find a solution to subset sum using dynamic programming

Question

What I want to do

I want to find a subset of an array that sums to a target T. I also want to use to a dynamic programming approach (and a bottom-up solution at that) to do this.

What I currently have

Currently I only found a way to see if amongst all subsets of size N, whether or not there is at least one subset that has the desired sum. See code below.

public boolean solve(int[] numbers, int target) {

    //Safeguard against invalid parameters
    if ((target < 0) || (sum(numbers) < target)){
        return false;
    }

    boolean [][] table = new boolean [target + 1] [numbers.length + 1]  ;

    for (int i = 0; i <= numbers.length; ++i) {
        table[0][i] = true;
    }


    /* Base cases have been covered. 
     * Now look set subsets [1..n][target] to be true or false.
     * n represents the number of elements from the start that have a subset 
     * that sums to target
     */
    for (int i = 1; i <= target; ++i){
        for (int j = 1; j <= numbers.length; ++j){
            /* Mark index j as one of the numbers in the array
             *  which is part of the solution with the given subtarget */
            table [i][j] = table[i][j-1];
            if (i >= numbers[j-1])
                table[i][j] = table[i][j] || table[i - numbers[j-1]] [j-1];
        }
    }

    return table[target][numbers.length];
}

Where I am stuck

Right now, I know if there is a solution, but I can't think of a way to actually output a solution.

I am not looking for anyone to provide me specific code, but pseudocode is welcome as are hints to how a solution may be saved.

Solution 1

The algorithm you provided can stay the same, you don't need to store anything else besides the DP-table table[][]. You just need an additional post-processing phase in which you step "backwards" through table[][] to get the solution set.

Just to recall:

You've computed the table table[i][j], which stores for every value 0<=i<=t(:=target) and every 0<=j<=n(:=numbers.length) whether there is a subset of numbers in numbers[0..j-1] that sum to i.

Consider the subset S corresponding to table[i][j] (, which is true). Note that:

• The subset S contains the number numbers[j] only if table[ i-numbers[j] ][j-1] is true.

  (Proof: recursively take the solution subset S' for table[ i-numbers[j] ][j-1], and add numbers[j])

• On the other hand, this subset S does not contain the number numbers[j] only if table[ i-numbers[j] ][j-1] is false.

  (Proof: assume S contains numbers[j], trow numbers[j] out of S, this implies table[ i-numbers[j] ][j-1], contradiction)

numbers[n-1]

• If so, recursively compute whether numbers[n-2] is in the subset summing to t-numbers[n-1],
• else recursively compute whether numbers[n-2], is in the subset summing to t

Solution 2

Here are the two Java solutions for the subset sum problem.
First using Recursive Approach.
Second using Dynamic Programming Approach.

/*
Question: Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set 
with sum equal to given sum.

Examples: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output:  True  //There is a subset (4, 5) with sum 9.
Let isSubSetSum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with 
sum equal to sum. n is the number of elements in set[].


*/
package SubsetSumProblem;

import java.util.Scanner;

public class UsingResursiveAndDPApproach {
public static void main(String[] args) {
    Scanner in = new Scanner(System.in);
    try{
        System.out.println("Enter the number of elements in the array");
        int n =in.nextInt();
        System.out.println("Enter the elements of the array");
        int[] a=new int[n];
        for(int i=0;i<n;i++)
            a[i]=in.nextInt();
        System.out.println("Enter the sum, which you need to find");
        int sum = in.nextInt();
        System.out.println("Using recursion, the result is: "+usingRecursion(a,a.length,sum));
        System.out.println("Using Dynamic Programming, the result is: "+usingDP(a,sum));
    }
    finally{
        in.close();
    }
}


private static boolean usingRecursion(int[] a,int length, int sum) {

    // 1. Base Cases
    if(sum==0)
        return true;
    if(length==0 && sum!=0)
        return false;

    // 2. To avoid unnecessary steps, we will optimize the recursion method by avoiding 
    //    recursive calls to areas where we are definite that we can SAFELY ignore the case since
    //    the SOLUTION does not exist there.

    // If last element is greater than sum, then ignore it
    if(a[a.length-1]>sum)
        return usingRecursion(a,length-1,sum);

    // 3. This is the recursion step where we will call the method again and again
    /* else, check if sum can be obtained by any of the following
    (a) including the last element
    (b) excluding the last element   */
    return (usingRecursion(a, length-1, sum-a[length-1])|| usingRecursion(a, length-1, sum));

}
/*
Analysis:
    Time Complexity = O(2^n)
    Space Complexity =       // Don't know
*/

private static boolean usingDP(int[] a, int sum) {
    // using boolean matrix for DP
    boolean dp[][] = new boolean[a.length+1][sum+1];  // +1 in row and column


    // if the length of the array is variable (and sum is 0) then fill TRUE, since the SUM=0 
    for(int row=0;row<dp.length;row++){
        dp[row][0] = true;    // NOTE: dp[length=VARIABLE][sum=0], thus we satisfy the condition where length is VARIABLE
                              // and the SUM=0
    }

    // if the SUM is variable and length is 0 then FALSE, since (sum=variable && length=0)
    for(int column=1;column<dp[0].length;column++){
        dp[0][column] = false;  // NOTE: dp[length=0][sum=VARIABLE], thus we satisfy the condition where 
                                // (length=0 && sum=variable)
    }

    for(int i=1;i<dp.length;i++){
        for(int j=1;j<dp[0].length;j++){


            /* Check if sum can be obtained by any of the following
              (a) including the last element
              (b) excluding the last element   */


            // VERY VERY IMP: This is same as "excluding the last element" which is represented in DP 
            dp[i][j] = dp[i-1][j]; // the current position[i][j] would be same as previous position.
                                   // the previous position means that SUM is ACHIEVED OR NOT-ACHIEVED
                                   // int the previous position then it will ofcourse be ACHIEVED or NOT-ACHIEVED
                                   // in the current position.


            // VERY VERY IMP: This is same as "including the last element" which is represented in DP 
            // if the column[ sum is represented in column of the matrix i.e this sum exist] > = sum-a[last_index]
            // then decrease the sum
            if(j>=a[i-1])   // i.e sum >= array[last index element]. If it is true then include this last element by
                            // deducting it from the total sum
                dp[i][j] = dp[i][j] || dp[i-1][j-a[i-1]];  // VERY VERY IMP NOTE: Here dp[i][j] on R.H.S represent
                            // dp[i-1][j] which we have assigned in the previous step

        }
    }
    return dp[a.length][sum];
}
/*
Analysis:
    Time Complexity = O(a.length*sum)
    Space Complexity = O(a.length*sum)
*/
}

Solution 3

Here is my solution is an iterative dp, but with only one dimension: Hope it can help you.

#include <iostream>
#include <cstring>

using namespace std;

const int maxN=1000;
int memo[maxN];
int pi[maxN];

int main(){
    int a[]={7,8,5,1,4};
    memset(memo,-1,sizeof memo);
    memset(pi,-1,sizeof pi);
    int n;
    cin>>n;
    memo[0]=0;
    pi[0]=0;
    for(int i=0;i<(int)sizeof(a)/4;i++){
        for(int num=n;num>=0;num--){
            if(num-a[i]>=0 and memo[num-a[i]]!=-1 and (memo[num]==-1 or memo[num]>1+memo[num-a[i]])){
                memo[num]=1+memo[num-a[i]]; 
                pi[num]=num-a[i];           
            }
        }
    }   
    int N=n;
    while(N!=0){
        cout<<N-pi[N]<<" ";
        N=pi[N];
    }
    cout<<endl;
    cout<<memo[n]<<endl;
    return 0;
}