<!-- doc/src/sgml/planstats.sgml -->

<chapter id="planner-stats-details">
 <title>How the Planner Uses Statistics</title>

  <para>
   This chapter builds on the material covered in <xref
   linkend="using-explain"> and <xref linkend="planner-stats"> to show some
   additional details about how the planner uses the
   system statistics to estimate the number of rows each part of a query might
   return. This is a significant part of the planning process,
   providing much of the raw material for cost calculation.
  </para>

  <para>
   The intent of this chapter is not to document the code in detail,
   but to present an overview of how it works.
   This will perhaps ease the learning curve for someone who subsequently
   wishes to read the code.
  </para>

 <sect1 id="row-estimation-examples">
  <title>Row Estimation Examples</title>

  <indexterm zone="row-estimation-examples">
   <primary>row estimation</primary>
   <secondary>planner</secondary>
  </indexterm>

  <para>
   The examples shown below use tables in the <productname>PostgreSQL</>
   regression test database.
   The outputs shown are taken from version 8.3.
   The behavior of earlier (or later) versions might vary.
   Note also that since <command>ANALYZE</> uses random sampling
   while producing statistics, the results will change slightly after
   any new <command>ANALYZE</>.
  </para>

  <para>
   Let's start with a very simple query:

<programlisting>
EXPLAIN SELECT * FROM tenk1;

                         QUERY PLAN
-------------------------------------------------------------
 Seq Scan on tenk1  (cost=0.00..458.00 rows=10000 width=244)
</programlisting>

   How the planner determines the cardinality of <structname>tenk1</structname>
   is covered in <xref linkend="planner-stats">, but is repeated here for
   completeness. The number of pages and rows is looked up in
   <structname>pg_class</structname>:

<programlisting>
SELECT relpages, reltuples FROM pg_class WHERE relname = 'tenk1';

 relpages | reltuples
----------+-----------
      358 |     10000
</programlisting>

    These numbers are current as of the last <command>VACUUM</> or
    <command>ANALYZE</> on the table.  The planner then fetches the
    actual current number of pages in the table (this is a cheap operation,
    not requiring a table scan).  If that is different from
    <structfield>relpages</structfield> then
    <structfield>reltuples</structfield> is scaled accordingly to
    arrive at a current number-of-rows estimate.  In the example above, the value of
    <structfield>relpages</structfield> is up-to-date so the rows estimate is
    the same as <structfield>reltuples</structfield>.
  </para>

  <para>
   Let's move on to an example with a range condition in its
   <literal>WHERE</literal> clause:

<programlisting>
EXPLAIN SELECT * FROM tenk1 WHERE unique1 &lt; 1000;

                                   QUERY PLAN
--------------------------------------------------------------------------------
 Bitmap Heap Scan on tenk1  (cost=24.06..394.64 rows=1007 width=244)
   Recheck Cond: (unique1 &lt; 1000)
   -&gt;  Bitmap Index Scan on tenk1_unique1  (cost=0.00..23.80 rows=1007 width=0)
         Index Cond: (unique1 &lt; 1000)
</programlisting>

   The planner examines the <literal>WHERE</literal> clause condition
   and looks up the selectivity function for the operator
   <literal>&lt;</literal> in <structname>pg_operator</structname>.
   This is held in the column <structfield>oprrest</structfield>,
   and the entry in this case is <function>scalarltsel</function>.
   The <function>scalarltsel</function> function retrieves the histogram for
   <structfield>unique1</structfield> from
   <structname>pg_statistic</structname>.  For manual queries it is more
   convenient to look in the simpler <structname>pg_stats</structname>
   view:

<programlisting>
SELECT histogram_bounds FROM pg_stats
WHERE tablename='tenk1' AND attname='unique1';

                   histogram_bounds
------------------------------------------------------
 {0,993,1997,3050,4040,5036,5957,7057,8029,9016,9995}
</programlisting>

