best thing in this course is that if u didn't understand anything from the lecture u gonna enjoy the outro my Greetings from Egy best instructor ever
@hritiksoni6930
4 жыл бұрын
Andy! You are the best PERIOD.
@pb25193
4 жыл бұрын
At 58:00 the sum is supposed to be 2114. Andy forgets to add 100 to B in the end. The error is $6 and not $106.
@prathibhapb
3 жыл бұрын
I was thinking of exactly the same!
@yenzyhebron5278
Жыл бұрын
Yeah I'm so mad at this, took me a while to figure out it was just a typo since I was only reading the slides
@shrivats83
4 жыл бұрын
May be I am missing some subtleties here: At kzitem.info/news/bejne/zo981WWXfmZoq3o, Prof Pavlo says that if there are no cycles in the graph, then the schedules are conflict serializable. However, if we go back to the write-read conflict (dirty reads) example, where T1 writes a value of $12 for object A and T2 reads $12 and writes a value of $14 for the same object (T2->T1), commits the transaction T2 after which T1 aborts, I don't believe there is a cycle. But clearly, there is a conflict. Could someone clarify this?
@pb25193
4 жыл бұрын
Think of the abort as another "write" which undo's the first write, and there you will have your cycle :D
@hemanthaugust7217
Жыл бұрын
1:10:34, Assume that these 3 transactions were issued to RDBMS, when it received T3 Commit, would it just wait until T1 & T2 issue commit statements? how does DB make use of what has been just explained? Thanks
@happythoughts67
3 жыл бұрын
love your courses. Thank you so much. A small question, at time around 1:16:52, if w(A) = A +1, which depend on previous value of A, then it isn't safe? So the order also matters. Am I right? Thanks!
@ruiyangxu5101
3 жыл бұрын
Can precedence graph detect write-read conflict? For example, t1 = [write(A), abort]. t2 = [read(A), commit]. Schedule is [write(A), read(A), abort T1, commit T2]. The precedence graph is from t1 to t2, which does not have a cycle. However, T2 should never get the value written by T1 if they were executed in sequence.
@shangruyi5439
3 жыл бұрын
Probably not? They are committed at the end, without an abort.
@shangruyi5439
3 жыл бұрын
One thing comes to mind. Abort is not a single operation. It contains the reverse back of the object version, which is a read-write operation in the end. Given this, the precedence graph is completed. There will be a cycle and it is not conflict-serializable.
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