этот запрос получает доминирующие множества в сети. так, например, учитывая сеть
A<----->B<br>
B<----->C<br>
B<----->D<br>
C<----->E<br>
D<----->C<br>
D<----->E<br>
F<----->E
возвращает
B, E
B, F
A, E
но это не работает для больших данных, потому что я использую строковые методы в моем результате. я пытался удалить строковые методы и вернуть представление или что-то еще, но безрезультатно
With t as (select 'A' as per1, 'B' as per2 from dual union all
select 'B','C' from dual union all
select 'B','D' from dual union all
select 'C','B' from dual union all
select 'C','E' from dual union all
select 'D','C' from dual union all
select 'D','E' from dual union all
select 'E','C' from dual union all
select 'E','D' from dual union all
select 'F','E' from dual)
,t2 as (select distinct least(per1, per2) as per1, greatest(per1, per2) as per2 from t union
select distinct greatest(per1, per2) as per1, least(per1, per2) as per1 from t)
,t3 as (select per1, per2, row_number() over (partition by per1 order by per2) as rn from t2)
,people as (select per, row_number() over (order by per) rn
from (select distinct per1 as per from t union
select distinct per2 from t)
)
,comb as (select sys_connect_by_path(per,',')||',' as p
from people
connect by rn > prior rn
)
,find as (select p, per2, count(*) over (partition by p) as cnt
from (
select distinct comb.p, t3.per2
from comb, t3
where instr(comb.p, ','||t3.per1||',') > 0 or instr(comb.p, ','||t3.per2||',') > 0
)
)
,rnk as (select p, rank() over (order by length(p)) as rnk
from find
where cnt = (select count(*) from people)
order by rnk
) select distinct trim(',' from p) as p from rnk where rnk.rnk = 1`