Join Cardinality Estimation Methods_in_Oracle12c.pdf

© 2016 Chinar A. Aliyev Hotsos Symposium March 6-10 1
Join Cardinality Estimation Methods
Chinar Aliyev
chinaraliyev@gmail.com
As it is known the Query Optimizer tries to select best plan for the query and it does that based on
generating all possible plans, estimates cost each of them, and selects cheapest costed plan as optimal
one. Estimating cost of the plan is a complex process. But cost is directly proportionate to the
number of I/O s. Here is functional dependence between number of the rows retrieved from
database and number of I/O s. So the cost of a plan depends on estimated number of the rows
retrieved in each step of the plan – cardinality of the operation. Therefore optimizer should accurately
estimate cardinality of each step in the execution plan. In this paper we going to analyze how oracle
optimizer calculates join selectivity and cardinality in different situations, like how does CBO
calculate join selectivity when histograms are available (including new types of histograms, in 12c)?,
what factors does error (estimation) depend on? And etc. In general two main join cardinality
estimation methods exists: Histogram Based and Sampling Based.
Thanks to Jonathan Lewis for writing “Cost Based Oracle Fundamentals” book. This book actually
helped me to understand optimizer`s internals and to open the “Black Box”. In 2007 Alberto
Dell`Era did an excellent work, he investigated join size estimation with histograms. However there
are some questions like introduction of a “special cardinality” concept. In this paper we are going to
review this matter also.
For simplicity we are going to use single column join and columns containing no null values. Assume
we have two tables t1, t2 corresponding join columns j1, j2 and the rest of columns are filter1 and
filter2. Our queries are
(Q0)
SELECT COUNT (*)
FROM t1, t2
WHERE t1.j1 = t2.j2
AND t1.filter1 ='value1'
AND t2.filter2 ='value2'
(Q1)
SELECT COUNT (*)
FROM t1, t2
WHERE t1.j1 = t2.j2;
(Q2)
SELECT COUNT (*)
FROM t1, t2;

Histogram Based Estimation
Understanding Join Selectivity and Cardinality
As you know the query Q2 is a Cartesian product. It means we will get Join Cardinality -
𝐶𝑎𝑟𝑑𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 for the join product as:
𝐶𝑎𝑟𝑑𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 =num_rows(𝑡1)*num_rows(𝑡2)
Here num_rows(𝑡𝑖) is number of rows of corresponding tables. When we add join condition into the
query (so Q1) then it means we actually get some fraction of Cartesian product. To identify this
fraction here Join Selectivity has been introduced.
Therefore we can write this as follows
𝐶𝑎𝑟𝑑𝑄1 ≤ 𝐶𝑎𝑟𝑑𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛
𝐶𝑎𝑟𝑑𝑄1 = 𝐽𝑠𝑒𝑙*𝐶𝑎𝑟𝑑𝑐𝑎𝑟𝑡𝑒𝑠𝑖𝑎𝑛 = 𝐽𝑠𝑒𝑙 ∗num_rows(𝑡1)*num_rows(𝑡2) (1)
𝐽𝑠𝑒𝑙 = 𝐶𝑎𝑟𝑑𝑄1/(𝑛𝑢𝑚_𝑟𝑜𝑤𝑠(𝑡1) ∗ 𝑛𝑢𝑚_𝑟𝑜𝑤𝑠(𝑡2))
Definition: Join selectivity is the ratio of the “pure”-natural cardinality over the Cartesian product.
I called 𝐶𝑎𝑟𝑑𝑄1 as “pure” cardinality because it does not contain any filter conditions.
Here 𝐽𝑠𝑒𝑙 is Join Selectivity. This is our main formula. You should know that when optimizer tries to
estimate JC- Join Cardinality it first calculates 𝐽𝑠𝑒𝑙 . Therefore we can use same 𝐽𝑠𝑒𝑙 and can write
appropriate formula for query Q0 as
𝐶𝑎𝑟𝑑𝑄0 = 𝐽𝑠𝑒𝑙 ∗Card(𝑡1)*Card(𝑡2) (2)
Here Card (𝑡𝑖) is final cardinality after applying filter predicate to the corresponding table. In other
words 𝐽𝑠𝑒𝑙 for both formulas (1) and (2) is same. Because 𝐽𝑠𝑒𝑙 does not depend on filter columns,
unless filter conditions include join columns. According to formula (1)
𝐽𝑠𝑒𝑙 = 𝐶𝑎𝑟𝑑𝑄1/(num_rows(𝑡1) ∗ num_rows(𝑡2)) (3)
or 𝐶𝑎𝑟𝑑𝑄0 =
𝐶𝑎𝑟𝑑𝑄1∗Card(𝑡1)∗Card(𝑡2)
num_rows(𝑡1)∗num_rows(𝑡2)
(4)
Based on this we have to find out estimation mechanism of expected cardinality - 𝑪𝒂𝒓𝒅𝑸𝟏. Now
consider that for 𝑗𝑖 join columns of 𝑡𝑖 tables here is not any type of histogram. So it means in this

case optimizer assumes uniform distribution and for such situations as you already know 𝐽𝑐𝑎𝑟𝑑 and
𝐽𝑠𝑒𝑙 are calculated as
𝐽𝑠𝑒𝑙 = 1/max(𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗1
), 𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗2
)) (5)
𝐽𝑐𝑎𝑟𝑑 = 𝐽𝑠𝑒𝑙 ∗ num_rows(𝑡1) ∗ num_rows(𝑡2)
The question now is: where does formula (5) come from? How do we understand it?
According to (3) in order to calculate 𝐽𝑠𝑒𝑙 we first have to estimate “pure” expected cardinality -
𝐶𝑎𝑟𝑑𝑄1. And it only depends on Join Columns. For 𝑡1 table, based on uniform distribution the
number of rows per distinct value of the 𝑗1 column will be num_rows(𝑡1)/𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗1) and for
𝑡2 table it will be num_rows(𝑡2)/𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗2). Also here will be
min(𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗1), 𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗2)) common distinct values. Therefore expected “pure” cardinality
is
𝐶𝑎𝑟𝑑𝑄1 = min (𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗1), 𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗2)) ∗
num_rows(𝑡1)
num_dist(𝑗1)
∗
num_rows(𝑡2)
num_dist(𝑗2)
(6)
Then according to formula (3) Join Selectivity will be:
𝐽𝑠𝑒𝑙 =
𝐶𝑎𝑟𝑑𝑄1
num_rows(𝑡1)∗num_rows(𝑡2)
=
min (𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗1),𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗2))
𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗1)∗ 𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗2)
=
1
max (𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗1),𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗2))
As it can be seen we have got formula (5). Without histogram optimizer is not aware of the data
distribution, so in dictionary of the database here are not “(distinct value, frequency)” – this pairs
indicate column distribution. Because of this, in case of uniform distribution, optimizer actually
thinks and calculates “average frequency” as 𝑛𝑢𝑚_𝑟𝑜𝑤𝑠(𝑡1)/𝑛𝑢𝑚_𝑑𝑖𝑠𝑡(𝑗1). Based on “average
frequency” optimizer calculates “pure” expected cardinality and then join selectivity. So if a table
column has histogram (depending type of this) optimizer will calculates join selectivity based on
histogram. In this case “(distinct value, frequency)” pairs are not formed based on “average
frequency”, but are formed based on information which are given by the histogram.
Case 1. Both Join columns have frequency histograms
In this case both join columns have frequency histogram and our query(freq_freq.sql) is
SELECT COUNT (*)
FROM t1, t2
WHERE t1.j1 = t2.j2 AND t1.f1 = 13;
Corresponding execution plan is

