Talk:The Complexity of Songs

Latest comment: 11 years ago by Altenmann in topic further results

S1

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  is undefined. —The preceding unsigned comment was added by 216.232.222.122 (talkcontribs) .

No, S1 is defined in terms of S0. —Keenan Pepper 21:09, 27 May 2006 (UTC)Reply
Actually, it wasn't, since that definition was stated as being for values of k strictly greater than 1. I've fixed it now. --Tango 22:17, 8 June 2006 (UTC)Reply

\epsilon

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The first part of the definition should use an epsilon ε rather than the letter e. Wcbarksdale 14:27, 9 July 2007 (UTC)Reply

Gee, I am glad people are actually reading this article! Fixed. `'Miikka 15:44, 9 July 2007 (UTC)Reply
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Does anyone know if it would be allowed to re-TeX Knuth’s original paper (the fonts in the linked one are a little bit ‘dirty’) or to translate it into a other language? --Frakturfreund (talk) 16:19, 1 December 2009 (UTC)Reply

Everything published on paper is copyrighted to the publisher. The author retains the rights as well. Only expressly released in public domain or expired may be used without asking permissions. While this does not answer your question, I'd say that changes are slim that you may re-publish/re-post it in any form without permission. - Altenmann >t 18:24, 1 December 2009 (UTC)Reply
Thank you very much for your rapid clarification! And you are right, the copyright has to be respected. Probably it would make more sense to translate this wikipedia article into the german language first. --Frakturfreund (talk) 04:13, 2 December 2009 (UTC)Reply
If it's so, how come there is a full version linked here? Is it not a violation of copyright?
My other question is, is Kurt Eisemann's letter available for reading somewhere? Marczellm (talk) 23:17, 29 December 2011 (UTC)Reply

further results

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This blog post, "The descriptive complexity of songs" notes that the song "  botthels on a wall" has complexity O(log log n) and makes other interesting remarks. Unfortunately refs to blogs are invalid for wikipedia. - Altenmann >t 20:05, 7 January 2013 (UTC)Reply