   Next the fraction of the histogram occupied by <quote>&lt; 1000</quote>
   is worked out. This is the selectivity. The histogram divides the range
   into equal frequency buckets, so all we have to do is locate the bucket
   that our value is in and count <emphasis>part</emphasis> of it and
   <emphasis>all</emphasis> of the ones before. The value 1000 is clearly in
   the second bucket (993-1997).  Assuming a linear distribution of
   values inside each bucket, we can calculate the selectivity as:

<programlisting>
selectivity = (1 + (1000 - bucket[2].min)/(bucket[2].max - bucket[2].min))/num_buckets
            = (1 + (1000 - 993)/(1997 - 993))/10
            = 0.100697
</programlisting>

   that is, one whole bucket plus a linear fraction of the second, divided by
   the number of buckets. The estimated number of rows can now be calculated as
   the product of the selectivity and the cardinality of
   <structname>tenk1</structname>:

<programlisting>
rows = rel_cardinality * selectivity
     = 10000 * 0.100697
     = 1007  (rounding off)
</programlisting>
  </para>

  <para>
   Next let's consider an example with an equality condition in its
   <literal>WHERE</literal> clause:

<programlisting>
EXPLAIN SELECT * FROM tenk1 WHERE stringu1 = 'CRAAAA';

                        QUERY PLAN
----------------------------------------------------------
 Seq Scan on tenk1  (cost=0.00..483.00 rows=30 width=244)
   Filter: (stringu1 = 'CRAAAA'::name)
</programlisting>

   Again the planner examines the <literal>WHERE</literal> clause condition
   and looks up the selectivity function for <literal>=</literal>, which is
   <function>eqsel</function>.  For equality estimation the histogram is
   not useful; instead the list of <firstterm>most
   common values</> (<acronym>MCV</acronym>s) is used to determine the
   selectivity. Let's have a look at the MCVs, with some additional columns
   that will be useful later:

<programlisting>
SELECT null_frac, n_distinct, most_common_vals, most_common_freqs FROM pg_stats
WHERE tablename='tenk1' AND attname='stringu1';

null_frac         | 0
n_distinct        | 676
most_common_vals  | {EJAAAA,BBAAAA,CRAAAA,FCAAAA,FEAAAA,GSAAAA,JOAAAA,MCAAAA,NAAAAA,WGAAAA}
most_common_freqs | {0.00333333,0.003,0.003,0.003,0.003,0.003,0.003,0.003,0.003,0.003}

</programlisting>

   Since <literal>CRAAAA</> appears in the list of MCVs, the selectivity is
   merely the corresponding entry in the list of most common frequencies
   (<acronym>MCF</acronym>s):

<programlisting>
selectivity = mcf[3]
            = 0.003
</programlisting>

   As before, the estimated number of rows is just the product of this with the
   cardinality of <structname>tenk1</structname>:

<programlisting>
rows = 10000 * 0.003
     = 30
</programlisting>
  </para>

  <para>
   Now consider the same query, but with a constant that is not in the
   <acronym>MCV</acronym> list:

<programlisting>
EXPLAIN SELECT * FROM tenk1 WHERE stringu1 = 'xxx';

                        QUERY PLAN
----------------------------------------------------------
 Seq Scan on tenk1  (cost=0.00..483.00 rows=15 width=244)
   Filter: (stringu1 = 'xxx'::name)
</programlisting>

   This is quite a different problem: how to estimate the selectivity when the
   value is <emphasis>not</emphasis> in the <acronym>MCV</acronym> list.
   The approach is to use the fact that the value is not in the list,
   combined with the knowledge of the frequencies for all of the
   <acronym>MCV</acronym>s:

<programlisting>
selectivity = (1 - sum(mvf))/(num_distinct - num_mcv)
            = (1 - (0.00333333 + 0.003 + 0.003 + 0.003 + 0.003 + 0.003 +
                    0.003 + 0.003 + 0.003 + 0.003))/(676 - 10)
            = 0.0014559
</programlisting>

