This chapter builds on the material covered in and 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. 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. row estimation planner The examples shown below use tables in the 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 ANALYZE uses random sampling while producing statistics, the results will change slightly after any new ANALYZE. Let's start with a very simple query: EXPLAIN SELECT * FROM tenk1; QUERY PLAN ------------------------------------------------------------- Seq Scan on tenk1 (cost=0.00..458.00 rows=10000 width=244) How the planner determines the cardinality of tenk1 is covered in , but is repeated here for completeness. The number of pages and rows is looked up in pg_class: SELECT relpages, reltuples FROM pg_class WHERE relname = 'tenk1'; relpages | reltuples ----------+----------- 358 | 10000 These numbers are current as of the last VACUUM or 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 relpages then reltuples is scaled accordingly to arrive at a current number-of-rows estimate. In the example above, the value of relpages is up-to-date so the rows estimate is the same as reltuples. Let's move on to an example with a range condition in its WHERE clause: EXPLAIN SELECT * FROM tenk1 WHERE unique1 < 1000; QUERY PLAN -------------------------------------------------------------------&zwsp;------------- Bitmap Heap Scan on tenk1 (cost=24.06..394.64 rows=1007 width=244) Recheck Cond: (unique1 < 1000) -> Bitmap Index Scan on tenk1_unique1 (cost=0.00..23.80 rows=1007 width=0) Index Cond: (unique1 < 1000) The planner examines the WHERE clause condition and looks up the selectivity function for the operator < in pg_operator. This is held in the column oprrest, and the entry in this case is scalarltsel. The scalarltsel function retrieves the histogram for unique1 from pg_statistic. For manual queries it is more convenient to look in the simpler pg_stats view: 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} Next the fraction of the histogram occupied by < 1000 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 part of it and all 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: selectivity = (1 + (1000 - bucket[2].min)/(bucket[2].max - bucket[2].min))/num_buckets = (1 + (1000 - 993)/(1997 - 993))/10 = 0.100697 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 tenk1: rows = rel_cardinality * selectivity = 10000 * 0.100697 = 1007 (rounding off) Next let's consider an example with an equality condition in its WHERE clause: 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) Again the planner examines the WHERE clause condition and looks up the selectivity function for =, which is eqsel. For equality estimation the histogram is not useful; instead the list of most common values (MCVs) is used to determine the selectivity. Let's have a look at the MCVs, with some additional columns that will be useful later: 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,&zwsp;JOAAAA,MCAAAA,NAAAAA,WGAAAA} most_common_freqs | {0.00333333,0.003,0.003,0.003,0.003,0.003,&zwsp;0.003,0.003,0.003,0.003} Since CRAAAA appears in the list of MCVs, the selectivity is merely the corresponding entry in the list of most common frequencies (MCFs): selectivity = mcf[3] = 0.003 As before, the estimated number of rows is just the product of this with the cardinality of tenk1: rows = 10000 * 0.003 = 30 Now consider the same query, but with a constant that is not in the MCV list: 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) This is quite a different problem: how to estimate the selectivity when the value is not in the MCV 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 MCVs: 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 That is, add up all the frequencies for the MCVs and subtract them from one, then divide by the number of other 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: rows = 10000 * 0.0014559 = 15 (rounding off) The previous example with unique1 < 1000 was an oversimplification of what scalarltsel 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 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 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 scalarltsel directly applies the condition (e.g., < 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 EXPLAIN SELECT * FROM tenk1 WHERE stringu1 < 'IAAAAA'; QUERY PLAN ------------------------------------------------------------ Seq Scan on tenk1 (cost=0.00..483.00 rows=3077 width=244) Filter: (stringu1 < 'IAAAAA'::name) We already saw the MCV information for stringu1, and here is its histogram: SELECT histogram_bounds FROM pg_stats WHERE tablename='tenk1' AND