GATE IT 2008 | Question: 63

Question

Consider the following three schedules of transactions T1, T2 and T3. [Notation: In the following NYO represents the action Y (R for read, W for write) performed by transaction N on object O.]$$\begin{array}{llll}\hline (S1) & 2RA & 2WA & 3RC & 2WB & 3WA & 3WC & 1RA & 1RB & 1WA & 1WB  \\\hline (S2) & 3RC & 2RA & 2WA & 2WB & 3WA & 1RA & 1RB & 1WA & 1WB & 3WC  \\\hline (S3) & 2RA & 3RC & 3WA & 2WA & 2WB & 3WC & 1RA & 1RB & 1WA & 1WB  \\\hline \end{array}$$Which of the following statements is TRUE?

• A. S1, S2 and S3 are all conflict equivalent to each other
• B. No two of S1, S2 and S3 are conflict equivalent to each other
• C. S2 is conflict equivalent to S3, but not to S1
• D. S1 is conflict equivalent to S2, but not to S3

Comments

Is there a typo error in this question or not ?? If yes please correct it or it create unnecessary confusion!!

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Similar question:

https://gateoverflow.in/333194/gate-2020-cse-question-37

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Its not Wrong- Two Schedules are conflict equivalent when there (i)Precedence graphs are isomorphic AND (ii) Graph should be Acyclic.

Answer 1

As the order of conflicting operations should be same. But in S3 it is not same as S1 and S2. I am attaching the picture for the reference.

P.S: Please watch the following video for more clarity

 

Answer 2

in S3 , there is typo , it should be 2RA instead of 2RZ
becoz all action on an objects should be equal
R(A) 2 times    W(A) 3 times
R(B) 1 time    W(B) 2 times
R(C) 1 time     W(C) 1 time
so in S3 it shoould be only 2RA
Now can draw precedence graph of each schedule
S1 and S2 dont hv cycle and arc also same so conflict
equivalent to each other
but S3 hv cycle so not conflict equivalent
so ans should be D only

Answer 3

Ans:D

Answer 4

Option D

So, from the above Conflict Diagram, we can conclude that S1 is CONFLICT EQUIVALENT to S2, but not to S3 and option D is the right answer.