kgyrtkirk commented on a change in pull request #1439:
URL: https://github.com/apache/hive/pull/1439#discussion_r482119043
##########
File path:
ql/src/java/org/apache/hadoop/hive/ql/optimizer/calcite/cost/HiveOnTezCostModel.java
##########
@@ -89,22 +89,23 @@ public RelOptCost getAggregateCost(HiveAggregate aggregate)
{
} else {
final RelMetadataQuery mq = aggregate.getCluster().getMetadataQuery();
// 1. Sum of input cardinalities
- final Double rCount = mq.getRowCount(aggregate.getInput());
- if (rCount == null) {
+ final Double inputRowCount = mq.getRowCount(aggregate.getInput());
+ final Double rowCount = mq.getRowCount(aggregate);
+ if (inputRowCount == null || rowCount == null) {
return null;
}
// 2. CPU cost = sorting cost
- final double cpuCost = algoUtils.computeSortCPUCost(rCount);
+ final double cpuCost = algoUtils.computeSortCPUCost(rowCount) +
inputRowCount * algoUtils.getCpuUnitCost();
Review comment:
maybe...I'm trying to catch the case when `inputRowCount >>
outputRowCount`; we are also grouping - so it will not be a full sort at all ;
I was using the above to achieve:
```
log(outputRowCount)*outputRowCount + inputRowCount*COST
```
the rational behind this is that it needs to really sort `oRC` and read
`iRC` rows - this could be an underestimation...but `log(iRC)*iRC` was highly
overestimating the cost
one alternative for the above could be to use:
```
log(outputRowCount) * inputRowCount
```
the rational behind this:
we will need to find the place for every input row; but we also know that
the output will be at most `outputRowCount` - so it shouldn't take more time to
find the place for the actual row than `log(outputRowCount)`
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