Re: Planner chose a much slower plan in hashjoin, using a large table as the inner table.

2019-11-28 Thread Andy Fan
On Thu, Nov 28, 2019 at 7:19 PM Jinbao Chen  wrote:

> 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?
>

Sorry for the noise..  you are right.  I thought I load the data but and
run the query immediately without running the analyzing.

now it is using big table as inner table.


> 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%10 from generate_series(1, 1)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  wrote:
>
>>
>>
>> On Fri, Nov 22, 2019 at 6:51 PM Jinbao Chen  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  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 1 distinct values.
 We can assume that all the 100 distinct values of the inner table are
 included in the
 1 distinct values of the outer table. So (100/1)*outer_rows
 tuples will
 probe the buckets has chain. And (9900/1)*outer_rows tuples will
 probe
 all the 1000 buckets randomly. So (9900/1)*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 
 wrote:

> On Mon, Nov 18, 2019 at 7:48 PM Jinbao Chen 
> 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 (100/10)* tuple_num tuples will search the whole chain. The
> other tuples
> > (number = (98900/10)*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 

Re: Planner chose a much slower plan in hashjoin, using a large table as the inner table.

2019-11-28 Thread Jinbao Chen
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%10 from generate_series(1, 1)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  wrote:

>
>
> On Fri, Nov 22, 2019 at 6:51 PM Jinbao Chen  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  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 1 distinct values.
>>> We can assume that all the 100 distinct values of the inner table are
>>> included in the
>>> 1 distinct values of the outer table. So (100/1)*outer_rows
>>> tuples will
>>> probe the buckets has chain. And (9900/1)*outer_rows tuples will
>>> probe
>>> all the 1000 buckets randomly. So (9900/1)*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 
>>> wrote:
>>>
 On Mon, Nov 18, 2019 at 7:48 PM Jinbao Chen  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 (100/10)* tuple_num tuples will search the whole chain. The
 other tuples
 > (number = (98900/10)*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 

Re: Planner chose a much slower plan in hashjoin, using a large table as the inner table.

2019-11-28 Thread Andy Fan
On Fri, Nov 22, 2019 at 6:51 PM Jinbao Chen  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  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 1 distinct values.
>> We can assume that all the 100 distinct values of the inner table are
>> included in the
>> 1 distinct values of the outer table. So (100/1)*outer_rows
>> tuples will
>> probe the buckets has chain. And (9900/1)*outer_rows tuples will probe
>> all the 1000 buckets randomly. So (9900/1)*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 
>> wrote:
>>
>>> On Mon, Nov 18, 2019 at 7:48 PM Jinbao Chen  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 (100/10)* tuple_num tuples will search the whole chain. The
>>> other tuples
>>> > (number = (98900/10)*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,

Re: Planner chose a much slower plan in hashjoin, using a large table as the inner table.

2019-11-22 Thread Jinbao Chen
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  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 1 distinct values.
> We can assume that all the 100 distinct values of the inner table are
> included in the
> 1 distinct values of the outer table. So (100/1)*outer_rows tuples
> will
> probe the buckets has chain. And (9900/1)*outer_rows tuples will probe
> all the 1000 buckets randomly. So (9900/1)*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 
> wrote:
>
>> On Mon, Nov 18, 2019 at 7:48 PM Jinbao Chen  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 (100/10)* tuple_num tuples will search the whole chain. The
>> other tuples
>> > (number = (98900/10)*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 

Re: Planner chose a much slower plan in hashjoin, using a large table as the inner table.

2019-11-19 Thread Jinbao Chen
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 1 distinct values.
We can assume that all the 100 distinct values of the inner table are
included in the
1 distinct values of the outer table. So (100/1)*outer_rows tuples
will
probe the buckets has chain. And (9900/1)*outer_rows tuples will probe
all the 1000 buckets randomly. So (9900/1)*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  wrote:

> On Mon, Nov 18, 2019 at 7:48 PM Jinbao Chen  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 (100/10)* tuple_num tuples will search the whole chain. The
> other tuples
> > (number = (98900/10)*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.
>


Re: Planner chose a much slower plan in hashjoin, using a large table as the inner table.

2019-11-18 Thread Thomas Munro
On Mon, Nov 18, 2019 at 7:48 PM Jinbao Chen  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 (100/10)* tuple_num tuples will search the whole chain. The other 
> tuples
> (number = (98900/10)*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.




Planner chose a much slower plan in hashjoin, using a large table as the inner table.

2019-11-17 Thread Jinbao Chen
Hi Hackers,

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. This
problem occurs on master
branch without parallel scan.

For example

create table t_small(a int);
create table t_big(b int);
insert into t_small select i%100 from generate_series(0, 3000);
insert into t_big select i%10 from generate_series(1, 1)i ;
analyze t_small;
analyze t_big;
set max_parallel_workers_per_gather = 0;

and the plan made by planner is
demo2=# explain select * from t_small, t_big where a = b;
  QUERY PLAN
---
 Hash Join  (cost=3083104.72..3508073.65 rows=3045990 width=8)
   Hash Cond: (t_small.a = t_big.b)
   ->  Seq Scan on t_small  (cost=0.00..44.01 rows=3001 width=4)
   ->  Hash  (cost=1442478.32..1442478.32 rows=10032 width=4)
 ->  Seq Scan on t_big  (cost=0.00..1442478.32 rows=10032
width=4)

and it runs nearly 58s
demo2=# select * from t_small, t_big where a = b;
Time: 58544.525 ms (00:58.545)

But if we do some hack and use the small table as inner. It runs 19s.
demo2=# explain select * from t_small, t_big where a = b;
   QUERY PLAN
-
 Hash Join  (cost=81.52..1723019.82 rows=3045990 width=8)
   Hash Cond: (t_big.b = t_small.a)
   ->  Seq Scan on t_big  (cost=0.00..1442478.32 rows=10032 width=4)
   ->  Hash  (cost=44.01..44.01 rows=3001 width=4)
 ->  Seq Scan on t_small  (cost=0.00..44.01 rows=3001 width=4)

demo2=# select * from t_small, t_big where a = b;
Time: 18751.588 ms (00:18.752)


RCA:

The cost of the inner table mainly comes from creating a hash table.
startup_cost += (cpu_operator_cost * num_hashclauses + cpu_tuple_cost)
* inner_path_rows;

The cost of the outer table mainly comes from search the hash table.
Calculate the hash value:
run_cost += cpu_operator_cost * num_hashclauses * outer_path_rows;

Traverse the linked list in the bucket and compare:
run_cost += hash_qual_cost.per_tuple * outer_path_rows *
clamp_row_est(inner_path_rows * innerbucketsize) * 0.5;

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.

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 (100/10)* tuple_num tuples will search the whole chain. The other
tuples
(number = (98900/10)*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.