---------------------------------------------------------------
| Id | Operation | Name | Starts | E-Rows | A-Rows |
---------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | | 1 |
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 2272 | 2260 |
|* 3 | TABLE ACCESS FULL| T1 | 1 | 40 | 40 |
| 4 | TABLE ACCESS FULL| T2 | 1 | 1000 | 1000 |
---------------------------------------------------------------
Predicate Information (identified by operation id):
---------------------------------------------------
2 - access("T1"."J1"="T2"."J2")
3 - filter("T1"."F1"=13)
Estimation is good enough for this situation, but it has not been exactly estimated. And why? How
did optimizer calculate cardinality of the join as 2272?
If we enable SQL trace for the query then we will see oracle queries only histgrm$ dictionary table.
Therefore information about columns and tables is as follows.
Select table_name,num_rows from user_tables where table_name in (‘T1’,’T2’);
tab_name num_rows
T1 1000
T2 1000
(Freq_values1)
SELECT endpoint_value COLUMN_VALUE,
endpoint_number - NVL (prev_endpoint, 0) frequency,
endpoint_number ep
FROM (SELECT endpoint_number,
NVL (LAG (endpoint_number, 1) OVER (ORDER BY endpoint_number),
0
)
prev_endpoint,
endpoint_value
FROM user_tab_histograms
WHERE table_name = 'T1' AND column_name = 'J1')
ORDER BY endpoint_number
tab t1,
col j1
tab t2, col
j2
value frequency ep value frequency ep
0 40 40 0 100 100
1 40 80 2 40 140
2 80 160 3 120 260
3 100 260 4 20 280

4 160 420 5 40 320
5 60 480 6 100 420
6 260 740 8 40 460
7 80 820 9 20 480
8 120 940 10 20 500
9 60 1000 11 60 560
12 20 580
13 20 600
14 80 680
15 80 760
16 20 780
17 80 860
18 80 940
19 60 1000
Frequency histograms exactly express column distribution. So “(column value, frequency)” pair
gives us all opportunity to estimate cardinality of any kind of operations. Now we have to try to
estimate pure cardinality 𝐶𝑎𝑟𝑑𝑄1 then we can find out 𝐽𝑠𝑒𝑙 according to formula (3). Firstly we
have to find common data for the join columns. These data is spread between
max(min_value(j1),min_value(j2)) and min(max_value(j1),max_value(j2)). It means we are
not interested in the data which column value greater than 10 for j2 column. Also we have to take
equval values, so we get following table
tab t1, col
j1
tab t2, col
j2
value frequency value frequency
0 40 0 100
2 80 2 40
3 100 3 120
4 160 4 20
5 60 5 40
6 260 6 100
8 120 8 40
9 60 9 20
Because of this expected pure cardinality 𝐶𝑎𝑟𝑑𝑄1 Will be
100*40+80*40+100*120+160*20+60*40+260*100+120*40+60*20=56800 and Join selectivity
𝐽𝑠𝑒𝑙 =
𝐶𝑎𝑟𝑑𝑄1
𝑛𝑢𝑚_𝑟𝑜𝑤𝑠(𝑡1)∗𝑛𝑢𝑚_𝑟𝑜𝑤𝑠(𝑡2)
=
56800
1000∗1000
= 0.0568
And eventually our cardinality will be according to the formula (2)

𝐶𝑎𝑟𝑑𝑄0 = 𝐽𝑠𝑒𝑙 ∗Card (𝑡1)*Card (𝑡2) = 0.0568 ∗ 40 ∗ 1000 = 2272
Also if we enable 10053 event then in trace file we see following lines regarding on join selectivity.
Join Card: 2272.000000 = outer (40.000000) * inner (1000.000000) * sel
(0.056800)
Join Card - Rounded: 2272 Computed: 2272.000000
As we see same number as in above execution plan. Another question was why we did not get exact
cardinality – 2260? Although join selectivity by definition does not depend on filter columns and
conditions, but filtering actually influences this process. Optimizer does not consider join column
value range, max/min value, spreads, distinct values after applying filter – in line 3 of execution plan.
It is not easy to resolve. At least it will require additional estimation algorithms, then efficiency of
whole estimation process could be harder. So if we remove filter condition from above query we will
get exact estimation.
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 56800 | 56800 |
---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."J1"="T2"."J2")
It means optimizer calculates “average” join selectivity. I think it is not an issue in general. As result
we got the following formula for join selectivity.
𝐽𝑠𝑒𝑙 =
∑ 𝑓𝑟𝑒𝑞(𝑡1.𝑗1)∗𝑓𝑟𝑒𝑞(𝑡2.𝑗2))
min _𝑚𝑎𝑥
𝑖=max _𝑚𝑖𝑛
𝑛𝑢𝑚_𝑟𝑜𝑤𝑠(𝑡1)∗𝑛𝑢𝑚_𝑟𝑜𝑤𝑠(𝑡2)
(7)
Here freq is corresponding frequency of the column value.
Case 2. Join columns with height-balanced (equ-height) and frequency histograms
Now assume one of the join column has height-balanced(HB) histogram and another has
frequency(FQ) histogram (Height_Balanced_Frequency.sql) We are going to investiagte cardinality
estimation of the two queries here
select count(*)
from t1, t2 --- (Case2 q1)
where t1.j1 = t2.j2;

select count(*)
from t1, t2 --- (Case2 q2)
where t1.j1 = t2.j2 and t1.f1=11;
For the column J1 here is Height balanced histogram - HB and for the column j2 here is frequency
- FQ histogram avilable. The appropriate information from user_tab_histogrgrams dictionary view
shown in Table 3.
tatb t1, col
j1
tab t2 , col
j2
column
value
frequency ep column
value
frequency ep
1 0 0 1 2 2
9 1 1 7 2 4
16 1 2 48 3 7
24 1 3 64 4 11
32 1 4
40 1 5
48 2 7
56 1 8
64 2 10
72 2 12
80 3 15
Ferquency column for t1.j1 of Table 3 does not express real frequency for the column. It is actually
“frequency of the bucket”. First we have to identify common values. So we have to ignore HB
histogram buckets with endpoint number greater than 10. We have exact “value, frequency” pairs of
the t2.j2 column therefore our base source must be values of the t2.j2 column. But for the t1.j1 we
do not have exact frequencies. HB histogram cointains buckets which hold approximately same
number of rows. Also we can find number of the distinct values per bucket. Then for every value of
the frequency histogram we can identify appropriate bucket of the HB histogram. Within HB bucket
we aslo can assume uniform distrbution then we can estimate size of this disjoint subset – {value of
FQ and Bucket of HB} .
Although this approach gave me some approximation and estimation of the join cardinality but it did
not give me exact number(s) which oracle optimizer calculates and reports in 10053 trace file. We
have to find what information we need to improve this approach? ,
Firstly Alberto Dell'Era investigated joins based on the histograms in 2007- (Join Over histograms).
His approach was based on grouping values into three major categories:
- “populars matching populars”
- “populars not matching populars”

