How many lanes has a freeway? (Crossing free Kuperberg G2, that is.) EDITED For referencing, I keep the original title and post and ask
only about the simplest case (and forget the freeway with crossings
for now).
Consider a trivalent graph, e.g. the dodecahedron or cube net. I like to
draw it in braid fashion, i.e. (see pic below) with a closure (black),
lanes (red) and some elements (U yellow, H blue, T green - yes, the naming
is silly but I named H "H" long befor I "invented" T and U). The graph
below would correspond to the word U1H3T2.

Of course the very simple graph of the pic can be drawn with less than
4 lanes. I want to know if there is a number n such that any trivalent
graph can be drawn with maximally n lanes. E.g. surely n>2 because on
2 lanes you can't do pentagons with U,T,H.
If yes, are one or even two elements of the set {U,T,H} 
actually superfluous? 
Of course, my ultimative goal is to define an algebra in U,T,H 
(see old pic below) but these equations are NOT to be used for the
question above! (Still, it would be fortituous if the answer is n=3 
as all "natural" equations use maximally 3 lanes. Somehow, it must
relate to Hamilton circuits.)

#OLD#######OLD########OLD

The question is analogous to that on the braid index of a knot.
Here are again the allowed building blocks:  
 
For example, the net of a dodecahedron can be drawn as a closure
of a three lane freeway, just using "U" (and I, of course) pieces. 
(Using U+T+H probably can reduce the lane number to 2.)    
Actually, these are several questions rolled into one:
AFAIK, the braid index problem is still unsolved. Does this change
if (standard braid uses S+R) a) one allows H, b) one allows H but
forbids R, c) one forbids H and R (surely some links then can't 
be drawn at all)?
And before the freeway is tackled, better start with just the
trivalent graph problem: Can all trivalent graphs be drawn with
a finite lane number (or drawn at all), using only
a) U,T,H b) U,T c) U,H d) U?
 A: I think that the question is asking whether planar trivalent graphs have bounded width when they are put in thin position.  (This is "thin position" in the same sense as in knot theory, except in this case in 2-dimensional topology rather than 3-dimensional topology.)  As you might expect, the answer is no.  If the graph is a large mat made of hexagons, then you can show using the isoperimetric inequality that the width is at least $\Omega(\sqrt{v})$.  On the other hand, there is a classic theorem of Lipton and Tarjan that every planar graph has width $O(\sqrt{v})$.  So the correct answer is $\Theta(\sqrt{v})$ rather than $O(1)$.
A: If I understand the question the answer depends on whether or not you include the relations. The question has G2 in the title and you have simplified the words of length two which implicitly suggests you are using relations. On the other hand there is no explicit mention of relations in the question.
If you do not have relations the answer is No. Counterexamples are given by taking a word that begins and ends with (your) H_1. Then remove the top half of H from the top left corner and the bottom half of H from the bottom right corner.
If you do have relations, or rather rewrite rules, and you only consider irreducible diagrams with no crossings then the answer is Yes, with U, T, H.
Why do you introduce R and S?
