How do I solve the following as f(n)=n! does not as to my knowledge apply to any of the cases of master theorem. T (n) = 16T (n/4) + n!
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David Eisenstat is partially correct. Case 3 does apply, but T(n) = theta(n!), not O(n!).
T(n) = 16T(n/4) + n!
Case 3 of the Master Theorem (AKA Master Method) applies. a = 16, b = 4, f(n) = n!. n^(log [base(b)] a) = n^2. f(n) is n!. Since n! is omega(f(n)) i.e. n! omega n^2 AND af(n/b) <= cf(n) for a large n, T(n) is theta(n!).
For reference, consult #10 here : http://www.csd.uwo.ca/~moreno/CS433-CS9624/Resources/master.pdf
Jay ARe
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Hey @Jay, How can we prove that af(n/b) <= cf(n)? – Kumaravel Rajan Nov 20 '20 at 11:23