# Character selectivity

A recent OTN posting asked how the optimizer dealt with “like” predicates for character types quoting the DDL and a query that I had published some time ago in a presentation I had done with Kyle Hailey. I thought that I had already given a detailed answer somewhere on my blog (or even in the presentation) but found that I couldn’t track down the necessary working, so here’s a repeat of the question and a full explanation of the working.

The query is very simple, and the optimizer’s arithmetic takes an “obvious” strategy in the arithmetic. Here’s the sample query, with the equiavalent query that we can use to do the calculation:

```
select * from t1 where alpha_06 like 'mm%';

select * from t1 where alpha_06 >= 'mm' and alpha_06 < 'mn';

```

Ignoring the possible pain of the EBCDIC character set and multi-byte national-language character sets with “strange” collation orders, it should be reasonably easy to see that ‘mn’ is the first string in alphabetical order that fails to match ‘mm%’. With that thought in mind we can apply the standard arithmetic for range-based predicates assuming, to stick with the easy example, that there are no histograms involved. For a range closed at one end and and open at the other the selectivity is:

```
( ( 'mn' - 'mm') / (high_value - low_value) ) + 1/num_distinct

```

The tricky bits, of course, are how you subtract ‘mm’ from ‘mn’ and how you use the values stored in the low_value and high_value columns of view user_tab_cols. So let’s generate the orginal data set and see where we go (running on 12c, and eliminating redundant bits from the original presentation):

```
rem
rem     Script:         selectivity_like_char.sql
rem     Author:         Jonathan Lewis
rem     Dated:          Sep 2013
rem

execute dbms_random.seed(0)

create table t1 nologging as
with generator as (
select rownum id
from dual
connect by rownum <= 1000
)
select
cast(dbms_random.string('l',6) as char(6))      alpha_06
from
generator,
generator
where
rownum <= 1e6 -- > comment to avoid WordPress formatting issue
;

execute dbms_stats.gather_table_stats(user,'t1',method_opt=>'for all columns size 1')

column low_value  format a32
column high_value format a32

select
column_name,
num_distinct,
density,
low_value,
high_value
from
user_tab_cols
where
table_name = 'T1'
order by
column_name
;

select min(alpha_06), max(alpha_06) from t1;

set autotrace traceonly explain

select
*
from
t1
where
alpha_06 like 'mm%'
;

set autotrace off

```

It will probably take a couple of minutes to generate the data – it’s 1M random strings, lower-case, 6 characters fixed – and will take up about 12MB of space. Here are the results from the stats and min/max queries, with the execution plan for the query we are testing:

```
COLUMN_NAME          NUM_DISTINCT    DENSITY LOW_VALUE                  HIGH_VALUE
-------------------- ------------ ---------- -------------------------- --------------------------
ALPHA_06                  1000000    .000001 616161616E72               7A7A7A78747A

MIN(AL MAX(AL
------ ------
aaaanr zzzxtz

Execution Plan
----------------------------------------------------------
Plan hash value: 3617692013

--------------------------------------------------------------------------
| Id  | Operation         | Name | Rows  | Bytes | Cost (%CPU)| Time     |
--------------------------------------------------------------------------
|   0 | SELECT STATEMENT  |      |   157 |  1099 |   265  (20)| 00:00:01 |
|*  1 |  TABLE ACCESS FULL| T1   |   157 |  1099 |   265  (20)| 00:00:01 |
--------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------
1 - filter("ALPHA_06" LIKE 'mm%')

```

Given that there are power(26,6) = 308,915,776 different combinations available for lower-case strings of 6 charactgers it’s not too surprising that Oracle generated 1M different strings, nor is it particularly surprising that the lowest value string started with ‘aaa’ and the highest with ‘zzz’.

So how do we get 157 as the cardinality for the query or, to put it another way, how do we get 0.000157 as the selectivity of the predicate. We need to refer to a note I wrote a few years ago to help us on our way (with a little caveat due to a change that appeared in 11.2.0.4) – what number would Oracle use to represent ‘mm’ and the other three strings we need to work with ?

According to the rules supplied (and adjusted in later versions) we have to:

1. pad the strings with ASCII nulls (zeros) up to 15 bytes
2. treat the results as a hexadecimal number and convert to decimal
3. round off the last 21 decimal digits

We can model this in SQL with a statement like:

```
SQL> column dec_value format 999,999,999,999,999,999,999,999,999,999,999,999
SQL> select round(to_number(utl_raw.cast_to_raw(rpad('aaaanr',15,chr(0))),'xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx'),-21) dec_val from dual;

DEC_VAL
------------------------------------------------
505,627,904,294,763,000,000,000,000,000,000,000

1 row selected.

```

As an alternative, or possibly a cross-check, I created a table with a varchar2(6) column, inserted the four values I was interested in and created a histogram of 4 buckets on the column (there’s a suitable little demo at this URL) and got the following endpoint values:

```
ENDPOINT_NUMBER                                   ENDPOINT_VALUE
--------------- ------------------------------------------------
1  505,627,904,294,763,000,000,000,000,000,000,000
2  568,171,140,227,094,000,000,000,000,000,000,000
3  568,191,422,636,698,000,000,000,000,000,000,000
4  635,944,373,827,734,000,000,000,000,000,000,000

```

Once we’ve got these numbers we can slot them into the standard formula (not forgetting the 1/1,000,000 for the closed end of the predicate) – and to save typing I’m going to factor out 10^21 across the board in the division:

Selectivity = (568,191,422,636,698 – 568,171,140,227,094) / (635,944,373,827,734 – 505,627,904,294,763) + 1/1,000,000

Selectivity = 20,282,409,604 / 130,316,469,532,971 + 1/1,000,000

Selectivity = 0.00015564 + 0.000001 = 0.00015664

From which the cardinality = (selectivity * num_rows) = 156.64, which rounds up to 157. Q.E.D.