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:''SCA is not involved in the copyright dispute.''[http://docs.law.gwu.edu/facweb/claw/penrose.htm]
:''SCA is not involved in the copyright dispute.''[http://docs.law.gwu.edu/facweb/claw/penrose.htm]


OK... the ref cited here doesn't seem to meet Wikipedia standards either. It comes from a law dept. at a genuine school, but it's not a publshed source and appears to be just a handout to students or something. On top of that, claims of a single law professor in an informal way is not a reliable, primary source on a law case. There must be real sources, and ones that can better sort out what really happened. Until that happens the section is extremely misleading (so an expired US patent gives a UK copyright on something most courts say can't be copyrighted at all??) and can't be in the article.
OK... the ref cited here doesn't seem to meet Wikipedia standards either. It comes from a law dept. at a genuine school, but it's not a publshed source and appears to be just a handout to students or something. On top of that, claims of a single law professor in an informal way is not a reliable, primary source on a law case. There must be real sources, and ones that can better sort out what really happened. Until that happens the section is extremely misleading (so an expired US patent gives a UK copyright on something most courts say can't be copyrighted at all??) and can't be in the article. [[User:DreamGuy|DreamGuy]] ([[User talk:DreamGuy|talk]]) 14:57, 20 November 2008 (UTC)

:[http://news.google.com/archivesearch?q=kimberly-clark+penrose Google news archive search] provides a number of sources. The [http://www.time.com/time/magazine/article/0,9171,986297,00.html Time Magazine] one supports precisely the second sentence quoted: that Penrose and Pentaplex sued Kimberly-Clark for copyright infringement. And I don't see how a handout in an intellectual property law class at a law school fails [[WP:RS]]: it is a [[WP:SPS|self-published sources]], but falls into the "established expert" clause of that part of the verifiability policy. [[WP:OR|Your interpretation]] about whether using copyright law in such cases could be an "error" is not grounds for removing this sourced material from the article. [[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 15:16, 20 November 2008 (UTC)


=="Citation needed"? ==
=="Citation needed"? ==
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I lean toward the final option, but I don't know if my inclinations are at odds with the customs here. Would somebodt please help a brother out and chime in on this issue? Thanks. [[Special:Contributions/76.105.238.158|76.105.238.158]] ([[User talk:76.105.238.158|talk]]) 15:28, 6 January 2009 (UTC)
I lean toward the final option, but I don't know if my inclinations are at odds with the customs here. Would somebodt please help a brother out and chime in on this issue? Thanks. [[Special:Contributions/76.105.238.158|76.105.238.158]] ([[User talk:76.105.238.158|talk]]) 15:28, 6 January 2009 (UTC)


Removing the tag seems OK to me. The obvious problem is that the person who originally put it there may, without reading this discussion, put it back. Nevertheless I should try removing the tag, and, if it comes back, copying the references from the other article to here. [[User:JamesBWatson|JamesBWatson]] ([[User talk:JamesBWatson|talk]]) 15:41, 22 January 2009 (UTC)
:Removing the tag seems OK to me. The obvious problem is that the person who originally put it there may, without reading this discussion, put it back. Nevertheless I should try removing the tag, and, if it comes back, copying the references from the other article to here. [[User:JamesBWatson|JamesBWatson]] ([[User talk:JamesBWatson|talk]]) 15:41, 22 January 2009 (UTC)


::Thanks for your input. I just removed the one instance mentioned above, and will observe how things play out in the future. [[Special:Contributions/76.105.238.158|76.105.238.158]] ([[User talk:76.105.238.158|talk]]) 02:00, 28 January 2009 (UTC)
'''Refs'''
<references/>


[[User:DreamGuy|DreamGuy]] ([[User talk:DreamGuy|talk]]) 14:57, 20 November 2008 (UTC)


'''Refs'''
:[http://news.google.com/archivesearch?q=kimberly-clark+penrose Google news archive search] provides a number of sources. The [http://www.time.com/time/magazine/article/0,9171,986297,00.html Time Magazine] one supports precisely the second sentence quoted: that Penrose and Pentaplex sued Kimberly-Clark for copyright infringement. And I don't see how a handout in an intellectual property law class at a law school fails [[WP:RS]]: it is a [[WP:SPS|self-published sources]], but falls into the "established expert" clause of that part of the verifiability policy. [[WP:OR|Your interpretation]] about whether using copyright law in such cases could be an "error" is not grounds for removing this sourced material from the article. [[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 15:16, 20 November 2008 (UTC)
<references/>

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I find it odd that the whole article makes no mention of the significance of Penrose tiles as a quasi-counterexample of the crystallographic restriction, since they exhibit five-fold symmetry. Why is that? Was it deliberately decided to omit what looks like one of the most important facts about Penrose tiles, or am I confused about their nature? Swap (talk) 04:30, 7 March 2008 (UTC)[reply]

What is a Tiling?

What, then, is a "tiling"? Is there such a thing as "aperiodic tiling"? For that matter what is "the penrose tiling"?

Unfortunately, the terms, as used in this article, are thoroughly entrenched, even though they are not mathematically defined. (I've made several changes to this article, and the article on Aperiodic tiling to reflect this. Much more work is needed.)

A tiling is a covering of the plane by a set of tiles with non-overlapping interiors. A given set of tiles "admits" a tiling iff there is a tiling by congruent copies of tiles in the set. One can consider the collection of all tilings admitted by a given set of tiles.

