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August 9

Lower bound on Kolmogorov complexity

Show that $K(x)\geq l(x)$ for most x, i.e. all x except for $o(x)$ exceptions.--130.56.94.183 (talk) 04:28, 9 August 2012 (UTC)[reply]

This looks like homework, so:

If your question is homework, show that you have attempted an answer first, and we will try to help you past the stuck point. If you don't show an effort, you probably won't get help. The reference desk will not do your homework for you.

— Kieff | Talk 04:41, 9 August 2012 (UTC)[reply]