5 votes 5 votes A) Number of conflict serializable schedules possible R4(a) R2(a) R3(a) W1(b) W2(a) R3(b) W2(b) B) NUmber of conflict serializABLE SCHEDULES possible by T1 and T2 T1 : R(a) W(a) R(b) W(b) T2 : R(a) W(a) R(b) W(b) A_i_$_h asked Oct 10, 2017 A_i_$_h 2.0k views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Anu007 commented Dec 19, 2017 reply Follow Share For part a answer is 3. 0 votes 0 votes hs_yadav commented Dec 25, 2017 reply Follow Share a) 3 b) 2 0 votes 0 votes viratsh133 commented Aug 9, 2018 reply Follow Share qn( A)instead of conflict serializable what if the question is how many conflict equivalent schedules are possible? 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes i correct me if i done wrong Prateek Raghuvanshi answered Oct 11, 2017 • edited Oct 12, 2017 by Prateek Raghuvanshi Prateek Raghuvanshi comment Share Follow See all 17 Comments See all 17 17 Comments reply A_i_$_h commented Oct 12, 2017 reply Follow Share for 1st problem there is also a conflict between T1 and T3 right? please check @prateek 0 votes 0 votes Prateek Raghuvanshi commented Oct 12, 2017 reply Follow Share yeah u r right sorry for that i'll do that again for (b) solution is correct ?? 0 votes 0 votes Prateek Raghuvanshi commented Oct 12, 2017 reply Follow Share for (a) conflict serializable schedule =no. of topological order in precedence grapgh 0 votes 0 votes A_i_$_h commented Oct 12, 2017 reply Follow Share yea (b) is correct can u explain (a) along with the graph and the possible orders 0 votes 0 votes Prateek Raghuvanshi commented Oct 12, 2017 reply Follow Share correct me if i done wrong? 0 votes 0 votes A_i_$_h commented Oct 12, 2017 reply Follow Share @prateek this is where i want help i dont understand how the 3 topological ordering is formed 0 votes 0 votes Prateek Raghuvanshi commented Oct 13, 2017 reply Follow Share first choose that node which has zero indegree as in graph,T1 & T4 has indegree zero so we can start either from T1 or T4 if i choose T1 then remove all edges associated with T1 ,now again check which has indegree zero, now T3 and T4 has indegree zeo T1 T3 T1 T4 NOW if i goes with T1 T3 ,at this situation only t4 ha indegree zero so order will be T1 T3 T4 T2 same as T1 T4 T3 T2, these two order starting with T1 now starting with T4, then only T1 has indegree zero so order will T4 T1 after that T3 has indegree zero, finally order will be T4 T1 T3 T2 so there are 3 order 2 votes 2 votes A_i_$_h commented Oct 13, 2017 reply Follow Share got it :) thank u @prateeek 0 votes 0 votes vamp_vaibhav commented Dec 19, 2017 reply Follow Share Just a doubt out of curiosity.. The method we use for T1->T2 will give number of conflict serializable schedules or just number of serializable schedules..?? 0 votes 0 votes A_i_$_h commented Dec 19, 2017 reply Follow Share @vamp whats the difference between both ? 0 votes 0 votes vamp_vaibhav commented Dec 19, 2017 reply Follow Share I know I am asking the silly doubt here.. But as we know conflict serializable schedules are subset of view serializable.. I mean if we are talking about serializable schedules then we should focus on the bigger one isn't? 0 votes 0 votes Prateek Raghuvanshi commented Dec 20, 2017 reply Follow Share it depends on mode of question for eg. lets assume two transaction T1 and T2 (1)if question ask about total no. of serializable schedule = we have to find no. of serializable schedule from T1 to T2 and T2 to T1 (2)for given schedule , no. of conflict equal serial schedule =no.of topological order of precedence graph (3)no. of schedules conflict equal to schedule=serial as well as non serial schedule just look this example,this may help to u 2 votes 2 votes vamp_vaibhav commented Dec 20, 2017 reply Follow Share Hmm prateek I m getting your point.. But as you can see the second question is asking conflict serializable schedules possible by T1 and T2 here definitely we can't go for topological order.. It means by default We have to consider T1->T2 nd vice versa.. As it is giving conflict serializable schedules but not all the serializable schedules including view also..? 0 votes 0 votes Prateek Raghuvanshi commented Dec 20, 2017 reply Follow Share view serializable is bigger set than conflict serializable,so when we talk about serializability,by default we talk about view serializable ,so all serializable from T1 ->T2 and vice versa are view serializable which are also conflict serializable schedules 1 votes 1 votes A_i_$_h commented Dec 21, 2017 reply Follow Share @prateek thank you :) but do bear with me for a few doubts why r2(a) cannot be between w2(a) and w2(b) ? 0 votes 0 votes Prateek Raghuvanshi commented Dec 24, 2017 reply Follow Share because transaction order should be preserved 0 votes 0 votes Likun swain commented Nov 30, 2018 reply Follow Share @Prateek Raghuvanshi nice explanation 0 votes 0 votes Please log in or register to add a comment.