In $T_{1}\rightarrow T_{2}$
$T_{1}$ |
$T_{2}$ |
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$r(A)$ |
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$w(A)$ |
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$r(B)$ |
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$w(B)$ |
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${\color{DarkRed} {r(C)}}$ |
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${\color{DarkRed} {w(C)}}$ |
${\color{DarkRed} {r(B)}}$ |
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${\color{DarkRed} {w(B)}}$ |
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$r(C)$ |
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$w(C)$ |
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$r(D)$ |
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$w(D)$ |
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Here those colored $4$ transaction has no conflict
So, Number of conflict will be $=\frac{4!}{2!2!}=6$
We divided by $2!$ because $r(C),w(C)$ cannot combine among them
Similarly $r(B),w(B)$ cannot arrange among them.
Now for $T_{2}\rightarrow T_{1}$
Case $1:$
$r1(A),w1(A)$ has no conflict with any transaction of $T_{2}$
So, total conflict serializable schedule$=\frac{8!}{2!6!}=28$
Now for $r1(B),w1(B)$ has no conflict with $r2(C),w2(C)$,$r2(D),w2(D)$ of $T_{2}$
So, total conflict serializable schedule$=\frac{6!}{2!4!}=15$
Now for $r1(C),w1(C)$ has no conflict with $r2(D),w2(D)$
So, total conflict serializable schedule$=\frac{4!}{2!2!}=6$
Total number of conflict serializable schedule$=6+28+15+6=55$