Arrange the following functions in decreasing asymptotic order:
f(n)= 32^n
g(n)= n!n3
(a) f(n),g(n)
(b) g(n),f(n)
Apply log on both functions
log(32^n) = 2n log3 = O(2n)
log(n!n3) = log n! + 3logn = O(nlogn) [ we know log n! = O(nlogn) ]
O(2n) > O(nlogn)
So, option (a) is correct.
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