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Let $C$ be a convex shape in the plane. Your task is to cover the plane with copies of $C$, each under any rigid motion. My question is essentially: What is the worst $C$, the shape that forces the most wasteful overlap?

To be more precise, assume $C$ has unit area. Let $n_C(A)$ be the fewest copies of $C$ (under any rigid motions) that suffice to cover a disk of area $A$. I seek the $C$ that maximizes the "waste": $$ w(C) = \lim_{A \to \infty} n_C(A) / A \;.$$ So if $C$ is a perfect tiler of the plane, then $\lim_{A \to \infty} n_C(A) = A$ (because $C$ has area $1$) and $w(C)=1$, i.e., no waste.

Consider a regular pentagon $P$, which cannot tile the plane. Here is one way to cover the plane with regular pentagons:

          TilerPentagon
If I've calculated correctly, this arrangement shows that $w(P) \le 1.510$. So one could cover an area $A=100$ with about $151$ unit-area regular pentagons, a $51$% waste. I doubt this is the best way to cover the plane with copies of $P$ (Q3 below), but it is one way.

Three questions.

Q1. Is it known that the disk is the worst shape $C$ to cover the plane? My understanding is that L.F.Tóth's paper[1], which I have not accessed, establishes this for lattice tilings/coverings. Is it known for arbitrary coverings?

Q1 Answered. Thanks to several, and especially Yoav Kallus, for pointing me in the right direction. Q1 remains an open problem. In [2,p.15], what I call the waste of a convex body $C$ is called $\theta(C)$. It is about $1.209$ for a disk. The best upperbound is $\theta(C) \le 1.228$ due to Dan Ismailescu, based on finding special tiling "p-hexagons" in $C$. A p-hexagon has two opposite, parallel edges of the same length.

Q2. Since every triangle, and every quadrilateral, tiles the plane, the first interesting polygonal shape is pentagons. What is the most wasteful pentagon?

Q3. More specifically, what is the waste $w(P)$ for the regular pentagon?

[1] L. Fejes Tóth, "Lagerungen in der Ebene auf der Kugel und im Raum." Die Grundlehren der mathematischen Wissenschaften Vol. 65. Springer-Verlag, 1972. doi:10.1007/978-3-642-65234-9

[2] Brass, Peter, William OJ Moser, and János Pach. Research Problems in Discrete Geometry. Springer Science & Business Media, 2006.

