ok, and when n is not even, you need to test explicitly the last element
with the whole cards.

2008/12/10 Miroslav Balaz <[EMAIL PROTECTED]>

> 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.
>>
>> >>
>>
>

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