TITLE: Lower bound in algorithmic puzzle
QUESTION [20 upvotes]: Puzzle: there are $n$ computers most of which are good; the others may be bad ("most" in the strict sense: there are strictly more good computers than bad ones). You may ask any computer $A$ about the good/bad status of another computer $B$. if $A$ is good it will correctly indicate $B$'s status, but otherwise it may answer whatever. 
Your goal is to locate a good computer using the minimum number of questions in the worst case. In other words, devise an algorithm that requires no more than $N$ questions regardless of the outcome and is guaranteed to pinpoint a good computer, and make $N$ as small as possible. 
The original puzzle asks for the optimal $N$ when $n=100$. 
Warning: spoilers below. Stop here if you wish to think about this fun puzzle. 
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I, and everyone else I know who solved this, can do the $n=100$ case with $97$ questions in the worst case. I'm pretty sure this is optimal but I do really miserably on lower bounds. The simplest case where I can't match the bounds is $n=7$ (at most $3$ bad computers): this is doable with $5$ questions and I can rule out $3$ but I can't rule out $4$. 
More generally, if the number of bad computers is at most $k$ (so $n=2k+1$ or $n=2k+2$), I can show that at least $k+1$ questions are needed while $2k-1$ questions suffice. Can anyone narrow that gap?
EDIT: starting a bounty, looking for improvements to the lower bound (or the upper bound, though I'd be surprised if the latter is possible) for general $n$. A transparent argument for why 7 computers require 5 questions is also good, but a computer-assisted case-by-case enumeration is not.

REPLY [3 votes]: Several related problems are discussed in papers of Martin Aigner (and his coauthors). See for example M. Aigner: Two Colors and More, In: Csiszár, Katona, Tardos (eds.): Entropy, Search, Complexity, Springer 2007, p. 9-26. 
This paper mentions the Bleher paper in which the Knights and Spies problem (that Aigner call the liar problem) is solved. The version where we only have to find one knight was posed as an open problem in the 2001 COSSAC meeting in Ischia (http://www.di.unisa.it/COSSAC/Program.html) and was solved independently in 2002 by Aigner (see the paper above)  and myself (G. Wiener: A Solution to the Liar Problem, In: Proceedings of the 3rd Japanese-Hungarian Workshop on Discrete Mathematics and its Applications. Tokyo, Japan, 21/01/2003-24/01/2003. pp. 232-239.). Aigner solved a somewhat generalized version of the  problem, where it is known that there are at least k>n/2 knights and proved that the number of questions needed is exactly 2(n-k)-H(n-k). He also proved a similar result for the original problem, which is just sketched in the Bleher paper: if there are at least k>n/2 knights, then exactly 2n-k-1 questions are needed.