Answers to Normalization Exercises

### Answer P.1

1NF |
---|

3NF |

### Answer P.2

1NF |
---|

3NF |

### Answer P.3

- A, C, [A, B], F
- [[A, B]]
- Answers:
- There are three separate transitive dependencies: 1) A —> C —> D, 2) A —>C —> E, and 3) [A, B] —> F —> G.
- A partial key dependency: A —> C.

- Answers:
- A —> C
- C —> D, E
- [A, B] —> F
- F —> G

### Answer P.4

I say “apparently” because you are only shown data which might be used to *disprove* the assertion that one set of attributes determines another set. It may just be an artifact of the data that you are shown or it may actually be the case that the determinancy *does* exist. The least you can say is that the determinancy *apparently* exists — it may actually exist but you can't prove that it does given only data in a table.

- False. There is one input (Lee) which gives two outputes (3, 6).
- Apparently true. Each input (i.e., each A) produces just one output (i.e., an F). Each time A takes on the value of 3, F takes on the value of Lee. That is, for one particular input there is one (and only one) particular output. So far, so good. Each time A takes on the value of 6, F takes on the value of Lee. Again, for one particular input there is one (and only one) particular output. It does not matter that different inputs produce the same output — all that matters is that one input always produces the same output.
- Apparently true. You look to disprove the existence of a determinancy by looking for instances where the attributes on the left have the same value but produce different values on the right. Here, [A, B] take on the values of [3, big] twice (in the 3rd and 4th rows).

### Answer P.5

1NF |
---|

3NF (with conditional dependencies) |

4NF (with conditional dependencies) |

### Answer P.6

1NF |
---|

3NF |

4NF (with conditional dependencies) |

page revision: 1, last edited: 29 Jul 2008 00:47