Rod Cutting Problem: Dynamic Programming Deep Dive With Java
- Rod cutting finds the maximum revenue by choosing how to cut a rod into pieces.
- dp[i] = max revenue for rod of length i. Recurrence: max over all first-cut lengths j.
- This is equivalent to unbounded knapsack where cut lengths are reusable items.
Imagine you have a chocolate bar with 10 squares and a price list: 1 square sells for $1, 2 squares sell for $5, 3 squares sell for $8, and so on. You want to break the bar and sell the pieces to make the most money possible. The rod cutting problem is exactly that — you have a rod of length N, a price for every possible piece length, and your job is to figure out the most profitable way to cut it up (or not cut it at all).
Every compiler that packs instructions into cache lines, every stock trader slicing a large order into smaller trades, every furniture manufacturer deciding how to cut timber to minimize waste — they're all solving variations of the same fundamental optimization question: given a resource you can divide, what's the most valuable partition? The rod cutting problem is the canonical formulation of that question, and it sits at the intersection of combinatorics and optimization in a way that makes it a perfect vehicle for understanding dynamic programming at depth.
What is Rod Cutting? — Plain English
The Rod Cutting problem: given a rod of length n and a price table where price[i] is the value of a piece of length i, determine the maximum revenue obtainable by cutting the rod and selling the pieces. Pieces can be sold in any combination. You choose how many cuts to make and where.
Example: a rod of length 4 with prices [0, 2, 5, 7, 8] (price[1]=2, price[2]=5, price[3]=7, price[4]=8). Selling as two pieces of length 2 yields 5+5=10, better than the unsplit price of 8.
How Rod Cutting Works — Step by Step
Define dp[i] = maximum revenue from a rod of length i.
- dp[0] = 0 (zero-length rod has zero value).
- For each length i from 1 to n:
- For each first cut at length j from 1 to i:
- dp[i] = max(dp[i], price[j] + dp[i - j])
- (price[j] is the value of selling a piece of length j; dp[i-j] is the optimal value of the remainder)
- Return dp[n].
Time: O(n^2). Space: O(n).
Worked Example — Tracing the Algorithm
Rod length n=4. Prices: price[1]=2, price[2]=5, price[3]=7, price[4]=8.
dp[0]=0.
dp[1]: try j=1: price[1]+dp[0]=2+0=2. dp[1]=2.
dp[2]: try j=1: price[1]+dp[1]=2+2=4. try j=2: price[2]+dp[0]=5+0=5. dp[2]=max(4,5)=5.
dp[3]: try j=1: price[1]+dp[2]=2+5=7. try j=2: price[2]+dp[1]=5+2=7. try j=3: price[3]+dp[0]=7+0=7. dp[3]=7.
dp[4]: try j=1: price[1]+dp[3]=2+7=9. try j=2: price[2]+dp[2]=5+5=10. try j=3: price[3]+dp[1]=7+2=9. try j=4: price[4]+dp[0]=8+0=8. dp[4]=max(9,10,9,8)=10.
Maximum revenue for length-4 rod = 10 (cut into two pieces of length 2).
Implementation
The following implementation demonstrates Rod Cutting with clear comments.
def rod_cutting(prices, n): """ prices: list where prices[i] = value of piece of length i (1-indexed). n: total rod length. """ dp = [0] * (n + 1) for i in range(1, n + 1): for j in range(1, i + 1): dp[i] = max(dp[i], prices[j] + dp[i - j]) return dp[n] prices = [0, 2, 5, 7, 8] # prices[0] unused, prices[1]=2, ..., prices[4]=8 for length in range(1, 5): print(f'Length {length}: max revenue = {rod_cutting(prices, length)}')
Length 2: max revenue = 5
Length 3: max revenue = 7
Length 4: max revenue = 10
| Concept | Use Case | Example |
|---|---|---|
| Rod Cutting Problem | Core usage | See code above |
🎯 Key Takeaways
- Rod cutting finds the maximum revenue by choosing how to cut a rod into pieces.
- dp[i] = max revenue for rod of length i. Recurrence: max over all first-cut lengths j.
- This is equivalent to unbounded knapsack where cut lengths are reusable items.
- Time: O(n^2). Space: O(n).
- Track a 'cuts' array to reconstruct which cuts produce the optimal revenue.
⚠ Common Mistakes to Avoid
Interview Questions on This Topic
- QWhat is the recurrence relation for the rod cutting DP?
- QHow is rod cutting similar to the unbounded knapsack problem?
- QHow would you reconstruct the actual cuts from the DP table?
Frequently Asked Questions
How is rod cutting related to the unbounded knapsack problem?
Rod cutting is identical to unbounded knapsack where each possible cut length is an 'item' that can be reused. The rod length is the capacity, the cut length is the item weight, and the price is the value. The DP formulation and iteration direction are the same.
How do you find the actual cuts, not just the maximum value?
Track a 'cuts' array where cuts[i] stores the first cut length that achieves dp[i]. To reconstruct: start at length n, read cuts[n] to get the first piece, subtract it, then repeat until length 0.
What if cut costs are involved?
Add a fixed cost for each cut. dp[i] = max over j from 1 to i of (prices[j] + dp[i-j] - cut_cost if i-j > 0 else prices[j]). The cut cost only applies when you actually make a cut (i.e., i-j > 0), not when you sell the whole rod unsplit.
Developer and founder of TheCodeForge. I built this site because I was tired of tutorials that explain what to type without explaining why it works. Every article here is written to make concepts actually click.