Find which items are taken in knapsack - c++

Question

solving the knapsack problem recursively, I want to know which items (item's weight) are taken in the bag that gives the maximum value.
so far I have this:
int MAX(int a, int b) { return (a > b) ? a : b ; }
int thief(int W, int weight[], int value[], int n)
{
int a,b,c;
//basecase:
if(n == 0 || weight <= 0) return 0;
// each item's weight can't be more than W:
if(weight[n-1] > W){
return thief(W, weight, value, n-1);}
a=value[n-1] + thief(W-weight[n-1], weight, value, n-1);// a: nth item included
b=thief(W, weight, value, n-1);// b:nth item not included
c= MAX(a,b);//answer is the maximum of situation a and b
if (c==a) { //if situation a occurs then nth item is included
cout<<weight[n]<<endl;
}
return c;
}
consider n=4 and maximum weight (W) = 30
let weights be : 30 10 20 5
and values : 100 50 60 10
but this code outputs: 20 5 20 10 5
I just want to output 10 and 20.
also I have tried to define a bool array with default values of false and its nth element changes to true if c==a occurs but this won't give the correct result as well.
I'm supposed to do it recursively.

Answer 1

Your basic algorithm doesn't work. You can't do the printing while you test different combinations.
However, first you must fix a bug:
cout<<weight[n-1]<<endl; // n-1 instead of n
Your algorithm does this:
a = value[3] + thief(30-weight[3], weight, value, 3); // Use item 3
b = thief(30, weight, value, 3); // Don't use item 3
The second line will lead to
a = value[2] + thief(30-weight[2], weight, value, 2); // Use item 2
b = thief(30, weight, value, 2); // Don't use item 2
The second line will lead to
a = value[1] + thief(30-weight[1], weight, value, 1); // Use item 1
b = thief(30, weight, value, 1); // Don't use item 1
The second line will lead to
a = value[0] + thief(30-weight[0], weight, value, 0); // Use item 0
b = thief(30, weight, value, 0); // Don't use item 0
This causes
a = 30
b = 0
so your code will select item 0 and print 30 but that is a bug!
So as I stated in the start: You can't do the printing while you test different combinations.
Instead you need to keep track of which elements you use in the different combinations and only keep the "best".
I haven't tested the code below but I think you can do it like this (assuming your code calculates the best combination correctly):
#include <vector>
// The vector v is used for holding the index of the items selected.
// The caller must supply a vector containing the items included so far.
// This function will test whether item "n-1" shall be included or
// excluded. If item "n-1" is included the index is added to the vector.
int thief(int W, int weight[], int value[], int n, vector<int>& v) // Added vector
{
vector<int> v1, v2; // Vector to hold elements of the two combinations
int a,b,c;
//basecase:
if(n == 0 || weight <= 0) return 0;
// each item's weight can't be more than W:
if(weight[n-1] > W){
return thief(W, weight, value, n-1, v2);}
v1.push_back(n-1); // Put n-1 in vector v1 and pass the vector v1
a=value[n-1] + thief(W-weight[n-1], weight, value, n-1, v1);// a: nth item included
// Don't put anything in v2 but pass the vector v2
b=thief(W, weight, value, n-1, v2);// b:nth item not included
c= MAX(a,b);//answer is the maximum of situation a and b
if (c==a) { //if situation a occurs then nth item is included
// cout<<weight[n-1]<<endl;
// Copy elements from v1 to v
for (auto e : v1)
{
v.push_back(e);
}
}
else
{
// Copy elements from v2 to v
for (auto e : v2)
{
v.push_back(e);
}
}
return c;
}
int main() {
vector<int> v;
int weight[4] = {30, 10, 20, 5};
int value[4] = {100, 50, 60, 10};
cout << "result=" << thief(30, weight, value, 4, v) << endl;
// Print the elements used
for (auto e : v)
{
cout << "elem=" << e << endl;
}
return 0;
}
Finally notice - your brute force method is very expensive in terms of execution time in the start value for n is high. There are much better ways to solve this problem.

Answer 2

can you write the code in c?
I wrote this but didn't work. (I think the difference is in the bold)
int knapSack(int W, int wt[], int val[], int n, int arr[])
{
int x, y, c, j, arr1[50], arr2[50];
// Base Case
if (n == 0 || W == 0)
return 0;
// If weight of the nth item is more than Knapsack capacity W, then
// this item cannot be included in the optimal solution
if (wt[n - 1] > W)
return knapSack(W, wt, val, n - 1, arr2);
// Return the maximum of two cases:
// x nth item included
// y not included
**arr1[n - 1] = val[n - 1];**
x = val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1, arr1);
//copyArr(arr, out, n);
y = knapSack(W, wt, val, n - 1, arr2);
if (x > y)
c = x;
else
c = y;
if (c == x)
for (j = 0; j < 50; j++)
arr[j] = arr1[j];
else
for (j = 0; j < 50; j++)
arr[j] = arr2[j];
return c;
}