- “not popular subtables”
Estimating each of them. The sum of cardinality of each group will give us join cardinality. But my
point of view to the matter is quite different:
- We have to identify “(distinct value, frequency)” pairs to approximate “pure” cardinality 𝐶𝑎𝑟𝑑𝑄1
- Our main data here is t2.j2 column`s data, because it gives us exact frequencies
- We have to walk t2.j2 columns (histograms) values and identify second part of “(distinct
value, frequency)” based on height balanced histogram.
- 𝐶𝑎𝑟𝑑𝑄1=∑ 𝐹𝑟𝑒𝑞𝑏𝑎𝑠𝑒𝑑 𝑡1(value=t2.j2)*𝐹𝑟𝑒𝑞𝑏𝑎𝑠𝑒𝑑 𝑡2(value=t2.j2)
- Then we can calculate join selectivity and cardinality
We have to identify (value, frequency) pairs based on HB histogram, then it is easy to calculate “pure”
cardinality so it means we can easily and more accurately estimate join cardinality. But when forming
(value, frequency) pairs based on HB histogram, we should not approach as uniform for the single
value which is locate within the bucket, because HB gives us actually “average” density –
NewDensity (actually the density term has been introduced to avoid estimation errors in
non-uniform distribution case and has been improved with new density mechanism) for un-
popular values and special approach for popular values. So let’s identify “(value, frequency)” pairs
based on the HB histogram.
tab_name num_rows (user_tables) col_name num_distinct
T1 11 T1.J1 30
T2 130 T2.J2 4
Number of buckets - num_buckets=15( as max(ep) from Table 3 )
Number of popular buckets – num_pop_bucktes=9(as sum(frequency) from table 3 where
frequency>1)
Popular value counts – pop_value_cnt=4(as count(frequency) from table 3 where frequency>1)
NewDensity=
𝑛𝑢𝑚_𝑢𝑛𝑝𝑜𝑝_𝑏𝑢𝑐𝑘𝑒𝑡𝑠
𝑢𝑛𝑝𝑜𝑝_𝑛𝑑𝑣∗𝑛𝑢𝑚_𝑏𝑢𝑐𝑘𝑒𝑡𝑠
=
𝑛𝑢𝑚_𝑏𝑢𝑐𝑘𝑒𝑡𝑠−𝑛𝑢𝑚_𝑝𝑜𝑝_𝑏𝑢𝑐𝑘𝑒𝑡𝑠
(𝑁𝐷𝑉−𝑝𝑜𝑝_𝑣𝑎𝑙𝑢𝑒_𝑐𝑛𝑡)𝑛𝑢𝑚_𝑏𝑢𝑐𝑘𝑒𝑡𝑠
=
15−9
(30−4)∗15
=0.015384615≈0.015385 (8)
And for popular values selectivity is:
𝑒𝑝_𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑛𝑢𝑚_𝑏𝑢𝑐𝑘𝑒𝑡𝑠
=
𝑒𝑝_𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
15
So “(value, frequency)” pairs based on the HB histogram will be:

column
value
popular frequency calculated
1 N 2.00005 130*0.015385 - (num_rows*density)
7 N 2.00005 130*0.015385 - (num_rows*density)
48 Y 17.33333333 130*2/15 - (num_rows*frequency/num_buckets)
64 Y 17.33333333 130*2/15 - (num_rows*frequency/num_buckets)
We have got all “(value, frequency)” pairs so according formula (7) we can calculate Join Selectivity.
tab t1 , col
j1
tab t2 , col
j2
column
value
frequency column
value
frequency freq*freq
1 2.00005 1 2 4.0001
7 2.00005 7 2 4.0001
48 17.33333333 48 3 52
64 17.33333333 64 4 69.33333
Sum 129.3335
And finally 𝐽𝑠𝑒𝑙 =
129.3335
numrows(t1)∗numrows(t2)
=
129.3335
11∗130
=0.090443
So our “pure” cardinality is 𝐶𝑎𝑟𝑑𝑄1 = 129. Execution plan of the query is as follows
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 129 | 104 |
---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."J1"="T2"."J2")
And the corresponding information from 10053 trace file :
Join Card: 129.333333 = outer (130.000000) * inner (11.000000) * sel (0.090443)
It means we were able to figure out exact estimation mechanism in this case. Execution plan of the
second query (Case2 q2) as follows

---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 5 | 7 |
|* 3 | TABLE ACCESS FULL| T1 | 1 | 5 | 5 |
---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."J1"="T2"."J2")
3 - filter("T1"."F1"=11)
According our approach join cardinality should computed as
𝐽𝑠𝑒𝑙*card(t1)*card(t2)= 0.090443*card(t1)*card(t2)
Also from optimizer trace file we will see the following:
It actually confirms our approach. However execution plan shows cardinality of the single table t1 as
5, it is correct because it must be rounded up but during join estimation process optimizer consider
original values rather than rounding.
Reviewing Alberto Dell'Era`s – complete formula (join_histogram_complete.sql)
We can list column information from dictionary as below:
tatb t1, col value tatb t2, col value
column value frequency column value frequency
20 1 10 1
40 1 30 2
50 1 50 1
60 1 60 4
70 2 70 2
80 2
90 1
99 1
So we have to find common values, as you see min(t1.value)=20 due to we must ignore t2.value=10
also max(t1.val)=70 it means we have to ignore column values t2.value>70. In addition we do not
have the value 40 in t2.value therefore we have to delete it also. Because of this we are getting
following table

tatb t1, col j1 tab t2 , col
j2
column value frequency column
value
frequency
20 1 30 2
50 1 50 1
60 1 60 4
70 2 70 2
Num_rows(t1)=12;num_buckets(t1.value)=6;num_distinct(t1.value)=8,=>
newdensity=
𝑛𝑢𝑚_𝑢𝑛𝑝𝑜𝑝_𝑏𝑢𝑐𝑘𝑒𝑡𝑠
𝑢𝑛𝑝𝑜𝑝_𝑛𝑑𝑣∗𝑛𝑢𝑚_𝑏𝑢𝑐𝑘𝑒𝑡𝑠
=
6−2
(8−1)∗6
=0.095238095, so appropriate column values
frequency based on HB histogram will be :
t1.value freq calculated
30 1.142857143 num_rows*newdensity
70 4 num_rows*freq/num_buckets
And finally cardinality will be.
t1.value t2.value
column
value
frequency column
value
frequency freq*freq
30 1.142857143 30 2 2.285714286
50 1.142857143 50 1 1.142857143
60 1.142857143 60 4 4.571428571
70 4 70 2 8
sum 16
Optimizer also estimated exactly 16 as we see it in execution plan of the query.
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 16 | 13 |
---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."VALUE"="T2"."VALUE")

