Can I use Master Theorem if the partitions are not fractions?

AI Thread Summary
The discussion centers on the application of the Master Theorem to recurrences that do not involve fractional partitions, specifically when the recurrence is of the form 2C(n-1) + f(n). It concludes that the Master Theorem cannot be applied in this scenario, and instead, the recursion tree method should be used to analyze the recurrence. The exponential growth of the function is highlighted, as each level of the recursion tree doubles the number of nodes. The closed form expression for the recurrence is suggested to be 2^n, contingent on the values of C(0) and f(n). Understanding the recursion tree method is recommended for further clarity.
zeion
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Homework Statement



If I had a recurrence expression that recurs on partitions of size n - 1 each time, (as opposed some fraction of the original size ie. n/2), how can I apply the Master Theorem? I don't know what the "b" value is?


Homework Equations





The Attempt at a Solution



ie. If I had 2 C(n-1) + f(n), what is b?
 
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You can not use the Master Theorem in this case, just use the recursion tree method. In this case it is easy to see that it will be exponential, since each level has twice the number of nodes of the previous level.
 
Max.Planck said:
You can not use the Master Theorem in this case, just use the recursion tree method. In this case it is easy to see that it will be exponential, since each level has twice the number of nodes of the previous level.

I'm not sure what a recursion tree method is..
I see that each recursion will have twice the calls to the function as the last, but they are also on one less element than the last, does that matter?

Actually nevermind, so if I find a closed form expression it will be 2^n, right?
 
zeion said:
Actually nevermind, so if I find a closed form expression it will be 2^n, right?

Depends on C(0) and f(n). The recursion tree method should be described in your textbook.
 
Oh I see. It says that f(n) is constant. But it doesn't say what C(0) is.
 
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