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 with the arithmetical complexity $$O(T_{LP} \cdot (1/\varepsilon)^{m+3} \cdot (2m)^{2m + 6} \cdot \Delta),$$ 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. The goal of the paper is only to prove the existence of an FPTAS, and more accurate analysis can give better constants in exponents. Moreover, we are not worry to much about memory usage, it can be roughly estimated as $O( n \cdot (1/\varepsilon)^{m+3} \cdot (2m)^{2m+6} \cdot \Delta)$.
翻译:已知的是,除非W[1]美元=FPT3}\cdot (2m)\2m)\\2m\2m\2m+6d}\cdot\Delta,否则美元=2美元=2美元=2美元=2美元=2美元。在本注释中,我们显示,美元=Delta$-modual 限制矩阵中,美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=cdot。美元=2美元=2美元=2美元=2美元=3m+6美元=6美元=6美元=2美元=2美元=2美元=3美元=3m+6美元。美元=2美元=3美元=2美元=6美元=2美元=2美元=3美元=2美元=3美元=2美元=2美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=30美元=3美元=3美元=30=3美元=30美元=3美元=30美元=30=30=30=30=30美元=30=30=30=30=30=30=30=30美元=3美元=3美元=3美元=3美元=3美元=3美元=3美元=10美元=10美元=3美元=3美元=10美元=30=30=30=30=30=30=30美元=30美元=30美元=3美元=30美元=30美元=30美元=30=30=30=30=30=