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Unit Distance Heawood Graph

Posted 10 years ago

At MathWorld, there's the Heawood graph. Noted there and over at Wikipedia there are some unit distance embeddings, where all edges have length 1. But none has symmetry. Let's fix that.

pts=N[{ {-1/2,-(1/2) (-1+Sqrt[3-4 x-4 x^2])}, {x,1/2}, {x,-1/2}, {-1/2,1/2 (-1+Sqrt[3-4 x-4 x^2])}, {1/2,1/2 (-1+Sqrt[3-4 x-4 x^2])}, {-x,-1/2}, {-x,1/2}, {1/2,-(1/2) (-1+Sqrt[3-4 x-4 x^2])}, {1/2 (-1+Sqrt[3-4 x (1+x)]) Sqrt[(7+8 Sqrt[3-4 x (1+x)]-8 x (-1+x (1+2 x (2+x))))/(13+8 x (1+x) (-3+2 x (1+x)))],-(1/2) Sqrt[(7+8 Sqrt[3-4 x (1+x)]-8 x (-1+x (1+2 x (2+x))))/(13+8 x (1+x) (-3+2 x (1+x)))]},{Sqrt[3-4 x^2]/(2 Sqrt[1+4 x^2]),-((x Sqrt[3-4 x^2])/Sqrt[1+4 x^2])}, {-Sqrt[3-4 x^2]/(2 Sqrt[1+4 x^2]),-((x Sqrt[3-4 x^2])/Sqrt[1+4 x^2])}, {-(1/2) (-1+Sqrt[3-4 x (1+x)]) Sqrt[(7+8 Sqrt[3-4 x (1+x)]-8 x (-1+x (1+2 x (2+x))))/(13+8 x (1+x) (-3+2 x (1+x)))],-(1/2) Sqrt[(7+8 Sqrt[3-4 x (1+x)]-8 x (-1+x (1+2 x (2+x))))/(13+8 x (1+x) (-3+2 x (1+x)))]} , {1/2,-((x Sqrt[3-4 x^2])/Sqrt[1+4 x^2])-Sqrt[(8 x^2+Sqrt[3+8 x^2-16 x^4])/(2+8 x^2)]}, {-1/2,-((x Sqrt[3-4 x^2])/Sqrt[1+4 x^2])-Sqrt[(8 x^2+Sqrt[3+8 x^2-16 x^4])/(2+8 x^2)]} }/.x->Root[-153027+1353114 #1-1986024 #1^2+2763312 #1^3+10497808 #1^4+31063072 #1^5-8457728 #1^6-40590336 #1^7+44468736 #1^8+164975616 #1^9+119705600 #1^10-494067712 #1^11-763912192 #1^12+414924800 #1^13+1154613248 #1^14+48496640 #1^15-864485376 #1^16-376307712 #1^17+253231104 #1^18+271581184 #1^19+89128960 #1^20+10485760 #1^21&,1],300];

points = pts[[#]] & /@ {13, 12, 4, 5, 9, 14, 11, 3, 2, 1, 8, 7, 6, 10};

edges = {{1, 2}, {1, 6}, {1, 14}, {2, 3}, {2, 11}, {3, 4}, {3, 8}, {4, 5}, {4,13}, {5, 6}, {5, 10}, {6, 7}, {7, 8}, {7, 12}, {8, 9}, {9, 10}, {9,14}, {10, 11}, {11, 12}, {12, 13}, {13, 14}}

These are the same edges as GraphData["HeawoodGraph", "EdgeIndices"]

Graphics[{Red, Disk[#, .02] & /@ points, Black, Line[{points[[#[[1]]]], points[[#[[2]]]]}] & /@ edges}]

Unit Heawood

With this code, we can verify that all the edges are 1.

EuclideanDistance[points[[#[[1]]]], points[[#[[2]]]]] & /@ edges

So, how did I find this? I originally had the 8 points at the top in a regular octagon, but that made the bottom edge longer than 1. I needed an 8 cycle where all the edges had length 1.

MyEuc[{a, b}, {c, d}] := Sqrt[(a - c)^2 + (b - d)^2]; Solve[MyEuc[{x, 1/2}, {-1/2, -y}] == 1]

That gives the first eight pts. Next we need pts 9 at a distance of 1 from pts 1 and 5.

