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Merge branch '10.6' into 10.8

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Oleksandr Byelkin
2023-05-03 11:33:57 +02:00
parent 7d967423fe 043d69bbcc
commit f0f1f2de0e
370 changed files with 2088 additions and 826 deletions
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44
sql/sql_statistics.cc
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@@ -4024,50 +4024,16 @@ double Histogram_binary::point_selectivity(Field *field, key_range *endpoint,
}
else
{
/*
/*
The value 'pos' fits within one single histogram bucket.
Histogram_binary buckets have the same numbers of rows, but they cover
different ranges of values.
We assume that values are uniformly distributed across the [0..1] value
range.
*/
/*
If all buckets covered value ranges of the same size, the width of
value range would be:
We also have avg_sel which is per-table average selectivity of col=const.
If there are popular values, this may be larger than one bucket, so
cap the returned number by the selectivity of one bucket.
*/
double avg_bucket_width= 1.0 / (get_width() + 1);
/*
Let's see what is the width of value range that our bucket is covering.
(min==max currently. they are kept in the formula just in case we
will want to extend it to handle multi-bucket case)
*/
double inv_prec_factor= (double) 1.0 / prec_factor();
double current_bucket_width=
(max + 1 == get_width() ? 1.0 : (get_value(max) * inv_prec_factor)) -
(min == 0 ? 0.0 : (get_value(min-1) * inv_prec_factor));
DBUG_ASSERT(current_bucket_width); /* We shouldn't get a one zero-width bucket */
/*
So:
- each bucket has the same #rows
- values are unformly distributed across the [min_value,max_value] domain.
If a bucket has value range that's N times bigger then average, than
each value will have to have N times fewer rows than average.
*/
sel= avg_sel * avg_bucket_width / current_bucket_width;
/*
(Q: if we just follow this proportion we may end up in a situation
where number of different values we expect to find in this bucket
exceeds the number of rows that this histogram has in a bucket. Are
we ok with this or we would want to have certain caps?)
*/
sel= MY_MIN(avg_bucket_width, avg_sel);
}
return sel;
}