[math-fun] computing Penrose tilings
I'm having trouble finding explicit recipes on the web for having a computer generate a finite excerpt of a Penrose tiling (in any of its equivalent forms; personally I like the rhombus-tilings). One sort of solution would be a prescription of the form "To generate the set of vertices of the rhombus-tiling that lie within distance d of the origin in C, let zeta = exp(2 Pi i / 5) and take all sums of the form a_1 zeta + a_2 zeta^2 + a_3 zeta^3 + a_4 zeta^4 with a_1, a_2, a_3, a_4 in Z satisfying the following inequalities: ..." Another sort of recipe would be an algebraic version of the inflation rule, where one starts with a finite set of colored points in R^2 and then replaces it by a larger set of colored points, applying some linear maps to get new points from old (where the choice of map depends on the color of the point, and new points get colored according to some rule). Surely something of this sort is on the web, but I haven't been able to find it. For that matter, I recall that there's a nice way to characterize one of the versions of Penrose tilings by means of the algebraic conjugacy of Q(zeta) (with zeta as above) that sends zeta^k to zeta^2k: we take those points in Q(zeta) such that the image of the point under the conjugacy lies in some finite window (a pentagon of some size centered on 0, as I recall). I learned about this from Veit Elser at an Oberwolfach meeting many years ago, and I believe it may be due independently to Veit and Rick Kenyon; was a description of this point of view ever published? Jim Propp
http://merganser.math.gvsu.edu/david/reed05/projects/halbert/discussion.html On Thu, Apr 7, 2011 at 9:54 AM, James Propp <jpropp@cs.uml.edu> wrote:
I'm having trouble finding explicit recipes on the web for having a computer generate a finite excerpt of a Penrose tiling (in any of its equivalent forms; personally I like the rhombus-tilings).
One sort of solution would be a prescription of the form "To generate the set of vertices of the rhombus-tiling that lie within distance d of the origin in C, let zeta = exp(2 Pi i / 5) and take all sums of the form a_1 zeta + a_2 zeta^2 + a_3 zeta^3 + a_4 zeta^4 with a_1, a_2, a_3, a_4 in Z satisfying the following inequalities: ..."
Another sort of recipe would be an algebraic version of the inflation rule, where one starts with a finite set of colored points in R^2 and then replaces it by a larger set of colored points, applying some linear maps to get new points from old (where the choice of map depends on the color of the point, and new points get colored according to some rule).
Surely something of this sort is on the web, but I haven't been able to find it.
For that matter, I recall that there's a nice way to characterize one of the versions of Penrose tilings by means of the algebraic conjugacy of Q(zeta) (with zeta as above) that sends zeta^k to zeta^2k: we take those points in Q(zeta) such that the image of the point under the conjugacy lies in some finite window (a pentagon of some size centered on 0, as I recall). I learned about this from Veit Elser at an Oberwolfach meeting many years ago, and I believe it may be due independently to Veit and Rick Kenyon; was a description of this point of view ever published?
Jim Propp
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Jim, One of the nicest constructions was written up in the AMM by Richard Schwartz (he mentions you in the acknowledgements!): http://www.math.brown.edu/~res/Papers/ll.pdf I did this independently and have a lengthy paper that is cited as ref. 1 in the Schwartz paper. I recently completed the classification of all the crystallographic Riemann surfaces and there are exactly four that have the Penrose symmetry. These are specified by conformal maps of triangles (that map vertices to vertices). Here are the ones that give Penrose-like patterns: {1, 2, 7} -> {3, 6, 1} {1, 3, 6} -> {7, 1, 2} {1, 4, 5} -> {3, 2, 5} {2, 3, 5} -> {4, 1, 5} The numbers are the triangle angles in units of pi/10, and the 1st, 2nd, 3rd angle in the first bracket get mapped to the 1st, 2nd, 3rd angle in the second bracket. Veit On Apr 7, 2011, at 12:54 PM, James Propp wrote:
I'm having trouble finding explicit recipes on the web for having a computer generate a finite excerpt of a Penrose tiling (in any of its equivalent forms; personally I like the rhombus-tilings).
One sort of solution would be a prescription of the form "To generate the set of vertices of the rhombus-tiling that lie within distance d of the origin in C, let zeta = exp(2 Pi i / 5) and take all sums of the form a_1 zeta + a_2 zeta^2 + a_3 zeta^3 + a_4 zeta^4 with a_1, a_2, a_3, a_4 in Z satisfying the following inequalities: ..."
Another sort of recipe would be an algebraic version of the inflation rule, where one starts with a finite set of colored points in R^2 and then replaces it by a larger set of colored points, applying some linear maps to get new points from old (where the choice of map depends on the color of the point, and new points get colored according to some rule).
Surely something of this sort is on the web, but I haven't been able to find it.
For that matter, I recall that there's a nice way to characterize one of the versions of Penrose tilings by means of the algebraic conjugacy of Q(zeta) (with zeta as above) that sends zeta^k to zeta^2k: we take those points in Q(zeta) such that the image of the point under the conjugacy lies in some finite window (a pentagon of some size centered on 0, as I recall). I learned about this from Veit Elser at an Oberwolfach meeting many years ago, and I believe it may be due independently to Veit and Rick Kenyon; was a description of this point of view ever published?
Jim Propp
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I have uploaded the following to my public directory at googledocs https://docs.google.com/leaf?id=0B6QR93hqu1AhZGY1NGYzNTktMzAzOC00ZDRlLWEzMzM... (1) A short description of the thick-section construction of the Penrose tiling [thik-sect.pdf] ; (2) A user-modifiable (if sufficiently brave) Postscript program [dartkite.ps] ; (3) PDF output from (2) showing full-page canonical "cartwheel" kite&dart and sling&arrow (rhombic) variations [dartkite0.pdf]; (4) ditto showing the 7 Conway countries (regions forced by connected sets of tiles) at various stages of inflation [dartkite1.pdf]. Incidentally, a few years later Peter Pleasants informed me that he had a proof that modulo inflation there are a finite number of "empires" (regions forced by disconnected sets of tiles); but since 1988 I have not returned to this problem. Fred Lunnon On 4/7/11, James Propp <jpropp@cs.uml.edu> wrote:
I'm having trouble finding explicit recipes on the web for having a computer generate a finite excerpt of a Penrose tiling (in any of its equivalent forms; personally I like the rhombus-tilings).
One sort of solution would be a prescription of the form "To generate the set of vertices of the rhombus-tiling that lie within distance d of the origin in C, let zeta = exp(2 Pi i / 5) and take all sums of the form a_1 zeta + a_2 zeta^2 + a_3 zeta^3 + a_4 zeta^4 with a_1, a_2, a_3, a_4 in Z satisfying the following inequalities: ..."
Another sort of recipe would be an algebraic version of the inflation rule, where one starts with a finite set of colored points in R^2 and then replaces it by a larger set of colored points, applying some linear maps to get new points from old (where the choice of map depends on the color of the point, and new points get colored according to some rule).
Surely something of this sort is on the web, but I haven't been able to find it.
For that matter, I recall that there's a nice way to characterize one of the versions of Penrose tilings by means of the algebraic conjugacy of Q(zeta) (with zeta as above) that sends zeta^k to zeta^2k: we take those points in Q(zeta) such that the image of the point under the conjugacy lies in some finite window (a pentagon of some size centered on 0, as I recall). I learned about this from Veit Elser at an Oberwolfach meeting many years ago, and I believe it may be due independently to Veit and Rick Kenyon; was a description of this point of view ever published?
Jim Propp
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participants (4)
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Fred lunnon -
James Propp -
Mike Stay -
Veit Elser