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March 21

Wikipedia Page Traffic Statistic V3

hi im student and im working on my new project about "Wikipedia Page Traffic Statistic V3" is there any resource to help me on this project?i don`t have any idea about how to start thank you — Preceding unsigned comment added by Sheal-bero (talk • contribs) 05:03, 21 March 2016 (UTC)[reply]

Well, all of the statistics that Wikipedia gathers about itself are here Wikipedia:Statistics. There is a ton of useful material linked from there. Wikipedia_article_traffic may also have some useful links - although the article itself is little more than a collection of anecdotes. If you need more help than this, you're going to have to tell us more about the requirements of your project - although, note that we have a policy that you Wikipedia:Do your own homework. SteveBaker (talk) 14:01, 21 March 2016 (UTC)[reply]

Pumping length of DFA

Could anyone tell what's really the pumping length of a DFA in Pumping lemma for regular languages.I'm confused whether it's the number of states or is it the length of a string that is needed to pump a string.For example consider this dfa.Could anyone tell what's the pumping length of this DFA.Is it 4? — Preceding unsigned comment added by JUSTIN JOHNS (talk • contribs) 10:26, 21 March 2016 (UTC)[reply]

Well, in Pumping lemma for regular languages it's defined as the length of the shortest string in a language for which we can guarantee (addition after Ben's helpful comment) that it can be pumped, i.e. it is not defined for an automaton, but for a language. If you have a given DFA for a language, then the number of states of the automaton is an upper bound for p (because for every accepted string of greater length the DFA must loop, and you can then repeat the loop). But there may, of course, be other DFAs. Also, its not a sharp bound - your example DFA has 4 states, but the language has a pumping lengths of no more than 3. --Stephan Schulz (talk) 12:15, 21 March 2016 (UTC)[reply]

It's not the length of the shortest string that can be pumped, it's one more than the length of the longest string that can't be pumped. For example, in the language a*|b, there is a pumpable string of length 1 ("a"), but also a non-pumpable string of that length ("b"), so the minimum pumping length is not 1 (it's 2). In the language a(bc)*d, the longest non-pumpable string is "ad", so the minimum pumping length is 3 (even though there are no strings of length 3). -- BenRG (talk) 19:17, 21 March 2016 (UTC)[reply]

Yes, of course you are right. Fixed it above. --Stephan Schulz (talk) 20:13, 21 March 2016 (UTC)[reply]

Could you tell how the pumping length of language of the given dfa is 3?Since I think that we can pump any string in the language only if it is of the form abcd from which we could get abcbcd,abcbcbcd etc.JUSTIN JOHNS (talk) 06:24, 22 March 2016 (UTC)[reply]

Any string of length 4 or higher can be pumped (by the pigeonhole principle, as explained in the article, or just because the only such strings are abcd, abcbcd, etc.), so 4 is a valid pumping length. Any string of length 3 can be pumped (vacuously, since there are no strings of length 3), so 3 is also a valid pumping length. The string "ad" can't be pumped, so 2 isn't a valid pumping length. Therefore the minimum pumping length is 3. To put it another way, any string of length 3 or more can be pumped because the only strings of length 3 or more are abcd, abcbcd, etc. -- BenRG (talk) 07:11, 22 March 2016 (UTC)[reply]

Do you mean to say that we should also consider cases where there aren't any strings to pump?JUSTIN JOHNS (talk) 08:37, 22 March 2016 (UTC)[reply]

Yes. Any finite language is regular, but has no pumpable words (otherwise it would be infinite). In this case, p is the length of the longest word in L plus 1. --Stephan Schulz (talk) 14:19, 22 March 2016 (UTC)[reply]

Could you give an example to illustrate how 'p' becomes the length of the longest word in L plus 1.I think it's hard to find the longest word in a infinite language.JUSTIN JOHNS (talk) 06:21, 23 March 2016 (UTC)[reply]

Yes, that's why that applies to finite languages, as I wrote above. Infinite languages do not have a longest word (the alphabet is finite, after all). E.g. for L={ab, ac, abc, abcd} p is 5. All words in L of length 5 or greater (which are none) can be pumped. --Stephan Schulz (talk) 08:59, 23 March 2016 (UTC)[reply]

rpmfusion U.S. mirror

The rpmfusion website is apparently hosted in Germany. My workplace blocks Germany. Is there such thing as a U.S. mirror of rpmfusion? I can't Google much for it because all the results end up being German websites that are, obviously, blocked. 199.15.144.250 (talk) 12:18, 21 March 2016 (UTC)[reply]

I don't know if that helps, but both the registration and the ip address locate to France, not Germany. --Stephan Schulz (talk) 12:35, 21 March 2016 (UTC)[reply]

I would not be surprised if the IT people also block France. They operate on the idea that if nobody can use the computers, then the computers will be secure. 199.15.144.250 (talk) 12:44, 21 March 2016 (UTC)[reply]

You can ask the IT people to whitelist that website, you can use an existing proxy, or you can set up a proxy at home. The Quixotic Potato (talk) 14:43, 21 March 2016 (UTC)[reply]

Your workplace blocks Germany?! Is this a common thing in normal business (not counting sensitive government networks etc)? It seems overkill to ban entire countries from a network, unless you're talking North Korea and the like. Fgf10 (talk) 19:04, 21 March 2016 (UTC)[reply]

Actually, our IT department blocks entire continents. If it isn't the United States, it must be nothing but hackers. I'm not saying that I agree with them. I just have to deal with them. What I did was set up my own mirror on my personal server at home and then set up my server at work to use that as the rpmfusion server. 199.15.144.250 (talk) 13:38, 22 March 2016 (UTC)[reply]