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March 13

Vectors

Suppose we take ${F_{2}}^{n}$ , i.e. the n dimensional vector space over the field $F_{2}$ of two elements. Now suppose we wish to construct a maximal set of these vectors such that each pair of vectors differ in at least d of their entries, if not more, by taking an initial vector and then continually adding another that is at least d away from all the others until we can no longer iterate the procedure. Can we place a bound on the size of this set? Or, better still, determine precisely how big it will be? Thanks. 131.111.216.115 (talk) 12:31, 13 March 2012 (UTC)[reply]

I should add that at each stage, including the initial one, we want the vectors to be randomly generated and only added to the set if they satisfy the given criterion. Otherwise, we reject the vector and try a new one. Aside from that, we have no control over what the vector is. 131.111.216.115 (talk) 12:32, 13 March 2012 (UTC)[reply]

Singleton bound gives an upper bound of 2^n-d+1. Hamming bound gives another upper bound. Those pages also links to some other bounds that may be useful. Not sure about the expected size for this random algorithm. Rckrone (talk) 15:45, 13 March 2012 (UTC)[reply]

The expected size for the random algorithm iirc is what you would hope for (this figures into the proof of the noisy channel coding theorem) but of course the worst case is, uh, pessimal. 67.117.144.57 (talk) 08:01, 17 March 2012 (UTC)[reply]

Using Euler's theorem to find the last digits of a tetration

I am trying to solve a problem on Project Euler, which deals with finding some of the last digits of a tetration ${^{n}p}$ , where p is a specific prime number given in the problem and n is another specific integer. To get some ideas on how to solve it, I skimmed thorugh the article and stumbled on this sentence at the end of Tetration#Iterated powers

When a and n are coprime, we can compute the last m decimal digits of $\scriptstyle \underbrace {a^{a^{\cdot ^{\cdot ^{a}}}}} _{n}$ using Euler's theorem.

But is that true? I can see for myself that if a and 10 (but not n) are coprime (and they are in my case since a is prime), then any tetration of ${^{k}p},k=1,\ldots ,n$ and 10^m will also be coprime, and I can use Euler's theorem recursively starting from ${^{2}p}$ and then eat the tower of exponentials (mod 10^m) top-down. So I am not asking for help to solve the problem, but questioning the statement in the article. --Slaunger (talk) 19:04, 13 March 2012 (UTC). My solution was accepted, confirming my own observations, but I still question the validity of the statement in the article.--Slaunger (talk) 19:55, 13 March 2012 (UTC)[reply]

I agree, it doesn't seem right. a and n being coprime doesn't seem to do anything. You only care if a and 10 are coprime. I changed the statement in the article, although maybe it should be removed entirely. Rckrone (talk) 03:17, 14 March 2012 (UTC)[reply]

Ok, thanks. A small victory for a number theory noob :-) --Slaunger (talk) 08:27, 14 March 2012 (UTC)[reply]

A week in Planck times

How many Planck times is a week? I guessed that it was 1,600,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 Planck times (1.6e+48) because Orders of magnitude (time) says that one is equal to approximately 10^-44 seconds, but I'm not at all sure that I've done the math right. Nyttend (talk) 20:38, 13 March 2012 (UTC)[reply]

A week is 7*24*60*60=604800 seconds. The Planck time is 5.39106*10^-44 seconds. If we divide 1 week by 1 Planck time, we get 1.1218573*10⁴⁹. So you were close, just less than one order of magnitude out (I think you did the maths right, but you use 10^-445.39106*10^-44 not 5.39106*10^-44 and must have had an approximation for the length of a week as well. --Tango (talk) 21:02, 13 March 2012 (UTC)[reply]

I thought that I put -44 into my computer calculator, clicked the 10^x button, multiplied by 5.4, put the result into memory, typed 604800, clicked the divide key, selected memory recall, and finished with the equals key. However, that gives me the number you tell me; I'm not sure where I went wrong. Thanks for the help; I've fixed some statements that I made at WP:RFPP, where I was using a different method of expressing the fact that I'd protected a page for a week :-) Nyttend (talk) 21:21, 13 March 2012 (UTC)[reply]

http://www.wolframalpha.com/input/?i=1+week+in+planck+times does the job. Staecker (talk) 14:45, 14 March 2012 (UTC)[reply]