(And here's the trouble-- everything in the paragraph above is often collectively referred to as "tiling"; The Penrose Tiling, approximately, seems to refer to a system consisting of the Penrose tiles (in one of their variations), and all of the tilings they produce)

A given "set of tiles" is aperiodic iff it admits only non-periodic tilings. So aperiodicity, formally, is a property of a set of tiles, not a tiling. This is an important distinction--- mere non-periodicity, a property of tilings, is not so special.

I guess one could mathematically define tiling and Tiling, but as tempting as that might be, I don't think there is any precedent for that kind of nonsense. The fact is, we have a well established, but inadequate, informal term, and a less-known, but more precise mathematical term.

In casual conversation, to be frank, I'm as likely as the next guy to refer to The Penrose Tiling; but it is not a completely meaningful term in itself. This is subtle, but essential. — C. Goodman-Strauss

If 'aperiodic tiling' is a misnomer then we cannot have the genus-species explanation, i.e. tiling>aperiodic tiling>Penrose tiling.
This suggests to discuss a renaming of the aperiodic tiling page.
For mathematicians, apparently, sets of tiles come before tilings, so finite precedes infinite. But there is the awkward point about the (aperiodic) sets of tiles which do not admit a tiling - why call such shapes 'tiles'?al 12:01, 23 September 2007 (UTC)[reply]
The genus-species definition is possible as
tiling>nonperiodic tiling>Penrose tiling
which avoids the term 'aperiodic tiling' and also the quaint wording that presents a tiling as set of tiles. 195.96.229.83 —Preceding comment was added at 12:04, 18 October 2007 (UTC)[reply]

L-systems

Several images were not Penrose tilings, but were outputs from L-systems. I've moved them to L-system. Now someone needs to explain in that article what those images actually are, and how the L-system generated them.

Now I too wonder how the L-system generates the tilings. I have one guess: it might use deflation. Is this correct? —Sverdrup(talk) 15:31, 27 Dec 2003 (UTC)

A free Microsft Windows program to generate and explore rhombic Penrose tiling is available at http://www.jkssoftware.com/penrose. The software was written by Stephen Collins of JKS Software, in collaboration with the Universities of York, UK and Tsuka, Japan.

LSystems can be generated using the free Software http://jlsystem.sourceforge.net. helohe 12:10, 1 October 2005 (UTC)[reply]

Clark Richert

Yeah, I know that a reference to a random guy posting at Slashdot is exactly what Wikipedia needs. :) But his post mentions tome interesting things about the tiles, including that Clark Richert has figured at least a part of that at the same time as Penrose. Paranoid 15:05, 12 May 2005 (UTC)[reply]

A very important distinction is that Richert does not claim he had invented the matching rules; as I understand it, he only is claiming to have discovered the pattern they force. This is the crux of the matter: the Penrose tiles are interesting because they can only form this particular non-periodic structure. Of course, Kimberly-Clark was only printing the pattern on it's tissues, not making any infringement, I think, on the matching rules and tiles themselves. --69.152.216.28 18:53, 17 September 2007 (UTC)[reply]

Projection

Note that the Penrose tiling is a projection of a five dimensional lattice (which has cubic symmetry) down to two dimensions; thus, the readily apparent symmetry in five dimensions is rather hidden and obfuscated when seen in two.

I heard that it is not a projection of the entire five dimensional lattice, but just of the part of the lattice within a slab between two parallel four-dimensional flats. Is that right? JRSpriggs 07:19, 28 February 2007 (UTC)[reply]

No Matching Rules Discussed

There is a serious-- indeed critical!-- flaw in this article. The rhombs shown CAN tile periodically. (indeed, of course, any quadrilateral can). The essential aspect of the Penrose tiles is that they are marked in such a way that they can ONLY tile non-periodically. With no discussion or illustration of the matching rules that drive the construction, unfortunately, the article is nonsense.

(The illustrations do show the structure the Penrose tiles are forced to assume, but not the actual tiles themselves)

Here is a reference, chosen by Google: www2.spsu.edu/math/tile/aperiodic/penrose/penrose2.htm --—Preceding unsigned comment added by 69.151.118.199 (talk)

The article currently reads (after describing the shapes):

The tiles are put together with one rule: no two tiles can be touching so as to form a single parallelogram. The tiles can be modified with bumps and dents around the perimeter to enforce this rule, but the tiling looks better if they have smooth sides.

Perhaps this should be more emphasized to avoid confusion by other future readers. --C S (Talk) 03:26, 11 March 2006 (UTC)[reply]
That rule is erroneous. The actual matching rule can be made by placing an arrow on each side of each rhomb, using two colors as in the diagram. The directions of the arrows can be assigned by choosing a direction for one edge in the diagram and propagating it. The rule is that when two rhombs are placed together, the arrows and colors must match. This seems to boil down to the above rule plus the color constraint when placing two identical rhombs together, but the actual rule also constrains putting different rhombs together. –Dan Hoey 02:28, 24 April 2007 (UTC)

Thank you; I did overlook this. But of course I wouldn't have been the only one. And the standard stripes that are often drawn on the rhombs are quite attractive!

True. This needs to be covered in the article, preferably with a picture. Reyk YO! 22:08, 20 May 2006 (UTC)[reply]
Is there any reference or better description for that strange rule

The tiles are put together with one rule: no two tiles can be touching so as to form a single parallelogram. The tiles can be modified with bumps and dents around the perimeter to enforce this rule, but the tiling looks better if they have smooth sides.

