Since geeky science questions seem to be today's fashion...
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That would be a fractal, such as this Sierpinski triangle[^]. ADDED Although not many would agree they have 2 or 3 (or any integer) number of dimensions... :)
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modified on Thursday, May 6, 2010 6:06 PM
Very good! I missed your response when I posted my answer below. You are also correct, I should have said that the shape could be rendered geometrically in a plane or a space. But that might have given away too much.
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Luc Pattyn wrote:
you mean a hyperbole
NOW who's using hyperbole. ;P
Luc Pattyn wrote:
takes up a finite amount of space
Depends on how you define space. If you consider bounding rectangle, it takes up infinite space. If you consider the bound between the equation and the axes, I'm pretty sure that area is finite (if I felt like polishing my calculus skills, I could probably calculate exactly how much that area is). If you consider the amount of volume the curve itself displaces, it would take up no space at all. :)
I don't think you need polish, a little push will suffice, hence: The integral of
1/xisln(x) + some constant, and your function is symmetrical around the first diagonal, so the integral from 1 to infinity would cover one quarter of the total area (ignoring signs), and that quarter is infinite as it equalsln(infinity)Sorry, I can't store it without damaging it. :)Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]
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Very good! I missed your response when I posted my answer below. You are also correct, I should have said that the shape could be rendered geometrically in a plane or a space. But that might have given away too much.
you also silently switched from edge to perimeter in your solution message; the edges get smaller and smaller after each iteration, it is the perimeter that grows to infinity. But that didn't stop me :laugh:
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you also silently switched from edge to perimeter in your solution message; the edges get smaller and smaller after each iteration, it is the perimeter that grows to infinity. But that didn't stop me :laugh:
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The solution is correct whether you use "edge" or "perimeter" as both approach infinity as the number of iterations increases. As I corrected myself below, not all edges diminish in size ;P
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The solution is correct whether you use "edge" or "perimeter" as both approach infinity as the number of iterations increases. As I corrected myself below, not all edges diminish in size ;P
Hmm. When you start with a finite triangle, and all you do is cut some edges in half, none of the edges will ever grow, let alone grow to infinity; the perimeter yes, the edges no. :)
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Hmm. When you start with a finite triangle, and all you do is cut some edges in half, none of the edges will ever grow, let alone grow to infinity; the perimeter yes, the edges no. :)
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Not so. You always have the outside edges of the starting triangle; those remain the same size. What you are adding are the edges created by cutting out the middle triangle, which are half the size of the triangle's outer edges.
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Q: Describe a 2 or 3 dimensional shape with an infinite edge and zero area, which takes up a finite amount of space.
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Hmm. When you start with a finite triangle, and all you do is cut some edges in half, none of the edges will ever grow, let alone grow to infinity; the perimeter yes, the edges no. :)
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As the number of edges approaches infinity, the sizes of those edges decreases toward 0, so they approach being a single edge. Or so, that's how I interpreted it.
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I'm surprised that no one has suggested fractals. The one I had in mind is called the Sierpinski Triangle. Observe: 1. Draw an equilateral triangle. 2. For each triangle, find the mid-point of its sides and draw lines connecting them, creating four triangles. 3. Remove the triangle in the middle, leaving three equilateral triangles connected at their vertices, each having three edges that are half of the starting triangle. 4. Go to step 2. The triangle starts with three edges of length x, so its total perimeter is 3x. After the first iteration, the shape has nine edges -- three on each of the three triangles -- each of which have a length of x/2, meaning the shape's total perimeter is 9x/2, longer than what we started with. After the first iteration, the area is 3/4 what it was before. As you continue with more iterations, the number of edges increases without bounds, and so does the shape's perimeter. As the number of iterations approaches infinity, so do the number of edges and, consequently, the length of its perimeter. Also, each iteration decreases the area geometrically: as the number of iterations approaches infinity, the area bound by the perimeter approaches zero. The shape itself never exceeds the bounds set by the starting triangle, which makes it finite. With a variant called the Sierpinski Carpet, you start with a single square, divide it into nine squares, remove the center one and repeat. Again, the number of edges and the perimeter approach infinity while the area bound by the perimeter approaches zero. There are also 3-d versions of these shapes, called sponges, which start with a tetrahedron and a cube respectively. Added: Oops, a bit of a screw up. The number of edges is doubled, not tripled; I was counting the starting edges of the triangle twice. The increase in the length of the perimeter is still 9x/2, however, as you have the three starting edges (x + x + x = 3x or 6x/2) plus the three edges of the now empty center triangle (x/2 + x/2 + x/2 = 3x/2)
modified on Thursday, May 6, 2010 8:31 PM
Here's another way to visualize it. A fractal that starts as a circle. Make a slightly smaller circle that is inside the larger circle. Line them up so the top of the smaller circle is touching the top of the larger circle. Now, the edge of that shape includes both circles (if you were to trace a pencil around the outer circle then continue on, fluidly to the inner circle). Continue making smaller circles in the smallest circle, and you end up with an infinite edge.