Joseph O'Rourke
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    How about a house? (Square plus triangle) Gerhard "Welcome To The Suburban Future" Paseman, 2016.12.03. – Gerhard Paseman Dec 04 '16 at 02:30
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    Oh wait. That tiles. Maybe with a weird roof? Gerhard "Not Yet A City Planner" Paseman, 2016.12.03. – Gerhard Paseman Dec 04 '16 at 02:34
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    Any pentagon with two parallel sides tiles the plane. (See for instance the first picture in http://3.bp.blogspot.com/-2YKhXVT62rc/Vd2KGHpYwnI/AAAAAAAAGWU/6Rw5Dn_tTIY/s1600/type-1-15-byEd%2BPegg.jpg) – Noam D. Elkies Dec 04 '16 at 02:36
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    However, you can use these good tilers to estimate the ratio. Pick a good tiler nearest the shape of the bad tiler and embed a scaled down copy into the bad tiler. The best scale factor among many choices should allow a good estimate of the waste factor. Gerhard "It Is Like Urban Renewal" Paseman, 2016.12.03. – Gerhard Paseman Dec 04 '16 at 02:52
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    @GerhardPaseman: Yes. And the lack of understanding of which are the perfect pentagonal tilers is an impediment to answering Q2. – Joseph O'Rourke Dec 04 '16 at 02:55
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    I imagine so. However, a general version of Q3 seems to be equivalent to "What is the area of the largest good (maybe convex) tiler contained entirely inside a bad convex tiler?". Even if it is not equivalent, I think it would make a good Q4. Gerhard "Consider It A Holiday Gift" Paseman, 2016.12.03. – Gerhard Paseman Dec 04 '16 at 03:03
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    As an example of Gerhard Paseman's recent comments, we get a waste ratio of only $w = (5+\sqrt{80})/11 < 1.2676611$ (if I computed right) for regular pentagons $P$ by finding a pentagon $P' \subset P$ of area $|P|/w$ that has two parallel sides and thus tiles the plane: fix a side $s$ of $P$, and let $P'$ consist of all points of $P$ that project to a point of $s$. – Noam D. Elkies Dec 04 '16 at 03:03
  • Indeed. One can use Noam Elkies's projection method to establish a quick upper bound for the waste. Perhaps by perturbing good tilers in known ways, one can carve out a selection of so-so tilers and reduce the search space for bad tilers significantly. Not a Q2 answer, but definitely clearing away the examples that won't work. Maybe asking Q2 for a Pentagon three of whose sides match a good tiler? Gerhard "Using This Mostly For Good" Paseman, 2016.12.03. – Gerhard Paseman Dec 04 '16 at 03:26
  • Great question - I wonder if it's equivalent to the one of what convex shape(s) pack the plane (in the sphere sense) with the lowest density. Obviously they're equivalent at the other end - zero waste is equal to perfect density - but that wouldn't a priori imply equivalence for the worst-case scenarios. – Steven Stadnicki Dec 04 '16 at 03:48
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    I want to edit the reference to give more info (L. Fejes Tóth, "Lagerungen in der Ebene auf der Kugel und im Raum.", Die Grundlehren der mathematischen Wissenschaften Vol. 65. Springer-Verlag, 1972. http://dx.doi.org/10.1007/978-3-642-65234-9) but you wrote 2013, which makes me pause. Is that a second edition? EDIT hmm, no 2nd ed was apparently 1972... – David Roberts Dec 04 '16 at 08:15
  • ...And you wrote 'paper' not 'book', dated after he died! – David Roberts Dec 04 '16 at 08:21
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    Here is an article about the dual problem, densest packing of regular pentagons in the plane: http://blogs.ams.org/visualinsight/2014/12/01/packing-regular-pentagons/ – Gerry Myerson Dec 04 '16 at 11:16
  • @StevenStadnicki: I asked earlier a question that addressed the relationship between tiling and covering (Optimal sphere packings ==> Thinnest ball coverings?). The conclusion was that they are rather different. – Joseph O'Rourke Dec 04 '16 at 12:11
  • @DavidRoberts: Thanks for the correction. I relied (mindlessly) on Google scholar. – Joseph O'Rourke Dec 04 '16 at 12:33
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    If 1) the disc is the worst shape for lattice coverings and 2) the best covering for the disc is a lattice covering, then the disc is the worst shape overall (because the lattice covering of a disc is worse than the lattice covering for each other shape, hence worse than the best covering for each other shape). Presumably 2 is known? – Will Sawin Dec 04 '16 at 12:48
  • Could some results on this page be useful (this is p. 29 in Pach, Agarwal, Combinatorial Geometry)? – Watson Dec 04 '16 at 13:17
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    What is known about the most economic covering of the plane by regular pentagons? That is, bounding $w(P )$ from below. – Pietro Majer Dec 04 '16 at 15:51
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    Sorry, I had to edit to fix the link syntax for the Toth book. What was in my comment was missing the http://! – David Roberts Dec 04 '16 at 15:58
  • @PietroMajer: That is essentially my Q3. I don't know. Likely explored in the past, but I haven't located bounds. – Joseph O'Rourke Dec 04 '16 at 16:27
  • To get an upper bound on $w(C )$, one could consider any shape $T\subset C$ that can form a tassellation of the plane, like the square inside the pentagon in the picture, and argue analogously. Is there e.g. a good hexagon prototile in $P$? The square seems close to optimal, though. – Pietro Majer Dec 04 '16 at 17:12
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    @PietroMajer: See Noam Elkies remarks. He fit a "house" pentagon inside a regular pentagon, knowing that the former tiles the plane. – Joseph O'Rourke Dec 04 '16 at 17:14
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    So a variation could be: let's remove the roof of Noam's house $P'$, but restore the two lateral pieces in $P\setminus P'$. The resulting hexagons make a tassellation of the plane as well. – Pietro Majer Dec 04 '16 at 17:34
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    This way I got exactly $w(P )\le5/4=1.25$ – Pietro Majer Dec 04 '16 at 17:56
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    re:Q1, Brass, Moser, and Pach have a good summary of results in their book https://books.google.com/books?id=cT7TB20y3A8C&lpg=PR1&pg=PA17#v=onepage&q=least%20economical%20convex%20sets%20for%20covering&f=false – Yoav Kallus Dec 04 '16 at 18:17
  • Question Q1 is indeed answered by L. Fejes Tóth in [1] (btw it's a book, not a paper) in the positive for all crossing-free coverings, and it is conjectured that the "crossing-free" assumption is not needed. – Wlodek Kuperberg Dec 05 '16 at 22:40
  • @WlodekKuperberg: I'm not sure I understand your comment, and would like you to clarify. As I read it, you are saying that every domain permits a crossing-free covering no denser than $1.209$. It follows that every domain permits a covering no denser than $1.209$. So, the "crossing-free" qualification is unnecessary. What did I get wrong? – Yoav Kallus Dec 05 '16 at 23:18
  • @YoavKallus: Of course, you're right, I made a mistake. The circle is known to be the worst for covering among centrally symmetric convex disks. In general, it is still an open problem. – Wlodek Kuperberg Dec 06 '16 at 16:59

