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CSE 344 -- Spring 2011
Lecture 09: Datalog and Relational Calculus
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Reading assignment: 5.3, 5.4


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A datalog rule is:

  Q(args)  :- R1(args1), R2(args2), ...

sometimes also written as:

  Q(args)  :- R1(args1) AND R2(args2) AND ...

Terminology:

  the rule head = Q(args)
  the rule body = R1(args1), R2(args2), ...
  atom, or subgoal = any one of Ri(argsi)
  head variables  = the variables occurring the head
  existential variables = all the other variables

A datalog program is:

  rule1
  rule2
  ...

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Examples of single rules:

Schema:
/*****************************************************/
/*   Purchase(pid, product, price, quantity)         */
/*   Product(pname, manufacturer)                    */
/*     (Purchase.product refers to Product.pname)    */
/*****************************************************/

Find all products under 9.99:

  A(y) :- Purchase(x,y,z,u),  z < 9.99

Find all manufacturers that manufacture some products under 9.99

  A(v) :- Purchase(x,y,z,u),  z < 9.99, Product(y, v)

Find manfuacturers that manufacture products both < 9.99 and > 999.99

  A(v) :- Purchase(x1,y1,z1,u1), z1 < 9.99, Product(y1, v),
          Purchase(x2,y2,z2,u2), z2 < 9.99, Product(y2, v),

Change this query so it returns manfuacturer that sold THE SAME
PRODUCT both under 9.99 and over 999.99


Examples of multiple rules:

/************************/
/* Friend(name1, name2) */
/* Enemy(name1, name2)  */
/************************/

Find Joe's friends, and Joe's friends' friends.

  A(x) :- Friend('Joe', x)
  A(x) :- Friend('Joe', z), Friend(z, x)

What does this compute ?

  Mystery(x,y) :- Friend(x,z), Friend(z,y)

Whata does this compute ?

  M1(x,y) :- Friend(x,z), Friend(z,y)
  M2(x,y) :- M1(x,z), M1(z,y)
  A(x,y) :- M2(x,z), M2(z,y)

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We allow only *non-recursive* datalog in ths course, meaning the
following:

   the relation in the head of rule k cannot appera in the body of rules

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Meaning of a datalog rule =  a logical statement !

  M(x,y) :- Friend(x,z), Friend(z,y)

Means:

forall x. forall y. forall z. (Friend(x,z) and Friend(z,y) ==> M(x,y))

and M is the smallest relation that has this property

Note: logically equivalent to:

forall x. forall y. (exists z. Friend(x,z) and Friend(z,y)) ==> M(x,y)

That's why the variables not in the head are called "existential
variables".


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Can we express everythign in the Relational Algebra in datalog ?

In class:

  1. Union  R(A,B) U S(A,B)
  2. Selection
  3. Join
  4. Projections
  5. Difference

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To express difference, we add negation (in class)

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Find Joe's friends' friends that are not Joe's friends.

  A(x) :- Friend('Joe',y), Friend(y,x), not Friend('Joe',x)

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what's wrong ?

  A(x,y) :- Friend('Joe',y), Enemy(y,z)

  A(z) :- Friend('Joe', x), x != z

  A(x) :- Friend('Joe', x), not (Friend(x,y))

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"Safe datalog rule": every variable occurs in a positive atom.

Every rule must be safe !

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**** In Class

Find all x such that all their enemies' enemies are their friends

Find all x having some friend all of whose enemies are x's enemies.

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**** In class

Express the two queries above in SQL.

Note: this seems quite straighforward.
      datalog with negation = simple formalism for SQL

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**** Can you write these two SQL queries WITHOUT subqueries ?