There Alberto also has introduced “Contribution 4: special cardinality”, but it seems it is not
necessary.
Reviewing Alberto Dell'Era`s essential case (join_histogram_essentials.sql)
This is quite interesting case, firstly because in oracle 12c optimizer calculates join cardinality as 31
but not as 30, and second in this case old and new densities are same. Let’s interpret the case.
The by corresponding information from user_tab_histograms.
tatb t1, col
value
tab t2 , col
value
column
value
frequency ep column
value
frequency ep
10 2 2 10 2 2
20 1 3 20 1 3
30 2 5 50 3 6
40 1 6 60 1 7
50 1 7 70 4 11
60 1 8
70 2 10
And num_rows(t1)=20,num_rows(t2)=11,num_dist(t1.value)=11,num_dist(t2.val)=5,
Density (t1.value)=(10-6)/((11-3)*10)= 0.05. Above mechanism does not give us exact number as
expected as optimizer estimation. Because in this case to estimate frequency for un-popular values
oracle does not use density it uses number of distinct values per bucket and number of rows per
distinct values instead of the density. To prove this one we can use join_histogram_essentials1.sql.
In this case t1 table is same as in join_histogram_essentials.sql . The column T2.value has only one
value 20 with frequency one.
t1.value freq EP t2.value freq EP
10 2 2 20 1 1
20 1 3
30 2 5
40 1 6
50 1 7
60 1 8
70 2 10
In this case oracle computes join cardinality 2 as rounded up from 1.818182. We can it from trace
file

It looks like the estimation totally depend on t1.value column distribution. So
num_rows_bucket(number of rows per bucket) is 2 and num_rows_distinct(number of distinct
value per bucket) is 20/11=1.818182. Every bucket has 1.1 distinct value and within bucket every
distinct value has 2/1.1=1.818182 rows. And this is our cardinality. But if we increase frequency of
the t2.value - join_histogram_essentials2.sql. The (t2.value, frequency)=(20,5) and t1 table is same
as in previous case.
t1.value freq EP t2.value freq EP
10 2 2 20 5 5
20 1 3
30 2 5
40 1 6
50 1 7
60 1 8
70 2 10
Corresponding lines from 10053 trace file:
Tests show that in such cases cardinality of the join computed as frequency of the t2.value. So it
means frequency of the popular value will be:
Frequency (non-popular t1.value) = {
𝑛𝑢𝑚_𝑟𝑜𝑤𝑠_𝑏𝑢𝑐𝑘𝑒𝑡
𝑛𝑢𝑚_𝑑𝑖𝑠𝑡_𝑏𝑢𝑐𝑘𝑒𝑡𝑠
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡2. 𝑣𝑎𝑙𝑢𝑒 = 1
1 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑡2. 𝑣𝑎𝑙𝑢𝑒 > 1
or
Cardinality = max (frequency of t2.val, number of rows per distinct value within bucket)
Question is why? In such cases I think optimizer tries to minimize estimation errors. So

tab t1,col
val
column
value
frequency calculated
10 4 - (num_rows*frequency/num_buckets)
20 1.818181818 - (num_rows_bucket/num_dist_buckets)
50 1 -frequency of t2.value
60 1.818181818 - (num_rows_bucket/num_dist_buckets)
70 4 - (num_rows*frequency/num_buckets)
Therefore
tab t1,col
value
tab t2 , col
value
column value frequency column value frequency freq*freq
10 4 10 2 8
20 1.818181818 20 1 1.818181818
50 1 50 3 3
60 1.818181818 60 1 1.818181818
70 4 70 4 16
sum 30.63636364
We get 30.64≈31 as expected cardinality. Let`s see trace file and execution plan
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 31 | 29 |
---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."VALUE"="T2"."VALUE")

Case 3. Join columns with hybrid and frequency histograms
In this case we are going to analyze how optimizer calculates join selectivity when there are hybrid
and frequency histograms available on the join columns (hybrid_freq.sql). Note that the query is
same -(Case2 q1).The corresponding information from dictionary view.
SELECT endpoint_value COLUMN_VALUE,
endpoint_number - NVL (prev_endpoint, 0) frequency,
ENDPOINT_REPEAT_COUNT,
endpoint_number
FROM (SELECT endpoint_number,
ENDPOINT_REPEAT_COUNT,
NVL (LAG (endpoint_number, 1) OVER (ORDER BY
endpoint_number),0)
prev_endpoint,
endpoint_value
FROM user_tab_histograms
WHERE table_name = 'T3' AND column_name = 'J3')
ORDER BY endpoint_number
Tab t1, col j1 tab t2 , col j2
column value frequency endpoint_rep_cnt column value frequency
0 6 6 0 3
2 9 7 1 6
4 8 5 2 6
6 8 5 3 8
7 7 7 4 11
9 10 5 5 3
10 6 6 6 3
11 3 3 7 9
12 7 7 8 6
13 4 4 9 5
14 5 5
15 5 5
16 5 5
17 7 7
19 10 5
As it can be seen common column values are between 0 and 9. So we are not interested in buckets
which contain column values greater than or equival 10. Hybrid histogram gives us more information
to estimate single table and also join selectivity than height balanced histogram. Specially endpoint
repeat count column are used by optimizer to exactly estimate endpoint values. But how does
optimizer use this information to estimate join? Principle of the estimation “(value,frequency)” pairs
based on hybrid histogram are same as height based histogram. So it depends on popularity of the
value, if value is popular then frequency will be equval to the corresponding endpoint repeat count,

otherwise it will be calculated based on the density. If we enable dbms_stats trace when gathering
hybrid histogram. We get the following
DBMS_STATS:
SELECT SUBSTRB (DUMP (val, 16, 0, 64), 1, 240) ep,
freq,
cdn,
ndv,
(SUM (pop) OVER ()) popcnt,
(SUM (pop * freq) OVER ()) popfreq,
SUBSTRB (DUMP (MAX (val) OVER (), 16, 0, 64), 1, 240) maxval,
SUBSTRB (DUMP (MIN (val) OVER (), 16, 0, 64), 1, 240) minval
FROM (SELECT val,
freq,
(SUM (freq) OVER ()) cdn,
(COUNT ( * ) OVER ()) ndv,
(CASE
WHEN freq > ( (SUM (freq) OVER ()) / 15) THEN 1
ELSE 0
END)
pop
FROM (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 */
"ID"
val,
COUNT ("ID") freq
FROM "SYS"."T1" t
WHERE "ID" IS NOT NULL
GROUP BY "ID"))
ORDER BY valDBMS_STATS: > cdn 100, popFreq 28, popCnt 4, bktSize 6.6, bktSzFrc .6
DBMS_STATS: Evaluating hybrid histogram: cht.count 15, mnb 15, ssize 100, min_ssize
2500, appr_ndv TRUE,
ndv 20, selNdv 0, selFreq 0, pct 100, avg_bktsize 7, csr.hreq TRUE, normalize TRUE
Average bucket size is 7. Oracle considers value as popular when correspoindg endpoint repeat count
is greater than or equval average bucket size. Also in our case density is (crdn- popfreq)/((NDV-
popCnt)*crdn)=(100-28)/((20-4)*100)= 0.045. If we enable 10053 trace event you can clearly see
columns and tables statistics. Therefore “(value,frequency)” will be as
t1.j1 popular frequency calculated
0 N 4.5 density*num_rows
2 Y 7 endpoint_repeat_count
7 Y 7 endpoint_repeat_count

