CODEWITHSUNDEEP
algorithmrecurrence
FutureTrillionaire
FutureTrillionaire

Reputation: 39

How to solve T(n) = T(0.2*n^0.5) + 1?

I know how to solve T(n) = T(n^0.5) + 1. Let m = lg n and S(m) = T(2^m). We then get S(m) = S(m/2) + 1. And we know S(m) = Θ(lg m). So T(n) = Θ(lg lg n).

However I'm not sure how to solve T(n) = T(0.2*n^0.5) + 1. The 0.2 is throwing me off. If I use the same method, I'm not sure how to figure out what S(m) is.

Upvotes: 3

Views: 648

Answers (1)

David Eisenstat
David Eisenstat

Reputation: 65458

On the new one, you get

S(lg n) = S(lg 0.2 + 0.5 lg n) + 1
S(m) = S(lg 0.2 + 0.5 m) + 1.

The trick is to substitute R(x) = S(m + 2 lg 0.2).

R(x) = S(m + 2 lg 0.2)
     = S(lg 0.2 + 0.5 (m + 2 lg 0.2)) + 1
     = S(0.5 m + 2 lg 0.2) + 1
     = R(0.5 x) + 1

Then unravel the substitutions and conclude that T(n) = Theta(lg lg n), as before.

Upvotes: 3

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