finding all the keys

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Department of Computer Science and Engineering, HKUST Slide 1 Finding All the Keys Computationally, finding all the keys can be done by exhaustive search: Given a table with 6 attributes, the number of all possible combinations of attributes is: 63 6 6 6 5 6 4 6 3 6 2 6 1 C C C C C C Since testing if a set of attributes is a candidate key or not is not difficult, trying out all 63 possibilities is a breeze for a computer

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Finding All the Keys. Computationally, finding all the keys can be done by exhaustive search: Given a table with 6 attributes, the number of all possible combinations of attributes is:. - PowerPoint PPT Presentation

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Page 1: Finding All the Keys

Department of Computer Science and Engineering, HKUST Slide 1

Finding All the KeysFinding All the Keys

• Computationally, finding all the keys can be done by exhaustive search:– Given a table with 6 attributes, the number of all possible

combinations of attributes is:

6366

65

64

63

62

61 CCCCCC

• Since testing if a set of attributes is a candidate key or not is not difficult, trying out all 63 possibilities is a breeze for a computer

Page 2: Finding All the Keys

Department of Computer Science and Engineering, HKUST Slide 2

Heuristics to Reduce the PossibilitiesHeuristics to Reduce the Possibilities

• Of course, students have to do it by hand (in exams)!• Go back to the example in chap6.ppt: R = (A, B, C, G, H, I)

F = A BA CCG HCG IB H

• Heuristics can cut down the total combinations from 63 to a few:– Attributes that no other attributes determine must be part of ANY

candidate key (i.e., A and G)– Attributes that don’t determine any other attributes but are

determined by other attributes should not belong to ANY candidate key (i.e., H and I, which does not conflict with our previous conclusion)

– Only 4 possibilities remain: AG, AGB, AGC, AGBC– Since A->B and A-> C, so AG is the only key for R

Page 3: Finding All the Keys

Department of Computer Science and Engineering, HKUST Slide 3

FDs require just Logical ReasoningFDs require just Logical Reasoning

• The above deductions are just logical reasoning, which are not unique to database design

• Other “obvious” results can be deduced by reasoning:– R(A,B) is always in BCNF, regardless of what FDs are given– Consider all 4 possibilities:

• Case 1: no FDs, R must be in BCNF (since no FD can violate BCNF definition!)• Case 2: Only A->B, then A is a candidate key, FD does not violate BCNF• Case 3: Only B->A, then B is a candidate key, FD does not violate BCNF• Case 4: Both A->B and B->A, both A and B are candidate keys; neither FD

violates BCNF

– There are other “interesting” properties that can be proven by reasoning

• In real life, FDs may not be very complicated but knowing why FDs affects data redundancy and updates, how to reason on FDs and how to decompose a table help to get a better database design

Page 4: Finding All the Keys

Department of Computer Science and Engineering, HKUST Slide 4

To Normalize or Not to Normalize, That is the Question

To Normalize or Not to Normalize, That is the Question

• Good practice: All tables must be in 3NF, in BNCF if possible• Problems with having a lot of tables:

– Computational cost is high (a lot of joins, but can create a physical view)– Insertion cost COULD BE high, inserting a set of values into the database

may cause several tables to be updated (likewise for deletion)

• When NOT to normalize (i.e., use a big table)?– High retrieval speed is required – When a table is never updated (e.g., access log for a website),

inconsistency and update anomaly due to data redundancy are not concerns

– Each transaction generates a large group of data (e.g., IP address, cookies, time, date, URL, etc., in an access log table), appending all the data into a table is more efficient than updating several tables

• Although a non-normalized table is much bigger than the normalized tables, searching a large table is still much faster than doing joins