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Cutting a Rod | DP-13
  • Difficulty Level : Medium
  • Last Updated : 27 May, 2021
 

Given a rod of length n inches and an array of prices that includes 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 the length of the rod is 8 and the values of different pieces are given as the 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
 

A naive solution to this problem is to generate all configurations of different pieces and find the highest-priced configuration. This solution is exponential in terms of time complexity. Let us see how this problem possesses both important properties of a Dynamic Programming (DP) Problem and can efficiently be solved using Dynamic Programming.
1) Optimal Substructure: 
We can get the best price by making a cut at different positions and comparing the values obtained after a cut. We can recursively call the same function for a piece obtained after a cut.
Let cutRod(n) be the required (best possible price) value for a rod of length n. cutRod(n) can be written as follows.
cutRod(n) = max(price[i] + cutRod(n-i-1)) for all i in {0, 1 .. n-1}
2) Overlapping Subproblems 
The following is a 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<stdio.h>
#include<limits.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 %dn", cutRod(arr, size));
    getchar();
    return 0;
}

Java




// // A Naive recursive solution for Rod cutting problem
class RodCutting
{
    /* 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 = Integer.MIN_VALUE;
 
        // 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 program to test above functions */
    public static void main(String args[])
    {
        int arr[] = new int[] {1, 5, 8, 9, 10, 17, 17, 20};
        int size = arr.length;
        System.out.println("Maximum Obtainable Value is "+
                            cutRod(arr, size));
 
    }
}
/* This code is contributed by Rajat Mishra */

Python




# A Naive recursive solution
# for Rod cutting problem
import sys
 
# A utility function to get the
# maximum of two integers
def max(a, b):
    return a if (a > b) else b
     
# Returns the best obtainable price for a rod of length n
# and price[] as prices of different pieces
def cutRod(price, n):
    if(n <= 0):
        return 0
    max_val = -sys.maxsize-1
     
    # Recursively cut the rod in different pieces 
    # and compare different configurations
    for i in range(0, n):
        max_val = max(max_val, price[i] +
                      cutRod(price, n - i - 1))
    return max_val
 
# Driver code
arr = [1, 5, 8, 9, 10, 17, 17, 20]
size = len(arr)
print("Maximum Obtainable Value is", cutRod(arr, size))
 
# This code is contributed by 'Smitha Dinesh Semwal'

C#




// 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

PHP




<?php
// A Naive recursive solution for
// Rod cutting problem
 
/* Returns the best obtainable
   price for a rod of length n and
   price[] as prices of different
   pieces */
function cutRod( $price, $n)
{
 
    if ($n <= 0)
        return 0;
    $max_val = PHP_INT_MIN;
     
    // Recursively cut the rod in different
    // pieces and compare different
    // configurations
    for ($i = 0; $i < $n; $i++)
            $max_val = max($max_val, $price[$i] +
                    cutRod($price, $n - $i - 1));
     
    return $max_val;
}
 
    // Driver Code
    $arr = array(1, 5, 8, 9, 10, 17, 17, 20);
    $size =count($arr);
    echo "Maximum Obtainable Value is "
               , cutRod($arr, $size);
 
// This code is contributed anuj_67.
?>

Javascript




<script>
 
    // A Naive recursive solution for
    // Rod cutting problem
     
    /* Returns the best obtainable
       price for a rod of length
       n and price[] as prices of
       different pieces */
    function cutRod(price, n)
    {
        if (n <= 0)
            return 0;
        let max_val = Number.MIN_VALUE;
   
        // Recursively cut the rod in
        // different pieces and compare
        // different configurations
        for (let i = 0; i < n; i++)
            max_val = Math.max(max_val, price[i] +
                        cutRod(price, n - i - 1));
   
        return max_val;
    }
     
    let arr = [1, 5, 8, 9, 10, 17, 17, 20];
    let size = arr.length;
    document.write("Maximum Obtainable Value is "+ cutRod(arr, size));
                                
</script>
Output



Maximum Obtainable Value is 22n

Considering the above implementation, the following is the 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 solved twice. We can see that there are many subproblems that are solved again and again. Since the same subproblems are called again, this problem has the 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 the same subproblems can be avoided by constructing a temporary array val[] in a bottom-up manner. 

