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How many Conflict Serializable and View serializable schedules for the Schedule given below

$S:r_1(A),w_1(B),w_1(C),r_2(A),w_2(B),w_2(C)$

To both my answer comes to be 7.

Is it correct?
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How to proceed for view equivalent, all I am understanding that final writes on B and C  must be fixed by T2 by any view equivalent schedules 

My ques is

1.first read has to be fixed for T1 or not ?pls someone clear this doubt?

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5 CS and 4 VS ??
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edited by

@ sir,

How many Conflict Serializable and View serializable schedules for the Schedule given below

‘serializable’ doesn’t make sense here. It should be ‘equivalent’ as in the heading of the question.

And yes we are getting both as 7.

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