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mobius
---------------------------------------------------------- Lorem ipsum dolor sit amet.
That has an area.
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As the number of edges approaches infinity, the sizes of those edges decreases toward 0, so they approach being a single edge. Or so, that's how I interpreted it.
That is what I suggested here[^] but I didn't expect you to buy it. :)
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Not so. You always have the outside edges of the starting triangle; those remain the same size. What you are adding are the edges created by cutting out the middle triangle, which are half the size of the triangle's outer edges.
yes the number of edges increases; none of the new edges is larger than the three original ones. Or are we discussing something other than the Sierpinski Triangle now? You said the shape remained finite as it was limiting itself to the bounds of the original triangle. :)
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Hmm. When you start with a finite triangle, and all you do is cut some edges in half, none of the edges will ever grow, let alone grow to infinity; the perimeter yes, the edges no. :)
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Hmmm, after drawing it out, I think you are correct. While the edges get smaller, all the angles stay the same. They never really blend into a single edge. On the other hand, I still haven't been able to poke any holes in my circle fractal proposal. Feel free to do that for me. ;P
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Hmmm, after drawing it out, I think you are correct. While the edges get smaller, all the angles stay the same. They never really blend into a single edge. On the other hand, I still haven't been able to poke any holes in my circle fractal proposal. Feel free to do that for me. ;P
aspdotnetdev wrote:
I still haven't been able to poke any holes in my cirlce fractal proposal
The way I see it, it has: - no straight parts at all; - a single edge, infinitely long; - a real, non-zero, area (using the even/odd rule for inside/outside), intuitively I would say half that of the original circle. So I wouldn't call it a fractal, I wouldn't call it a solution to the problem in the OP, and frankly I wouldn't call it very pretty either. It is original though, I haven't seen it before. :)
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I'm surprised that no one has suggested fractals. The one I had in mind is called the Sierpinski Triangle. Observe: 1. Draw an equilateral triangle. 2. For each triangle, find the mid-point of its sides and draw lines connecting them, creating four triangles. 3. Remove the triangle in the middle, leaving three equilateral triangles connected at their vertices, each having three edges that are half of the starting triangle. 4. Go to step 2. The triangle starts with three edges of length x, so its total perimeter is 3x. After the first iteration, the shape has nine edges -- three on each of the three triangles -- each of which have a length of x/2, meaning the shape's total perimeter is 9x/2, longer than what we started with. After the first iteration, the area is 3/4 what it was before. As you continue with more iterations, the number of edges increases without bounds, and so does the shape's perimeter. As the number of iterations approaches infinity, so do the number of edges and, consequently, the length of its perimeter. Also, each iteration decreases the area geometrically: as the number of iterations approaches infinity, the area bound by the perimeter approaches zero. The shape itself never exceeds the bounds set by the starting triangle, which makes it finite. With a variant called the Sierpinski Carpet, you start with a single square, divide it into nine squares, remove the center one and repeat. Again, the number of edges and the perimeter approach infinity while the area bound by the perimeter approaches zero. There are also 3-d versions of these shapes, called sponges, which start with a tetrahedron and a cube respectively. Added: Oops, a bit of a screw up. The number of edges is doubled, not tripled; I was counting the starting edges of the triangle twice. The increase in the length of the perimeter is still 9x/2, however, as you have the three starting edges (x + x + x = 3x or 6x/2) plus the three edges of the now empty center triangle (x/2 + x/2 + x/2 = 3x/2)
modified on Thursday, May 6, 2010 8:31 PM
Whoa, whoa wait a second. Peano curve[^] beats serpinski triangle for your criteria by a wide margin.
I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon
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aspdotnetdev wrote:
I still haven't been able to poke any holes in my cirlce fractal proposal
The way I see it, it has: - no straight parts at all; - a single edge, infinitely long; - a real, non-zero, area (using the even/odd rule for inside/outside), intuitively I would say half that of the original circle. So I wouldn't call it a fractal, I wouldn't call it a solution to the problem in the OP, and frankly I wouldn't call it very pretty either. It is original though, I haven't seen it before. :)
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Luc Pattyn wrote:
a real, non-zero, area (using the even/odd rule for inside/outside)
Are you saying that the outer crescent (for want of a better term) is solid, the one inside that is not, the one inside that is, and so on? Not what I was thinking... I was thinking more like a spool of wire... or a hose that you are wrapping up by looping it around your hand and the part of the hose you've already wrapped. Suppose the hose is infinitely thin and you are wrapping in reverse (starting with the largest circle).
Luc Pattyn wrote:
So I wouldn't call it a fractal
By what definition? Start by viewing the whole thing: you see a circle with smaller nested circles. Now, zoom in with top of the "camera" staying focused on the top of the circle. You now see a circle with smaller nested circles. Continue zooming in and it looks pretty much the same, no matter how far you zoom in.
