Counting distinct entries in a column using relational algebra

Question

So I have a table similar to this one. Each user has posted a review about one or more hotels(A,B,C,D) but on different dates so there are no duplicate tuples even though a person might have reviewed the same hotel more than once.

I need to count the number of DISTINCT hotels every user has reviewed using RELATIONAL ALGEBRA only. How can I do that?

example to show notations I use:

R = ƔUser,COUNT(Hotel_reviewed)->Num_Reviews (InitialRelation- table 1) 

would give the number of reviews by each user

The result should be the following table:

example to show notations I use:

R = ƔUser,COUNT(Hotel_reviewed)->Num_Reviews (InitialRelation- table 1) 

would give the number of reviews by each user

Answer 1

Besides the more compact syntax (from @McNets' answer):

select   User,
         count(distinct Hotel_Reviewed) HotelsReviewed
from     InitialRelation
group by User;

we can also do a projection first to find distinct User, Hotel_Reviewed pairs and then aggregate:

select   User,
         count(Hotel_Reviewed) as Hotels_Reviewed
from     
    ( select distinct
               User,
               Hotel_Reviewed
      from     InitialRelation
    ) as D
group by User ;

This leads us to the relational algebra notation:

R = Ɣ User, COUNT(Hotel_Reviewed) -> Hotels_Reviewed 
        (π User, Hotel_Reviewed (InitialRelation)) -> D

Answer 2

You can get it by counting distinct hotels, grouped by user.

select   User,
         count(distinct Hotel_Reviewed) HotelsReviewed
from     your_table
group by User;

create table reviews([user] varchar(20), date_review date, hotel_reviewed varchar(10) );
insert into reviews values
('Joe',   '20170101', 'A'),
('Joe',   '20170201', 'A'),
('Tom',   '20170101', 'C'),
('Bryan', '20170101', 'B'),
('Bryan', '20170201', 'A'),
('Mike',  '20170101', 'B'),
('Mike',  '20170201', 'C'),
('Mike',  '20170301', 'D');
GO

select   [User],
         count(distinct Hotel_Reviewed) HotelsReviewed
from     reviews
group by [User];
GO

User  | HotelsReviewed
:---- | -------------:
Joe   |              1
Tom   |              1
Bryan |              2
Mike  |              3

dbfiddle here