0-1 Knapsack algorithm

Question

Is the following 0-1 Knapsack problem solvable:

• \'float\' positive values and
• \'float\' weights (can be positive or negative)
• \'float\' capacity

Answer 1

public class KnapSackSolver {

public static void main(String[] args) {
    int N = Integer.parseInt(args[0]); // number of items
    int W = Integer.parseInt(args[1]); // maximum weight of knapsack

    int[] profit = new int[N + 1];
    int[] weight = new int[N + 1];

    // generate random instance, items 1..N
    for (int n = 1; n <= N; n++) {
        profit[n] = (int) (Math.random() * 1000);
        weight[n] = (int) (Math.random() * W);
    }

    // opt[n][w] = max profit of packing items 1..n with weight limit w
    // sol[n][w] = does opt solution to pack items 1..n with weight limit w
    // include item n?
    int[][] opt = new int[N + 1][W + 1];
    boolean[][] sol = new boolean[N + 1][W + 1];

    for (int n = 1; n <= N; n++) {
        for (int w = 1; w <= W; w++) {

            // don't take item n
            int option1 = opt[n - 1][w];

            // take item n
            int option2 = Integer.MIN_VALUE;
            if (weight[n] <= w)
                option2 = profit[n] + opt[n - 1][w - weight[n]];

            // select better of two options
            opt[n][w] = Math.max(option1, option2);
            sol[n][w] = (option2 > option1);
        }
    }

    // determine which items to take
    boolean[] take = new boolean[N + 1];
    for (int n = N, w = W; n > 0; n--) {
        if (sol[n][w]) {
            take[n] = true;
            w = w - weight[n];
        } else {
            take[n] = false;
        }
    }

    // print results
    System.out.println("item" + "\t" + "profit" + "\t" + "weight" + "\t"
            + "take");
    for (int n = 1; n <= N; n++) {
        System.out.println(n + "\t" + profit[n] + "\t" + weight[n] + "\t"
                + take[n]);
    }
}

}