Mohamed Houri’s Oracle Notes

January 20, 2016

Natural and Adjusted Hybrid Histogram

Filed under: Oracle,Statistics — hourim @ 7:16 pm

I was going to write a comment in this Jonathan Lewis article and have finally decided to write a blog article because it turned to be a long comment.  In the above mentioned article a reader was wondering why the bucket size of his modified data set is not obeying the minimum bucket size explained by Jonathan. Coincidentally I am writing a second article on Hybrid histogram for allthingsOracle where I have used my proper terminology to define two types of Hybrid histogram: a first type, which I have named ‘’Natural Hybrid’, is close to Jonathan’s original data set. And a second type, which I have named ‘’Adjusted Hybrid’’ is of the same vein as the reader modified data set. By  ‘’Natural Hybrid’’ type  I refer to a data set a data set that doesn’t qualify for a TOP-Frequency histogram because the threshold is greater than the naturally non-adjusted TopNRows. By ‘Adjusted Hybrid’type, I am refering to a data set that initially satisfies the TOP-Frequency threshold but  fails to qualify for a TOP-Frequency histogram because Oracle finds at the middle of the process that the Adjusted TopNRows is greater than the threshold .

Let’s explain with example. Firstly here’s a model that qualify for a “Natural Hybrid” histogram:

SQL> create table t1 (id number, n1 number);
SQL> start InsT1.sql (see downloadable script at the end)

If we gather histogram for this data set of 37 distinct values using 20 buckets we will obtain a HYBRID histogram because the TOP-Frequency threshold accounts for 95 rows while the TOP-20 rows account for 80 rows as shown below:

SQL> select round ((20-1)/20 * 100) threshold from dual;

SQL> select
         sum (cnt) TopNRows
    from (select
           ,count(*) cnt
         from t1
         group by n1
          order by count(*) desc
   where rownum <= 20;


In order to compute the Hybrid histogram information, Oracle will, in this case, use the bucket-frequency method explained by Jonathan Lewis. This method uses the minimum bucket size of 5 (not valid at the end of the data set though) and the unchanged initial number of bucket of 20 condition is respected. This is what I prefer labelling a Natural Hybrid histogram.

Let’s now use the reader model

create table t_jar (id number, n1 number);
insert into t_jar values (1,5 );
insert into t_jar values (1,5 );
insert into t_jar values (1,7 );
insert into t_jar values (1,7 );
insert into t_jar values (1,7 );
insert into t_jar values (1,7 );
insert into t_jar values (1,10 );
insert into t_jar values (1,12 );
insert into t_jar values (1,15 );
insert into t_jar values (1,15 );
insert into t_jar values (1,15 );
insert into t_jar values (1,20 );

This new data set of 6 distinct values over 12 rows will normally qualify for a TOP-3 Frequency histogram as it satisfies the threshold formula:

SQL> select round ((3-1)/3 * 12) threshold from dual;

SQL> select
          sum (cnt) TopNRows
     from (select
            ,count(*) cnt
          from t_jar
          group by n1
          order by count(*) desc
     where rownum >= 3;

However, Oracle will not accept this at face value. It has to check if the low and high values are among the TOP-3 distinct values. If one of these values (or both) are not in the TOP-3, oracle will force it into the histogram, exclude the least repetitive value from the TOP-3, adjust the TopNRows and check again whether theses modifications have not altered the data set so that it still qualify or not for a TOP-Frequency histogram.

Here’s below a snippet of a corresponding dbms_stats trace file

SQL> exec dbms_stats.set_global_prefs ('TRACE', to_char (1+16));

          (user, 't_jar', method_opt =>; 'for columns n1 size 3');
SQL> exec dbms_stats.set_global_prefs('TRACE', null);
DBMS_STATS:   Y    Y    Y    Y    Y    Y    Y                        Y    N1
DBMS_STATS: Approximate NDV Options
DBMS_STATS: start processing top n values for column N1
DBMS_STATS: topn sql (len: 415):
DBMS_STATS: +++ select /*+
                       no_parallel(t) no_parallel_index(t)
			   dbms_stats cursor_sharing_exact use_weak_name_resl
                       dynamic_sampling(0) no_monitoring
			 xmlindex_sel_idx_tbl no_substrb_pad
		   substrb(N1,16,0,64),1,240) val,
                 rowidtochar(rowid) rwid
		  from "XXXX&".T_JAR& t
		  where rowid in(chartorowid('AAAJ4MAAEAACN4EAAA')
		order by N1;
DBMS_STATS: remove last bucket: Typ=2 Len=2: c1,10 add: Typ=2 Len=2: c1,15
DBMS_STATS: removal_count: 1 total_nonnull_rows: 12 mnb:  3
DBMS_STATS: adjusted coverage: .667

