AI 中文总结
该研究改进了d维向量背包问题的近似算法运行时间，给出二维情况的细粒度下界，确定其最优指数，且二维特殊情况的算法达到双准则最优。
AI 中文摘要
本文重新研究d维向量背包问题（d-Knapsack）：给定一个d维容量向量和一组物品，每个物品具有一个d维权重向量和一个利润，目标是选择一组物品，使得在不超过任何维度容量的前提下最大化总利润。对于任意d≥2，目前已知的d-Knapsack最佳近似方案的运行时间为O(n^{⌈d/ε⌉−d})[Caprara、Kellerer、Pferschy、Pisinger '00]。我们将该运行时间改进为$\tilde{O}_{d,ε,ρ}(n^{⌈(d−1)/(2ε)−1/2+ρ⌉}+n^d)$，其中ε∈(0,1)且ρ∈(0,1)。我们通过设计首个针对d-Knapsack的中途相遇算法实现了这一加速，该算法需要基于LP结构论证，用高效的动态规划算法替代先前算法中使用的LP求解器来生成代表性解。这是25多年来的首次改进，也是首个将指数以常数因子改进的结果。我们补充了基于k-SUM的细粒度下界，表明2-Knapsack需要时间$n^{⌈1/(2ε)−1/2⌉−o(1)}$，这确定了2-Knapsack的最优指数为$1/(2ε)±O(1)$，其精度误差仅为加性O(1)。据我们所知，这是首个针对任何具有PTAS但无EPTAS的问题，确定最优指数精度优于O(1)因子的结果。对于2-Knapsack的特殊情况，我们还在时间$\tilde{O}_{δ,ε}(n^{⌈1/(2ε)−1/2⌉})$内实现了(1−ε−δ)近似，该结果几乎匹配我们的下界，因为对于稍好的近似比，不可能有稍好的运行时间，因此我们的算法是双准则最优的。
英文摘要
We revisit the $d$-dimensional Vector Knapsack problem ($d$-Knapsack): Given a $d$-dimensional capacity vector and a set of items, each with a $d$-dimensional weight vector and a profit, the goal is to select a set of items that maximizes the total profit without exceeding the capacity in any dimension. For any $d\ge2$, the best known approximation scheme for $d$-Knapsack runs in time $O(n^{\lceil d/\varepsilon\rceil-d})$ [Caprara, Kellerer, Pferschy, Pisinger '00]. We improve this running time to $\widetilde O_{d,\varepsilon,ρ}(n^{\lceil\frac{d-1}{2\varepsilon}-\frac12+ρ\rceil}+n^d)$ for any $\varepsilon\in(0,1)$ and every parameter $ρ\in(0,1)$. We achieve this speedup by designing the first meet-in-the-middle algorithm for $d$-Knapsack. This requires replacing the LP solver used in prior algorithms by a highly efficient dynamic programming algorithm to generate representative solutions, building on an LP-based structural argument. This is the first improvement in over 25 years, and the first result that improves the exponent by a constant factor. We complement this by a fine-grained lower bound based on $k$-SUM showing that 2-Knapsack requires time $n^{\lceil\frac1{2\varepsilon}-\frac12 \rceil-o(1)}$. This establishes the optimal exponent of 2-Knapsack as $\frac1{2\varepsilon}\pm O(1)$, which is precise up to an additive $O(1)$. To the best of our knowledge, this is the first result that determines the optimal exponent more precisely than up to a factor $O(1)$, for any problem that admits a PTAS but no EPTAS. For the special case of 2-Knapsack we further attain a $(1-\varepsilon-δ)$-approximation in time $\widetilde O_{δ,\varepsilon}(n^{\lceil\frac1{2\varepsilon}-\frac12\rceil})$. This nearly matches our lower bound, as for a slightly better approximation ratio a slightly better running time is impossible -- so our algorithm is bicriteria-optimal.
CommentsAbstract shortened to fit ArXiV requirements
