Talk:De Bruijn sequence

to create a de-bruijn sequance, "find an Eularian cycle on a de-bruijn graph" is wrong, because Eularian cycle has to use every "edge" once, but to create a de-bruijn sequance need to use every "node" once on a de-bruijn graph.

the article is correct and you are wrong, 'to create a de-bruijn sequance need to use every "node" once on a de-bruijn graph' is incorrect.

I think that the article, which now says "by taking an Hamiltonian cycle of a complete graph of order $k^{n}$", is wrong. A Hamiltonian cycle over the complete graph is just an ordering of the $k^{n}$ sequences, but since only some orderings give a de Bruijn sequence, this is meaningless. It should rather say "by taking a Hamiltonian cycle of an $n$-dimensional de Bruijn graph over $k$ symbols. Am I wrong? --fudo 14:29, 10 July 2006 (UTC)

I was wondering the same thing. Since no-one else seems to have objected here over the past 2+ months, I'll be bold and commit it. —JLD 20:31, 28 September 2006 (UTC)

As I understand it, you do use every EDGE exactly once. Michael Hardy 22:26, 28 September 2006 (UTC)

... and now I've added a diagram and an accompanying explanation of why it's Eulerian and not Hamiltonian. Michael Hardy 23:43, 28 September 2006 (UTC)

Everything you said is correct, except for that "not". In fact, by following an Eulerian cycle over an $n$-dimensional de Bruijn graph, you obtain a de Bruijn sequence of order $n+1$. Since it's easy to prove (by the definition) that the $n+1$-dimensional de Bruijn graph is just the line graph of the $n$-dimensional one, an Eulerian cycle over the first corresponds exactly to a Hamiltonian cycle over the second one. Try it by yourself. I'll modify slightly the article, leaving your example (which is perfect) as is. --fudo 13:05, 4 October 2006 (UTC)

The early on sentence: "Taking A = {0,1}, there are two distinct B(2,3): 00010111 and 00011101, one being the reverse of the other." Those two sequences are not the reverse of each other. Either a) this should say there are 4 distinct, 00010111 and 00011101 and these two reversed, which appears correct to me or b.) the part about "one being the reverse of the other" should be taken out. 17:18, 5 November 2006 (UTC)

They are reverses of each other if you mod out by cyclic shifts. Michael Hardy 20:15, 6 November 2006 (UTC)

The initial example given (A={0,1}) does not contain the subsequence 110 (although its claimed "reverse" does). So what's going on?

Oh never mind -- it is cyclic, so the sequence is "properly" viewed including the wrap-around from back to front. So the initial B(2,3) contains 110 via the "final" ones + the "initial" zero.

I think the use of the term "subsequence" in the first paragraph is wrong. It should be changed into "substring" or something (Mathworld sais "subrange"), since the elements of a subsequence need not be consecutive in the original sequence. —Preceding unsigned comment added by 85.73.195.2 (talk) 12:27, 16 December 2007 (UTC)

Capitals

Why is "de Bruijn sequence" not written as "De Bruijn sequence", with capital D? At least in Dutch, the first character of a name has to be a capital, even if it is something like "de" or "van" (so meneer (mister) De Bruijn, but Nicolaas Govert de Bruijn) and I never learned that that's different in English. And after all, De Bruijn was a Dutchman.DaanAlberga 13:15, 8 June 2006 (UTC)

Then why is the initial d in lower case at [[1]]? Michael Hardy 23:52, 12 June 2006 (UTC)

The initial d only is in lower case when it is _not_ the first letter of his name. In "a de Bruijn sequence", the d _is_ the initial letter of his name. In your reference, the Dutch page about De Bruijn, one can see that as well. When it says "Nicolaas de Bruijn" or something like that, the d is in lower case, and when it just says "De Bruijn", the d is a capital. DaanAlberga 09:31, 13 June 2006 (UTC)

Possitional encoding improvements

For a flat (not wraped application of possitional detection. Arrange for the sequence of zeros to be at one end and ones at the other. i.e. for the sequence from the article 0 0 0 0 1 1 1 1 0 1 1 0 0 1 0 1 change the point where it would but does not wrap so the sequence reads. 1 1 1 1 0 1 1 0 0 1 0 1 0 0 0 0

Now you can extend the sequence with e.g. (1 1 1 1 1...) 1 1 1 1 0 1 1 0 0 1 0 1 0 0 0 0 (0 0 0 0 0...) Now if the senser detects it is in an all zero or all one regon it knows it is off the map and in which direction. Note: not worked things out for a 2d map but, if in corners it has a 50% chance of getting one axis incorrect, the corect axis will get it out of the corner where it will head back to the map.

—Preceding unsigned comment added by 204.193.45.69 (talk) 16:50, 11 January 2008 (UTC)

Use in fMRI

GKA pointed out my info on its use with fMRI was a bit too complicated for the page, but I thought I'd put the info here, in case someone wanted to expand on it in future.

In an fMRI experiment, activity at certain frequencies is most easy to detect. By selecting the correct order of stimuli, the researcher can ensure the activity changes of interest vary at the most detectable frequency. A subset of the possible De Bruijn cycles will have this feature. In general the de Bruijn cycle allows the experimenter to maximise the statistical power of the experiment, without compromising on counterbalancing. The method also allows the signal caused by carry-over effects to be maximised, if required.

Thanks, Lionfish0 (talk) 10:26, 22 March 2012 (UTC)

I should also say that I'm happy with the changes GKA made, very nicely summarised :) Lionfish0 (talk) 10:28, 22 March 2012 (UTC)

Misunderstanding

I read "Taking A = {0, 1}, there are two distinct B(2, 3): 00010111 and 11101000, one being the reverse or negation of the other." Maybe I do not understand De Bruijn sequence's definition, but 000, 001, 010, 011, 100, 101, 110 and 111 should all be substrings of 00010111 and 11101000, right? But I cannot find 100 in the 1^st one, and I cannot find 001 in the 2^nd one. Where is the mistake ? --Gloumouth1 12:29, 5 December 2013 (UTC)

Ok I think I understood. The sequence is cyclical. --Gloumouth1 12:32, 5 December 2013 (UTC)