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. > .. > > > > > > > > -- > > > > > > You received this message because you are subscribed to the > Google > > > Groups > > > > > > "Algorithm Geeks" group. > > > > > > To post to this group, send email to algogeeks@googlegroups.com. > > > > > > To unsubscribe from this group, send email to > > > > > > algogeeks+unsubscr...@googlegroups.com. > > > > > > For more options, visit this group at > > > > > >http://groups.google.com/group/algogeeks?hl=en. > > > > > -- > > > You received this message because you are subscribed to the Google > Groups > > > "Algorithm Geeks" group. > > > To post to this group, send email to algogeeks@googlegroups.com. > > > To unsubscribe from this group, send email to > > > algogeeks+unsubscr...@googlegroups.com. > > > For more options, visit this group at > > >http://groups.google.com/group/algogeeks?hl=en. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. 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