Consider a schema for a picture tagging website:

Member(mid, name, age) 
Picture(pid, year) 
Tagged(mid, pid)


1. Return the names of all members that were tagged in both 2011 and 2014 sorted in alphabetic order

2. Return the name of all users who were never tagged in 2015.

3. For each query below indicate if it is the correct solution to question 2 above.

	a. select distinct x.name 
		from Member x, Tagged y 
		where x.mid = y.mid 
		and not exists 
			(select * 
		 	from Picture z 
		 	where y.pid = z.pid 
		 	and z.year = 2015);

	b. select distinct x.name 
		from Member x 
		where not exists 
			(select * 
		 	from Tagged y, Picture z 
		 	where x.mid = y.mid 
		 	and y.pid = z.pid and z.year = 2015);

	c. select distinct x.name 
	 	from Member x 
	 	where not exists 
			(select * 
		 	from Tagged y 
		 	where x.mid = y.mid 
		 	and not exists 				(select * 
			 	from Picture z 
			 	where y.pid = z.pid 
			 	and z.year = 2015));

	d. select distinct x.name 
	 	from Member x, Tagged y, Picture z 
	 	where x.mid = y.mid and y.pid = z.pid 
	 	and z.year = 2015 
	 	group by x.name 
	 	having count(z.pid) = 0;


4. Write a Relational Algebra Expression (draw a tree) for the following query:

select w.year, max(w.c) as m 
from 
	(select x.name, z.year, count(*) as c 
	 from Member x, Tagged y, Picture z 
	 where x.mid = y.mid 
	 and y.pid = z.pid 
	 and age < 20 
	 group by x.name, z.year) w 
group by w.year 
having sum(w.c) > 100;


5. Write a query in datalog with negation that returns the mids and names of all members that were tagged only in pictures were Alice was also tagged.