attname='stringu1'; histogram_bounds -------------------------------------------------------------------&zwsp;------------- {AAAAAA,CQAAAA,FRAAAA,IBAAAA,KRAAAA,NFAAAA,PSAAAA,SGAAAA,VAAAAA,&zwsp;XLAAAA,ZZAAAA} Checking the MCV list, we find that the condition stringu1 < 'IAAAAA' is satisfied by the first six entries and not the last four, so the selectivity within the MCV part of the population is selectivity = sum(relevant mvfs) = 0.00333333 + 0.003 + 0.003 + 0.003 + 0.003 + 0.003 = 0.01833333 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 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 IAAAAA. We then combine the estimates for the MCV and non-MCV populations: selectivity = mcv_selectivity + histogram_selectivity * histogram_fraction = 0.01833333 + 0.298387 * 0.96966667 = 0.307669 rows = 10000 * 0.307669 = 3077 (rounding off) 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. Now let's consider a case with more than one condition in the WHERE clause: EXPLAIN SELECT * FROM tenk1 WHERE unique1 < 1000 AND stringu1 = 'xxx'; QUERY PLAN -------------------------------------------------------------------&zwsp;------------- Bitmap Heap Scan on tenk1 (cost=23.80..396.91 rows=1 width=244) Recheck Cond: (unique1 < 1000) Filter: (stringu1 = 'xxx'::name) -> Bitmap Index Scan on tenk1_unique1 (cost=0.00..23.80 rows=1007 width=0) Index Cond: (unique1 < 1000) The planner assumes that the two conditions are independent, so that the individual selectivities of the clauses can be multiplied together: selectivity = selectivity(unique1 < 1000) * selectivity(stringu1 = 'xxx') = 0.100697 * 0.0014559 = 0.0001466 rows = 10000 * 0.0001466 = 1 (rounding off) 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. Finally we will examine a query that involves a join: EXPLAIN SELECT * FROM tenk1 t1, tenk2 t2 WHERE t1.unique1 < 50 AND t1.unique2 = t2.unique2; QUERY PLAN -------------------------------------------------------------------&zwsp;------------------- Nested Loop (cost=4.64..456.23 rows=50 width=488) -> Bitmap Heap Scan on tenk1 t1 (cost=4.64..142.17 rows=50 width=244) Recheck Cond: (unique1 < 50) -> Bitmap Index Scan on tenk1_unique1 (cost=0.00..4.63 rows=50 width=0) Index Cond: (unique1 < 50) -> Index Scan using tenk2_unique2 on tenk2 t2 (cost=0.00..6.27 rows=1 width=244) Index Cond: (unique2 = t1.unique2) The restriction on tenk1, unique1 < 50, 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 unique1 histogram: 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) The restriction for the join is t2.unique2 = t1.unique2. The operator is just our familiar =, however the selectivity function is obtained from the oprjoin column of pg_operator, and is eqjoinsel. eqjoinsel looks up the statistical information for both tenk2 and tenk1: 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 | In this case there is no MCV information for unique2 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: selectivity = (1 - null_frac1) * (1 - null_frac2) * min(1/num_distinct1, 1/num_distinct2) = (1 - 0) * (1 - 0) / max(10000, 10000) = 0.0001 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: rows = (outer_cardinality * inner_cardinality) * selectivity = (50 * 10000) * 0.0001 = 50 Had there been MCV lists for the two columns, eqjoinsel 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. Notice that we showed inner_cardinality as 10000, that is, the unmodified size of tenk2. It might appear from inspection of the 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 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. For those interested in further details, estimation of the size of a table (before any WHERE clauses) is done in src/backend/optimizer/util/plancat.c. The generic logic for clause selectivities is in src/backend/optimizer/path/clausesel.c. The operator-specific selectivity functions are mostly found in src/backend/utils/adt/selfuncs.c. row estimation multivariate Multivariate correlation can be demonstrated with a very simple data set — a table with two columns, both containing the same values: CREATE TABLE t (a INT, b INT); INSERT INTO t SELECT i % 100, i % 100 FROM generate_series(1, 10000) s(i); ANALYZE t; As explained in , the planner can determine cardinality of t using the number of pages and rows obtained from pg_class: SELECT relpages, reltuples FROM pg_class WHERE relname = 't'; relpages | reltuples ----------+----------- 45 | 10000 The data distribution is very simple; there are only 100 distinct values in each column, uniformly distributed. The following example shows the result of estimating a WHERE condition on the a column: EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1; QUERY PLAN -------------------------------------------------------------------&zwsp;------------ 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 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 WHERE condition to use the b column, an identical plan is generated. But observe what happens if we apply the same condition on both columns, combining them with AND: EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1 AND b = 1; QUERY PLAN -------------------------------------------------------------------&zwsp;---------- 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 The planner estimates the selectivity for each condition individually, arriving at the same 1% estimates as above. Then it assumes that the conditions are independent, and so it multiplies their selectivities, producing a final selectivity estimate of just 0.01%. This is a significant underestimate, as the actual number of rows matching the conditions (100) is two orders of magnitude higher. This problem can be fixed by creating a statistics object that directs ANALYZE to calculate functional-dependency multivariate statistics on the two columns: CREATE STATISTICS stts (dependencies) ON a, b FROM t; ANALYZE t; EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1 AND b = 1; QUERY PLAN -------------------------------------------------------------------&zwsp;------------ 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 A similar problem occurs with estimation of the cardinality of sets of multiple columns, such as the number of groups that would be generated by a GROUP BY clause. When GROUP BY lists a single column, the n-distinct estimate (which is visible as the estimated number of rows returned by the HashAggregate node) is very accurate: EXPLAIN (ANALYZE, TIMING OFF) SELECT COUNT(*) FROM t GROUP BY a; QUERY PLAN -------------------------------------------------------------------&zwsp;---------------------- HashAggregate (cost=195.00..196.00 rows=100 width=12) (actual rows=100 loops=1) Group Key: a -> Seq Scan on t (cost=0.00..145.00 rows=10000 width=4) (actual rows=10000 loops=1) But without multivariate statistics, the estimate for the number of groups in a query with two columns in GROUP BY, as in the following example, is off by an order of magnitude: EXPLAIN (ANALYZE, TIMING OFF) SELECT COUNT(*) FROM t GROUP BY a, b; QUERY PLAN -------------------------------------------------------------------&zwsp;------------------------- HashAggregate (cost=220.00..230.00 rows=1000 width=16) (actual rows=100 loops=1) Group Key: a, b -> Seq Scan on t (cost=0.00..145.00 rows=10000 width=8) (actual rows=10000 loops=1) By redefining the statistics object to include n-distinct counts for the two columns, the estimate is much improved: DROP STATISTICS stts; CREATE STATISTICS stts (dependencies, ndistinct) ON a, b FROM t; ANALYZE t; EXPLAIN (ANALYZE, TIMING OFF) SELECT COUNT(*) FROM t GROUP BY a, b; QUERY PLAN -------------------------------------------------------------------&zwsp;------------------------- HashAggregate (cost=220.00..221.00 rows=100 width=16) (actual rows=100 loops=1) Group Key: a, b -> Seq Scan on t (cost=0.00..145.00 rows=10000 width=8) (actual rows=10000 loops=1) As explained in , functional dependencies are very cheap and efficient type of statistics, but their main limitation is their global nature (only tracking dependencies at the column level, not between individual column values). This section introduces multivariate variant of MCV (most-common values) lists, a straightforward extension of the per-column statistics described in . These statistics address the limitation by storing individual values, but it is naturally more expensive, both in terms of building the statistics in ANALYZE, storage and planning time. Let's look at the query from again, but this time with a MCV list created on the same set of columns (be sure to drop the functional dependencies, to make sure the planner uses the newly created statistics). DROP STATISTICS stts; CREATE STATISTICS stts2 (mcv) ON a, b FROM t; ANALYZE t; EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1 AND b = 1; QUERY PLAN -------------------------------------------------------------------&zwsp;------------ 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 The estimate is as accurate as with the functional dependencies, mostly thanks to the table being fairly small and having a simple distribution with a low number of distinct values. Before looking at the second query, which was not handled by functional dependencies particularly well, let's inspect the MCV list a bit. Inspecting the MCV list is possible using pg_mcv_list_items set-returning function. SELECT m.