Well that is true...and at the outset the sum may appear to be 198+49 but the truth is that if we eliminated 49 tl groups then we would not have had to ask the 198 questions in the first place remember. The worst case calculation of 100+50+25+(12+1)+... assumes that no group gets eliminated in each round. If a group were a "tl" and it got eliminated then that means in all the subsequent rounds from there on the group would not contribute.
I am not sure how this works with the even number of groups problem since there eliminating a group may not affect the next round but with odd number of groups even reducing a group a question. The sum i guess (still guessing not sure for sure) is still 198 therefore. Its like If I eliminate a group in say the second round then I will have to only ask 12 questions in the next round hence i can afford to ask a question in the end to the eliminated group. Similarly If a group was eliminated in round 1 then I would have 24+1 in second round but I think there is a work around If we move that single person to the next round. The majority reasoning would still not be affected because in the worst case there could be 51 and 49 in first round and after elimination we could have 25 and 24 in the next round. Between them If we choose the liar to be kept alone then we still have majority in this round 25 vs 23 and therefore we will give atleast 2 true ppl to the next round and still carry a majority. If the true person was kept apart we would have 24 and 24 so in the worst case only the true person would get through. --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