And then final cardinality.
t1.j1 t2.j2
value frequency value frequency freq*freq
0 4.5 0 3 13.5
1 4.5 1 6 27
2 7 2 6 42
3 4.5 3 8 36
4 4.5 4 11 49.5
5 4.5 5 3 13.5
6 4.5 6 3 13.5
7 7 7 9 63
8 4.5 8 6 27
9 4.5 9 5 22.5
sum 307.5
Join sel 0.05125
Lets now check execution plan
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 308 | 293 |
---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."J1"="T2"."J2")
And trace from 10053 trace file :
Case 4. Both Join columns with Top frequency histograms
In this case join columns have top-frequency histogram (TopFrequency_hist.sql). We are going to
use same query as above – (Case2 q1). Corresponding column information is.
Table Stats::
Table: T2 Alias: T2
#Rows: 201 SSZ: 0 LGR: 0 #Blks: 1 AvgRowLen: 3.00 NEB: 0 ChainCnt: 0.00
SPC: 0 RFL: 0 RNF: 0 CBK: 0 CHR: 0 KQDFLG: 1

#IMCUs: 0 IMCRowCnt: 0 IMCJournalRowCnt: 0 #IMCBlocks: 0 IMCQuotient: 0.000000
Column (#1): J2(NUMBER)
AvgLen: 4 NDV: 21 Nulls: 0 Density: 0.004975 Min: 1.000000 Max: 200.000000
Histogram: Top-Freq #Bkts: 192 UncompBkts: 192 EndPtVals: 12 ActualVal: yes
***********************
Table Stats::
Table: T1 Alias: T1
t1.j1 freq t2.j2 freq
4 10 1 14
5 16 2 18
6 17 3 18
8 12 4 17
100 1 5 15
6 19
7 19
8 22
9 17
10 18
11 13
200 2
By definition of the Top-Frequency histogram, we can say that here are two types of buckets.
Oracle placed high frequency values into some buckets (appropriate) and rest of the values of the
table oracle actually “placed” into another “bucket”. So we actually have “high frequency” and
“low frequency” values. Therefore for “high frequency” values we also have exact frequencies, but
for “low frequency” values we can approach by using “Uniform distribution”. Firstly we have to
build high frequency pairs based on common values. The max(min(t1.j1),min(t2.j2))=4 and also
max(max(t1.j1),max(t2.j2))=100. In principle we have to see and gather common values which are
between 4 and 100. So after identifying common values, for popular values we are going to use
exact frequency and for non-popular values new density. Therefore we could create following table:

common value j2.freq j1.freq freq*freq
4 17 10 170
5 15 16 240
6 19 17 323
7 19 1.000025 19.000475
8 22 12 264
9 17 1.000025 17.000425
10 18 1.000025 18.00045
11 13 1.000025 13.000325
100 1 1.000025 1.000025
sum 1065.0017
For t1 table we have num_rows-popular_rows=65-56=9 unpopular rows and ndv-
popular_value_count=14-5=9 also for t2 table we have 201-192=9 unpopular rows and 21-12=9
unpopular distinct values. Frequency for each unpopular rows of the t2.j2 is
num_rows(t2)*density(t2)= 201*0.004975= 0.999975 also for t1.j1 it is num_rows(t1)*density(t1)=
65*0.015385 = 1.000025. Due to cardinality for each individual unpopular rows will be:
CardIndvPair=unpopular_freq(t1.j1)* unpopular_freq(t2.j2)= 0.999975*1.000025=1.
Test cases show that oracle considers all low frequency (unpopular rows) values during join when
top frequency histograms are available it means cardinality for “low frequency” values will be
Card(Low frequency values)=max(unpopular_rows(t1.j1),unpopular_rows(t2.j2))* CardIndvPair=9
Therefore final cardinality of the our join will be CARD(high freq values)+ CARD(low freq
values)=1065+9=1074. Lets see execution plan.
-----------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)|
-----------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 6 | 4 (0)|
| 1 | SORT AGGREGATE | | 1 | 6 | |
|* 2 | HASH JOIN | | 1074 | 6444 | 4 (0)|
| 3 | TABLE ACCESS FULL| T1 | 65 | 195 | 2 (0)|
| 4 | TABLE ACCESS FULL| T2 | 201 | 603 | 2 (0)|
-----------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."J1"="T2"."J2")
And the trace file

Case 5. Join columns with Top frequency and frequency histograms
Now consider that we have tables which for join columns there are top frequency and frequency
histogram (TopFrequency_Frequency.sql). The columns distribution from dictionary is as below
(Freq_values1)
1 3 0 4
2 3 1 7
3 5 2 2
4 5 4 3
5 5
6 4
7 6
8 4
9 5
25 1
In this case there is a frequency histogram for the column t2.j2 and we have exact common {1, 2, 4}
values. But test cases show that optimizer also considers all the values from top frequency histogram
which are between max(min(t1.j1),min(t2.j2)) and min(max(t1.j1),max(t2.j2)). It is quite interesting
case. Because of this we have frequency histogram and it should be our main source and this case
should have been similar to the case 3.
Table Stats::
Table: T2 Alias: T2
#IMCUs: 0 IMCRowCnt: 0 IMCJournalRowCnt: 0 #IMCBlocks: 0 IMCQuotient:
0.000000
Histogram: Freq #Bkts: 4 UncompBkts: 16 EndPtVals: 4 ActualVal: yes
***********************
Table Stats::
Table: T1 Alias: T1
0.000000
Considered values and their frequencies:

Considered
values
j1.freq j2.freq freq*freq
1 3 7 21
2 3 2 6
3 5 1 5
4 5 3 15
sum 47
Here for the value 3 j2.freq calculated as num_rows(t2)*density=16*0.0625=1. And in 10053 file
Let see another example-TopFrequency_Frequency2.sql
1 3 0 4
2 3 1 7
3 5 2 2
4 5 4 3
5 5 10 2
6 4
7 6
8 4
9 5
25 1
Tables and columns statistics:
Table Stats::
Table: T3 Alias: T3
#Rows: 18 SSZ: 0 LGR: 0 #Blks: 1 AvgRowLen: 3.00 NEB: 0 ChainCnt: 0.00 SPC:
0 RFL: 0 RNF: 0 CBK: 0 CHR: 0 KQDFLG: 1
Histogram: Freq #Bkts: 5 UncompBkts: 18 EndPtVals: 5 ActualVal: yes
***********************
Table Stats::
Table: T1 Alias: T1
#Rows: 42 SSZ: 0 LGR: 0 #Blks: 1 AvgRowLen: 3.00 NEB: 0 ChainCnt: 0.00 SPC:

Considered column values and their frequencies are:
Considered
val
j1.freq calculated j3.freq calculated freq*freq
1 3 freq 7 freq 21
2 3 freq 2 freq 6
3 5 freq 1.000008 num_rows*density 5.00004
4 5 freq 3 freq 15
10 1.00002 num_rows*density 2 freq 2.00004
sum 73.000272
And from trace file
But if we compare estimated cardinality with actual values then we will see:
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 73 | 42 |
---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."J1"="T3"."J3")
As we see here is significant difference. 73 vs 42, error estimation is enough big. That is why we said
before its quite interesting case, so optimizer should consider only values from frequency histogram,
these values should be main source of the estimation process – as similar to the case3. So if consider
and walk on the values of the frequency histogram as common values then we will get the following
table:
common val j1.freq calculated j3.freq calculated freq*freq
1 3 freq 7 freq 21
2 3 freq 2 freq 6

