How to solve: T(n) = T(n - 1) + n

Question

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?

Answer 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.

Answer 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.

Answer 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).

Answer 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).