? While the second (well-known) method implies the first, the converse does not hold, i.e. the "one rule" is not sufficient to guarantee non-periodicity: One big rhomb and two small ones can be put together to form a hexagon and these tile the plane periodically and obeying that rule (smeone produced some piece ASCII art of this on the talk page of the german article).--Hagman-de 08:58, 10 March 2007 (UTC)[reply]
Periodic tiling showing that the parallelogram rule is insufficient
Here is an example of the parallelogram-free hexagon tiling. –Dan Hoeytalk 23:04, 30 April 2007 (UTC)[reply]
bump-and-color-coding for rhombs
I see that Commons:User:Hagman has put up an image of a real set of matching rules. I prefer these rules to the arrow rules by User_talk:Ael 2. Actually, it would look even better without the bumps. Getting the colored arcs into a larger picture for the beginning of the article would require more work, unless we can find something in one of the other wikipedia articles--both the French and German versions seem to have better content than this one. –Dan Hoeytalk 23:42, 30 April 2007 (UTC)[reply]

Proposal for matching rules

I made a christmas and new-year card for my friends using penrose tilings. The program that I wrote for this card can also easily generate other figures, and I believe they are usefull for explaining the matching rules. I have split up the section "Drawing the penrose tiling" two subsections: "L-Systems" and "Deflation". Under the section "Deflation" I have put some figures and explanations about how to generate a penrose tiling based on the matching rule, using of the deflation principle. It is certainly instructive to have some extra words about these matching rules in the introduction. —The preceding unsigned comment was added by Tovrstra (talkcontribs) 14:35, 30 December 2006 (UTC).[reply]

Explanation?

How about a section explaining why the tiling is aperiodic, how it works, and why this is interesting? Torokun 22:05, 28 March 2006 (UTC) And also, why insist (as many others do) that ' given a bounded region of the pattern, no matter how large, that region will be repeated an infinite number of times within the tiling' which is a feature of random systems? 85.187.217.182 23:17, 13 November 2006 (UTC)[reply]

More penrose tiles needed?

See The colossal book of mathematics, Gardner M., Penrose tiles. The tiles featured there are more interesting and should be added.Doomed Rasher 18:19, 2 September 2006 (UTC)[reply]

Ref: Kepler/Penrose tiling problem...i am an independent artist/designer and about 20 yrs ago i painted a picture depicting a periodic pattern using the Kepler/Penrose tiles (derived from the dissection of a pentagon)...and this image can be perused on my web-site at:

http://www.peterhugomcclure.com/colour%20images/36.htm

Best regards pete mcclure.--81.86.8.62 13:00, 22 November 2006 (UTC)[reply]

Hi Pete. Are you willing to allow use of the image directly on the wikipedia? There's a choice of copyright tags to go with an image here; see WP:TAG.--Niels Ø 09:08, 23 November 2006 (UTC)[reply]
Jos Leys has also made some imagery using Penrose Tiles. Jos Leys's website is here, while the Penrose specific images can be perused here. Still, I think these images should only considered if they can add something to the article. Or maybe in a section on the use of Penrose Tiles by artists. Scribblesinmindscapes 17:04, 23 January 2007 (UTC)[reply]
I think you misunderstand the significance of Pete's work. He has created a periodic Penrose tiling having translational symmetry in both dimensions (wich I verified in a paint program). This contradicts a statement in the article that the "parallelogram rule" results in an aperiodic tiling. =Axlq 19:21, 27 January 2007 (UTC)[reply]
Aaah, okay, I initially just took Pete's comment as artistic, as being indicative of an artistic work inspired by Penrose Tiling. The tiling in his picture is not the Penrose tiling. It has translational symmetry and it uses the constituents of a Penrose tiling to form a tiling with translational symmetry - but the real Penrose tiling must use the constituent tiles in a certain manner as discussed in the wikipedia article. Penrose's original paper also starts by saying: "'kites' and 'darts', which, when matched according to certain simple rules, could tile the entire plane, but only in a non-periodic way". Penrose's article can be found here (and should perhaps be included as an external link): [1] Pete's work does not agree with the pictures in the wikipedia article and I can therefore only assume that it doesn't match the tiles correctly. I think the usage of the words Penrose Tiles for the constituents are a bit misleading as indicated they need to be tiled in a certain way to be aperiodic. I'm not a researcher in the field of aperiodic tilings or quasicrystals so disagreement welcomed. Scribblesinmindscapes 09:24, 28 January 2007 (UTC)[reply]
Pete's work falls under the section of this article titled "Rhombus tiling" – not kites and darts – and follows the rhombus tiling rule exactly.
I just confirmed that the picture now shown in that section also has translational symmetry in both dimensions — the region outline itself is tile-able. There is a disconnect between the text in that section and the picture; now that I know the picture has translational symmetry, the text now seems flawed or unclear. =Axlq 19:23, 28 January 2007 (UTC)[reply]
- —Preceding unsigned comment added by JamesBWatson (talkcontribs) 15:49, 22 January 2009 (UTC)[reply]

Kite or Kile?

Most (all?) of the times that the word appears in the text, it's 'kite.' However, the word appears in some images (and the name of the images) as 'kile.' I expect that 'kite' is correct, though I have no idea. It'd be a good thing for someone knowledgeable in Penrose tiling to make consistent.

—The preceding unsigned comment was added by Stomv (talkcontribs) 12:45, 23 February 2007 (UTC).[reply]

Probably some typing error; never heard about 'kile', definitely.al 20:39, 26 February 2007 (UTC)[reply]
"Kile" in norwegian means a wedge (or to tickle) Cuddlyable3 07:57, 15 May 2007 (UTC)[reply]

Some more external links which I thought was quite good concerning Penrose Tiles (see below). I'm adding this here so that someone more familiar with the topic can review it and see if it might be useful re the article.

