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Extreme Orchards for Gardner

GROUPS:

Plant 11 trees to get 16 lines of 3 trees. (Friedman, Tree Planting Problems)
Plant 19 trees to get 20 lines of 4 trees. (Friedman, Tree Planting Problems)
Plant 25 trees to get 18 lines each having 5 trees. (Elkies, On some points-and-lines problems )
Plant 62 trees on the Earth to get 15 lines of 12 trees. (Answer)

The modern father of recreational math Martin Gardner would have turned 102 today, so I thought I'd take a look at the orchard-planting problem. I'm never content to just quote previous work, so I looked closely at that "25 trees to get 18 lines each having 5 trees" configuration. The points were at simple barycentric coordinates. What are those? Consider a triangle with vertices A, B, C and an area of 1. Add an interior point D. The barycentric coordinates of D are the areas of triangles ABD, ADC, and DBC. The sum of the coordinates equals one since the original triangle had area 1. I used barycentric coordinates in my previous discussion Zak's Triangle.

Here's a picture. A =(1,0,0), B =(0,1,0), C =(0,0,1). For the interior points, you'll need to add a step -- divide by the total. The point marked 210 is (2/3, 1/3, 0). The point marked 123 is (1/6, 1/3, 1/2). Showing them as fractions would be a much messier picture.

25 trees to get 18 lines each having 5 trees

Code for that. First a triangle, then code for converting to and from barycentric coordinates.

tri = {{Sqrt[3], -1}, {0, 2}, {-Sqrt[3], -1}};

ToBarycentrics[{{x1_, y1_}, {x2_, y2_}, {x3_, y3_}}, {p_, q_}] := Solve[{m*x1 + n*x2 + (1 - m - n)*x3, m*y1 + n*y2 + (1 - m - n)*y3} == {p, q}];

FromBarycentrics[{m_, n_}, {{x1_, y1_}, {x2_, y2_}, {x3_, y3_}}] := {m*x1 + n*x2 + (1 - m - n)*x3, m*y1 + n*y2 + (1 - m - n)*y3};  

Next, it's handy to have barycentric lines. A barycentric point a b c is on a barycentric line A B C if their dot product A a + B b + C c equals 0. A canonical barycentric line can have three forms:

  1. (a, b, c) with total 1.
  2. (a, -1-a, 1) with total 0, going through the center point (1/3, 1/3, 1/3).
  3. (1,-1,0), the vertical line through the center point.

Here's code for finding a barycentric line and finding the intersection point of two barycentric lines:

BaryLiner[{pt1_, pt2_}] := Module[{a, b},
  If[Det[{pt1, pt2, {1, 1, 1}/3}] != 0,
   {a, b, 1 - a - b} /. 
    Solve[{a, b, 1 - a - b}.# == 0 & /@ {pt1, pt2}][[1]],
   If[pt1[[1]] != pt1[[2]] ,
    {a, -1 - a, 1} /. 
     Solve[{a, -1 - a, 1}.# == 0 & /@ {pt1, pt2}][[1]],
    {1, -1, 0}
    ]]];

lineintersect[{{a1_, b1_, c1_}, {a2_, b2_, c2_}}] := 
  With[{init = {b1 c2 - b2 c1, a2 c1 - a1 c2, a1 b2 - a2 b1}}, 
   If[Total[init] == 0, init, init/Total[init]]];

With those pieces of code, I was able to find some more extreme orchards. Here's 72 trees making 57 rows of 6 or more trees. There are also 27 lines of 4 trees.

peggpoints = Sort[#/Total[#] & /@ Flatten[(Permutations /@ {{0, 0, 1}, {0, 1, 1}, {0, 1, 2}, {0, 4, 5}, {1, 1, 2}, {1, 2, 2}, {1, 2, 3}, {1, 2, 6}, {1, 4, 4}, {2, 2, 3}, {2, 2, 5}, {2, 3, 4}, {2, 3, 5}, {2, 5, 5}, {2, 6, 7}, {4, 5, 6}}), 1]];  
pegglines = First /@ Select[SortBy[Tally[BaryLiner[#] & /@ Subsets[peggpoints, {2}]], Last], Last[#] > 5 &];  
linecoords = Table[FromBarycentrics[{#[[1]], #[[2]]}, tri] & /@ Select[peggpoints, pegglines[[n]].# == 0 &], {n, 1, Length[pegglines]}];  
Graphics[{AbsoluteThickness[1], {{White, White, White, Cyan, Red, Black, Green, Magenta}[[Length[#]]], Line[#]} & /@ linecoords, 
  With[{coord = FromBarycentrics[{#[[1]], #[[2]]}, tri]}, {Black, Disk[coord, .057], White, Disk[coord, .054], Black, Style[Text[StringJoin[ToString /@ (# (Max[Denominator[#]]))], coord], 8]}] & /@ peggpoints}, ImageSize -> {600, 520}]

72 trees in 57 lines of 6 or more trees

After finding that, I started thinking about something -- all the coordinates had sum one, and all the lines not through the center had sum one. Could I find a really nice configuration where the point set and the line set were the same? Here's one base set. See the notebook for full code.

base = {{-5, 3, 4}, {-4, 3, 5}, {-2, -1, 6}, {-2, 1, 2}, {-2, 1, 3}, {-1, 0, 2}, {-1, 0, 3}, {-1, 1, 1}, {-1, 1, 2}, {-1, 1, 3}, {-1, 2, 2}, {-1, 2, 3}, {-1, 3, 3}, {-1, 3, 6}, {0, 0, 1}, {0,1, 1}, {0, 1, 2}, {0, 1, 3}, {1, 1, 2}, {1, 1, 4}, {1, 2, 2}, {1,3, 3}, {1, 5, 7}};  

The resulting set of 111 lines and 111 points looks like the following:

Every line is a point. Every point is a line.

