CODEWITHSUNDEEP
recursioncomplexity-theorybig-o
gpol
gpol

Reputation: 966

Solution of T(n)=2T(n/2) + n*log(logn)

I am struggling to solve this equation. I concluded that the Master Theorem does not apply in this situation so I tried to successively substitute the terms in order to solve this equation but cannot proceed. Could somebody tell me what would be the best way to solve this?

Thanx

Upvotes: 0

Views: 2322

Answers (1)

phoenyx
phoenyx

Reputation: 119

As you deduced we can't use the master theorem because of failure of a condition in case three.

but for a start we have n^(log_b a) = n(log_2 2) = n

now n < n* log(log(n))

So the contribution of the recursion term 2T(n/2) keeps on decreasing, as compared to the other term n*(log(log n))....Hence we can forget that term for now...

Now consider the term n(log(log n))

Expanding the recurrence but we give importance to this term, we have,

T(n) = 2^k * T(n/2^k) + n [ log(log(n/2^(k-1))) + log(log(n/2^(k-2)) ..... ]

putting 2^k as n we have,

T(n) = ignore first term + n [ log(log(2)) + log(log(4)) +.....+ log(log(2^(k-1))) ]

T(n) = ignore first term + n [ log(1) + log(2) + log(3) +......+ log(k-1) ]

T(n) = ignore first term + n [ log((k-1)!) ]

We just need an upper bound, hence we can make use of Stirling's approxiamation,

n! < (n/e)^n

hence (k-1)! ~ ((k-1)/e)^(k-1)

hence log((k-1)!) = (k-1)[(log(k-1) - log (e)]

but k = log n => [ log((k-1)!) ] = (log(n) - 1) [log(log(n)-1 - log(e)]

hence our term n [ log((k-1)!) ] = n*log(n)*log(log(n)) and some lower order terms which we can ignore..

Hence answer is: O(n*log(n)*log(log(n)))

PS: I am not sure of this answer, do check and tell if I have made some errors!

Upvotes: 1

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