Here’s the data set

SQL> select *
       (select n1, count(1) cnt
         from t_jar
        group by n1
         order by n1);

        N1        CNT
---------- ----------
         5          2 -> low value
         7          4
        10          1
        12          1
        15          3
        20          1 -> high value
6 rows selected.

And here are the TOP-3 rows of the same data set:

SQL> select
     from (select
            ,count(*) cnt
          from t_jar
          group by n1
          order by count(*) desc)
     where rownum <= 3;

        N1        CNT
---------- ----------
         7          4
        15          3
         5          2
3 rows selected.

Since the high value is not in the TOP-3 it will be forced into the histogram to the cost of the exclusion of the least repetitive TOP-3 values which is 5 in this case (frequency =2). But, before doing this task, Oracle has to check if, after this high value forcing, the data set is still qualifying for a Top-Frequency using the AdjustedTopNRows

 AdjustedTopNRows = TopNRows – 2 + 1 = 9-2+1 = 8

The AdjustedTopNRows is not any more greater than the threshold of 8 which signifies that Oracle will finally stop collecting TOP-Frequency and transform what it has already gathered into  a HYBRID histogram as shown in the same trace file (Trying to convert frequency histogram to hybrid)

DBMS_STATS: adjusted coverage: .667
DBMS_STATS: hist_type in exec_get_topn: 2048 ndv:6 mnb:3
DBMS_STATS: Evaluating frequency histogram for col: N1
DBMS_STATS:  number of values = 4, max # of buckects = 3, pct = 100, ssize = 12
DBMS_STATS:  Trying to convert frequency histogram to hybrid
DBMS_STATS:  > cdn 10, popFreq 4, popCnt 1, bktSize 4, bktSzFrc 0
DBMS_STATS:  Evaluating hybrid histogram:  cht.count 3, mnb 3, ssize 10, min_ssize 12, appr_ndv  TRUE, ndv 4, selNdv 1,
selFreq 4,
pct 100, avg_bktsize 3, csr.hreq TRUE, normalize TRUE
DBMS_STATS:   Histogram gathering flags: 7
DBMS_STATS:  Accepting histogram
DBMS_STATS: done_hist in process_topn: TRUE  csr.ccnt: 1
DBMS_STATS: Mark column N1 as top N computed
DBMS_STATS: Skip topn computed column N1 numhist: 0

This is what I have labelled an ‘’Adjusted Hybrid’’ histogram which is collected using a method that seems not involving the bucket size and which tends to be a top-frequency to hybrid conversion.

        (row_number() over(order by ept_nbr)-1) NumBucket
        ,ept_nbr - (lag(ept_nbr,1,0) over(order by ept_nbr)) "new bucket size"
        ,bucket_size "original bucket_size"
             ah.endpoint_number            ept_nbr
            ,ah.endpoint_actual_value      ept_act_val
            ,lag(ah.endpoint_number,1,0) over(order by ah.endpoint_number) ept_lag
            ,ah.endpoint_repeat_count rpt_cnt
            ,at.sample_size/at.num_buckets bucket_size
            user_tab_histograms      ah
           ,user_tab_col_statistics  at
         WHERE ah.table_name  = at.table_name
         AND ah.column_name = at.column_name
         AND ah.table_name  = 'T_JAR'
         AND ah.column_name = 'N1'
       ) ORDER BY ept_nbr;

NUMBUCKET    EPT_NBR EPT_ACT_VA    RPT_CNT new bucket size original bucket_size
---------- ---------- ---------- ---------- --------------- --------------------
         0          2 5                   2               2                    4
         1          6 7                   4               4                    4
         2         10 20                  1               4                    4

In passing, the adjusted coverage mentioned in the above trace file is nothing than

AdjustedTopNRows/num_rows = 8/12 = 0.667

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