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Chapter 2: Problem 40

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

Short Answer

Expert verified
a. The recurrence relation for the number of rounds in the tournament is \( t(k) = t(k-1) + 1 \) with \( t(0) = 0 \). b. There are 6 rounds in the tournament when there are 64 teams. c. The solution to the recurrence relation is \( t(k) = k \).

Step by step solution

01

Formulate the Recurrence Relation

In an elimination tournament with \( n = 2^k \) teams, half of the teams get eliminated in each round. So in the first round, there are \( n/2 = 2^{k-1} \) games. Since the winner of each game moves to the next round, the number of games becomes half in the next round. This keeps repeating till one team is left and it is declared the winner. Hence, the number of games is halved at each step till we reach 1. This leads to the recurrence relation: \[ t(k) = t(k-1) + 1, \] with initial condition \( t(0) = 0 \), where \( t(k) \) represents the number of rounds needed for \( 2^k \) teams.
02

Calculate the Total Rounds for 64 teams

Given \( n = 64 \) teams, convert it to the form of \( 2^k \), where \( k \) is the value to be found in the recurrence relation. We have \( 64 = 2^6 \), so \( k = 6 \). Using the recurrence relation established in Step 1, we get the total rounds equal to \( k \), so there are 6 rounds when 64 teams are participating.
03

Solve the Recurrence Relation

To solve the recurrence relation \( t(k) = t(k-1) + 1 \), we replace \( t(k-1) \) with its equivalent from the recurrence relation until we ultimately express everything in terms of \( t(0) \). Plugging \( k-1 \) into our recurrence relation, we obtain: \[ t(k-1) = t(k-2) + 1. \] Replacing \( t(k-1) \) in the initial recurrence relation, we get: \[ t(k) = t(k-2) + 1 + 1. \] Repeating this process \( k \) times, we get \( t(k) = t (0) + k = k \). This shows that the number of rounds in the tournament is given by \( k \), where \( k \) is such that \( n = 2^k \), the number of teams.

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