On 04/12/2018 10:08 AM, Vladimir Sementsov-Ogievskiy wrote:
I propose, not to say that bitmap represents a checkpoint. It is simpler to say (and it reflects the reality) that bitmap is a difference between two consecutive checkpoints. And we can say, that active state is some kind of a checkpoint, current point in time.
So, we have checkpoints (5* is an active state) which are points in time:
1 2 3 4 5*
Oh -- the most recent checkpoint there doesn't belong to a ***specific time*** yet. It's a floating checkpoint -- it always represents the most current version. It's not really a checkpoint at all. 1, 2, 3, and 4 however are associated with a specific timestamp though.
And bitmaps, first three are disabled, last is enabled:
"1->2", "2->3", "3->4", "4->5*"
OK; so 1->2, 2->3 and 3->4 define deltas between two ***defined*** points in time. 4->5* however is only anchored by one specific point in time, and is floating just like the most recent checkpoint is floating.
So, remove first checkpoint: just remove bitmap "A->B".
I assume you mean "1->2" here. And... yes, I agree -- if you don't care about your very first checkpoint anymore, you can just delete the first bitmap, too.
Remove any other checkpoint N: create new bitmap "(N-1)->(N+1)" = merge("(N-1)->N", "N->(N+1)"), drop bitmaps "(N-1)->N" and "N->(N+1)".
err, okay, so let's say we want to drop checkpoint 3: create: "2->4" merge: "2->3", "3->4" [and presumably store in "2->4"] drop: 2->3, 3->4 OK, that makes more sense to me. In essence; (1) We could consider this 2->3 absorbing 3->4, or (2) 3->4 absorbing 2->3 and in either case it's the same, really.
If the latter was active, the new one becomes active. And we cant remove 5* checkpoint, as it is an active state, not an actual checkpoint.
OK, crystal. --js