Solve Induction Problem: cn <= n log2n

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SUMMARY

The discussion centers on solving the induction problem defined by the recurrence relation for the number of comparisons, cn, in an algorithm for a list of n elements. The relation states that c0 = c1 = 0, c2 = 1, c3 <= 3, and cn <= (2cn/2) + (n - 1) for n >= 4. The goal is to prove that cn <= n log2n for any n >= 2. The solution involves applying mathematical induction, which is confirmed to be a suitable approach for this problem.

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nano1
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Hey I've been given an equality to solve as a bonus question with a strong hint that a like one would appear on my midterm.
However, I am stumped by it, it appears quite complex to me. Any insight into how to solve this would be greatly appreciated!
I'll try to type it as best I can:

A student is studying a list with n elements. The student determines that the number of comparisons c sub n when the list has n elements satisfies the following relation:

c0 = c1 = 0
c2 = 1
c3 <= 3
cn <= (2cn/2) + (n - 1) if n >= 4

Prove that for any n >= 2 the value cn satisfies the equation cn <= n log2n

I think I have to use induction to solve this but the question appears tricky and weirdly worded. Any help would be awesome, thank you !
 
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Hi,
The attachment answers your question. The induction is just a little tricky. Presumably, this is the number of comparisons in some algorithm for a list of length n.

View attachment 1715
 

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