1. Overview
The Rod Cutting Problem is a classic optimization problem that involves finding the best way to cut a rod into pieces to maximize the total revenue.
In this tutorial, we’ll understand the Rod Cutting Problem and explore various approaches to solving it in Java.
2. Understanding the Problem
Imagine we’ve got a rod of length n. We’ve got the flexibility to cut this rod into various lengths and sell those cut segments. Furthermore, we possess a table of prices for different lengths of cut rods. Our goal is to maximize the total revenue.
For example, consider a rod of length n=4 with prices Pi = [1,5,8,9]. The Pi array denotes the price of a rod piece of length i. It means:
P1 = 1 shows the price of a rod piece of length 1 is 1 unit.
P2 = 5 shows the price of a rod piece of length 2 is 5 units.
Similarly, P3 = 8 shows the price of a rod piece of length 3 is 8 units, and
P4 = 9 shows the price of a rod piece of length 4 is 9 units.
We can make various cuts to the rod, and each cut will result in a certain revenue based on the prices of the rod pieces. Here are the possible cuts for a rod of length 4:
- Cut into 4 pieces of length 1m each [1,1,1,1], resulted in revenue = 1+1+1+1 = $4
- Cut into 2 pieces of length 2m each [2,2], resulted in revenue = 5+5 = $10
- Cut into 1 piece of length 4m, resulted in revenue = $9
- Cut into 3 pieces of length 1m and 2m [1,1,2], resulted in revenue = 1+1+5 = $7
- Cut into 3 pieces of length 1m and 2m [1,2,1], resulted in revenue = 1+5+1 = $7
- Cut into 3 pieces of length 1m and 2m [2,1,1], resulted in revenue= 5+1+1 = $7
Here, we can see that by cutting our rod into 2 pieces of length 2m each, we get a total price of $10. Every other combination gives a lower price, so we can see that this is the optimal answer.
3. Recursive Solution
The recursive solution involves decomposing the problem into subproblems and solving them recursively. The recurrence relation used to calculate the maximum revenue for a rod of length n is Rn = max1<=i<=n(Pi+Rn-i).
In the above relation, Rn represents the maximum revenue for a rod of length n, and Pi is the price of a rod piece of length i. Let’s implement the recursive solution:
int usingRecursion(int[] prices, int n) {
if (n <= 0) {
return 0;
}
int maxRevenue = Integer.MIN_VALUE;
for (int i = 1; i <= n; i++) {
maxRevenue = Math.max(maxRevenue, prices[i - 1] + usingRecursion(prices, n - i));
}
return maxRevenue;
}
In our example, we check the base case to determine if the rod’s length is equal to or less than 0, resulting in a revenue of 0. We systematically explore all potential cuts using recursion to calculate the maximum revenue for each cut. The result is the highest revenue achieved among all the cuts.
Now, let’s test this approach:
@Test
void givenARod_whenUsingRecursion_thenFindMaxRevenue() {
int[] prices = {1, 5, 8, 9};
int rodLength = 4;
int maxRevenue = RodCuttingProblem.usingRecursion(prices, rodLength);
assertEquals(10, maxRevenue);
}
The recursive approach works and is relatively easy to understand. However, its exponential time complexity O(2n) makes it impractical for larger instances, as recursive calls will rapidly increase. So, if we’ve got a 4m rod, this approach will take 24 = 16 iterations.
This solution doesn’t directly store the state in a separate data structure like an array or a map. However, it utilizes the call stack provided by the language to keep track of function classes and their respective parameters and local variables.
The lack of memorization in this approach results in redundant computations, as identical subproblems are solved multiple times during the recursive exploration. This inefficiency becomes a significant concern for realistic scenarios with rods of considerable lengths. Recursive solutions can also lead to stack overflow errors if the recursion depth is too large.
4. Memoized Recursive Solution
We can improve our recursive solution by the use of memoization. In this approach, we’ll use a memoization table to store and reuse the results of previously solved subproblems:
int usingMemoizedRecursion(int[] prices, int n) {
int[] memo = new int[n + 1];
Arrays.fill(memo, -1);
return memoizedHelper(prices, n, memo);
}
int memoizedHelper(int[] prices, int n, int[] memo) {
if (n <= 0) {
return 0;
}
if (memo[n] != -1) {
return memo[n];
}
int maxRevenue = Integer.MIN_VALUE;
for (int i = 1; i <= n; i++) {
maxRevenue = Math.max(maxRevenue, prices[i - 1] + memoizedHelper(prices, n - i, memo));
}
memo[n] = maxRevenue;
return maxRevenue;
}
The idea is to avoid redundant computations by checking whether a subproblem has already been solved before solving it again. Our memoization table takes the form of an array, where each index corresponds to the rod length, and the associated value holds the maximum revenue for that specific length.
