On Tue, Jun 18, 2024 at 11:54 AM Dilip Kumar <dilipbalaut@gmail.com> wrote:
>
> On Mon, Jun 17, 2024 at 8:51 PM Robert Haas <robertmhaas@gmail.com> wrote:
> >
> > On Mon, Jun 17, 2024 at 1:42 AM Amit Kapila <amit.kapila16@gmail.com> wrote:
> > > The difference w.r.t the existing mechanisms for holding deleted data
> > > is that we don't know whether we need to hold off the vacuum from
> > > cleaning up the rows because we can't say with any certainty whether
> > > other nodes will perform any conflicting operations in the future.
> > > Using the example we discussed,
> > > Node A:
> > > T1: INSERT INTO t (id, value) VALUES (1,1);
> > > T2: DELETE FROM t WHERE id = 1;
> > >
> > > Node B:
> > > T3: UPDATE t SET value = 2 WHERE id = 1;
> > >
> > > Say the order of receiving the commands is T1-T2-T3. We can't predict
> > > whether we will ever get T-3, so on what basis shall we try to prevent
> > > vacuum from removing the deleted row?
> >
> > The problem arises because T2 and T3 might be applied out of order on
> > some nodes. Once either one of them has been applied on every node, no
> > further conflicts are possible.
>
> If we decide to skip the update whether the row is missing or deleted,
> we indeed reach the same end result regardless of the order of T2, T3,
> and Vacuum. Here's how it looks in each case:
>
> Case 1: T1, T2, Vacuum, T3 -> Skip the update for a non-existing row
> -> end result we do not have a row.
> Case 2: T1, T2, T3 -> Skip the update for a deleted row -> end result
> we do not have a row.
> Case 3: T1, T3, T2 -> deleted the row -> end result we do not have a row.
>
In case 3, how can deletion be successful? The row required to be
deleted has already been updated.
--
With Regards,
Amit Kapila.