vishal , what will we do to count when both   p[i-1][j] and
p[i-1][j-a[i]] is true .

On Thu, Dec 30, 2010 at 12:36 PM, Ankur Khurana
<ankur.kkhur...@gmail.com> wrote:
> 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 <vishal.ge...@gmail.com> 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 <vishal.ge...@gmail.com>
>> wrote:
>>>
>>> yeah, My bad.
>>> Missed that.
>>>
>>> On Wed, Dec 29, 2010 at 10:52 PM, Wladimir Tavares <wladimir...@gmail.com>
>>> 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 <ankur.kkhur...@gmail.com>
>>>> 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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