Solving a relational query: part 5

Found: the relational algebra to tackle the problem outlined in part 3.

This is based on a solution to a similar problem found in Database Management Systems, Third Edition, though, unfortunately, that book prints a solution that is not relational algebra at all, as it relies on the order in which results are returned (relational algebra is a form of set algebra and sets have no order).

$\rho (TEMP1, \rho (BIDDER\_ID \rightarrow USER\_ID, \sigma _{amount > 100} (BID)) \Join USER \Join LOT)$

(Rename the attributes in TEMP1 – this is the step that is broken in the book)

$\rho (name \rightarrow TEMP1\_name, TEMP1)$

$\rho(TEMP2, TEMP1 \times ( \rho(BIDDER\_ID \rightarrow USER\_ID, BID) \Join USER \Join LOT ))$

$\Pi _{NAME, EMAIL} ( \sigma _{TEMP1\_LOT\_ID=LOT\_ID \wedge TEMP1\_USER\_ID=USER\_ID \wedge TEMP1\_BID\_ID \neq BID\_ID} (TEMP2))$

Solving a relational query: part 4 (cartesianproduct.wordpress.com)
Solving a relational query: part 3 (cartesianproduct.wordpress.com)
Solving a relational query: part 1 (cartesianproduct.wordpress.com)
Solving a relational query: part 2 (cartesianproduct.wordpress.com)
Relational algebra problem (cartesianproduct.wordpress.com)
Types of Database Management Systems (brighthub.com)

cartesian product … stuff about computing mostly

Solving a relational query: part 5

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Solving a relational query: part 5

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