tag:blogger.com,1999:blog-3829599.post113851492699376383..comments2017-03-12T10:07:49.832-07:00Comments on Agoraphilia: Fractal Coordinates, Stretching Einstein, and Shorting EuclidGlen Whitmannoreply@blogger.comBlogger11125tag:blogger.com,1999:blog-3829599.post-1139251947069690752006-02-06T10:52:00.000-08:002006-02-06T10:52:00.000-08:00Jeff Brown: I think I've grok your account of the...Jeff Brown: I think I've grok your account of the Peano curve. You say that either it is volume-filling or self-avoiding, but not both. It depends on its iteration. At any iteration short of infinity, it is not volume-filling, but it is space-avoiding. After infinite interations, it fills the volume--but at the cost of touching itself.<BR/><BR/>I should think, though I am not certain, that the same would hold true of the Hilbert curve or any volume-filling fractal curve. It is *not* equivalent to Godel's Theorem or the Heisenberg Principle. But it *is* attractively reminescent of them.<BR/><BR/>But, ok, so what? I still think I might get the fractal coordinate system to work with this hack. Define each point in the fractal volume by the coordinates (s, i), where s=arc lenth of curve from origin and i=iteration-1/iteration.<BR/><BR/>Each point in the fractal volume can now (I hypothesize) be defined by a unique coordinate (s, i). The uncertainty of defining a point lying on the infinite iteration of the curve decreases to zero as you near it. Again, I emphasize, it's only a hypothesis--not a theory!<BR/><BR/>Cool blog, by the way. Loved that stuff about perfect tunings.Tom W. Bellhttp://www.blogger.com/profile/05244942671829268273noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138727433670027412006-01-31T09:10:00.000-08:002006-01-31T09:10:00.000-08:00Jadagul wrote: "Depends on what kind of calculus y...Jadagul wrote: "Depends on what kind of calculus you want to do with it ..." Could you do any calculus at all on it?<BR/><BR/>Tom's original post: "Assume a volume-filling, self-avoiding, fractal curve, like Peano curve or Hilbert curve. In one or more of its iterations, the curve will touch each point in the volume." That's not quite right. Most points will never be on any finite iteration of the Peano curve; you have to define the "ultimate" Peano curve, as the limit of the sequence of Peano curves (i.e. a function that is the limit of a series of functions).<BR/><BR/>The finite iterations are self-avoiding but not space-filling. The ultimate Peano curve is space-filling but not self-avoiding: it sends multiple points to the same point. I said that earlier, but I didn't explain. Here's one example of a violation: if a sequence of finite Peano curves is filling in four quadrants of a square (or eight octants of a cube), then any point on the boundary between two quadrants (octants) can be approached from either side. In the limit, multiple points will map to the same point.<BR/><BR/>I don't know if the Peano curve is more than 4-to-1 (8-to-1) anywhere.Jeff Brownhttp://www.blogger.com/profile/10798023289599415591noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138701815449905042006-01-31T02:03:00.000-08:002006-01-31T02:03:00.000-08:00Jeff Brown: depends on what kind of calculus you w...Jeff Brown: depends on what kind of calculus you want to do with it, I suppose; if you just want to compress information, then you don't need the reals and you don't need continuity, and the conversion function I just gave will work (incidentally, the return function—from the integers to Q3—is continuous because the integers are a discrete space). If you want to work in the reals, there's an analysis result that says that, essentially, a bounded subset of R2 has the same cardinality as a bounded subset of R (which has the same cardinality of R. I'm pretty sure R2 would have that same cardinality, and likewise R3, but my reference didn't mention that specifically). Thus there's a bijection from any bounded subset of R3 into a bounded subset of R, if you really want one. Of course, a computer can't actually store a real number, as far as I know, and most concrete computations work with approximations anyway; seems like you're fine just sticking with a rational approximation of the point.Jadagulhttp://www.blogger.com/profile/15781108432873815972noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138697205737507142006-01-31T00:46:00.000-08:002006-01-31T00:46:00.000-08:00There exists no 1-to-1 continuous function between...There exists no 1-to-1 continuous function between Euclidean spaces of different dimension, although there exist plenty of functions with one characteristic or the other. In particular, the Peano curves are continuous but not bijective. I feel like that would make their use in physics difficult.Jeff Brownhttp://www.blogger.com/profile/10798023289599415591noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138682646751657362006-01-30T20:44:00.000-08:002006-01-30T20:44:00.000-08:00If anyone cares, I took a break from my math homew...If anyone cares, I took a break from my math homework today to (and yes, this <I>does</I> qualify as a break) find the formula for converting rational three-space to an integer. Actually pretty staightforward: If I have a point expressed by (a/b,c/d,e/f), I compress it into n=(2^a)*(3^b)*(5^c)*(7^d)*(11^e)*(13^f). Leave aside for the moment that it's almost certainly shorter to send the sequence (a,b,c,d,e,f), you can see that this number n can be uniquely decomposed into its prime factorization, which gives a unique ordered sextuplet. I once again question the utility of this method, especially since unwrapping it—going from n back to the rational three-space—is hard and long, even on a computer; but it's certainly possible.