First time here? Checkout the FAQ!
+6 votes

What is the optimized version of the relation algebra expression $\pi_{A1}(\pi_{A2}(\sigma_{F1}(\sigma_{F2}(r))))$, where $A1, A2$ are sets of attributes in $r$ with  $A1 \subset A2$ and $F1,F2$ are Boolean expressions based on the attributes in $r$?

  1. $\pi_{A1}(\sigma_{(F1 \wedge F2)}(r))$
  2. $\pi_{A1}(\sigma_{(F1 \vee F2)}(r))$
  3. $\pi_{A2}(\sigma_{(F1 \wedge F2)}(r))$
  4. $\pi_{A2}(\sigma_{(F1 \vee F2)}(r))$
asked in Databases by Veteran (77.2k points)  
edited by | 441 views

2 Answers

+17 votes
Best answer

(A) πA1(σ(F1∧F2)(r))

since A1 is subset of A2 will get only A1 attributes as it is in the outside, so we can remove project A2.

Two Selects with boolean expression can be combined into one select with And of two boolean expressions.

answered by Loyal (3.2k points)  
edited by
0 votes

The Relational Algebra expression in the question above, does 4 operations, step by step ( innermost braces first ) .

1. Select those tuples from relation r which satisfies
   expression/condition F1, say the result of this 
   operation is set A.

2. Select those tuples from set A which satisfies
   expression/condition F2, say the result of this
   operation is set B.

3. Select attrributes set A2 from set B, say the 
   result of this operation is set C.

4. Select attrributes set A1 from set C, say the 
   result is set D which is the final result.

Now to optimize this expression, we can combine operations/steps 1 and 2 by AND operator between F1 and F2 condition, like F1 ^ F2, and instead of selecting first attribute set A2, we can directly select attribute set A1 from the result of the combined operation, which is represented by expression in Option A .

answered by Loyal (3.2k points)  

Top Users May 2017
  1. akash.dinkar12

    3292 Points

  2. pawan kumarln

    1652 Points

  3. sh!va

    1650 Points

  4. Arjun

    1424 Points

  5. Bikram

    1372 Points

  6. Devshree Dubey

    1272 Points

  7. Debashish Deka

    1142 Points

  8. Angkit

    1044 Points

  9. LeenSharma

    904 Points

  10. srestha

    718 Points

Monthly Topper: Rs. 500 gift card
Top Users 2017 May 22 - 28
  1. Bikram

    458 Points

  2. Arnab Bhadra

    402 Points

  3. pawan kumarln

    278 Points

  4. Ahwan

    236 Points

  5. bharti

    194 Points

22,786 questions
29,121 answers
27,661 users