Comments on: Laying train tracks

By: bancuri de groaza

bancuri de groaza — Fri, 05 Apr 2013 09:37:11 +0000

I do not know if it’s just me or if perhaps everybody else experiencing issues with your blog. It seems like some of the written text within your content are running off the screen. Can someone else please comment and let me know if this is happening to them as well? This may be a problem with my internet browser because I’ve had this happen previously.
Thank you

By: wjtgpf

wjtgpf — Tue, 25 Dec 2012 20:45:50 +0000

Here’s a post from another blog that might be a good starting point for a track layout script.

http://blog.jgc.org/2010/01/more-fun-with-toys-ikea-lillabo-train.html

By: Danny Calegari

Danny Calegari — Sat, 11 Aug 2012 15:19:36 +0000

Hi Jeff – I think it would be very interesting to try to enumerate (eg by computer) all possible *embedded* (i.e. without self-intersections) tracks of a given length. Even with only two pieces as above this is already very complicated I think. One way to adjust the parameters of the problem is to change the “angle” through which a piece of track turns from 2\pi/8 to 2\pi/n for some other integer n. Please let me know if you decide to write your program, and how it turns out.

By: Jeff

Jeff — Mon, 04 Jun 2012 15:33:55 +0000

Interesting post even for a layman like myself. I was thinking about this problem, to write a program to generate all track layouts for a given set of pieces.

How does having more track piece types change things? It appears to not fundamentally change anything other than make the problem more messy.

By: Group Theoretic Origin of the Domino Height Functions « monsieurcactus

Group Theoretic Origin of the Domino Height Functions « monsieurcactus — Thu, 12 Apr 2012 15:32:48 +0000

[...] Laying Train Tracks [...]

By: Danny Calegari

Danny Calegari — Mon, 13 Feb 2012 08:12:37 +0000

I’m not on twitter, so @lamington is definitely not me.

best,

Danny

By: human mathematics

human mathematics — Mon, 13 Feb 2012 06:20:35 +0000

Are you on twitter? The name @lamington doesn’t look like you … if you could tweet @isomorphisms that would make it easy for me to find you.

By: Sheldon

Sheldon — Wed, 28 Dec 2011 05:19:27 +0000

Proof that geometry is applicable in all arenas of life!

By: Swiat Gal

Swiat Gal — Wed, 07 Dec 2011 22:06:10 +0000

Oops! I forgot to copy one more relation LRRL=RLLR.

By: Swiat Gal

Swiat Gal — Wed, 07 Dec 2011 22:01:06 +0000

Just for fun I wrote down presentation for Γ_0: R^8, (R^5L)^4, (R^7L^3)^4. Of course the third relator is obtained from the second by substituting R^3 and L^3 for R and S respectively (3.5≡7(8)). So,
the two track you have mention at the beggining are almosl all you would like to know about that group.

BTW. you should have cusps (switches?) in your tracks. R^2L^{-2}R^2L^{-2} is a nice closed track!

By: Swiat Gal

Swiat Gal — Wed, 07 Dec 2011 16:46:51 +0000

Actually, I have chosen a dumb reference point At the beginning of the track. If I take it to be the centre of the R-circle the group Γ_0 is generated by the rotation R and the translation RL. And the relations for R and RL are the same as for O and T.

By: Swiat Gal

Swiat Gal — Wed, 07 Dec 2011 15:12:41 +0000

-1. Would you mind posting the postscript code you are so proud of? What tools do you use?

0. I think that declaring groups by generators and relations is some misidea. I think that we should investigate more groups defined by action (of generators). Alas, using presentations as tools.

1. The group generated by L and S is virtually abelian, so the word problem is easy to solve. I did not derive the representattion of it, but it is a finite index subgroup of the group generated by a rotation O by π/4 and a translation T. Then R=OT, L=TO^{-1}. (T is the vector between the ends of the track rotated by π/8.)

The relations in the group generated by T and O are: O^8, (O^4T)^2, [O^iTO^{-1},T]. Then if you want to know if a LR-word closes you write it in terms of O and T, you group conjugates of T, commute (is it the right verb for moving commuting elements?) them
to cancel (or not). There are only few left if and only if you could close the track if few steps.

2. What is the index of Γ_0:= in Γ:=? What is the presentation of Γ_0? I know tat I need to take representatives of cosets of Γ_0 together with L and R and play with Tietze moves…

By: Danny Calegari

Danny Calegari — Sun, 04 Dec 2011 12:14:54 +0000

This is quite a difficult question in general, but I can try to give a few vague ideas. Part of the problem is that it is possible to have an embedded train track which can’t be extended to a closed up embedded train track. This is actually quite easy to arrange: if the embedded train track has a long narrow cul-de-sac, then a path that enters won’t have enough room to turn around without intersecting itself or the walls of the dead end. So imagine someone laying out track more or less randomly, but trying to look ahead a bit to avoid cul-de-sacs. Conformally speaking, the size of a cul-de-sac depends (roughly) only on the narrowest bottleneck, not on how big it is past that bottleneck. So by avoiding bottlenecks and self-intersections and otherwise laying track “randomly”, the track that’s laid out will, on the large scale, resemble some conformally invariant random embedded path, which is to say SLE at some parameter. A segment of path of length n will tend to have diameter about sqrt(n) (it can’t be smaller than this, since it has area n and is embedded). Moreover, the distance between the endpoints measured in the translation group A (which is abstractly isomorphic to Z^4) will be usually of order sqrt(n). Now, providing neither end is trapped in a cul-de-sac, the path can be closed up to an embedded track by adding an additional O(sqrt(n)) pieces. First, give yourself a bit of room by extending the two ends outwards in “spikes” from what’s laid out so far. Then the trick is to add a big embedded “loop” of the form (LR)^{n_1}R(LR)^{n_2}R . . . (LR)^{n_8}R. If the n_i are all reasonably large (say, of order twice the diameter) and are chosen so that this new loop represents the inverse in A of what is laid out so far, then it will close up the track in an embedded way.

By: Anonymous

Anonymous — Sun, 04 Dec 2011 08:36:16 +0000

This is really nice! Given an embedded train track, is there a nice way to tell whether there is a way to close the loop so that the closed loop is also embedded?

By: ACME Science » The Internet Maths Aperiodical –

ACME Science » The Internet Maths Aperiodical – — Sat, 03 Dec 2011 13:05:41 +0000

[...] last bit of real maths is something I found via Jeff Erickson. It’s an essay on the topic of wooden train-track sets, answering the question of which layouts of left- and right-turning pieces construct a closed [...]