-            \State $i \gets \argmax_{1\leq j\leq n}\frac{V(j)}{c_j}$
-	    \State $S_G \gets \emptyset$
-            \While{$c_i\leq \frac{B}{2}\frac{V(S_G\cup\{i\})-V(S_G)}{V(S_G\cup\{i\})}$}
-                \State $S_G \gets S_G\cup\{i\}$ 
- \State $i \gets \argmax_{j\in\mathcal{N}\setminus S_G}
-                \frac{V(S_G\cup\{j\})-V(S_G)}{c_j}$
-            \EndWhile
-            \State \textbf{return} $S_G$
-        \EndIf
-    \end{algorithmic}
-\end{algorithm}
-
-
-We present here the proof of Theorem~\ref{thm:main}. 
-Our mechanism for \EDP{} is composed of
-(a) the allocation function presented in Algorithm~\ref{mechanism}, and 
-(b) the payment function which pays each allocated agent $i$ her threshold
-payment as described in Myerson's Theorem (see Lemma~\ref{thm:myerson-variant}).  In the case where $\{i^*\}$ is the
-allocated set, her threshold payment is $B$. %(she would be have been dropped on
-%line 1 of Algorithm~\ref{mechanism} had she reported a higher cost).
-A closed-form formula for threshold payments when $S_G$ is the allocated set can
-be found in~\cite{singer-mechanisms}. 
-
-
-We use the notation $OPT_{-i^*}$ to denote the optimal value of \EDP{} when the maximum value element $i^*$ is  excluded. We also use $OPT'_{-i^*}$ to denote the approximation computed by the $\delta$-decreasing, $\epsilon$-accurate approximation of $L^*_{c_{-i^*}}$, as defined in Algorithm~\ref{mechanism}.
-
-
-The properties of $\delta$-truthfulness and
-individual rationality follow from $\delta$-monotonicity and threshold
-payments. $\delta$-monotonicity and budget feasibility follow  similar steps as the
-analysis of \citeN{chen}, adapted to account for $\delta$-monotonicity:
-\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
-    a cost $c_i'\leq c_i-\delta$ while all  other costs stay the same.
-    Suppose that when $i$ reports $c_i$, $OPT'_{-i^*} \geq C V(i^*)$; then, as $s_i(c_i,c_{-i})=1$, $i\in S_G$.
-     By reporting cost $c_i'$, $i$ may be selected at an earlier iteration of the greedy algorithm.
-    %using the submodularity of $V$, we see that $i$ will satisfy the greedy
-    %selection rule:
-    %\begin{displaymath}
-    %    i = \argmax_{j\in\mathcal{N}\setminus S} \frac{V(S\cup\{j\})
-    %    - V(S)}{c_j}
-    %\end{displaymath}
-    %in an earlier iteration of the greedy heuristic.
- Denote by $S_i$
-    (resp. $S_i'$) the set to which $i$ is added when reporting cost $c_i$
-    (resp. $c_i'$). We have $S_i'\subseteq S_i$; in addition, $S_i'\subseteq S_G'$, the set selected by the greedy algorithm under $(c_i',c_{-i})$; if not, then greedy selection would terminate prior to selecting $i$ also when she reports $c_i$, a contradiction. Moreover, we have
-    \begin{align*}
-        c_i' & \leq c_i \leq
-        \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  $i\in S_G'$. By
-    $\delta$-decreasingness of
-    $OPT'_{-i^*}$, under $c'_i\leq c_i-\delta$ 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.
-
-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 $OPT'_{-i^*} \geq C V(i^*)$. 
-Denote by $S_G$ the set selected by the greedy algorithm, and for $i\in S_G$,  denote by
-$S_i$ the subset of the solution set that was selected by the greedy algorithm just prior to the addition of $i$---both sets determined for the present cost vector $c$. 
-%Chen \emph{et al.}~\cite{chen} show that, 
-Then for any submodular function $V$, and for all $i\in S_G$:
-%the reported cost of an agent selected by the greedy heuristic, and holds for
-%any submodular function $V$:
-\begin{equation}\label{eq:budget}
-   \text{if}~c_i'\geq \frac{V(S_i\cup\{i\}) - V(S)}{V(S_G)} B~\text{then}~s_i(c_i',c_{-i})=0
-\end{equation}
-In other words, if $i$ increases her cost to a value higher than $\frac{V(S_i\cup\{i\}) - V(S)}{V(S_G)}$, she will cease to be in the selected set $S_G$. As a result,
-\eqref{eq:budget}
-implies that the threshold payment of user $i$ is bounded by the above quantity.
-%\begin{displaymath}
-%\frac{V(S_i\cup\{i\}) - V(S_i)}{V(S_G)} =  B
-%\end{displaymath}
-Hence, the total payment is bounded by the telescopic sum:
-\begin{displaymath}
-    \sum_{i\in S_G} \frac{V(S_i\cup\{i\}) - V(S_i)}{V(S_G)} B = \frac{V(S_G)-V(\emptyset)}{V(S_G)} B=B\qedhere
-\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
-    $O\big(poly(n, d, \log\log\frac{B}{b\varepsilon\delta})\big)$
-\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 \ref{relaxexec} of Algorithm~\ref{mechanism}
-    can be computed in time
-    $O(\text{poly}(n, d, \log\log \frac{B}{b\varepsilon\delta}))$. 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 Lemma~\ref{lemma:greedy-bound} we
-can complete the proof of the approximation ratio of our mechanism
-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 $L^*_{c_{-i^*}}$ be the  maximum value of $L$ subject to
-$\lambda_{i^*}=0$, $\sum_{i\in \mathcal{N}\setminus{i^*}}c_i\leq B$. From line
-\ref{relaxexec} of Algorithm~\ref{mechanism}, we have
-$L^*_{c_{-i^*}}-\varepsilon\leq OPT_{-i^*}' \leq L^*_{c_{-i^*}}+\varepsilon$.
-
-If the condition on line \ref{c} of the algorithm holds then, from the lower bound in Proposition~\ref{prop:relaxation},
-\begin{displaymath}
-    V(i^*) \geq \frac{1}{C}L^*_{c_{-i^*}}-\frac{\varepsilon}{C} \geq
-    \frac{1}{C}OPT_{-i^*} -\frac{\varepsilon}{C}.
-\end{displaymath}
-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 on line \ref{c} does not hold, by observing that $L^*_{c_{-i^*}}\leq L^*_c$ and
-the upper bound of Proposition~\ref{prop:relaxation}, we get
-\begin{displaymath}
-    V(i^*)\leq \frac{1}{C}L^*_{c_{-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}
-Note that $C$ satifies $C(e-1) -6e  +2 > 0$, hence
-\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 \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}
-Our choice of $C$, namely,
-\begin{equation}\label{eq:constant}
-    C =  \frac{8e-1 + \sqrt{64e^2-24e + 9}}{2(e-1)},
-\end{equation}
- is precisely to minimize the maximum among the coefficients of $V_{i^*}$  and $V(S_G)$  in \eqref{eq:bound1}
-and \eqref{eq:bound2}, respectively. Indeed, consider:
-\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 precisely \eqref{eq:constant}.
-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
-
-Finally, we prove the lower bound stated in Theorem~\ref{thm:main}.
-Suppose, for contradiction, that such a mechanism exists. From Myerson's Theorem \cite{myerson}, a single parameter auction is truthful if and only if the allocation function is monotone and agents are paid theshold payments. 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