The 0-1 Knapsack Problem in C

scvalex (Alexandru Scvortov) — Tue, 20 Nov 2007 17:17:50 GMT

This is a dynamic-programming algorithm implementation for solving the the 0-1 Knapsack Problem in C.

Further explanation is given here.



#include 



#define MAXWEIGHT 100



int n = 3; /* The number of objects */

int c[10] = {8, 6, 4}; /* c[i] is the *COST* of the ith object; i.e. what

				YOU PAY to take the object */

int v[10] = {16, 10, 7}; /* v[i] is the *VALUE* of the ith object; i.e.

				what YOU GET for taking the object */

int W = 10; /* The maximum weight you can take */ 



void fill_sack() {

	int a[MAXWEIGHT]; /* a[i] holds the maximum value that can be obtained

				using at most i weight */

	int last_added[MAXWEIGHT]; /* I use this to calculate which object were

					added */

	int i, j;

	int aux;



	for (i = 0; i <= W; ++i) {

		a[i] = 0;

		last_added[i] = -1;

	}



	a[0] = 0;

	for (i = 1; i <= W; ++i)

		for (j = 0; j < n; ++j)

			if ((c[j] <= i) && (a[i] < a[i - c[j]] + v[j])) {

				a[i] = a[i - c[j]] + v[j];

				last_added[i] = j;

			}



	for (i = 0; i <= W; ++i)

		if (last_added[i] != -1)

			printf("Weight %d; Benefit: %d; To reach this weight I added object %d (%d$ %dKg) to weight %d.\n", i, a[i], last_added[i] + 1, v[last_added[i]], c[last_added[i]], i - c[last_added[i]]);

		else

			printf("Weight %d; Benefit: 0; Can't reach this exact weight.\n", i);



	printf("---\n");



	aux = W;

	while ((aux > 0) && (last_added[aux] != -1)) {

		printf("Added object %d (%d$ %dKg). Space left: %d\n", last_added[aux] + 1, v[last_added[aux]], c[last_added[aux]], aux - c[last_added[aux]]);

		aux -= c[last_added[aux]];

	}



	printf("Total value added: %d$\n", a[W]);

}



int main(int argc, char *argv[]) {

	fill_sack();



	return 0;

}

The Fractional Knapsack Problem in C

scvalex (Alexandru Scvortov) — Tue, 20 Nov 2007 13:39:13 GMT

This is the classic Greedy algorithm implementation for solving the Fractional Knapsack Problem in C.

Further explanations here



#include 



int n = 5; /* The number of objects */

int c[10] = {12, 1, 2, 1, 4}; /* c[i] is the *COST* of the ith object; i.e. what

				YOU PAY to take the object */

int v[10] = {4, 2, 2, 1, 10}; /* v[i] is the *VALUE* of the ith object; i.e.

				what YOU GET for taking the object */

int W = 15; /* The maximum weight you can take */



void simple_fill() {

	int cur_w;

	float tot_v;

	int i, maxi;

	int used[10];



	for (i = 0; i < n; ++i)

		used[i] = 0; /* I have not used the ith object yet */



	cur_w = W;

	while (cur_w > 0) { /* while there's still room*/

		/* Find the best object */

		maxi = -1;

		for (i = 0; i < n; ++i)

			if ((used[i] == 0) &&

				((maxi == -1) || ((float)v[i]/c[i] > (float)v[maxi]/c[maxi])))

				maxi = i;



		used[maxi] = 1; /* mark the maxi-th object as used */

		cur_w -= c[maxi]; /* with the object in the bag, I can carry less */

		tot_v += v[maxi];

		if (cur_w >= 0)

			printf("Added object %d (%d$, %dKg) completly in the bag. Space left: %d.\n", maxi + 1, v[maxi], c[maxi], cur_w);

		else {

			printf("Added %d%% (%d$, %dKg) of object %d in the bag.\n", (int)((1 + (float)cur_w/c[maxi]) * 100), v[maxi], c[maxi], maxi + 1);

			tot_v -= v[maxi];

			tot_v += (1 + (float)cur_w/c[maxi]) * v[maxi];

		}

	}



	printf("Filled the bag with objects worth %.2f$.\n", tot_v);

}



int main(int argc, char *argv[]) {

	simple_fill();



	return 0;

}

DZone Snippets: knapsack code

The 0-1 Knapsack Problem in C

The Fractional Knapsack Problem in C