4 5 freq 3 freq 15
10 1.00002 num_rows*density 2 freq 2.00004
sum 44.00004
You can clearly see that, such estimation is very close to the actual rows.
Case 6. Join columns with Hybrid and Top frequency histograms
It is quite hard to interpret when one of join column has top frequency histogram
(Hybrid_topfreq.sql). For example here is hybrid histogram for t1.j1 and top frequency histogram
for t2.j2. Column information from dictionary
t1.j1 t1.freq ep_rep_cnt t2.j2 j2.freq
1 3 3 1 5
3 10 6 2 3
4 6 6 3 4
6 5 2 4 5
9 8 5 5 4
10 1 1 6 3
11 2 2 7 1
13 5 3 26 1
30 1
And from dbms_stats trace file
SELECT SUBSTRB (DUMP (val, 16, 0, 64), 1, 240) ep,
freq,
cdn,
ndv,
(SUM (pop) OVER ()) popcnt,
(SUM (pop * freq) OVER ()) popfreq,
SUBSTRB (DUMP (MAX (val) OVER (), 16, 0, 64), 1, 240) maxval,
SUBSTRB (DUMP (MIN (val) OVER (), 16, 0, 64), 1, 240) minval
FROM (SELECT val,
freq,
(SUM (freq) OVER ()) cdn,
(COUNT ( * ) OVER ()) ndv,
(CASE WHEN freq > ( (SUM (freq) OVER ()) / 8) THEN 1 ELSE 0 END)
pop

FROM (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 */
"J1"
val,
COUNT ("J1") freq
FROM "T"."T1" t
WHERE "J1" IS NOT NULL
GROUP BY "J1"))
ORDER BY val
DBMS_STATS: > cdn 40, popFreq 12, popCnt 2, bktSize 5, bktSzFrc 0
DBMS_STATS: Evaluating hybrid histogram: cht.count 8, mnb 8, ssize 40, min_ssize 2500,
appr_ndv TRUE, ndv 13, selNdv 0, selFreq 0, pct 100, avg_bktsize 5, csr.hreq TRUE,
normalize TRUE
High frequency common values are located between 1 and 7. Also we have two popular values for
t1.j1 column :{3,4}.
Table Stats::
Table: T2 Alias: T2
#Rows: 30 SSZ: 0 LGR: 0 #Blks: 5 AvgRowLen: 3.00 NEB: 0 ChainCnt: 0.00 SPC
#IMCUs: 0 IMCRowCnt: 0 IMCJournalRowCnt: 0 #IMCBlocks: 0
IMCQuotient: 0.000000
AvgLen: 3 NDV: 12 Nulls: 0 Density: 0.033333 Min:
1.000000 Max: 30.000000
Histogram: Top-Freq #Bkts: 27 UncompBkts: 27
EndPtVals: 9 ActualVal: yes
***********************
Table Stats::
Table: T1 Alias: T1
#Rows: 40 SSZ: 0 LGR: 0 #Blks: 5 AvgRowLen: 3.00 NEB: 0 ChainCnt: 0.00 SPC
#IMCUs: 0 IMCRowCnt: 0 IMCJournalRowCnt: 0 #IMCBlocks: 0
IMCQuotient: 0.000000
AvgLen: 3 NDV: 13 Nulls: 0 Density: 0.063636 Min:
1.000000 Max: 13.000000
Histogram: Hybrid #Bkts: 8 UncompBkts: 40
EndPtVals: 8 ActualVal: yes
Therefore common values and their frequencies are:
Common value t1.freq t2.freq freq*freq
1 2.54544 5 12.7272
2 2.54544 3 7.63632
3 6 4 24

4 6 5 30
5 2.54544 4 10.18176
6 2.54544 3 7.63632
7 2.54544 1 2.54544
sum 94.72704
Moreover we have num_rows-top_freq_rows=30-27=3 infrequency rows and NDV-
top_freq_count=12-9=3 unpopular NDV. I have done several test cases and I think cardinality of
the join in this case consists two parts: High frequency values and low frequency values
(unpopular). In different cases estimating cardinality for low frequency values was different for me.
In current case I think based on the uniform distribution. It means for t1.j1 “average frequency” is
number for rows(t1)/NDV(j1)=40/13= 3.076923 . Also we have 3 unpopular (low frequency
values) rows and 3 unpopular NDV. For each “low frequency” value we have
num_rows(t1)*density(j1)=2.54544≈3 frequency and we have 3 low frequency-unpopular rows
therefore unpopular cardinality is 3*3=9 so final cardinality will be
CARD(popular rows)+CARD(un popular rows)= 94.72704+ 9= 103.72704.
Lines from 10053 trace file
And execution plan
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 104 | 101 |
---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."J1"="T2"."J2")
The above test case was a quite simple because popular values of the hybrid histogram also are located
within range of high frequency values of the top frequency histogram. I mean popular values {1, 5,
6} of the hybrid histogram actually located 1-6 range of top frequency histogram.
Let see another example
CREATE TABLE t1(j1 NUMBER);
INSERT INTO t1 VALUES(6);

COMMIT;
/* execute dbms_stats.set_global_prefs('trace',to_char(512+128+2048+32768+4+8+16)); */
execute dbms_stats.gather_table_stats(null,'t1',method_opt=>'for all columns size 8');
/*exec dbms_stats.set_global_prefs('TRACE', null);*/
---Creating second table
CREATE TABLE t2(j2 number);
COMMIT;
execute dbms_stats.gather_table_stats(null,'t2',method_opt=>'for all columns size 4');
Lets enable 10053 trace file
ALTER SESSION SET EVENTS '10053 trace name context forever';
EXPLAIN PLAN
FOR
SELECT COUNT ( * )
FROM t1, t2

WHERE t1.j1 = t2.j2;
SELECT *
FROM table (DBMS_XPLAN.display);
ALTER SESSION SET EVENTS '10053 trace name context off';
Corresponding information from the dictionary:
t1.j1 freq ep_rep ep_num t2.j2 freq ep_num
1 1 1 1 1 3 3
2 2 2 3 3 4 7
4 3 1 6 4 7 14
6 4 3 10 20 1 15
7 4 4 14
17 3 1 17
18 1 1 18
20 2 1 20
And lines from dbms_stats trace
DBMS_STATS: > cdn 20, popFreq 7, popCnt 2, bktSize 2.4, bktSzFrc .4
DBMS_STATS: Evaluating hybrid histogram: cht.count 8, mnb 8, ssize 20,
min_ssize 2500, appr_ndv TRUE, ndv 13, selNdv 0, selFreq 0, pct 100, avg_bktsize
3, csr.hreq TRUE, normalize TRUE
DBMS_STATS: Histogram gathering flags: 527
DBMS_STATS: Accepting histogram
DBMS_STATS: Start fill_cstats - hybrid_enabled: TRUE
So we our average bucket size is 3 and we have 2 popular values {6, 7}. These values are not a part
of high frequency values in top frequency histogram. Table and column statistics from optimizer
trace file:
Table Stats::
Table: T2 Alias: T2
0.000000
***********************
Table Stats::
Table: T1 Alias: T1
0.000000

