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README.md
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README.md
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# Cut Rod Problem — Dynamic Programming
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A complete implementation of the classic **Rod Cutting Problem** from *Introduction to Algorithms* (CLRS), with three algorithmic approaches and an interactive browser-based visualizer.
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---
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## The Problem
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You have a rod of length **n** inches and a price table that tells you how much each cut length sells for. You want to cut the rod into pieces (or keep it whole) to **maximize total revenue**.
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```
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Length: 1 2 3 4 5 6 7 8 9 10
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Price: 1 5 8 9 10 17 17 20 24 30
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```
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For a rod of length 4, the best strategy is two pieces of length 2 (5 + 5 = **$10**), not selling it whole for $9.
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This is a classic **optimal substructure** problem: the best way to cut a rod of length n depends on the best ways to cut shorter rods — making it a perfect fit for dynamic programming.
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---
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## Algorithms
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### 1. Naive Recursion — `simple_cut_rod(p, n)`
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Tries every possible first cut at position i and recurses on the remainder.
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```python
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def simple_cut_rod(p, n):
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if n == 0:
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return 0
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q = float('-inf')
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for i in range(1, n + 1):
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q = max(q, p[i] + simple_cut_rod(p, n - i))
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return q
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```
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**Time complexity:** O(2ⁿ) — exponential. Each subproblem is recomputed from scratch.
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---
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### 2. Top-Down DP with Memoization — `memoized_cut_rod(p, n)`
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Same recursion as above, but caches results in an array `r`. If `r[n]` has already been computed, return it immediately.
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```python
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def memoized_cut_rod(p, n):
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r = [float('-inf')] * (n + 1)
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return _memoized_cut_rod_aux(p, n, r)
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```
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**Time complexity:** O(n²) — each of the n subproblems is solved exactly once.
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---
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### 3. Bottom-Up DP — `bottom_up_cut_rod(p, n)`
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Fills a table `r[0..n]` iteratively from the smallest subproblem up. For each rod length j, tries every first cut i from 1 to j and stores the best result.
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```python
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def bottom_up_cut_rod(p, n):
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r = [0] * (n + 1)
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for j in range(1, n + 1):
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q = float('-inf')
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for i in range(1, j + 1):
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q = max(q, p[i] + r[j - i])
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r[j] = q
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return r[n]
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```
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**Time complexity:** O(n²) | **Space complexity:** O(n)
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This is the approach animated in the visualizer.
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---
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## Complexity Comparison
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| Approach | Time | Space | Notes |
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|---|---|---|---|
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| Naive recursion | O(2ⁿ) | O(n) call stack | Impractical for n > 30 |
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| Top-down (memoized) | O(n²) | O(n) | Natural recursive structure |
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| Bottom-up | O(n²) | O(n) | No recursion overhead, cache-friendly |
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---
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## Project Files
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```
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.
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├── cut_rod.py # Python implementation (all 3 algorithms)
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├── visualization.html # Interactive browser visualizer
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└── Cut Rod Problem/
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└── Cut Rod Problem/
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├── Program.cs # Original C# implementation
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└── Data.txt # Price table (33 prices, space-separated)
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```
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### `Data.txt` — Price Table
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```
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1 5 8 9 10 17 17 20 24 30 33 31 31 31 40 39 20 45 42 46 70 76 77 80 85 90 95 100 110 120 111 200 678
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```
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Prices for lengths 1 through 33 inches.
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---
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## Running the Python Version
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```bash
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python3 cut_rod.py
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```
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You will be prompted for a rod length. Switch between algorithms by uncommenting the desired line in `main()`:
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```python
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best_price = memoized_cut_rod(prices, n) # default
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# best_price = bottom_up_cut_rod(prices, n)
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# best_price = simple_cut_rod(prices, n) # slow for n > 25
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```
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**Example output for n = 10:**
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```
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Enter number of Inches: 10
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Best Revenue is upto : 30
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Time Elapsed : 0.000021875s
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```
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---
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## Running the Visualizer
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Open `visualization.html` directly in any modern browser — no server or dependencies needed.
