4 ms·

On a less trollish note, I had always been puzzled by the related question you discuss: why Manhattan Distance does not converge to the Euclidean distance. I'm

by neilk 16y ago

On a less trollish note, I had always been puzzled by the related question you discuss: why Manhattan Distance does not converge to the Euclidean distance. I'm pretty uneducated about math, so please let me know what trap I'm falling into. Your answer isn't yet helping me because you just note that Manhattan Distance remains the same no matter how many twists and turns there are. Yes, this is the premise of the question! But turn up the twistiness to infinity. Now the Manhattan Distance line is identical to the hypotenuse. There is no point on the Manhattan distance line isn't also on the hypotenuse and vice versa. They both have an infinite number of points, of the same aleph-number, I think. You mention a hypothetical path that doubled back on itself -- it is easy to see why that would result in a different answer. But the Manhattan Distance line is not doubling back on itself. Every point is a step towards the goal, and it doesn't cross itself. All of its deviations from the simple hypotenuse are infinitely small. So why are the distances still different?

simcop2387 16y ago
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.
RiderOfGiraffes 16y ago
Perhaps I could more easily answer your question if you tried to explain why you think the length of the zig-zag line should approach the length of the straight line. It's zig-zagging - why should it end up the same length? You say that in the Manhattan Distance every point is a step towards the goal, but it's not directly towards the goal, therefore you always get a bit more than you want. The number of points isn't relevant - the Cantor Set (also known as Cantor Dust) is an uncountably large subset of the real line, and yet its length is zero. I guess the only insight I can offer at this point is that length has nothing to do with points.
- jules 16y ago
  I think he is not asking what he wants to know. He wants to know how we know why Archimedes' solution does converge to the right length. Couldn't there be a problem with his solution that we're just not seeing yet? i.e. what makes Archimedes right and the troll wrong?
  - caf 16y ago
    It's because the tangents to Archimedes' polygons also approach the tangents to the circle. (This is the "velocity vector" convergence condition mentioned by cousin_it above).
    - jules 16y ago
      That doesn't explain why. Why is it that if tangents approach the circle (whatever that means precisely) that the length of the approximation approaches the length of the circle?
      - caf 16y ago
        Loosely speaking, the total path length is defined as the sum of a bunch of infintesimal segments that have both position and direction. For the sum (integral) of those infintesimals to approach the sum of the similar infintisemals around the actual circle, those of the approximation path have to approach those of the circle in the limit, and of course this means in both position and direction.
ced 16y ago
Playing with infinities is dangerous business, and it's very easy to yield paradoxes like the OP. But note that before the 20th century, mathematicians like Gauss would avoid playing with such ill-defined objects, for precisely this reason[1]. Their rule was that you should first derive an expression in the finite case, then take the limit to infinity at the end. In this case, it's mathematically simple that a Manhattan line should have the same length, no matter what N (the number of steps) is. [1] http://en.wikipedia.org/wiki/Russell%27s_paradox http://en.wikipedia.org/wiki/Russell%27s_paradox for instance. The Banach-Tarski paradox is also worthwhile: http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox I think nowadays most mathematicians are "comfortable" with such aberrant results, but that was not always the case. And there's still a chance that future generations of mathematicians will dismiss this work.
deleted 16y ago
[deleted]
cousin_it 16y ago
There's no such thing as "infinitely small" deviation. The Manhattan Distance limit line is the hypotenuse, and its length is the length of the hypotenuse. The question is not about the limit line, which does not exhibit any weird behavior. The question is about the sequence of approximations to that line: why the length of the approximations doesn't converge to the length of the limit. And the answer is that it doesn't have to, because even though the approximations are very similar to the limit line in one respect (geometric closeness), they are all very different from it in another respect (directions and angles of travel). If we had a sequence of approximations whose direction of travel converged correctly, the length would converge correctly too.

by neilk 16y ago

simcop2387 16y ago

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.

RiderOfGiraffes 16y ago

Perhaps I could more easily answer your question if you tried to explain why you think the length of the zig-zag line should approach the length of the straight line. It's zig-zagging - why should it end up the same length? You say that in the Manhattan Distance every point is a step towards the goal, but it's not directly towards the goal, therefore you always get a bit more than you want. The number of points isn't relevant - the Cantor Set (also known as Cantor Dust) is an uncountably large subset of the real line, and yet its length is zero. I guess the only insight I can offer at this point is that length has nothing to do with points.

jules 16y ago
I think he is not asking what he wants to know. He wants to know how we know why Archimedes' solution does converge to the right length. Couldn't there be a problem with his solution that we're just not seeing yet? i.e. what makes Archimedes right and the troll wrong?
- caf 16y ago
  It's because the tangents to Archimedes' polygons also approach the tangents to the circle. (This is the "velocity vector" convergence condition mentioned by cousin_it above).
  - jules 16y ago
    That doesn't explain why. Why is it that if tangents approach the circle (whatever that means precisely) that the length of the approximation approaches the length of the circle?
    - caf 16y ago
      Loosely speaking, the total path length is defined as the sum of a bunch of infintesimal segments that have both position and direction. For the sum (integral) of those infintesimals to approach the sum of the similar infintisemals around the actual circle, those of the approximation path have to approach those of the circle in the limit, and of course this means in both position and direction.

ced 16y ago

Playing with infinities is dangerous business, and it's very easy to yield paradoxes like the OP. But note that before the 20th century, mathematicians like Gauss would avoid playing with such ill-defined objects, for precisely this reason[1]. Their rule was that you should first derive an expression in the finite case, then take the limit to infinity at the end. In this case, it's mathematically simple that a Manhattan line should have the same length, no matter what N (the number of steps) is. [1] http://en.wikipedia.org/wiki/Russell%27s_paradox http://en.wikipedia.org/wiki/Russell%27s_paradox for instance. The Banach-Tarski paradox is also worthwhile: http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox I think nowadays most mathematicians are "comfortable" with such aberrant results, but that was not always the case. And there's still a chance that future generations of mathematicians will dismiss this work.

deleted 16y ago

[deleted]

cousin_it 16y ago

There's no such thing as "infinitely small" deviation. The Manhattan Distance limit line is the hypotenuse, and its length is the length of the hypotenuse. The question is not about the limit line, which does not exhibit any weird behavior. The question is about the sequence of approximations to that line: why the length of the approximations doesn't converge to the length of the limit. And the answer is that it doesn't have to, because even though the approximations are very similar to the limit line in one respect (geometric closeness), they are all very different from it in another respect (directions and angles of travel). If we had a sequence of approximations whose direction of travel converged correctly, the length would converge correctly too.