Suppose that you decompose a schema R (A, B, C, D, E) into two schemas as below:
R1 (A, B, C)
R2 (C, D, E)
Show that this decomposition is not a lossless join decomposition.
Hint: Take a suitable Relation, r, for the Schema R and show that r is a lossy join decomposition.
A decomposition {R1, R2} is a lossless-join decomposition
if R1 ∩ R2 → R1 or R1 ∩ R2 → R2 where THis after intersection result must be a candidate key
so
R1 ∩ R2 = Φ
and Φ canot be a candidate key so it is not a lossless join decomposition.
Suppose that you decompose a schema R (A, B, C, D, E) into two schemas as...
consider the schema R-(A,B.C,D,E) and the following set F of functional dependencies holds on R ABC CD-E B- D E-A Problem 2. Suppose that we decompose the relation schema R into R, -(A, B, C) and R, (C, D,E). Show that this decomposition is not a lossless-join decomposition.
Find the decopmosition of R into R1(A, B, C), R2(B, C,D ) and R3(C, D, E) Let R(A,B,C,D,E) be a relation with FDs F = {AB-CDAE, C-D, D-E} The decomposition of R into R1(A, B, C), R2(B, C, D) and RP(C, D, E) is (2 Points) Select one: Lossy and Not Dependency Preserving. Lossless and Not Dependency Preserving. Lossy and Dependency Preserving. Lossless and Dependency Preserving.
can you answer part e of the question Q.31 Answer the following questions (a) Explain difference between lossy decomposition and lossless decomposition (b) If you write a SQL statement to inner join the following two tables based on Plocation value as inner join condition, is the result table lossy or lossless? (Explain why?) EMP LOCS P.K EMP PROJ1 Hours Pname Plocation Р.К. (c) Given a relation schema R ={SSN, Ename, Pnumber, Pname, Plocation, Hours) R is decomposed to R1, R2,...
Let R(A,B,C,D,E) be a relation with FDs F = {AB-CD, A-E, C-D, D-E} The decomposition of Rinto R1(A, B, C), R2(B, C, D) and R3(C, D, E) is 2 Points) Select one: Lossless and Dependency Preserving. Lossy and Not Dependency Preserving. Lossless and Not Dependency Preserving. Lossy and Dependency Preserving.
Let R(A,B,C,D,E) be a relation with FDs F = {AB-CD, A-E, C-D, D-E} The decomposition of Rinto R1(A, B, C), R2(B, C, D) and R3(C, D, E) is 2 Points) Select one: Lossless and Dependency Preserving. Lossy and Not Dependency Preserving. Lossless and Not Dependency Preserving. Lossy and Dependency Preserving.
Let R(A,B,C,D,E) be a relation with FDs F = {AB-CD, A-E, C-D, DE} The decomposition of R into R1(A, B, C), R2(B, C, D) and R3(C, D, E) is (2 Points) Select one: Lossy and Dependency Preserving. Lossless and Not Dependency Preserving. Lossy and Not Dependency Preserving. Lossless and Dependency Preserving.
Consider the schema R=(A, B, C, D, E) and let the following set F of functional dependencies hold for R: F= {A → BC, CD → E, B D } Problem 3 Suppose that the schema R=(A, B, C, D, E) is decomposed into R/ - (A, B, C) and R=(A, D, E). Show if this decomposition is a lossless decomposition with respect to the given set of functional dependencies F.
Given the following Schema S = (R, FD) where R = (A, B, C, D, E, F) and FD contains the following dependencies: A -> BC B ->C C -> D D ->E C -> E E -> F DE -> F C -> F 1. Find a minimal cover of F 2. Find a key for the schema 3. Find a 3N decomposition of the schema that satisfies the lossless join decomposition and dependency preservation properties 4. Find a...
Given the following relation schemas and the sets of FD's: a- R(A,B,C,D) F={ABẠC,C7D, D´A, BC+C} b- R(A,B,C,D) F={BẠC, BD, AD>B} C- R(A,B,C,D) F={AB-C, DC+D, CD+A, AD+B} d- R(A,B,C,D) F={AB=C, C+D, D™B, DE} e- R(A, B, C, D, E) F= {AB+C, DB+E, AE>B, CD+A, ECD} In each case, (i) Give all candidate keys (ii) Indicate the BCNF violation Give the minimal cover and decompose R into a collection of relations that are BCNF. Is it lossless? Does it preserve the dependencies?...
Help ASAP! 1. (10 pts) Given a relation schema R = (A, B, C, D, G, H) and a set of functional dependencies F = {D -> G, CD -> G, D -> C, H -> G} Find FC, a canonical cover of F. Please show all the steps to get your answer. 2. (30 pts) A Hollywood movie studio uses a relation called Movie to keep track of information about movie stars, what fee a star charges for...