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```bash
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open visualization.html # macOS
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start visualization.html # Windows
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xdg-open visualization.html # Linux
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```
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### Controls
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| Control | Action |
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|---|---|
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| Rod Length | Set the rod length (1–25). Resets the animation. |
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| Speed | Adjust playback speed (1× slow → 10× fast). |
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| ▶ Play | Run the animation automatically. |
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| ⏸ Pause | Pause at the current step. |
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| ⏭ Step | Advance one step forward. |
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| ⏮ Back | Go one step backward. |
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| ↺ Reset | Restart from the beginning. |
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---
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## Example: Rod of Length 10
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The algorithm finds that the maximum revenue is **$30**, achieved by keeping the rod uncut (selling the full 10-inch piece at $30).
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For a rod of length 7, the optimal is **3 + 4 = $17** (8 + 9 = $17, compared to selling whole for $17 — same either way, but the s-table reveals the first cut chosen).
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---
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## References
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- Cormen, T. H., Leiserson, C. E., Rivest, R. L., & Stein, C. (2009). *Introduction to Algorithms* (3rd ed.), Section 15.1 — Rod cutting.
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@ -314,6 +314,81 @@ input[type=range] { width: 110px; accent-color: var(--purple); }
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.r { color: var(--red); }
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.p { color: var(--purple); }
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.o { color: var(--orange); }
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/* ── LEGEND ──────────────────────────────────── */
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.legend {
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display: flex;
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flex-wrap: wrap;
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gap: 10px 20px;
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justify-content: center;
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max-width: 1120px;
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margin: 0 auto 20px;
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padding: 12px 20px;
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background: var(--surf);
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border: 1px solid var(--border);
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border-radius: 10px;
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font-size: .78rem;
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}
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.legend-item {
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display: flex;
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align-items: center;
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gap: 7px;
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color: var(--dim);
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}
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.swatch {
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width: 14px;
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height: 14px;
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border-radius: 3px;
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flex-shrink: 0;
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}
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.swatch-outline {
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width: 14px;
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height: 14px;
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border-radius: 3px;
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flex-shrink: 0;
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border: 2px solid;
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}
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/* ── EXPLAINER ───────────────────────────────── */
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.explainer {
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max-width: 1120px;
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margin: 20px auto 0;
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display: grid;
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grid-template-columns: repeat(3, 1fr);
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gap: 14px;
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}
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@media (max-width: 800px) {
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.explainer { grid-template-columns: 1fr; }
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}
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.exp-card {
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background: var(--surf);
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border: 1px solid var(--border);
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border-radius: 12px;
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padding: 18px 20px;
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}
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.exp-card h3 {
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font-size: .8rem;
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text-transform: uppercase;
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letter-spacing: .08em;
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color: var(--dim);
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margin-bottom: 12px;
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}
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.exp-card p, .exp-card li {
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font-size: .83rem;
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line-height: 1.65;
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color: #a0aab4;
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}
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.exp-card ul { padding-left: 16px; }
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.exp-card li { margin-bottom: 5px; }
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.exp-card code {
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font-family: 'SFMono-Regular', 'Consolas', monospace;
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font-size: .8rem;
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background: var(--bg);
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padding: 1px 5px;
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border-radius: 4px;
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color: var(--orange);
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}
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.exp-card .accent { color: var(--text); font-weight: 600; }
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</style>
