Reputation: 3159
I solved my homework problem, I used recursion tree.
But solution says that this recurrence relation can be solved by master's theorem!
T(N) = 49T(N/25) + n^(3/2)log(n)
I solved n^(3/2) log^2(n)
But solution said n^(3/2) log(n)
I don't know why this case can use master's theorem and it is correct.
Upvotes: 0
Views: 6336
Reputation: 1445
We can see that a=49
and b=25
. Note that log_b(a) ~ 1.2
and that 3/2 = 1.5
. Hence, log_b(a) < 3/2
. Thus, we can see that f(n) = n^{3/2}log(n) = Omega(n^{log_b(a) + epsilon})
for some epsilon, so that Case 3 of the master theorem applies. Thus, the run time is
T(n) = Theta(f(n)) = n^{3/2}log(n)
Note: You also have to check that
af(n/b) <= cf(n)
for some constant c
. Of course
49 (n/25)^{3/2} log(n/25) <= c n^{3/2}log(n)
which can be checked by dividing both sides by n^{3/2}
and then subtracting c log(n)
from both sides which gives
(49/25^{3/2} - c) log n - 49/25^{3/2} log(25) <= 0
This is certainly true at least for c > 49/25^{3/2}
(no need to make this tight).
Upvotes: 3