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CS422
Multiversion Lock

Author	Message
lishenyu Posts: 103	Posted 13:37 Dec 03, 2015 \| ~T1: w1(b1), w1(b2) T2: w2(b1), w2(b2) T3: r3(b1), r3(b2) T4: w4(b2) After watching the video I still didn't hear clearly at last about r3(b2) reads Tx1 or Tx2 ? The protocol says "~it reads from the block that was most recently committed at the time when this transaction began" , 'this transaction' refers to which transaction ? Many thanks!
lishenyu Posts: 103	Posted 17:59 Dec 03, 2015 \| As far as I know if a sequence has no cycle in its precedence graph then it's serializable. So what's the relationship between multiversion and serializable? thanks!
gbkprajapati Posts: 8	Posted 18:03 Dec 03, 2015 \| r3(b2) reads T1 . as per protocol says. Last edited by gbkprajapati at 18:03 Dec 03, 2015.
cysun Posts: 2935	Posted 20:01 Dec 03, 2015 \| lishenyu wrote: As far as I know if a sequence has no cycle in its precedence graph then it's serializable. So what's the relationship between multiversion and serializable? thanks! Multiversion Locking is a protocol that ensures serializability.
lishenyu Posts: 103	Posted 20:40 Dec 03, 2015 \| cysun wrote: lishenyu wrote: As far as I know if a sequence has no cycle in its precedence graph then it's serializable. So what's the relationship between multiversion and serializable? thanks! Multiversion Locking is a protocol that ensures serializability. for this sequence: ~w1(b1),w1(b2),c1,w2(b1),r3(b1),w4(b2),c4,r3(b2),c3,w2(b2),c2 the most recently committed transaction when T3 began is T4, so r3(b2) should read b2 from T4, is this understanding correct ?if not ,why ? thanks!
cysun Posts: 2935	Posted 08:45 Dec 04, 2015 \| lishenyu wrote: cysun wrote: lishenyu wrote: As far as I know if a sequence has no cycle in its precedence graph then it's serializable. So what's the relationship between multiversion and serializable? thanks! Multiversion Locking is a protocol that ensures serializability. for this sequence: ~w1(b1),w1(b2),c1,w2(b1),r3(b1),w4(b2),c4,r3(b2),c3,w2(b2),c2 the most recently committed transaction when T3 began is T4, so r3(b2) should read b2 from T4, is this understanding correct ?if not ,why ? thanks! That's incorrect. Because r3(b1) is before c4, T4 didn't commit before T3 began.