If you take a replacement sequence L -> LS, S -> L, then you generate a fibonacci ruler. That is, you get LSL LS LSL LSL LS LSL LS LSL LSL LS LSL ...

If instead of making L = 1.618033&c, you put it as -0.618033... then the ruler is _bounded_. That is, it varies from -0.618033 to +1, and gives an L segment whenever the number is positive, and an S segment when it's negative. This binding restricts what kinds of tiles might fit together in the quasicrystal, makes the quasicrystal as 'iso-bound'.

When you plot the stuff out on a plane of 1, Ø, the effect is that the infinite part goes from edge to edge, but the bound part is restricted by the two parallel lines iv=-0.618 and iv=1.

By moving the origin along the band, one can effectively change the sequence as given above, so instead of starting at 0, one is starting somewhere else.

The entire line then changes according to where you are standing on it, in that it will oscillate between two values separated by Ø, but at different points.

If one were now to consider instead of penrose tiles, a tiling of pentagons, then this would be wound infinitely, but it's much like the reduction of the 1,Ø plane onto a single line. The effect of iso-bound is caused by the fact that molecules can not occupy the same position, and one is left with a zigzag that is roughly twice the line width, caused by binding rules. If you were to stand on a pentagon in a {5,10/3}, you would only see those pentagons, and parts of pentagons, that a straight ray without turning would lead. This means that when you come to a vertex, you don't see the full 10/3, but only 3 1/3 of the pentagons around it.

Moving around the pentagon you are standing in, would mean that different parts of the 10/3 come into view, and different segments of remote pentagons are seen. It is rather similar to drawing a 2-space through a 4d lattice, and mapping only the nearest points.

Because the construction is iso-bounded, one can not have an infinite repeating pattern, but the amount that can repeat depends on how small the iso-image is to the bounding circle. That is, if the isomorph is only 1/10 of the circle, you can not repeat the image more than ten times before a spacer is called for.

The heptagon example you give in your thesis is governed by the same rules in two separate images, and it is sufficient to show that _any_one_ of the images is bounded. This is because an infinite repeating pattern is infinite in _every_ image. So the heptagonal system with L, M, S, gives rise to two images one in L', M', S', and the second in L", M", S". Since the first is necessarily infinite, then either the second or third might be finite.

The same arguments are used when one divvies up the hyperbolic tilings to see what goes into what. For example, {5,3,4} is hyperbolic, but {5/2,3,4} is discrete (real, infinitely dense). On the other hand, {5,3,6} and {5/2,3,6} are both hyperbolic. In the first pair, {5,3,4} and {5/2,3,4} are the projections of a thin section of six-space, rather as the pentagonal zigzag thrice over. So if we find on isomorphism, any other real:hyperbolic pairs, there is a chance of joining.

With the heptagon series, the cycle is 7, 7/2, 7/3 and 14, 14/5, 14/3. So we use the usual curvature rule to show that {7,3}, and {7,4} can have the same symmetry, but {7,6} can't. This is because the first pair give H,S,S as their coordinate, but {7,6} goes H,H,S (only 7/3,6 can be started in spherical space). If one gets, eg S,H,S, it is fair to rotate these so that the first isomorph becomes prime. Note that the 7's and 14's must both be rotated in step. So {7,14} gives isomorphs {7/2, 14,5} and {7/3, 14/3}, which is the same HSS as the {7,3} set, and this can exist as a subgroup.