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########OLDThe 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?