Chapter 7: Normalization Flashcards

(44 cards)

1
Q

What does decomposition do?

A

reduce repetition

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

lossy decomposition

A

loss of information

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

lossless Decomposition

A

includes all attributes of R

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

Functional Dependencies

A

generalization of the notion of a key

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

legal instance

A

relation satisfies all real world constraints

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

B can hold A but…

A

sometimes A cannot hold B

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

trivial

A

A gives B and B is a subset of A

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

Trivial Examples:

A

ID, NAME gives ID
name gives name

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

Closure

A

when all functional dependencies can be implied

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

SuperKey

A

sufficient to identify a unique tuple

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

Candidate Key

A

it is minimal, meaning just one key alone

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

satisfies

A

relation r is legal under set F functional dependency

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

holds on

A

all legal relations on R satisfies F functional dependency

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

Visual of lossless Decomposition

A

R1 and R2 gives R1

R2 and R1 gives R2

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

First Normal Form

A

Domain is atomic

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Atomic

A

indivisible units
ex: names, or ID numbers

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
17
Q

non-atomic numbers can..

A
  • complicate storage
  • cause redundancy
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
18
Q

When decomposing a relation, to do testing for dependency preservation you cannot use …

A

Cartesian product or joins

19
Q

Boyce - Codd Normal Form

A

at least One of the following holds true…

a to beta is trivial
a is a superkey of R

20
Q

Decomposing into BCNF

A

R1 = a or B
R2 = R - (B - a)

21
Q

Not always possible to achieve both BCNF and…

A

Dependency Preservation

22
Q

Third Normal Form is only if…

A
  • a gives B is trivial
  • a is a superkey for R
  • attributes in B-a is in the candidate key
23
Q

Advantages of 3NF versus BCNF

A

3NF always has dependency preservation

24
Q

Disadvantages of 3NF

A

can have null values
redundancy

25
Armstrong Axiom's
* Reflexive Rule is B is part of A then A gives B * Augmentation rule is a gives b then ya gives yb *Transitivity rule is a gives b and b gives y then a gives y
26
These rules are...
sound complete
27
sound
generates only functional dependencies
28
Complete
generate all functional dependencies
29
Union Rule
a gives b and a gives y then a gives BY
30
decomposition rule
if a gives by then a gives b and a gives y
31
pseudotransitivity rule
if a gives b and yb gives c then ay gives c
32
Computing Closure of F
* apply reflexivity and augmentation rule * add dependencies in f1 and f2 only if they can combine through transivity
33
test a is a superkey
compute closure of a and see if it contains all the attributes
34
Canonical Cover
testing simplified set of functional dependencies that has the same closure as the set
35
extraneous attributes
attribute can be removed without changing the closure of F
36
Extraneous from the left makes it stronger but...
A gives C implies AB gives C but it cant be the otherway around
37
Extraneous from the right makes it weaker
AB gives CD cannot imply that AB only gives D
38
restriction
all functional dependencies include only attributes of R
39
Dependency Preservation Time
it takes polynomial time instead of exponential
40
Simplified Test
only check dependencies in set F for a violation
41
However simplified test using only F is incorrect when...
testing a relation in decomposition of R
42
for every attribute of A subgroup of R, check that closure of a includes...
no attributes of R-a or includes all attributes of R
43
nothing
44