The time complexities of some standard graph algorithms are given. Match each algorithm with its time complexity ? (n and m are no. of nodes and edges respectively)
a. Bellman Ford algorithm 1. O (m log n)
b. Kruskals algorithm 2. O (n3)
c. Floyd Warshall algorithm 3. O(mn)
d. Topological sorting 4. O(n + m)
| A. | 3 1 2 4 |
| B. | 2 4 3 1 |
| C. | 3 4 1 2 |
| D. | 2 1 3 4 |
|
Option: A Explanation : Click on Discuss to view users comments. |
Let V1 = 2I – J + K and V2 = I + J – K, then the angle between V1 & V2 and a vector perpendicular to both V1 & V2 shall be :
| A. | 90o and (–2I + J – 3K) |
| B. | 60o and (2I + J + 3K) |
| C. | 90o and (2I + J – 3K) |
| D. | 90o and (–2I – J + 3K) |
|
Option: D Explanation : Click on Discuss to view users comments. Anuradha said: (3:20pm on Sunday 12th January 2014)
V1.V2 = (2I - J K).(I J - K) = 2 - 1 -1 = 0Hence angle between V1 and V2 is 90º.|i j k |V1 X V2 = |2 -1 1 ||1 1 -1 |= i(1-1)-j(-2-1) k(2 1) = i.0 j.3 k.3= 3j 3kHence, no answer is matched.
|
| A. | {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} |
| B. | {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} |
| C. | {2, 3, 4, 5, 6, 7, 8, 9, 10} |
| D. | { } |
|
Option: C Explanation : Click on Discuss to view users comments. Anuradha said: (2:13pm on Sunday 12th January 2014)
A ={0,1, 2, 3, 4, 5, 6, 7, 8, 9,10}u^a={0,1/3,2/4,3/5,4/6,5/7,7/8,8/9,9/10,10/11}Hence, Alpha cut where alpha=0.5 is membership value greater or equal to 0.5.So {0,1/3,2/4,3/5,4/6,5/7,7/8,8/9,9/10,10/11} => {0,0.3,0.5,0.6,0.6,0.7,0.8,0.9,0.9}Hence, {2, 3, 4, 5, 6, 7, 8, 9, 10}Hence, (c) is correct
Then the α cut corresponding to α = 0.5 will be
|
