Thread: Planner chose a much slower plan in hashjoin, using a large table asthe inner table.

On Mon, Nov 18, 2019 at 7:48 PM Jinbao Chen <jinchen@pivotal.io> wrote:
> In the test case above, the small table has 3000 tuples and 100 distinct values on column ‘a’.
> If we use small table as inner table.  The chan length of the bucket is 30. And we need to
> search the whole chain on probing the hash table. So the cost of probing is bigger than build
> hash table, and we need to use big table as inner.
>
> But in fact this is not true. We initialized 620,000 buckets in hashtable. But only 100 buckets
> has chains with length 30. Other buckets are empty. Only hash values need to be compared.
> Its costs are very small. We have 100,000 distinct key and 100,000,000 tuple on outer table.
> Only (100/100000)* tuple_num tuples will search the whole chain. The other tuples
> (number = (98900/100000)*tuple_num*) in outer
> table just compare with the hash value. So the actual cost is much smaller than the planner
> calculated. This is the reason why using a small table as inner is faster.

So basically we think that if t_big is on the outer side, we'll do
100,000,000 probes and each one is going to scan a t_small bucket with
chain length 30, so that looks really expensive.  Actually only a
small percentage of its probes find tuples with the right hash value,
but final_cost_hash_join() doesn't know that.  So we hash t_big
instead, which we estimated pretty well and it finishes up with
buckets of length 1,000 (which is actually fine in this case, they're
not unwanted hash collisions, they're duplicate keys that we need to
emit) and we probe them 3,000 times (which is also fine in this case),
but we had to do a bunch of memory allocation and/or batch file IO and
that turns out to be slower.

I am not at all sure about this but I wonder if it would be better to
use something like:

  run_cost += outer_path_rows * some_small_probe_cost;
  run_cost += hash_qual_cost.per_tuple * approximate_tuple_count();

If we can estimate how many tuples will actually match accurately,
that should also be the number of times we have to run the quals,
since we don't usually expect hash collisions (bucket collisions, yes,
but hash collisions where the key doesn't turn out to be equal, no*).

* ... but also yes as you approach various limits, so you could also
factor in bucket chain length that is due to being prevented from
expanding the number of buckets by arbitrary constraints, and perhaps
also birthday_problem(hash size, key space) to factor in unwanted hash
collisions that start to matter once you get to billions of keys and
expect collisions with short hashes.

Hi hackers,

I have made a patch to fix the problem. 

Added the selection rate of the inner table non-empty bucket

The planner will use big table as inner table in hash join
if small table have fewer unique values. But this plan is
much slower than using small table as inner table.

In general, the cost of creating a hash table is higher
than the cost of querying a hash table. So we tend to use
small tables as internal tables. But if the average chain
length of the bucket is large, the situation is just the
opposite.

If virtualbuckets is much larger than innerndistinct, and
outerndistinct is much larger than innerndistinct. Then most
tuples of the outer table will match the empty bucket. So when
we calculate the cost of traversing the bucket, we need to
ignore the tuple matching empty bucket.

So we add the selection rate of the inner table non-empty
bucket. The formula is:
(1 - ((outerndistinct - innerndistinct)/outerndistinct)*
((virtualbuckets - innerndistinct)/virtualbuckets))

On Tue, Nov 19, 2019 at 5:56 PM Jinbao Chen <jinchen@pivotal.io> wrote:

I think we have the same understanding of this issue.

Sometimes use smaller costs on scanning the chain in bucket like below would

be better. 

run_cost += outer_path_rows * some_small_probe_cost;
run_cost += hash_qual_cost.per_tuple * approximate_tuple_count();

In some version of GreenPlum(a database based on postgres), we just disabled

the cost on scanning the bucket chain. In most cases, this can get a better query

plan. But I am worried that it will be worse in some cases.

Now only the small table's distinct value is much smaller than the bucket number,

and much smaller than the distinct value of the large table, the planner will get the

wrong plan. 

For example, if inner table has 100 distinct values, and 3000 rows. Hash table 

has 1000 buckets. Outer table has 10000 distinct values.

We can assume that all the 100 distinct values of the inner table are included in the

10000 distinct values of the outer table. So (100/10000)*outer_rows tuples will

probe the buckets has chain. And (9900/10000)*outer_rows tuples will probe

all the 1000 buckets randomly. So (9900/10000)*outer_rows*(900/1000) tuples will

probe empty buckets. So the costs on scanning bucket chain is

hash_qual_cost.per_tuple*innerbucketsize*outer_rows*

(1 - ((outer_distinct - inner_distinct)/outer_distinct)*((buckets_num - inner_disttinct)/buckets_num))

Do you think this assumption is reasonable?

