T1 |
T2 |
T3 |
R(x) |
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R(y) |
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W(x) |
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R(y) |
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W(y) |
W(x) |
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R(y) |
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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 |
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R(y) |
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R(y) |
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W(y) |
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R(y) |
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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.