O(N*K)
On Nov 28, 1:04 pm, Nitin Garg <nitin.garg.i...@gmail.com> wrote:
> @Sourabh - whats the running time?
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> On Mon, Nov 28, 2011 at 3:28 AM, Ankur Garg <ankurga...@gmail.com> wrote:
> > Cool Solution...I was thinking of DP but wasnt clear on the recurrence...
>
> > Nice thinking man and thanks :)
>
> > On Mon, Nov 28, 2011 at 2:47 AM, sourabh <sourabhd2...@gmail.com> wrote:
>
> >> Consider the example that you have given:
> >> [0,0,1,1,0,0,1,1,0,1,1,0] , here n = 12 and k=3
>
> >> Now we need to partition the array into 3 contiguous sub arrays such
> >> that :
> >> a) The expected sum value is maximum
> >> b) and the size of each sub array should be between 2 and 6, both
> >> inclusive. In case, this constraint is not satisfied then its not a
> >> valid candidate for the solution even if the partition produces the
> >> max value.
>
> >> 2 = ceil (n / 2k) = ceil (12/6)
> >> 6 = floor (3n / 2k) = floor (36/6)
> >> ---------------------------------------------
> >> As mentioned above the following equation :
> >> F(n,k) = MAX for all r such that ceil(n/2k) <= r <= floor(3n/2k)
> >> { (expected value of array elems from A[n] to A[n-r+1]) + F(n-r,
> >> k-1) }
>
> >> /**
> >> For understanding how partitioning of the array is represented by the
> >> above equation:
>
> >> Say there is an array denoted by A[i] and it needs to be divided into
> >> 3 contiguous parts, one of the ways to do so would be to take the
> >> following steps :
>
> >> Let K(partition no.) be initialized to 3.
> >> Let array size N be 12.
>
> >> a) If N is 0, the goto step 'f'
> >> b) If K == 1 then call it as partition K and goto step 'e'.
> >> c) Lets take X no. of elements from the end of array A of size N and
> >> call it partition K.
> >> d) Decrement K by 1 and N by X { --K; and N-=X;}
> >> d) Goto step 'a'
> >> e) Valid partition and End.
> >> f) Not a valid partition and End.
>
> >> Now if the above set of steps is run for all values of X such that
> >> 2<=X<=6 , then it will generate all possible candidates (partitions)
> >> as per the given problem statement. And for all the valid
> >> partitions(the ones that will hit step 'e') we need to calculate the
> >> expected sum value.
> >> **/
>
> >> can be translated into,
> >> // I am using 1-based array indexing for better clarity
> >> // A[x .. y] means all elements from A[y] to A[x]..
>
> >> F(12, 3) = MAX
> >> {
> >> ExpVal (A[12 .. 11]) + F(10, 2) ,
> >> ExpVal (A[12 .. 10]) + F(9, 2) ,
> >> ExpVal (A[12 .. 9]) + F(8, 2) , //
> >> this will yield the maximum sum..
> >> ExpVal (A[12 .. 8]) + F(7, 2) ,
> >> ExpVal (A[12 .. 7]) + F(6, 2)
> >> }
>
> >> which is nothing but,
> >> F(12, 3) = MAX
> >> {
> >> 1/2 + F(10, 2) ,
> >> 2/3 + F(9, 2) ,
> >> 2/4 + F(8, 2) , // this will yield the
> >> maximum sum..
> >> 3/5 + F(7, 2) ,
> >> 4/6 + F(6, 2)
> >> }
>
> >> Trace the above equation and you should get it..
>
> >> On Nov 28, 12:57 am, Ankur Garg <ankurga...@gmail.com> wrote:
> >> > Hey Sourabh
>
> >> > Could you please explain the solution in a bit detail perhaps using an
> >> > example or so..It wud be really helpful ..Just logic not code
>
> >> > On Mon, Nov 28, 2011 at 1:03 AM, sourabh <sourabhd2...@gmail.com>
> >> wrote:
> >> > > Looks like a dynamic programming problem....
>
> >> > > Say F(n,k) denotes the maximum expected sum value for an array of n
> >> > > elements and partition k , then
>
> >> > > F(n,k) = MAX for all r such that ceil(n/2k) <= r <= floor(3n/2k)
> >> > > { (expected value of array elems from A[n] to A[n-r+1]) + F(n-r,
> >> > > k-1) }
>
> >> > > Base condition:
> >> > > 1) F(N, 1) = expected value for array A[n] such that ceil(n/2k) <= N
> >> > > <= floor(3n/2k)
> >> > > 2) If any of the sub problems where the array size is not between
> >> > > ceil(n/2k) and floor(3n/2k) , both inclusive, then its not a valid
> >> > > candidate for the final solution. This is can be handled by giving
> >> > > initial value to all such combination a value of -1.
>
> >> > > To store that the intermediate computations take an array Max[N][K],
> >> > > F(N,K) = Max[N][K]
>
> >> > > On Nov 28, 12:17 am, sourabh <sourabhd2...@gmail.com> wrote:
> >> > > > Because in the previous example k = 3.
>
> >> > > > On Nov 27, 10:46 pm, Piyush Grover <piyush4u.iit...@gmail.com>
> >> wrote:
>
> >> > > > > Optimal split: [0,0][1,1][0,0][1,1][0,1][1,0]
> >> > > > > Expected value of optimal split: 0 + 1 + 0 + 1 + 1/2 + 1/2 = 3
> >> > > > > why this is not the optimal split???
>
> >> > > > > On Sun, Nov 27, 2011 at 6:58 PM, Ankur Garg <ankurga...@gmail.com
>
> >> > > wrote:
> >> > > > > > You have an array with *n* elements. The elements are either 0
> >> or 1.
> >> > > You
> >> > > > > > want to *split the array into kcontiguous subarrays*. The size
> >> of
> >> > > each
> >> > > > > > subarray can vary between ceil(n/2k) and floor(3n/2k). You can
> >> > > assume that
> >> > > > > > k << n. After you split the array into k subarrays. One element
> >> of
> >> > > each
> >> > > > > > subarray will be randomly selected.
>
> >> > > > > > Devise an algorithm for maximizing the sum of the randomly
> >> selected
> >> > > > > > elements from the k subarrays. Basically means that we will
> >> want to
> >> > > split
> >> > > > > > the array in such way such that the sum of all the expected
> >> values
> >> > > for the
> >> > > > > > elements selected from each subarray is maximum.
>
> >> > > > > > You can assume that n is a power of 2.
>
> >> > > > > > Example:
>
> >> > > > > > Array: [0,0,1,1,0,0,1,1,0,1,1,0]
> >> > > > > > n = 12
> >> > > > > > k = 3
> >> > > > > > Size of subarrays can be: 2,3,4,5,6
>
> >> > > > > > Possible subarrays [0,0,1] [1,0,0,1] [1,0,1,1,0]
> >> > > > > > Expected Value of the sum of the elements randomly selected
> >> from the
> >> > > subarrays: 1/3 + 2/4 + 3/5 = 43/30 ~ 1.4333333
>
> >> > > > > > Optimal split: [0,0,1,1,0,0][1,1][0,1,1,0]
> >> > > > > > Expected value of optimal split: 1/3 + 1 + 1/2 = 11/6 ~
> >> 1.83333333
>
> >> > > > > > Source ->
> >> > >http://stackoverflow.com/questions/8189334/google-combinatorial-optim.
> >> ..
>
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> Nitin Garg
>
> "Personality can open doors, but only Character can keep them open"
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