Question

Consider a database with the following schema. 
        LIKES(drinker,beer);  /* key: all columns */
        FREQUENTS(drinker,pub); /* key: all columns */
        SERVES(pub,beer,cost); /* key: (pub,beer)  */

Write the following queries in relational algebra.  You can use the
math notation (greek letters sigma, pi, etc.)  or the ASCii "linear"
notation I used in class.  In order to make things more clear, please
use intermediate results defined with the assignment notation in the
algebra: R(a,b) := <rel-alg expression>. Try to give meaningful names
for R. (The textbook sometimes does this using the rho operator as
    rho(    R(a,b), <rel-alg expression> )


Consider the following queries:

1.   Find pubs that serve some beer that Joe likes.
2.   Find drinkers who frequent pubs where they can get a drink for less 
       than $3
3.   Find drinkers who like at least one expensive (over $8) beer that Joe likes. 
4.   Find drinkers who like some beers but do not frequent any pubs.
5.   Find drinkers who frequent pubs that serve either ’Stella Artois’ or ’Molsons’.
6.  Find pubs that serve every beer that Joe likes.
7.  Find all drinkers who frequent a pub that serves at least 2 beers they like, 
      and one of them for at most $3.

(i) Give the numbers of the non-monotone queries (and give an example
    of why you think they are non-monotone)

(ii) Also, write Datalog queries for 1, 2, 4, 6. Again, use
intermediate predicates in Datalog to make things clearer.

Answer 1

i)

.The queries in which there will not be any lose with the tuples when some other tuples get added it. This is said to be known as monotone queries. This process will be displayed in the output already. Example of monotone queries: A query which is having selection-projection joins with equality condition is said to be a monotone query are not only equality condition, there are other conditions too satisfied The following non-monotone queries are in the (Consider given queries) as follows Drinkers who frequent pubs where they can get a drink for less than $3. . Drinkers who like at least one expensive (over S8) beer that Joe likes Drinkers who like some beers but do not frequent any pubs . Drinkers who frequent pubs that serve either 'Stella Artois' or 'Molsons'. . Pubs that serve every beer that Joe likes.

ii)

Consider the datalog queries for 1, 2, 4, 6 from the question. The following intermediate predicate queries in datalog are: :find 2pubs where[?:LIKES/drinker-FREQUENT/drinker ?d LIKES/drinker "joe This query joins the table LIKES and FREQUENT on the basis of attributes of the table It compares the drinker attributes with string "joe" :find drinker :where [?1LIKES/drinker-FREQUENT/drinker] 2dSERVES/cost3 This query is joining LIKES and DRINKER tables and compares the cost attribute of SERVES table with 3 to display the value ;ess than 3

[find ?drinker :where NOT[?! :LIKES/drinker=FREQUENT/drinker 4 It is same as first query since the serves table has not been used in this question. [?! [?d LIKE/drinker "joe" find ?pub :where :LIKES/drinker=FREQUENT/drinker]