American Mathematical Society article: [2]

Clay Mathematics Institute article: [3]

Scribblesinmindscapes 16:40, 10 January 2007 (UTC)[reply]

Another Penrose Tiles article at AMS: [4]
Scribblesinmindscapes 18:48, 27 January 2007 (UTC)[reply]

Media Hype

Rewrote the passage about the Steinhardt & Lu paper and trimmed irrelevant links. The idea is not really new, but the name of Steinhardt, an acknowledged expert, gives it now more weight. Artisanal practices suggest examples, but do not produce mathematical objects; an ellipse does not prove that its daughtsman had a theory of conic sections.195.96.229.83 13:28, 27 February 2007 (UTC)[reply]

I think that this paper is important, not just for the link to Islamic architecture, but because it gives tiles which allow you to achieve 10-fold rotational symmetry rather than the mere 5-fold symmetry achievable with the Penrose's own tilings. In both cases there is also an additional reflectional symmetry. So the pattern can be generated by reflections of an 18 degree (for Lu's scheme) or 36 degree (for Penrose's scheme) section between two mirrors. JRSpriggs 07:19, 28 February 2007 (UTC)[reply]
If you believe that tenfold symmetry is somehow superior to fivefold perhaps you should appreciate the pinwheel tiling which has symmetry of infinite order. The Radin paper [1] explains that the Penrose tiling has indeed a tenfold statistical symmetry but the local is just fivefold (added this just the other day and meant to clean up references and notes).

A place for links to art could be the end section ('Triva') which we should perhaps rename. I agree that Jos Leys' site mentioned somewhere above should not be missed. 195.96.229.83

I am not talking about some kind of average or statistical symmetry. I am talking about a perfect rotational symmetry around a single point. JRSpriggs 10:39, 28 February 2007 (UTC)[reply]
A nice symmetry around a central point can be easily achieved by arranging mirror pairs of identical pie-slices from an aperiodic tiling. Their angle should be pi/n. A more elegant solution would be to find an aperiodic decomposition of a triangle and to arrange (pairs of) triangles to form a vertex. Tubingen triangles, Robinson triangles and the pinwheel substitution illustrate this idea.
The main interst in an aperiodic tiling is a lack of symmetry: aperiodic means lacking translational symmetry. The surprise is that only with an aperiodic tiling you can achieve otherwise 'forbidden' symmetries.
Roger Penrose was searching to decrease the number of tiles needed for an aperidic tiling and he arrived at two. The medieval islamic decorators needed some more.
And their interest was to obtain a complicated and elegant motif. In the Mediterranean world interlaced motifs are believed to be a charm against the evil eye, which is apparently one of the sources for interest in them. One might speculate that aperiodic tilings have a particular effect on the perceptive apparatus which could explain their psychological appeal.195.96.229.83
Peter J. Lu has combined inflation rules for two tiles defined by the Islamic architects with one inflation rule of his own devising and gotten a set of rules for three tiles (all used by the architects along with two other tiles) which can tile the plane with perfect ten-fold rotational symmetry. In my opinion, this is an improvement on the five-fold symmetry of the Penrose tiling. I hope to get a link to a file with all three of Lu's tiles. JRSpriggs 09:26, 7 March 2007 (UTC)[reply]

Removed ref to Islamic art from lead as it is out of place and incorrect: see above and also reactions to the Steinhardt-Lou paper. One can say that something equivalent to a Penrose tiling might have been obtained as it is done later in the article.195.96.229.104 (talk) 09:35, 13 December 2007 (UTC)[reply]

Substitution Matrix

The word "Substitution Matrix" is used without reference and introduction. Is it related to the substitutions used to build a penrose tiling and if so, how is it defined? —The preceding unsigned comment was added by 80.202.238.117 (talk) 20:24, 6 March 2007 (UTC).[reply]

Substitutions are transforms which for simple 'linear' cases are represented a matrix, hence the name. The usual notation is New=Matrix.Old, e.g for the penrose tiling:

If the eigenvalues of the substitution matrix are pisot numbers the substitution generates a quasicrystal and the physicists say that it produces Bragg diffraction.91.92.179.156 23:36, 7 March 2007 (UTC)[reply]

use of the word "uncountable"

the article says "there are many ways (infact, uncountably many).." This is kinda vague and might lead someone to think uncountable is a synonym of infinite. In fact, countability/uncountability really has nothing to do with the size of a set. It should definitely be mentioned that the ways to form a penrose tiling is uncountable, but it should not be confused with an implication about the size of the set. 164.76.162.135 16:56, 5 December 2006 (UTC)[reply]

i just decided to go for it and made this change myself 164.76.162.135 17:05, 5 December 2006 (UTC)[reply]

Uncountable/countable are characteristics of the two basic infinite sets and the first is more 'powerful' than the second. This funny talk tries to avoid the paradoxes and confusions when dealing with the infinite. The assertion that there are uncountable ways to arrange a Penrose tiling sounds plausible as different tilings within the same perimeter are possible.

