Database-Management-System
|
Question 475
|
Suppose a database schedule S involves transactions T1, T2, ………….,Tn. Consider the precedence graph of S with vertices representing the transactions and edges representing the conflicts. If S is serializable, which one of the following orderings of the vertices of the precedence graph is guaranteed to yield a serial schedule ?
|
Topological order
|
|
|
Depth – first order
|
|
|
Breadth – first order
|
|
|
Ascending order of transaction indices
|
Question 475 Explanation:
If a schedule is conflict serializable then no cycle in precedence graph should be present.
But BFS and DFS are also possible for cyclic graphs.
And topological sort is not possible for cyclic graph.
Moreover option (D) is also wrong because in a transaction with more indices might come before lower one.
But BFS and DFS are also possible for cyclic graphs.
And topological sort is not possible for cyclic graph.
Moreover option (D) is also wrong because in a transaction with more indices might come before lower one.
Correct Answer: A
Question 475 Explanation:
If a schedule is conflict serializable then no cycle in precedence graph should be present.
But BFS and DFS are also possible for cyclic graphs.
And topological sort is not possible for cyclic graph.
Moreover option (D) is also wrong because in a transaction with more indices might come before lower one.
But BFS and DFS are also possible for cyclic graphs.
And topological sort is not possible for cyclic graph.
Moreover option (D) is also wrong because in a transaction with more indices might come before lower one.
Subscribe
Login
0 Comments