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Chapter 2: Problem 40

Suppose that there are \(n=2^{k}\) teams in an elimination tournament, where there are \(n / 2\) games in the first round, with the \(n / 2=2^{2-1}\) winners playing in the second round, and so on. (a) Develop a recurrence equation for the number of rounds in the tournament. (b) How many rounds are there in the tournament when there are 64 teams? (c) Solve the recurrence cquation of part (a).

Short Answer

Expert verified
The recurrence equation that represents the number of tournament rounds for \(n = 2^k\) teams is \(R(n) = R(n / 2) + 1\) with \(R(1) = 0\). For a 64-team tournament, there would be 6 rounds. The solution to this recurrence equation is \(R(n) = log_2(n)\).

Step by step solution

01

Develop a recurrence relation

Assuming there are \(n = 2^k\) teams in a tournament, and each match occurs between two teams and eliminates one, then the number of rounds \(R(n)\) needed to determine a champion is \(R(n) = R(n/2) + 1\). This is because after each round, half of the teams are eliminated, so we recursively define the number of rounds based on the number of teams left, which is \(n/2\), and add one to count the current round.
02

Apply the recurrence relation for a specific n

To determine how many rounds are needed for a 64-team tournament, we substitute \(n = 64 = 2^6\), which means \(R(64) = R(2^6) = 6+R(0)\). As \(R(1) = 0\) (base case: when there is only one team, no tournament is needed), so \(R(64) = 6\). So there will be 6 rounds in this tournament.
03

Solve the recurrence relation

The recurrence equation \(R(n) = R(n/2) + 1\) with base case \(R(1) = 0\) is a logarithmic relation. Since \(n = 2^k\), we can rewrite the relation as \(R(2^k) = R(2^{k-1}) + 1\). The solution of this relation is \(R(2^k) = k + R(1) = k\), which is a logarithm base 2 function.

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