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Topic: Uncountable Copies (Read 840 times) |
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Icarus
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Below are the 5 standard symbols used in "ESP" experiments. (Sorry for the crude "wavy lines" - I created them using a spline curve generator freehand.) For each of these shapes, an infinite number of similar (in the strict geometric sense) shapes can be fit inside a finite rectangle so that no two shapes overlap. For which of these 5 shapes can an uncountable number of similar figures fit in a finite space without overlapping? (Consider all lines to be of zero width.)
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« Last Edit: Nov 5th, 2003, 8:23pm by Icarus » |
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
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BNC
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Re: Uncountable Copies
« Reply #1 on: Nov 5th, 2003, 11:19pm » |
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Does the space between the wavy lines count as part of the shape, or as free space?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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towr
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Re: Uncountable Copies
« Reply #2 on: Nov 6th, 2003, 12:36am » |
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::Unless I misunderstand the question, I'd say at least the circle, and the square and I think also the pentagram The results would look like a filled version.::
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BNC
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Re: Uncountable Copies
« Reply #3 on: Nov 6th, 2003, 5:31am » |
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on Nov 6th, 2003, 12:36am, towr wrote:The results would look like a <hidden>. |
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« Last Edit: Nov 6th, 2003, 6:36am by BNC » |
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Icarus
wu::riddles Moderator Uberpuzzler
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Re: Uncountable Copies
« Reply #4 on: Nov 6th, 2003, 10:03am » |
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Only the lines count as part of the shape, and the 3 wavy lines are all congruent (they don't look like it in the drawing to me, but they are). And yes, you are correct about the 3 closed figures, towr - you can get an uncountable number of similar copies of them just by scaling. What about the other two?
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
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towr
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Re: Uncountable Copies
« Reply #5 on: Nov 6th, 2003, 10:34am » |
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::the wavy lines might also work, but not the plus I think.. the plus would be countably infinite, since you divide the plane in 4, and each subsquare repeats the pattern.::
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aero_guy
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Re: Uncountable Copies
« Reply #6 on: Nov 6th, 2003, 12:39pm » |
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wow, I wish I knew what a countable infinity was.
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towr
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Re: Uncountable Copies
« Reply #7 on: Nov 6th, 2003, 12:44pm » |
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0,1,2,3,4,5,6,7,8,9.... = [bbn] or 0,1,-1,2,-2,3,-3... = [bbz] basicly if you can give some method which can bring you to any element in a finite number of steps.. [bbr] is an uncountable set
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« Last Edit: Nov 6th, 2003, 12:45pm by towr » |
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Eigenray
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Re: Uncountable Copies
« Reply #8 on: Nov 6th, 2003, 10:16pm » |
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I'd have to say for all but the cross. For the pentagon, square, or circle, we can just scale by any r in (0,1), which is uncountable. For the wavy lines, you can take uncountably many translates. To see why the cross doesn't work, let the rectangle by WxH, W[ge]H, and specify a cross by the coordinates of its center and its size, so each cross is a point in (0,W)x(0,H)x(0,H), which has finite measure. Every time we put down a cross, we eliminate a certain set of points representing more allowable crosses. The idea is that we can't put down an uncountable number of crosses without eliminating an infinite volume of cross space: If we put down a cross C=(x,y,2r), that eliminates, among others, the set of crosses that intersect C and lie within the r-by-r square centered at (x,y) (the "center square"): Uc = { (a,b,2s) : 0 < s < min(x+r/2-a,a-x+r/2,y+r/2-b,b-y+r/2) }, which has volume r3/6 > 0. It actually eliminates a larger set, but this way the Uc's will be disjoint. They're clearly disjoint(?): given two disjoint crosses C and C', if the center square of C intersects the center square of C', then you can't have the center square of C intersecting C' and the center square of C' intersecting C (as I wave my hands about furiously). Then, if we have an uncountable collection of disjoint crosses {C}, we get an uncountable collection of disjoint sets {UC}, each with positive measure, which is impossible since their union would have infinite measure*, while being contained in [0,W]x[0,H]x[0,H]. *Lemma: An uncountable set {xj : j [in] J} of positive numbers has an infinite sum. (The sum of an infinite set is the supremum of all sums of finitely many elements.) Proof: Let Sn = { j | xj > 1/n }; then J = [cup]Sn, a countable union, so some Sn must be uncountable, so for any N, we can take at least N*n elements from there, giving a sum > N.
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« Last Edit: Nov 7th, 2003, 12:01am by Eigenray » |
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Eigenray
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Re: Uncountable Copies
« Reply #9 on: Nov 7th, 2003, 12:53am » |
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The reason that was so messy was because I didn't have a positive lower bound on the radii. With that in mind, I present a much simpler, and more rigorous, proof: Take an uncountable collection of disjoint crosses S; let Sn = { C [in] S : radius(C) > 1/n }. Then S = [cup]Sn, a countable union, so some Sn must be uncountable; let r=1/n. We may therefore assume that all the crosses have radius greater than r. Now, to each cross C, associate the open ball BC with the same center as C and radius r/2. These balls are disjoint: If there's a point in BC and BC', then the centers of C and C' are less than r apart, by the triangle inequality, but it's impossible to have two disjoint crosses of radii > r with centers < r apart. (If we fix one cross, the center p of the second cross must lie within the circle of radius r centered at the first cross, but drawing the perpendiculars from p to the two arms of the first cross that p lies between, those two lines extend an arc of 90 degrees, which means there must be an arm of the second cross between them, but that arm must intersect one of the two arms of the first cross. Or just draw it and it's obvious.) Now we have an uncountable union of disjoint open balls, which is impossible (each one contains a rational point, say).
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« Last Edit: Nov 7th, 2003, 1:13am by Eigenray » |
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Icarus
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Re: Uncountable Copies
« Reply #10 on: Nov 7th, 2003, 3:24pm » |
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That proves it! In the book I got this from, Martin Gardner said that there was no known criterion for deciding whether a given figure could be copied uncountably many times or not.
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rmsgrey
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Re: Uncountable Copies
« Reply #11 on: Nov 14th, 2003, 7:23am » |
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I think I have that book somewhere... As I recall, the proof given for uncountable numbers was by ::establishing a one-one mapping between shapes and (real) points on a finite line segment - radii for the closed figures and horizontal across a gap for the wavy lines::
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Icarus
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Re: Uncountable Copies
« Reply #12 on: Nov 14th, 2003, 3:39pm » |
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By a "criterion", I was refering to a quick test one could apply to determine if a given figure could be uncountably copied or not. Instead, it must be examined on a case-by-case basis. Any figure that can be magnified/contracted or translated over even a small range without intersection clearly is in the uncountable category.
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? " - Anonymous
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