Talk:Biggest little polygon

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it takes the form of a regular pentagon with an obtuse isosceles triangle attached to one of its sides, with the apex of the triangle at unit distance from the opposite pentagon vertex.[4]

I think this description is not correct. Where can it bou found in the given source? Mo271 (talk) 20:05, 2 August 2014 (UTC)

We had a discussion by email and I agree that this is incorrect. The pentagon has equal diagonals but is not regular. —David Eppstein (talk) 03:17, 3 August 2014 (UTC)

Bad graph

On Aug. 3 the old biggest little hexagon graph was deleted on the grounds of its not being accurate. Unfortunately it was replaced by something, also allegedly the biggest little hexagon, which is not even a hexagon and indeed not even a polygon, since it is not a closed loop. And even if the edges were redrawn to make it a hexagon, it would still not be right as the pentagon that it is based on is supposed to be symmetric. See Graham's Figure 5a for the correct image. Can someone with graphics skills put that in here? In the meantime I'm reverting the non-hexagon. 208.50.124.65 (talk) 17:26, 29 August 2014 (UTC)

Incorrect edits

Moreover, I think the article was entirely correct before the edits of August 3. It said:

it takes the form of a regular pentagon with an obtuse isosceles triangle attached to one of its sides, with the apex of the triangle at unit distance from the opposite pentagon vertex.

As far as I can see this is an accurate description of Graham's Figure 5a, while the replacement text is not. And the original graph here was, as far as I can see, an exact replica of Graham's Figure 5a (rotated 180°). Thus our original wording above referring to the "apex of the triangle" refers to the bottom point in Graham's Figure 5a and equivalently the top point in our original graph.

So I'm reverting to the Aug.3 status quo ante. 208.50.124.65 (talk) 20:26, 29 August 2014 (UTC)

No, the pentagon in the solution is not regular, as we have established here already. It has equal diagonals and is bilaterally symmetric but does not have equal sides or angles and does not have 5-fold rotational symmetry. —David Eppstein (talk) 20:59, 29 August 2014 (UTC)

Okay, I recommend that instead of calling the pentagon simply "irregular equidiagonal" we call it "irregular but equidiagonal and with bilateral symmetry." Also, why does the current version say this?:

Graham conjectured that the optimal solution for the general case of even values of n consists in the same way of a regular (n − 1)-gon with an isosceles triangle attached

All I can find in Graham is

It is conjectured that the optimal solution has the diameter graph D(X_2m) which consists of a circuit of length 2m-1 together with a single edge from one vertex....

208.50.124.65 (talk) 23:32, 29 August 2014 (UTC)

Presumably it's the same mistake as in the hexagon case, not yet corrected. —David Eppstein (talk) 23:40, 29 August 2014 (UTC)

Even numbers of sides

Comment by Charles Audet: The current version of the page gives the exact solution to the hexagon, mentions a numerical approximation for the octagon, and states that Graham's conjecture on the optimal polygon is true. The proposed modification is to restructure the presentation, and include recent development for the optimal axially symmetric octagon. This is a proposition to revise the paragraph. Also, here is a link to a picture showing the optimal polygons: https://www.dropbox.com/s/kberl477yth2x5r/OptimalPolygons.pdf?dl=0 (I was not able to include the picture in the Wikipedia entry).

PS: This is my first Wikipedia edition. Please inform me constructively if this is not the proper way to propose modifications to a Wikipedia entry.

-- Begin Modifications --

In the case n = 4, the optimal quadrilateral is not unique. With an area of 1/2, the square with unit length diagonals is optimal. More generally, any quadrilateral with perpendicular unit-length diagonals and with side-lengths not exceeding one is also optimal.

