How to determine equaivalent groupings (recursively) in SQL? - mysql

Question

I have a list of products identified by their SKUs. To simplify it, I just name them as A, B, C, D,... here. Each of these SKUs has been assigned by default an already existing GroupID, for simplicity I just number them as 1, 2, 3,... here.
The same GroupID would mean "These SKUs are equivalent, so it is ok to use/buy either one of them, as it makes no difference".
The problem is, some SKUs show up more than once as they come from a different buying source, but as they come from a different source, they have a different grouping.
The goal is therefore to consolidate the grouping and make sure they have the same groupings.
I already apologize if my illustration may not be super pretty, but I'm trying. Here's a small data table sample on how the raw data looks like (first line is the column names):
Source SKU GroupID
Seller1 A 1
Seller1 B 1
Seller1 C 1
Seller2 B 2
Seller2 D 2
Seller2 E 2
Seller3 A 3
Seller3 B 3
Seller4 F 4
Seller4 G 4
Seller4 H 4
The result should be like:
Source SKU GroupID
Seller1 A 1
Seller1 B 1
Seller1 C 1
Seller2 B 1
Seller2 D 1
Seller2 E 1
Seller3 A 1
Seller3 B 1
Seller4 F 4
Seller4 G 4
Seller4 H 4
Basically, if Any SKU in GroupID X is a subset of GroupID Y, then GroupID Y = GroupID X. But that should be applied on all GroupIDs, so it appears to be recursive.
I wish I could show the code that I tried already and I tried already for a few days, but I literally only managed to produce garbage.
In C# I'd know how to deal with this, but I can't seem to wrap my head around SQL as I am not that experienced and unfortunately I would need this in SQL.
I would be thankful for any kind of help, even if it's just a hint or direction you guys would suggest I should try. Thanks a lot!

Answer 1

You want a correspondence between groups, which you can calculate with a recursive CTE:
with recursive tt as (
select distinct t1.groupid as groupid1, t2.groupid as groupid2
from t t1 join
t t2
on t1.sku = t2.sku
),
cte as (
select tt.groupid1, tt.groupid2, concat_ws(',', tt.groupid1, tt.groupid2) as visited
from tt
union all
select cte.groupid1, tt.groupid2, concat_ws(',', visited, tt.groupid2)
from cte join
tt
on cte.groupid2 = tt.groupid1
where find_in_set(tt.groupid2, cte.visited) = 0
)
select groupid1, min(groupid2) as overall_group
from cte
group by groupid1;
You can then join this back to the original table to get the "overall group":
with recursive tt as (
select distinct t1.groupid as groupid1, t2.groupid as groupid2
from t t1 join
t t2
on t1.sku = t2.sku
),
cte as (
select tt.groupid1, tt.groupid2, concat_ws(',', tt.groupid1, tt.groupid2) as visited
from tt
union all
select cte.groupid1, tt.groupid2, concat_ws(',', visited, tt.groupid2)
from cte join
tt
on cte.groupid2 = tt.groupid1
where find_in_set(tt.groupid2, cte.visited) = 0
)
select t.*, g.overall_group
from t join
(select groupid1, min(groupid2) as overall_group
from cte
group by groupid1
) g
on t.groupid = g.groupid1;
Here is a db<>fiddle.
Note: Your sample data is rather "complete" so you don't need a recursive CTE for that particular data. However, I am guessing that your real groups have a bit less overlap in which case recursion is necessary.

Answer 2

First is to get all those sellers with subsets based on count. then filter using Group By
select table1.Source, SKU, case when table1.Source = t6.Source and t6.cnt > 1 then 1 else 2 end as GroupID
from table1
left join
(select t5.Source, count(t5.cnt) as cnt from (
select distinct t4.Source, t4.cnt from (
select t3.Source, count(t3.SKU) as cnt from (
select t1.Source, t1.SKU from table1 t1
left join table1 t2 on t2.SKU = t1.SKU ) t3
group by t3.Source, t3.SKU
order by t3.Source) t4) as t5
group by t5.Source) t6 on t6.Source = table1.Source