   That is, add up all the frequencies for the <acronym>MCV</acronym>s and
   subtract them from one, then
   divide by the number of <emphasis>other</emphasis> distinct values.
   This amounts to assuming that the fraction of the column that is not any
   of the MCVs is evenly distributed among all the other distinct values.
   Notice that there are no null values so we don't have to worry about those
   (otherwise we'd subtract the null fraction from the numerator as well).
   The estimated number of rows is then calculated as usual:

<programlisting>
rows = 10000 * 0.0014559
     = 15  (rounding off)
</programlisting>
  </para>

  <para>
   The previous example with <literal>unique1 &lt; 1000</> was an
   oversimplification of what <function>scalarltsel</function> really does;
   now that we have seen an example of the use of MCVs, we can fill in some
   more detail.  The example was correct as far as it went, because since
   <structfield>unique1</> is a unique column it has no MCVs (obviously, no
   value is any more common than any other value).  For a non-unique
   column, there will normally be both a histogram and an MCV list, and
   <emphasis>the histogram does not include the portion of the column
   population represented by the MCVs</>.  We do things this way because
   it allows more precise estimation.  In this situation
   <function>scalarltsel</function> directly applies the condition (e.g.,
   <quote>&lt; 1000</>) to each value of the MCV list, and adds up the
   frequencies of the MCVs for which the condition is true.  This gives
   an exact estimate of the selectivity within the portion of the table
   that is MCVs.  The histogram is then used in the same way as above
   to estimate the selectivity in the portion of the table that is not
   MCVs, and then the two numbers are combined to estimate the overall
   selectivity.  For example, consider

<programlisting>
EXPLAIN SELECT * FROM tenk1 WHERE stringu1 &lt; 'IAAAAA';

                         QUERY PLAN
------------------------------------------------------------
 Seq Scan on tenk1  (cost=0.00..483.00 rows=3077 width=244)
   Filter: (stringu1 &lt; 'IAAAAA'::name)
</programlisting>

   We already saw the MCV information for <structfield>stringu1</>,
   and here is its histogram:

<programlisting>
SELECT histogram_bounds FROM pg_stats
WHERE tablename='tenk1' AND attname='stringu1';

                                histogram_bounds
--------------------------------------------------------------------------------
 {AAAAAA,CQAAAA,FRAAAA,IBAAAA,KRAAAA,NFAAAA,PSAAAA,SGAAAA,VAAAAA,XLAAAA,ZZAAAA}
</programlisting>

   Checking the MCV list, we find that the condition <literal>stringu1 &lt;
   'IAAAAA'</> is satisfied by the first six entries and not the last four,
   so the selectivity within the MCV part of the population is

<programlisting>
selectivity = sum(relevant mvfs)
            = 0.00333333 + 0.003 + 0.003 + 0.003 + 0.003 + 0.003
            = 0.01833333
</programlisting>

   Summing all the MCFs also tells us that the total fraction of the
   population represented by MCVs is 0.03033333, and therefore the
   fraction represented by the histogram is 0.96966667 (again, there
   are no nulls, else we'd have to exclude them here).  We can see
   that the value <literal>IAAAAA</> falls nearly at the end of the
   third histogram bucket.  Using some rather cheesy assumptions
   about the frequency of different characters, the planner arrives
   at the estimate 0.298387 for the portion of the histogram population
   that is less than <literal>IAAAAA</>.  We then combine the estimates
   for the MCV and non-MCV populations:

<programlisting>
selectivity = mcv_selectivity + histogram_selectivity * histogram_fraction
            = 0.01833333 + 0.298387 * 0.96966667
            = 0.307669

rows        = 10000 * 0.307669
            = 3077  (rounding off)
</programlisting>

