[Solved]: How to go from a recurrence relation to a final complexity

Problem Detail: 

I have an algorithm, shown below, that I need to analyze. Because it's recursive in nature I set up a recurrence relation.

//Input: Adjacency matrix A[1..n, 1..n]) of an undirected graph G   //Output: 1 (true) if G is complete and 0 (false) otherwise   GraphComplete(A[1..n, 1..n]) {   if ( n = 1 )     return 1 //one-vertex graph is complete by definition     else       if not GraphComplete(A[0..n − 1, 0..n − 1])        return 0       else        for ( j ← 1 to n − 1 ) do           if ( A[n, j] = 0 )            return 0         end       return 1 } 

Here is what I believe is a valid and correct recurrence relation:

$\qquad \begin{align} T(1) &= 0 \\ T(n) &= T(n-1) + n - 1 \quad \text{for } n \geq 2 \end{align}$

The "$n - 1$" is how many times the body of the for loop, specifically the "if A[n,j]=0" check, is executed.

The problem is, where do I go from here? How do I convert the above into something that actually shows what the resulting complexity is?

Asked By : Chris Bond

Answered By : jmad

From what you wrote it seems that for all $k$ you have $T(k)=k-1+T(k-1)$ and $T(1)=0$. Therefore you can get it directly:

$$T(n)=(n-1)+(n-2)+\dots+1+0 = \sum_{k=1}^{n}{(k-1)}=\frac{n(n-1)}{2}$$

So $T(n)$ is in $Θ(n^2)$.

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Question Source : http://cs.stackexchange.com/questions/1959