Need a way to find matches between two many-to-many-relationships

Question

Given the following tables:

   ------------     ------------
   |     BA   |     |    CA    | 
   ------+-----     ------+-----
   | BId|| AId|     | CId|| AId|
         


        

Answer 1

One way to do this is to use JOIN and COUNT. You first count the CIds (how many times each one of them is repeated). Then, you do the same for BIds. Then, you JOIN BA and link both (AIds) and COUNT. Your target is to match the count of ids and their AIds.

Example :

DECLARE 
    @a TABLE(id INT)

INSERT INTO @a VALUES 
(1),
(2),
(3),
(4),
(5),
(6)

DECLARE 
    @b TABLE(id CHAR(2))

INSERT INTO @b VALUES 
('B1'),
('B2'),
('B3')

DECLARE 
    @c TABLE(id CHAR(2))

INSERT INTO @c VALUES 
('C1'),
('C2'),
('C3')


DECLARE 
    @ba TABLE(BId CHAR(2), AId INT)

INSERT INTO @ba VALUES 
('B1',2),
('B1', 3),
('B2', 2),
('B2', 4),
('B2', 5)

DECLARE 
    @ca TABLE(CId CHAR(2), AId INT)

INSERT INTO @ca VALUES 
('C1',2),
('C2',2),
('C2',3),
('C3',4),
('C3',5)




SELECT DISTINCT CId 
FROM (
SELECT *
,   COUNT(*) OVER(PARTITION BY CId) cnt
FROM @ca ca
) c
LEFT JOIN (
    SELECT *
    ,   COUNT(*) OVER(PARTITION BY BId) cnt
    FROM @ba ba
) b ON b.AId = c.AId AND b.cnt = c.cnt 
WHERE 
    b.cnt IS NOT NULL 

So, in the example C2 has repeated 2 times, and in BA, B1 has repeated also 2 times. This is the first condition, second one is to match both AIds, if they're the same, then you have a group match.

Answer 2

I think the simplest solution uses window functions:

select ca.cid, ba.bid
from (select ca.*, count(*) over (partition by cid) as cnt
      from ca
     ) ca join
     (select ba.*, count(*) over (partition by bid) as cnt
      from ba
     ) ba
     on ca.aid = ba.aid and ca.cnt = ba.cnt
group by ca.cid, ba.bid, ca.cnt
having ca.cnt = count(*)  -- all match

Here is a db<>fiddle.

The result set is all matching cid/bid pairs.

The logic here is pretty simple. For each cid and bid, the subqueries calculate the count of aids. This number has to match.

Then the join is on aid -- this is an inner join, so only matching pairs are produced. The final group by is used to generate the count of matches to see if this tallies up with all the aids.

This particular version assumes that the rows are unique in each table, although the query can easily be adjusted if this is not the case.

Answer 3

The number of elements can be calculated using a CTE for CA and BA. Then you can get at the full rows via:

SQL

with ca_info as (
      select
           cid
         , count(*) as ccount
        from ca
      group by cid
  ),
ba_info as (
      select
           bid
         , count(*) as bcount
        from ba
      group by bid
  )
select
*
from
    ba
    join ca on (ba.aid = ca.aid)
    join ba_info on ba.bid=ba_info.bid
    join ca_info on ca.cid=ca_info.cid
where ccount = bcount

SQL Fiddle

Results

bid     aid     cid     aid     bid     bcount  cid     ccount
B1      2       C2      2       B1      2       C2      2
B1      3       C2      3       B1      2       C2      2

If you are just interested in C2 itself, you can restrict the result set more:

with ca_info as (
      select
           cid
         , count(*) as ccount
        from ca
      group by cid
  ),
ba_info as (
      select
           bid
         , count(*) as bcount
        from ba
      group by bid
  )
select
distinct ca.cid
from
    ba
    join ca on (ba.aid = ca.aid)
    join ba_info on ba.bid=ba_info.bid
    join ca_info on ca.cid=ca_info.cid
where ccount = bcount

Supersets

To also get sub/supersets, the condition enforcing the set-equality can be changed:

where ccount <= bcount

This returns all sets where Bx as at least as many elements as Cy:

bid    aid    cid    aid    bid    bcount    cid    ccount
B1     2      C1     2      B1     2          C1    1
B1     2      C2     2      B1     2          C2    2
B1     3      C2     3      B1     2          C2    2
B2     2      C1     2      B2     3          C1    1
B2     2      C2     2      B2     3          C2    2
B2     4      C3     4      B2     3          C3    2
B2     5      C3     5      B2     3          C3    2