{a, b} /. FullSimplify[Solve[{MyEuc[pts[[1]], {a, b}] == 1, MyEuc[pts[[5]], {a, b}] == 1}]]

Turns out we want the second of those. With symmetry, we have pts 1-12. For the last 2,

b /. FullSimplify[Solve[{MyEuc[{Sqrt[3 - 4 x^2]/(2 Sqrt[1 + 4 x^2]), -((x Sqrt[3 - 4 x^2])/Sqrt[1 + 4 x^2])}, {1/2, b}] == 1}]]

From that, we get the last two points.

{1/2, -((x Sqrt[3 - 4 x^2])/Sqrt[1 + 4 x^2]) - Sqrt[(8 x^2 + Sqrt[3 + 8 x^2 - 16 x^4])/(2 + 8 x^2)]}

To solve for x, we need a solution for a nasty equation. Solve[] wasn't working, so I tried a trick.

NMinimize[{Abs[ Sqrt[(((x Sqrt[3 - 4 x^2])/Sqrt[1 + 4 x^2] + Sqrt[(8 x^2 + Sqrt[3 + 8 x^2 - 16 x^4])/(2 + 8 x^2)] - 1/2 Sqrt[(7 + 8 Sqrt[3 - 4 x (1 + x)] - 8 x (-1 + x (1 + 2 x (2 + x))))/(13 + 8 x (1 + x) (-3 + 2 x (1 + x)))])^2 + (-(1/2) - 1/2 (-1 + Sqrt[3 - 4 x (1 + x)]) Sqrt[(7 + 8 Sqrt[3 - 4 x (1 + x)] - 8 x (-1 + x (1 + 2 x (2 + x))))/(13 + 8 x (1 + x) (-3 + 2 x (1 + x)))])^2)] - 1], .1 < x < .3}, x, AccuracyGoal -> 2000, WorkingPrecision -> 4000]

Then I took that and put it into RootApproximant[] -- and got the solution.

Root[-153027 + 1353114 #1 - 1986024 #1^2 + 2763312 #1^3 + 10497808 #1^4 + 31063072 #1^5 - 8457728 #1^6 - 40590336 #1^7 + 44468736 #1^8 + 164975616 #1^9 + 119705600 #1^10 - 494067712 #1^11 - 763912192 #1^12 + 414924800 #1^13 + 1154613248 #1^14 + 48496640 #1^15 - 864485376 #1^16 - 376307712 #1^17 + 253231104 #1^18 + 271581184 #1^19 + 89128960 #1^20 + 10485760 #1^21 &, 1]

And so, we get an exact unit-distance embedding with the Heawood graph, with symmetry.

Attachments:
POSTED BY: Ed Pegg
4 Replies
Posted 10 years ago

...

POSTED BY: Ion Dodon

enter image description here

POSTED BY: Simon Cadrin
Posted 10 years ago

NMinimize[{Abs[ Sqrt[(((x Sqrt[3 - 4 x^2])/Sqrt[1 + 4 x^2] + Sqrt[(8 x^2 + Sqrt[3 + 8 x^2 - 16 x^4])/(2 + 8 x^2)] - 1/2 Sqrt[(7 + 8 Sqrt[3 - 4 x (1 + x)] - 8 x (-1 + x (1 + 2 x (2 + x))))/(13 + 8 x (1 + x) (-3 + 2 x (1 + x)))])^2 + (-(1/2) - 1/2 (-1 + Sqrt[3 - 4 x (1 + x)]) Sqrt[(7 + 8 Sqrt[3 - 4 x (1 + x)] - 8 x (-1 + x (1 + 2 x (2 + x))))/(13 + 8 x (1 + x) (-3 + 2 x (1 + x)))])^2)] - 1], .1 < x < .3}, x, AccuracyGoal -> 2000, WorkingPrecision -> 4000]

It was right in the notebook. The big Sqrt[] doesn't copy over correct as input text.

POSTED BY: Ed Pegg

enter image description here

POSTED BY: Simon Cadrin
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