On 08/03/2015 11:36 PM, Robert Haas wrote:
On Mon, Aug 3, 2015 at 3:33 PM, Peter Geoghegan <p...@heroku.com> wrote:
When it's time to drain the heap, in performsort, divide the array into two
arrays, based on the run number of each tuple, and then quicksort the arrays
separately. The first array becomes the in-memory tail of the current tape,
and the second array becomes the in-memory tail of the next tape.

You wouldn't want to actually allocate two arrays and copy SortTuples
around, but keep using the single large array, just logically divided into
two. So the bookkeeping isn't trivial, but seems doable.

You're talking about a new thing here, that happens when it is
necessary to dump everything and do a conventional merging of on-tape
runs. IOW, we cannot fit a significant fraction of overall tuples in
memory, and we need much of the memtuples array for the next run
(usually this ends as a TSS_FINALMERGE). That being the case

I don't think that's what Heikki is talking about.  He can correct me
if I'm wrong, but what I think he's saying is that we should try to
exploit the fact that we've already determined which in-memory tuples
can be part of the current run and which in-memory tuples must become
part of the next run.  Suppose half the tuples in memory can become
part of the current run and the other half must become part of the
next run.  If we throw all of those tuples into a single bucket and
quicksort it, we're losing the benefit of the comparisons we did to
figure out which tuples go in which runs.  Instead, we could quicksort
the current-run tuples and, separately, quick-sort the next-run
tuples.  Ignore the current-run tuples completely until the tape is
empty, and then merge them with any next-run tuples that remain.

Yeah, something like that. To paraphrase, if I'm now understanding it correctly, Peter's idea is:

When all the tuples have been fed to tuplesort, and it's time to perform the sort, quicksort all the tuples currently in the heap, ignoring the run numbers, and turn the resulting array into another tape. That tape is special: it's actually stored completely in memory. It is merged with the "real" tapes when tuples are returned from the tuplesort, just like regular tapes in TSS_FINALMERGE.

And my idea is:

When all the tuples have been fed to tuplesort, and it's time to perform the sort, take all the tuples in the heap belonging to currentRun, quicksort them, and make them part of the current tape. They're not just pushed to the tape as usual, however, but attached as in-memory tail of the current tape. The logical tape abstraction will return them after all the tuples already in the tape, as if they were pushed to the tape as usual. Then take all the remaining tuples in the heap (if any), belonging to next tape, and do the same for them. They become an in-memory tail of the next tape.

I'm not sure if there's any reason to believe that would be faster
than your approach.  In general, sorting is O(n lg n) so sorting two
arrays that are each half as large figures to be slightly faster than
sorting one big array.  But the difference may not amount to much.

Yeah, I don't think there's a big performance difference between the two approaches. I'm not wedded to either approach. Whichever approach we use, my main point was that it would be better to handle this in the logical tape abstraction. In my approach, you would have those "in-memory tails" attached to the last two tapes. In Peter's approach, you would have one tape that's completely in memory, backed by the array. In either case, the tapes would look like normal tapes to most of tuplesort.c. There would be no extra TSS state, it would be TSS_SORTEDONTAPE or TSS_FINALMERGE as usual.

The logical tape abstraction is currently too low-level for that. It's just a tape of bytes, and tuplesort.c determines where a tuple begins and ends. That needs to be changed so that the logical tape abstraction works tuple-at-a-time instead. For example, instead of LogicalTapeRead(N) which reads N bytes, you would have LogicalTapeReadTuple(), which reads next tuple, and returns its length and the tuple itself. But that would be quite sensible anyway.

- Heikki

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