Thanks everybody for wonderful support and special thanks to Vishal
raja. . But i was bit apprehensive about your last solution . . i will
test it :) and let you know as well . Thanks . . . .


On Thu, Dec 30, 2010 at 11:52 AM, vishal raja <[email protected]> wrote:
> But the same solution I've given above can give you the solution for this
> problem .
> In the formed table of P[i][j] , you can take another variable attached to
> it as count[i][j] for how many items we have selected yet.
> So you gotta find , the max. value of j which has count = 50.
> count[i][j] = count[i-1][j]                       if P(i-1,j) ==1
> count[i][j] = count[i-1][j-a[i]]                  if P(i-1,j-a[i]) ==1
> else count[i][j] = 0
>
>
>
>
> On Thu, Dec 30, 2010 at 11:42 AM, vishal raja <[email protected]>
> wrote:
>>
>> yeah, My bad.
>> Missed that.
>>
>> On Wed, Dec 29, 2010 at 10:52 PM, Wladimir Tavares <[email protected]>
>> wrote:
>>>
>>> Sum up all the number and divide by 2
>>>
>>> Using the algorithm subset problem to find a number close to median
>>>
>>>
>>> Wladimir Araujo Tavares
>>> Federal University of Ceará
>>>
>>>
>>>
>>>
>>>
>>>
>>> On Wed, Dec 29, 2010 at 2:07 PM, Ankur Khurana <[email protected]>
>>> wrote:
>>>>
>>>> How will you divide and array of approx 100 elements into two sub sets
>>>> of 50 each such that the difference between both the subsets is the
>>>> minimum possible one . .
>>>>
>>>>  Thanks in advance .
>>>> Ankur
>>>>
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