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Merge branch 'bb-10.4-release' into bb-10.5-release
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@ -3901,50 +3901,16 @@ double Histogram::point_selectivity(double pos, double avg_sel)
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}
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else
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{
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/*
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/*
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The value 'pos' fits within one single histogram bucket.
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Histogram buckets have the same numbers of rows, but they cover
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different ranges of values.
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We assume that values are uniformly distributed across the [0..1] value
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range.
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*/
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/*
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If all buckets covered value ranges of the same size, the width of
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value range would be:
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We also have avg_sel which is per-table average selectivity of col=const.
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If there are popular values, this may be larger than one bucket, so
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cap the returned number by the selectivity of one bucket.
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*/
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double avg_bucket_width= 1.0 / (get_width() + 1);
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/*
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Let's see what is the width of value range that our bucket is covering.
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(min==max currently. they are kept in the formula just in case we
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will want to extend it to handle multi-bucket case)
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*/
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double inv_prec_factor= (double) 1.0 / prec_factor();
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double current_bucket_width=
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(max + 1 == get_width() ? 1.0 : (get_value(max) * inv_prec_factor)) -
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(min == 0 ? 0.0 : (get_value(min-1) * inv_prec_factor));
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DBUG_ASSERT(current_bucket_width); /* We shouldn't get a one zero-width bucket */
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/*
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So:
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- each bucket has the same #rows
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- values are unformly distributed across the [min_value,max_value] domain.
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If a bucket has value range that's N times bigger then average, than
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each value will have to have N times fewer rows than average.
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*/
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sel= avg_sel * avg_bucket_width / current_bucket_width;
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/*
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(Q: if we just follow this proportion we may end up in a situation
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where number of different values we expect to find in this bucket
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exceeds the number of rows that this histogram has in a bucket. Are
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we ok with this or we would want to have certain caps?)
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*/
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sel= MY_MIN(avg_bucket_width, avg_sel);
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}
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return sel;
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}
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