# Recent questions tagged decomposition

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Relation R (A,B,C,D,E,F) Functional dependency: ABC -->DEF, BC-->EF, EF-->D. Decomposition: R1 (ABCDF), R2 (BCDE). Is this decomposition lossy? Is this decomposition preserve functional dependency? Plz explain how to check functional dependency is preserved or not in decomposition.
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Is minimal set of functional dependency for a functional dependency set is always unique???
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Consider a relation R(A,B,C,D,E,F,H) with A as the only key. Assume that the dependencies E->F and C-> DEH hold.on R. 1. Is R in 2NF? If not, decompose to 2NF. 2. is R in 3NF? if not, decompose to 3NF. //What does " with A as the only key" mean ?
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For relation R=(L, M, N, O, P), the following dependencies hold and R is decomposed into R1 = (L, M, N, P) and R2 = (M, O). How to calculate what candidate keys are there for relation R1 and R2. Please tell me the algorithm. FD's are $M\rightarrow O, NO\rightarrow P, P\rightarrow L, L\rightarrow MN$
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R is divided into R1 and R2 ,but since there is no common attribute in R1 and R2, so it should form lossy join,as for loseless join the common attribute has to be key in one of the table.But here the image above,it is comming as loseless join.Can someone please check?
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$R(X,Y,Z,W) is\ decomposed\ into \\ R_1(X,Y)\\ R_2(Y,Z)\\ R_3(Y,W).\\The\ FDs\ are\ :\\ X -> Y,\\ Z->Y,\\ Y->W \\ Find\ whether\ the\ decomposition\ is\ lossless\ or\ lossy\ ?$
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Q.Let R (X, Y, Z, W) be a relational schema with the following functional dependencies: X→ Y, Y → Z, Z → W and W → Y. The decomposition of R into (X, Y), (Y, Z), (Y, W) it is asked to find whether it is lossless join and dependency preserving or not ? I am sure it is lossless join,but couldn't figure how it is dependency preserving.
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How is it option D??
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Relation R(ABCDEF) with the following set of functional dependencies {AB→CDEF, C→A, D→B, E→F, B→E}. The decomposed relationslations of R into 3NF are __________________________.
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I know the condition for a 2 table to be lossless, but i cannot visualize it. How is 1 table which is decomposed, and if common attribute is key in any one of table, then its lossless decomposition. Can anyone explain me this with different analogy or help me visualize it.
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Which of the following statement is true while decomposition ? 3NF guaranteed dependency preserving 3NF guaranteed lossless property Both of the above None of the above
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State true or false:- Symbol " ^ " stands for an intersection AND letters in bold are the candidate key of the respective table. 1) R1(A, B, C) ^ R2(B, C) = {B,C} IS LOSSLESS JOIN. 2) R1(A, B, C) ^ R2(B, D) = {B} IS LOSSY JOIN. 3) R1(A, B, C) ^ R2(B, C) = {B ... R1(A, B, C), R2(D, E), R3(F, B), R4(D, G, A) Also, tell what will be the candidate key after performing join each time and at the end.
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$X$ is an entity set. $E$ and $F$ are multivalued attributes. How many minimum tables which satisfy 1NF ? How many minimum tables which satisfy 2NF ? How many minimum tables which satisfy 3NF ? How many minimum tables which satisfy BCNF ? How many minimum tables which satisfy 4NF ?
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Consider a schema R(A, B, C, D) and functional dependencies {AB → C, C → D, D→ A}. Then the decomposition of R into R1 (CD), R2 (AC) and R3 (BC) is in a. 1NF but not 2NF b. 2NF but not 3NF c. 3NF but not BCNF d. BCNF
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1 vote
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Tell whether the following decomposition of relations lossless and dependency preserving or not. 1. R(ABCDEFGHIJ) and FD sets AB->C, A->DE, B->F, F->GH, D->IJ a) D1"={ DIJ, ACE, FGH, BF, ADC} b) D2={ FGH, DIJ, ADEBF, ABC} 2.R(ABCDEG) and FD sets AB->C, AC->B, AD->E, B->D, BC->A, E->G a) D=( ABC, ACDE, ADG)
1 vote
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Consider the table $R$ with attributes $A, B$ and $C$. The functional dependencies that hold on $R$ are : $A \rightarrow B, C \rightarrow AB$. Which of the following statements is/are True ? I. The decomposition of $R$ into $R1(C, A)$ and $R2(A, B)$ is lossless. II. The decomposition of $R$ into $R1(A, B)$ and $R2(B, C)$ is lossy. Only $I$ Only $II$ Both $I$ and $II$ Neither $I$ nor $II$
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The decomposition of relation R with FD set F into R1 and R2 has lossless join property iff R1 $\cap$ R2 $\rightarrow$ R1 $\in$ F$^+$ OR R1 $\cap$ R2 $\rightarrow$ R2 $\in$ F$^+$ If R is decomposed into more than 2 relations, can we apply ... join property ? Or is that table drawing thing the only way to check for lossless join property ? Answers with reference(s) will be much appreciated. Thanks.
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Decomposition help in eliminating some of the problems of bad design Redundancy Inconsistencies Anomalies All of the above
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The relation schemas $R_1$ and $R_2$ form a Lossless join decomposition of $R$ if and only if $R_1 \cap R_2 \twoheadrightarrow (R_1-R_2)$ $R_1 \rightarrow R_2$ $R_1 \cap R_2 \twoheadrightarrow (R_2-R_1)$ $R_2 \rightarrow R_1 \cap R_2$ a and b happen a and d happen a and c happen b and c happen
1 vote
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If a relation with a Schema R is decomposed into two relations $R_1$ and $R_2$ such that $(R_1 \cup R_2) = R_1$ then which one of the following is to be satisfied for a lossless joint decomposition ($\rightarrow$ ... $R_1 \cap R_2 \rightarrow R_2$ $(R_1 \cap R_2) \rightarrow R_1 \text{ and } R_1 \cap R_2 \rightarrow R_2$
1 vote
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Consider the following set of functional dependency on the scheme (A, B,C) A-->BC, B-->C, A--> B, AB-->C The canonical cover for this set is: (A) A-->BC and B--> C B. A-->BC and AB--> C C. A--> BC and A--> B D. A--> B and B--> C
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Consider a relation R= {M, N, O,P, Q, R, S, T} with the following set of dependencies: MN--> Q M--> RQ N--> R R--> ST Next consider the following set of decompositions for the relation schema R: D1= {R1,R2,R3,R4}: R1={M,N,O,P}, R2={M,P,Q}, R3= {N,R}, ... R4= {R,S,T} Which of the above decomposition (s) has/ have lossless join property? A. Only D1 B. Only D2 C. Both D1 and D2 D. Neither D1 nor D2
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For which of the following set of functions dependencies does the relation $R(A, B, C, D)$ has $AB, CD$ as closed sets? $A \rightarrow B, B \rightarrow A, C \rightarrow D$ $A \rightarrow B, B \rightarrow C, C \rightarrow D, D \rightarrow A$ $A \rightarrow B, B \rightarrow A, C \rightarrow A, D \rightarrow A$ $A \rightarrow B, B \rightarrow A, C \rightarrow D, D \rightarrow C$