There is however a serious problem here: if the Penrose tiling is in fact an infinite set of variants, what is the meaning of the definite article? 'The' Penrose tiling is the one that Roger Penrose first proposed, but how to define the rest? In the recent hype an attempt to escape from this ambiguity has been made by speaking about the quasicrystalline Penrose tiling. 'Perfect' would have been a better choice of adjective as the variants are seen to be produced by local 'defects' and thus considered to be 'imperfect'. I suspect that the second illustration in the article has been the cause for the unfortunate word choice but it replaced an older picture of something that was not a Penrose tiling.91.92.179.156 21:08, 7 March 2007 (UTC)[reply]
The word uncountable has a special meaning in Mathematics; have a look at the article about the countable set. In short, something can be finite or it can be infinite. The latter can be distinguished into countably infinite and uncountably infinite. In my opinion, you can find a bijection to the set of the natural numbers, so the set of Penrose tilings is countably infinite. This bijection is an algorithm (not really, because it does not terminate, but otherwise it is) creating all possible penrose tilings in an infinite number of steps. It starts with a single tile. Then it enters an endless loop (could be implemented using a width-search) in which it branches a finite number of times to place a neighbouring tile to all possible positions on the current tiling. This way each given possible tiling is generated in a specific amount of steps. This is a bijection to the set of the natural numbers (the numbers of steps are members of this set), so we have the necessary bijection. qed. So it should be countably infinite. Alfe 07:38, 10 March 2007 (UTC)[reply]
The article says "It is easy to check that some of the compact patches consisting of three tiles admit two different arrangements within the same perimeter and thus variations are possible.". If that is true and there are a countably infinite number of such patches (as there must be), then the number of variations which can be achieved this way is uncountable in the technical mathematical sense used in set theory. JRSpriggs 11:10, 10 March 2007 (UTC)[reply]
That's not a proof, though. The proof of uncountability that I know of is by deflation. It's easier to prove that there are uncountable choices of a single tile in a tiling; since each tiling only has countably many tiles, it amounts to the same thing. To choose a single tile in a tiling, first choose whether it is a kite or a dart, second choose what kind of patch this tile is part of at the second level of deflation, etc. Thus one can specify the whole tiling by an infinite sequence of finite choices of this type. The number of sequences of choices that can be made is the number of the continuum. —David Eppstein 16:05, 10 March 2007 (UTC)[reply]
Yes, David. My argument was for the rhombus tilings. Yours works for the kite and dart tilings (and would work also for the rhombus tiles with a different choice of words). What would you like me to clarify or justify to make my proof complete in your eyes? JRSpriggs 09:54, 11 March 2007 (UTC)[reply]
The algorithm I presented above produces a countably infinite number of finite tilings. So it proves that the number of finite tilings is countably infinite. It seems to be of more importance, though, what the number of tilings is which fill the whole infinite plane. That number may likely be uncountably infinite. The text could point out that difference more clearly. Alfe 01:47, 15 March 2007 (UTC)[reply]
Sorry, Alfe, you're mistaken. You said you branch a finite number of times, but that only tiles a finite part of the plane. If you want to tile the entire plane, then you have to branch infinitely many times, and that makes the number of tilings uncountable. If we take them modulo rotations and translations (placing a vertex at the origin and including a horizontal edge) that's still uncountable. So there are uncountably many different tilings of the plane, but you only can tell if you look at the whole plane. If you look at a finite patch, all the tilings agree (and each agrees in a set translation that is not only infinite, but has positive density.) Only two of the tilings (up to symmetry) have a fivefold center of symmetry. –Dan Hoeytalk 14:59, 5 June 2007 (UTC)[reply]

Multiple tilings and rotational symmetry

There are certainly finitely many connected tilings given any finite number N of tiles, but there are uncountably many tilings of the plane, using the deflation argument. However, it is important to note that only two of the tilings possess five-fold rotational symmetry. This renders most of the statements about five-fold symmetry false. It should be mentioned that these two, and uncountably many others, also possess mirror symmetry; only the two rotationally-symmetric ones possess mirror symmetry through more than one line. The distinction between finite and infinite tilings is crucial here, since a finite subtiling cannot be used to determine which infinite tiling you are in, nor even where you are in that infinite tiling.

Statements about a "rule" that no two rhombs can form a parallelogram are also incorrect, as noted above. The true rule can be seen in the diagram; color the edges of the rhombs as in that diagram, and only allow matching-colored edges to be adjacent.

There doesn't seem to be a rating for an article that has quite a bit of good stuff and some glaring falsehoods. –Dan Hoey 02:28, 24 April 2007 (UTC)

I removed the statement of the bogus rule about two rhombs being forbidden to form a parallelogram. I also removed the statement It is easy to check that some of the compact patches consisting of three tiles admit two different arrangements within the same perimeter and thus variations are possible which I believe to be incorrect (not in that it is not "easy", but in that it is "impossible"). Perhaps it was referring to the bogus rule, which permits reorientation of the convex hexagon formed of two thin and one thick rhomb. That cannot be made with correct matching rules.
I'm about ready to dike out the discussion of image:VarPenrT.jpg since it does not seem to be relevant to Penrose tilings. I don't know what it means to say it is "not a quasicrystal" Does it represent an aperiodic tiling rule or not?
The statement that there are 23 ways that these rhombs can meet at a vertex is apparently wrong; I don't know how it was determined. I count 49 ways; as multiples of 36 degrees, these are 1111111111, 111111112, 11111113, 11111122, 1111114, 11111212, 1111123, 11112112, 1111213, 111124, 111133, 111121112, 1111213, 111124, 111133, 11121112, 1112113, 1112122, 111214, 111223, 111313, 11134, 1121122, 112114, 1121212, 112123, 112213, 112222, 11224, 11233, 11242, 113113, 11314, 11323, 1144, 121222, 121213, 12124, 12133, 122122, 12214, 12223, 12313, 1234, 1324, 1333, 1414, 22222, 2224, 2233, 2323, 244, and 334. This was done by hand and may not be entirely complete, so I changed the wording to "over twenty".
Indeed, there are 54 ways: My list omitted 1111222, 111232, 112132, 12232, and 1243. Of course, to write "54" would be OR. Perhaps the OEIS has a sequence for "necklaces from {1,2,3,4} summing to n". Or I could send them that sequence, if find the time to compute it (and its relatives) properly. Or is there a combinatorial objects server that would cough up the answer in a citable way?–Dan Hoeytalk 15:38, 12 June 2007 (UTC)[reply]
The statement did not originally say 23 ways (which is vague), it said 23 combinations (which is specific). I believe your 54 figure refers to permutations. For example you count 11233, 11323, 12133, 12313 separately, these are all permutations of the 11233 combination. I wouldn't mind if the article says "there are X combinations" or "there are Y permutations", but it should be precise; "there are at least Z ways" is vague in two ways - the use of the phrase "at least", and the use of the term "way". There are, as far as I know, 23 combinations of vertex. Until you can precisely say how many permutations there are, we should stick with this statement.