Now, let’s test this approach:
@Test
void givenARod_whenUsingMemoizedRecursion_thenFindMaxRevenue() {
int[] prices = {1, 5, 8, 9};
int rodLength = 4;
int maxRevenue = RodCuttingProblem.usingMemoizedRecursion(prices, rodLength);
assertEquals(10, maxRevenue);
}
By avoiding the redundant calculations through memoization, the time complexity is significantly reduced compared to the naive recursive solution. Now, the time complexity is determined by the number of unique subproblems solved, and for the Rod Cutting problem, it is often O(n2) due to the two nested loops.
This means for a rod of length 10m, we’ll have a time complexity of 100 here, compared to 1024 before, because we’re doing notably less work due to the trimming of nodes. However, this comes at a cost in space complexity. The algorithm must store up to one value for each rod length, giving it a space complexity of O(n).
5. Dynamic Programming Solution
Another approach to solving this problem is via dynamic programming. Dynamic programming solves a problem by dividing it into smaller subproblems. This is very similar to the divide-and-conquer algorithm-solving technique. The major difference, however, is that dynamic programming solves a subproblem only once.
5.1. Iterative (Bottom-Up) Approach
In this approach, avoiding recursion and using an iterative process eliminates the function call overhead associated with the recursive solutions, contributing to enhanced performance:
int usingDynamicProgramming(int[] prices, int n) {
int[] dp = new int[n + 1];
for (int i = 1; i <= n; i++) {
int maxRevenue = Integer.MIN_VALUE;
for (int j = 1; j <= i; j++) {
maxRevenue = Math.max(maxRevenue, prices[j - 1] + dp[i - j]);
}
dp[i] = maxRevenue;
}
return dp[n];
}
Let’s test this approach as well:
@Test
void givenARod_whenUsingDynamicProgramming_thenFindMaxRevenue() {
int[] prices = {1, 5, 8, 9};
int rodLength = 4;
int maxRevenue = RodCuttingProblem.usingDynamicProgramming(prices, rodLength);
assertEquals(10, maxRevenue);
}
Our dynamic programming table, represented by the dp array, stores the optimal solutions for each subproblem, where the index i corresponds to the rod length, and dp[i] holds the maximum revenue. The final entry in the dp array, dp[n], contains the maximum revenue for the original problem of a rod with length n.
This approach is also memory-efficient as it doesn’t require an additional memoization table.
5.2. Unbounded Knapsack Solution
The unbounded knapsack approach is a dynamic programming technique used to solve optimization problems where an item can be chosen unlimited times. This means that the same cut length can be selected and included in the rod multiple times if it contributes to maximizing the revenue. Let’s take a look at the implementation:
int usingUnboundedKnapsack(int[] prices, int n) {
int[] dp = new int[n + 1];
for (int i = 1; i <= n; i++) {
for (int j = 0; j < prices.length; j++) {
if (j + 1 <= i) {
dp[i] = Math.max(dp[i], dp[i - (j + 1)] + prices[j]);
}
}
}
return dp[n];
}
Similar to the iterative (bottom-up) approach, this algorithm computes the maximum revenue for each rod length. Let’s test this approach:
@Test
void givenARod_whenUsingGreedy_thenFindMaxRevenue() {
int[] prices = {1, 5, 8, 9};
int rodLength = 4;
int maxRevenue = RodCuttingProblem.usingUnboundedKnapsack(prices, rodLength);
assertEquals(10, maxRevenue);
}
The drawback of this approach is the potential for increased space complexity.
6. Conclusion
In this tutorial, we’ve understood the Rod Cutting Problem and discussed various ways to solve it.
The recursive approach offers simplicity but suffers from exponential time complexity. The memoization method addresses this by reusing solutions and introducing slight space complexity. The iterative approach further improves efficiency, eliminating recursion overhead.
The choice between these approaches depends on specific problem characteristics, involving trade-offs in time, space, and implementation complexity.
As always, the code used in the examples is available over on GitHub.