does incremented column makes the b-tree index on the column unbalanced?

问题

I have been thinking about two questions. Couldn't find any resources on the internet about this. How do dbms handle it ? Or do they ? Especially Oracle.

Before the questions, here is an example: Say I have a master table "MASTER" and slave table "SLAVE". Master table has an "ID" column which is the primary key and index is created by Oracle.Slave table has the foreign key "MASTER_ID" which refers to master table and "SLAVE_NO". These two together is the primary key of slave table, which is again indexed.

 **MASTER**  |  **SLAVE**
     (P) ID <------> (P)(F) MASTER_ID 
                     (P) SLAVE_NO

Now the questions;

1- If MASTER_ID is an autoincremented column, and no record is ever deleted, doesn't this get the table's index unbalanced ? Does Oracle rebuilds indexes periodically ? As far as i know Oracle only balances index branches at build time. Does Oracle re-builds indexes Automatically ever ? Say if the level goes up to some high levels ?

2- Assuming Oracle does not rebuild automatically, apart from scheduling a job that rebuilds index periodically, would it be wiser to order SLAVE table's primary key columns reverse ? I mean instead of "MASTER_ID", "SLAVE_NO" ordering it as "SLAVE_NO", "MASTER_ID"i, would it help the slave table's b-tree index be more balanced ? (Well each master table might not have exact number of slave records, but still, seems better than reverse order)

Anyone know anything about that ? Or opinions ?

回答1:

If MASTER_ID is an autoincremented column, and no record is ever deleted, doesn't this get the table's index unbalanced ?

Oracle's indexes are never "unbalanced": every leaf in the index is at the same depth as any other leaf.

No page split introduces a new level by itself: a leaf page does not become a parent for new pages like it would be on a non-self-balancing tree.

Instead, a sibling for the split page is made and the new record (plus possibly some of the records from the old page) go to the new page. A pointer to the new page is added to the parent.

If the parent page is out of space too (can't accept the pointer to the newly created leaf page), it gets split as well, and so on.

These splits can propagate up to the root page, whose split is the only thing which increases the index depth (and does it for all pages at once).

Index pages are additionally organized into double-linked lists, each list on its own level. This would be impossible if the tree were unbalanced.

If master_id is auto-incremented this means that all splits occur at the end (such called 90/10 splits) which makes the most dense index possible.

Would it be wiser to order SLAVE table's primary key columns reverse?

No, it would not, for the reasons above.

If you join slave to master often, you may consider creating a CLUSTER of the two tables, indexed by master_id. This means that the records from both tables, sharing the same master_id, go to the same or nearby data pages which makes a join between them very fast.

When the engine found a record from master, with an index or whatever, this also means it already has found the records from slave to be joined with that master. And vice versa, locating a slave also means locating its master.

回答2:

The b-tree index on MASTER_ID will remain balanced for most useful definitions of "balanced". In particular, the distance between the root block and any child block will always be the same and the amount of data in any branch will be at least roughly euqal to the amount of data in any other sibling branch.

As you insert data into an index on a sequence-generated column, Oracle will perform 90-10 block splits on the leaves when the amount of data in any particular level increases. If, for example, you have a leaf block that can hold 10 rows, when it is full and you want to add an 11th row, Oracle creates a new block, leaves the first 9 entries in the first block, puts 2 entries in the new block, and updates the parent block with the address of the new block. If the parent block needs to split because it is holding addresses for too many children, a similar process takes place. That leaves the indexes relatively balanced throughout their life. Richard Foote (the expert on Oracle indexes) has an excellent blog on When Does an Oracle Index Increase in Height that goes into much more detail about how this works.

The only case that you potentially have to worry about an index becoming skewed is when you regularly delete most but not all blocks from the left-hand side of the index. If, for example, you decide to delete 90% of the data from the left-hand side of the index leaving one row in every leaf node pointing to data, your index can become unbalanced in the sense that some paths lead to vastly more data than other paths. Oracle doesn't reclaim index leaf nodes until they are completely empty so you can end up with the left-hand side of the index using a lot more space than is really necessary. This doesn't really affect the performance of the system much-- it is more of a space utilization issue-- but it can be fixed by coalescing the index after you've purged the data (or structuring your purges so that you don't leave lots of sparse blocks).

来源：https://stackoverflow.com/questions/10901558/does-incremented-column-makes-the-b-tree-index-on-the-column-unbalanced