I thik it is wll known classical problem,or there is some similar well known problem, you need O(n) comparisons. Randomized pick eight random cards and compare then to each other. and you have very low probabilty that you always choose bad card. Deterministic. just compare 1 and second, third and fourth... 2k-1 and 2k. And pick the smaller elemen from ecah pair that the result is true. now how many representants will have the good set? let m be the number of good pairs you compared, the result for them ist true, aleso 1<=m<=n/2; within the representatns we must have m good, and at most n/2-m-(n/2+x-m*2) bad, because each good card when compares with bad card results in false; so we dont pic any of them in representants. So we are to solve the same problem with n/2 cards, of course the x means how much more than n/2 cards it is.
I am not sure about boundary cases 2008/12/9 Tower <[EMAIL PROTECTED]> > > A bank has a collection of n bank cards that they've confiscated, > suspecting them of being used in a fraud. Each bank card corresponds > to a unique account in the bank. Each account can have many cards > corresponding to it, and we'll say that two bank cards are equivalent > if they correspond to the same account. The only way to say 2 cards > are equivalent is by using a high-tech "equivalence-tester" that takes > in 2 cards, and after performing some computations, determines whether > they are equivalent. > Their question is the following: among the collection of n cards, is > there a set of more than n/2 of them that are all equivalent to one > another? Assume that the only feasible operations you can do with the > cards are to pick two of them and plug them in to the equivalence > tester. > > > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---
