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Consider the given schedule and choose the suitable option S = T1:R(x), T1:R(y), T1:W(x), T2:R(y), T3:W(y), T1:W(x), T2:R(y)

  1. Schedule is view serializable
  2. Schedule is conflict serializable but not view serializable
  3. Schedule is view serializable but not conflict serializable
  4. Neither view serializable nor conflict serializable

1 Answer

Best answer
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T1 T2 T3
R(x)    
R(y)    
W(x)    
  R(y)  
    W(y)
W(x)    
  R(y)  

Here $T1$ is only doing operations on Data item $x$ so we can ignore the $R(x)$ and $W(x)$ as they will not produce any conflicts.

T1 T2 T3
     
R(y)    
     
  R(y)  
    W(y)
     
  R(y)  

Since there is a cycle $T2\rightarrow T3\rightarrow T2$ produced due to $R(y)W(y)$ and $W(y)R(y)$ conflicts

$\Rightarrow$ The schedule is $not$ conflict serializable.

For view serializable,

Since the graph contains a blind write at $T3$ so view serializable schedule is not equal to conflict serializable schedule.

$T1:R(y)$ and $T2:R(y)$ are initial reads and $T2:R(y)$ is at last and also we have only 1 Write operation by T3:W(y) so we can't make a serial schedule by executing in the any order.

So the sequence is not view serializable.

$\therefore$ Option $D.$ is the correct answer.

edited by

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