1. CS 480: Database Systems
Lecture 15
February 18, 2013

2. Example { x |  y  order ((y[buyer_name] = ‘Jewel’)   z  supplies
ORDER(order_id,buyer_name,item,quantity), SUPPLIES(supp_name,item)
Query: Retrieve names of all the suppliers that supply all the items that ‘Jewel’ ordered.
{ x |  y  order
((y[buyer_name] = ‘Jewel’) 
 z  supplies
(z[supp_name] = x[supp_name] 
y[item] = z[item] )) }

3. Example { x |  y  order ((y[buyer_name] = ‘Jewel’)   z  supplies
ORDER(order_id,buyer_name,item,quantity), SUPPLIES(supp_name,item)
Query: Retrieve names of all the suppliers that supply all the items that ‘Jewel’ ordered.
{ x |  y  order
((y[buyer_name] = ‘Jewel’) 
 z  supplies
(z[supp_name] = x[supp_name] 
y[item] = z[item] )) }

4. Beer Drinking Example #3
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent every
bar that serves ‘Bud’?
{ t |  d  drinker ( d[d_name] = t[d_name] 
 s  serves ( s[be_name] = ‘Bud’ 
 f  frequents ( f [ba_name] = s[ba_name] 
f [d_name] = t[d_name])))}

5. Beer Drinking Example #3
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent every
bar that serves ‘Bud’?
{ t |  d  drinker ( d[d_name] = t[d_name] 
 s  serves ( s[be_name] = ‘Bud’ 
 f  frequents ( f [ba_name] = s[ba_name] 
f [d_name] = t[d_name])))}
If the bolded formula is untrue, it doesn’t matter. That’s exactly what we want.

6. Domain Relational Calculus (DRC)
Similar to tuple relational calculus, except that each variable is defined on an attribute, rather than a tuple.
An expression in the DRC is of the form:
{ < x1,x2,…,xn > | P(x1,x2,…,xn) }
x1,x2,…,xn are domain variables. P is a formula composed of atoms (as in TRC).

7. Example #1 student1(STUDENT) student2(STUDENT)
STUDENT = {id,name,major,GPA}
RA Query: student1  student2
Domain Relational Calculus:
{ < i,n,m,g > | < i,n,m,g >  student1 
< i,n,m,g >  student2 }

8. Example #2 student(STUDENT) STUDENT = {id,name,major,GPA}
RA Query: Πid(student)
Domain Relational Calculus:
“ n,m,g” is shorthand for  n ( m ( g (…)))
{ < i > |  n,m,g ( < i,n,m,g >  student )}

9. Example #3 student(STUDENT), enrol(ENROL)
STUDENT = {id,name,major,GPA}
ENROL = {id,course,grade}
RA Query: Πname,grade(student enrol)
Domain Relational Calculus:
{ < n,g2 > |  i,m,g1,c
(< i,n,m,g1 >  student 
< i,c,g2 >  enrol }

10. { < x1,x2,…,xn > | P(x1,x2,…,xn) }
Formal Definition
A domain-relational-calculus expression is of the form:
{ < x1,x2,…,xn > | P(x1,x2,…,xn) }
Where P(x1,x2,…,xn) is a formula (predicate) in which x1,x2,…,xn are free (unbound) domain variables.
A free variable will be one that is not quantified by  or  .
Go back one slide to show which are the free variables and which are the bound variables in the previous example.

11. Formal Definition
Let x1,x2,…,xn be domain variables, r a relation. Then, the following are atoms:
< x1,x2,…,xn >  r
xi θ xj, where θ is a comparison operator
xi θ c, where c is a constant
An atom is a formula.

12. Formal Definition Same as with TRC:
If P1 and P2 are formulas, then the following are also formulas:
(P1), P1
P1  P2, P1  P2, P1  P2
If P1(s) contains a free domain variable s and r is a relation then the following are also formulas:
 s (P1(s))
 s (P1(s))
There exists s such that P1 is true
For all s, P1 is true.

