here,nothing is given about referential integrity so,we apply some case
CASE 1 (default): All values of R1(R) and R2(R) are same then 20*10=200
CSAE 2: suppose R2(R) references R1(R) (foreign key) and all the values are same in R2(R) then,
1 tuple of R1(R) matches with 10 tuples of R2(R) then no of combination=10
and 19 tuples left in R1 which also come in resultant table with NULL values for R2 tables attributes value
so,19 no of more tuples
so, total =10+19=29
CASE 3: if we change RI make R1(R) references R2(R) then if all values R1(R) is same
then, also as in case 2, i.e. 20+9=29
CASE 4: if r be act as primary key (unique) for both the table then for EX
R1
P |
Q |
R |
S |
a |
d |
1 |
o |
b |
e |
2 |
u |
c |
f |
3 |
g |
R2
R |
G |
T |
U |
4 |
k |
n |
q |
5 |
l |
o |
r |
6 |
m |
p |
s |
R1(leftouter join)R2
P |
Q |
R |
S |
R |
G |
T |
U |
a |
d |
1 |
o |
n |
n |
n |
n |
b |
e |
2 |
u |
n |
n |
n |
n |
e |
f |
3 |
g |
n |
n |
n |
n |
so,all left part of table present with null values is right part of table
so,cardinality of left table will be maxm no of tuples so, 20 will be maxm