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nomadreid
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in Appendix D, pp. 179 ff, of Alain Conne's "Noncommutative Geometry", he outlines a process. He looks at Penrose tilings of the plane which are composed of two types of triangles
L [with 2 longer sides of length [itex]\phi[/itex]n+1, and one shorter side of [itex]\phi[/itex]n) for integers n, whereby [itex]\phi[/itex] is the golden ratio (1+[itex]\sqrt{5}[/itex])/2] and
S [with 2 shorter sides of length [itex]\phi[/itex]m, and one longer side of [itex]\phi[/itex]m+1) for integers m]
put together by certain rules (among which include that a common side to two triangles must have the same endpoints for both sides). [The familiar darts and kites are combinations of these triangles that help in putting the triangles together, but are not important for this question.]
The process [the purpose of which is to code the tiling into a series of 0's and 1's, but that is also not important for this question] is to transform a tiling into another tiling by eliminating each side which fulfills two conditions:
(a) constitutes a border between two different kinds of triangles
(b) is a longer side of L and a shorter side of S.
This works fine, as long as n≠m (specifically, if n+1=m). But he himself gives an example (Figure 3, p. 94) of a tiling where n=m. In this case, I don't see how this process could work. Can anyone please explain this?
In alternative but similar process [which is to achieve the same set of codings] in pp. 412-413 of Appendix F, in the book "In Search of the Riemann Zeros" by Michel L. Lapidus, which uses the same triangles (labeling them L and R instead of L and S, resp.), the author eliminates sides that fulfill (a) and
(c) is the longer side of S (=R) and the shorter side of L (that is, the reverse of (b)).
Again, unless m+1=n, I don't see how this could work either. (Lapidus gives the example of a tiling with m=n in Figure 2 on page 413.)
(Lapidus' method is a bit nicer in that it is more definite: in Conne's method, you have to choose between two sides of each triangle to look at, whereas Lapidus has only one side of each triangle that has to be considered. That is seen in the difference of the indefinite and definite articles in conditions (b) and (c) resp.)
I would be grateful for help on this step.
L [with 2 longer sides of length [itex]\phi[/itex]n+1, and one shorter side of [itex]\phi[/itex]n) for integers n, whereby [itex]\phi[/itex] is the golden ratio (1+[itex]\sqrt{5}[/itex])/2] and
S [with 2 shorter sides of length [itex]\phi[/itex]m, and one longer side of [itex]\phi[/itex]m+1) for integers m]
put together by certain rules (among which include that a common side to two triangles must have the same endpoints for both sides). [The familiar darts and kites are combinations of these triangles that help in putting the triangles together, but are not important for this question.]
The process [the purpose of which is to code the tiling into a series of 0's and 1's, but that is also not important for this question] is to transform a tiling into another tiling by eliminating each side which fulfills two conditions:
(a) constitutes a border between two different kinds of triangles
(b) is a longer side of L and a shorter side of S.
This works fine, as long as n≠m (specifically, if n+1=m). But he himself gives an example (Figure 3, p. 94) of a tiling where n=m. In this case, I don't see how this process could work. Can anyone please explain this?
In alternative but similar process [which is to achieve the same set of codings] in pp. 412-413 of Appendix F, in the book "In Search of the Riemann Zeros" by Michel L. Lapidus, which uses the same triangles (labeling them L and R instead of L and S, resp.), the author eliminates sides that fulfill (a) and
(c) is the longer side of S (=R) and the shorter side of L (that is, the reverse of (b)).
Again, unless m+1=n, I don't see how this could work either. (Lapidus gives the example of a tiling with m=n in Figure 2 on page 413.)
(Lapidus' method is a bit nicer in that it is more definite: in Conne's method, you have to choose between two sides of each triangle to look at, whereas Lapidus has only one side of each triangle that has to be considered. That is seen in the difference of the indefinite and definite articles in conditions (b) and (c) resp.)
I would be grateful for help on this step.
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