Tuesday, 1 December 2015

Rectangular Circles - Yet Another Response to Mathematician

This is yet another response to Mathematician.  Here's what he wrote, interspersed with my responses (minor editing for format, and I've excised the final part that I've already addressed back in the comments section which can be read in its entirety here):

> At N=100, the 1/2 method does not have gaps or clumping.

The whole point of my "subintervals in intervals" example, was to show you that the problem of "gaps or clumping" is only a problem if you think it is. In the interval example, if you require that there is no "gaps or clumping" when N is finite, then the only possible answer is 0.

I think we agree that this is not reasonable at all. And that the most obvious way to select a subinterval will give gaps and clumping.

So how do you choose A PRIORI, in which contexts "gaps or clumping" are problematic, and in which contexts they are not?

It seems to me that you are blending the discussion about the disc and the discussion about the interval.

The "1/2 method" that I am talking about refers to the disc (and only the disc).  I am pretty certain that you are clear on this, but I want to be as totally certain as I can be.

I'm not as convinced as you are that the only possible proportion of subintervals greater than L/2 is zero when using a method that eliminates gaps and clumping.  What I am pretty sure of, however, is that if we looked at the distribution of subintervals and found that they were clustered around the ends of the interval and their lengths clustered around what could be described as "very very short" (much less than L/2), then we'd have reason to doubt how fair this distribution was.

Perhaps there's a good mathematical reason to not care about such clumping (and the implied "gap" between the ends of the interval in which the density of subintervals would dip), but don't you agree that using such a distribution would not meet the general understanding of "at random" - perhaps not even your own understanding of what "at random" would mean in this context?

> if distribution continued towards an as yet unknown value, or whether it still approaches zero

As far as I understand what you are trying to do, I'm pretty sure that it will approach 0.

I'm not sure that what you are doing proves anything at all, but that's another problem.

I wasn't trying to prove anything.  I was just pondering the puzzle that you presented.

> Perhaps it was not clear to you, but the "corrected" 1/3 method, ends up being the 1/2 method

No it was clear.

> And, no, I don't agree that it's the same thing

Let me repeat something for sake of clarity:
For any (c,θ) in [-1,1]x[0,pi], there exists a unique chord that is at distance c from the center, in direction θ.

The "1/2 method of selecting a chord", amounts to pick a couple (c,θ) uniformly in the rectangle [-1,1]x[0,pi]. Do we agree on that?

When you draw your picture to show "granularity", what you are doing is that you choose a θ, once and for all, and then you take 100 values of c that are evenly spaced in [-1,1].

What I'm suggesting is that you do the opposite: Choose a c, once and for all, and then take 100 values of θ that are evenly spaced in [0,pi].

In the end, this is exactly the same method, but you're not drawing the same picture. (In mathematical terms, you are just doing a projection on one of the coordinates)

This is possibly where the meat of the issue is.

I agree that for any (c,θ) in [-1,1]x[0,π], there exists a unique chord that is at distance c from the centre, in direction θ.  To be absolutely clear, I am interpreting this to mean that you are talking about a chord that is offset from the locus by c at its midpoint and that, therefore, the direction mentioned is the direction from the locus to that midpoint.

This is not what I thought you meant before.  I thought you meant to pick a point at c from the locus (direction irrelevant), and then consider the chords that pass through that point with gradients defined by θ.  You'd agree that such a scheme, picking a single value of c, won't give you ALL the chords (certainly not if you pick any value of c less than R, being the radius of the disc), right?

However, you seem to misunderstand my intention.  I made clear (somewhere, I can dig it up if you insist) that I was notionally selecting a single value of θ (direction from locus to midpoint) only because that single value can represent all possible values of θ.  The same applies when selecting a single Point 1 on the circumference in the 1/3 method.

I fully expect that, to get the ALL the chords, you’d have to consider all possible values of θ - in no way was I suggesting that I should "choose a value of θ, once and for all".

So, I understand that if someone foolishly suggested that we select a value of c, "once and for all", and then look at the chords at c given all possible values of θ (as a direction from the locus to the midpoint of a chord), then you'll never get ALL chords.  You'll get an infinite number of chords with precisely the same length but different gradients.

Perhaps I am still misunderstanding your point.  I think I must be, because I do not believe that you are this foolish (insert smile here to minimise any unintended offense).

I want to step back a bit to your question:

The "1/2 method of selecting a chord", amounts to pick a couple (c,θ) uniformly in the rectangle [-1,1]x[0,pi]. Do we agree on that?

I agree, with a minor reservation.  I'm a bit uncomfortable calling [-1,1]x[0,π] a "rectangle": that space represents a circle (hence my little joke in the title of this article).  However, I think I get what you mean - it's a useful way to visualise things for the purpose of considering a uniform distribution of values of c and θ.

What occurs to me is that this can be used in association with the 1/4 method.

My "fix" involved selecting a midpoint from this space (precisely like you seem to be suggesting), while the standard 1/4 method involves selecting from a reduced space.  I think it might be, notionally, a bit like this (think density rather than direct correspondence with values of θ):


I'm not suggesting that these are accurate representations of the shapes corresponding to the 1/3 and 1/4 methods, I just used a triangle for 1/3 and cut out circular chunks for 1/4 because it was convenient.  However, the concept does point towards the notion that the 1/2 and 1/4 methods are missing chords - and where they are missing from.

> because I am not focussed on how we select chords, I am focussed on ensuring that we have ALL chords (and where N is less than infinity, a representative sample of ALL chords).

Can you provide a single example of a chord that you can get with the 1/2 method, but that you cannot get with the 1/3 method?

See above.  Of course I can't point to a single example, which you would clearly realise, but I can (at least conceptually) show that there are fewer chords near the locus with the standard 1/3 and 1/4 methods than there are with the 1/2 method.

> Between -R and -R/2 and R/2 and R, there will be a decrease in the proportion of chords greater than sqrt(3)

You are apparently thinking that "c" should be taken in a predefined direction, and then choose another direction θ. It's not what I said. Just fix some c, once and for all, and then choose a bunch of θ, and then draw the chords corresponding to (c,θ).

So, when c is between, R/2 and R (and between -R and -R/2), the proportion of chords greater than sqrt(3) is 0. So the final answer is 1/2. (Which is absolutely not surprising because it's exactly the same method)

See above.  I think I've already addressed your "once and for all" objection, perhaps once and for all (but I am not holding my breath).

I don't know how you end up with 1/2 with what you've said here, but I do agree that all my methods - the standard 1/2 method, the "corrected" 1/3 method and the " corrected " 1/4 method - are effectively (and exactly) the same method.

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I note that there might have been confusion about my use of the word "fixed" when I mean "corrected".  When I used "fixed" previously, I did not mean "never to be changed" as in "fixed in stone".  I meant "fixed" as in "my keyboard is broken, I am going to have to get it either fixed or replaced".