Finding Closure of Set of Functional Dependencies Examples / Solved Examples - Closure of Set F of Functional Dependencies / Example for finding F+ / What is meant by the closure of a set of functional dependencies illustrate with an example
Closure of Set F of Functional Dependencies can be found from the given set of functional dependencies by applying the Armstrong's axioms. Closure is denoted as F+.
Recall the axioms;
Reflexivity rule
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If X is a set of attributes, and Y is subset of X, then we would say,
X →
Y.
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Augmentation rule
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If X →
Y, and Z is another set of attributes, then we write XZ →
XY.
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Transitivity rule
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If X →
Y, and Y →
Z, then X →
Z is true.
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Union rule
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If X →
Y and X →
Z, then X →
YZ is true.
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Decomposition rule
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Its reverse of Rule 4. If X →
YZ, then X →
Y, and X →
Z are true
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Pseudotransitivity rule
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If X →
Y and ZY →
A are true, then XZ →
A is also true.
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Example 1 - Finding F+ of given set F of functional dependencies:
Consider a relation R with the relation schema R = (A, B, C, D, E). Assume that the relation holds the following set of functional dependencies. That is, F is given as follows;
F = { A → BC, CD → E, B → D, E → A}
Find the F+ (Closure).
From A → BC and B → D, we can deduce A → D (By applying Transitivity rule)
Through A → BC, A → D, and CD → E, we can derive A → E (By applying Transitivity rule)Through above deductions, we can write A → ABCDE.
From E → A, and A → ABCDE, we can derive E → ABCDE
From CD → E and E → ABCDE, then CD → ABCDE
From B → D, we can derive BC → CD using Augmentaion rule. It would further mean, BC → ABCDE
Like this, we can continue to derive all the possible functional dependencies.
We would write all the functional dependencies as follows to get F+;
F+ = { A → BC, CD → E, B → D, E → A, A → D, A → E, A → ABCDE, BC → CD, CD → ABCDE }