For completeness the 23 combinations are: 1111111111, 111111112, 11111113, 11111122, 1111114, 1111123, 1111222, 111124, 111133, 111223, 11134, 112222, 11224, 11233, 1144, 12223, 1234, 1333, 22222, 2224, 2233, 244, 334. Chris 10:53, 21 June 2007 (UTC)[reply]

"Combinations" may be ordered or not, so it is nearly as ambiguous as "way"; "permutations" usually refers to bijections between sets of distinct objects, so is incongruous here. I claim that cyclically ordered combinations of rhombus angles are the appropriate statistic, because the rules of the tiling are based on tiles that share an edge, not a vertex. Thus 11233, 11323, 12133, 12313 are distinguished by rules that specify whether or not a "2" angle may abut a "1" or "3" and whether a "1" may abut "3"s on both sides. So counting the number of cyclically ordered combinations,
the unordered angle multiset 111223 appears in six distinct cyclically ordered combinations (namely 111223, 111232, 112123, 112213, 112312, and 121213),
the multisets 11111122, 1111222, 11224, and 11233 appear in four ordered combinations each,
the multisets 1111123, 111124, 111133, 112222, and 1234 appear in three ordered combinations each, and
the multisets 11134, 1144, 12223, and 2233 appear in two ordered combinations each.
Thus your count of 23 is short by 5 + 4×3 + 5×2 + 4×1 = 31, and the total of ordered combinations is 54.
I appreciate that you have now clarified your language on the article, but I dispute your contention that "combinations" may be ordered or not: in combinatorial mathematics combination refers to an unordered collection of elements, whereas permutation refers to distinguishable sequences. If you dispute this, you'd better sort out the pages for these terms. Taking this definition of combination, my count of 23 combinations is spot on. My original point was that when the article said there were over 20 ways to add up to 360 degrees at the vertex, this was unnecessarily vague. The current wording is better, although I'd still prefer the word permutation since it has more specific meaning in this context (as long as you are sure there are 54 permutations - do you have a citation? - you shouldn't really be working the combinations/permutations out yourself since that would constitute original research). Chris 15:44, 21 June 2007 (UTC)[reply]
I am still concerned that there is no citable reference for either of these numbers, nor for the number of vertices of these types that appear in a Penrose tiling.–Dan Hoeytalk 14:37, 21 June 2007 (UTC)[reply]
Agreed - I've seen both numbers used in text books before, but unfortunately can't remember them: doesn't really count as a proper citation does it?Chris 15:44, 21 June 2007 (UTC)[reply]
The vague wording '23 combination' being replaced by '54 cyclically ordered', what about the 7 allowed combinations? Are they ordered? In fact some people prefer to say 8 as they count two types of 5-stars due to the matching rules.

If counting is OR and unacceptable, I guess that here at least two citations are needed.al 21:36, 28 July 2007 (UTC)[reply]

Also, the statement that the frequencies of the two rhombs are equal is incorrect: there are more thick rhombs than thin ones, in the golden ratio. –Dan Hoeytalk 14:43, 12 June 2007 (UTC)[reply]
I figured out that what was meant was that the frequency of each orientation of a rhomb is equal, so I clarified that. This whole discussion should be moved to a section that treats both rhombus tiles and kite and dart tiles together, since much of its content is applicable to both.–Dan Hoeytalk 14:58, 21 June 2007 (UTC)[reply]

The Pentagonal Penrose Tiling

Perhaps something should be said about the earlier steps, when Penrose proposed a 'Keplerian' tilings built with more than two tiles. Apparently the original Penrose tiling was built with pentagons, rhombuses, stars and boats (3/5 of a star). The derivation can be seen on Savard's page [5]. Here is a good introduction [6] which includes all of this and could be linked somewhere. In professional jargon the Penrose tiling(s) are just P1, P2 and P3, which correspond to the pentagonal, kite and dart and rhombus variants. al 22:22, 30 June 2007 (UTC)[reply]

Importance

Just added a section about the Decagonal covering which seems important in physics. Removed the high-rating tag on this page as I believe that the Penrose tiling is mathematically trivial and pertains more to recreational maths. If people believe that tenfold symmetry is somehow important, here is perhaps an important link [7].al 17:49, 14 March 2007 (UTC)[reply]

