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The time complexity of computing the transitive closure of a binary relation on a set of n elements is known to be:
- a)O(n)
- b)O(nLogn)
- c)O(n(3/2))
- d)O(n3)
Correct answer is option 'D'. Can you explain this answer?
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The time complexity of computing the transitive closure of a binary re...
Explanation:
The transitive closure of a binary relation on a set of n elements is a matrix that represents the reachability information between pairs of elements in the set.
To compute the transitive closure of a binary relation, we can use the Floyd-Warshall algorithm. This algorithm iteratively considers all pairs of elements and checks if there exists a path between them through any intermediate element. If such a path exists, the algorithm updates the matrix to reflect this reachability information.
The time complexity of the Floyd-Warshall algorithm is O(n^3), where n is the number of elements in the set. Let's break down the complexity of each step of the algorithm:
1. Initialization: Initializing the matrix takes O(n^2) time as we need to set the initial reachability information for all pairs of elements.
2. Iterations: The algorithm iterates n times, considering each element as an intermediate element. For each iteration, it checks all pairs of elements. Since there are n^2 pairs, this step takes O(n^2) time.
3. Update: For each pair of elements, the algorithm checks if there exists a path through the intermediate element. This check takes constant time, O(1).
Overall, the time complexity of the algorithm is O(n^2) for initialization and O(n^2) for each iteration, resulting in a total time complexity of O(n^2 + n^2) = O(n^2).
However, in the worst case, the algorithm may need to iterate n times, resulting in a time complexity of O(n^3).
Therefore, the correct answer is option D: O(n^3).
The transitive closure of a binary relation on a set of n elements is a matrix that represents the reachability information between pairs of elements in the set.
To compute the transitive closure of a binary relation, we can use the Floyd-Warshall algorithm. This algorithm iteratively considers all pairs of elements and checks if there exists a path between them through any intermediate element. If such a path exists, the algorithm updates the matrix to reflect this reachability information.
The time complexity of the Floyd-Warshall algorithm is O(n^3), where n is the number of elements in the set. Let's break down the complexity of each step of the algorithm:
1. Initialization: Initializing the matrix takes O(n^2) time as we need to set the initial reachability information for all pairs of elements.
2. Iterations: The algorithm iterates n times, considering each element as an intermediate element. For each iteration, it checks all pairs of elements. Since there are n^2 pairs, this step takes O(n^2) time.
3. Update: For each pair of elements, the algorithm checks if there exists a path through the intermediate element. This check takes constant time, O(1).
Overall, the time complexity of the algorithm is O(n^2) for initialization and O(n^2) for each iteration, resulting in a total time complexity of O(n^2 + n^2) = O(n^2).
However, in the worst case, the algorithm may need to iterate n times, resulting in a time complexity of O(n^3).
Therefore, the correct answer is option D: O(n^3).
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The time complexity of computing the transitive closure of a binary relation on a set of n elements is known to be:a)O(n)b)O(nLogn)c)O(n(3/2))d)O(n3)Correct answer is option 'D'. Can you explain this answer?
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