1 Answers1

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Here is a better (possibly best) way of covering the plane with congruent regular pentagons:

  PentCover
The density of this covering is ${\sqrt5}/2 = 1.1180...$. This covering is generated by the maximum-area $p$-hexagon contained in $P$, and is the thinnest among all coverings with $P$ that are of the double-lattice type. I conjecture that this density is minimum among all coverings with $P$, not just double-lattice ones.

For the notion of double-lattice see: Kuperberg, G.; Kuperberg, W. (1990), Double-lattice packings of convex bodies in the plane, Discrete and Computational Geometry, 5 (4): 389–397, MR 1043721

This covering is obtained by a continuous transition starting with the best packing with regular pentagons, see http://www.auburn.edu/~kuperwl/pent_movie.mp4

Joseph O'Rourke
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Wlodek Kuperberg
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  • Do you know if it is known---or considered likely---that the regular pentagon is the worst pentagon for covering? (Of course the regular hexagon tiles, so regularity is no universal guide.) – Joseph O'Rourke Dec 06 '16 at 00:01
  • If you "raise the roof", so that the overlapping triangle is enlarged when the top two sides are made longer (but leave the other three as is), you get more waste for this type of packing of the new Pentagon (which goes back to my house comment). You might even get halfway to the waste ratio for the circle . Gerhard "Has To Come Full Circle" Paseman, 2016.12.05. – Gerhard Paseman Dec 06 '16 at 02:01
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    Joe: As far as I know - no, but yes, I do think that the regular pentagon is the worst pentagon for covering. Some of its affine images may be equally bad, though. – Wlodek Kuperberg Dec 06 '16 at 04:17
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    No. "Raising the roof" does not make the pentagon worse; it makes it better. Namely, in the modified pentagon one can use a different corner for the overlap; most efficiently the one at which the angle is greatest. – Wlodek Kuperberg Dec 06 '16 at 04:28
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    For some people orange and green colors, of similar darkness, flash and get confused with each other... Otherwise great answer! – Yaakov Baruch Dec 06 '16 at 07:16
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    @YaakovBaruch: colors changed in the picture; working on the animation. – Wlodek Kuperberg Dec 18 '16 at 03:17
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    Actually this is precisely the covering I suggested in my last comment to the question; my computation was $w(P )\le5/4$ – Pietro Majer Dec 18 '16 at 17:38
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    @PietroMajer: that's correct. But my computation was $\sqrt{5/4}$, and in the picture the clipped off tip looks indeed smaller than $1/5$ of the pentagon's area. – Wlodek Kuperberg Dec 18 '16 at 23:27
  • yes, 5/4 is a quick bound for w(P ), based on Area(clipped off tip)<1/5 :) – Pietro Majer Dec 19 '16 at 09:12