@Lakhan
Why are you not considering sets of size 2 ? Because two sets of size two
cannot have both of the elements as same.




On Sat, Jan 15, 2011 at 9:39 PM, Lakhan Arya <[email protected]> wrote:

> @bittu
> I don't think answer of 6th question to be a)
> No. of vertices of degree 0 will be those who didnot intersect with
> any set i exactly 2 points. All sets of size greater than equal 2 must
> intersect with any other set having exactly 2 common elements between
> them in exactly 2 points. e.g if a set is (1,2) then it will be
> adjacent to (1,2,3) , (1,2,3,4) etc..
> The sets of size 0 and 1 cannot intersect in 2 points so they all will
> be of degree 0.
> Number of Sets of size 0 --- 1
> Number of Sets of size 1 --- n
> so Total number of vertices n+1.
>
> In the similar way number of connected components will be n+2.
>
>
> On Jan 15, 8:44 pm, bittu <[email protected]> wrote:
> > 1.c U Can verify by putting n =I where I is positive integer value say
> > n=5 & try it out its so easy
> >
> > 2 a...what i have understood.
> >     as we know that  formal grammar is defined as (N, Σ, P, S)
> >     so  For instance, the grammar G with N = {S, A}, Σ = {a, b}, P
> > with start symbol S and rules
> >
> >     S → aA
> >     A → Sb
> >     S → ε
> >
> >   generates { a^ib^i : >=0}   so answer is A.
> >
> > 3  expected value doe discrete distributional is defined as
> > E(i)=sum(pi * xi);  so from my points of view ans is 1/n ...Really Gud
> > Question one has think..still thinking
> >
> > 4.b -Explaination
> >
> > Informally the NP-complete problems are the "toughest" problems in NP
> > in the sense that they are the ones most likely not to be in P. NP-
> > complete problems are a set of problems that any other NP-problem can
> > be reduced to in polynomial time, but retain the ability to have their
> > solution verified in polynomial time. In comparison, NP-hard problems
> > are those at least as hard as NP-complete problems, meaning all NP-
> > problems can be reduced to them, but not all NP-hard problems are in
> > NP, meaning not all of them have solutions verifiable in polynomial
> > time.
> >
> > (A) is incorrect because set NP includes both P(Polynomial time
> > solvable) and NP-Complete .
> > (B) is incorrect because X may belong to P (same reason as (A))
> > (C) is correct because NP-Complete set is intersection of NP and NP-
> > Hard sets.
> > (D) is incorrect because all NP problems are decidable in finite set
> > of operations.
> >
> > 5. The Most Typical..Still Need Time....
> > 6 a   zero degree means vertex is not connected from any other vertex
> > in graph
> > 7.a
> > 8.No Answer  Answer Comes to Be 252
> >     15c10,14c9,10c5,10*9*8*7*6 all are greater then from output so say
> > No Answer
> >
> >   Correct Me if I am Wrong
> >
> >   Next Time I will Try to provide the solution of 2nd, 5th
> > problem ..explanations from-others are appreciated
> >
> > Thanks & Regards
> > Shashank Mani "Don't B Evil U Can Earn while U Learn"
> > Computer Science & Engg.
> > BIT Mesra
>
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-- 
regards,
chinna.

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