Histogram: Hybrid #Bkts: 8 UncompBkts: 20 EndPtVals: 8 ActualVal: yes
---Join Cardinality
SPD: Return code in qosdDSDirSetup: NOCTX, estType = JOIN
Firstly lets calculate cardinality for the high frequency values.
High freq values j2.freq j1.freq freq*freq
1 3 1.18182 3.54546
3 4 1.18182 4.72728
4 7 1.18182 8.27274
20 1 1.18182 1.18182
sum 17.7273
So our cardinality for high frequency values is 17.7273. And we also have num_rows(t1)-
popular_rows(t1)=20-15=5 unpopular rows. But as you see oracle computed final cardinality as
31. In my opinion popular rows of the hybrid histogram here play role. Test cases show that
optimizer in such situations also tries to take advantage of the popular values. In our case the value
6 and 7 are popular values and popular frequency is 7 (sum of popular frequency). If we try find
out frequencies of these values based on the top frequency histogram then we have to use density.
So cardinality for popular values will be:
Popular frequency*num_rows(t1)*density(j2)=7*20*0.0625=8.75. Moreover for every “low
frequency” values we have 1.18182≈1 frequency and we have 5 “low frequency” values (or
unpopular rows of the j2 column) therefore cardinality for “low frequency” could be consider as 5.
Eventually we can figure out final cardinality.
CARD = CARD (High frequency values) + CARD (Low frequency values) + CARD (Unpopular
rows) = 17.7273+8.75+5=31.4773.
And execution plan
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 31 | 26 |
---------------------------------------------------------------

---------------------------------------------------
2 - access("T1"."J1"="T2"."J2")
So it is an expected cardinality.
But in general here could be estimation or approximation errors which are related with rounding.
Sampling based estimation
As we know in Oracle Database 12c new dynamic sampling feature has been introduced. The
dynamic sampling level=11 is designed for the operations like single table, group by and join
operations for which oracle automatically defines sample size and tries to estimate cardinality of the
operations. Lets see following example and try to understand sampling mechanism in the join size
estimation.
CREATE TABLE t1
AS SELECT * FROM dba_users;
CREATE TABLE t2
AS SELECT * FROM dba_objects;
EXECUTE dbms_stats.gather_table_stats(user,'t1',method_opt=>'for all columns size 1');
EXECUTE dbms_stats.gather_table_stats(user,'t2',method_opt=>'for all columns size 1');
SELECT COUNT (*)
FROM t1, t2
WHERE t1.username = t2.owner;
Without histogram and in default sampling mode execution plan is:
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 92019 | 54942 |
---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."USERNAME"="T2"."OWNER")
Without histogram and automatic sampling mode execution plan is:
---------------------------------------------------------------
---------------------------------------------------------------
| 1 | SORT AGGREGATE | | 1 | 1 | 1 |
|* 2 | HASH JOIN | | 1 | 58728 | 54942 |

---------------------------------------------------------------
---------------------------------------------------
2 - access("T1"."USERNAME"="T2"."OWNER")
Note
-----
- dynamic statistics used: dynamic sampling (level=AUTO)
As we see without histogram there is significant difference between actual and estimated rows but in
case when automatic (adaptive) sampling is enabled estimation is good enough. The Question is how
did optimizer actually get cardinality as 58728? How did optimizer calculate it? To give the
explanation we could use 10046 and 10053 trace events. So in SQL trace file we could see following
lines.
SQL ID: 1bgh7fk6kqxg7
Plan Hash: 3696410285
SELECT /* DS_SVC */ /*+ dynamic_sampling(0) no_sql_tune no_monitoring
optimizer_features_enable(default) no_parallel result_cache(snapshot=3600)
*/ SUM(C1)
FROM
(SELECT /*+ qb_name("innerQuery") NO_INDEX_FFS( "T2#0") */ 1 AS C1 FROM
"T2" SAMPLE BLOCK(51.8135, 8) SEED(1) "T2#0", "T1" "T1#1" WHERE
("T1#1"."USERNAME"="T2#0"."OWNER")) innerQuery
call count cpu elapsed disk query current rows
------- ------ -------- ---------- ---------- ---------- ---------- ----------
Parse 1 0.00 0.00 0 0 0 0
Execute 1 0.00 0.00 0 0 0 0
Fetch 1 0.06 0.05 0 879 0 1
------- ------ -------- ---------- ---------- ---------- ---------- ----------
total 3 0.06 0.05 0 879 0 1
Misses in library cache during parse: 1
Optimizer mode: CHOOSE
Parsing user id: SYS (recursive depth: 1)
Rows Row Source Operation
------- ---------------------------------------------------
1 SORT AGGREGATE (cr=879 pr=0 pw=0 time=51540 us)
30429 HASH JOIN (cr=879 pr=0 pw=0 time=58582 us cost=220 size=1287306 card=47678)
42 TABLE ACCESS FULL T1 (cr=3 pr=0 pw=0 time=203 us cost=2 size=378 card=42)
51770 TABLE ACCESS SAMPLE T2 (cr=876 pr=0 pw=0 time=35978 us cost=218 size=858204
card=47678)
During parsing oracle has executed this SQL statement and result has been used to estimate size of
the join. The SQL statement used sampling (undocumented format) actually read 50 percent of the
T2 table blocks. Sampling was not applied to the T1 table because its size is quite small when
compared to the second table and 100% sampling of the T1 table does not consume “lot of” time
during parsing. It means oracle first identifies appropriate sampling size based on the table size and

then execute specific SQL statement. So we get 30429 rows based on the 51.8135 percent therefore
our estimated cardinality is 30429/51.8135*100=58727.94≈58728. Now let’s check optimizer trace
file:
>> Join Card adjusted from 92019.000000 to 58727.970000 due to adaptive dynamic sampling,
prelen=2
Adjusted Join Cards: adjRatio=0.638216 cardHjSmj=58727.970000 cardHjSmjNPF=58727.970000
cardNlj=58727.970000 cardNSQ=58727.970000 cardNSQ_na=92019.000000
Let see what will happen if we increase sizes of both tables – using multiple
insert into t select * from t
table name blocks row nums size mb
T1 3186 172032 25
T2 6158 368076 49
In this case oracle completely ignores adaptive sampling and uses uniform distribution to estimate
join size.
Table Stats::
Table: T2 Alias: T2
#Rows: 368076 SSZ: 0 LGR: 0 #Blks: 6158 AvgRowLen: 115.00 NEB: 0 ChainCnt:
0.00 SPC: 0 RFL: 0 RNF: 0 CBK: 0 CHR: 0 KQDFLG: 1
Column (#1): OWNER(VARCHAR2)
AvgLen: 6 NDV: 31 Nulls: 0 Density: 0.032258
***********************
Table Stats::
Table: T1 Alias: T1
Column (#1): USERNAME(VARCHAR2)
(0.023810)
In addition if you see SQL trace file then
SQL ID: 0ck072zj5gf73
Plan Hash: 3774486692
*/ SUM(C1)
FROM
("T1#1"."USERNAME"="T2#0"."OWNER")) innerQuery