   In this particular example, the correction from the MCV list is fairly
   small, because the column distribution is actually quite flat (the
   statistics showing these particular values as being more common than
   others are mostly due to sampling error).  In a more typical case where
   some values are significantly more common than others, this complicated
   process gives a useful improvement in accuracy because the selectivity
   for the most common values is found exactly.
  </para>

  <para>
   Now let's consider a case with more than one
   condition in the <literal>WHERE</literal> clause:

<programlisting>
EXPLAIN SELECT * FROM tenk1 WHERE unique1 &lt; 1000 AND stringu1 = 'xxx';

                                   QUERY PLAN
--------------------------------------------------------------------------------
 Bitmap Heap Scan on tenk1  (cost=23.80..396.91 rows=1 width=244)
   Recheck Cond: (unique1 &lt; 1000)
   Filter: (stringu1 = 'xxx'::name)
   -&gt;  Bitmap Index Scan on tenk1_unique1  (cost=0.00..23.80 rows=1007 width=0)
         Index Cond: (unique1 &lt; 1000)
</programlisting>

   The planner assumes that the two conditions are independent, so that
   the individual selectivities of the clauses can be multiplied together:

<programlisting>
selectivity = selectivity(unique1 &lt; 1000) * selectivity(stringu1 = 'xxx')
            = 0.100697 * 0.0014559
            = 0.0001466

rows        = 10000 * 0.0001466
            = 1  (rounding off)
</programlisting>

   Notice that the number of rows estimated to be returned from the bitmap
   index scan reflects only the condition used with the index; this is
   important since it affects the cost estimate for the subsequent heap
   fetches.
  </para>

  <para>
   Finally we will examine a query that involves a join:

<programlisting>
EXPLAIN SELECT * FROM tenk1 t1, tenk2 t2
WHERE t1.unique1 &lt; 50 AND t1.unique2 = t2.unique2;

                                      QUERY PLAN
--------------------------------------------------------------------------------------
 Nested Loop  (cost=4.64..456.23 rows=50 width=488)
   -&gt;  Bitmap Heap Scan on tenk1 t1  (cost=4.64..142.17 rows=50 width=244)
         Recheck Cond: (unique1 &lt; 50)
         -&gt;  Bitmap Index Scan on tenk1_unique1  (cost=0.00..4.63 rows=50 width=0)
               Index Cond: (unique1 &lt; 50)
   -&gt;  Index Scan using tenk2_unique2 on tenk2 t2  (cost=0.00..6.27 rows=1 width=244)
         Index Cond: (unique2 = t1.unique2)
</programlisting>

   The restriction on <structname>tenk1</structname>,
   <literal>unique1 &lt; 50</literal>,
   is evaluated before the nested-loop join.
   This is handled analogously to the previous range example.  This time the
   value 50 falls into the first bucket of the
   <structfield>unique1</structfield> histogram:

<programlisting>
selectivity = (0 + (50 - bucket[1].min)/(bucket[1].max - bucket[1].min))/num_buckets
            = (0 + (50 - 0)/(993 - 0))/10
            = 0.005035

rows        = 10000 * 0.005035
            = 50  (rounding off)
</programlisting>

   The restriction for the join is <literal>t2.unique2 = t1.unique2</>.
   The operator is just
   our familiar <literal>=</literal>, however the selectivity function is
   obtained from the <structfield>oprjoin</structfield> column of
   <structname>pg_operator</structname>, and is <function>eqjoinsel</function>.
   <function>eqjoinsel</function> looks up the statistical information for both
   <structname>tenk2</structname> and <structname>tenk1</structname>:

<programlisting>
SELECT tablename, null_frac,n_distinct, most_common_vals FROM pg_stats
WHERE tablename IN ('tenk1', 'tenk2') AND attname='unique2';

tablename  | null_frac | n_distinct | most_common_vals
-----------+-----------+------------+------------------
 tenk1     |         0 |         -1 |
 tenk2     |         0 |         -1 |
</programlisting>

