Decomposition algorithms that convert to 3NF hold function dependencies:

Decomposition algorithms that convert to 3NF hold function dependencies:

Ρ={r1<u1,f1>,r2<u2,f2>,..., Rk<uk,fk>} is a decomposition of the relational pattern r<u,f>, u={a1,a2,..., an},f={fd1,fd2,..., FDp}, and set F is a minimum dependency set, note FDI as XI→ALJ, the steps are as follows:

① the function dependency set F of the r<u,f> is minimized (the processed result is still recorded as f);

② find properties that do not appear in F, and make such attributes a relational pattern. Remove these attributes from U, and the remaining attributes are still remembered as u;

③ If there is x→a€f, and Xa=u, then ρ={r}, the algorithm terminates;

④ Otherwise, the f is grouped by a principle with the same left part (assuming a K-group), and each set of functions relies on all the attributes involved in fi to form an attribute set UI. If Ui¢uj (i≠j), remove the UI. As a result of the step ②, so the composition of a hold function dependent decomposition. Also, each ri (Ui,fi) belongs to 3NF and maintains the function dependency.

Example 1: The relational pattern R<u,f>, where u={c,t,h,i,s,g},f={cs→g,c→t,th→i,hi→c,hs→i}, decomposes it into 3NF and keeps the function dependent.

(i) Calculating the minimum function dependency set of f

① uses decomposition rules to turn all function dependencies into a function dependency of a single property on the right. Since the right side of all function dependencies of f is a single attribute, no decomposition is necessary.

② Remove redundant function dependencies in F

A Set Cs→g as a redundant function dependency, then remove the cs→g to:

F1={c→t,th→i,hi→c,hs→i}

Calculation (CS) f1+:

Set X (0) =cs

Calculate x (1): Scan F1 for function dependencies, find a function dependency on left CS or CS subset, find a c→t function dependency. So there are X (1) =x (0) ∪t=cst.

Calculation x (2): Scans the individual function dependencies in the F1, finds the function dependencies on the left side of the CST or CST subset, and finds no function dependencies. So there are X (2) =x (1). The algorithm terminates.

(CS) f1+= CST does not contain G, so cs→g is not a redundant function dependency and cannot be removed from F1.

B Set C→t as a redundant function dependency, then remove the c→t to:

F2={cs→g,th→i,hi→c,hs→i}

Calculation (C) f2+:

Set X (0) =c

Calculation x (1): Scans the F2 for each function dependency, and does not find the function dependency of C on the left. So there are X (1) =x (0). The algorithm terminates. Therefore, c→t is not a redundant function dependency and cannot be removed from the F2.

C Set Th→i as a redundant function dependency, then remove the th→i to:

F3={cs→g,c→t,hi→c,hs→i}

Calculation (TH) f3+:

Set X (0) =th

Calculation x (1): Scans the individual function dependencies in the F3, and does not find the function dependency of the left part as th or th subset. So there are X (1) =x (0). The algorithm terminates. Therefore, Th→i is not a redundant function dependency and cannot be removed from the F3.

D Set Hi→c as a redundant function dependency, then remove the hi→c to:

F4={cs→g,c→t,th→i,hs→i}

Calculation (HI) f4+:

Set X (0) =hi

Calculation x (1): Scans the individual function dependencies in the F4, and does not find a function dependency on the left side as a subset of hi or hi. So there are X (1) =x (0). The algorithm terminates. Therefore, Hi→c is not a redundant function dependency and cannot be removed from the F4.

E. Set Hs→i as a redundant function dependency, then remove the hs→i to:

F5={CS→G,C→T,TH→I,HI→C}

Calculation (HS) f5+:

Set X (0) =hs

Calculation x (1): Scans the F5 for individual function dependencies, and does not find a function dependency on the left part of the HS or HS subset. So there are X (1) =x (0). The algorithm terminates. Therefore, Hs→i is not a redundant function dependency and cannot be removed from the F5. That is: F5={cs→g,c→t,th→i,hi→c,hs→i}

③ remove each function in the F5 depends on the left superfluous attribute (check only the left is not a single property of the function dependency)

There is no dependency on the function on the left that has extra attributes. Therefore, the minimum function dependency set is:

F={cs→g,c→t,th→i,hi→c,hs→i}

(b) Since all the properties in R appear in F, go to the next step.

(iii) by the principle of having the same left part of F is divided into: r1=csg,r2=ct,r3=thi,r4=hic,r5=hsi. So Ρ={r1 (CSG), R2 (CT), R3 (THI), R4 (HIC), R5 (HSI)}.

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Decomposition algorithms that convert to 3NF hold function dependencies: