*Why the Closure Algorithm Works*

On May 25, 2013, In The Relational Data Model by Admin

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In this section, we shall show why the closure algorithm correctly decides whether or not a FD A_{1}A_{2} . . . A_{n} → B follows from a given set of FD's S. There are two parts to the proof:

1. We must prove that the closure algorithm does not claim too much. That is, we must show that if A_{1}A_{2} . . . A_{n} → B is asserted by the closure test (i.e., B is in {A_{1}, A_{2}, . . . , A_{n}}^{+}), then A_{1}A_{2} . . . A_{n} → B holds in any relation that satisfies all the FD's in S.

2. We must prove that the closure algorithm does not fail to discover a FD that truly follows from the set of FD's S.

### Why the Closure Algorithm Claims only True FD's

We can prove by induction on the number of times that we apply the growing operation of step 2 that for every attribute D in X, the FD A_{1}A

_{2}. . . A

_{n}→ D holds (in the special case where D is among the As, this FD is trivial). That is, every relation R satisfying all of the FD's in S also satisfies A

_{1}A

_{2}. . . A

_{n}→ D.

**BASIS:**The basis case is when there are zero steps. Then D must be one of A

_{1}, A

_{2}. . . , A

_{n}, and surely A

_{1}A

_{2}. . . A

_{n}→ D holds in any relation, because it is a trivial FD.

**INDUCTION:**For the induction, suppose D was added when we used the FD B

_{1}B

_{2}. . . B

_{m}→ D. We know by the inductive hypothesis that R satisfies A

_{1}A

_{2}. . . A

_{n}→ B

_{i}for all i = 1,2, . . . , m. Put another way, any two tuples of R that agree on all of A

_{1}, A

_{2}, . . . , A

_{n}also agree on all of B

_{1}, B

_{2}, . . , B

_{m}. Since R satisfies B

_{1}B

_{2}. . . B

_{m}→ D, we also know that these two tuples agree on D. Thus, R satisfies A

_{1}A

_{2}. . . A

_{n}→ D.

### Why the Closure Algorithm Discovers All True FD's

Suppose A_{1}A

_{2}. . . A

_{n}→ B were a FD that the closure algorithm says does not follow from set S. That is, the closure of {A

_{1}, A

_{2}, . . . , A

_{n}} using set of FD's S does not include B. We must show that FD A

_{1}A

_{2}. . . A

_{n}→ B really doesn't follow from S. That is, we must show that there is at least one relation instance that satisfies all the FD's in S, and yet does not satisfy A

_{1}A

_{2}. . . A

_{n}→ B.

This instance I is actually quite simple to construct; it is shown in the following figure. I has only two tuples t and s. The two tuples agree in all the attributes of {A

_{1}, A

_{2}, . . . , A

_{n}}

^{+}, and they disagree in all the other attributes. We must show first that I satisfies all the FD's of S, and then it does not satisfy A

_{1}A

_{2}. . . A

_{n}→ B.

_{1}C

_{2}. . . C

_{k}→ D in set S that instance I does not satisfy. Since I has only two tuples, t and s, those must be the two tuples that violate C

_{1}C

_{2}. . . C

_{k}→ D. That is, t and s agree in all the attributes of {C

_{1}, C

_{2}, . . . , C

_{k}}, yet disagree on D. If we examine the above figure, we see that all of C

_{1}, C

_{2}, . . . , C

_{k}must be among the attributes of {A

_{1}, A

_{2}, . . . , A

_{n}}

^{+}, because those are the only attributes on which t and s agree. Likewise, D must be among the other attributes, because only on those attributes do t and s disagree.

But then we did not compute the closure correctly. C

_{1}C

_{2}. . . C

_{k}→ D should have been applied when X was {A

_{1}, A

_{2}, . . . , A

_{n}} to add D to X. We conclude that C

_{1}C

_{2}. . . C

_{k}→ D cannot exist; i.e., instance I satisfies S.

Second, we must show that I does not satisfy A

_{1}A

_{2}. . . A

_{n}→ B. However, this part is easy. Surely, A

_{1}, A

_{2}, . . . , A

_{n}are among the attributes on which t and s agree. Also, we know that B is not in {A

_{1}, A

_{2}, . . . , A

_{n}}

^{+}, so B is one of the attributes on which t and s disagree. Thus, I does not satisfy A

_{1}A

_{2}. . . A

_{n}→ B. We conclude that the closure algorithm asserts neither too few nor too many FD's; it asserts exactly those FD's that do follow from S.

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