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Blatm wrote:Spoiler:
Qaanol wrote:The Sierpinski Carpet is the set of points (x, y) such that in base 3, for each integer k, the k'th digit of x and the k'th digit of y are not both 1.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
afarnen wrote:EDIT: Although I guess since, some points would require an infinite amount of iterations to be proven not on the carpet, such as (1/3, 1/3) (the top-left corner of the largest square), there is a clearer notion of points "on" the carpet. Is that what is meant? Am I totally off-track here?
jaap wrote:afarnen wrote:There is no iteration where (1/3,1/3) is removed, therefore the carpet (the limit of those iterations, or really the intersection of all those iterations) will still contain that point.
afarnen wrote:So that mean only points which are on the edges of square holes are actually on the carpet? Otherwise it can be removed in a definite amount of iterations.
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
EricSeverson wrote:Is there any way of identifying the points (such as 1/4) that stay in the carpet even though they are not on an edge?
Is there a way to mathematically prove that 1/4 will always stay on the carpet?
You need either that the kth digit of x and y are never both 1, or that they are both one and one of them then has every digit subsequent digit be 0.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
doogly wrote:Oh yea, obviously they wouldn't know Griffiths from Sakurai if I were throwing them at them.
EricSeverson wrote:But are there any other diagonal lines contained on the carpet that have slopes other than +/- 1/2 or 2 ?
EricSeverson wrote:Also, are there any points contained on the carpet that are not contained on a line through the carpet?
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
phlip wrote:(1/8,3/8) is on the carpet, but isn't on any horizontal or vertical line segment... that's (0.\overline{01}_3, 0.\overline{10}_3). I'm not sure about diagonal line segments, though... I don't really have a good grasp of what they look like.
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
phlip wrote:Do we know these are the only gradients we need to concern ourselves with? Or is there still the possibility of others?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
quintopia wrote:I highly doubt there are others, almost as much as I doubt I could prove it. I think the reason 1/2 works is because it is what the sum of powers of 1/3 converge to.
But if there are no others then, surprise surprise, this puzzle is solved.
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
jestingrabbit wrote:Gradient 1 or -1 starting from the midpoints are also good.
quintopia wrote:But I don't understand how if a point does not fall on any connected line segment, there could still be a path between any two points in the set. Perhaps there are some other gradients we missed.
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
phlip wrote:And anyway, we haven't found a point that doesn't fall on any connected line segment... just specifically the few gradients we've tried.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
jestingrabbit wrote:In general, I expect that we have to have an infinite number of straight segments connecting any two points, with a big straight bit in the middle, followed by smaller and smaller straight bits, getting closer and closer to the points in question.
quintopia wrote:jestingrabbit wrote:In general, I expect that we have to have an infinite number of straight segments connecting any two points, with a big straight bit in the middle, followed by smaller and smaller straight bits, getting closer and closer to the points in question.
Just like road trips in the real world. . .
However, it's almost deceiving when you put it this way, since for some points the straight bits will get infinitely short. And at this point, we aren't really looking at a straight bit at all, but rather a curve of some sort. I suspect it will be a smooth curve whose tangent approaches one of the gradients that works, and at the precise moment it reaches that gradient, it will have arrived at some point that's on a line of that gradient and so we straighten out and follow a piecewise linear function like you said for some time, and then optionally we follow a curve again at the end.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
doogly wrote:Oh yea, obviously they wouldn't know Griffiths from Sakurai if I were throwing them at them.
skeptical scientist wrote:Can anyone make anything of my above proof that the carpet is path-connected? Should I try explaining it better? I'm worried that it's completely incomprehensible. Would that I had you all in a room with me and a blackboard...
quintopia wrote:Given a point's single-number representation, what is the shortest distance from it to the lower left corner?
dedalus wrote:Seeing as for most of these rectangles the width would come very close to zero for the full carpet, wouldn't it be fairly accurate to say that the shortest path IS that path?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
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