2 votes 2 votes Consider the following schedule: The number of serial schedules which are view equal to schedule (S) ___________. Answer is 10 as per the given solution. Please explain how to solve these type of questions Databases databases transaction-and-concurrency madeeasy-testseries-2018 made-easy-test-series + – ashish pal asked Jan 12, 2018 • edited Mar 5, 2019 by ajaysoni1924 ashish pal 880 views answer comment Share Follow See all 14 Comments See all 14 14 Comments reply hs_yadav commented Jan 12, 2018 reply Follow Share i'm getting 15 0 votes 0 votes Mk Utkarsh commented Jan 12, 2018 reply Follow Share according to the rules initial read must be same can someone explain the meaning of this through this quesiton? 0 votes 0 votes Anu007 commented Jan 12, 2018 reply Follow Share i getting 10 only. 0 votes 0 votes hs_yadav commented Jan 12, 2018 reply Follow Share @Anu i 'm getting like this... 1->2->5 and 1->4 1 2 5 now 4 may arrage in 3 way and t3 could be anywhere (5 way..) so i think 3*5 =15 ? 0 votes 0 votes ashish pal commented Jan 12, 2018 reply Follow Share @Anu007 this is the polygraph I am getting 0 votes 0 votes Anu007 commented Jan 12, 2018 reply Follow Share T4 cannot come after T5 since T5 must be last b/c of final write of B 0 votes 0 votes hs_yadav commented Jan 12, 2018 reply Follow Share i missed a point i got it....10 is the answer..... 2->4 because initial value of a must be read by 2 not 4... so....1 2 5 ....4 may arranged in 2 way and 3 could be anywhere (5)...total 5*2=10 0 votes 0 votes vishal chugh commented Jan 12, 2018 reply Follow Share Can anyone please tell the complete solution,i.e. how to approach such kind of problems. 0 votes 0 votes ashish pal commented Jan 12, 2018 reply Follow Share @vishal chugh there are 3 types of restrictions that we need to find out 1.) the first read on all the data items must be done by same transaction in order to be view serializable with given transaction. 2.) the final write on all the data items must be done by same transaction. 3.) the write read dependency must follow the same order as in given schedule. this is what i know as of now 0 votes 0 votes ashish pal commented Jan 12, 2018 reply Follow Share @hs_yadav why there are 2 possibilities for T4 I mean the initial read is to be done by T2 and. T4 is also reading on B. So T2 must come before T4 in order to be the first one to read data item B please explain 0 votes 0 votes hs_yadav commented Jan 12, 2018 reply Follow Share ye sure..... 1 2 _5_ at place of dot T4 could be placed.... now if u think that T5 must be last ...but why if there are any other write on B then may be problem but for this case ..T5 is not neccessory at last...becoz it is only write on B...(according to view seri. defination last write on all account sho 0 votes 0 votes hs_yadav commented Jan 12, 2018 reply Follow Share ye sure..... 1 2 _5_ at place of dot T4 could be placed.... now if u think that T5 must be last ...but why if there are any other write on B then may be problem but for this case ..T5 is not neccessory at last...becoz it is only write on B...(according to view seri. defination last write on all account should be same and that is) 0 votes 0 votes ashish pal commented Jan 12, 2018 reply Follow Share @hs_yadav but T5 must occur after T4 so that is does not mess with the W-R dependency of T1-T4 please tell me where i am going wrong :( 0 votes 0 votes vishal chugh commented Jan 18, 2018 reply Follow Share Yes, I think 5 should be the answer. Can't understand why 10,i.e. how there are 2 ways of arranging 1,2,4 and 5? 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes let me know if done any mistake- Prateek Raghuvanshi answered Aug 3, 2018 Prateek Raghuvanshi comment Share Follow See all 0 reply Please log in or register to add a comment.