[algogeeks] Re: how to tell honest people
yes..thats the reasoning that i had too but let us see if anyone else sees some thing fishy in our reasoning --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
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[algogeeks] Re: Interesting Probability Question
On 3/9/07, Nat (Padmanabhan Natarajan) [EMAIL PROTECTED] wrote: 1. We cannot bound the triangle if we don't bound the space...thats the reason why I choose a unit square I think we don't really need to bound anything. I think the question as is phrased can only yield what I guess should be called _limiting_ probabilities. ;-) 2. It is true that there are a lot of points outside the triangle that you cannot choose but they all lie in a finite set of lines Again, I think this is incorrect. Please refer to my argument in my previous post. -- Regards, Rajiv Mathews --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
[algogeeks] Re: how to tell honest people
I think I made a mistake. After we found out the truth teller, we need 2 questions for each of the tl and ll groups. This is because we know at least one of them is a liar and nothing else. Say the answer is yn then the group could be tl or ll. Similarly if the answer is nn then also the group could be tl or ll. We need two questions to tell who's who and not one. So my previous argument was wrong. Here's another argument. Say there are 'a' number of tt groups, 'b' number of ll who answer yy, 'c' number of tl and 'd' number of ll whose answer has a no in it. Then total number of questions asked so far for grouping is 2(a+b+c+d) = N the number of people we started with. Now with induction we'll prove that for 'p' people we need at most 2p-2 questions. In the next iteration we have a+b people so we need at most 2(a+b)-2 questions to identify all the truth tellers and liars among a+b. Once we have that we need 2 questions each for c,d groups so we need to ask 2(c+d) questions. So total questions asked are 2(a+b+c+d) + 2(a+b) -2 + 2(c+d) = 4(a+b+c +d)-2 = 2N-2. Hence proved. For 100 people we need at most 198. The -2 is coming from the fact that for the case of N=2 we don't need to ask any questions as both have to be truth tellers. I hope I am right this time. --~--~-~--~~~---~--~~ 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 -~--~~~~--~~--~--~---