C++




// A Dynamic Programming solution for Rod cutting problem
#include<stdio.h>
#include<limits.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 %dn", cutRod(arr, size));
    getchar();
    return 0;
}

Java




// A Dynamic Programming solution for Rod cutting problem
class RodCutting
{
    /* 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 = Integer.MIN_VALUE;
            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 program to test above functions */
    public static void main(String args[])
    {
        int arr[] = new int[] {1, 5, 8, 9, 10, 17, 17, 20};
        int size = arr.length;
        System.out.println("Maximum Obtainable Value is " +
                            cutRod(arr, size));
    }
}
/* This code is contributed by Rajat Mishra */

Python




# A Dynamic Programming solution for Rod cutting problem
INT_MIN = -32767
 
# Returns the best obtainable price for a rod of length n and
# price[] as prices of different pieces
def cutRod(price, n):
    val = [0 for x in range(n+1)]
    val[0] = 0
 
    # Build the table val[] in bottom up manner and return
    # the last entry from the table
    for i in range(1, n+1):
        max_val = INT_MIN
        for j in range(i):
             max_val = max(max_val, price[j] + val[i-j-1])
        val[i] = max_val
 
    return val[n]
 
# Driver program to test above functions
arr = [1, 5, 8, 9, 10, 17, 17, 20]
size = len(arr)
print("Maximum Obtainable Value is " + str(cutRod(arr, size)))
 
# This code is contributed by Bhavya Jain

C#




// 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

PHP




<?php
// A Dynamic Programming solution for
// Rod cutting problem
 
/* Returns the best obtainable price
for a rod of length n and price[] as
prices of different pieces */
function cutRod( $price, $n)
{
    $val = array();
    $val[0] = 0;
    $i; $j;
     
    // Build the table val[] in bottom
    // up manner and return the last
    // entry from the table
    for ($i = 1; $i <= $n; $i++)
    {
        $max_val = PHP_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
$arr = array(1, 5, 8, 9, 10, 17, 17, 20);
$size = count($arr);
echo "Maximum Obtainable Value is ",
                      cutRod($arr, $size);
     
// This code is contributed by anuj_67.
?>

Javascript




<script>
    // A Dynamic Programming solution
    // for Rod cutting problem
     
    /* Returns the best obtainable
       price for a rod of length n
       and price[] as prices of
       different pieces */
    function cutRod(price, n)
    {
        let val = new Array(n + 1);
        val[0] = 0;
  
        // Build the table val[] in
        // bottom up manner and return
        // the last entry from the table
        for (let i = 1; i<=n; i++)
        {
            let max_val = Number.MIN_VALUE;
            for (let j = 0; j < i; j++)
                max_val = Math.max(max_val, price[j] + val[i - j - 1]);
            val[i] = max_val;
        }
  
        return val[n];
    }
     
    let arr = [1, 5, 8, 9, 10, 17, 17, 20];
    let size = arr.length;
    document.write("Maximum Obtainable Value is " + cutRod(arr, size) + "n");
</script>
Output
Maximum Obtainable Value is 22n

The Time Complexity of the above implementation is O(n^2), which is much better than the worst-case time complexity of Naive Recursive implementation.

3) Using the idea of Unbounded Knapsack.

This problem is very similar to the Unbounded Knapsack Problem, where there are multiple occurrences of the same item. Here the pieces of the rod.

Now I will create an analogy between Unbounded Knapsack and the Rod Cutting Problem. 


C++






// CPP program for above approach
#include <iostream>
using namespace std;
 
// Global Array for
// the purpose of memoization.
int t[9][9];
 
// A recursive program, using ,
// memoization, to implement the
// rod cutting problem(Top-Down).
int un_kp(int price[], int length[],
                    int Max_len, int n)
{
 
    // The maximum price will be zero,
    // when either the length of the rod
    // is zero or price is zero.
    if (n == 0 || Max_len == 0)
    {
        return 0;
    }
 
    // If the length of the rod is less
    // than the maximum length, Max_lene will
    // consider it.Now depending
    // upon the profit,
    // either Max_lene  we will take
    // it or discard it.
    if (length[n - 1] <= Max_len)
    {
        t[n][Max_len]
            = max(price[n - 1]
                      + un_kp(price, length,
                           Max_len - length[n - 1], n),
                  un_kp(price, length, Max_len, n - 1));
    }
 
    // If the length of the rod is
    // greater than the permitted size,
    // Max_len we will  not consider it.
    else
    {
        t[n][Max_len]
            = un_kp(price, length,
                              Max_len, n - 1);
    }
 
