Solving Cheryl's Birthday in SQL

cheryls-bday.sql

create table bdays (m, d) as
values
  ('May', 15), ('May', 16), ('May', 19),
  ('June', 17), ('June', 18),
  ('July', 14), ('July', 16),
  ('August', 14), ('August', 15), ('August', 17)
;

with x as (
  select
    m, 
    (count(*) over (partition by d) = 1) as m_decides
  from bdays
),
s1 as (
  select
    m,
    not(bool_or(m_decides)) as s1
  from x
  group by m
),
s2 as (
    select 
      bdays.*,
      s1,
      (count(*) over (partition by s1, bdays.d) = 1) as s2
    from
      bdays
      inner join s1 on s1.m = bdays.m
),
s3 as (
  select
    s2.*,
    (count(*) over (partition by s1, s2, m) = 1) as s3
  from s2
)
select *
from s3
where s1 and s2 and s3
;