[Translated from mos] explanation of histogram information and mos Article histogram Information

Interpreting histogram Information

Source:

Interpreting Histogram Info (Doc ID 72539.1)

Applicable:

Oracle Database-Enterprise Edition-Version 7.3.0.0 and later

Oracle Database-Standard Edition-Version 7.3.0.0 and later

Oracle Database-Personal Edition-Version 7.3.0.0 and later

Information in this document applies to any platform.

Purpose:

How is the histogram information stored and explained.

Range:

For other useful histograms, refer:

Document 1445372.1 Histograms: An Overview (10g and Above)

Details:

A histogram is a mechanism used to store detailed information about column data. This data is used by CBO to determine the optimal access path for a query statement ).

Without a histogram, the optimizer relies on all the information: the high and low values of a column, the number of different values of the column, the number of null values of the column, and the total number of records of the table.

(In fact, the high and low values of a column are stored in raw format, so they are not particularly useful). Other information can be found in dictionary views.

When no column statistics are available, the optimizer assumes that the data is evenly distributed. For equivalent predicates, a column selectivity is generated. The selection rate is calculated as follows: 1/NVD (Number of Distinct Values)

When a histogram is available, you can access more distribution information of row data.

When the data distribution of a column is not balanced (that is, the data distribution of the column is highly skewed-the data distribution is very skewed), Oracle can store the column histogram to provide a better choice rate. this will produce a better execution plan than using standard statistics (high and low values plus Number of Distinct Values)

For specific implementation (In terms of implementation), we can choose to save each different value together with the number of records of this value, which is effective for the number of records with few values, 'width balanced' histograms is used.

As the number of different values increases, the amount of stored data becomes too high. We need to use a different method to store the histogram data. In this case, we can select height balanced histograms.

Using the above two methods, the column histogram provides an effective and centralized method to display data distribution. When a histogram is created, the stored information depends on "whether the number of different values is less than or equal to the number of buckets (75 by default, up to 254.

If the number of different values is less than or equal to the number of Histogram buckets (up to 254 buckets), Frequency Histogram is created

If the number of different values is greater than the number of buckets in the Histogram, Height Balanced Histogram is created.

Frequency Histogram

Frequency Histogram uses a bucket to record the number of records with different values

Height Balanced Histogram

Height Balanced Histogram is implemented by dividing data into different buckets. Each bucket contains the same number of column values. The maximum value (or END_POINT) and minimum value in each bucket are recorded in the zero bucket.

Once the data is stored in the bucket, we can identify two types of data value --- Non-popular values and popular values

Non-popular values -- are those that do not occur multiple times as end points. will not appear multiple times

Popular values -- occur multiple times as end points. Appears multiple times.

We can use Popular and Non-Popular Values to provide use with varous statistics. since we know how many values there are in a bucket we can use this information to estimate the number of rows in total that are covered by Popular and Non-Popular values.

• The selectivity for popular values can be obtained by calculation the proportion of bucket endpoints filled by that popular value.

• The selectivity for non popular values can now be calculated as 1/number non-popular bucket endpoints, so we can now be more accurate about selectivities than the original 1/Newcastle, because we have removed the popular values from the equation.

How histograms are used

The histogram is used to obtain a better selectivity Estimation for column predicate.

Where there are fewer distinct values than buckets, the selectivity is simply calculated as we have accurate row information for each value. for the case where we have more distinct values than buckets, the following outlines how these selectivities are obtained.

Equality Predicate selecti1_calculated from:

• Popular Value:

Number of buckets for value/Total Number of buckets

• Non-Popular Value:

Density see:

Document 43041.1 Query Optimizer: What is Density?

Less than <(Same principle applies for >&>=)

• All Values:

Buckets with endpoints <value/Total No. of buckets

Histogram Examples

Table TAB1SQL> desc tab1 Name Null? Type ------------------------------- -------- ---- A NUMBER(6) B NUMBER(6)

Column A contains unique values from 1 to 10000.

Column B contains 10 distinct values.

The value '5' occurs 9991 times.

Values '1, 2, 3, 4, 9996,999 7, 9998,999 9, 10000 'occur only once.

I. e.

select distinct B , count(*)from HTAB1group by Border by B; B COUNT(*)---------- ---------- 1 1 2 1 3 1 4 1 5 9991 9996 1 9997 1 9998 1 9999 1 10000 110 rows selected.

There is an index on Column B.