Luc Pattyn wrote:
I wouldn't call it very pretty either
Pft, whatever... I'm going to draw it and make it my desktop background. ;P
Luc Pattyn wrote:
I wouldn't call it a solution to the problem in the OP
Seems to satisfy all the requirements to me. How about a spiral (like the kind you get hypnotized with)? Only with a different function to determine the rate of shrinkage. If you still aren't convinced, just look here and I think you'll come to your senses eventually. You will come to your senses... you will come to your senses... you will come to your senses... ;)
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I don't think you need polish, a little push will suffice, hence: The integral of
1/xisln(x) + some constant, and your function is symmetrical around the first diagonal, so the integral from 1 to infinity would cover one quarter of the total area (ignoring signs), and that quarter is infinite as it equalsln(infinity)Sorry, I can't store it without damaging it. :)Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]
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We're gonna need a bigger box.
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Luc Pattyn wrote:
a real, non-zero, area (using the even/odd rule for inside/outside)
Are you saying that the outer crescent (for want of a better term) is solid, the one inside that is not, the one inside that is, and so on? Not what I was thinking... I was thinking more like a spool of wire... or a hose that you are wrapping up by looping it around your hand and the part of the hose you've already wrapped. Suppose the hose is infinitely thin and you are wrapping in reverse (starting with the largest circle).
Luc Pattyn wrote:
So I wouldn't call it a fractal
By what definition? Start by viewing the whole thing: you see a circle with smaller nested circles. Now, zoom in with top of the "camera" staying focused on the top of the circle. You now see a circle with smaller nested circles. Continue zooming in and it looks pretty much the same, no matter how far you zoom in.
Luc Pattyn wrote:
I wouldn't call it very pretty either
Pft, whatever... I'm going to draw it and make it my desktop background. ;P
Luc Pattyn wrote:
I wouldn't call it a solution to the problem in the OP
Seems to satisfy all the requirements to me. How about a spiral (like the kind you get hypnotized with)? Only with a different function to determine the rate of shrinkage. If you still aren't convinced, just look here and I think you'll come to your senses eventually. You will come to your senses... you will come to your senses... you will come to your senses... ;)
OK, I understood your concoction as a two-dimensional drawing where you have N "top-connected" circles of decreasing diameter (with N increasing without bound), and you travel to the next circle every time you reach the top, until you reached the smallest one, then you step to the outer one again. So that is a closed line, there is nothing to zoom that would keep the overall impression, and the area is a half circle. Not a fractal. If you want to visualize it as a spiral, i.e. each next circle moves you a bit in the third dimension, then you have somewhat of a fractal effect as you can move forward over the pitch of the spiral, and zoom in a bit to compensate for the decreasing diameter. But now it is just a spiral, it spans an infinite z-axis. So it is not contained in a finite space. (In fact it resembles a worm hole in Stargate-1). All my senses and I agree yours is not a space-limited fractal, and not the right answer to the OP. But I agree you might still like it as a wall paper. You, not me. :laugh: This[^] might offer some consolation.
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Whoa, whoa wait a second. Peano curve[^] beats serpinski triangle for your criteria by a wide margin.
I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon
In terms of area, remember that
null != 0;:laugh: -
OK, I understood your concoction as a two-dimensional drawing where you have N "top-connected" circles of decreasing diameter (with N increasing without bound), and you travel to the next circle every time you reach the top, until you reached the smallest one, then you step to the outer one again. So that is a closed line, there is nothing to zoom that would keep the overall impression, and the area is a half circle. Not a fractal. If you want to visualize it as a spiral, i.e. each next circle moves you a bit in the third dimension, then you have somewhat of a fractal effect as you can move forward over the pitch of the spiral, and zoom in a bit to compensate for the decreasing diameter. But now it is just a spiral, it spans an infinite z-axis. So it is not contained in a finite space. (In fact it resembles a worm hole in Stargate-1). All my senses and I agree yours is not a space-limited fractal, and not the right answer to the OP. But I agree you might still like it as a wall paper. You, not me. :laugh: This[^] might offer some consolation.
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Luc Pattyn wrote:
OK, I understood your concoction as a two-dimensional drawing where you have N "top-connected" circles of decreasing diameter (with N increasing without bound), and you travel to the next circle every time you reach the top
Yep.
Luc Pattyn wrote:
until you reached the smallest one, then you step to the outer one again
There is no "smallest" one. It keeps going, forever.
Luc Pattyn wrote:
there is nothing to zoom that would keep the overall impression
I used my excellent skills as an artist to make this rendition of what I was thinking. Suppose you started out zoomed to view the full shape. Then, you zoom so that you can only view the part of the shape composed of light grey circles. Then you keep zooming in that fashion. You'll always see circles within circles, all intersecting at the top of the view. It is this self-similarity that I used to define this as a fractal.
Luc Pattyn wrote:
not the right answer to the OP
Nonsense!
Luc Pattyn wrote:
This[^] might offer some consolation.
Sorry it took so long to respond... I just woke up from a seizure induced by that crazy image. ;P
