Question

Let R(A,B,C,D) be a relation with FDs F = {A-B, A-C, C-A, B-C, ABC—D} Which of the following statements is correct ? (2 Points) Select one: G= {A-B, B-C, C-A, AC-D } is a canonical cover of F O H = {A-C, C-A,B,C,B-D) is a canonical cover of F. o F is a canonical cover of itself. G and Hare canonical covers of F. None of the above.

Answer 1

Canonical cover It is the set minimal means Version of previous of E Functional dependency on R (A, B, C, D) functional iwhich dependency simplified functional dependency steps right 0 Remoning dependency redundant dependency We canonical can cover fund m redundant attributes im present side FA @ Removing whole in Removing redundant attributes present right left dependency FD's E = {A» B, AC, CA, B>C, ABC D} Relation R. o Firstly decompose defendenses defendencies such that only so attribute is present no reduction 2 right side that that attribute can be done oner

& v is But F's dependenties there 1 21 → ßr eng & ß in fortunately already i attribute en decomposed forne only side remanat rensual of full dependency dependency that is determinant present at meyl beje i Now Check each we one ty Find the closure of attributes side Now hide that dependency and and again find the closure that determinant g closure changes then removed otherwise dependency cannot be remone it AB At - ABC we By looking at all dependences dependencies uniquely fund that closure as attributes which van be determined from A Now Hide it

Again find closure At AC [ B is not is hidden present in closure ute AB As closure changes this defendency that means me cannot remoue AC 2 At ABC Closure Hule Find closure again ) determine A-C by looking looking at all dependencies) again =) Aty ABC B from A ('AB) (B+C) as me can and G from As losure remams same so this AC is redundant and be removed so can remoue it F from immediately • B Updated Fc = {278, CPA, B 3C, ABC D}

C> et CAB [cap Agd A-s 8] Hide (-A ct c las c A is hidden Again find closure that means (→ A cannot closure changes be removed В с B C Bt B CA [ CA & B c] Hide B t B [ By C is hidden] so B - C cannot be removed closure changes AGC D ABCt ABCD Hide it ABCt ABC As closure can't remoue A B C D so changes redundant left 3 Removing There only ow dependency sule attributes which present at м 1 attributes are more than left side Settim which is ABC D

possible that one one Take all possible subsets ABC A, B, C, AB, BC, CA Hide the FD = ABC Find closure od those subsets and with closure ABC which then following =) compare it A B CD the

that means closure remains same present in ABC but not is redundant си that the attributes particular subset it otherwise not we van remove Abct - ABCD At ABC A B C & ABCD et CAB ABT ABC Bt BCA B CA Z A B CD CAB FAB CD A B C & ABCD Acts BCA ACT ACB ACB & ABCD be removed BCA & AB CD A B C D cannot Also D is present only in this, FD FD making will that subset none of any making it clear have same closure

cover So the canone al off F is =) Fe = { AB, [A, BC, A B C D } Now couch of D cover F is incorrect option I G = {AB, BC, CTA, ACD] cannonical incorrect option 2 - & AAC, B3(, (A, BD} is canonical options canonical which is also incorrect F different that option G and H canonical ge also incorrect P F is cover of itsey as than are covers This s option's None the about US the the F of 15 This correct option Fc canonical cover which mentioned in "any of the options Correct option option option (3) not