How to solve the recurrence A(n) = A(n-1) + n*log(n)?

Question

Given the recurrence:
A(n) = A(n-1) + n*log(n).
How do I find the time complexity of A(n)?

@Shantanu did my answer solve your problem? Could you somehow respond? I'm open to give you more explanation if needed. — xenteros, Nov 07 '16 at 12:54
@xenteros yes it does answer my question. I have marked it as the correct answer. Thanks very much! I understood the big-O part. How would you approach finding the tight bound (big-theta)? — awareeye, Nov 07 '16 at 14:56
@LukeRahne sorry for not making that explicit in the question. It does terminate at A(1) — awareeye, Nov 07 '16 at 14:56

xenteros · Accepted Answer · 2016-11-07T12:01:30.790

6

A(n) = A(n-1) + nlog(n)

Look, your recurence says: take the previous value and add nlogn to it.

So... assuming that A(1) = log(1) is the first element of the sequence: A(n) = SUM from i = 1 to n (ilog(i)).

Why?

A(1) = log(1).
A(2) = log(1) + 2log(2).
A(3) = A(2) + 3log(3) = 1log(1) + 2log(2) + 3log(3).
.
.
.
A(n) = 1log(1) + 2log(2) + 3log(3) + ... + nlog(n)

This way of solving recursion can be always used when F(n) - F(n-1) is a non recursive function. In our case it's nlogn so it worked.

Similar rule can be used when F(n)/F(n-1) is a non recursive function. Then we use PI instead of SIGMA.

If I were asked to give it an upper bound I would suggest to try the following:

  log(n)   + log(n)   + log(n)   + log(n)   + ...
+            log(n-1) + log(n-1) + log(n-1) + ...
+                       log(n-2) + log(n-2) + ...
.
.
.

so

Now you have a very clear upper bound, so the big-o is there for free (O(nlog(n!))). In case you're looking for a big-theta you need to struggle a bit longer.

edited Nov 07 '16 at 12:01

answered Nov 07 '16 at 11:34

xenteros

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Nope, it's not. It's a sum. – xenteros Nov 07 '16 at 11:53
Question was what is complexity for calculating recurrence relation, not what result is. – Luka Rahne Nov 07 '16 at 11:53
@xenteros oh yes I see, but it should be A(n) = ½ • n(n-1) • sum_{i=1}^n{log(i)} then. – ToxiCore Nov 07 '16 at 11:53
@ToxiCore if you find my answer satisfying feel free to mark it as solution – xenteros Nov 07 '16 at 12:11
@xenteros I find it very satisfying and interesting - sadly I am not the one who asked the question elseways I already would have done that :/ but you've got an upvote! – ToxiCore Nov 07 '16 at 12:20
@ToxiCore haha, suprising. Anyway, you made me to struggle a bit and build a better answer. – xenteros Nov 07 '16 at 12:22
3

`O(N log N!)` is the same as `O(N^2 log N)`, via Stirling's approximation: https://en.wikipedia.org/wiki/Stirling%27s_approximation – Matt Timmermans Nov 07 '16 at 13:04
@xenteros do you happen to know how the big-theta (tight bound) can be found for this recurrence? Thanks! – awareeye Nov 07 '16 at 14:59
@Shantanu not yet – xenteros Nov 07 '16 at 15:03

score 3 · Answer 2 · answered Nov 07 '16 at 11:37

Let A(0)=c. Find A(n) as a function of n and c, defined by sum.
How many terms are in the sum?
What is a minimal value of the sum? Try to find an estimation.
What is a maximal value of the sum? Try to find an estimation.
If you can estimate (3) and (4) such that they one of them is constant times bigger than other, are you done? Why?

How to solve the recurrence A(n) = A(n-1) + n*log(n)?

2 Answers2