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"A knock-out tournament is being held with 2^n tennis players..."
a) 2^n tennis players play \frac{2^n}{2}=2^{n-1} matches each round. Each match has 1 winner. Therefore the next round will have \frac{2^{n-1}}{2}=2^{n-2} etc until n-k=0 as 2^0=1 (final match), where k is the number of rounds. n rounds for 2^n players. Idk why there is a log identity in the solution.
Author uses big N here to refer to the little n as stated by the problem, yet for part C uses it to refer to the total number of players N=2^n (???)
Not to mention the N_{r) from part B.
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"A knock-out tournament is being held with 2^n tennis players..."
a) 2^n tennis players play \frac{2^n}{2}=2^{n-1} matches each round. Each match has 1 winner. Therefore the next round will have \frac{2^{n-1}}{2}=2^{n-2} etc until n-k=0 as 2^0=1 (final match), where k is the number of rounds. n rounds for 2^n players. Idk why there is a log identity in the solution.
Author uses big N here to refer to the little n as stated by the problem, yet for part C uses it to refer to the total number of players N=2^n (???)
Not to mention the N_{r) from part B.
This passed the Turing test with flying colors.
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