On Tue, Nov 19, 2019 at 3:46 PM Thomas Munro <thomas.munro@gmail.com> wrote:

On Mon, Nov 18, 2019 at 7:48 PM Jinbao Chen <jinchen@pivotal.io> wrote:
> In the test case above, the small table has 3000 tuples and 100 distinct values on column ‘a’.
> If we use small table as inner table.  The chan length of the bucket is 30. And we need to
> search the whole chain on probing the hash table. So the cost of probing is bigger than build
> hash table, and we need to use big table as inner.
>
> But in fact this is not true. We initialized 620,000 buckets in hashtable. But only 100 buckets
> has chains with length 30. Other buckets are empty. Only hash values need to be compared.
> Its costs are very small. We have 100,000 distinct key and 100,000,000 tuple on outer table.
> Only (100/100000)* tuple_num tuples will search the whole chain. The other tuples
> (number = (98900/100000)*tuple_num*) in outer
> table just compare with the hash value. So the actual cost is much smaller than the planner
> calculated. This is the reason why using a small table as inner is faster.

So basically we think that if t_big is on the outer side, we'll do
100,000,000 probes and each one is going to scan a t_small bucket with
chain length 30, so that looks really expensive.  Actually only a
small percentage of its probes find tuples with the right hash value,
but final_cost_hash_join() doesn't know that.  So we hash t_big
instead, which we estimated pretty well and it finishes up with
buckets of length 1,000 (which is actually fine in this case, they're
not unwanted hash collisions, they're duplicate keys that we need to
emit) and we probe them 3,000 times (which is also fine in this case),
but we had to do a bunch of memory allocation and/or batch file IO and
that turns out to be slower.

I am not at all sure about this but I wonder if it would be better to
use something like:

  run_cost += outer_path_rows * some_small_probe_cost;
  run_cost += hash_qual_cost.per_tuple * approximate_tuple_count();

If we can estimate how many tuples will actually match accurately,
that should also be the number of times we have to run the quals,
since we don't usually expect hash collisions (bucket collisions, yes,
but hash collisions where the key doesn't turn out to be equal, no*).

* ... but also yes as you approach various limits, so you could also
factor in bucket chain length that is due to being prevented from
expanding the number of buckets by arbitrary constraints, and perhaps
also birthday_problem(hash size, key space) to factor in unwanted hash
collisions that start to matter once you get to billions of keys and
expect collisions with short hashes.

Hi Andy,

I just test the query on 12.1. But pg use big_table as inner.

demo=# explain (costs off) select * from t_small, t_big where a = b;
             QUERY PLAN
------------------------------------
 Hash Join
   Hash Cond: (t_small.a = t_big.b)
   ->  Seq Scan on t_small
   ->  Hash
         ->  Seq Scan on t_big

Do you insert data and  set max_parallel_workers_per_gather to 0 like above?

create table t_small(a int);
create table t_big(b int);
insert into t_small select i%100 from generate_series(0, 3000)i;
insert into t_big select i%100000 from generate_series(1, 100000000)i ;
analyze t_small;
analyze t_big;
set max_parallel_workers_per_gather = 0;

On Thu, Nov 28, 2019 at 5:46 PM Andy Fan <zhihui.fan1213@gmail.com> wrote:

On Fri, Nov 22, 2019 at 6:51 PM Jinbao Chen <jinchen@pivotal.io> wrote:

Hi hackers,

I have made a patch to fix the problem. 

Added the selection rate of the inner table non-empty bucket

The planner will use big table as inner table in hash join
if small table have fewer unique values. But this plan is
much slower than using small table as inner table.

In general, the cost of creating a hash table is higher
than the cost of querying a hash table. So we tend to use
small tables as internal tables. But if the average chain
length of the bucket is large, the situation is just the
opposite.

If virtualbuckets is much larger than innerndistinct, and
outerndistinct is much larger than innerndistinct. Then most
tuples of the outer table will match the empty bucket. So when
we calculate the cost of traversing the bucket, we need to
ignore the tuple matching empty bucket.

So we add the selection rate of the inner table non-empty
bucket. The formula is:
(1 - ((outerndistinct - innerndistinct)/outerndistinct)*
((virtualbuckets - innerndistinct)/virtualbuckets))

On Tue, Nov 19, 2019 at 5:56 PM Jinbao Chen <jinchen@pivotal.io> wrote:

I think we have the same understanding of this issue.

