@Nithin: Sorry I did not understand your logic!! If a person looses a match
he should be knocked out of the tournament. Could you please explain why 2
matches to knock out a person??

On Mon, Sep 19, 2011 at 2:47 PM, praveen raj <[email protected]> wrote:

> For question 2 see ashima link.
>
> On 19-Sep-2011 1:43 PM, "Nitin Garg" <[email protected]> wrote:
> >
> > Can someone tell answers to question 2 and 5 with explanation??
> >
> >
> >
> >
> > On Mon, Sep 19, 2011 at 1:40 PM, Nitin Garg <[email protected]>
> wrote:
> >>
> >> In Question 4 i just kept counting new processes that are being added in
> every iteration.
> >> No. of new processes being created is equal to the already running no.
> of even pid processes.
> >>
> >> Time - PId
> >> 0  - 0                          1
> >> 1 - 0,1                        2
> >> 2, - 0,1,2                    3
> >> 3, - 0,1,2,3,4              5
> >> 4 - 0,1,2,3,4,5,6,7       8
> >> .
> >> .
> >> .
> >>
> >>
> >>
> >> 1,2,3,5,8,11,17,25,38,57,86,129,194
> >>
> >> I kept counting, got 194.
> >> Don't know of any shortcut.
> >>
> >>
> >>
> >>
> >>
> >> On Mon, Sep 19, 2011 at 1:35 PM, Nitin Garg <[email protected]>
> wrote:
> >>>
> >>> Question 6 -
> >>> Intuitively you can see that the greater the sum is, the greater the
> favorable events in sample space.
> >>>
> >>> e.g. - sum = 1 .. cases {(1)}   Pr = 1/6
> >>> sum = 2             cases {(2),(1,1)}   Pr = 1/6 + 1/36
> >>> sum = 3            cases {(3),(2,1)(1,2)(1,1,1)}  Pr = 1/6 + 1/36 +1/36
> + 1/216
> >>>
> >>>
> >>> for a more formal proof, look at the recursion -
> >>>
> >>>
> >>> P(k) = (P(k-6) + P(k-5) + P(k-4)... P(k-1)))/6
> >>>
> >>> where P(0) = 1, P(i) = 0  for i<0
> >>>
> >>> Base case -
> >>> P(2) > P(1)
> >>>
> >>> Hypothesis -
> >>>
> >>> P(i) > P(i-1) for  all i <= k
> >>>
> >>> To prove
> >>> P(k+1)  > P(k)
> >>>
> >>> Proof
> >>> P(k+1) - P(k) = (P(k) - P(k-6))/6 > 0
> >>>
> >>>
> >>>
> >>>
> >>>
> >>>
> >>>
> >>>
> >>> On Mon, Sep 19, 2011 at 1:04 PM, Nitin Garg <[email protected]>
> wrote:
> >>>>
> >>>> Question 3 -
> >>>> To eliminate one player, you need to host atleast 2 matches and make
> him loose in both 2. These 2 matches can not contribute to elimination of
> any other player.
> >>>> So, min 2 matches for every player who is to be eliminated, hence 100.
> >>>>
> >>>>
> >>>> On Mon, Sep 19, 2011 at 11:54 AM, Bhanu Chowdary <
> [email protected]> wrote:
> >>>>>
> >>>>> @Nitin: Answer to question 3 is 50.
> >>>>>
> >>>>>
> >>>>> On Mon, Sep 19, 2011 at 11:44 AM, praveen raj <[email protected]>
> wrote:
> >>>>>>
> >>>>>> @nitin Plz explain how u have reached answer of question no. 4 and 6
> >>>>>>
> >>>>>>
> >>>>>> On 19-Sep-2011 12:26 AM, "Nitin Garg" <[email protected]>
> wrote:
> >>>>>> >
> >>>>>> > Answer  3 - 100
> >>>>>> > Answer 6 - 103
> >>>>>> > Answer 4 - 194 total processes including the parent
> >>>>>> > Answer 7 - 12 km south, 12 km east
> >>>>>> >
> >>>>>> >
> >>>>>> > On Sun, Sep 18, 2011 at 11:53 PM, Ashima . <[email protected]>
> wrote:
> >>>>>> >>
> >>>>>> >> @malay: how cm n+logn-2?
> >>>>>> >> cn u explain the logic ?
> >>>>>> >>
> >>>>>> >> Ashima
> >>>>>> >> M.Sc.(Tech)Information Systems
> >>>>>> >> 4th year
> >>>>>> >> BITS Pilani
> >>>>>> >> Rajasthan
> >>>>>> >>
> >>>>>> >>
> >>>>>> >>
> >>>>>> >>
> >>>>>> >> On Sun, Sep 18, 2011 at 11:07 AM, Ashima . <
> [email protected]> wrote:
> >>>>>> >>>
> >>>>>> >>> rite! 62.5%
> >>>>>> >>>
> >>>>>> >>> Ashima
> >>>>>> >>> M.Sc.(Tech)Information Systems
> >>>>>> >>> 4th year
> >>>>>> >>> BITS Pilani
> >>>>>> >>> Rajasthan
> >>>>>> >>>
> >>>>>> >>>
> >>>>>> >>>
> >>>>>> >>>
> >>>>>> >>> On Sat, Sep 17, 2011 at 9:04 PM, malay chakrabarti <
> [email protected]> wrote:
> >>>>>> >>>>
> >>>>>> >>>> create a tournament tree.in each round one value is eliminated
> to obtain in the process the winner or the highest value in n-1 comparisons.
> Then check the queue of the winner which contains log(n) entries of the
> values beaten by the winner which implicitly will contain the runners
> up.Then log(n)-1 comparisons to find the highest among all the losers whom
> the winner had beaten. So all over complexity will be n-1 +log(n) -1 =
> n+log(n)-2. Hp that answers ur query. nice question btw :)
> >>>>>> >>>>
> >>>>>> >>>>
> >>>>>> >>>> On Sun, Sep 18, 2011 at 8:02 AM, VIHARRI <
> [email protected]> wrote:
> >>>>>> >>>>>
> >>>>>> >>>>> hey i'm also thinking n + logn -2.. but couldnt able to figure
> out
> >>>>>> >>>>> how??? can you please explain the logic
> >>>>>> >>>>>
> >>>>>> >>>>> --
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> >>>>>> >>>>>
> >>>>>> >>>>
> >>>>>> >>>> --
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> >>>>>> >>>
> >>>>>> >>>
> >>>>>> >>
> >>>>>> >> --
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> >>>>>> >
> >>>>>> >
> >>>>>> >
> >>>>>> >
> >>>>>> > --
> >>>>>> > Nitin Garg
> >>>>>> >
> >>>>>> > "Personality can open doors... but only Character can keep them
> open"
> >>>>>> >
> >>>>>> > --
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> >>>>>
> >>>>>
> >>>>>
> >>>>>
> >>>>> --
> >>>>> Bhanu Chowdary
> >>>>>
> >>>>> --
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> >>>>
> >>>>
> >>>>
> >>>>
> >>>> --
> >>>> Nitin Garg
> >>>>
> >>>> "Personality can open doors... but only Character can keep them open"
> >>>
> >>>
> >>>
> >>>
> >>> --
> >>> Nitin Garg
> >>>
> >>> "Personality can open doors... but only Character can keep them open"
> >>
> >>
> >>
> >>
> >> --
> >> Nitin Garg
> >>
> >> "Personality can open doors... but only Character can keep them open"
> >
> >
> >
> >
> > --
> > Nitin Garg
> >
> > "Personality can open doors... but only Character can keep them open"
> >
> > --
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>



-- 
Bhanu Chowdary

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