Question

For each of the following, give the Big-O time and explain your answer:
a. Breadth-first search/traversal using an adjacency matrix.
b. Breadth-first search/traversal using an adjacency list.
c. Depth-first search/traversal using an adjacency matrix.
d. Depth-first search/traversal using an adjacency list.

Answer 1

a. Breadth-first search/traversal using an adjacency matrix.

in adjacency matrix graph is represented by using n*n Matrix where n is number of vertex in graph.

for loop runs two times in the logic. So complexity is O(V²)

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b. Breadth-first search/traversal using an adjacency list.

Complexity using an adjacency list is O(V+E) where V is vertices of the graph and E is edges of the graph.

BFS using adjacency list takes O(V) time to initialize the distance and predecessor for each vertex. then it traverse only the vertices reachable from a given source vertex so it takes max O(E) . so we get O(V+E).

if |E| >= |V| then  ∣V∣ + ∣E∣ <= ∣E∣+ ∣E∣ <= 2.|E| igonre constant then O(V+E) = O(E)

if |V| >= |E| then  ∣V∣ + ∣E∣ <= ∣V∣+ ∣V∣ <= 2.|V| igonre constant then O(V+E) = O(V)

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c. Depth-first search/traversal using an adjacency matrix.

in adjacency matrix graph is represented by using n*n Matrix where n is number of vertex in graph.

for loop runs two times in the logic. So complexity is O(V²)

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d. Depth-first search/traversal using an adjacency list.

Complexity using an adjacency list is O(V+E) where V is vertices of the graph and E is edges of the graph.

if |E| >= |V| then  ∣V∣ + ∣E∣ <= ∣E∣+ ∣E∣ <= 2.|E| igonre constant then O(V+E) = O(E)

if |V| >= |E| then  ∣V∣ + ∣E∣ <= ∣V∣+ ∣V∣ <= 2.|V| igonre constant then O(V+E) = O(V)

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