In the Two-dimensional Bin Packing (2BP) problem, we are given a set of rectangles of height and width at most one and our goal is to find an axis-aligned nonoverlapping packing of these rectangles into the minimum number of unit square bins. The problem admits no APTAS and the current best approximation ratio is $1.406$ by Bansal and Khan [SODA'14]. A well-studied variant of the problem is Guillotine Two-dimensional Bin Packing (G2BP), where all rectangles must be packed in such a way that every rectangle in the packing can be obtained by recursively applying a sequence of end-to-end axis-parallel cuts, also called guillotine cuts. Bansal, Lodi, and Sviridenko [FOCS'05] obtained an APTAS for this problem. Let $\lambda$ be the smallest constant such that for every set $I$ of items, the number of bins in the optimal solution to G2BP for $I$ is upper bounded by $\lambda\operatorname{opt}(I) + c$, where $\operatorname{opt}(I)$ is the number of bins in the optimal solution to 2BP for $I$ and $c$ is a constant. It is known that $4/3 \le \lambda \le 1.692$. Bansal and Khan [SODA'14] conjectured that $\lambda = 4/3$. The conjecture, if true, will imply a $(4/3+\varepsilon)$-approximation algorithm for 2BP. According to convention, for a given constant $\delta>0$, a rectangle is large if both its height and width are at least $\delta$, and otherwise it is called skewed. We make progress towards the conjecture by showing $\lambda = 4/3$ for skewed instance, i.e., when all input rectangles are skewed. Even for this case, the previous best upper bound on $\lambda$ was roughly 1.692. We also give an APTAS for 2BP for skewed instance, though general 2BP does not admit an APTAS.
翻译:在二维的 Bin 包装 (2BP) 问题中, 我们得到的是一组以高度和宽度为单位的直径矩形。 最多一个, 我们的目标是找到一个以轴为主的不重叠包装这些正方形的最小数量 。 问题没有接受 APTAS, 而目前的最佳近似比率是 Bansal 和 Khan [SODA'14] 的 1.406 美元。 问题的一个得到很好研究的变式是 吉洛丁 双维本包( G2BP) 。 所有正方块的平面矩形必须以这样的方式打包, 使每个正方块的正方块 $ 2. 0. 3 美元 的正方格可以再次使用端对端轴轴轴轴轴的切分解 。 当 以美元为正值的平面的平面的平面的平面是 。 以美元平面的平面的平面的平面的平面的平面的平面是 。 当 以美元平面的平面的平面的平面的平面的平面的平面的平面的平面是 。