4 ms·

The reason it won't converge is that a very large number of the points will ALWAYS lie off of the line. While you might be able to make something that looks li

The reason it won't converge is that a very large number of the points will ALWAYS lie off of the line. While you might be able to make something that looks like a line on a computer screen when you make it smaller and smaller segments of manhattan distance. it will in fact only ever be able to intersect the real line (they are infinitely thin remember) at N+1 points at most (where N is the number of turns you make when following it along the grid in manhattan distance). because of this you will always have a very large portion of your approximation lying off of the line.

ewjordan 16y ago
Put another way, the right angles approximation as N->infinity puts a countable infinity of points exactly on the line. The true line has an uncountable infinity of points on the line, and the difference between an uncountable infinity and a countable one is an uncountable infinity, so we're still way the hell off.
- RiderOfGiraffes 16y ago
  That's irrelevant. + The limit has all the points on the line. + Uncountable sets can be of length zero + There are families of curves where at no stage do any of the points lie on the circle, and yet the limit is the circle, and the limit of the length is the length of the circle. In short, your comment isn't really explaining anything. ADDED IN EDIT: Whoa - cool - a downvote! Please, let me know where I'm wrong. I thought I'd explained clearly why the comment was wrong, and I have explained elsewhere more about what's really going on. No doubt you, the downvoter, have moved on now, but whoever you are, I'd love more information as to what you think I've got wrong. Thanks.
RiderOfGiraffes 16y ago
What you (simcop2387) say doesn't help, as I can produce curves that do as you describe and yet they do converge. It's possible to have convergence even when all the points lie off the line (in the only sense I can understand you to mean) so your comment isn't really an explanation.
neilk 16y ago
> The reason it won't converge is that a very large number of the points will ALWAYS lie off of the line. No they won't. It moves zero distance horizontally and zero distance vertically. All the points of the infinitely zigzagging line are on the hypotenuse. Of course, now we're back to Zeno; if it's not moving off the line, and it only makes progress when it's off the line, how can it get anywhere?
- RiderOfGiraffes 16y ago
  For every fractally-type enclosing shape, yes they will. And when you talk about the limit, then you have a circle. You can't talk about moving "zero distance horizontally and zero distance vertically" becuase then, as you rightly say, we're back to Zeno. You don't have an infinitely zigzagging line. That way of trying to think about things is a dead end, and unhelpful. Better answers are given elsewhere in this thread.

ewjordan 16y ago

Put another way, the right angles approximation as N->infinity puts a countable infinity of points exactly on the line. The true line has an uncountable infinity of points on the line, and the difference between an uncountable infinity and a countable one is an uncountable infinity, so we're still way the hell off.

RiderOfGiraffes 16y ago
That's irrelevant. + The limit has all the points on the line. + Uncountable sets can be of length zero + There are families of curves where at no stage do any of the points lie on the circle, and yet the limit is the circle, and the limit of the length is the length of the circle. In short, your comment isn't really explaining anything. ADDED IN EDIT: Whoa - cool - a downvote! Please, let me know where I'm wrong. I thought I'd explained clearly why the comment was wrong, and I have explained elsewhere more about what's really going on. No doubt you, the downvoter, have moved on now, but whoever you are, I'd love more information as to what you think I've got wrong. Thanks.

RiderOfGiraffes 16y ago

What you (simcop2387) say doesn't help, as I can produce curves that do as you describe and yet they do converge. It's possible to have convergence even when all the points lie off the line (in the only sense I can understand you to mean) so your comment isn't really an explanation.

neilk 16y ago

> The reason it won't converge is that a very large number of the points will ALWAYS lie off of the line. No they won't. It moves zero distance horizontally and zero distance vertically. All the points of the infinitely zigzagging line are on the hypotenuse. Of course, now we're back to Zeno; if it's not moving off the line, and it only makes progress when it's off the line, how can it get anywhere?

RiderOfGiraffes 16y ago
For every fractally-type enclosing shape, yes they will. And when you talk about the limit, then you have a circle. You can't talk about moving "zero distance horizontally and zero distance vertically" becuase then, as you rightly say, we're back to Zeno. You don't have an infinitely zigzagging line. That way of trying to think about things is a dead end, and unhelpful. Better answers are given elsewhere in this thread.