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Penrose Tiling.Pdf Andrew Glassner’s Notebook http://www.research.microsoft.com/research/graphics/glassner Penrose Tiling ______________________________________ Andrew heme and variation are part of nature. Birdwatch- translated copies of a single, larger piece. If you can do Glassner Ters identify the species of a bird by its distinctive this, then the overall pattern—and the set of tiles used markings, even though the specific colors and shapes to make it— are periodic. Microsoft vary from one bird to the next. Every thunderclap If you can’t find such a single, big piece that replicates Research sounds roughly like thunder, but each one is different. the pattern by translation, the pattern is called nonpe- And of course snowflakes are beloved for the hexagonal riodic. Many tiles can create both periodic and nonperi- symmetry they all share, as well as the delicate patterns odic patterns, depending on how they’re laid down. unique to each one. If you happen to have a set of tiles that can only make In the May/June 1998 issue of IEEE CG&A I discussed nonperiodic patterns, and you can prove that’s the case, the topic of aperiodic tiling for the plane. This technique then you have an aperiodic set of tiles. helps us create patterns with lots of theme and varia- The quest for aperiodic tiles has gone on for several tion, like the leaves on a tree. I’ll continue discussing the decades. The first set discovered involved 26,000 tiles. subject, so let’s first briefly summarize the main ideas In the last issue I discussed two far smaller sets of ape- from last time. riodic tiles, named after their creators. These Robinson The basic approach is to take a bunch of 2D shapes and Ammann tiles are based on squares, which make and impose rules on how they can connect, like the them very convenient for computer graphics applica- pieces of a jigsaw puzzle. Suppose that you have an infi- tions like texture mapping and sampling. Here I’ll look nite supply of these shapes, or tiles, and you cover the at perhaps the most famous set of aperiodic tiles, dis- plane with them, out to infinity in all directions. You covered by Roger Penrose. Actually, Penrose found two might be able to find a region of the pattern that you distinct sets of aperiodic tiles, one called “kites and could pick up and use as a rubber stamp, and by stamp- darts” and the other called “rhombs.” Each set is made ing it out an infinite number of times (without rotating of only two tiles. Let’s look at the kites and darts first. or scaling it), fill the plane with the identical design that you started with. So in essence you’ve reduced the orig- Aperiodicity inal set of pieces and their interlocking rules to just many Figure 1 shows Penrose’s kite and dart. The “kite” (the larger tile) and the “dart” (the smaller, pointed one) may be assembled only by placing them so that edges of sim- φ 1 The geome- ilar length are adjacent and the colored bands are con- θ θ θ 2 1 try of the 2 1 φ tinuous. (The names kite and dart, and this decoration Penrose kite 1 of the tiles, are due to John Conway.) Figure 2 shows an 6θ 1 and dart (a), φ 4θ φ2 example of a pattern created by these tiles. φ and a decora- φ 1 As you can see, these tiles are indeed aperiodic. A full 1 1 φ tion that forces θ proof would take more room than I have, but a gener- the matching 2θ 2θ al outline of the approach conveys most of the key ideas. φ rules (b) . I’ll use a construction technique that’s based on substi- φ=(1+√5 )/2. (a) (b) tution rules. Like the rules of a formal grammar in com- θ=36 degrees. puter science, or an L system used in botanical simulations, I’ll take each tile and replace it, in position, with another set of tiles. 