Dynamic Programming - Rod Cutting Problem

Objective: Given a rod of length n inches and a table of prices p_i, i=1,2,…,n, write an algorithm to find the maximum revenue r_n obtainable by cutting up the rod and selling the pieces.

This is very good basic problem after fibonacci sequence if you are new to Dynamic programming. We will see how the dynamic programming is used to overcome the issues with recursion(Time Complexity).

Given: Rod lengths are integers and For i=1,2,…,n we know the price p_i of a rod of length i inches

Example:

Length	1	2	3	4	5	6	7	8	9	10
Price	1	5	8	9	10	17	17	20	24	30

for rod of length: 4
Ways to sell :
•	selling length : 4  ( no cuts) , Price: 9
•	selling length : 1,1,1,1  ( 3 cuts) , Price: 4
•	selling length : 1,1,2  ( 2 cuts) , Price: 7
•	selling length : 2,2  ( 1 cut) , Price: 10
•	selling length : 3, 1  ( 1 cut) , Price: 9

Best Price for rod of length 4: 10

Approach:

Naive Approach : Recursion

There can be n-1 cuts can be made in the rod of length n, so there are 2^n-1ways to cut the rod.
So for every length we have 2 options either we cut it or not. we will consider both the options and choose the optimal out of it.

Code:

	public class RodCuttingRecursion {
	public static int profit(int[] value, int length) {
	if (length <= 0)
	return 0;
	// either we will cut it or don't cut it
	int max = –1;
	for(int i=0;i<length;i++)
	max = Math.max(max, value[i] + profit(value, length–(i+1)));
	return max;
	}
	public static void main(String[] args) {
	int[] value = { 2, 3, 7, 8, 9 };
	int len = 5;
	System.out.println("Max profit for length is " + len + ":"
	+ profit(value, len));
	}
	}

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RodCuttingRecursion.java

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Time Complexity: O(2^n-1)
But this time complexity is very high since we are solving many sub problems repeatedly.

Rod Cutting Problem - Overlapping Sub problems — Rod Cutting Problem – Overlapping Sub problems

Dynamic Programming: Bottom-Up

Instead of solving the sub problems repeatedly we can store the results of it in an array and use it further rather than solving it again.

See the Code for better explanation:

Code:

	public class RodCutting {

	public static int profitDP(int[] value, int length) {
	int[] solution = new int[length + 1];
	solution[0] = 0;

	for (int i = 1; i <= length; i++) {
	int max = –1;
	for (int j = 0; j < i; j++) {
	max = Math.max(max, value[j] + solution[i – (j + 1)]);
	solution[i] = max;
	}
	}
	return solution[length];
	}

	public static void main(String[] args) {
	int[] value = { 2, 3, 7, 8, 9 };
	int len = 5;
	System.out.println("Max profit for length is " + len + ":"
	+ profitDP(value, len));

	}

	}

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RodCutting.java

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Result: Max profit for length is 5:11

Reference: Link

Dynamic Programming – Rod Cutting Problem

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