Consider the following schedules involving three transaction :
$S_{1} : r_{1} (x), r_{2} (z), r_{1} (z), r_{3} (x), r_{3} (y), w_{1} (x), w_{3} (y), r_{2} (y), w_{2} (z), w_{2} (y)$

$S_{2} : r_{1} (x), r_{2} (z), r_{3} (x), r_{1} (z), r_{2} (y), r_{3} (y), w_{1} (x), w_{2} (z), w_{3} (y), w_{2} (y)$

Which of the following schedules is conflict serializable?

B o t h S_{1} a n d S_{2}

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O n l y S_{1}

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O n l y S_{2}

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None of the above

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Solution

The correct option is B $O n l y S_{1}$
A schedule S is contact serializable if the precedence graph of S does not contain any cycle.
Precedence graph of
$S_{1}$ .

It does not contain any cycle so conflict serializable,
Precedence graph of
$S_{2}$

It contain cycle so not conflict serializable.
So option (b) correct.

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S_{1} : W_{2} (x), W_{1} (x), R_{3} (x), W_{2} (y), R_{3} (y), R_{3} (z), R_{2} (x), R_{1} (y)

S_{2} : R_{3} (z), W_{2} (x), W_{2} (y), R_{1} (x), R_{3} (x), R_{2} (z), R_{3} (y), W_{1} (x)

S : r_{1} (x), r_{2} (x), ω_{2} (x), ω_{3} (x), r_{1} (z), ω_{1} (x)

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