User talk:RobinK

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Computational complexity of mathematical operations

re: recent revert. I overlooked the definition of M(n) because it's buried in a specific table, yet used throughout the article. Maybe it should be moved out of the table in question? -kj (talk) 18:38, 30 January 2010 (UTC)[reply]

Yes, feel free to do so. I just reverted because what you wrote was incorrect; I don't object to changing the placement of that text. --Robin (talk) 23:34, 30 January 2010 (UTC)[reply]

Positive definiteness

Your comments are requested on a discussion about whether or not a particular page is a disambiguation page or a stub here. Neelix (talk) 20:06, 14 February 2010 (UTC)[reply]

DSPACE, etc.

Thanks for several corrections and a {{dn}} fix. I don't know why I didn't see Alternation (complexity)...

CRGreathouse (t | c) 05:56, 5 March 2010 (UTC)[reply]

No problem. You didn't see it because I fixed the disambiguation page for Alternation to list Alternation (complexity) instead of (algorithms), and created a redirect from (complexity) -> Alternating TM, after you were done disambiguating. --Robin (talk) 13:54, 5 March 2010 (UTC)[reply]

AM,MA

Hi Robin, thanks for the fix in Arthur–Merlin_protocol. I saw that the current condition for AM reads as follows (and is correct)

if x is not in L, then $\Pr \nolimits _{y\in \{0,1\}^{p(n)}}(\forall z\in \{0,1\}^{q(n)}\,M(x,y,z)=0)\geq 2/3.$

I wanted to note that it can be equally written as

if x is not in L, then $\Pr \nolimits _{y\in \{0,1\}^{p(n)}}(\exists z\in \{0,1\}^{q(n)}\,M(x,y,z)=1)\leq 1/3.$

However, the second condition seems to go well together with the x is in L condition, as 2/3 and 1/3 add up to 1.