Please don't remove the rating template. If you feel that "High" is the wrong importance, that can be changed, but changing it is very different from ripping out the whole rating block. As for mathematical triviality, I disagree but more importantly I think that mathematical depth is far from the only thing determining importance. —David Eppstein 20:57, 14 March 2007 (UTC)[reply]
Sorry for the tag; please explain why the importance is high.Imho it is very low, the Penrose tiling being just an example.I believe that the quality is also low and a lot of rewriting has to be done (Be bold).al 18:32, 16 March 2007 (UTC)[reply]
My feeling is that in terms of mathematical depth the importance of Penrose tilings should be mid, not low. The concept of a periodic tiling itself is nontrivial, as is the construction of Penrose tilings by inflation and deflation, as is the relation to duals of pentagrids, as is the connection to sections of higher dimensional lattices. However, since Penrose tilings are so well known popularly, and have some relation to physical quasicrystals as well, I think it's enough to boost the importance to high. As a similar example, I've been editing regular number recently; in terms of mathematical content it is much more trivial (merely the numbers which have only 2, 3, or 5 in their prime factorization) so in mathematical depth I would rate it as low, but again it is a concept with several important applications outside of mathematics, due to which I boosted the rating to mid. —David Eppstein 18:51, 16 March 2007 (UTC)[reply]
There is no doubt that the PT is the paradigmatic case, but it is just an instance of an aperiodic tiling. Everything that is known about the PT, has been found to be valid mutatis mutandis for the Ammann-Beenker tiling. As I see it, aperiodic to periodic is just as rational to irrational. The binary Fibonacci word can be inflated, deflated and projected and it offers a more obvious access to understanding the properties of the PT, which is more or less its generalization. Socolar's multigrid approach is perhaps 'deep', but it is also a much more general topic than PT. Briefly: popular is not the same as important. Shall we lower the rating with one or two notches? al 17:03, 17 March 2007 (UTC)[reply]

Let's try a different tack. My feeling of the importance rating is, again, not so much importance within mathematics, but importance as a contribution to an encyclopedia. How embarrassed should we be if this topic were missing? Not embarrassed at all, it's so trivial as to be unimportant, and there is room for legitimate debate about whether it even meets WP's standards of notability: low importance. The article makes a solid positive contribution to the encyclopedia's overall depth, is on a clearly notable topic, has some applications and connections to other topics, but could be removed without causing us significant embarrassment: mid importance. The topic is notable enough that any encyclopedia worthy of the name should carry it, and it would be a clear embarrassment to us not to carry it: high importance. The topic is central to human knowledge and any educated person should be embarrassed not to know a little about what it is: top importance. That's my own calibration, anyway, and I think it's more conservative than the calibration described at Wikipedia:WikiProject_Mathematics/Wikipedia_1.0. Now, where does Penrose tiling fit on this scale? I think mid or high are both defensible choices, but I'd pick high because I'd be embarrassed to be working on an encyclopedia that doesn't carry an article on a mathematical topic that is so well known in popular culture. —David Eppstein 18:33, 17 March 2007 (UTC)[reply]

I would go for mid importance for this article, but I'd accept high. Low is really for the very obscure stuff, see Category:Low-importance mathematics articles (although there are a few there which I think deserve mid). I do think this is mathematically important as it opened up a new area of investigation, it what was beleived to be a closed book. It further prompted investigation into real world crystals which had five-fold symmetry where none were thought to exist. Hence is has implication outside of mathematics, which is one reason for for an increased rating. --Salix alba (talk) 22:00, 17 March 2007 (UTC)[reply]

OK, I see I got it wrong, but I was mislead by the Physics project which explicitely says to rate importance 'within physics'. The Maths project, being tied with the W_1.0, suggests the oposite. Sorry for the trouble.al 18:00, 25 March 2007 (UTC)[reply]

Alain Connes considers the space of Penrose tilings a "very interesting `noncommutative' space" and makes it a main example in his book, Noncommutative geometry. So I think calling the topic "mathematically trivial" is a rather narrow viewpoint and one that would presumably change with more experience and knowledge. --C S (Talk) 22:39, 11 June 2007 (UTC)[reply]

Early history

I think that the content of 'Early history' paragraph did not match its title and I have modified it. Penrose's name is still buried among technical details but I have tried to connect it with a larger context (not just mathematics). Perhaps Dürer and girih tiles should also be moved here. The discarded ending could be moved to Wang tiles. 91.92.179.156 09:33, 21 September 2007 (UTC)[reply]

Surprising

The opening section states "* any finite region in a tiling appears infinitely many times in that tiling and, in fact, in any other tiling. This property would be expected if the tilings had translational symmetry so it is not a surprising fact given their lack of translational symmetry." Why does the fact that it is expected given translational symmetry (isn't that a tautology anyway?) mean that it isn't surprising given no translational symmetry. I believe the statement due to the largeness of infinite tilings, but I don't see the link to the statement made. -- SGBailey (talk) 22:48, 22 June 2008 (UTC)[reply]

Is there a better L-System rule set?

As can be seen by this SVG file with ECMAScript animation of a Penrose tiling using an L-System (essentially following the given axiom/rules in this article), this particular rule set covers many rhombus edges MUCH more then twice.

This leads to MANY edges overlapping each other, especially near the center.

Also some of the pathes are closed (and then can be filled with a color), but others are open and can't be filled.

Does anybody know of a better rule set, only producing closed pathes for the neccessary rhombuses?

Every single rule in the given set forms essentially a narrow or a wide rhombus (2 of them each, the narrow ones have an additional entry/exit line, which can't be removed without breaking the over all tiling. I can provide an SVG file demonstrating the single rules on request).

Would a better rule set even be possible?

Deerwood (talk) 06:20, 7 July 2008 (UTC)[reply]

Contradiction between Penrose and Aperiodic articles?

