"Where am I Quiz" with a Twist!

[Starkey7]

Heh I knew that HÃ¥kan would start getting technical at some point. Nice question. Do you think that, by doing this, there's a way of halving the possibilities each time?

[hwolge]

Well, Starkey7, had it not been for the 1-lie rule, it would obviously had been a good strategy to half the remaining options every time and solve it in 9 questions at the most (if there are 512 stations at the most). With the 1-lie rule, it's probably still the best strategy to try half the set of possible stations every time, but it's not as easy, since you don't really know what set to half - so you'd better go for halfing the entire set every time, which makes it harder to come up with "good" questions - just listing a random half of the stations in every question is hardly an interesting way to play the game. I have to admit, I've not come up with any easily described strategy or even a complete analysis of the 1-lie rule. Maybe I'll spend some time on it - it can't really be all that complicated, can it?

PS Starkey7, you are reading your PMs aren't you?

[snudge27]

Are you in an interchange (more than one LU line) station OR does the station name start with a vowel?
(I know, I'm making it a tiny bit complicated here, but this is stiall a YES/NO-question! )

No

[hwolge]

Sorry, I just have to make a general comment here (inspired by Starkey7's comment above):

I looked further into the theory of the 1-lie rule and I realised it's almost identical to the theory of Hamming codes http://en.wikipedia.org/wiki/Hamming_code which is, among other things, used when construction ECC (Error Correction Code) RAM for computers.

I say "Almost", because if there are an even power of 2 (e.g. 256 or 512) possible stations, then it's "exact", in that a series of questions can be posed so that it's guaranteed to be solved in the theoretical number of questions (in the case of 256, exactly 11 questions, but with 512 it would be 12 OR 13 questions), but I suppose there must be a simple way that is not a pure use of Hamming codes that solves the 257 problem in exactly 12 questions (rather than 12 or 13 using the pure Hamming code).

In fact, 12 questions could possibly cover cases up to 315 stations, but maybe a little less...