# Why does nature prefer hexagons?

The best ratio of surface to volume in three dimensional space is the ball.

This can be easily observed with soap-bubbles, rain-drops and so on. They "choose" this shape naturally.

Given restricted space, soap-bubbles "choose" the hexagon shape. The hexagon can be observed on other examples, too, like honeycombs.

Is there a mathematical reason for that? Why are there not seven, eight, nine edges? (anything that comes closer to a circle)

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Maybe related: math.stackexchange.com/questions/10648/… –  Jonas Meyer Sep 29 '13 at 18:21
This might be true for honeycombs, but not for soap-bubbles. There seems to be an intrinsic reason for the hexagon shape. –  Mare Infinitus Sep 29 '13 at 18:22
Can't comment on the "intelligence" of things that choose hexagon but Hexagonal Close Packing has the maximum Packing fraction in a lattice packing scenario. –  Sudarsan Sep 29 '13 at 18:22
I never wanted to ascribe intelligence to things. It is just because I have no better description on how those things come to the shape. –  Mare Infinitus Sep 29 '13 at 18:25
The basalt columns at the Giants' Causeway are mostly hexagonal also, probably for the same reason: the columns start out circular, and as they grow they tend to pack into a hexagonal lattice because that's the lowest-energy packing. Then they continue to grow into the spaces left over between the circles, thus becoming hexagons. –  MJD Sep 29 '13 at 18:51

We can ignore vertices of degree 1 (dead ends) and of degree 2 (not distinguished from a point of an edge). We can also ignore the case of degree $\ge 4$ as so many edges incident with one vertex would be highly coincidental. Thus all vertices have degree $3$. Now if we cut out some large but finite portion of this infinite graoh with $v$ vertices, $e$ edges and $f$ faces, then Euler says that $v+f=e+2$. The cutting will turn about $\sqrt v$ vertices (say $c\sqrt v$ for some small $c$) into degree $2$ vertices. By counting edge-vertex incidences, we find $3v-c\sqrt v=2e$. The cutting produced one outer face that is a $c'\sqrt v$-gon for some small $c'\ge c$. For $\nu=3,4,\ldots$, let $f_\nu$ be the number of $\nu$-gonal faces apart from that outer face. Then $1+\sum f_\nu=f$ and $c'\sqrt v+\sum\nu f_\nu=2e$. Plug this into Euler to obtain $$12 = 6f+6v-6e=\sum(6-\nu)f_\nu+6+(2c-c')\sqrt v.$$ Especially, $f\approx \frac12v$ as $v$ gets large and each $\nu$-gon with $\nu>6$ must be "cancelled" by a $5$-gon or lower. In fact, any $\nu>10$ requires at least two small-gons and thus should be somewhat unusual.