问题
I have the following worked out:
T(n) = T(n - 1) + n = O(n^2)
Now when I work this out I find that the bound is very loose. Have I done something wrong or is it just that way?
回答1:
Think of it this way:
In each "iteration" of the recursion you do O(n) work.
Each iteration has n-1 work to do, until n = base case. (I'm assuming base case is O(n) work)
Therefore, assuming the base case is a constant independant of n, there are O(n) iterations of the recursion.
If you have n iterations of O(n) work each, O(n)*O(n) = O(n^2).
Your analysis is correct. If you'd like more info on this way of solving recursions, look into Recursion Trees. They are very intuitive compared to the other methods.
回答2:
You need also a base case for your recurrence relation.
T(1) = c
T(n) = T(n-1) + n
To solve this, you can first guess a solution and then prove it works using induction.
T(n) = (n + 1) * n / 2 + c - 1
First the base case. When n = 1 this gives c as required.
For other n:
T(n)
= (n + 1) * n / 2 + c - 1
= ((n - 1) + 2) * n / 2 + c - 1
= ((n - 1) * n / 2) + (2 * n / 2) + c - 1
= (n * (n - 1) / 2) + c - 1) + (2 * n / 2)
= T(n - 1) + n
So the solution works.
To get the guess in the first place, notice that your recurrence relationship generates the triangular numbers when c = 1:
T(1) = 1:
*
T(2) = 3:
*
**
T(3) = 6:
*
**
***
T(4) = 10:
*
**
***
****
etc..
Intuitively a triangle is roughly half of a square, and in Big-O notation the constants can be ignored so O(n^2) is the expected result.
回答3:
The solution is pretty easy for this one. You have to unroll the recursion:
T(n) = T(n-1) + n = T(n-2) + (n - 1) + n =
= T(n-3) + (n-2) + (n-1) + n = ... =
= T(0) + 1 + 2 + ... + (n-1) + n
You have arithmetic progression here and the sum is 1/2*n*(n-1). Technically you are missing the boundary condition here, but with any constant boundary condition you see that the recursion is O(n^2).
回答4:
Looks about right, but will depend on the base case T(1). Assuming you will do n steps to get T(n) to T(0) and each time the n term is anywhere between 0 and n for an average of n/2 so n * n/2 = (n^2)/2 = O(n^2).
来源：https://stackoverflow.com/questions/2752977/how-to-solve-tn-tn-1-n