2 A pattern Let’s simplify the problem for a moment and look just created by at 1D patterns. Penrose kites Suppose that you had an infinite string, and you want- and darts. (a) ed to fill it up with an aperiodic pattern of white and With kites and black beads. I’ll use two production rules to do the job. darts outlined. First, every white bead will be replaced with one white (b) Just the and two black beads in that order. I write this W→WBB, decoration. as shown in Figure 3a. Similarly, we’ll replace each black (a) (b) bead with the rule B→BWW, as in Figure 3b. 78 July/August 1998 . Let’s start with a single white bead and then apply these rules to start filling up the string. Figure 3c shows 3 Creating an (a) the first few steps. Each time you apply the rules, you aperiodic bead take the original string of beads and apply all the rules sequence. at once. So the first generation is simply WBB. Now the (a) The rule (b) new W goes to WBB, and each of the two Bs indepen- W→WBB, dently go to BWW, giving us the new string WBBB- (b) the rule WWBWW. Let’s run the process forever, so that we have B→BWW, and (theoretically) filled up the infinite string with beads. (c) three steps If the pattern generated this way is periodic, then we starting with a could find some chunk of beads—perhaps short, perhaps white bead. enormously long—that repeats. We could then take one (c) of these chunks and glue it together into a super bead, and then match the original pattern by simply filling another string with an infinite number of copies of this single super bead. If we can’t find such a super bead, the pattern is either nonperiodic or aperiodic. You 4 (a) Deflating might be able to convince yourself a kite turns it that in this example we would never (a) into two kites be able to find such a super bead. and a dart. We follow a similar process in 2D (b) Deflating a for Penrose tiles. The technique is dart turns it into typically called deflation since the a kite and two new tiles are smaller (or deflated) half-darts. versions of their predecessors. Figure 4 shows the deflation rules for the kite and the dart. Notice that two of the newly created darts in the dart (b) rule fall off the side of the tile. The matching rules come to the rescue here—you can prove that each half- tile is exactly completed by each of its legal neighbors when that neighbor 5 A Penrose tile is deflated. If you run the process “Ace” deflated back the other way and reduce the four times. number of tiles in a tiling, it’s called inflation, since each tile gets bigger. Figure 5 shows a few deflation steps applied to a small starting pat- tern of Penrose tiles. You may notice that some config- urations of tiles seem to appear sev- eral times in these tilings. This is because you can only assemble tiles around a vertex in a limited number of ways. For example, take the ver- 6 The atlas for tex at the tip of the dart and enu- kites and darts. merate all the ways that other kites Sun and Star and darts can be assembled around (top); Ace and that vertex while obeying the Deuce (middle); matching rules. Then repeat the and Jack, process for the other vertices on the Queen, and dart, and then for the kite. King (bottom). If you run through this process and eliminate duplicates, you’ll find that only seven different kinds of clusters can be formed. These clus- ters taken together are called the atlas of the tiling. Figure 6 shows the atlas for the Penrose kites and darts. IEEE Computer Graphics and Applications 79 . Andrew Glassner’s Notebook C C this limerick (the third line is kind K G K G of clunky, but let’s overlook that). The major problem here is that you J F J F just can’t rhyme with orange— B B you’re stuck and it’s nobody’s fault. I E H D You just have to back up a line or two and try again. H D I E Analytic It’s the same thing with building 7 A A expressions for Penrose tilings. You can follow all vertices in the Sun: Jack: the rules and find yourself unable to Sun and Jack. In A (0,0) A (0,0) add any more tiles. You need to take these expres- B (0, φ) B (0, k) some tiles away and try again, in a sions, k2 = 1 + φ2 C (0, φ + k) C (0, φ +k) trial-and-error process. Obviously if − 2φcos(2π/5). D (cos (π/10), sin (π/10)) D φ (cos (3π/10), sin (3π/10)) we want to create and use Penrose E φ (sin (2π/5), 1 − cos (2π/5)) E k(cos (π/10), sin (π/10)) tilings regularly we need a more reli- F k(sin (2π/5), φ + cos (2π/5)) F (φ cos (π/10), k + φsin (π/10)) able approach for generating the G (cos (π/10), φ + k −sin (π/10)) G (cos (π/10), φ+k −sin (π/10)) patterns. H (−cos (π/10), sin (π/10)) H φ (−cos (3π/10), sin (3π/10)) We saw one approach above: start I φ (−sin (2π/5), 1− cos (2π/5)) I k(−cos (π/10), sin (π/10)) with a single tile and deflate it over J k(−sin (2π/5), φ + cos (2π/5)) J (−φ cos (π/10), k + φsin (π/10)) and over, producing an ever denser K (−cos (π/10), φ +k −sin (π/10)) K (−cos (π/10), φ+ k − sin (π/10)) (or ever larger) tiling.
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