How atomicity is maintained in this scheme(check pointing) because the operations which appears before checkpoint address being done permanently and operations after checkpoint of active transaction are undone. So some part of active transactions is done and some doesn't. Please clarify.
In this explanation, 8:33 it is provided that all the transactions in the undo list will be undone in the reverse order but while solving the problem it was not taken into consideration.
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Thanks
thanks !
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Excellent sir
Really easily understood.
Thank u sir
How atomicity is maintained in this scheme(check pointing) because the operations which appears before checkpoint address being done permanently and operations after checkpoint of active transaction are undone. So some part of active transactions is done and some doesn't. Please clarify.
did anyone got any answer for this doubt?
T2 was added because it was never committed
Thank you so much, very well explained, to the point.
Thank you 👍
At 12:36 for T3, why A = 9? Shouldn't it be A = 1 ?
Because T3 has committed after checkpoint so need to be redo.
Can someone tell does it matter if undo list is done before redo list or vice versa?
You have to do undo list first
miny miny miny miny thanks :P
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Jai Shree Ram
Good explanation. Thanks.
Wanna have idea about check points characteristics
amazing. best explained
T2 and then T4. Isnt it??
In this explanation, 8:33 it is provided that all the transactions in the undo list will be undone in the reverse order but while solving the problem it was not taken into consideration.
@@biginepallinagasai4403 Reverse with respect to log records order not the list order. There is no wrong in the solution.
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