------- ------ -------- ---------- ---------- ---------- ---------- ----------
Parse 1 0.00 0.00 0 2 0 0
Execute 1 0.00 0.00 0 0 0 0
Fetch 1 1.70 1.91 0 885 0 0
------- ------ -------- ---------- ---------- ---------- ---------- ----------
total 3 1.70 1.91 0 887 0 0
------- ---------------------------------------------------
4049738 HASH JOIN (cr=885 pr=0 pw=0 time=2696440 us cost=1835 size=5288231772
card=195860436)
card=47817)
6468 TABLE ACCESS FULL T1 (cr=124 pr=0 pw=0 time=28902 us cost=866 size=1548288
card=172032)
It is obvious that oracle stopped execution of this SQL during parsing, we can see it from rows
column of the execution statistics and also from row source statistics. Oracle did not complete HASH
JOIN operation in this SQL, we can be confirm that with result of the above SQL and row source
statistics. Sizes of the tables are not big actually but why did optimizer ignore and decided to continue
with previous approach? In my opinion here could be two factors, although sample size is not small
but in our case sample SQL actually took quite long time during parsing (1.8 sec elapsed time)
therefore oracle stopped it. I have added one filter predicate to the query:
SELECT COUNT (*)
FROM t1, t2
WHERE t1.username = t2.owner AND t2.object_type = 'TABLE';
In this case we will get the following lines
SQL ID: 8pu5v8h0ghy1z
Plan Hash: 3252009800
*/ SUM(C1)
FROM
(SELECT /*+ qb_name("innerQuery") NO_INDEX_FFS( "T2") */ 1 AS C1 FROM "T2"
SAMPLE BLOCK(12.9912, 8) SEED(1) "T2" WHERE ("T2"."OBJECT_TYPE"='TABLE'))
innerQuery
------- ------ -------- ---------- ---------- ---------- ---------- ----------
Parse 1 0.00 0.00 0 2 0 0
Execute 1 0.00 0.00 0 0 0 0
Fetch 1 0.01 0.01 0 761 0 1
------- ------ -------- ---------- ---------- ---------- ---------- ----------

total 3 0.01 0.01 0 763 0 1
------- ---------------------------------------------------
card=1018)
********************************************************************************
SQL ID: 9jv79m9u42jps
Plan Hash: 3525519047
OPT_ESTIMATE(@"innerQuery", TABLE, "T2#0", ROWS=5819.31) */ SUM(C1)
FROM
("T2#0"."OBJECT_TYPE"='TABLE') AND ("T1#1"."USERNAME"="T2#0"."OWNER"))
innerQuery
------- ------ -------- ---------- ---------- ---------- ---------- ----------
Parse 1 0.01 0.00 0 2 0 0
Execute 1 0.00 0.00 0 0 0 0
Fetch 1 0.74 1.02 0 6283 0 0
------- ------ -------- ---------- ---------- ---------- ---------- ----------
total 3 0.76 1.02 0 6285 0 0
------- ---------------------------------------------------
1412128 HASH JOIN (cr=6283 pr=0 pw=0 time=1243755 us cost=1908 size=215466084
card=5985169)
9880 TABLE ACCESS FULL T2 (cr=6167 pr=0 pw=0 time=20665 us cost=1674 size=87285
card=5819)
card=43197)
It means oracle firstly tried to estimate size of T2 table, because it has filter predicate and optimizer
thinks using ADS could be very efficient. If we should have added predicate like t2.owner=’HR’ then
optimizer would tried to estimate also T1 table cardinality. But the mechanism of estimating subset
of the join and then estimate whole join principle in this case actually ignored. However in this case
only T2 table has been estimated. We can easily see this fact from the trace file:
BASE STATISTICAL INFORMATION
***********************
Table Stats::

Table: T2 Alias: T2
Column (#1): OWNER(VARCHAR2)
***********************
Table Stats::
Table: T1 Alias: T1
Column (#1): USERNAME(VARCHAR2)
SINGLE TABLE ACCESS PATH
Single Table Cardinality Estimation for T1[T1]
SPD: Return code in qosdDSDirSetup: NOCTX, estType = TABLE
*** 2016-02-13 19:59:49.416
** Performing dynamic sampling initial checks. **
** Not using old style dynamic sampling since ADS is enabled.
Table: T1 Alias: T1
Card: Original: 172032.000000 Rounded: 172032 Computed: 172032.000000 Non
Adjusted: 172032.000000
Scan IO Cost (Disk) = 865.000000
Scan CPU Cost (Disk) = 48493707.840000
Total Scan IO Cost = 865.000000 (scan (Disk))
= 865.000000
Total Scan CPU Cost = 48493707.840000 (scan (Disk))
= 48493707.840000
Access Path: TableScan
Cost: 866.262422 Resp: 866.262422 Degree: 0
Cost_io: 865.000000 Cost_cpu: 48493708
Resp_io: 865.000000 Resp_cpu: 48493708
Best:: AccessPath: TableScan
Cost: 866.262422 Degree: 1 Resp: 866.262422 Card: 172032.000000 Bytes:
0.000000
Access path analysis for T2
***************************************
SINGLE TABLE ACCESS PATH
Single Table Cardinality Estimation for T2[T2]
SPD: Return code in qosdDSDirSetup: NOCTX, estType = TABLE
*** 2016-02-13 19:59:49.417
** Performing dynamic sampling initial checks. **
** Not using old style dynamic sampling since ADS is enabled.
Column (#6): OBJECT_TYPE(VARCHAR2)
Table: T2 Alias: T2
Card: Original: 368144.000000 >> Single Tab Card adjusted from 7832.851064 to
5819.310000 due to adaptive dynamic sampling
Rounded: 5819 Computed: 5819.310000 Non Adjusted: 7832.851064
Best NL cost: 5031394.059186
resc: 5031394.059186 resc_io: 5022741.000000 resc_cpu: 332391893348
resp: 5031394.059186 resp_io: 5022741.000000 resc_cpu: 332391893348

(0.023810)
*** 2016-02-13 19:59:51.055
In last case I have increased of both table sizes as
table name blocks row nums size mb
T1 101950 5505024 800
T2 196807 11780608 1600
In this case oracle completely ignored ADS and used statistics from dictionary to estimate size of
tables and join cardinality.
Summary
In this paper has explained the mechanism of the oracle optimizer to calculate join selectivity and
cardinality. We learned that firstly optimizer calculates join selectivity based on “pure” cardinality. To
estimate the “pure” cardinality optimizer identifies “distinct value, frequency” pairs for each column,
based on the column distribution. The column distribution information is identified by the histogram.
And as we know that, frequency histogram gives us completely whole data distribution of the column.
Also top frequency histogram gives us enough information for high frequency values. However for
less significantly values we can approach “uniform distribution”. Moreover if here are hybrid
histograms for the join columns in the dictionary then optimizer can use endpoint repeat count to
formulate frequency. In addition optimizer has chance to estimate join cardinality via sampling.
Although this process influenced by time restriction and size of the tables. As a result optimizer can
completely ignore adaptive dynamic sampling.
References
• Lewis, Jonathan. Cost-Based Oracle: Fundamentals Based Oracle: Fundamentals. Apress.
2006
• Alberto Dell'Era. Join Over Histograms. 2007
• http://www.adellera.it/investigations/join_over_histograms/JoinOverHistograms.pdf
• Chinar Aliyev. Automatic Sampling in Oracle 12c. 2014
• https://www.toadworld.com/platforms/oracle/w/wiki/11036.automaticadaptive-dynamic-
sampling-in-oracle-12c-part-2
• https://www.toadworld.com/platforms/oracle/w/wiki/11052.automaticadaptive-dynamic-
sampling-in-oracle-12c-part-3

Join Cardinality Estimation Methods_in_Oracle12c.pdf

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