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Q: Divide & Conquer ( No Answer,   0 Comments )
Question  
Subject: Divide & Conquer
Category: Computers
Asked by: spark_82-ga
List Price: $5.00
Posted: 16 Oct 2005 22:53 PDT
Expires: 19 Oct 2005 22:41 PDT
Question ID: 581165
Suppose you're consulting for a bank that's concerned about fraud
detection, and they come to you with the following problem. They have
a collection of n bank-cards that they've confiscated, suspecting them
of being used in fraud. Each bank-card is a small plastic object,
containing a magnetic stripe with some encrypted data, and it
corresponds to a unique account in the bank. Each account can have
many bank-cards corresponding to it, and we'll say that two bank-cards
are equivalent if they correspond to the same account.

It's very difficult to read the account number off a bank-card
directly, but the bank has a high-tech "equivalence tester" that takes
two bank-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. Show how to decide the answer to their question with only O(n
log n) invocations of the equivalence tester.

please give Pseudo-Code and a prove this algorithm is correct
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