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authorThibaut Horel <thibaut.horel@gmail.com>2015-03-06 18:16:03 -0500
committerThibaut Horel <thibaut.horel@gmail.com>2015-03-06 18:16:03 -0500
commit73a47224ff0c5825efe0435db600774b037fb5bf (patch)
tree0e89700c57ed651f1955668b7a22dbd978f1226c /paper/sections
parent96f5a11aec4e9a6c94eedc3167ac42b4f62c6f93 (diff)
downloadfast-seeding-73a47224ff0c5825efe0435db600774b037fb5bf.tar.gz
Final version
Diffstat (limited to 'paper/sections')
-rw-r--r--paper/sections/adaptivity.tex10
-rw-r--r--paper/sections/algorithms.tex13
-rw-r--r--paper/sections/experiments.tex1
-rw-r--r--paper/sections/model.tex2
4 files changed, 12 insertions, 14 deletions
diff --git a/paper/sections/adaptivity.tex b/paper/sections/adaptivity.tex
index d590408..8e2e52f 100644
--- a/paper/sections/adaptivity.tex
+++ b/paper/sections/adaptivity.tex
@@ -149,10 +149,10 @@ adaptive policy is upper bounded by the optimal non-adaptive policy:
% \end{displaymath}
%\end{proposition}
-The proof of this proposition can be found in the full version of the
-paper~\cite{full} and relies on the following fact: the optimal adaptive policy
-can be written as a feasible non-adaptive policy, hence it provides a lower
-bound on the value of the optimal non-adaptive policy.
+The proof of this proposition can be found in Appendix~\ref{sec:ad-proofs} and
+relies on the following fact: the optimal adaptive policy can be written as
+a feasible non-adaptive policy, hence it provides a lower bound on the value of
+the optimal non-adaptive policy.
\subsection{From Non-Adaptive to Adaptive Solutions}\label{sec:round}
@@ -212,7 +212,7 @@ $S$ that we denote by $A(S)$.
More precisely, denoting by $\text{OPT}_A$ the optimal value of the adaptive
problem~\eqref{eq:problem}, we have the following proposition whose proof can
-be found in the full version of this paper~\cite{full}.
+be found in Appendix~\ref{sec:ad-proofs}.
\begin{proposition}\label{prop:cr}
Let $(S,\textbf{q})$ be an $\alpha$-approximate solution to the
non-adaptive problem \eqref{eq:relaxed}, then $\mathrm{A}(S) \geq \alpha
diff --git a/paper/sections/algorithms.tex b/paper/sections/algorithms.tex
index 810653f..7dd22b5 100644
--- a/paper/sections/algorithms.tex
+++ b/paper/sections/algorithms.tex
@@ -49,7 +49,7 @@ An optimal solution to the above problem can be found in polynomial time using
standard LP-solvers. The solution returned by the LP is \emph{fractional}, and
requires a rounding procedure to return a feasible solution to our problem,
where $S$ is integral. To round the solution we use the pipage rounding
-method~\cite{pipage}. We defer the details to the full version of the paper~\cite{full}.
+method~\cite{pipage}. We defer the details to Appendix~\ref{sec:lp-proofs}.
\begin{lemma}
For \mbox{\textsc{AdaptiveSeeding-LP}} defined in \eqref{eq:lp}, any fractional solution $(\boldsymbol\lambda, \mathbf{q})\in[0,1]^m\times[0,1]^n$ can be rounded to an integral solution $\bar{\boldsymbol\lambda} \in \{0,1\}^{m}$ s.t. $(1-1/e) F(\mathbf{p}\circ\mathbf{q}) \leq A(\bar{\lambda})$ in $O(m + n)$ steps.
@@ -109,7 +109,7 @@ $\mathcal{O}(T,b)$.
non-decreasing in $b$.
\end{lemma}
-The proof of this lemma can be found in the full version of the paper~\cite{full}. The main
+The proof of this lemma can be found in Appendix~\ref{sec:comb-proofs}. The main
idea consists in writing:
\begin{multline*}
\mathcal{O}(T\cup\{x\},c)-\mathcal{O}(T\cup\{x\},b)=\int_b^c\partial_+\mathcal{O}_{T\cup\{x\}}(t)dt
@@ -137,10 +137,10 @@ y\in\neigh{X}\setminus T$, we need to show that:
This can be done by partitioning the set $T$ into ``high value
items'' (those with weight greater than $w_x$) and ``low value items'' and
carefully applying Lemma~\ref{lemma:nd} to the associated subproblems.
- The proof is in the full version of the paper~\cite{full}.
+ The proof is in Appendix~\ref{sec:comb-proofs}.