I'm not satisfied with that one yet. It needs a smaller set, and ideally there should be 6 lines through each point and 6 points on each line. Whether that's possible is actually an unsolved question in the realm of [configurations](https://en.wikipedia.org/wiki/Configuration_(geometry)). If every point has n lines and every line has n points it's known as an n-configuration. Here are some 4-configurations.

4-configurations

Whether a 5-configuration existed wasn't settled until 2007, when Leah Berman found the following:

5-configuration

Ideally, I should fold those configurations into barycentric coordinates and see if I could make improvements. Here are two more extreme orchards I found while searching in this space.

extreme orchard 1

extreme orchard 2

Can these be cleaned up to make the first 6-configuration, or beyond? Do all of the known best orchard solutions work out nicely in barycentric coordinates? With 30 trees, what is the maximal number of 3-lines, 4-lines, 5-lines, or 6-lines possible? All of these are unsolved questions. Based on my playing around here, I think that barycentric coordinates might lead to some new solutions in the world of extreme orchards.

Attachments:
POSTED BY: Ed Pegg
Answer
1 year ago

enter image description here - another post of yours has been selected for the Staff Picks group, congratulations! We are happy to see you at the top of the "Featured Contributor" board. Thank you for your wonderful contributions, and please keep them coming!

POSTED BY: Moderation Team
Answer
1 year ago

As someone with 200+ apple trees in my 7 acre yard, this is fun to think about. My trees are planted in rows, which is practical if a bit boring.

But there are certain pragmatic constraints involved in planting apple trees.

The orchardist needs access to the trees from two sides both for maintenance (pruning, spraying, etc.) and also to harvest apples. Assuming semi-dwarf trees, this involves aisles with a minimum width of about 22 ft (ca. 6.7 m) starting from the center of each trunk. The trees should be planted no closer than intervals of 16 ft (ca. 4.9 m) to give them enough air and light.

So how big does an orchard need to be for each of these configurations? Eyeballing it, only configurations in which there is a small variation in the segments connecting trees could realistically be planted as something that would, on the ground, resemble an orchard. Most of the configurations would require an enormous amount of land and so are mostly mathematical abstractions rather than something one could really implement.

How can we modify this to accommodate real trees? The simplest way might be to treat it as packing problem involving circles 8 ft. in diameter that also need 6 ft. of breathing space on two sides.

One advantage I see in the configurations with a small variation in segment length is that planting the orchard to look like a simple mandala reduces the amount of grass under the trees to be maintained, thus significantly reducing mowing and therefore labor and gasoline costs. So it is not completely foolish to consider planting at least a small orchard this way. I am somewhat attracted to the 25 tree pentagon configuration because of its center empty circle, creating a private grove space. Taking into account an air gap around the outside, my guess is that a circle in the field of about 125 ft in diameter should be big enough. That center circle could, for example, hold a very nice circle of wildflowers 20 ft. across for bee forage, maybe some bee hives in the center, and still leave room for equipment to navigate. Another advantage: this would be a good layout for planting five types of trees in groups of five. The could then be easily identified in their mini-groves and harvested together.

So what would the Wolfram Language code be that would tell me how much land I needed and the placement of the trees? (I might actually try this.)

A further thought: if one were considering using the pentagon configuration, one might also want to consider the Orchard Visibility Problem.

POSTED BY: Kathryn Cramer
Answer
1 year ago

For the penta-orchard with 25 trees, you might guess that the central 10 trees made a regular decagon. We can check that, and then add a variable if it doesn't work. First, some line intersection code and a pointset.

 lineIntersect[{pointset_, {a_, b_}, {c_, d_}}] := Module[{m, n, kk}, kk = ({m, n} /. Solve[ m pointset[[a]] + (1 - m) pointset[[b]] == n pointset[[c]] + (1 - n) pointset[[d]]][[1]]);kk[[1]] pointset[[a]] + (1 - kk[[1]]) pointset[[b]]] 

 p = RootReduce[Table[{Sin[Pi (2 n + 1)/10], Cos[Pi (2 n + 1)/10]}, {n, 0, 9}]]; 

 Graphics[MapIndexed[Text[#2[[1]], #1] &, p]] 

ten points

Then the four points on the left can be calculated.

left = RootReduce[{lineIntersect[{p, {7, 10}, {1, 4}}],  lineIntersect[{p, {5, 10}, {1, 2}}],  lineIntersect[{p, {1, 10}, {9, 4}}], lineIntersect[{p, {2, 9}, {5, 8}}]}];

Are these all on a line? RootReduce[Det[Append[#, 1] & /@ Take[left, 3]]] gives 0, so yes they are.

 new = Join[p, Flatten[Table[RootReduce[RotationMatrix[2 Pi n/5].# & /@ Take[left, 3]], {n, 0,4}], 1]];  

 Graphics[MapIndexed[Text[#2[[1]], #1] &, new]]

25 points

That points the closest trees 0.618034 apart, so just divide by that and multiply by the required distance between trees.

Attachments:
POSTED BY: Ed Pegg
Answer
1 year ago

Great! Thanks!

POSTED BY: Kathryn Cramer
Answer
1 year ago

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