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algorithmsubstitutionrecurrenceproofinduction
Sudocode
Sudocode

Reputation: 79

Recurrence with logs T(n) = T(logn)+log(log(n))

How to find T(n)=Θ(f(n)) for the recurrence T(n) = T(logn)+log(log(n))?

I think T(n)= Θ(log(n)) but the proof is the hard part for me. I'm going to attempt to prove by substitution but please help me with that. I also tried a proof by induction but that got out of hand quickly...

Assume T(n) = logn such that

proof of big o:

T(n+1) = T(log(n+1))+loglog(n+1)
       = loglog(n+1) + loglog(n+1) < c*log(n) (check)

proof of big ohmega:

T(n-1) = T(log(n-1))+loglog(n-1)
       = loglog(n-1) + loglog(n-1) > c*log(n) (not good)

Thoughts on proving this via sub. or via induction? ... wish I could just plug and chug into master theorem ~

Upvotes: 0

Views: 1654

Answers (1)

Gassa
Gassa

Reputation: 8846

Direct approach seems the best fit.

T (n) = log log n + T (log n) =
        log log n + log log log n + T (log log n) =
        log log n + log log log n + log log log log n + T (log log log n) =
        log log n + log log log n + log log log log n + log log log log log n + ...

The first term is log log n, and every term dominates the next one.
So the function is Omega (log log n).

Upvotes: 3

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