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C Program for Cutting a Rod | DP-13

  • Last Updated : 25 Jun, 2021

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)

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

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



C++




// A Naive recursive solution for Rod cutting problem
#include <limits.h>
#include <stdio.h>
 
// A utility function to get the maximum of two integers
int max(int a, int b) { return (a > b) ? a : b; }
 
/* Returns the best obtainable price for a rod of length n and
   price[] as prices of different pieces */
int cutRod(int price[], int n)
{
    if (n <= 0)
        return 0;
    int max_val = INT_MIN;
 
    // Recursively cut the rod in different pieces and compare different
    // configurations
    for (int i = 0; i < n; i++)
        max_val = max(max_val, price[i] + cutRod(price, n - i - 1));
 
    return max_val;
}
 
/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 5, 8, 9, 10, 17, 17, 20 };
    int size = sizeof(arr) / sizeof(arr[0]);
    printf("Maximum Obtainable Value is %d", cutRod(arr, size));
    getchar();
    return 0;
}


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 subproblems are called again, this problem has Overlapping Subproblems 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.

C++




// A Dynamic Programming solution for Rod cutting problem
#include <limits.h>
#include <stdio.h>
 
// A utility function to get the maximum of two integers
int max(int a, int b) { return (a > b) ? a : b; }
 
/* Returns the best obtainable price for a rod of length n and
   price[] as prices of different pieces */
int cutRod(int price[], int n)
{
    int val[n + 1];
    val[0] = 0;
    int i, j;
 
    // Build the table val[] in bottom up manner and return the last entry
    // from the table
    for (i = 1; i <= n; i++) {
        int max_val = INT_MIN;
        for (j = 0; j < i; j++)
            max_val = max(max_val, price[j] + val[i - j - 1]);
        val[i] = max_val;
    }
 
    return val[n];
}
 
/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 5, 8, 9, 10, 17, 17, 20 };
    int size = sizeof(arr) / sizeof(arr[0]);
    printf("Maximum Obtainable Value is %d", cutRod(arr, size));
    getchar();
    return 0;
}


Output:

Maximum Obtainable Value is 22

 

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




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