Test: Graph Based Algorithms- 2 - Computer Science Engineering (CSE) MCQ

There are total n*(n-1)/2 possible edges. For every edge, there are to possible options, either we pick it or don't pick. So total number of possible graphs is 2^n(n-1)/2.

In a graph with three nodes, r u and v, with edges (r; u) and (r; v), and r is the starting point for the DFS, u and v are siblings in the DFS tree, neither as the ancestor of the other.

n * (n - 1) / 2 when cyclic. But acyclic graph with the maximum number of edges is actually a spanning tree and therefore, correct answer is n-1 edges.

In option D, 1 appears after 2 and 3 which is not possible in Topological Sorting. In the given DAG it is directly visible that there is an outgoing edge from vertex 1 to vertex 2 and 3 hence 2 and 3 cannot come before vertex 1 so clearly option D is incorrect topological sort. But for questions in which it is not directly visible we should know how to find topological sort of a DAG.

Background : Given a connected and undirected graph, a spanning tree of that graph is a subgraph that is a tree and connects all the vertices together.

• A spanning tree has exactly V - 1 edges.
• A single graph can have many different spanning trees. A minimum spanning tree (MST) or minimum weight spanning tree for a weighted, connected and undirected graph is a spanning tree with weight less than or equal to the weight of every other spanning tree. The weight of a spanning tree is the sum of weights given to each edge of the spanning tree.
• There can be more that one possible spanning trees of a graph. For example, the graph in this question has 6 possible spanning trees.
• MST has lightest edge of every cutset. Remember Prim's algorithm which basically picks the lightest edge from every cutset.

Choices of this question :

a) There exists a cutset in G having all edges of maximum weight : This is correct. If there is a heaviest edge in MST, then there exist a cut with all edges with weight equal to heavies edge. See point 4 discussed in above background.

b) There exists a cycle in G having all edges of maximum weight : Not always TRUE. The cutset of heaviest edge may contain only one edge. In fact there may be overall one edge of heavies weight which is part of MST (Consider a graph with two components which are connected by only one edge and this edge is the heavies)

c) Edge e cannot be contained in a cycle. Not Always True. The curset may form cycle with other edges. d) All edges in G have the same weight Not always True. As discussed in option b, there can be only one edge of heaviest weight.

Depth First Search of a graph takes O(m+n) time when the graph is represented using adjacency list. In adjacency matrix representation, graph is represented as an "n x n" matrix. To do DFS, for every vertex, we traverse the row corresponding to that vertex to find all adjacent vertices (In adjacency list representation we traverse only the adjacent vertices of the vertex). Therefore time complexity becomes O(n²)

Option (C) is correct, because the cyclomatic complexity of the flow graph of a program provides an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at least once.

BFS always produces shortest path from source to all other vertices in an unweighted graph.

A graph is connected iff all nodes can be traversed from each node. For a graph with n nodes, there will be n-1 minimum number of edges.
Given that there are n edges, that means a cycle is there in the graph.
The simplex graph with these conditions may be:

Now we can make a different spanning tree by removing one edge from the cycle, one at a time.
Minimum cycle length can be 3, So, there must be atleast 3 spanning trees in any such Graph.

The following diagram shows the worst case situation where the recursion tree has maximum depth.

So the recursion depth is 19 (including the first node).

In prims algorithm we start with any node and keep on exploring minimum cost neighbors of nodes already covered.

Below example shows that A and B are FALSE:

Below example shows C is false:

It’s a special case in which all edge weights are equal, so dfs would work and resultant depth first tree would be the spanning tree. So the answer would be O(m+n).

In DFS, if 'v' is visited
after 'u', then one of the following is true.
1) (u, v) is an edge.
     u
   /   \
  v     w
 /     / \
x     y   z

2) 'u' is a leaf node.
     w
   /   \
  x     v
 /     / \
u     y   z 
In DFS, after visiting a node, we first recur for all unvisited children. If there are no unvisited children (u is leaf), then control goes back to parent and parent then visits next unvisited children.

It is by definition that a graph is bipartite iff it does not contain odd length cycles. So the answer is O(1).

Tree edges are the edges that are part of DFS tree.  If there are x tree edges in a tree, then  x+1 vertices in the tree. The output of DFS is a forest if the graph is disconnected.  Let us see below simple example where graph is disconnected.

 The above example matches with D option More Examples:

1) All vertices  of Graph are connected.  k must be n-1.  We get number of connected components  = n- k =  n - (n-1) = 1

2) No vertex is connected. k must be 0.  We get number of connected components  = n- k =  n - 0 = n

Case 1: If the vertex we are deleting from G is an isolated vertex, which is a component by itself, then number of components in G becomes 7.
Case 2: If G is a start Graph, then by deleting the cut vertex of G, we get 19 components.
Hence, number of components in G should lie between 7 and 19.

Following are different 6 Topological Sortings
a-b-c-d-e-f
a-d-e-b-c-f
a-b-d-c-e-f
a-d-b-c-e-f
a-b-d-e-c-f
a-d-b-e-c-f

When vertices are arranged with difference of 8 there are 8 components as shown by 8 columns in the image below:

When vertices are arranged withdifference of 12 the number of components is reduced to 4 as first column will be connected with fifth column, second column will be connected with sixth column, third column will be connected with seventh column and fourth column will be connected with eighth column. No other form of connection exists so total 4 connected components are there. So, option (B) is correct. This explanation is contributed by Pradeep Pandey.

• As T is a minimum spanning tree and we need to add a new edge to existing spanning tree.
• Later we need to check still T is a minimum spanning tree or not, So we need to check all vertices whether there is any cycle present after adding a new edge.
• All vertices need to traverse to confirm minimum spanning tree after adding new edge then time complexity is O(V).

Option (D) is correct.