<BR/><BR/>My apologies for the math-nerd-out.Jadagulhttp://www.blogger.com/profile/15781108432873815972noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138664433594471432006-01-30T15:40:00.000-08:002006-01-30T15:40:00.000-08:00Thanks, Chris. My error. I thought he meant to d...Thanks, Chris. My error. I thought he meant to describe a continous line rather than successive sweeps in the same direction.Tom W. Bellhttp://www.blogger.com/profile/05244942671829268273noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138655055850664602006-01-30T13:04:00.000-08:002006-01-30T13:04:00.000-08:00Tom, either you have a typo in your response to Ja...Tom, either you have a typo in your response to Jadagul, or you haven't understood his point. He's listing successive diagonals (upper right-to lower left). The leftmost diagonal has one element (0,0), the second two (1,0),(0,1), and third has three (2,0),(1,1),(0,2). When you list them all you get the sequence he gave: ((0,0),(1,0),(0,1),(2,0),(1,1),(0,2)...) rather than the one you gave, which repeats (2,0).Chris Hibberthttp://www.blogger.com/profile/12235621011708498622noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138647531012086002006-01-30T10:58:00.000-08:002006-01-30T10:58:00.000-08:00Joel: Thanks for the cogent critique. I'm not su...Joel: Thanks for the cogent critique. I'm not sure about the length of the infinitely-iterated curve, either, but I'll bet it's infinite. It has to fill the volume, after all, and each curve cross-section is infintely thin. You would thus end up, were you to use a Peano curve, with infinite points at the coordinates (infinite, infinite). That hardly makes for a completely descriptive coordinate system!<BR/><BR/>It might still make for a useful one, though. Note that the fractal coordinate system I described leaves space cut into distinct chunks. Sound familiar? Right: our own atomic-scale space (at least under quantum theory).<BR/><BR/>You also hit the mark with the approximation strategy, I think. Cresson appears to use a similar strategy.<BR/><BR/>"Real analysis isn't my strong suit,"? I like that. How's your surreal analysis? Hardy har har.<BR/><BR/>dgm: I approach math somewhat hat-in-hand. But, granted, the prospect of listening to me beg cannot appeal. It's just not my strong suit.<BR/><BR/>jadagul: Thanks for that argument. (I think though, you meant to write, "Now pass through the grid in a crosswise order, so you get the sequence ((0,0),(1,0),(0,1),(2,0),(1,1),(2,0)...)." I think I grok it. I certainly undestand the translation function problem. I'd recognized that problem in my own proposed coordinate system.<BR/><BR/>A fractal coordinate system would prove useful untranslated, though, if it described curves that exist in nature. I've got some more crazy ideas up my sleeve on that front. Please check back, later. I need good math critics.Tom W. Bellhttp://www.blogger.com/profile/05244942671829268273noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138589378506894912006-01-29T18:49:00.000-08:002006-01-29T18:49:00.000-08:00On some level, it's possible to take any n-space a...On some level, it's possible to take any n-space and specify it with one number, at least in theory. That's actually the core of Cantor's proof that the integers and the rational numbers are equal in cardinality (have a one-to-one correspondence): any rational number can be expressed as one or more ordered pair of integers, so if you can set up a one-to-one correspondence between ordered pairs and the integers, you show the integers and rationals are the same size. This method can be extended to show that the integers are the same size as the set of integer n-tuples for any n, and I believe it would do the same for the reals and real n-tuples.<BR/><BR/>The method, in brief: create a grid of ordered pairs.<BR/><BR/>(0,0) (1,0) (2,0) (3,0)...<BR/>(0,1) (1,1) (2,1) (3,1)...<BR/>(0,2) (1,2) (2,2) (3,2)...<BR/>(0,3) (1,3) (2,3) (3,3)...<BR/>....<BR/><BR/>Now pass through the grid in a crosswise order, so you get the sequence ((0,0),(1,0),(0,1),(2,0),(1,1),(0,2)...). Then any point in this 2-space can be specified by a single integer ((0,2), for instance, is the number 5). You can use the same principle on the rationals, and thus could ultimately specify any point in rational n-space with a single integer. But writing the translation function would be a huge pain in the rear, and the only way I could think of it possibly being useful is if bandwidth is extremely expensive and computing power extremely cheap. And that would have to be a pretty huge discrepancy.Jadagulhttp://www.blogger.com/profile/15781108432873815972noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138575715229036232006-01-29T15:01:00.000-08:002006-01-29T15:01:00.000-08:00you lost me when you begged for my indulgence.you lost me when you begged for my indulgence.dgmhttp://www.blogger.com/profile/09180130771290773700noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1138572724997542722006-01-29T14:12:00.000-08:002006-01-29T14:12:00.000-08:00The biggest problem I see with this scheme is that...The biggest problem I see with this scheme is that certain points will *only* lay on the curve when the number of iterations is infinite. Specifically, the Peano curves won't touch any irrational numbers at a finite number of iterations, and the length of the infinitely-iterated curve may very well be unmeasurable (I haven't checked). I suspect the problem is that irrational numbers are not countable.<BR/><BR/>Of course, you could use this as an *approximation* of any point in 3d space, but there are much easier methods than this. You could divide a finite 3d space into a finite number of small cubes with sides of length x, line them up, and approximate the location of any point by specifying x and which cube it's in.<BR/><BR/>Keep in mind that real analysis isn't my strong suit, and I may be completely wrong.Joel Bernsteinhttp://www.blogger.com/profile/07230973094188840754noreply@blogger.com