it means we r considring d value for 2*n.......so need total n passes...

On Mon, Sep 5, 2011 at 1:33 PM, aditya kumar
<[email protected]>wrote:

> @dheeraj : u are right . i dint read the question properly . i thot there
> are total n glasses so i answerd : n/4 swaps . bt since we have 2n glass so
> we need n/2 swaps . i guess gopi did the same mistake
> @sachin : yes u are also correct since the glasses are identical the order
> of swapping doesnt matter . yes u can do that way also . the onli thing is
> dat u will require n/2 swaps .
>
>
>
> On Mon, Sep 5, 2011 at 12:12 PM, Dheeraj Sharma <
> [email protected]> wrote:
>
>> i thnk n/2 swaps are required... 1/4 is of the total glasses present..
>> i.e. 2n/4 ..=n/2
>>
>>
>> On Mon, Sep 5, 2011 at 11:52 AM, sachin goyal <[email protected]>wrote:
>>
>>> PLEASE CORRECT ME ADITYA
>>>
>>>
>>> On Mon, Sep 5, 2011 at 11:52 AM, sachin goyal <[email protected]>wrote:
>>>
>>>> 12345|12345
>>>> FFFFF|EEEEE
>>>> WE HAVE TO CREATE THE PATTERN LIKE
>>>> F E F E F E F E F E
>>>> ACCORDING TO MY UNDERSTANDING WE CAN DIRECTLY REPLACE THE 2ND WITH
>>>> SECOND AND 4TH WITH 4TH
>>>> TELL CORRECT ME IF I AM WRONG AND I AM TRATING WRONG???????
>>>>
>>>>
>>>>
>>>> On Sun, Sep 4, 2011 at 8:04 PM, *$* <[email protected]> wrote:
>>>>
>>>>> yes n/4 swaps are required. +1 aditya kumar
>>>>>
>>>>>
>>>>> On Sun, Sep 4, 2011 at 6:32 PM, aditya kumar <
>>>>> [email protected]> wrote:
>>>>>
>>>>>> swap n/2-1 with n/2+1 , and then n/2-3 with n/2+3 till we reach n-1 .
>>>>>> so we need n/4 swaps .
>>>>>>
>>>>>>
>>>>>> On Sun, Sep 4, 2011 at 5:28 PM, Anup Ghatage <[email protected]>wrote:
>>>>>>
>>>>>>> That's interesting.
>>>>>>>
>>>>>>> when n = 3
>>>>>>>
>>>>>>> We have been given this :F F F | E E E
>>>>>>>
>>>>>>> Swap the middle element and it becomes: F E F | E F E
>>>>>>>
>>>>>>> Which is what you want, but when n = 5
>>>>>>>
>>>>>>> F F F F F | E E E E E
>>>>>>>
>>>>>>> And you swap the middle element it becomes: F F E F F | E E F E E
>>>>>>>
>>>>>>> So the same startegy doesn't apply.
>>>>>>>
>>>>>>> But if you do with from the center for every alternate element, it
>>>>>>> works
>>>>>>>
>>>>>>> for n = 3
>>>>>>>
>>>>>>> F F F | E E E > F F E | F E E > E F E | E F E
>>>>>>>
>>>>>>> It also works for n = 5 etc. So, It is a more uniform solution if you
>>>>>>> may.
>>>>>>>
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>>>>>
>>>>>
>>>>>
>>>>> --
>>>>> Thx,
>>>>> --Gopi
>>>>>
>>>>>
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>>>>
>>>>
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>>
>>
>>
>> --
>> *Dheeraj Sharma*
>> Comp Engg.
>> NIT Kurukshetra
>> +91 8950264227
>>
>>
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>
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