It is known that there is no EPTAS for the $m$-dimensional knapsack problem unless $W[1] = FPT$. It is true already for the case, when $m = 2$. But, an FPTAS still can exist for some other particular cases of the problem. In this note, we show that the $m$-dimensional knapsack problem with a $\Delta$-modular constraints matrix admits an FPTAS, whose complexity bound depends on $\Delta$ linearly. More precisely, the proposed algorithm complexity is $$O(T_{LP} \cdot (1/\varepsilon)^{m+3} \cdot (2m)^{2m + 6} \cdot \Delta),$$ where $T_{LP}$ is the linear programming complexity bound. In particular, for fixed $m$ the arithmetical complexity bound becomes $$ O(n \cdot (1/\varepsilon)^{m+3} \cdot \Delta). $$ Our algorithm is actually a generalisation of the classical FPTAS for the $1$-dimensional case. Strictly speaking, the considered problem can be solved by an exact polynomial-time algorithm, when $m$ is fixed and $\Delta$ grows as a polynomial on $n$. This fact can be observed combining previously known results. In this paper, we give a slightly more accurate analysis to present an exact algorithm with the complexity bound $$ O(n \cdot \Delta^{m + 1}), \quad \text{ for $m$ being fixed}. $$ Note that the last bound is non-linear by $\Delta$ with respect to the given FPTAS.
翻译:已知的是,除非W[1]美元=FPT$,否则美元维度 knapsack 问题就没有EPTAS 。 更确切地说, 提议的算法复杂性是美元O( T ⁇ LP}) 美元, 美元=2美元。 但是, 美元=2美元=3}\cdot 仍然可以存在。 在本说明中, 我们显示, 美元=2美元=6美元为线性程序复杂度的美元问题。 特别是, 以美元为单位的算术复杂性取决于 美元=Delta$ 线性。 更确切地说, 拟议的算法复杂性是美元=O( T ⁇ LP} $$ (cdo) 美元=cdot 美元=cdot 美元 (1/\\\\\\\\\\\\\\\\\qlsl) 美元 美元 美元=xxxxxxxxxxxxxxxxxxal =xal- 美元=xnalalal- 美元=xal- 美元=xal- 美元=xlation 美元=xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx=xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx