how to solve the recurrence equation . T(n) = T(n/2) + T(n/4) + T(n/8) + Ω(n). I can solve it if instead of Ω(n), we had (n), but now I can't solve it. please help me!
Asked
Active
Viewed 177 times
1 Answers
2
It seems a little bit unusual to have a lower bound like that, but I believe we can use strong induction to give a lower bound on T(n). I start with an educated guess that the recursion will add a factor of O(lg n) to the Ω(n) bound, and use strong induction to verify this guess. Let's consider the case in which Ω(n) is minimised, to achieve a lower bound on the whole recurrence. That is, we assume that the function of n hiding in the Omega notation is actually Θ(n):
Assume true for all values of n up to k:
I.H. : T(n) < cn lg n - dn (for n < k)
Prove that it is true for k also:
T(k) = T(k/2) + T(k/4) + T(k/8) + Θ(n)
(IH) T(k) = (k/2)lg(k/2) + (k/4)lg(k/4) + (k/8)lg(k/8) - 3dn + Θ(n)
T(k) = ck lg k - 3dn + Θ(n)
< ck lg k - dn as required
Since we can choose the constant d large enough to outweigh the constant hidden in the Theta-notation. Since cn lg n is Ω(n lg n), we can give this as a lower bound on the recurrence. Owing to the Ω(n) term in the original, I believe this is the tightest asymptotic bound we can give.
soothsooth
- 173
- 8
-
If you don't think there's any problem with it, you can accept the answer so that it's marked as answered – soothsooth Nov 20 '20 at 13:19