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 <nitin.garg.i...@gmail.com>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 <nitin.garg.i...@gmail.com>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 <bhanuchowd...@gmail.com
>> > wrote:
>>
>>> @Nitin: Answer to question 3 is 50.
>>>
>>>
>>> On Mon, Sep 19, 2011 at 11:44 AM, praveen raj <praveen0...@gmail.com>wrote:
>>>
>>>> @nitin Plz explain how u have reached answer of question no. 4 and 6
>>>>
>>>> On 19-Sep-2011 12:26 AM, "Nitin Garg" <nitin.garg.i...@gmail.com>
>>>> 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 . <ashima.b...@gmail.com>
>>>> 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 . <ashima.b...@gmail.com>
>>>> 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 <
>>>> m1234...@gmail.com> 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 <viharri....@gmail.com>
>>>> 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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>>>> >
>>>> >
>>>> >
>>>> >
>>>> > --
>>>> > 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"
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