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Let r and s be two relations over the relation schemes R and S respectively, and let A be an attribute in R.  The relational algebra expression $\sigma_{A=a}(r \bowtie s)$ is always equal to

1. $\sigma_{A=a}(r)$
2. $r$
3. $\sigma_{A=a}(r) \bowtie s$
4. None of the above
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0
here selection is based on constant value ,so its better to filter r first

,if selection is based on compaison of attributes it is better to first cartesian then apply condition
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why this question tagged as difficult?
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A and B will be false when S is empty.

C is just the better form of query, more execution friendly because requires less memory while joining. query, given in question takes more time and memory while joining.

edited
+9
thanks to early use of selection
+1
If we consider answer C means, tables will be joined using column 'A'.There may be other similar columns in two tables.Suppose column 'B' is in both.

PLZ explain.
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According to the equation in the question, we take a join of r and s and then select the rows where attribute A has the value a. In option C, we first apply the filter, i.e. select those rows from r where A attribute has value a and then apply join with s. Hence, C is optimised as the number of rows used in the join is less.
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@sumaiya mam very nice

It is an efficient way to write the query to select first the tuples then cross with other relation.

option c 0
what is "relation schemes R and S respectively" in this question??
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why you have taken B attribute in relation r and how come B & C attribute
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What if B dont have entry

20,Y?

It is an efficient way to write the query.First apply selection condition to reduce the number of tuples then apply cross product.