Graphs: find a sink in less than O(|V|) - or show it can't be done

Question

I have a graph with n nodes as an adjacency matrix.

Is it possible to detect a sink in less than O(n) time?

If yes, how? If no

Answer 1

I've been working on this problem and I believe this does it:

int graph::containsUniversalSink() {
/****************************************************************
 Returns: Universal Sink, or -1 if it doesn't exist
 Paramters: Expects an adjacency-matrix to exist called matrix
****************************************************************/

//a universal sink is a Vertex with in-degree |V|-1 and out-degree 0
//a vertex in a graph represented as an adjacency-matrix is a universal sink
//if and only if its row is all 0s and its column is all 1s except the [i,i] entry - path to itself (hence out-degree |V|-1)
//for i=0..|V|-1, j=0..|V|-1
//if m[i][j]==0 then j is not universal sink (unless i==j) - column is not all 1s
//if m[i][j]==1 then i is not universal sink - row is not all 0s
int i=0,j=1;
while (ii && matrix[i][j]==true) {
        //we found a 1, disqualifying vertex i, and we're not in the last row (j>i) so we move to that row to see if it's all 0s
        i=j;
        if (j

}