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In computer science the concept of a Lossless-Join Decomposition is central in removing redundancy safely from databases while preserving the original data.

Lossless-join Decomposition

Can also be called Nonadditive. If you decompose a relation $R$ into relations $R_{1}$ and $R_{2}$ you will guarantee a Lossless-Join if $R_{1}$ ⋈ $R_{2}$ = $R$ .

If R is split into R1 and R2, for the decomposition to be lossless then at least one of the two should hold true.

Projecting on R1 and R2, and joining back, results in the relation you started with.^[1] Let $R$ be a relation schema.

Let $F$ be a set of functional dependencies on $R$ .

Let $R_{1}$ and $R_{2}$ form a decomposition of $R$ .

The decomposition is a lossless-join decomposition of R if at least one of the following functional dependencies are in $F$ ⁺ (where $F$ ⁺ stands for the closure for every attribute in $F$ ):^[2]

$R_{1}$ ∩ $R_{2}$ → $R_{1} - R_{2}$
$R_{1}$ ∩ $R_{2}$ → $R_{2} - R_{1}$

Example

Let $R = (A, B, C, D)$ be the relation schema, with $A$ , $B$ , $C$ and $D$ attributes.
Let $F = {A \to B C}$ be the set of functional dependencies.
Decomposition into $R_{1} = (A, B, C)$ and $R_{2} = (A, D)$ is lossless under $F$ because $R_{1} \cap R_{2} = (A)$ and $A \to B C$ so $R_{1} \cap R_{2} \to R_{1} - R_{2}$ .

References

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[1] ttp://stackoverflow.com/questions/5771810/lossless-join-property

[2] Template:Cite news

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+{{Orphan|date=October 2013}}
+In [[computer science]] the concept of a '''Lossless-Join Decomposition''' is central in removing redundancy safely from [[database]]s while preserving the original data.
+==Lossless-join Decomposition==
+Can also be called Nonadditive.
+If you decompose a relation <math>R</math> into relations <math>R_1</math> and <math>R_2</math> you will guarantee a Lossless-Join if <math>R_1</math>⋈<math>R_2</math> = <math>R</math>.
+If R is split into R1 and R2, for the decomposition to be lossless then at least one of the two should hold true.
+Projecting on R1 and R2, and joining back, results in the relation you started with.<ref>http://stackoverflow.com/questions/5771810/lossless-join-property</ref>
+Let <math>R</math> be a relation schema.
+Let <math>F</math> be a set of [[Functional dependency|functional dependencies]] on <math>R</math>.
+Let <math>R_1</math> and <math>R_2</math> form a decomposition of <math>R</math>.
+The decomposition is a lossless-join decomposition of R if at least one of the following functional dependencies are in <math>F</math><sup>+</sup> (where <math>F</math><sup>+</sup> stands for the closure for every attribute in <math>F</math>):<ref>{{cite news | first= | last= | coauthors= | title=Lossless Join Decomposition | date= | publisher=Jan Chomicki | url =http://www.cse.buffalo.edu/~chomicki/560/handout-design.pdf | work =[[University at Buffalo]] | pages = | accessdate = 2012-02-08 | language = }}</ref>
+* <math>R_1</math>&nbsp;∩&nbsp;<math>R_2</math> → <math>R_1 - R_2</math>
+* <math>R_1</math>&nbsp;∩&nbsp;<math>R_2</math> → <math>R_2 - R_1</math>
+==Example==
+* Let <math>R = (A, B, C, D)</math> be the relation schema, with <math>A</math>, <math>B</math>, <math>C</math> and <math>D</math> attributes.
+* Let <math>F = \{ A \rightarrow BC \}</math> be the set of functional dependencies.
+* Decomposition into <math>R_1 = (A, B, C)</math> and <math>R_2 = (A, D)</math> is lossless under <math>F</math> because <math>R_1 \cap R_2 = (A)</math> and <math>A \rightarrow BC</math> so <math>R_1 \cap R_2 \rightarrow R_1 - R_2</math>.
+==References==
+{{Reflist}}
+[[Category:Databases]]
+[[Category:Data modeling]]
+[[Category:Database constraints]]
+[[Category:Database normalization| ]]
+[[Category:Relational algebra]]

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Lossless-join Decomposition

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