Knapsack Calculator | 0-1 Knapsack with DP Table & Fractional Comparison
Solve the 0-1 knapsack optimization problem using dynamic programming. Enter items with weights and values and a capacity; the calculator builds the complete DP table, traces back the optimal item selection, reports total value and weight, and compares the result with the fractional knapsack greedy solution.
PRESETS
ITEMS (UP TO 10)
| # | Name | Weight | Value | V/W ratio | Action |
|---|---|---|---|---|---|
| 1 | 3.00 | ||||
| 2 | 5.00 | ||||
| 3 | 4.00 | ||||
| 4 | 2.60 |
What Is the Knapsack Calculator | 0-1 Knapsack with DP Table & Fractional Comparison?
Solve the 0-1 knapsack problem with dynamic programming. Enter items with integer weights and values and a capacity; the calculator builds the full DP table, traces back the optimal item selection, and compares the result with the fractional knapsack greedy solution.
Formula
dp[i][w] = max(dp[i−1][w], dp[i−1][w−wᵢ] + vᵢ) if w ≥ wᵢ, else dp[i−1][w]
How to Use
- 1
Click a preset to load a classic example, or clear the table and enter your own items.
- 2
For each item, type a name, integer weight, and integer value into the table cells.
- 3
Use the + Add item button to add more items (up to 10) or × to remove any row.
- 4
Set the capacity W (integer 1–100) in the capacity field.
- 5
Results update automatically: optimal value, selected items highlighted in colour, weight used vs capacity.
- 6
Toggle Show DP table to see the full dynamic programming table with values updated by each item row.
- 7
Compare the 0-1 optimal value with the fractional knapsack (greedy) value shown in the result cards.
Enter items (name, weight, value) and a capacity, then read the optimal value, selected items, and DP table.
Example Calculation
Simple 4-item problem (capacity 10): A(w=2, v=6), B(w=2, v=10), C(w=3, v=12), D(w=5, v=13). DP gives optimal = 22 by selecting B+C (w=2+3=5, v=10+12=22) or B+D (w=7, v=23). Fractional would take all of B, all of C, and 5/5 of D for the same integer result here.
Understanding Knapsack | 0-1 Knapsack with DP Table & Fractional Comparison
DP Recurrence Explanation
The 0-1 knapsack builds a table dp[i][w] = maximum value achievable using the first i items with weight capacity w. Each cell is filled in O(1) by considering two choices.
| Condition | Formula | Meaning |
|---|---|---|
| w < weight[i] | dp[i][w] = dp[i−1][w] | Item i is too heavy to include at capacity w — skip it |
| w ≥ weight[i] | dp[i][w] = max(dp[i−1][w], dp[i−1][w−weight[i]] + value[i]) | Take the better of: skip item i, or include item i |
| Base case i=0 | dp[0][w] = 0 for all w | No items → zero value regardless of capacity |
| Traceback | dp[i][w] ≠ dp[i−1][w] → item i was selected | Walk back from dp[n][W] to identify chosen items |
Time complexity: O(n × W). Space complexity: O(n × W) for full table (can be reduced to O(W) for value-only output).
0-1 Knapsack vs Fractional Knapsack
| Property | 0-1 Knapsack | Fractional Knapsack |
|---|---|---|
| Item divisibility | Items must be taken whole (0 or 1) | Items can be taken in fractions |
| Algorithm | Dynamic programming | Greedy (sort by v/w ratio) |
| Time complexity | O(n × W) | O(n log n) |
| Optimal value | Always ≤ fractional solution | Always ≥ 0-1 solution |
| Real-world analogy | Indivisible goods: laptops, paintings | Divisible goods: gold dust, grain |
| Solution approach | Exact optimal via DP table | Greedy gives exact optimal |
| NP-hardness | NP-hard in general case | Solvable in polynomial time |
Applications of Knapsack Optimization
- ›Resource allocation: assign a fixed budget or CPU time to projects, each with a cost and expected return, to maximise total return.
- ›Portfolio optimisation: select a subset of investment opportunities within a capital constraint to maximise expected value or return.
- ›Cargo loading: choose which freight containers to load onto a truck or ship with a weight limit to maximise revenue.
- ›Cloud computing: pack virtual machines onto physical servers with limited RAM and CPU to maximise utilisation or revenue.
- ›Cutting stock problem: cut raw materials (pipes, fabric rolls) to minimise waste — related to bin packing and knapsack variants.
- ›Cryptography: the knapsack problem (with superincreasing sequences) was an early public-key cryptosystem proposed by Merkle and Hellman in 1978.
- ›Bioinformatics: fragment assembly and sequence alignment problems share structural similarity with knapsack and edit-distance DP.
Frequently Asked Questions
Why is the 0-1 knapsack NP-hard?
The problem is NP-hard because there is no known polynomial-time algorithm for all inputs. The DP solution is pseudo-polynomial — it runs in O(n×W) time, which is polynomial in the numeric value of W but exponential in the number of bits needed to represent W.
How does the DP traceback work?
Start at cell dp[n][W]. If dp[n][W] ≠ dp[n−1][W], item n was selected: mark it, subtract its weight, and move to dp[n−1][W−wₙ]. If dp[n][W] = dp[n−1][W], item n was not selected: move to dp[n−1][W]. Repeat until reaching row 0.
Why is the fractional knapsack value always ≥ the 0-1 value?
The fractional problem relaxes the integrality constraint, allowing partial items. This larger feasible region can only produce a higher or equal optimal value. The gap between the two solutions measures how much "integrality cost" exists in your specific instance.
What happens when two items have the same value/weight ratio?
The greedy fractional algorithm breaks ties arbitrarily (first encountered in the sorted order). The 0-1 DP is unaffected — it considers all combinations exactly. Both solutions are still optimal within their respective problem formulations.
Can the DP handle large capacities?
This calculator supports capacities up to 100 and up to 10 items for interactive performance. In production systems with large W, a space-optimised rolling array (O(W) space) or branch-and-bound algorithms are used. For very large W, approximation schemes (FPTAS) provide (1−ε)-optimal solutions in polynomial time.
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