The introduction here says that Penrose tilings are not correctly described as aperiodic tilings. Yet the introduction of Aperiodic tiling article says "the various Penrose tiles are best known examples of an aperiodic set of tiles" and "the Penrose tiles are an aperiodic set of tiles." Unless I'm missing some subtle semantics, this is confusing to a reader. --Ds13 (talk) 07:48, 11 July 2008 (UTC)[reply]

I noticed that contradiction as well, and you seem to be correct. Under the citation I just added ([8]), which distinguishes between aperiodic and nonperiodic tiling, the Penrose tiling is indeed aperiodic, since it is not also possible to construct a periodic tiling from Penrose tiles. I've updated the article accordingly. --ArthurDenture (talk) 05:05, 26 August 2008 (UTC)[reply]

I removed the following from the article, as it has multiple errors of sourcing and fact. Since someone reverted the removal with claim that it is "notable" I am taking the time to explain in more detail. The content was so screwed up it was dangerously misleading and could not stay. Perhaps with some work it can be reworded to be accurate and adequately sourced.

Pentaplex Ltd., a company in Yorkshire, England controlled by Penrose, owns the licensing rights to Penrose tilings.[1]

This sentence is severely flawed. First up, ownership of intellectual property is always contentious issue, and Wikipedia itself should not take sides on claims of ownership. If Pentaplex asserts ownership to some sort of legally enforceable intellectual property, then we need to be specific in what they claim to own and in what jurisdictions. The ref that was provided here says they have a US patent -- but it also says it's expired. Thus the only reliable source provided shows no ownership rights at all, and only former potential rights (patents can be disputed, ownership comes after it's been tested and prevailed in court), and only in the United States... which is a problem, as Pentaplex is in the UK. If they assert any ownership then it must be something other than this patent. We need a reliable source about their claims if we are going to list these claims... and then they are only claims, not findings of law.

Penrose and Pentaplex filed a lawsuit against Kimberly-Clark for breach of copyright.

Is there a reliable source about the lawsuit and what exactly the grounds were? It sounds dubious to me, as copyrights are only for fixed forms, not theoretical constructs like tilings.

Kimberly-Clark had allegedly embossed Penrose tilings on Kleenex quilted toilet paper in the UK.

This is probably the least bothersome sentence. It could use a real source, but it's not exactly confusing or misleading.

SCA Hygiene Products later came to control Kleenex products and reached an agreement with Penrose and Pentaplex on the Penrose tiling issue.

In which case no ownership rights were legally established, just that a settlement of some sort was made. And a source would be needed here

SCA is not involved in the copyright dispute.[9]

OK... the ref cited here doesn't seem to meet Wikipedia standards either. It comes from a law dept. at a genuine school, but it's not a publshed source and appears to be just a handout to students or something. On top of that, claims of a single law professor in an informal way is not a reliable, primary source on a law case. There must be real sources, and ones that can better sort out what really happened. Until that happens the section is extremely misleading (so an expired US patent gives a UK copyright on something most courts say can't be copyrighted at all??) and can't be in the article. DreamGuy (talk) 14:57, 20 November 2008 (UTC)[reply]

Google news archive search provides a number of sources. The Time Magazine one supports precisely the second sentence quoted: that Penrose and Pentaplex sued Kimberly-Clark for copyright infringement. And I don't see how a handout in an intellectual property law class at a law school fails WP:RS: it is a self-published sources, but falls into the "established expert" clause of that part of the verifiability policy. Your interpretation about whether using copyright law in such cases could be an "error" is not grounds for removing this sourced material from the article. David Eppstein (talk) 15:16, 20 November 2008 (UTC)[reply]

"Citation needed"?

I'm unsure of the protocol, so rather than make the (trivial) edit I'm putting this on the Talk page. My concern relates to the usage of "citation needed" tags, and it's bothered me in many articles... perhaps someone can set me straight on this one, so I'll know how to deal with it elsewhere? In this article I'm unsure about the state of the following sentence:

"Robert Ammann independently discovered the tiling at approximately the same time as Penrose.[citation needed]"

Given that Ammann's name is a link, and that following that link leads you to a Wikipedia article that addresses the statement in some detail and cites some sources of its own (which, unfortunately, don't include links to web-accessible information), is the "citation needed" tag truly needed here? It seems to me there are a few possibilities:

  • Cite the same sources as Ammann's Wikipedia page. This seems a bit pedantic.
  • Cite Ammann's Wikipedia page itself. This seems redundant, since that link is already in the sentence.
  • Find newer and more useful citation sources, and update the article, Ammann's article, or both. While this would solidify things a bit, it also just adds a layer of abstraction to the question: the placement and propagation of the citations would still need to be determined.
  • Remove the "citation needed" tag. Assume that the link at the beginning of the sentence is a sufficient guide for interested readers.

I lean toward the final option, but I don't know if my inclinations are at odds with the customs here. Would somebodt please help a brother out and chime in on this issue? Thanks. 76.105.238.158 (talk) 15:28, 6 January 2009 (UTC)[reply]

Removing the tag seems OK to me. The obvious problem is that the person who originally put it there may, without reading this discussion, put it back. Nevertheless I should try removing the tag, and, if it comes back, copying the references from the other article to here. JamesBWatson (talk) 15:41, 22 January 2009 (UTC)[reply]
Thanks for your input. I just removed the one instance mentioned above, and will observe how things play out in the future. 76.105.238.158 (talk) 02:00, 28 January 2009 (UTC)[reply]


Refs

  1. ^ Penrose, Roger, U.S. patent 4,133,152 "Set of tiles for covering a surface", patent issued January 9, 1979 (expired)