   In this case there is no <acronym>MCV</acronym> information for
   <structfield>unique2</structfield> because all the values appear to be
   unique, so we use an algorithm that relies only on the number of
   distinct values for both relations together with their null fractions:

<programlisting>
selectivity = (1 - null_frac1) * (1 - null_frac2) * min(1/num_distinct1, 1/num_distinct2)
            = (1 - 0) * (1 - 0) / max(10000, 10000)
            = 0.0001
</programlisting>

   This is, subtract the null fraction from one for each of the relations,
   and divide by the maximum of the numbers of distinct values.
   The number of rows
   that the join is likely to emit is calculated as the cardinality of the
   Cartesian product of the two inputs, multiplied by the
   selectivity:

<programlisting>
rows = (outer_cardinality * inner_cardinality) * selectivity
     = (50 * 10000) * 0.0001
     = 50
</programlisting>
  </para>

  <para>
   Had there been MCV lists for the two columns,
   <function>eqjoinsel</function> would have used direct comparison of the MCV
   lists to determine the join selectivity within the part of the column
   populations represented by the MCVs.  The estimate for the remainder of the
   populations follows the same approach shown here.
  </para>

  <para>
   Notice that we showed <literal>inner_cardinality</> as 10000, that is,
   the unmodified size of <structname>tenk2</>.  It might appear from
   inspection of the <command>EXPLAIN</> output that the estimate of
   join rows comes from 50 * 1, that is, the number of outer rows times
   the estimated number of rows obtained by each inner index scan on
   <structname>tenk2</>.  But this is not the case: the join relation size
   is estimated before any particular join plan has been considered.  If
   everything is working well then the two ways of estimating the join
   size will produce about the same answer, but due to round-off error and
   other factors they sometimes diverge significantly.
  </para>

  <para>
   For those interested in further details, estimation of the size of
   a table (before any <literal>WHERE</> clauses) is done in
   <filename>src/backend/optimizer/util/plancat.c</filename>. The generic
   logic for clause selectivities is in
   <filename>src/backend/optimizer/path/clausesel.c</filename>.  The
   operator-specific selectivity functions are mostly found
   in <filename>src/backend/utils/adt/selfuncs.c</filename>.
  </para>
 </sect1>

 <sect1 id="multivariate-statistics-examples">
  <title>Multivariate Statistics Examples</title>

  <indexterm>
   <primary>row estimation</primary>
   <secondary>multivariate</secondary>
  </indexterm>

  <sect2>
   <title>Functional dependencies</title>
   <para>
    Multivariate correlation can be seen with a very simple data set &mdash; a
    table with two columns, both containing the same values:

<programlisting>
CREATE TABLE t (a INT, b INT);
INSERT INTO t SELECT i % 100, i % 100 FROM generate_series(1, 10000) s(i);
ANALYZE t;
</programlisting>

    As explained in <xref linkend="planner-stats">, the planner can determine
    cardinality of <structname>t</structname> using the number of pages and
    rows obtained from <structname>pg_class</structname>:

<programlisting>
SELECT relpages, reltuples FROM pg_class WHERE relname = 't';

 relpages | reltuples
----------+-----------
       45 |     10000
</programlisting>

    The data distribution is very simple; there are only 100 distinct values
    in each column, uniformly distributed.
   </para>

   <para>
    The following example shows the result of estimating a <literal>WHERE</>
    condition on the <structfield>a</> column:

<programlisting>
EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1;
                                 QUERY PLAN                                  
-------------------------------------------------------------------------------
 Seq Scan on t  (cost=0.00..170.00 rows=100 width=8) (actual rows=100 loops=1)
   Filter: (a = 1)
   Rows Removed by Filter: 9900
</programlisting>

