Social Network Extra Problems:
 
1
Select H1.Name, H1.Grade, H2.Name, H2.Grade, H3.Name, H3. Grade
From Highschooler As H1, Highschooler As H2, Highschooler As H3,
        Likes as L1, Likes as L2
Where H1.ID = L1.ID1
    AND H2.ID = L1.ID2
    AND H2.ID = L2.ID1
    AND H3.ID = L2.ID2
    AND L1.ID1 <> L2.ID2;
 
2
Select Distinct H1.name, H1.Grade
From Friend As F, Highschooler As H1, Highschooler As H2
Where F.ID1 = H1.ID AND F.ID2 = H2.ID AND H1.grade <> H2.grade
EXCEPT
Select Distinct H1.name, H1.Grade
From Friend As F, Highschooler As H1, Highschooler As H2
Where F.ID1 = H1.ID AND F.ID2 = H2.ID AND H1.grade = H2.grade;  
 
3
Select (1.0 * Count(*))/
(Select Count(*)
From Highschooler)
From Friend;
 
4
Select count(*)
From Highschooler Join Friend ON (Highschooler.ID = Friend.ID1)
Where Name <> 'Cassandra' AND ID1 in
    (Select Friend.ID1
     From Highschooler Join Friend on
     (Highschooler.ID = Friend.ID2 AND Highschooler.Name = 'Cassandra')
);
 
5
Select Name, Grade
From (Friend Join Highschooler ON Friend.Id1 = Highschooler.ID) As supertable
Group By ID1
Having Count(ID1) =
  (SELECT max(count.c)
    FROM
      (SELECT count(ID1) as c
      FROM friend
      GROUP BY ID1)
    as count)
;