Suppose you have a supply of infinite-length, opaque, unit-radius cylinders, and you would like to block all visibility from a point $p \in \mathbb{R}^3$ to infinity with as few cylinders as possible. (The cylinders are infinite length in both directions.) The cylinders may touch but not interpenetrate, and they should be disjoint from $p$, leaving a small ball around $p$ empty. (Another variation would insist that cylinders be pairwise disjoint, i.e., not touching one another.)

A collection of parallel cylinders arranged to form a "fence" around $p$ do not suffice, leaving two line-of-sight $\pm$ rays to infinity. Perhaps a grid of cylinders in the pattern illustrated left below suffice, but at least if there are not many cylinders, there is a view from an interior point to infinity (right below).

Crossed Cylinders

I feel like I am missing a simple construction that would obviously block all rays from $p$. Perhaps crossing the cylinders like the poles of a tipi (teepee) could help, but it seems this would at best lead to inefficient blockage. Suggestions welcome—Thanks!

Addendum1. Perhaps if the weaving above is rendered irregular by displacing the cylinders slightly by different amounts, so that cracks do not align, then a sufficient portion of the weaving will block all visibility.

Here (left below) is the start of Gerhard's first suggested construction (a portion of the weaving above), which I don't see how to complete. But perhaps seeing this depiction will aid intuition.
Crossed Tangent to sphere Cylinder forest

Addendum2. To the right above I added (three-quarters of) a forest along the lines (but not exactly as) Yaakov suggested.

  • $\begingroup$ can you arrange $3$ cylinders vertically with their centers forming a equilateral triangle around $p$, and then seal off the top and bottom holes with (I'd guess) $4$ other cylinders for each hole? $\endgroup$ – Olivier Bégassat May 9 '11 at 14:47
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    $\begingroup$ @Olivier: But I intended that all cylinders must be infinite in length. Likely my figure misled in that regard. Edited to make clearer. $\endgroup$ – Joseph O'Rourke May 9 '11 at 15:05
  • $\begingroup$ I posted a bad comment and deleted it. I think the question is interesting for dounbly infinite cylinders. If cylinders are allowed to be singly infinite take a hexagon of seven doubly infinite cylinders eg vertically and cut the middle one to allow space for p. (previous comment was six in a triangle - but until you get ten in a triangle there is no central one, and when you have ten it is only the central hexagon which counts) $\endgroup$ – Mark Bennet May 9 '11 at 15:17
  • $\begingroup$ You can mimic larger cylinders by groups of small cylinders. Should you not then be able to place large groups far enough away to patch any holes that a close and densely packed arrangement leaves? Gerhard "The Forest Is The Trees" Paseman, 2011.05.09 $\endgroup$ – Gerhard Paseman May 9 '11 at 17:34
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    $\begingroup$ Perhaps my visualization skills could use some assistance. What about 6 cylinders tangent (or near tangent) to a unit sphere (say a certain subset of your red, blue, and green in your picture) and then 8 small groups of 2 or 3 cylinders close but not touching to block the remaining lines of sight? Gerhard "Ask Me About System Design" Paseman, 2011.05.09 $\endgroup$ – Gerhard Paseman May 9 '11 at 18:24

Here is one construction. On the horizontal xy plane place a forest of vertical cylinders of radius r<1/2 (or =1/2 if we allow contacts) centered at each point in $(\mathbb{Z} \backslash{\lbrace0\rbrace})\times(\mathbb{Z}\backslash{\lbrace0\rbrace})$; moreover place 2 similar cylinders parallel to x centered at (y=+/-1, z=0), 2 more parallel to y centered at (x=0, z=+/-1) and the last 2 parallel to z and centered at (x=+/-1, y=0). Then (0,0,0) is blocked by the forest in all directions except those in the xz and yz planes, which are blocked by the other 6 cylinders. The forest clearly does not need to be infinite and it should be easy to find an upper limit on its size.

${\bf UPDATE}$ As pointed out by Mark in a comment, the forest should be based on $(\mathbb{Z} \backslash{\lbrace0\rbrace})\times(\mathbb{Z}\backslash{\lbrace-1,0,1\rbrace})$.

  • $\begingroup$ I think this construction is very close to what Gerhard meant with his second comment to the question. $\endgroup$ – Yaakov Baruch May 10 '11 at 13:07
  • $\begingroup$ I realize now, I simply added a forest of green cylinders to the last picture. $\endgroup$ – Yaakov Baruch May 10 '11 at 13:11
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    $\begingroup$ @Yaakov: I'm afraid I don't follow. The cylinders parallel to $x$ seem to intersect with the trees in your forest with centres at $(x,\pm 1)$? $\endgroup$ – Mark Grant May 10 '11 at 13:14
  • $\begingroup$ @Mark: you are right - I need to move the 4 islands of the forest a bit further away from the x-axis - I'm editing my answer accordingly. $\endgroup$ – Yaakov Baruch May 10 '11 at 13:28
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    $\begingroup$ CLARIFICATION: a finite forest suffices because the projection of any open cylinder on the surface of the unit sphere is open and the sphere is compact. $\endgroup$ – Yaakov Baruch May 10 '11 at 16:21

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