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Author | Message |
Priya
Groupie ![]() Joined: 04Jan2007 Online Status: Offline Posts: 82 |
![]() ![]() ![]() Posted: 15Feb2007 at 10:45am |
Q13. consider the following premises
P->Q,Q->not R,R,PV(J^S) which of the following conclusion is true?? ans is J^S plz explain..// Q11. A hash table implementation uses function of(n mod7) and linear probing to resolve collision,what is the ratio of number in the following without collision and with collision if 7 buckets are used? 32,56,87,23,65,26,93 A.2,5 B.3,4 c.5,2 D.4,3 plz tell me how I proceed??? 72. Let 'R' be the relation in natural numbers N={1,2,3,....} defined as "x+2y=10" Find the range of R A. {2468} B.{1236} C {4321} D{4368} ans. is C. in C. if we take y=1 ....them what would be the value of X??...it must be 8 but 8 is missing in set C.in this way if we take it the other three no options are fit for the answer..plz. discuss. question no 38 i think the answer should be B.3(A,BD,E) plz. confirm. Question number 10.the number of spanning tree of a complete graph(Kn) is n^n-2....plz explain(i think it should be2^n-1) In Q45..how i get min cost? |
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Priya
Groupie ![]() Joined: 04Jan2007 Online Status: Offline Posts: 82 |
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Hii Priya, for Q 72: the relation is xRy.. thus range is all possible values of 'y' for x ={1,2,3...} so compute values of y considering each value of x.. x y --------------------- 1 no value possible from N 2 4 3 not possible 4 3 5 not possible 6 2 7 not possible 8 1 9.. not possible thus range is {4,3,2,1} Q.13 I didnt find any comprehensive material for this type of problems.. but here is approch that i follow: In such a question the only statements true are the one given as premises. Next, all the statements that can be concluded from premises are true. As a formula use "Modus Ponens" & "Modus Tollens" Modus Ponens: If P->Q & Q is true that implise : P is true Modus tollens: If ~Q (read Q is false) & P->Q that implise : ~P (P is false) Well now apply these rules to given set of premises: given : P->Q,Q->~R,R,PV(J^S) R is true Q->~R therefore Q is false P->Q therefore P is false P V (J^S) since above premise is true & we just deduced that P is false therefore J^S must be true. hence the Answer. |
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