* FROM pg_statistic_ext join pg_statistic_ext_data on (oid = stxoid), pg_mcv_list_items(stxdmcv) m WHERE stxname = 'stts2'; index | values | nulls | frequency | base_frequency -------+----------+-------+-----------+---------------- 0 | {0, 0} | {f,f} | 0.01 | 0.0001 1 | {1, 1} | {f,f} | 0.01 | 0.0001 ... 49 | {49, 49} | {f,f} | 0.01 | 0.0001 50 | {50, 50} | {f,f} | 0.01 | 0.0001 ... 97 | {97, 97} | {f,f} | 0.01 | 0.0001 98 | {98, 98} | {f,f} | 0.01 | 0.0001 99 | {99, 99} | {f,f} | 0.01 | 0.0001 (100 rows) This confirms there are 100 distinct combinations in the two columns, and all of them are about equally likely (1% frequency for each one). The base frequency is the frequency computed from per-column statistics, as if there were no multi-column statistics. Had there been any null values in either of the columns, this would be identified in the nulls column. When estimating the selectivity, the planner applies all the conditions on items in the MCV list, and then sums the frequencies of the matching ones. See mcv_clauselist_selectivity in src/backend/statistics/mcv.c for details. Compared to functional dependencies, MCV lists have two major advantages. Firstly, the list stores actual values, making it possible to decide which combinations are compatible. EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a = 1 AND b = 10; QUERY PLAN -------------------------------------------------------------------&zwsp;-------- Seq Scan on t (cost=0.00..195.00 rows=1 width=8) (actual rows=0 loops=1) Filter: ((a = 1) AND (b = 10)) Rows Removed by Filter: 10000 Secondly, MCV lists handle a wider range of clause types, not just equality clauses like functional dependencies. For example, consider the following range query for the same table: EXPLAIN (ANALYZE, TIMING OFF) SELECT * FROM t WHERE a <= 49 AND b > 49; QUERY PLAN -------------------------------------------------------------------&zwsp;-------- Seq Scan on t (cost=0.00..195.00 rows=1 width=8) (actual rows=0 loops=1) Filter: ((a <= 49) AND (b > 49)) Rows Removed by Filter: 10000 Access to the table pg_statistic is restricted to superusers, so that ordinary users cannot learn about the contents of the tables of other users from it. Some selectivity estimation functions will use a user-provided operator (either the operator appearing in the query or a related operator) to analyze the stored statistics. For example, in order to determine whether a stored most common value is applicable, the selectivity estimator will have to run the appropriate = operator to compare the constant in the query to the stored value. Thus the data in pg_statistic is potentially passed to user-defined operators. An appropriately crafted operator can intentionally leak the passed operands (for example, by logging them or writing them to a different table), or accidentally leak them by showing their values in error messages, in either case possibly exposing data from pg_statistic to a user who should not be able to see it. In order to prevent this, the following applies to all built-in selectivity estimation functions. When planning a query, in order to be able to use stored statistics, the current user must either have SELECT privilege on the table or the involved columns, or the operator used must be LEAKPROOF (more accurately, the function that the operator is based on). If not, then the selectivity estimator will behave as if no statistics are available, and the planner will proceed with default or fall-back assumptions. If a user does not have the required privilege on the table or columns, then in many cases the query will ultimately receive a permission-denied error, in which case this mechanism is invisible in practice. But if the user is reading from a security-barrier view, then the planner might wish to check the statistics of an underlying table that is otherwise inaccessible to the user. In that case, the operator should be leak-proof or the statistics will not be used. There is no direct feedback about that, except that the plan might be suboptimal. If one suspects that this is the case, one could try running the query as a more privileged user, to see if a different plan results. This restriction applies only to cases where the planner would need to execute a user-defined operator on one or more values from pg_statistic. Thus the planner is permitted to use generic statistical information, such as the fraction of null values or the number of distinct values in a column, regardless of access privileges. Selectivity estimation functions contained in third-party extensions that potentially operate on statistics with user-defined operators should follow the same security rules. Consult the PostgreSQL source code for guidance.