Sometimes use smaller costs on scanning the chain in bucket like below would

be better. 

run_cost += outer_path_rows * some_small_probe_cost;
run_cost += hash_qual_cost.per_tuple * approximate_tuple_count();

In some version of GreenPlum(a database based on postgres), we just disabled

the cost on scanning the bucket chain. In most cases, this can get a better query

plan. But I am worried that it will be worse in some cases.

Now only the small table's distinct value is much smaller than the bucket number,

and much smaller than the distinct value of the large table, the planner will get the

wrong plan. 

For example, if inner table has 100 distinct values, and 3000 rows. Hash table 

has 1000 buckets. Outer table has 10000 distinct values.

We can assume that all the 100 distinct values of the inner table are included in the

10000 distinct values of the outer table. So (100/10000)*outer_rows tuples will

probe the buckets has chain. And (9900/10000)*outer_rows tuples will probe

all the 1000 buckets randomly. So (9900/10000)*outer_rows*(900/1000) tuples will

probe empty buckets. So the costs on scanning bucket chain is

hash_qual_cost.per_tuple*innerbucketsize*outer_rows*

(1 - ((outer_distinct - inner_distinct)/outer_distinct)*((buckets_num - inner_disttinct)/buckets_num))

Do you think this assumption is reasonable?

On Tue, Nov 19, 2019 at 3:46 PM Thomas Munro <thomas.munro@gmail.com> wrote:

On Mon, Nov 18, 2019 at 7:48 PM Jinbao Chen <jinchen@pivotal.io> wrote:
> In the test case above, the small table has 3000 tuples and 100 distinct values on column ‘a’.
> If we use small table as inner table.  The chan length of the bucket is 30. And we need to
> search the whole chain on probing the hash table. So the cost of probing is bigger than build
> hash table, and we need to use big table as inner.
>
> But in fact this is not true. We initialized 620,000 buckets in hashtable. But only 100 buckets
> has chains with length 30. Other buckets are empty. Only hash values need to be compared.
> Its costs are very small. We have 100,000 distinct key and 100,000,000 tuple on outer table.
> Only (100/100000)* tuple_num tuples will search the whole chain. The other tuples
> (number = (98900/100000)*tuple_num*) in outer
> table just compare with the hash value. So the actual cost is much smaller than the planner
> calculated. This is the reason why using a small table as inner is faster.

So basically we think that if t_big is on the outer side, we'll do
100,000,000 probes and each one is going to scan a t_small bucket with
chain length 30, so that looks really expensive.  Actually only a
small percentage of its probes find tuples with the right hash value,
but final_cost_hash_join() doesn't know that.  So we hash t_big
instead, which we estimated pretty well and it finishes up with
buckets of length 1,000 (which is actually fine in this case, they're
not unwanted hash collisions, they're duplicate keys that we need to
emit) and we probe them 3,000 times (which is also fine in this case),
but we had to do a bunch of memory allocation and/or batch file IO and
that turns out to be slower.

I am not at all sure about this but I wonder if it would be better to
use something like:

  run_cost += outer_path_rows * some_small_probe_cost;
  run_cost += hash_qual_cost.per_tuple * approximate_tuple_count();

If we can estimate how many tuples will actually match accurately,
that should also be the number of times we have to run the quals,
since we don't usually expect hash collisions (bucket collisions, yes,
but hash collisions where the key doesn't turn out to be equal, no*).

* ... but also yes as you approach various limits, so you could also
factor in bucket chain length that is due to being prevented from
expanding the number of buckets by arbitrary constraints, and perhaps
also birthday_problem(hash size, key space) to factor in unwanted hash
collisions that start to matter once you get to billions of keys and
expect collisions with short hashes.

FYI:  I tried this on 12.1, and find it use small_table as inner table already.  I didn't look into the details so far.

postgres=# explain (costs off) select * from join_hash_t_small, join_hash_t_big where a = b;
                       QUERY PLAN
--------------------------------------------------------
 Hash Join
   Hash Cond: (join_hash_t_big.b = join_hash_t_small.a)
   ->  Seq Scan on join_hash_t_big
   ->  Hash
         ->  Seq Scan on join_hash_t_small
(5 rows)

postgres=# select version();
                                                     version
-----------------------------------------------------------------------------------------------------------------
 PostgreSQL 12.1 on x86_64-apple-darwin18.7.0, compiled by Apple LLVM version 10.0.1 (clang-1001.0.46.4), 64-bit
(1 row)