planner: zero estimation result if all values are in TopN

[qw4990]

Enhancement

create table t (a int, key(a));
insert into t values (1), (3);
analyze table t with 2 topn;
insert into t values (2);

explain select * from t where a=2;
+------------------------+---------+-----------+---------------------+-------------------------------+
| id                     | estRows | task      | access object       | operator info                 |
+------------------------+---------+-----------+---------------------+-------------------------------+
| IndexReader_6          | 0.00    | root      |                     | index:IndexRangeScan_5        |
| └─IndexRangeScan_5     | 0.00    | cop[tikv] | table:t, index:a(a) | range:[2,2], keep order:false |
+------------------------+---------+-----------+---------------------+-------------------------------+
2 rows in set (0.00 sec)

After v6.5, in the case above, the estimation result for a=2 is zero, which seems dangerous.

If no TopN, it'll be 1:

create table t (a int, key(a));
insert into t values (1), (3);
analyze table t with 0 topn;
insert into t values (2);

mysql> explain select * from t where a=2;
+------------------------+---------+-----------+---------------------+-------------------------------+
| id                     | estRows | task      | access object       | operator info                 |
+------------------------+---------+-----------+---------------------+-------------------------------+
| IndexReader_6          | 1.00    | root      |                     | index:IndexRangeScan_5        |
| └─IndexRangeScan_5     | 1.00    | cop[tikv] | table:t, index:a(a) | range:[2,2], keep order:false |
+------------------------+---------+-----------+---------------------+-------------------------------+

Should we avoid using 0 in this case (this value is not in TopN and there is no Histogram)?

[AilinKid]

the related code about this is as follows

	// 3. use uniform distribution assumption for the rest (even when this value is not covered by the range of stats)
	histNDV := float64(c.Histogram.NDV - int64(c.TopN.Num()))
	if histNDV <= 0 {
		return 0, nil
	}
	return c.Histogram.notNullCount() / histNDV, nil

in first case, since topN is collected, and it is overlapped with the NDV indication in Histogram, so the computation here is meaningless, leading histNDV := float64(c.Histogram.NDV - int64(c.TopN.Num())) = 2-2 = 0

in the second case, since topN is not collected, histNDV := float64(c.Histogram.NDV - int64(c.TopN.Num())) = 2-0 = 2, and the Histogram.notNullCount() is 2 as well , causing the returned value as 2/2 = 1

[seiya-annie]

/report customer