When compaction detects a parallel checkpoint, EBUSY is returned from an internal call and compaction is retried. See the corresponding code here.

Even though this is only a theory so far - a checkpoint could always be processing the same table that compaction is trying to work on and this would happen repeatedly. Currently, we try to compact a table up to 100 times which means we could have this scenario 100 times which could be undesirable.

This ticket should validate the theory and come up with a reproducer if this conclusion of the first step comes back as possible. Finally, we could detect N (<100) interruptions before exiting compaction.