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A farmer has a 10m x 10m field that has fences around the perimeter. What is the least number of 1m fences he needs to add to divide the field into 5 regions of equal area?

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    $\begingroup$ Can we use fractions of fence? Or must every section have integer length? $\endgroup$
    – fljx
    Commented Dec 12, 2022 at 8:53
  • $\begingroup$ @fljx I am going to say yes as it opens up extra solutions that I didn't consider. $\endgroup$ Commented Dec 12, 2022 at 10:33
  • $\begingroup$ It doesn't make a better sense to keep both the original problem and the one that replaced it visible, does it? Or maybe post a separate one. A generalization of the original one to arbitrary curved fences is very interesting. Both are indeed very interesting, just very different. $\endgroup$ Commented Dec 12, 2022 at 23:49
  • $\begingroup$ feel free to post a new question where curved fences are allowed $\endgroup$ Commented Dec 13, 2022 at 14:15
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    $\begingroup$ “feel free to post”—Thanks for the permission. The thing is, it is the exact original formulation of your puzzle (it didn't explicitly mention straight or curved fences). I cannot promise to post it, tho, too busy. If I do, you'll get a proper attribution, of course. $\endgroup$ Commented Dec 15, 2022 at 6:33

5 Answers 5

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Edit/Note: I gave this answer assuming fractional fences where not allowed.
As such I did not carefully think about the second picture.

Obviously that answer is not optimal if fractional fence solutions are allowed! Considering the talk if that should be another question: I won't give my better solution here. (evil grin).

Since farmers tend not to care about unit squares, 26 is doable.
If the farmer does not mind setting the saw into his fences, the optimal solution is around 25 1/4 fence (probably not worth it for the farmer, and probably not the intent of this question). But this suggests 25 is not possible. enter image description here

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  • $\begingroup$ Cool second solution! I didn't think of that. $\endgroup$ Commented Dec 12, 2022 at 10:34
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    $\begingroup$ Can you think of a proof that the solution 2 is optimal (a) considering only straight line fences, and (b) considering any non-intersecting, possibly curved lines? My head explodes... I'st one of the simplest and at the same time one of the most disccombobulating tiling problems I ever seen. Delightful! $\endgroup$ Commented Dec 12, 2022 at 23:46
  • $\begingroup$ @kkm: see edit. PS: only straight lines does not mean anything if fences of any length are allowed. $\endgroup$
    – Retudin
    Commented Dec 13, 2022 at 7:53
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    $\begingroup$ Re: 2nd solution... I saw this in the 'Hot Network Questions' next to the 'Puzzling' Icon and exclaimed 'The solution is right there!' XD $\endgroup$
    – TesseractE
    Commented Dec 13, 2022 at 17:36
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This is an attempt to locally optimize the solution by Retudin.

Some theory suggests that when you have segments meeting in a problem like this one, they should meet at 120 degree angles. And then to compensate for the fact that it's not possible to use straight line segments anymore, we use circular arcs instead. If my calculations are correct, then this achieves 25.0211 units of fence.

circular arcs and line segments meeting at 120 degrees

Some detail if someone wants to check my calculations: if the square has vertices at coordinates (0,0), (1,0), (1,1), and (0,1); then one of the points where three segments meet is (0.5, 0.791588112238442); the center of one of the arcs is (0.10168323124074008, 0.10168323124074008) and the radius of the circular arc is 0.7966335375185202. The radius squared is exactly one-fifth the reciprocal of 1-sqrt(3)+pi/3. The circular arcs are each one-twelfth of a full circle (that is, 30 degrees).

But perhaps the optimal solution has a different global structure?

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    $\begingroup$ With my own calculations I also get 25.0211293 for the fence length. Looks good. And according to this blog, it is the optimal structure. lama.univ-savoie.fr/pagesmembres/bogosel/anisotropic.html $\endgroup$
    – Florian F
    Commented Dec 12, 2022 at 21:48
  • $\begingroup$ Since the arcs are 1/12th of a circle, would 3 of them would fit in this square as a quarter of the circle and intersect at the vertical & horizontal lines? Or is the origin of this larger circle outside this square? $\endgroup$
    – Jim
    Commented Dec 12, 2022 at 21:54
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    $\begingroup$ By the way, this answer shows that the 26-unit solution must be optimal with unit fences, regardless how you arrange the fences. $\endgroup$
    – Florian F
    Commented Dec 12, 2022 at 22:09
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    $\begingroup$ Nice. It's basically a special case of a soap-bubble optimisation/simulation. I happen to write one last summer (screenshot: dropbox.com/s/6rpnh5xtm9vhily/…) $\endgroup$ Commented Dec 13, 2022 at 14:04
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    $\begingroup$ I find that the 'divide a square in 4 equal parts' solution also uses 120degree angles and (thus) is not 4way symmetrical even more counterintuitive at first sight. $\endgroup$
    – Retudin
    Commented Dec 13, 2022 at 14:50
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The best I can achieve with orthogonal fences is:

26

enter image description here

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This answer is good enough if the total length of the new fences is an integer (26): enter image description here

Why a better integer answer is not possible:

If it can be 25 or lower, the total fence length is $25*2+40 = 90$ with the average perimeter of a region being $18$. If so, the region's shorter dimension will be $4$ (with the shape being a $4$x$5$ rectangle), but we need different perimeters than $18$ too. Even the shortest distances through the shapes decrease less than they increase to keep the same area, so this is impossible.

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I think a lateral answer can be given if farmer thinks out of the box

or better, use the box.

So the farmer can achieve

zero meters of additional fences. Farmer can reposition the fences that are on the border of the field "box"

case 1) Fractional fences are allowed

Given that there would be no lower limit on the the fence size, we can have infinitesimal small, shapes. I do not think that this is a very interesting case.

case 2) Fractional fences are not allowed

The farmer would use at least 1m of fence each time. So the farmer can build 5 squares that shares one or more side each. A good configuration using squares is good 5 square configuration

Not sure that is the best general configuration, but the additional fences is still 0, as it requires 15 meters of fences and it can take them from the 40m of the border.

A better configuration could be achieved using equilateral triangles of 1m side equilateral so you can use only 11m of fences. Still, I am not sure that is the best configuration.

Basically this is a try to answer to the question "Minimize the non-fractional 1m fences that the farmer need to create 5 equal area regions". I think other answers that are not exploiting the border fences could be re-adapted to answer the question and achieve even a better minimization!

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