Question:
Consider the following three relations:
Car (model, year, serial, color)
Make (maker, model)
Own (owner, serial)
A tuple in Car represents a specific car of a given model, made in a given year, with a serial number and a color. A tuple in Make specifies that a maker company makes cars of a certain model. A tuple in Own specifies that an owner owns the car with a given serial number. Keys are underlined; (owner, serial) together form key for Own. (\(\bowtie\) denotes natural join)
\[
\pi_{owner} \left( Own \bowtie \sigma_{color="red"} \left( Car \bowtie \sigma_{maker="ABC"} Make \right) \right)
\]
Which one of the following options describes what the above expression computes?
Consider the following three relations:
Car (model, year, serial, color)
Make (maker, model)
Own (owner, serial)
A tuple in Car represents a specific car of a given model, made in a given year, with a serial number and a color. A tuple in Make specifies that a maker company makes cars of a certain model. A tuple in Own specifies that an owner owns the car with a given serial number. Keys are underlined; (owner, serial) together form key for Own. (\(\bowtie\) denotes natural join)
\[ \pi_{owner} \left( Own \bowtie \sigma_{color="red"} \left( Car \bowtie \sigma_{maker="ABC"} Make \right) \right) \] Which one of the following options describes what the above expression computes?
Car (model, year, serial, color)
Make (maker, model)
Own (owner, serial)
A tuple in Car represents a specific car of a given model, made in a given year, with a serial number and a color. A tuple in Make specifies that a maker company makes cars of a certain model. A tuple in Own specifies that an owner owns the car with a given serial number. Keys are underlined; (owner, serial) together form key for Own. (\(\bowtie\) denotes natural join)
\[ \pi_{owner} \left( Own \bowtie \sigma_{color="red"} \left( Car \bowtie \sigma_{maker="ABC"} Make \right) \right) \] Which one of the following options describes what the above expression computes?
Show Hint
To understand relational algebra expressions, break down each operation: selection (\(\sigma\)), projection (\(\pi\)), and joins (\(\bowtie\)) to determine the result.
Updated On: Apr 4, 2025
- All owners of a red car, a car made by ABC, or a red car made by ABC
- All owners of more than one car, where at least one car is red and made by ABC
- All owners of a red car made by ABC
- All red cars made by ABC
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The Correct Option is C
Solution and Explanation
The given expression is:
\[ \pi_{owner} \left( Own \bowtie \sigma_{color="red"} \left( Car \bowtie \sigma_{maker="ABC"} Make \right) \right) \]
1. Step 1: Analyze the components of the expression.
The first part \( \sigma_{maker="ABC"} Make \) selects the rows from the Make relation where the maker is "ABC."
The second part \( Car \bowtie \) natural joins the result with the Car relation, combining information about cars that are made by "ABC."
The third part \( \sigma_{color="red"} \) selects only those cars from the result where the color is "red."
The final part \( Own \bowtie \) joins the result with the Own relation, combining the data about car owners with the cars that are red and made by "ABC."
Finally, \( \pi_{owner} \) projects (selects) only the "owner" field, which gives the owners of the selected cars.
2. Step 2: Interpret the result.
The result is the list of owners of the cars that are red and made by ABC.
3. Conclusion:
This corresponds to Option (C), which states "All owners of a red car made by ABC."
\[ \pi_{owner} \left( Own \bowtie \sigma_{color="red"} \left( Car \bowtie \sigma_{maker="ABC"} Make \right) \right) \]
1. Step 1: Analyze the components of the expression.
The first part \( \sigma_{maker="ABC"} Make \) selects the rows from the Make relation where the maker is "ABC."
The second part \( Car \bowtie \) natural joins the result with the Car relation, combining information about cars that are made by "ABC."
The third part \( \sigma_{color="red"} \) selects only those cars from the result where the color is "red."
The final part \( Own \bowtie \) joins the result with the Own relation, combining the data about car owners with the cars that are red and made by "ABC."
Finally, \( \pi_{owner} \) projects (selects) only the "owner" field, which gives the owners of the selected cars.
2. Step 2: Interpret the result.
The result is the list of owners of the cars that are red and made by ABC.
3. Conclusion:
This corresponds to Option (C), which states "All owners of a red car made by ABC."
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