Statistics are gathered without Histograms using:

exec DBMS_STATS.GATHER_TABLE_STATS (NULL,'HTAB1', method_opt => 'FOR ALL COLUMNS SIZE 1');

Setup:

drop table HTAB1;create table HTAB1 (a number, b number); Insert into HTAB1 ( A,B) values ( 1,1); Insert into HTAB1 ( A,B) values ( 2,2); Insert into HTAB1 ( A,B) values ( 3,3); Insert into HTAB1 ( A,B) values ( 4,4); Insert into HTAB1 ( A,B) values ( 9996,9996); Insert into HTAB1 ( A,B) values ( 9997,9997); Insert into HTAB1 ( A,B) values ( 9998,9998); Insert into HTAB1 ( A,B) values ( 9999,9999); Insert into HTAB1 ( A,B) values ( 10000,10000);commit;begin for i in 5 .. 9995 loop Insert into HTAB1 ( A,B)values ( i,5); if (mod(i,100) = 0) then commit; end if; end loop; commit;end;/commit;create index HTAB1_B on HTAB1(b);exec DBMS_STATS.GATHER_TABLE_STATS (NULL,'HTAB1', method_opt => 'FOR ALL COLUMNS SIZE 1');alter session set OPTIMIZER_DYNAMIC_SAMPLING = 0;

Function to convert raw data in to numeric data:

create or replace function raw_to_number(my_input raw)return numberas my_output number;begin dbms_stats.convert_raw_value(my_input,my_output); return my_output;end;/

This results in statistics as follows:

column COLUMN_NAME format a5 heading COLcolumn NUM_DISTINCT format 99990column LOW_VALUE format 99990column HIGH_VALUE format 99990column DENSITY format 99990column NUM_NULLS format 99990column NUM_BUCKETS format 99990column SAMPLE_SIZE format 99990select COLUMN_NAME,NUM_DISTINCT,raw_to_number(LOW_VALUE) Low,raw_to_number(HIGH_VALUE) High,DENSITY,NUM_NULLS, NUM_BUCKETS,LAST_ANALYZED,SAMPLE_SIZE,HISTOGRAMfrom user_tab_columnswhere table_name = 'HTAB1';COL NUM_DISTINCT LOW HIGH DENSITY NUM_NULLS NUM_BUCKETS LAST_ANALYZED SAMPLE_SIZE HISTOGRAM----- ------------ ---------- ---------- ------- --------- ----------- -------------------- ----------- ---------------A 10000 1 10000 0 0 1 31-jan-2013 09:32:08 10000 NONEB 10 1 10000 0 0 1 31-jan-2013 09:32:08 10000 NONEselect lpad(TABLE_NAME,10) TAB, lpad(COLUMN_NAME, 10) COL, ENDPOINT_NUMBER, ENDPOINT_VALUEfrom user_histogramswhere table_name='HTAB1'order by COL, ENDPOINT_NUMBER;TAB COL ENDPOINT_NUMBER ENDPOINT_VALUE---------- ---------- --------------- -------------- HTAB1 A 0 1 HTAB1 A 1 10000 HTAB1 B 0 1 HTAB1 B 1 10000

In the above you can see that the statistics gathering has not created a histogram. there is a single bucket and high and a low ENDPOINT_NUMBER for each column value (you will always get 2 entries in USER_HISTOGRAMS for each column, for the high and low values respectively ).

Test queries:

To replicate the tests you will need to disable OPTIMIZER_DYNAMIC_SAMPLING

alter session set OPTIMIZER_DYNAMIC_SAMPLING = 0;

See:

Document 336267.1 maid (OPTIMIZER_DYNAMIC_SAMPLING)

Without Histograms, both queries do an index range scan because the optimizer believes that the data is uniformly distributed in column B and that each predicate with return 1/10th of the values because there are 10 distinct values:

---------------------------------------------------------------------------------------| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |---------------------------------------------------------------------------------------| 0 | SELECT STATEMENT | | 1111 | 6666 | 5 (0)| 00:00:01 || 1 | TABLE ACCESS BY INDEX ROWID| HTAB1 | 1111 | 6666 | 5 (0)| 00:00:01 ||* 2 | INDEX RANGE SCAN | HTAB1_B | 1111 | | 3 (0)| 00:00:01 |---------------------------------------------------------------------------------------

In fact it may be preferable to use a Full Table Scan for the select where B = 5 and index lookups for the others.

Gathering Histogram Statistics

If we collect histogram statistics with the recommended settings:

exec DBMS_STATS.GATHER_TABLE_STATS (NULL,'HTAB1', method_opt => 'FOR ALL COLUMNS SIZE AUTO');

The B = 5 query now does a Full Table Scan

select * from htab1 where b=5;---------------------------------------------------------------------------| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |---------------------------------------------------------------------------| 0 | SELECT STATEMENT | | 9991 | 69937 | 7 (0)| 00:00:01 ||* 1 | TABLE ACCESS FULL| HTAB1 | 9991 | 69937 | 7 (0)| 00:00:01 |---------------------------------------------------------------------------

The query where B is 3 still uses an index:

select * from htab1 where b=3;---------------------------------------------------------------------------------------| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |---------------------------------------------------------------------------------------| 0 | SELECT STATEMENT | | 1 | 7 | 2 (0)| 00:00:01 || 1 | TABLE ACCESS BY INDEX ROWID| HTAB1 | 1 | 7 | 2 (0)| 00:00:01 ||* 2 | INDEX RANGE SCAN | HTAB1_B | 1 | | 1 (0)| 00:00:01 |---------------------------------------------------------------------------------------

This is because a FREQUENCY Histogram has been created:

COL NUM_DISTINCT LOW HIGH DENSITY NUM_NULLS NUM_BUCKETS LAST_ANALYZED SAMPLE_SIZE HISTOGRAM----- ------------ ---------- ---------- ------- --------- ----------- -------------------- ----------- ---------------A 10000 1 10000 0 0 1 31-jan-2013 09:58:01 10000 NONEB 10 1 10000 0 0 10 31-jan-2013 09:58:01 10000 FREQUENCYTAB COL ENDPOINT_NUMBER ENDPOINT_VALUE---------- ---------- --------------- -------------- HTAB1 A 0 1 HTAB1 A 1 10000 HTAB1 B 1 1 HTAB1 B 2 2 HTAB1 B 3 3 HTAB1 B 4 4 HTAB1 B 9995 5 HTAB1 B 9996 9996 HTAB1 B 9997 9997 HTAB1 B 9998 9998 HTAB1 B 9999 9999 HTAB1 B 10000 1000012 rows selected.

On Column B there are 10 buckets matching up with the 10 distinct values.

The ENDPOINT_VALUE shows the column value and the ENDPOINT_NUMBER shows the cumulative number of rows. so the number of rows for ENDPOINT_VALUE 2, it has an ENDPOINT_NUMBER 2, the previous ENDPOINT_NUMBER is 1, hence the number of rows with value 2 is 1. another example is ENDPOINT_VALUE 5. its ENDPOINT_NUMBER is 9995. the previous bucket ENDPOINT_NUMBER is 4, so 9995-4 = 9991 rows containing the value 5.

Frequency histograms work fine with a low number of distinct values, but when the number exceeds the maximum number of buckets, you cannot create a bucket for each value. in this case the Optimizer creates Height balanced histograms.

Height Balanced Histograms

You can demonstrate this situation by forcing the optimizer to create fewer buckets than the Number of Distinct Values. I. e. using 8 buckets for 10 Distinct Values:

exec DBMS_STATS.GATHER_TABLE_STATS (NULL,'HTAB1', method_opt => 'FOR COLUMNS B SIZE 8');

So now we have gathered a height balanced histogram for Column B:

COL NUM_DISTINCT LOW HIGH DENSITY NUM_NULLS NUM_BUCKETS LAST_ANALYZED SAMPLE_SIZE HISTOGRAM----- ------------ ---------- ---------- ------- --------- ----------- -------------------- ----------- ---------------A 10000 1 10000 0 0 1 31-jan-2013 09:58:01 10000 NONEB 10 1 10000 0 0 8 31-jan-2013 09:59:09 10000 HEIGHT BALANCEDTAB COL ENDPOINT_NUMBER ENDPOINT_VALUE---------- ---------- --------------- -------------- HTAB1 A 0 1 HTAB1 A 1 10000 HTAB1 B 0 1 HTAB1 B 7 5 HTAB1 B 8 10000

Notice that there are 8 Buckets against B now.

Oracle puts the same number of values in each bucket and records the endpoint of each bucket.

With height balanced Histograms, the ENDPOINT_NUMBER is the actual bucket number and ENDPOINT_VALUE is the endpoint value of the bucket determined by the column value.

From the above, bucket 0 holds the low value for the column.

Because buckets 1-7 have the same endpoint, Oracle does not store all these rows to save space. but we have: bucket 1 with an endpoint of 5, bucket 2 with an endpoint of 5, bucket 3 with an endpoint of 5, bucket 4 with an endpoint of 5, bucket 5 with an endpoint of 5, bucket 6 with an endpoint of 5, bucket 7 with an endpoint of 5 AND bucket 8 with an endpoint of 10000 So bucket 1 contains values between 1 and 5, bucket 8 contains values between 5 and 10000.

All buckets contain the same number of values (which is why they are called height-balanced histograms), cannot the last bucket may have fewer values then the other buckets.

Storing Character Values in Histograms

For character columns, Oracle only stores the first 32 bytes of any string (there are also limits on numeric columns, but these are less frequently an issue since the majority of numbers are insufficiently large to encounter any problems ). see:

Document 212809.1 Limitations of the Oracle Cost Based Optimizer

Any predicates that contain strings greater than 32 characters will not use histogram information and the selectivity will be 1/Number of DISTINCT Values. Data in histogram endpoints is normalized to double precision floating point arithmetic.

For Example

SQL> select * from example;A----------abcdeeee

The table contains 5 distinct values. there is one occurence of 'A', 'B', 'C' and 'D' There are 4 occurrences of 'E '. if we create a histogram: Looking in user_histograms:

TABLE COL ENDPOINT_NUMBER ENDPOINT_VALUE---------- ----- --------------- -------------- EXAMPLE A 1 5.0365E+35 EXAMPLE A 2 5.0885E+35 EXAMPLE A 3 5.1404E+35 EXAMPLE A 4 5.1923E+35 EXAMPLE A 8 5.2442E+35

So:

ENDPOINT_VALUE 5.0365E+35 represents a5.0885E+35 represents b5.1404E+35 represents c5.1923E+35 represents d5.2442E+35 represents e

Then, if you look at the cumulative values for ENDPOINT_NUMBER, the corresponding ENDPOINT_VALUE's are correct.