    // Max_lene Max_lenill return the maximum
    // value obtained, Max_lenhich is present
    // at the nth roMax_len and Max_lenth column.
    return t[n][Max_len];
}
 
/* Driver program to
test above functions */
int main()
{
    int price[] = { 1, 5, 8, 9, 10, 17, 17, 20 };
    int n = sizeof(price) / sizeof(price[0]);
    int length[n];
    for (int i = 0; i < n; i++) {
        length[i] = i + 1;
    }
    int Max_len = n;
 
    // Function Call
    cout << "Maximum obtained value  is "
         << un_kp(price, length, n, Max_len) << endl;
}

C




// C program for above approach
#include <stdio.h>
#include <stdlib.h>
 
int max(int a, int b)
{
  return (a > b) ? a : b;
}
 
// Global Array for the
// purpose of memoization.
int t[9][9];
 
// A recursive program, using ,
// memoization, to implement the
// rod cutting problem(Top-Down).
int un_kp(int price[], int length[],
                     int Max_len, int n)
{
 
    // The maximum price will be zero,
    // when either the length of the rod
    // is zero or price is zero.
    if (n == 0 || Max_len == 0)
    {
        return 0;
    }
 
    // If the length of the rod is less
    // than the maximum length, Max_lene
    // will consider it.Now depending
    // upon the profit,
    // either Max_lene  we will take it
    // or discard it.
    if (length[n - 1] <= Max_len)
    {
        t[n][Max_len]
            = max(price[n - 1]
                      + un_kp(price, length,
                              Max_len - length[n - 1], n),
                  un_kp(price, length, Max_len, n - 1));
    }
 
    // If the length of the rod is greater
    // than the permitted size, Max_len
    // we will  not consider it.
    else
    {
        t[n][Max_len]
            = un_kp(price, length,
                             Max_len, n - 1);
    }
 
    // Max_lene Max_lenill return
    // the maximum value obtained,
    // Max_lenhich is present at the
    // nth roMax_len and Max_lenth column.
    return t[n][Max_len];
}
 
/* Driver program to test above functions */
int main()
{
    int price[] = { 1, 5, 8, 9, 10, 17, 17, 20 };
    int n = sizeof(price) / sizeof(price[0]);
    int length[n];
    for (int i = 0; i < n; i++)
    {
        length[i] = i + 1;
    }
    int Max_len = n;
 
    // Function Call
    printf("Maximum obtained value  is %d \n",
           un_kp(price, length, n, Max_len));
}

Javascript




<script>
 
// Javascript program for above approach
 
// Global Array for
// the purpose of memoization.
let t = new Array(9);
for (var i = 0; i < t.length; i++) {
    t[i] = new Array(2);
}
 
// A recursive program, using ,
// memoization, to implement the
// rod cutting problem(Top-Down).
function un_kp(price, length, Max_len, n)
{
 
    // The maximum price will be zero,
    // when either the length of the rod
    // is zero or price is zero.
    if (n == 0 || Max_len == 0)
    {
        return 0;
    }
 
    // If the length of the rod is less
    // than the maximum length, Max_lene will
    // consider it.Now depending
    // upon the profit,
    // either Max_lene  we will take
    // it or discard it.
    if (length[n - 1] <= Max_len)
    {
        t[n][Max_len]
            = Math.max(price[n - 1]
                      + un_kp(price, length,
                           Max_len - length[n - 1], n),
                  un_kp(price, length, Max_len, n - 1));
    }
 
    // If the length of the rod is
    // greater than the permitted size,
    // Max_len we will  not consider it.
    else
    {
        t[n][Max_len]
            = un_kp(price, length,
                              Max_len, n - 1);
    }
 
    // Max_lene Max_lenill return the maximum
    // value obtained, Max_lenhich is present
    // at the nth roMax_len and Max_lenth column.
    return t[n][Max_len];
}
     
    // Driver code
     
    let price = [ 1, 5, 8, 9, 10, 17, 17, 20 ];
    let n = price.length;
    let length = Array(n).fill(0);
    for (let i = 0; i < n; i++) {
        length[i] = i + 1;
    }
    let Max_len = n;
 
    // Function Call
    document.write("Maximum obtained value  is "
         + un_kp(price, length, n, Max_len));
 
</script>

 
 

 

 

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