i have already mentioned the source
it is famous shuffling algorithm..... u can seach some papers on In-Shuffle
and Out-Shuffle

the original question in the shuffling is like how many times we need to
shuffle the cards after which they will return back to initial sequence.

On Sat, Oct 15, 2011 at 5:35 PM, Ankur Garg <ankurga...@gmail.com> wrote:

> @Sunny ..Superb Algo ..but can you share some link where we can read abt it
>  :)..especially where the info abt the equation u used is given
>
>
> Thanks in Advance
>
>
> On Sat, Oct 15, 2011 at 12:37 PM, Kunal Patil <kp101...@gmail.com> wrote:
>
>> @Sunny: Thanks for the info !! That's gr8..& your logic also seems to be
>> working perfectly fine..
>>
>>
>> On Sat, Oct 15, 2011 at 12:16 PM, shady <sinv...@gmail.com> wrote:
>>
>>> u can always post the code.,... but before posting code, you must state
>>> your algorithm
>>>   else code becomes useless for other users
>>>
>>> On Sat, Oct 15, 2011 at 1:10 AM, Anika Jain <anika.jai...@gmail.com>wrote:
>>>
>>>> i have the code solution to this.. if m allowed to post it here then i
>>>> can i post it. m i allowed to post code here?
>>>>
>>>>
>>>> On Fri, Oct 14, 2011 at 9:40 PM, gaurav yadav <
>>>> gauravyadav1...@gmail.com> wrote:
>>>>
>>>>> @shiva...keep swapping the underline elements
>>>>>
>>>>> a1*a2*a3a4a5*b1*b2b3b4b5c1c2c3c4c5
>>>>> a1b1*a3*a4a5a2b2b3b4b5*c1*c2c3c4c5
>>>>> a1b1c1*a4*a5*a2*b2b3b4b5a3c2c3c4c5
>>>>> a1b1c1a2*a5*a4*b2*b3b4b5a3c2c3c4c5
>>>>> a1b1c1a2b2*a4*a5b3b4b5a3*c2*c3c4c5
>>>>> a1b1c1a2b2c2*a5*b3b4b5*a3*a4c3c4c5
>>>>> a1b1c1a2b2c2a3b3*b4*b5a5a4*c3*c4c5
>>>>> a1b1c1a2b2c2a3b3c3*b5*a5*a4*b4c4c5
>>>>> a1b1c1a2b2c2a3b3c3a4*a5*b5*b4*c4c5
>>>>> a1b1c1a2b2c2a3b3c3a4b4*b5*a5*c4*c5
>>>>> a1b1c1a2b2c2a3b3c3a4b4*c4*a5*b5*c5
>>>>> a1b1c1a2b2c2a3b3c3a4b4c4a5b5c5
>>>>>
>>>>>
>>>>>
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-- 
Sunny Aggrawal
B.Tech. V year,CSI
Indian Institute Of Technology,Roorkee

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