    The planner examines the condition and determines the selectivity
    of this clause to be 1%.  By comparing this estimate and the actual
    number of rows, we see that the estimate is very accurate
    (in fact exact, as the table is very small).  Changing the
    <literal>WHERE</> to use the <structfield>b</> column, an identical
    plan is generated.  Observe what happens if we apply the same
    condition on both columns combining them with <literal>AND</>:

<programlisting>
EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1 AND b = 1;
                                 QUERY PLAN                                  
-----------------------------------------------------------------------------
 Seq Scan on t  (cost=0.00..195.00 rows=1 width=8) (actual rows=100 loops=1)
   Filter: ((a = 1) AND (b = 1))
   Rows Removed by Filter: 9900
</programlisting>

    The planner estimates the selectivity for each condition individually,
    arriving to the 1% estimates as above, and then multiplies them, getting
    the final 0.01% estimate. The <quote>actual</quote> figures, however,
    show that this results in a significant underestimate, as the actual
    number of rows matching the conditions (100) is two orders of magnitude
    higher than the estimated value.
   </para>

   <para>
    This problem can be fixed by applying functional-dependency
    multivariate statistics on the two columns:

<programlisting>
CREATE STATISTICS stts WITH (dependencies) ON (a, b) FROM t;
ANALYZE t;
EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1 AND b = 1;
                                  QUERY PLAN                                   
-------------------------------------------------------------------------------
 Seq Scan on t  (cost=0.00..195.00 rows=100 width=8) (actual rows=100 loops=1)
   Filter: ((a = 1) AND (b = 1))
   Rows Removed by Filter: 9900
</programlisting>
   </para>
  </sect2>

  <sect2>
   <title>Multivariate N-Distinct coefficients</title>
   <para>
    A similar problem occurs with estimation of the cardinality of distinct
    elements, used to determine the number of groups that would be generated
    by a <command>GROUP BY</command> clause.  When <command>GROUP BY</command>
    lists a single column, the n-distinct estimate (which can be seen as the
    number of rows returned by the aggregate execution node) is very accurate:
<programlisting>
EXPLAIN (ANALYZE, TIMING OFF) SELECT COUNT(*) FROM t GROUP BY a;
                                       QUERY PLAN                                        
-----------------------------------------------------------------------------------------
 HashAggregate  (cost=195.00..196.00 rows=100 width=12) (actual rows=100 loops=1)
   Group Key: a
   -&gt;  Seq Scan on t  (cost=0.00..145.00 rows=10000 width=4) (actual rows=10000 loops=1)
</programlisting>
    But without multivariate statistics, the estimate for the number of
    groups in a query with two columns in <command>GROUP BY</command>, as
    in the following example, is off by an order of magnitude:
<programlisting>
EXPLAIN (ANALYZE, TIMING OFF) SELECT COUNT(*) FROM t GROUP BY a, b;
                                       QUERY PLAN                                        
--------------------------------------------------------------------------------------------
 HashAggregate  (cost=220.00..230.00 rows=1000 width=16) (actual rows=100 loops=1)
   Group Key: a, b
   -&gt;  Seq Scan on t  (cost=0.00..145.00 rows=10000 width=8) (actual rows=10000 loops=1)
</programlisting>
    By dropping the existing statistics and re-creating it to include n-distinct
    calculation, the estimate is much improved:
<programlisting>
DROP STATISTICS stts;
CREATE STATISTICS stts WITH (dependencies, ndistinct) ON (a, b) FROM t;
ANALYZE t;
EXPLAIN (ANALYZE, TIMING OFF) SELECT COUNT(*) FROM t GROUP BY a, b;
                                       QUERY PLAN                                        
--------------------------------------------------------------------------------------------
 HashAggregate  (cost=220.00..221.00 rows=100 width=16) (actual rows=100 loops=1)
   Group Key: a, b
   -&gt;  Seq Scan on t  (cost=0.00..145.00 rows=10000 width=8) (actual rows=10000 loops=1)
</programlisting>
   </para>

  </sect2>
 </sect1>
</chapter>
