Association loyalty

Composite fidelity (also called losslessness) means in the decomposition of a relation that the original relationship from the decomposed relation by means of natural composite (engl. Natural join can be restored).

Related to this is the dependency loyalty .

Simple criterion

A decomposition of a relation scheme R into two sub-schemes R1 and R2 is true to the network if the following applies:

• ${\displaystyle R1\cap R2\rightarrow R1}$ or
• ${\displaystyle R1\cap R2\rightarrow R2}$

Examples

Given a relation with a key candidate . ${\displaystyle R(A,B,C;\{A,B\})}$ ${\displaystyle \{A,B\}}$

Loyal to the association

The relation is divided into two relations and . The following applies and , simplified: ${\displaystyle R}$ ${\displaystyle R1(A,B;\{A,B\})}$ ${\displaystyle R2(B,C;\{B\})}$ ${\displaystyle R1\cap R2=\{A,B\}\cap \{B,C\}={B}}$ ${\displaystyle B\rightarrow BC}$ ${\displaystyle B\rightarrow C}$

R. A. B. C. 1 1 2 1 2 3 2 1 2

R1 A. B. 1 1 1 2 2 1

R2 B. C. 1 2 2 3

${\displaystyle R1\triangleright \!\!\triangleleft \,R2}$ A. B. C. 1 1 2 1 2 3 2 1 2

Not loyal to the association

R. A. B. C. 1 1 2 1 2 3 2 1 3

R1 A. B. 1 1 1 2 2 1

R2 B. C. 1 2 2 3 1 3

${\displaystyle R1\triangleright \!\!\triangleleft \,R2}$ A. B. C. 1 1 2 1 1 3 1 2 3 2 1 2 2 1 3

The reason for the lack of consistency is that in the second example there is neither a functional dependency between B → A nor between B → C. In the first example, however, B → C.

See also

• Relational algebra
• Relation (database)