Question

5. The in-degree of a vertex in a directed graph is the number of edges directed into it. Here is an algorithm for labeling each vertex with its in-degree, given an adjacency-list representation of the graph. for each vertex i: i.indegree = 0 for each vertex i: for each neighbor j of i: j.indegree = j.indegree + 1 Label each line with a big-bound on the time spent at the line over the entire run on the graph. Assume that there are no isolated vertices. The bounds should be expressed in terms of the number n of vertices and/or the number m of edges. Then give below the overall big-running time, omitting any terms from the expression that are dominated by others.

Answer 1

For simplicity, I will represent theta with 'O'

Number of vertices = n

Number of edges = m

Now for the first for loop
for each vertex i: ---> O(n)
i.indegrees = 0 ---> O(1)

and for the second for loop
for each vertex i: ----> O(n)
for each neighbor j of i: ----> O(2m)
j,indegree = j.indegree + 1 ----> O(1)

Why O(2m) ?
It's because we will traverse each edge twice. Let's say there are two nodes (- Node1 and Node2) and one edge (Edge1-2).
For i = Node1
edge = Edge1-2
For i = Node2
edge = Edge2-1
As you can see we traverse 2 nodes once and one edge twice(one for each side of the node connected by the edge)

Thus for the first for loop ---> O(n), and for the second for loop ---> O(n*2m) = O(2nm) = O(nm)

Finally ---> O(n + nm) = O(nm) since nm >>> n