Implement the Rod Cutting problem using dynamic programming.?

The Rod Cutting problem is a classic optimization problem in which we are given a rod of length N and a price list that contains the prices of all possible rod lengths up to N. Our task is to cut the rod into smaller pieces in such a way that we can maximize our profit.

Here’s an optimal substructure for the problem: suppose we know the optimal solution for a rod of length k, then we can easily compute the optimal solution for a rod of length N (N > k) by just considering every possible cut of length k and adding the price of that cut with the optimal solution for a rod of length (N-k).

Let’s use dynamic programming to solve the problem step by step:

1. Define the problem in terms of smaller subproblems:

For a given rod of length i, let’s say we are given an array ’price’ where price[j] represents the price for a rod of length j. We want to find the maximum revenue we can obtain by cutting up the rod and selling the pieces.

2. Define the optimal value function:

Let’s define R[i] as the maximum revenue we can obtain by cutting a rod of length i. Our task is to calculate R[N], which is the maximum revenue we can obtain by cutting a rod of length N.

3. Define the base case:

We want to start with a rod of length 0, so R[0] = 0.

4. Define the recursive function:

We can calculate R[i] for all values of i in the following way:

int[] price = {0, 1, 5, 8, 9, 10, 17, 17, 20};

int rodCutting(int N, int[] price) {
    int[] R = new int[N+1];
    R[0] = 0;

    for(int i=1; i<=N; i++) {
        int maxVal = Integer.MIN_VALUE;
        for(int j=1; j<=i; j++) {
            maxVal = Math.max(maxVal, price[j]+R[i-j]);
        }
        R[i] = maxVal;
    }

    return R[N];
}

System.out.println(rodCutting(8, price)); // Output: 22

In the above code, we are iterating through all possible cut lengths j for a rod of length i and calculating the maximum revenue we can obtain by using the optimal solution for a rod of length (i-j) and adding the revenue obtained by cutting a rod of length j.

5. Define the solution:

The solution to the problem is stored in R[N], which is the maximum revenue we can obtain by cutting a rod of length N.

In the above example, the optimal solution for a rod of length 8 is obtained by cutting it into two pieces of length 2 and 6, which gives us a revenue of 1+17 = 18 for the first piece and a revenue of 17+5 = 22 for the second. Therefore, the maximum revenue we can obtain is 22.