13. Beer Drinking Example #1
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which bars serve some beer
that ‘John Doe’ likes?
{ < a > |  d, e ( < a,e >  serves 
< d,e >  likes 
d = ‘John Doe’ ) }

14. Beer Drinking Example #1
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which bars serve some beer
that ‘John Doe’ likes?
{ < a > |  d, e ( < a,e >  serves 
< d,e >  likes 
d = ‘John Doe’ ) }

15. Beer Drinking Example #1
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which bars serve some beer
that ‘John Doe’ likes?
{ < a > |  d, e ( < a,e >  serves 
< d,e >  likes 
d = ‘John Doe’ ) }

16. Beer Drinking Example #1
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which bars serve some beer
that ‘John Doe’ likes?
{ < a > |  d, e ( < a,e >  serves 
< d,e >  likes 
d = ‘John Doe’ ) }

17. Beer Drinking Example #1
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which bars serve some beer
that ‘John Doe’ likes?
{ < a > |  d, e ( < a,e >  serves 
< d,e >  likes 
d = ‘John Doe’ ) }

18. Beer Drinking Example #2
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent at
least one bar that serves some
beer they like?
{ < d > |  a, e ( < a,e >  serves 
< d,e >  likes 
< d,a >  frequents ) }

19. Beer Drinking Example #3
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent every
bar that serves ‘Bud’?
{ < d > |  c ( < d,c >  drinker 
 a,e ( ( < a,e >  serves 
e = ‘Bud’ ) 
< d,a >  frequents ) }
I don’t need the  c…

20. Beer Drinking Example #3
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent every
bar that serves ‘Bud’?
{ < d > |  c ( < d,c >  drinker 
 a,e ( ( < a,e >  serves 
e = ‘Bud’ ) 
< d,a >  frequents ) ) }
We do need the  c because if it wasn’t there then if no bar served Bud then the query would return no drinkers which would be a mistake.

21. Beer Drinking Example #3
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent every
bar that serves ‘Bud’?
{ < d > |  c ( < d,c >  drinker 
 a,e ( ( < a,e >  serves 
e = ‘Bud’ ) 
< d,a >  frequents ) ) }
I don’t need the  c…

22. Beer Drinking Example #3
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent every
bar that serves ‘Bud’?
{ < d > |  c ( < d,c >  drinker 
 a,e ( ( < a,e >  serves 
e = ‘Bud’ ) 
< d,a >  frequents ) ) }
I don’t need the  c…

23. Beer Drinking Example #3
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent every
bar that serves ‘Bud’?
{ < d > |  c ( < d,c >  drinker 
 a,e ( ( < a,e >  serves 
e = ‘Bud’ ) 
< d,a >  frequents ) ) }
I don’t need the  c…

24. Beer Drinking Example #4
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent only
bars that serve some beer
they like?
{ < d > |  c ( < d,c >  drinker 
 a ( < d,a >  frequents 
 e ( < d,e >  likes 
< a,e >  serves ) ) ) }

25. Beer Drinking Example #4
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent only
bars that serve some beer
they like?
{ < d > |  c ( < d,c >  drinker 
 a ( < d,a >  frequents 
 e ( < d,e >  likes 
< a,e >  serves ) ) ) }

26. Beer Drinking Example #4
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent only
bars that serve some beer
they like?
{ < d > |  c ( < d,c >  drinker 
 a ( < d,a >  frequents 
 e ( < d,e >  likes 
< a,e >  serves ) ) ) }

27. Beer Drinking Example #4
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent only
bars that serve some beer
they like?
{ < d > |  c ( < d,c >  drinker 
 a ( < d,a >  frequents 
 e ( < d,e >  likes 
< a,e >  serves ) ) ) }

28. Beer Drinking Example #4
Relational Schema:
DRINKER(d_name,d_city)
BAR(ba_name,ba_city)
BEER(be_name,type)
FREQUENTS(d_name,ba_name)
SERVES(ba_name,be_name)
LIKES(d_name,be_name)
Query: Which drinkers frequent only
bars that serve some beer
they like?
{ < d > |  c ( < d,c >  drinker 
 a ( < d,a >  frequents 
 e ( < d,e >  likes 
< a,e >  serves ) ) ) }