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</head>
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<body>
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<button class="btn btn-secondary" id="btn-reset">↺ Reset</button>
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</div>
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<!-- LEGEND -->
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<div class="legend">
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<div class="legend-item">
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<div class="swatch" style="background:var(--red)"></div>
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<span>Cut piece — sells for <strong style="color:var(--text)">p[i]</strong></span>
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</div>
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<div class="legend-item">
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<div class="swatch" style="background:var(--blue)"></div>
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<span>Remainder — best revenue already known as <strong style="color:var(--text)">r[j−i]</strong></span>
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</div>
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<div class="legend-item">
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<div class="swatch-outline" style="border-color:var(--yellow);background:#2b2400"></div>
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<span><strong style="color:var(--yellow)">Yellow cell</strong> — currently being computed</span>
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</div>
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<div class="legend-item">
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<div class="swatch" style="background:#122620"></div>
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<span><strong style="color:var(--green)">Green cell</strong> — finalized value</span>
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</div>
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<div class="legend-item">
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<div class="swatch" style="background:#0d1f36"></div>
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<span><strong style="color:var(--blue)">Blue cell</strong> — value being read (r[j−i])</span>
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</div>
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<div class="legend-item">
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<div class="swatch" style="background:#1a1030"></div>
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<span><strong style="color:var(--purple)">Purple cell</strong> — optimal first cut stored in s[j]</span>
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</div>
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</div>
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<div class="layout">
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<!-- LEFT COLUMN -->
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@ -395,6 +498,47 @@ input[type=range] { width: 110px; accent-color: var(--purple); }
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</div>
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<!-- EXPLAINER -->
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<div class="explainer">
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<div class="exp-card">
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<h3>The Problem</h3>
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<p>
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You have a rod of length <span class="accent">n</span> inches and a price table
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<code>p[1..n]</code> where <code>p[i]</code> is what a piece of length <code>i</code> sells for.
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You can cut the rod into any combination of integer lengths. The goal is to
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<span class="accent">maximize total revenue</span>.
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</p>
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<br>
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<p>
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A rod of length 4 sold whole earns $9, but two pieces of length 2 earn $5 + $5 = <span class="accent">$10</span>.
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Finding the best combination by brute force takes O(2ⁿ) time — dynamic programming reduces this to <span class="accent">O(n²)</span>.
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</p>
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</div>
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<div class="exp-card">
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<h3>How the Algorithm Works</h3>
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<ul>
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<li><span class="accent">Build up from small rods.</span> Compute the best revenue for length 1, then 2, then 3 … up to n. Each answer reuses earlier answers.</li>
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<li><span class="accent">For each length j</span>, try every possible first cut at position <code>i = 1 … j</code>. The candidate revenue is <code>p[i] + r[j−i]</code>: sell the first piece and optimally cut the rest.</li>
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<li><span class="accent">Store the best.</span> <code>r[j]</code> holds the maximum over all cuts. <code>s[j]</code> records which cut produced that maximum — used later to reconstruct the actual pieces.</li>
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<li><span class="accent">Reconstruct.</span> Follow <code>s[n] → s[n−s[n]] → …</code> to read off the optimal cut sequence.</li>
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</ul>
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</div>
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<div class="exp-card">
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<h3>Reading the Visualization</h3>
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<ul>
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<li><span class="accent">Rod bar:</span> the red segment is the piece being sold now (length <code>i</code>); the blue segment is the remainder whose best revenue is already in the table.</li>
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<li><span class="accent">DP table row r[j]:</span> fills left-to-right. The yellow cell is being computed right now; blue cells are the ones being looked up.</li>
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<li><span class="accent">DP table row s[j]:</span> once a cell is finalized (purple), it tells you the optimal first cut for that rod length.</li>
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<li><span class="accent">Pseudocode:</span> the highlighted line shows exactly which line of the algorithm the current step corresponds to.</li>
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<li><span class="accent">Optimal Solution</span> (shown at the end): the colored bar divides the rod into its best pieces, each labeled with its length and sale price.</li>
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</ul>
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</div>
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</div>
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<script>
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// ── DATA ────────────────────────────────────────
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const PRICE_DATA = [0,1,5,8,9,10,17,17,20,24,30,33,31,31,31,40,39,20,45,42,46,70,76,77,80,85,90,95,100,110,120,111,200,678];
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