问题
We are to solve the recurrence relation through repeating substitution:
T(n)=T(n-1)+logn
I started the substitution and got the following.
T(n)=T(n-2)+log(n)+log(n-1)
By logarithm product rule, log(mn)=logm+logn,
T(n)=T(n-2)+log[n*(n-1)]
Continuing this, I get
T(n)=T(n-k)+log[n*(n-1)*...*(n-k)]
We know that the base case is T(1), so n-1=k -> k=n+1, and substituting this in we get
T(n)=T(1)+log[n*(n-1)*...*1]
Clearly n*(n-1)*...*1 = n! so,
T(n)=T(1)+log(n!)
I do not know how to solve beyond this point. Is the answer simply O(log(n!))? I have read other explanations saying that it is Θ(nlogn) and thus it follows that O(nlogn) and Ω(nlogn) are the upper and lower bounds respectively.
回答1:
This expands out to log (n!). You can see this because
T(n) = T(n - 1) + log n
= T(n - 2) + log (n - 1) + log n
= T(n - 3) + log (n - 2) + log (n - 1) + log n
= ...
= T(0) + log 1 + log 2 + ... + log (n - 1) + log n
= T(0) + log n!
The exact answer depends on what T(0) is, but this is Θ(log n!) for any fixed constant value of T(0).
A note - using Stirling's approximation, Θ(log n!) = Θ(n log n). That might help you relate this back to existing complexity classes.
Hope this helps!
回答2:
Stirling's formula is not needed to get the big-Theta bound. It's O(n log n) because it's a sum of at most n terms each at most log n. It's Omega(n log n) because it's a sum of at least n/2 terms each at least log (n/2) = log n - 1.
回答3:
Yes, this is a linear recurrence of the first order. It can be solved exactly. If your initial value is $T(1) = 0$, you do get $T(n) = \log n!$. You can approximate $\log n!$ (see Stirling's formula): $$ \ln n! = n \ln n - n + \frac{1}{2} \ln \pí n + O(\ln n) $$
[Need LaTeX here!!]
来源：https://stackoverflow.com/questions/22618778/calculating-the-recurrence-relation-tn-tn-1logn