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Before we perform this mass deletion, we have no blocks on the FREELIST and there are 1,043 blocks in the leafs of the index, holding data. Now, we ll perform the delete and measure the space utilization again: ops$tkyte@ORA11GR2> delete from big_table where id <= 250000; 250000 rows deleted. ops$tkyte@ORA11GR2> commit; Commit complete. ops$tkyte@ORA11GR2> declare 2 l_freelist_blocks number; 3 begin 4 dbms_space.free_blocks 5 ( segment_owner => user, 6 segment_name => 'BIG_TABLE_PK', 7 segment_type => 'INDEX', 8 freelist_group_id => 0, 9 free_blks => l_freelist_blocks ); 10 dbms_output.put_line( 'blocks on freelist = ' || l_freelist_blocks ); 11 dbms_stats.gather_index_stats 12 ( user, 'BIG_TABLE_PK' ); 13 end; 14 / blocks on freelist = 520 PL/SQL procedure successfully completed. ops$tkyte@ORA11GR2> select leaf_blocks from user_indexes 2 where index_name = 'BIG_TABLE_PK'; LEAF_BLOCKS ----------523 As we can see, over half of the index is on the FREELIST now (520 blocks) and there are only 523 leaf blocks. If we add 523 and 520, we get the original 1043. This means the blocks are totally empty and ready to be reused (blocks on the FREELIST for an index must be empty, unlike blocks on the FREELIST for a heap organized table). This demonstration highlights two points: Space is reused on index blocks as soon as a row comes along that can reuse it. When an index block is emptied, it can be taken out of the index structure and may be reused later. This is probably the genesis of this myth in the first place: blocks are not visible as having free space on them in an index structure as they are in a table. In a table, you can see blocks on the FREELIST, even if they have data on them. In an index, you will only see completely empty blocks on the FREELIST; blocks that have at least one index entry (and remaining free space) will not be as clearly visible.
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Myth: Most Discriminating Elements Should Be First
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This seems like common sense. If you are going to create an index on the columns C1 and C2 in a table T with 100,000 rows, and you find C1 has 100,000 distinct values and C2 has 25,000 distinct values, you would want to create the index on T(C1,C2). This means that C1 should be first, which is the commonsense approach. The fact is, when comparing vectors of data (consider C1, C2 to be a vector), it doesn t matter which you put first. Consider the following example. We will create a table based on ALL_OBJECTS and an index on the OWNER, OBJECT_TYPE, and OBJECT_NAME columns (least discriminating to most discriminating) and also on OBJECT_NAME, OBJECT_TYPE, and OWNER: ops$tkyte@ORA11GR2> create table t 2 as 3 select * from all_objects; Table created. ops$tkyte@ORA11GR2> create index t_idx_1 on t(owner,object_type,object_name); Index created. ops$tkyte@ORA11GR2> create index t_idx_2 on t(object_name,object_type,owner); Index created. ops$tkyte@ORA11GR2> select count(distinct owner), count(distinct object_type), 2 count(distinct object_name ), count(*) 3 from t; DISTINCTOWNER DISTINCTOBJECT_TYPE DISTINCTOBJECT_NAME COUNT(*) ------------- ------------------- ------------------- -------28 36 28537 48243 Now, to show that neither is more efficient space-wise, we ll measure their space utilization: ops$tkyte@ORA11GR2> analyze index t_idx_1 validate structure; Index analyzed. ops$tkyte@ORA11GR2> select btree_space, pct_used, opt_cmpr_count, opt_cmpr_pctsave 2 from index_stats; BTREE_SPACE PCT ----------- -----2702744 89.0 OPT_CMPR_COUNT OPT_CMPR_PCTSAVE -------------- ---------------2 28
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ops$tkyte@ORA11GR2> analyze index t_idx_2 validate structure; Index analyzed. ops$tkyte@ORA11GR2> select btree_space, pct_used, opt_cmpr_count, opt_cmpr_pctsave 2 from index_stats; BTREE_SPACE PCT ----------- -----2702744 89.0 OPT_CMPR_COUNT OPT_CMPR_PCTSAVE -------------- ---------------1 13
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They use exactly the same amount of space, down to the byte there are no differences there. However, the first index is a lot more compressible if we use index key compression, as evidenced by the OPT_CMP_PCTSAVE value. There is an argument for arranging the columns in the index in order from the least discriminating to the most discriminating. Now let s see how they perform, to determine if either index is generally more efficient than the other. To test this, we ll use a PL/SQL block with hinted queries (so as to use one index or the other): ops$tkyte@ORA11GR2> alter session set sql_trace=true; Session altered. ops$tkyte@ORA11GR2> declare 2 cnt int; 3 begin 4 for x in ( select /*+FULL(t)*/ owner, object_type, object_name from t ) 5 loop 6 select /*+ INDEX( t t_idx_1 ) */ count(*) into cnt 7 from t 8 where object_name = x.object_name 9 and object_type = x.object_type 10 and owner = x.owner; 11 12 select /*+ INDEX( t t_idx_2 ) */ count(*) into cnt 13 from t 14 where object_name = x.object_name 15 and object_type = x.object_type 16 and owner = x.owner; 17 end loop; 18 end; 19 / PL/SQL procedure successfully completed. These queries read every single row in the table by means of the index. The TKPROF report shows us the following: SELECT /*+ INDEX( t t_idx_1 ) */ COUNT(*) FROM T WHERE OBJECT_NAME = :B3 AND OBJECT_TYPE = :B2 AND OWNER = :B1 call count ------- -----Parse 1 Execute 48243 Fetch 48243 ------- -----total 96487 cpu elapsed disk query current -------- ---------- ---------- ---------- ---------0.00 0.00 0 0 0 10.63 10.78 0 0 0 1.90 1.77 0 145133 0 -------- ---------- ---------- ---------- ---------12.53 12.55 0 145133 0 rows ---------0 0 48243 ---------48243
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Rows Row Source Operation ------- --------------------------------------------------48243 SORT AGGREGATE (cr=145133 pr=0 pw=0 time=2334197 us) 57879 INDEX RANGE SCAN T_IDX_1 (cr=145133 pr=0 pw=0 time=1440672 us)(object ******************************************************************************** SELECT /*+ INDEX( t t_idx_2 ) */ COUNT(*) FROM T WHERE OBJECT_NAME = :B3 AND OBJECT_TYPE = :B2 AND OWNER = :B1
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