1. FDImplication: 1 Functional Dependencies (FDs) Let r(R) be a relation and let t  r, then the restriction of t to X  R, written t[X], is the projection of t onto X. R(A B C) 1 2 3 2 2 5 2 3 5 4 4 5 BC  ABC t[BC] = = {(B:2), (C:5)} Let R be a relational schema and let X  R and Y  R. X  Y is a functional dependency or FD. A relation r(R) satisfies the FD X  Y (or X  Y holds) if for any two tuples t 1 and t 2 in r, t 1 [X] = t 2 [X]  t 1 [Y] = t 2 [Y]; alternatively (and equivalently) if for every (sub)tuple s in  X r, |  X=s  XY r| = 1. A  C B  C AB  C AC  B t = r =

2. FDImplication: 2 FD Implication Let r(R) and let F be a set of FDs over R. Then r satisfies F if each FD in F holds for r. F may imply that other FDs also hold. F implies X  Y if X  Y holds for every relation that satisfies F. r = A B C D 1 2 3 4 5 6 7 8 5 6 7 9 0 1 2 3 F = {A  B, B  C} r satisfies F. F implies A  C. Proof: 1. Let s[A] = t[A] 2. s[A] = t[A]  s[B] = t[B] given, A  B 3. s[B] = t[B] 1 & 2, modus ponens 4. s[B] = t[B]  s[C] = t[C] given, B  C 5. s[C] = t[C] 3 & 4, modus ponens

3. FDImplication: 3 F+F+ Let S be a set of attributes. If F is a set of FDs over S, the set of all FDs implied by F is called the closure of F, denoted F +. When we assert an FD X  Y, we mean X  Y  F +. Rules for computing F + : (trivial implication) Y  X  X  Y e.g., Name City  Name (accumulation) X  Y, W  Z, W  Y  X  YZ e.g., GuestNr  Name City, Name  RoomNr  GuestNr  Name City RoomNr (projection) X  Y, Z  Y  X  Z e.g., GuestNr  Name City RoomNr  GuestNr  RoomNr We compute F + by a least-fixed-point process.

4. FDImplication: 4 Sound and Complete Rules The implication rules for F + are: sound: the derived FDs hold for any relation satisfying F; complete: repeated application of the rules derives all implied FDs. Proof of Soundness. (Y  X  X  Y): Let s[X] = t[X]. Then since Y  X, s[Y] = t[Y]. (X  Y, W  Z, W  Y  X  YZ): Let s[X] = t[X]. Then since X  Y, s[Y] = t[Y]. Then since W  Y, s[W] = t[W] and since W  Z, s[Z] = t[Z]. Now, since s[Y] = t[Y] and s[Z] = t[Z], s[YZ] = t[YZ]. (X  Y, Z  Y  X  Z): Let s[X] = t[X]. Then since X  Y, s[Y] = t[Y], and since Z  Y, s[Z] = t[Z]. Proof of Completeness (basic idea): Assume an FD X  Y holds but is not in F +. We can show, however, that X  Y can be computed by trivial implication, accumulation, and projection and thus contradict our assumption.

5. FDImplication: 5 Example of F + If the set of attributes is ABC and the set of FDs F = {A  B, B  C}, then F + = {A  B, B  C A  A, B  B, C  C, AB  AB, AC  AC, BC  BC, ABC  ABC, AB  A, AB  B, AC  A, AC  C, BC  B, BC  C, ABC  A, ABC  B, ABC  C, ABC  AB, ABC  AC, ABC  BC, A  BC, A  C, A  AB, A  AC, A  ABC, B  BC, AB  C, AB  AC, AB  BC, AB  ABC, AC  B, AC  AB, AC  BC, AC  ABC}

6. FDImplication: 6 Additional FD-Implication Rules (augmentation) X  Y  XZ  YZ RoomNr  Cost  RoomNr NrDays  Cost NrDays (transitivity) X  Y, Y  Z  X  Z GuestNr  Name, Name  RoomNr  GuestNr  RoomNr (union) X  Y, X  Z  X  YZ RoomNr  NrBeds, RoomNr  Cost  RoomNr  NrBeds Cost

7. FDImplication: 7 Checking for X  Y  F + Generate F + and see if X  Y is present (expensive) Derive X  Y from F, or determine that it’s not derivable –Derivation sequence for X  Y: sequence of FDs Each FD is given in F or follows by a sound derivation rule X  Y is the last FD in the sequence –Examples: R = ABCD, F = {A  B, B  C}, AD  C  F + ?, BC  A  F + ? 1. A  B given 2. B  C given 3. A  C transitivity, 1 & 2 4. AD  CD augmentation, 3 5. AD  C projection, 4 1. BC  B trivial implication 2. B  C given 3. BC  C transitivity, 1 & 2... How do we know we cannot derive BC  A?

8. FDImplication: 8 TAP Derivation Sequence A particular derivation sequence always works! –List the given FDs –T: Trivial Implication –A: Accumulation (repeated zero or more times with FDs in F) –P: Projection (if needed) Examples: R = ABCD, F = {A  B, B  C}, AD  C  F + ?, BC  A  F + ? 1. A  B given 2. B  C given 3. AD  AD T 4. AD  ADB A 5. AD  ADBC A 6. AD  C P 1. A  B given 2. B  C given 3. BC  BC T How do we know we cannot derive BC  A? Accumulation yields nothing more, and projection cannot yield A on the rhs, and BC  A  F + iff there is a TAP derivation sequence for BC  A.

9. FDImplication: 9 X + — Closure of a Set of Attributes X + = maximal accumulation in a TAP derivation sequence starting with X. Algorithm for X + given a set of FDs F: –1. Start with X + = X. –2. If Y  Z  F and Y  X +, X + becomes X + Z. –3. Repeat 2 until no more changes to X + (least fixed point). Examples: R = ABCD, F = {A  B, B  C} AD + = ABCD BC + = BC BD + = BCD D + = D A + = ABC

10. FDImplication: 10 X  Y  F + iff Y  X + Significant observation! –X  Y  F + looks like a problem requiring exponential time –BUT has a polynomial-time solution (linear with well-chosen data structures) This is an example of the essence of good computer science.

11. FDImplication: 11 X + and Hypergraph Reachability To test X  Y  F +, mark the vertices in X and see if the vertices in Y are reachable following directed edges. A  D  F + ? Yes A  G  F + ? No E  CD  F + ? Yes A  BH  F + ? No AE  HG  F + ? Yes...

12. FDImplication: 12 FD Equivalence Two sets of FDs F & G are equivalent, written F  G, if F implies each FD in G and conversely. If F  G, then F + = G +. F = {A  B, AB  C, C  D}  G = {A  BC, C  D, A  D} In F, A + = ABCD  A  BC & A  D and C + = CD  C  D. In G, A + = ABCD  A  B, AB + = ABCD  AB  C, and C + = CD  C  D. FG G+G+ F+F+ FG F+F+ G+G+ F  G G  F (G +  F +  F +  G + )  F + = G +

13. FDImplication: 13 Keys and FDs Let U be a set of attributes, and let F be a set of FDs over U. Let R  U be a relational schema. A subset K of R (K need not be a proper subset of R) is a superkey of R if K  R  F + and is a candidate key (minimal key) of R if there does not exist a proper subset K of K such that K  R  F +. Example: U = ABCDE and F = {A  B, B  A, AB  C, D  BC}. Schema Candidate Keys AB A, B CE CE ABCD D ABCDE DE