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OP-01106 Open

Slicing axis-parallel rectangles with few lines

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Call two axis-parallel rectangles in the plane independent if their projections to the $x$-axis are disjoint and their projections to the $y$-axis are also disjoint, and call a set of rectangles independent if its members are pairwise independent.

Vatter met the question in work on permutation classes and showed that the number $f(m)$ below is finite, with an exponential upper bound. Werner and Lenz proved $f(m) = O(m^2)$, a bound they report Tardos and Károlyi had found earlier, and Soberón pointed out that a theorem of Kaiser already gives $f(m) \le 2m$.

Problem. Let $f(m)$ be the least number such that every collection of axis-parallel rectangles containing no independent set of $m$ or more rectangles can be sliced by $f(m)$ horizontal and vertical lines, that is, every rectangle of the collection meets at least one of the lines. Determine $f(m)$.

Source: Vatter (2009); Problem 74 of The Open Problems Project (Demaine, Mitchell and O'Rourke).

Imported from topp.openproblem.net.

Classification

Primary
CO — Combinatorics
Secondary
MG
Keywords
axis-parallel rectangles, independent set, piercing number, The Open Problems Project

References

@misc{vatter2008,
  author        = {Vatter, Vincent},
  title         = {Small permutation classes},
  year          = {2008},
  eprint        = {0712.4006},
  archivePrefix = {arXiv},
  url           = {https://arxiv.org/abs/0712.4006v2},
}
@article{kaiser1997,
  author  = {Kaiser, T.},
  title   = {Transversals of $d$-intervals},
  journal = {Discrete Comput. Geom.},
  volume  = {18},
  number  = {2},
  year    = {1997},
  pages   = {195--203},
}
@misc{werner2010,
  author        = {Werner, Daniel and Lenz, Matthias},
  title         = {Polynomial bounds on the rectangle slicing number},
  year          = {2010},
  eprint        = {1004.3381},
  archivePrefix = {arXiv},
  url           = {https://arxiv.org/abs/1004.3381},
}

Citing this problem

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OP-01106, "Slicing axis-parallel rectangles with few lines", clopen.solutions.