I'm watching this video by Dr. Becky about the history of exoplanet discoveries and I was amused by the auto captions interpreting "Anglada-Escudé" (the last name of an astronomer) as "GLaDOS CUDA".

A little bit a mathematical question: Do we actually need the supremum in this definition, wouldn't the maximum be applicable? codegolf.meta.stackexchange.com/a/18190/24877

@flawr I suspect not, because the limit approaching the open end of a discontinuity seems to be smaller than the value at the dicontinuity.

Actually here is a simpler approach

I think we can always just iterate over all pairs of "gaps" between two points in the set {s_1,...,s_n} and find the maximum in there, and then take the maximum over all those

u and v will always be members of the sequence, if they were not then you could find a u and a v with a larger result by taking the convex hull of the members u and v capture.

This leaves a finite number of options.

For finite sets sup = max

@SriotchilismO'Zaic I don't agree: Assume [a,b] = [0,1] and s_1 = 1/3, s_2 = 2/3. Then the D_2 is at least 2/3 namely for [u,v] = [0,1]

Oh I was misunderstanding the definition.

Wait I am still misunderstanding the definition?

@SriotchilismO'Zaic I cannot answer that question:P

Wouldnit [u,v]=[1/3,2/3] be higher?

It was not a question, I just used ? to indicate rising tone.

@SriotchilismO'Zaic then N(u,v)/n = 2/2 = 1 and (v-u)/(b-a) = 1/3 and we again a value of 2/3

I think my previous calculation was wrong

for [u,v] = [0,1] we get a value of 0

actually for [u,v] = [a,b] we always get a value of 0

Ok that lines up with my expectation.

Ah now I think I see why we need the supremum:

Do you still disagree with that statement?

consider the example [a,b] = [0,1] and s_1 = 0, s_2 = 0.5, s_3 = 1

I'm not seeing the supremum being needed in that example. It looks to me like the maximum is attained.

then as u goes to 0 and simulatenously v goes to 1, the value goes to |1/3 - (v-u)| = 2/3

@SriotchilismO'Zaic for what [u,v]?

u=1/2, v=1/2

Oh wait that is wrong

I do not understand this metric.

@SriotchilismO'Zaic that should result in 1/3, right?

It would, I was optimizing a differnet metric

@SriotchilismO'Zaic well it actually only makes sense for infinite sequences, but it is harder to deal with those:)

It really makes less sense for infinite sequences

Since reordering them changes the result of the metric

I'm not convinced that this is the case, and even if it was the case I suspect then the sequence would also intuitively look less "uniformly distributed"

Here is an example: Take a sequence that has a discrepancy of zero

Let us have two subsequences x_i and y_i such that x_i is the members of the sequence less than 1/2 and y_i is the other ones.

Let us make a new sequence s' which is two mbembers of x_i and then one y_i etc. preserving the order of x_i and y_i.

This new sequence has a non-zero discrepancy. It should be 1/6.

@SriotchilismO'Zaic Why that?

(I would intuitively expect this sequence still to have D = 0.)

I got the number wrong it is 1/6

As n goes to infty the interval [1/2,1] will increasingly capture one third of the points.

But will always be half of the interval.

Oh wait I didn't read the two members, I have to rethink. In this case the asymptotic density would be twice as high in [0,0.5] as in [0.5,1].

Yes.

You can infact reorder discrepancy zero sequences to get discrepancy 1 sequences.

Ok yes, I agree with your argument. And in this case the sequence really does intuitively look less uniform.

So I'm still not convinced it is a bad measure for measuring uniformness

To me it doesn't seem to be measuring any intuitive concept.

I mean a measure can be useful regardless htough.

I don't know whether it is equivalent, but it seems quite closely related: Take some bin size epsilon: If you consider the sequences of histograms of bin size epsilon (cosidering these histograms as piecewise constant function), then a for a discrepancy zero sequence, the sequence of histograms should converge to an constant functoin (= pdf of uniform distribution U[a,b]).

For all epsilons

This histogram idea is more intuitive for me and is how I would go about measuring "uniformness" of a sequence.

(But as I said, I don't know whether it coincides with the discrepancy definition.)

This might be a little weaker. It should be the same on rational intervals, but I don't see how that bridges over to real intervals.

By rational intervals I mean intervals of rational numbers.

Rather than intervals of rational length or intervals with rational endpoints.

Actually, your histogram defnition works for infinite intervals.

Also I don't know why I was hung up on irrational numbers. This should work for real numbers.

You do need to worry about when epsilon does not divide |b-a| though.

A slight tweaking can likely be made.

I think you could extend it to the same way you define Rieman integrals: You take the limit over the finite partitions of the interval. (As the length largest element of the partition goes to zero)

Here you'd just have to scale the numbers in each bin with the inverse of the length of the bin

Ok here is maybe an unintuitive example.

Let us have all of the the set of all binary fractions in some discrepancy zero sequence x_i.

Let us also have all of the ternary fractions in some discrepancy zero sequence y_i.

can you elaborate, what are binary/ternary fractions?

Yeah sure,

a binary fraction is a fraction where the denominator is a power of two

ternary fractions are fractions where the denominator is a power of three

Oh and the numerator is an integer in both cases.

Both of these are dense on the reals.

Ok, so far I follow

Let me back track a bit to make things simpler, x_i is on the interval [0,1/2] and y_i is on the interval [1/2,1], all else is the same.

Now if we take a new sequence s_i which just alternates between x_i and y_i.

On the interval [0,1] this has discrepancy zero.

So far I think I agree

this is pretty unintuive

I mean binary fractions are more dense than ternary fractions it just seems like this should be skewed to the left.

(We would probably first have to assume that x_i and y_i without the restriction to [0,1/2] and [1/2,1] are discrepancy zero.)

@SriotchilismO'Zaic well they are equally dense:)

Yes in the sense that they are both dense

Thanks for the discussion, this gave me some more intuition about this definition!

I need to leave now:)

Have a good one

You too!

oh and I just found this thread about a new O(n * log(n)) integer multiplication algorithm: reddit.com/r/math/comments/dj5ygi/…

From this reddit comment.