Data Structures & Algorithms Multiple Choice Questions on “Dijkstra’s Algorithm”.
1. Dijkstra’s Algorithm is used to solve _____________ problems. Answer: b 2. Which of the following is the most commonly used data structure for implementing Dijkstra’s Algorithm? Answer: d 3. What is the time complexity of Dijikstra’s algorithm? Answer: c 4. Dijkstra’s Algorithm cannot be applied on ______________ 5. What is the pseudo code to compute the shortest path in Dijkstra’s algorithm? b) c) d) Answer: a 6. How many priority queue operations are involved in Dijkstra’s Algorithm? 7. How many times the insert and extract min operations are invoked per vertex? Answer: a 8. The maximum number of times the decrease key operation performed in Dijkstra’s algorithm will be equal to ___________ Answer: b 9. What is running time of Dijkstra’s algorithm using Binary min- heap method? Answer: d 10. The running time of Bellmann Ford algorithm is lower than that of Dijkstra’s Algorithm. Answer: b 11. Dijkstra’s Algorithm run on a weighted, directed graph G={V,E} with non-negative weight function w and source s, terminates with d[u]=delta(s,u) for all vertices u in V. Answer: a 12. Given pseudo code of Dijkstra’s Algorithm. What happens when “While Q != 0” is changed to “while Q>1”? Answer: b 13. Consider the following graph. Answer: d 14. In the given graph, identify the shortest path having minimum cost to reach vertex E if A is the source vertex. Answer: b 15. Dijkstra’s Algorithm is the prime example for ___________ Answer: a
a) All pair shortest path
b) Single source shortest path
c) Network flow
d) Sorting
Clarification: Dijkstra’s Algorithm is used for solving single source shortest path problems. In this algorithm, a single node is fixed as a source node and shortest paths from this node to all other nodes in graph is found.
a) Max priority queue
b) Stack
c) Circular queue
d) Min priority queue
Clarification: Minimum priority queue is the most commonly used data structure for implementing Dijkstra’s Algorithm because the required operations to be performed in Dijkstra’s Algorithm match with specialty of a minimum priority queue.
a) O(N)
b) O(N3)
c) O(N2)
d) O(logN)
Clarification: Time complexity of Dijkstra’s algorithm is O(N2) because of the use of doubly nested for loops. It depends on how the table is manipulated.
a) Directed and weighted graphs
b) Graphs having negative weight function
c) Unweighted graphs
d) Undirected and unweighted graphs
Answer: b
Clarification: Dijkstra’s Algorithm cannot be applied on graphs having negative weight function because calculation of cost to reach a destination node from the source node becomes complex.
a)if(!T[w].Known)
if(T[v].Dist + C(v,w) < T[w].Dist) {
Decrease(T[w].Dist to T[v].Dist +C(v,w));
T[w].path=v; }
if(T[w].Known)
if(T[v].Dist + C(v,w) < T[w].Dist) {
Increase (T[w].Dist to T[v].Dist +C(v,w));
T[w].path=v; }
if(!T[w].Known)
if(T[v].Dist + C(v,w) > T[w].Dist) {
Decrease(T[w].Dist to T[v].Dist +C(v,w);
T[w].path=v; }
if(T[w].Known)
if(T[v].Dist + C(v,w) < T[w].Dist) {
Increase(T[w].Dist to T[v].Dist);
T[w].path=v; }
Clarification: If the known value of the adjacent vertex(w) is not set then check whether the sum of distance from source vertex(v) and cost to travel from source to adjacent vertex is less than the existing distance of the adjacent node. If so, perform decrease key operation.
a) 1
b) 3
c) 2
d) 4
Answer: b
Clarification: The number of priority queue operations involved is 3. They are insert, extract-min and decrease key.
a) 1
b) 2
c) 3
d) 0
Clarification: Insert and extract min operations are invoked only once per vertex because each vertex is added only once to the set and each edge in the adjacency list is examined only once during the course of algorithm.
a) Total number of vertices
b) Total number of edges
c) Number of vertices – 1
d) Number of edges – 1
Clarification: If the total number of edges in all adjacency list is E, then there will be a total of E number of iterations, hence there will be a total of at most E decrease key operations.
a) O(V)
b) O(VlogV)
c) O(E)
d) O(ElogV)
Clarification: Time required to build a binary min heap is O(V). Each decrease key operation takes O(logV) and there are still at most E such operations. Hence total running time is O(ElogV).
a) True
b) False
Clarification: The number of iterations involved in Bellmann Ford Algorithm is more than that of Dijkstra’s Algorithm.
a) True
b) False
Clarification: The equality d[u]=delta(s,u) holds good when vertex u is added to set S and this equality is maintained thereafter by the upper bound property.//Initialise single source(G,s)
S=0
Q=V[G]
While Q != 0
Do u=extract-min(Q)
S=S union {u}
For each vertex v in adj[u]
Do relax(u,v,w)
a) While loop gets executed for v times
b) While loop gets executed for v-1 times
c) While loop gets executed only once
d) While loop does not get executed
Clarification: In the normal execution of Dijkstra’s Algorithm, the while loop gets executed V times. The change in the while loop statement causes it to execute only V – 1 times.
If b is the source vertex, what is the minimum cost to reach f vertex?
a) 8
b) 9
c) 4
d) 6
Clarification: The minimum cost to reach f vertex from b vertex is 6 by having vertices g and e as intermediates.
b to g, cost is 1
g to e, cost is 4
e to f, cost is 1
hence total cost 1+4+1=6.
a) a-b-e
b) a-c-e
c) a-c-d-e
d) a-c-d-b-e
Clarification: The minimum cost required to travel from vertex A to E is via vertex C
A to C, cost= 3
C to E, cost= 2
Hence the total cost is 5.
a) Greedy algorithm
b) Branch and bound
c) Back tracking
d) Dynamic programming
Clarification: Dijkstra’s Algorithm is the prime example for greedy algorithms because greedy algorithms generally solve a problem in stages by doing what appears to be the best thing at each stage.