self doubt(conflict serializable)

Question

Q.1 find total number of conflict serializable

T1=R(A)W(A)R(B)W(B)

T2=R(B)W(B)R(C)W(C)

Q.2 find total number of conflict serializable for

T1=R1(A)W1(A)R1(B)W1(B)R1(C)W1(C)

T2=R2(A)W2(A)R2(B)W2(B)R2(C)W2(C).

explanation??.

Comments

https://gateoverflow.in/118640/gate2017-2-44

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Actually, the answer provided in GATE-2017 question, didn't satisfy me.

I did with my approach, you may check it.

@BASANT KUMAR

Don't add two different questions, in one post from next time.

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@shaikh thnxx ver much for uploading these images

thesereally helped me

Answer 1

NEW APPROACH :-

For question 1 :-

see my answer at https://gateoverflow.in/118640/gate2017-2-44?show=289691#a289691

For question 2:-

Before coming to question 2, please understand the procedure which is used in the Q1, due to it is same concept

For clarity images :- https://drive.google.com/open?id=1dka-cqVm6ZppPI81Mm9WztqV_pG5iMMh

Recommended to see the same type of question  https://gateoverflow.in/272638/total-conflict-serializable-schedules

 

OLD APPROACH :-

For Question 1 :-

For clarity images https://drive.google.com/open?id=1o12YFKd8JLhlXWLB1D3tuph91ENJ0OCd

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For Question 2 :-

For Case 3 and Case 4 :-

Note that R2(B) should be after W1(B)

Comments

No it's not present

But we can use this DAG approach in  similar type of questions where we have to find no of conflict serializable right  ?

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thanks :)

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@Shaik Masthan

In the first method, second image –

 https://drive.google.com/file/d/1Z-tlThOEyVlBjV8Zky5_Wc3z7L-hvSGc/view?usp=sharing

Total number of possibilities such that T2->T1 = T1->T2 = 56 is because of the symmetricity of the DAG, right? i.e if we draw the graph for T2->T1 it will be same as that of T1->T2 with the corresponding labels interchanged(G for A, H for B etc) , right?