Question

No. Of serial schedules view equal to S

S: R1(A), R3(D), W1(B), R2(B), W3(B), R4(B), W2(C), R5(C), W4(E), R5(E), W5(B) .

If possible please provide detailed solutions.

Thank you in advance

Comments

https://gateoverflow.in/169297/serial-schedule

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let me explain :

dependecies : 

FOR FINAL WRITE

•  B = T5  here condition is (T1, T3)-> T5
• E = T4 NO RESTRICTION
• C = T2 NO RESTRICTION

FOR INITIAL READ 

• A= T1 NOT MATTER SINCE NO WRITE ON A
• D = T3 NO MATTER SINCE NO WRITE ON D

FOR WRITE READ 

• B = T1-> T2 ( NO WRITE COME IN BETWEEN I.E. T3 AND T5) 
• B = T3-> T4 ( NO WRITE COME IN BETWEEN I.E. T1 AND T5) 
• C = T2 -> T5
• E= T4-> T5

NOW CONDITIONS :

T5 MUST BE LAST 

T1-> T2 AND T3-> T4 COME ADJACENT ONLY

SO T1-> T2-> T3-> T4-> T5 AND T3-> T4->T1->T2-> T5

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@Ashwin here W3(B) extra present. It's totally different question.

@Anu007

Actually for ans. I am also confused about this I am getting only one. But ans is 2.

1,2,3,4,5. But ans says that one extra which is 3, 4, 1,2,5.

Please describe.

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CHECK NOW

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plz any one gives correct method for this

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@Anu007

hi! please check for T3..T1..T2..T4..T5

view eqiv rules are maintained as far as i see this...

 

Answer 1

LET ME EXPLAIN :

LOOK AT DEPENDENCIES : 

FOR FINAL WRITE

•  B = T5  here condition is (T1, T3)-> T5
• E = T4 NO RESTRICTION
• C = T2 NO RESTRICTION

FOR INITIAL READ 

• A= T1 NOT MATTER SINCE NO WRITE ON A
• D = T3 NOT MATTER SINCE NO WRITE ON D

FOR WRITE READ 

• B = T1-> T2 ( NO WRITE COME IN BETWEEN I.E. T3 AND T5) 
• B = T3-> T4 ( NO WRITE COME IN BETWEEN I.E. T1 AND T5) 
• C = T2 -> T5
• E= T4-> T5

NOW CONDITIONS :

T5 MUST BE LAST 

T1-> T2 AND T3-> T4 COME ADJACENT ONLY

SO T1-> T2-> T3-> T4-> T5 AND T3-> T4->T1->T2-> T5

Comments

When we use T3 --> T2 which generates new updated read (WR)problem. So how can we say it would be view serializable ?

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BEFORE T2 READ WE WRITE IT BY T2 SEE T1->T2

SO EFFECT OF T3 IS LOST

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Which of following is the Problem for data item B

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IN VIEW SERIAL ONLY LAST WRITE MATTER SO  T3 -> T5 NO ISSUE. SAME T3->T1 NO ISSUE

YOU ARE CHECKING CONFLICT SERIALIZABLE.

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That's means conflict problem (WW, WR RW for same data item) is not same as view problem or serializable problem ?

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YES SOME RESTRICTION ARE  REMOVED FROM VIEW , IN COMPARISION TO CONFLICT.

LIKE WW.

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For above image all are conflict pairs?

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Is is true or not?

(RW , WR conflict pair is same precedence for two schedule S and S') iff  (intial read and updated read same for S and S')

But only implication for 

(WW conflict pair) --> (final write)

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@ Anu007 If No. Of serial schedules conflict equal to S was asked then just this one right, T1-> T2-> T3-> T4-> T5