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  1. #1
    Join Date
    Feb 2006
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    18

    relational algebra

    there are 2 relations R(A,B) and S(B,C) where |R| = 3000,|R| = 1000 and B is the primary key of S.Then |R| natural join|S| <= 1000.

    Is this true or false?

    I think R Natjoin S is less than or equal to 1000,but the answer is false....why is this so? please answer.....

  2. #2
    Join Date
    Apr 2002
    Location
    Toronto, Canada
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    R has 3000 rows? then the answer is less than or equal to 3000, not 1000

    each row in R has a value for B, so joining R to S using B, every row in R will match at most one row in B

    suppose R is some set of people, A=name and B=gender (e.g. M/F)

    suppose S is the set of genders, B=gender (M/F), C=name (Male/Female)

    joining R to S you get one row for every person, right?
    rudy.ca | @rudydotca
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  3. #3
    Join Date
    Feb 2006
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    18

    Functional dependencies

    F={A →B, AC →D, AB →C}
    X=A
    Result should be X+=ABCD

    A->B.......AB
    AB->C (given) therefore ABC
    ABC......ABCD ....as AC->D


    Is this right?

    suppose for example A->B...then can we write AA->AB
    so AA is nothing but A...so A->AB..is this right..

  4. #4
    Join Date
    Feb 2006
    Posts
    18

    Functional Dependencies!

    F = { I -> B, IS -> Q, B -> O}
    • a. What are the keys of the relation?
    • b. Is the relation in BCNF?
    IS -> ISBQO IS is the only key...

    How is IS the key....can anyone please elaborate?

  5. #5
    Join Date
    Feb 2006
    Posts
    18

    functional dependencies

    Show that the following “rule” for functional dependencies is not sound: If
    A->B and C->B then A->B..??????????


    ans:for example 3 attributes:lets assume ::A is ssn ; C is empno ; B is name.

    ssn name empno
    100000 abc E4343
    200000 bhh E6456
    300000 njk E9079

    can the above rule be disproved...

    and one more doubt is

    in FD: A->B (A determines B )means A is the key...
    so

    A B
    1000 rr
    1000 rr

    this table satisfies A-> B

    but how can there be two 1000's as A is the key ???

  6. #6
    Join Date
    Apr 2002
    Location
    Toronto, Canada
    Posts
    20,002
    look, this is the 4th separate thread you've started, and, as with the other ones, i'm going to merge them into one thread

    please keep to one thread per topic, thanks
    rudy.ca | @rudydotca
    Buy my SitePoint book: Simply SQL

  7. #7
    Join Date
    Feb 2006
    Posts
    18

    fd's--urgent reply wanted please

    C = CourseNo, SN = SecNo, OD = OfferingDept, CH = CreditHours, CL = CourseLevel,
    I = InstructorSSN, S = Semester, Y = Year, D = Days_Hours, RM = RoomNo,
    NS = NoOfStudents
    Hence, R = {C, SN, OD, CH, CL, I, S, Y, D, RM, NS}, and the following functional
    dependencies hold:
    {C} -> {OD, CH, CL}
    {C, SN, S, Y} -> {D, RM, NS, I}
    {RM, D, S, Y} -> {I, C, SN}

    solution:
    (a) All attributes of R are in F. Attributes only in LHS: S,Y. Attributes only in RHS:
    OD,CH,CL,I,NS. Thus, attributes S and Y must be part of any candidate key, and the
    attributes OD,CH,CL,I,NS are not part of any candidate key. We combine S and Y
    with the remaining attributes: C, SN, D, RM. By looking at the LHS of the given FDs,
    we notice that {S,Y} is not a candidate key, and no combination of three attributes is
    a candidate key. The only sets of attributes that can generate the entire schema are:
    {S,Y,C,SN} and {S,Y, RM,D}, which are the candidate keys.
    why is this so...can anone please elaborate...from the bold....how can we decipher by looking at the LHS of the given fd's that s,y is not a candidate key.......

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