In the case n = 6, the unique optimal hexagon is not regular. The solution to this case was published in 1975 by Ronald Graham, answering a question posed in 1956 by Hanfried Lenz; it takes the form of an irregular equidiagonal pentagon with an obtuse isosceles triangle attached to one of its sides, with the distance from the apex of the triangle to the opposite pentagon vertex equal to the diagonals of the pentagon. Its area is 0.674981.... (sequence A111969 in the OEIS), a number that satisfies the equation

4096 x¹⁰ +8192x⁹ − 3008x⁸ − 30848x⁷ + 21056x⁶ + 146496x⁵ − 221360x⁴ + 1232x³ + 144464x² − 78488x + 11993 = 0.

Graham conjectured that the optimal solution for the general case of even values of n consists in the same way of an equidiagonal (n − 1)-gon with an isosceles triangle attached to one of its sides, its apex at unit distance from the opposite (n − 1)-gon vertex. This was numerically verified by a computer calculation in the case n = 8 by Audet et al. and in the case n ∈ {10,12} by Henrion and Messine. Graham's proof that his hexagon is optimal, and the computer proof of the n = 8 case, both involved a case analysis of all possible n-vertex thrackles with straight edges. The full conjecture of Graham, characterizing the solution to the biggest little polygon problem for all even values of n, was proven in 2007 by Foster and Szabo. However, that conjecture does not provide the exact optimal solution and does not even guarantee that the optimal polygons are axially symmetrical.

Assuming axial symmetry in the case n = 8, Audet, Hansen and Svrtan show that the area of the optimal octagon is the unique root (between √2/2 and π/4) of the degree 42 polynomial

147573952589676412928 x⁴² − 442721857769029238784 x⁴¹ + 2605602600411474165760 x⁴⁰

 + 7670386770149352931328 x³⁹ − 19803120195082488119296 x³⁸ − 90234644551552032833536 x³⁷

 − 5317091837915248694657024 x³⁶ − 17594041430635084655886336 x³⁵ + 29758395462703081578299392 x³⁴

 + 282207246119748476170403840 x³³ + 335103297887714904283021312 x³² − 1917928307706587784371240960 x³¹

 − 5240302758882335722850746368 x³⁰ + 4631615507099121446555746304 x²⁹ + 30114159874526648530622218240 x²⁸

 − 7175008161182179668028030976 x²⁷ − 148064818635686576530703515648 x²⁶ − 42551878829792132053254275072 x²⁵

 + 601318123428810231261639475200 x²⁴ + 332708870397989105275274002432 x²³ − 2358897389358876839124819509248 x²²

 − 680235061366055307103034146816 x²¹ + 7452392569346922858753860567040 x²⁰ − 1491865144134539091913264332800 x¹⁹

 − 15455347946546823025854527832064 x¹⁸ + 9574865040443004381891485761536 x¹⁷ + 20104198057699941048810876698624 x¹⁶

 − 20027080947914571766986403610624 x¹⁵ − 16192270866005062836001824866304 x¹⁴ + 23588130061203336356460301369344 x¹³

 + 8009206689639186621822611818496 x¹² − 17935820857956814364517526943744 x¹¹ − 2370238736752843325635609948160 x¹⁰

 + 9147034213711759916391887323136 x⁹ + 367361764236902187872898865664 x⁸ − 3078428637636379850280988117504 x⁷

 + 10555168880874361068013425792 x⁶ + 647330513128418259524157203072 x⁵ − 23523528029439955698746202488 x⁴

 − 76143004877906320975709476552 x³ + 5833707081723328603647313856 x² + 3773041038347596515021000956 x − 478425365462547737405343343 = 0.

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References

I think overwhelming the article with a paragraph-length polynomial is unhelpful and a bad idea, especially when it appears to need additional unproven assumptions for its correctness as the solution to this problem. —David Eppstein (talk) 20:45, 14 March 2022 (UTC)

I do not share your opinion. I find it very interesting to see that the optimal quadrilateral is trivial, that the optimal hexagon requires a degree 10 polynomial, and that the optimal octagon requires a degree 42 polynomial with large integer coefficients, even when assuming symmetry. This illustrates the extremely rapid growth in complexity of this class of problem, and it contrasts with the cassées with an even number of sides. Audet Charles (talk) 15:00, 15 March 2022 (UTC)