User talk:MCiura

Hello, and welcome to Wikipedia. Thank you for your contributions. I hope you like the place and decide to stay. Here are a few good links for newcomers:

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Hi. You've helped with the Wikipedia:WikiProject Wiki Syntax, so I thought it worth alerting you to the latest and greatest of Wikipedia fixing project, User:Yann/Untagged Images, which is seeking to put copyright tags on all of the untagged images. There are probably, oh, thirty thousand or so to do (he said, reaching into the air for a large figure). But hey: they're images ... you'll get to see lots of random pretty pictures. That must be better than looking for at at and the the, non? You know you'll love it. best wishes --Tagishsimon (talk)

You might want to know that there was a voting started at Wikipedia_talk:WikiProject_Geography_of_Poland#Vote. Halibutt 00:18, 8 February 2006 (UTC)[reply]

Cześć. Zapraszam cię do Wikipedia:WikiProject Silesia. Mam nadzieję że się wpiszesz na listę użytkowników i coś czasem poedytujesz związanego z naszym Śląskiem :) LUCPOL 23:07, 20 February 2007 (UTC)[reply]

Hello there. This is an automated message to tell you about the gradual phasing out of the preference entitled "Mark all edits minor by default", which you currently have (or very recently had) enabled.

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Thank you for your understanding and happy editing :) Editing on behalf of User:Jarry1250, LivingBot (talk) 21:29, 14 March 2011 (UTC)[reply]

I'll have some spare time tomorrow. I'll see what I can do. Thanks for your interest. --Simpsons contributor (talk) 18:18, 6 August 2011 (UTC)[reply]

No problem. I think that embossed orange bar/cornflower blue background colour scheme works better too. If you want the source code (the graphics/sorting is done with C#, animated gif creation with Java) then get back in touch. --Simpsons contributor (talk) 12:36, 7 August 2011 (UTC)[reply]

Hello MCiura. Thanks very much for making your expertise available to Wikipedia. I've copyedited Shellsort as you requested on the GOCE Requests page.

Discussing computational complexity in the 3-gap case, your original says "A complicated function ψ(h, g) from the last formula is asymptotically ...", and I think this does not quite make sense in English. I have changed it to "ψ(h, g) in the last formula is a complicated function asymptotically ...". I think this might be your meaning, but am not 100% sure. Please would you check this out and correct it if necessary, or we can discuss it if you wish.

By the way, I find the animation at the top very distracting and not really helpful. The walkthrough in the Description section is clear enough, I think.

To help you in future, please would you place <ref> tags after any punctuation. (See Wikipedia:MOS#Punctuation_and_footnotes). Thanks. --Stfg (talk) 14:52, 5 September 2011 (UTC)[reply]

We know that a lower bound on any sorting algorithm that uses comparison is ${\displaystyle \Omega (N\ln(N))}$. See Knuth, Vol. 3. You will notice that all of the other routines have a run time that are ${\displaystyle \omega (N\ln(N))}$.

Now, if ${\displaystyle \lim _{N\rightarrow \infty }{\frac {f(N)}{g(N)}}=0}$, we say that ${\displaystyle f(N)=o(g(N))}$.

In this case, the worst-case time complexity is given by ${\displaystyle O(e^{\sqrt {8\ln(5/2)\ln N}})}$. If we compare the given function and ${\displaystyle N}$, we get:

${\displaystyle \lim _{n\rightarrow \infty }{\frac {e^{\sqrt {8\ln(5/2)\ln(N)}}}{N}}=0}$

and you can either ask Maple,

limit( exp( sqrt( 8*ln(5/2)*ln(N) ) )/N, N = infinity );                                                                0 

or just try it yourself with increasingly larger numbers. Why not just pick your mathematical software of your choice and plot ${\displaystyle e^{\sqrt {8\ln(5/2)\ln N}}}$--its growth looks more like the logarithm.

Therefore, if the limit is 0, then the worst-case time complexity is being claimed as a function that is ${\displaystyle o(N)}$ which contradicts the above known lower bound of ${\displaystyle \Omega (N\ln(N))}$. In fact, if you try it out, you will find that ${\displaystyle e^{\sqrt {8\ln(5/2)\ln N}}=o(N^{d})}$ for any ${\displaystyle d>0}$ but ${\displaystyle e^{\sqrt {8\ln(5/2)\ln N}}=\omega (\ln(N))}$.

I don't know what the correct answer is--it might actually be ${\displaystyle O(Ne^{\sqrt {8\ln(5/2)\ln N}})}$, as this is at least plausible--but the answer that is there is wrong and it should not appear. — Preceding unsigned comment added by 129.97.10.179 (talk) 23:17, 16 December 2011 (UTC)[reply]

Hi,
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Hello, MCiura. Voting in the 2016 Arbitration Committee elections is open from Monday, 00:00, 21 November through Sunday, 23:59, 4 December to all unblocked users who have registered an account before Wednesday, 00:00, 28 October 2016 and have made at least 150 mainspace edits before Sunday, 00:00, 1 November 2016.

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Hello! Your submission of Pentagramma mirificum at the Did You Know nominations page has been reviewed, and some issues with it may need to be clarified. Please review the comment(s) underneath your nomination's entry and respond there as soon as possible. Thank you for contributing to Did You Know! Yoninah (talk) 21:42, 29 January 2019 (UTC)[reply]

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