add pdf

Author: wata-orz

README.md

@@ -1 +1,3 @@
-# santa17
+# santa17
+
+This is my 2nd place solution for the Kaggle 2017 Santa competition (https://www.kaggle.com/c/santa-gift-matching).

paper.pdf

@@ -0,0 +1,63 @@
+\documentclass{article}
+
+\usepackage{xspace,amsmath,amssymb,amsthm,fullpage}
+
+\newcommand{\R}{\mathbb{R}}
+
+
+\begin{document}
+
+\section{Overview}
+
+Instead of maximizing $(\sum \mathrm{ANCH})^3+(\sum \mathrm{ANSH})^3$, we maximized $M(\sum \mathrm{ANCH})+(\sum \mathrm{ANSH})$ for a very huge constant $M$.
+If we ignore the triple/twin constraints, the problem can be solved by using a minimum-cost flow algorithm.
+Using this relaxation lower bound, we solved the original problem by the branch-and-bound method.
+
+\section{Details}
+
+\subsection{Problem}
+
+The problem can be formulated as follows.
+Given a bipartite graph $G=(U\cup V,E)$, an edge-weight function $w:E\to\R$, a multiplicity function $d:U\to\R_{> 0}$, and a capacity function $c:V\to\R_{> 0}$,
+find a minimum-weight subset $M\subseteq E$ satisfying (i) $|M\cap \delta(u)|=1$ and (ii) $\sum_{uv\in M\cap\delta(v)}d(u)=c(v)$.
+
+\subsection{Relaxation}
+
+The LP relaxation of this problem is the minimum-cost flow problem, which can be solved in polynomial time.
+More particularly, we construct a network as follows.
+Each edge $uv\in E$ has capacity $d(u)$ and cost $w(uv)/d(u)$.
+Each vertex $u\in U$ has supply $d(u)$, and each vertex $v\in V$ has demand $c(v)$.
+In the original problem, we have additional constraints that, for every edge $uv\in E$, its flow value must be either $0$ or $d(u)$.
+
+\subsection{Branch and Bound}
+
+For solving the original problem, we use a branch-and-bound method.
+For every node of the search tree, we do as follows.
+
+First, we compute a relaxed solution by solving the minimum-cost flow problem (we implemented a cost-scaling algorithm by Goldberg and Tarjan).
+At the root node, we need to solve the relaxed problem from scratch, but for the other nodes, we can obtain a relaxed solution in $O(|E|\log |V|)$ time by modifying the relaxed solution for the parent node by searching an augmenting path.
+
+If the obtained relaxed solution is larger than the solution for the original problem we have found so far, we quit the subsequent search and backtrack to the parent node.
+Otherwise, we choose an edge $uv$ whose flow value $f(uv)$ is neither $0$ nor $d(u)$.
+If there are no such edges, we obtain an improved solution for the original problem and backtrack to the parent node.
+Otherwise, we branch into two cases: $f(uv)=0$ or $f(uv)=d(u)$.
+In order to make the search space smaller, we compute a relaxed solution for every unsatisfying edge and choose the one with the largest lower bound.
+
+\subsection{Reductions}
+
+Because the input is very large, we cannot solve it directly.
+First, we remove all the edges of weight zero and relax the constraints (i) and (ii) to (i') $|M\cap \delta(u)|\leq 1$ and (ii') $\sum_{uv\in M\cap\delta(v)}d(u)\leq c(v)$.
+From the solution for this relaxed problem, we construct a solution for the original problem by solving the knapsack problem.
+
+Second, we apply the reduced-cost fixing as follows.
+Let $D$ be the difference between a known upper bound and the current lower bound.
+Let $f$ be the minimum-cost flow and let $p:U\cup V\to\R$ be the dual variables (i.e., potentials).
+
+For edge $uv\in E$ with $f(uv)<d(uv)$, the increase of the cost for fixing $f(uv)=d(u)$ is at least $(w(uv)+p(u)-p(v))(d(u)-f(uv))$.
+Therefore, if this value is larger than $D$, we can fix $f(uv)=0$.
+
+For edge $uv\in E$ with $f(uv)>0$, the increase of the cost for fixing $f(uv)=0$ is at least $f(uv)\ell(u,v)$, where $\ell(u,v)$ is the shortest-path distance from $u$ to $v$ in the residual graph.
+Therefore, if this value is larger than $D$, we can fix $f(uv)=d(u)$.
+Because $|V|$ is small (1000), we can compute all the distances $\ell(u,v)$ by using the Dijkstra's algorithm against the reversed graph.
+
+\end{document}