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This is the problem:

You have two arrays A and B, of equal length. You have to partition them into two groups P and Q such that: (1) Their difference is minimized. (2) If A[i] goes into one of P or Q, B[i] should go into another.

Here is a link to the actual problem: http://opc.iarcs.org.in/index.php/problems/EQGIFTS

This is my logic (to solve the actual problem):

if the input is : a b c d e f g, a list of values and the index of a,b,c,d,e,f is 0,1,2,3,4,5 respectively

if t is a index of a,b,c,d,e,f,g the program checks for t and i such that: the value at [t] > value at [t-i] , beginning with t = 5, and i = 1, and increasing the value of i by 1 and decreasing the value of t by 1.

as soon as it finds a match, it swaps the values of both the indices and sorts the values beginning from [t-1]. the resulting list of values is the output.

I don't know what is wrong with this algorithm, but it produces a wrong answer for all the test cases. I know it can be solved using dynamic programming, and that it is a variation of the partition problem. But i don't know how to change the partition algorithm to solve this problem.

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  • Your problem statement differs from the one in your link in two ways (which may or may not be critical). In the original problem, the array elements are (1) positive and (2) bounded by a known constant (300 in the word problem). You have removed both of those constraints. However, I don't know if this changes the inherent complexity of the problem. – Ted Hopp Oct 29 '12 at 16:53

1 Answers1

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Reduce the problem to partition problem:
Create a third array D[i] = B[i] - A[i] for each i.

Now the problem is a classic partition problem on the array D, and you can use its DP solution to have a pseudo-polynomial time solution.

Correctness Proof:
If there is a solution on D (sum(D_1) = sum(D_2)) - then there are i_1,...,i_k chosen to D_1 and j_1,...,j_m chosen to D_2 (and each index is in i's or j's), such that:

sum(D[i's]) = sum(D[j's])

From the construction, it means:

sum(B[i]-A[i]) = sum(B[j]-A[j]) (for each relevant i's,j's)

and thus:

sum(B[i's]) - sum(A[i's]) = sum (B[j's]) - sum(A[j's])

From this:

sum(B[i's]) + sum(A[j's]) = sum(B[j's]) + sum(A[i's])

which exactly what we wanted, since each "index" is assigned to both parts, one part gets a B and the other gets A.
The other direction is similar.

QED

Complexity of the problem:
The problem is still NP-Hard with the simple reduction:

Given an instance of Partition Problem (S=[a_1,a_2,...,a_n]), create the instance of this problem:

A=S, B=[0,...,0]

It is easy to see that the same solution that gives optimal solution to this problem will be the needed partition to the original partition problem, and thus the problem is NP-Hard.

amit
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