# Iii if there are c 1 copies of tuple t 1 in r 1 and c

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iii. × If there are c 1 copies of tuple t 1 in r 1 and c 2 copies of tuple t 2 in r 2 , then there are c 1 c 2 copies of the tuple t 1 . t 2 in r 1 × r 2 . iv. a The output will be the same as a cross product followed by a selection. v. If there are c 1 copies of tuple t in r 1 and c 2 copies of t in r 2 , then there will be c 1 c 2 copies of t in r 1 r 2 , provided that c 1 c 2 is positive. vi.

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Chapter 13 Query Optimization If there are c 1 copies of tuple t in r 1 and c 2 copies of t in r 2 , then there will be c 1 + c 2 copies of t in r 1 r 2 . vii. If there are c 1 copies of tuple t in r 1 and c 2 copies of t in r 2 , then there will be min ( c 1 , c 2 ) copies of t in r 1 r 2 . b. All the equivalence rules 1 through 7.b of section 13.2.1 hold for the multiset version of the relational-algebra deFned in the Frst part. There exist equivalence rules which hold for the ordinary relational- algebra, but do not hold for the multiset version. ±or example con- sider the rule :- A B = A B ( A B ) ( B A ) This is clearly valid in plain relational-algebra. Consider a multiset instance in which a tuple t occurs 4 times in A and 3 times in B . t will occur 3 times in the output of the left hand side expression, but 6 times in the output of the right hand side expression. The reason for this rule to not hold in the multiset version is the asymmetry in the semantics of multiset union and intersection. 13.4 Consider the relations r 1 ( A , B , C ), r 2 ( C , D , E ), and r 3 ( E , F ), with primary keys A , C , and E , respectively. Assume that r 1 has 1000 tuples, r 2 has 1500 tuples, and r 3 has 750 tuples. Estimate the size of r 1 a r 2 a r 3 , and give an efFcient strategy for computing the join. Answer: The relation resulting from the join of r 1 , r 2 , and r 3 will be the same no matter which way we join them, due to the associative and com- mutative properties of joins. So we will consider the size based on the strategy of (( r 1 a r 2 ) a r 3 ). Joining r 1 with r 2 will yield a relation of at most 1000 tuples, since C is a key for r 2 . Likewise, joining that result with r 3 will yield a relation of at most 1000 tuples because E is a key for r 3 . Therefore the Fnal relation will have at most 1000 tuples. An efFcient strategy for computing this join would be to create an index on attribute C for relation r 2 and on E for r 3 . Then for each tuple in r 1 , we do the following: a. Use the index for r 2 to look up at most one tuple which matches the C value of r 1 . b. Use the created index on E to look up in r 3 at most one tuple which matches the unique value for E in r 2 . 13.5 Consider the relations r 1 ( A , B , C ), r 2 ( C , D , E ), and r 3 ( E , F ) of Practice Exercise 13.4. Assume that there are no primary keys, except the entire schema. Let V ( C , r 1 ) be 900, V ( C , r 2 ) be 1100, V ( E , r 2 ) be 50, and V ( E , r 3 ) be 100. Assume that r 1 has 1000 tuples, r 2 has 1500 tuples, and
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iii If there are c 1 copies of tuple t 1 in r 1 and c 2...

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