we need n/2 swaps

On Mon, Sep 5, 2011 at 2:34 PM, sachin goyal <[email protected]> wrote:

> 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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