Talk:Halton sequence

What is the distinction between pseudo-random number sequences and quasi-random sequences? It might be good to explain it. Andris 22:52, May 2, 2004 (UTC)

pseudo-random numbers perform like random numbers. quasi random numbers perform better than random numbers. Both are created deterministically.

Math images?

Why does this article use images instead of the <math></math> syntax? -- Cyrius|&#9998 01:08, May 3, 2004 (UTC)

Two Articles with the same content

Why do you start a new article on the Halton sequence ?

It is already mentioned in the article on the Van der Corput sequence. I do no claim that the article on the Van der Corput sequence is written well. I think, it should be considered as a starting point on this subject. However, I dont think that we should do the same work twice.

Best wishes, Klaus.

Rewrite needed

The following is an edited copy (some parts omitted) of a discussion on Wikipedia:Reference desk archive/Mathematics/2006 August 14#Binary interval-filling sequence:

Binary interval-filling sequence

Does the sequence 0, 1, 1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, 1/16, ... (or some sensible permutation thereof) have a name? Fredrik Johansson 16:30, 14 August 2006 (UTC)[reply]

...

This is known as the (binary) van der Corput sequence, after the Dutch mathematician J. G. van der Corput who described it in 1935. —Ilmari Karonen (talk) 11:09, 15 August 2006 (UTC)[reply]

Note that the binary vdC sequence is a "sensible permutation" of the original sequence. --Lambiam ^Talk 18:29, 15 August 2006 (UTC)[reply]

Hmm -- as far as I can tell, this is the same thing as the binary Halton sequence. Should the articles be merged? Or is there a distinction I'm missing? --Trovatore 20:15, 15 August 2006 (UTC)[reply]

I find that article quite incomprehensible. Where I come from, sequences are not particularly active; they do not "fill space" or "divide space into segments", nor do they "fill in empty spaces" or "use cycles" that in turn "place draws" (whatever these are) in segments. What I think I do understand is that there are several Halton sequences, all of which have finite lengths. There is an "original" Halton sequence, but it is not explained what it is and how it relates to the other ones. A small example might have worked wonders. In any case, the sequence we have here is infinite. --Lambiam ^Talk 23:23, 15 August 2006 (UTC)[reply]

This paper: (P. L'Ecuyer and C. Lemieux "Recent Advances in Randomized Quasi-Monte Carlo Methods", in Modeling Uncertainty: An Examination of Stochastic Theory, Methods, and Applications, M. Dror, P. L'Ecuyer, and F. Szidarovszki, eds., Kluwer Academic Publishers, 2002, 419–474.), accessed online from [1], states: "Halton sequence This sequence was introduced in 1960 by Halton [37] for constructing point sets of arbitrary length, and is a generalization of the one-dimensional van der Corput sequence [86]." In the definition given there, it is clear that there is an (infinite) Halton sequence of points in (0,1)ⁿ for all finite dimensions n > 0, and that the one-dimensional Halton sequence is the van der Corput sequence. I can't access the original Halton paper (J.H. Halton. "On the efficiency of certain quasi-random sequences of points in evaluating multi-dimensional integrals". Numerische Mathematik, 2:84--90, 1960.) to verify that this conforms to the original definition. In any case, there appears to be a distinction, and my advice is to keep the articles separate, although they should of course cross-refer. --Lambiam ^Talk 23:41, 15 August 2006 (UTC)[reply]

Ah, got it: So the vdC sequence is the one-d H sequence. The Halton sequence article needs a serious rewrite, and someone should write up Sobol sequence too (currently a redirect). --Trovatore 23:47, 15 August 2006 (UTC)[reply]

Right. I'm not really an expert on low-discrepancy sequences, but from what I gather the van der Corput sequence is more or less the granddaddy of them all, with the Halton and Hammersley sequences being generalizations of the van der Corput sequence to multiple dimensions. —Ilmari Karonen (talk) 00:40, 16 August 2006 (UTC)[reply]

Why a rewrite is needed:

The present article is incomprehensible.
It does not show how Halton sequences relate to the van der Corput sequences.

--Lambiam ^Talk 21:33, 16 August 2006 (UTC)[reply]

rewritten

i rewrote this and included an example. when i figure out a way to put in the references, i will do that. some things that should be added: (1) connections to other sequences, and (2) examples of application, i just used the one in the original article. any other suggestions? thanksEssap 16:18, 6 May 2007 (UTC)essap[reply]

Confusion

Is the binary sequence on this page really the Halton sequence ? I always thought that is is 1/2, 1/4, 3/4, 1/8,5/8,3/8,7/8, ... ?! In "your" sequence the monotonically increasing parts get longer and longer ... Regards, David Vonka —Preceding unsigned comment added by Vonkad (talk • contribs) 14:57, 29 June 2008 (UTC)[reply]

Update: At this moment, I'm almost sure that the example is wrong. if there is no reaction within a few days, I'll change it. —Preceding unsigned comment added by 137.56.140.22 (talk) 11:42, 1 July 2008 (UTC)[reply]

I've changed the examples in accordance with the references. Richard Pinch (talk) 05:54, 4 July 2008 (UTC)[reply]

Name origin needed

As a sequence obviously named after some person "Halton", one would expect to find information of that person as well. Who is he and when did he live etc. Jahibadkaret (talk) 18:23, 22 August 2010 (UTC)[reply]

The person Halton is Dr. John H Halton, who retired from the Comp Sci dept of UNC-Chapel Hill a number of years ago. https://www.researchgate.net/profile/John-Halton 2607:F280:3010:41A0:0:0:0:8156 (talk) 21:37, 14 January 2024 (UTC)[reply]

Prime number is needed?

The Halton sequence is constructed according to a deterministic method that uses a prime number as its base.(from article) Can we use s composite number as its base? I think it's possible to use two numbers a, b as its base where gcd(a,b)=1. —Preceding unsigned comment added by 121.190.39.217 (talk) 14:36, 28 January 2011 (UTC)[reply]

If you use a composite number, some of the fractions (eg. 2/4) will be equal to some of the previous fractions (1/2). You'd then have to filter those out. You may as well just use a prime. -naught101 (talk) 05:28, 20 March 2012 (UTC)[reply]