Finally, Lemma~\ref{lemma:sub} can be used to show Proposition~\ref{prop:sub}
-whose proof can be found in the full version~\cite{full}.
+whose proof can be found in Appendix~\ref{sec:comb-proofs}.
\begin{proposition}\label{prop:sub}
Let $b\in\mathbf{R}^+$, then $\mathcal{O}(\neigh{S},b)$ is monotone and
@@ -195,8 +195,7 @@ solution for the split with the highest value (breaking ties arbitrarily).
This process can be trivially parallelized across $k-1$ machines, each
performing a computation of a single split. With slightly more effort, for any
$\epsilon>0$ one can parallelize over $\log_{1+\epsilon}n$ machines at the cost
-of losing a factor of $\epsilon$ in the approximation guarantee (see full
-version of the paper~\cite{full} for details).\newline
+of losing a factor of $\epsilon$ in the approximation guarantee (see Appendix~\ref{sec:para} for details).\newline
\noindent \textbf{Implementation in MapReduce.} While the previous paragraph
describes how to parallelize the outer \texttt{for} loop of
@@ -206,7 +205,7 @@ applied to the function $\mathcal{O}\left(\neigh{\cdot}, t\right)$. The
\textsc{Sample\&Prune} approach successfully applied in \cite{mr} to obtain
MapReduce algorithms for various submodular maximizations can also be applied
to Algorithm~\ref{alg:comb} to cast it in the MapReduce framework. The details
-of the algorithm can be found in the full version of the paper~\cite{full}.
+of the algorithm can be found in Appendix~\ref{sec:mr}.
\newline
%A slightly more sophisticated approach is to consider only $\log n$ splits: $(1,k-1),(2,k-2),\ldots,(2^{\lfloor \log n \rfloor},1)$ and then select the best solution from this set. It is not hard to see that in comparison to the previous approach, this would reduce the approximation guarantee by a factor of at most 2: if the optimal solution is obtained by spending $t$ on the first stage and $k-t$ in the second stage, then since $t \leq 2\cdot2^{\lfloor \log t \rfloor}$ the solution computed for $(2^{\lfloor \log t \rfloor}, k - 2^{\lfloor \log t \rfloor})$ will have at least half that value.
diff --git a/paper/sections/experiments.tex b/paper/sections/experiments.tex
index 2faa011..650e41d 100644
--- a/paper/sections/experiments.tex
+++ b/paper/sections/experiments.tex
@@ -67,7 +67,6 @@ constitute our core set. We then crawled the social network of
these sets: for each user, we collected her list of friends, and the degrees
(number of friends) of these friends.\newline
-
\noindent\textbf{Data description.} Among the several verticals we collected,
we select eight of them for which we will report our results. We obtained
similar results for the other ones. Table~\ref{tab:data} summarizes statistics
diff --git a/paper/sections/model.tex b/paper/sections/model.tex
index ee24eec..5d64acf 100644
--- a/paper/sections/model.tex
+++ b/paper/sections/model.tex
@@ -77,6 +77,6 @@ the transition matrix of the Markov chain. This observation led to the
development of fast algorithms for influence maximization under the voter
model~\cite{even-dar}.\newline
-\noindent \textbf{\textsc{NP}-Hardness.} In contrast to standard influence maximization, adaptive seeding is already \textsc{NP}-Hard even for the simplest influence functions such as $f(S) = |S|$ and when all probabilities are one. We discuss this in the full version of the paper~\cite{full}.
+\noindent \textbf{\textsc{NP}-Hardness.} In contrast to standard influence maximization, adaptive seeding is already \textsc{NP}-Hard even for the simplest influence functions such as $f(S) = |S|$ and when all probabilities are one. We discuss this in Appendix~\ref{sec:alg-proofs}.
%In the case when $f(S)=|S|$ and all probabilities equal one, the decision problem is whether given a budget $k$ and target value $\ell$ there exists a subset of $X$ of size $k-t$ which yields a solution with expected value of $\ell$ using $t$ nodes in $\mathcal{N}(X)$. It is easy to see that this problem is \textsc{NP}-hard by reduction from \textsc{Set-Cover}.
%This is equivalent to deciding whether there are $k-t$ nodes in $X$ that have $t$ neighbors in $\mathcal{N}(X)$. To see this is \textsc{NP}-hard, consider reducing from \textsc{Set-Cover} where there is one node $i$ for each input set $T_i$, $1\leq i\leq n$, with $\neigh{i}= T_i$ and integers $k,\ell$, and the output is ``yes'' if there is a family of $k$ sets in the input which cover at least $\ell$ elements, and ``no'' otherwise.