C# Program for Cutting a Rod | DP-13

Given a rod of length n inches and an array of prices that contains prices of all pieces of size smaller than n. Determine the maximum value obtainable by cutting up the rod and selling the pieces. For example, if length of the rod is 8 and the values of different pieces are given as following, then the maximum obtainable value is 22 (by cutting in two pieces of lengths 2 and 6)


length   | 1   2   3   4   5   6   7   8  
--------------------------------------------
price    | 1   5   8   9  10  17  17  20

And if the prices are as following, then the maximum obtainable value is 24 (by cutting in eight pieces of length 1)

length   | 1   2   3   4   5   6   7   8  
--------------------------------------------
price    | 3   5   8   9  10  17  17  20

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Following is simple recursive implementation of the Rod Cutting problem. The implementation simply follows the recursive structure mentioned above.

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// A Naive recursive solution for
// Rod cutting problem
using System;
class GFG {
  
    /* Returns the best obtainable 
       price for a rod of length
       n and price[] as prices of 
       different pieces */
    static int cutRod(int[] price, int n)
    {
        if (n <= 0)
            return 0;
        int max_val = int.MinValue;
  
        // Recursively cut the rod in
        // different pieces and compare
        // different configurations
        for (int i = 0; i < n; i++)
            max_val = Math.Max(max_val, price[i] + cutRod(price, n - i - 1));
  
        return max_val;
    }
  
    // Driver Code
    public static void Main()
    {
        int[] arr = new int[] { 1, 5, 8, 9, 10, 17, 17, 20 };
        int size = arr.Length;
        Console.WriteLine("Maximum Obtainable Value is " + cutRod(arr, size));
    }
}
  
// This code is contributed by Sam007

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Output:

Maximum Obtainable Value is 22

Considering the above implementation, following is recursion tree for a Rod of length 4.

cR() ---> cutRod() 

                             cR(4)
                  /        /           
                 /        /              
             cR(3)       cR(2)     cR(1)   cR(0)
            /  |         /         |
           /   |        /          |  
      cR(2) cR(1) cR(0) cR(1) cR(0) cR(0)
     /        |          |
    /         |          |   
  cR(1) cR(0) cR(0)      cR(0)
   /
 /
CR(0)

In the above partial recursion tree, cR(2) is being solved twice. We can see that there are many subproblems which are solved again and again. Since same suproblems are called again, this problem has Overlapping Subprolems property. So the Rod Cutting problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by constructing a temporary array val[] in bottom up manner.

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// A Dynamic Programming solution
// for Rod cutting problem
using System;
class GFG {
  
    /* Returns the best obtainable 
       price for a rod of length n
       and price[] as prices of 
       different pieces */
    static int cutRod(int[] price, int n)
    {
        int[] val = new int[n + 1];
        val[0] = 0;
  
        // Build the table val[] in
        // bottom up manner and return
        // the last entry from the table
        for (int i = 1; i <= n; i++) {
            int max_val = int.MinValue;
            for (int j = 0; j < i; j++)
                max_val = Math.Max(max_val,
                                   price[j] + val[i - j - 1]);
            val[i] = max_val;
        }
  
        return val[n];
    }
  
    // Driver Code
    public static void Main()
    {
        int[] arr = new int[] { 1, 5, 8, 9, 10, 17, 17, 20 };
        int size = arr.Length;
        Console.WriteLine("Maximum Obtainable Value is " + cutRod(arr, size));
    }
}
  
// This code is contributed by Sam007

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Output:

Maximum Obtainable Value is 22

Please refer complete article on Cutting a Rod | DP-13 for more details!



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