Test: Graph Based Algorithms- 1 - Computer Science Engineering (CSE) MCQ

An array of lists is used. The size of the array is equal to the number of vertices. Let the array be an array[]. An entry array[i] represents the list of vertices adjacent to the ith vertex. This representation can also be used to represent a weighted graph. The weights of edges can be represented as lists of pairs. 

1) Breadth First Search uses Queue

2) Depth First Search uses Stack

3) Prim's Minimum Spanning Tree uses Priority Queue.

4) Kruskal' Minimum Spanning Tree uses Union Find.

In sequence IV, we have a vertex with degree 8 which is not possible in a simple graph (no self loops and no multiple edges) with total vertex count as 8. Maximum possible degree in such a graph is 7.

In sequence II, four vertices are connected to 6 other vertices, but remaining 4 vertices have degrees as 3, 3, 2 and 2 which are not possible in a simple graph (no self loops and no multiple edges).

Option (A) is MNOPQR. It cannot be a BFS as the traversal starts with M, but O is visited before N and Q. In BFS all adjacent must be visited before adjacent of adjacent. Option (B) is NQMPOR. It also cannot be BFS, because here, P is visited before O. (C) and (D) match up to QMNP. We see that M was added to the queue before N and P (because M comes before NP in QMNP). Because R is M's neighbor, it gets added to the queue before the neighbor of N and P (which is O). Thus, R is visited before O.

Connected components can be found in O(m + n) using Tarjan's algorithm. Once we have connected components, we can count them.

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.

d(r, u) and d(r, v) will be equal when u and v are at same level, otherwise d(r, u) will be less than d(r, v)

In an undirected graph, there can be maximum n(n-1)/2 edges. We can choose to have (or not have) any of the n(n-1)/2 edges. So, total number of undirected graphs with n vertices is 2^(n(n-1)/2).

We can check all DFSs for following properties.
In DFS, if a vertex '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.

P is true for undirected graph as adding an edge always increases degree of two vertices by 1. Q is true: If we consider sum of degrees and subtract all even degrees, we get an even number because every edge increases the sum of degrees by 2. So total number of odd degree vertices must be even.

Make can decide the order of building a software using topological sorting. Topological sorting produces the order considering all dependencies provide by makefile. See following for details. Topological Sorting

We can use both traversals to find if there is a path from s to t.

A cycle of length 3 can be formed with 3 vertices. There can be total 8C3 ways to pick 3 vertices from 8. The probability that there is an edge between two vertices is 1/2. So expected number of unordered cycles of length 3 = (8C3)*(1/2)^3 = 7

Consider the following graph. If we start from 'a', then there is one tree edge. If we start from 'b', then there is no tree edge. a→b

Since the given graph is undirected, every edge contributes as 2 to sum of degrees. So the sum of degrees is 2E.