minimum difference between sum of two subsets

Folks,

came across a problem... found this intersting... am modifying it a little bit just tu pep it up.

Given a set of integers (range 0-500), find the minimum difference between the sum of two subsets that can be formed by splitting them almost equally. (say count of int开发者_高级运维egers is n, if n is even, each set must have n/2 elements and if n is odd, one set has (n-1)/2 elements and other has (n+1)/2 elements)

sample imput : 1 2 3 4 5 6

minimal difference = 1 (subsets being 1 4 6 and 2 3 5 )

sample input 2 : [ 1 1 1 1 2 2 2 2 ]

minimal difference = 0 (subsets being 1 1 2 2 and 1 1 2 2 )

is there DP approach for this problem.

Thanks guys...

raj...

This problem looks almost like the "balanced partition".

You can use a DP approach to build a pseudo-polynomial time algorithm that solves the balanced partition. See problem 7 at http://people.csail.mit.edu/bdean/6.046/dp/

It sounds like you could have a similar approach.

I've solved this problem recently using Dynamic Programming in c++. I have not modified the code to answer your question. But changing some constants and little code should do.

The code below reads and solves N problems.Each problem has some people (in your case number of integers) and their weights (integer values). This code tries to split the set into 2 groups with difference being minimum.

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_PEOPLE 100
#define MAX_WEIGHT 450
#define MAX_WEIGHT_SUM MAX_PEOPLE*MAX_WEIGHT
using namespace std;

int weights[MAX_PEOPLE];
//bool table[MAX_PEOPLE + 1][MAX_WEIGHT_SUM + 1]; 

bool** create2D(int x, int y) {
    bool **array = new bool*[x];
    for (int i = 0; i < x; ++i) {
        array[i] = new bool[y];
        memset(array[i], 0, sizeof(bool)*y);
    }
    return array;
}

void delete2D(int x, int y, bool **array) {
    for (int i = 0; i < x; ++i) {
        delete[] array[i];
    }
    delete[] array;
}

void memset2D(int x, int y, bool **array) {
    for(int i = 0; i < x; ++i)
        memset(array[i], 0, sizeof(bool)*y);
}

int main(void) {
    int n, N, W, maxDiff, teamWeight, temp;
    int minWeight = MAX_WEIGHT, maxWeight = -1;
    cin >> N;
    while(N--) {
        cin >> n;
        W = 0;
        for(int i = 0; i < n; ++i) {
            cin >> weights[i];
            if(weights[i] < minWeight)
                minWeight = weights[i];
            if(weights[i] > maxWeight)
                maxWeight = weights[i];

            W += weights[i];
        }
        int maxW = maxWeight + (W>>1);
        int maxn = n>>1;
        int index = 0;
    /* 
       table[j][i] = 1 if a team of j people can form i weight 
                        from K people, where k is implicit in loop
       table[j][i] = table[j-1][i-weight[j]] if i-weight[j] >=0
     */
        bool **table = create2D(maxn+1, maxW+1);
        //memset2D(maxn+1, maxW+1, table);
        //memset(table, 0, sizeof(table));
        table[0][0] = true;
        /* for k people what can be formed?*/
        for(int k = 0; k < n; ++k) {
            /* forming team of size j out of k people*/
            for(int j = min(k, maxn) ; j >= 1; --j) { 
                /* using j people out of k, can I make weight i?*/
                for(int i = maxW; i >=minWeight ; --i) {
                    if (table[j][i] == false) {
                        /*do not consider k if more than allowable*/
                        index = i - weights[k];
                        if (index < 0) break;
                        /*if without adding k, we can make the weight
                          limit with less than one person then one can
                          also make weight limit by adding k.*/
                        table[j][i] = table[j-1][index];
                    } /*outer if ends here*/
                } /* ith loop */
            } /* jth loop */
        } /* kth loop */

        maxDiff = MAX_WEIGHT_SUM ;
        teamWeight = 0;
        for(int i = 0; i <= maxW; ++i) {
            if (table[n/2][i]) {
                temp = abs(abs(W - i) - i);
                if (temp < maxDiff) {
                    maxDiff = temp;
                    teamWeight = i;
                }
            }
        }
        delete2D(n+1, maxW+1, table);
        teamWeight = min(teamWeight, W-teamWeight);
            cout << teamWeight << " " << W - teamWeight << endl;
        if(N)
            cout << endl;
    }
        return 0;
}

One good way to think about it would be, if you had a DP solution to this problem, could you use it to answer subset sum in a P amount of time? If so then your DP solution probably is not correct.

This appears to be an instance of the Partition problem, which is NP-Complete.

According to the Wikipedia article, there is a pseudo-polynomial time dynamic programming solution.

I have written this program in C++ assuming that max sum could be 10000.

#include <iostream>
#include <vector>
#include <memory>
#include <cmath>

using namespace std;
typedef vector<int> VecInt;
typedef vector<int>::size_type VecSize;
typedef vector<int>::iterator VecIter;

class BalancedPartition {
public:
    bool doBalancePartition(const vector<int>*const & inList, int sum) {
        int localSum = 0,  j;
        bool ret = false;
        int diff = INT_MAX, ans=0;

        for(VecSize i=0; i<inList->size(); ++i) {
            cout<<(*inList)[i]<<"\t";
        }
        cout<<endl;
        for(VecSize i=0; i<inList->size(); ++i) {
            localSum += (*inList)[i];
        }
        M.reset(new vector<int>(localSum+1, 0));
        (*M)[0] = 1;
        cout<<"local sum "<<localSum<<" size of M "<<M->size()<<endl;

        for(VecSize k=0; k<inList->size(); ++k) {
            for(j=localSum; j>=(*inList)[k]; --j) {
                (*M)[j] = (*M)[j]|(*M)[j-(*inList)[k]];
                if((*M)[j]) {
                    if(diff > abs(localSum/2 -j)) {
                        diff = abs(localSum/2 -j);
                        ans = j;
                    }
                }
            }
        }
        mMinDiffSubSumPossible = abs(localSum - 2*ans);
        return ret;
    }

    friend ostream& operator<<(ostream& out, const BalancedPartition& bp) {
        out<<"Min diff "<<bp.mMinDiffSubSumPossible;
        return out;
    }
    BalancedPartition(): mIsSumPossible(false), mMinDiffSubSumPossible(INT_MAX) {

    }
private:
    shared_ptr<vector<int> > M;
    bool mIsSumPossible;
    int mMinDiffSubSumPossible;
    static const int INT_MAX = 10000;
};

int main(void) {
    shared_ptr<BalancedPartition> bp(new BalancedPartition());
    int arr[] = {4, 12, 13, 24, 35, 45};
    vector<int> inList(arr, arr + sizeof(arr) / sizeof(arr[0]));
    bp->doBalancePartition(&inList, 0);
    cout<<*bp;
    return 0;
}

    from random import uniform
    l=[int(uniform(0, 500)) for x in xrange(15)]
    l.sort()
    a=[]
    b=[]
    a.append(l.pop())
    while l:
        if len(a) > len(b):
            b.append(l.pop())
        elif len(b) > len(a):
            a.append(l.pop())
        elif sum(a) > sum(b):
            b.append(l.pop())
        else:
            a.append(l.pop())

    print a, b
    print sum(a), sum(b)
    print len(a), len(b)

Next step I would try to find a pair of numbers from opposite lists with difference half of the sums' difference (or close to that) and swap them.