How Many Rounds Are Needed in a 2^k Team Elimination Tournament?

[bfpri]

14. Suppose 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^(k-1) winners playing in the second round, and so on. Develop a recurrence relation for the number of rounds in the tournament.

So the recurrence relation would be f(n)=f(n/2)+1, with n=2^k. Not sure if that's right, seems a little too easy.

16. Solve the recurrence relation for the number of rounds in the tournament described in Exercise 14.

f(n)=f(1)+clog_bn. (Theorem 1) Since f(1)=0; (no rounds when there is a winner) and b=2 and c=1. then the answer is log_2n. did i do this right?

 

[KataKoniK]

I would like to provide some feedback and clarification on your approach to solving this problem.

Firstly, your recurrence relation for the number of rounds in the tournament is correct. However, it would be helpful to explain the meaning of each term in the relation. The term f(n) represents the number of rounds in a tournament with n teams, f(n/2) represents the number of rounds in the next round of the tournament (after the first round), and the constant 1 represents the additional round needed for the winners of the first round to compete in the second round. This helps to provide a better understanding of the problem and the recurrence relation.

Secondly, your solution for the recurrence relation is correct, but it would be beneficial to provide a step-by-step explanation of how you arrived at the solution. This would help others who may be struggling with solving recurrence relations to understand the process and apply it to similar problems.

Lastly, it is always good practice to check your solution by substituting values for n and verifying that the recurrence relation holds true. In this case, we can substitute n=2^k and see that f(2^k)=f(2^(k-1))+1 is indeed true.

Overall, your approach to solving this problem is correct, but it can be improved by providing more explanation and steps to your solution. Good job!