Greedy Algorithm - Knapsack Problem
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The document discusses the knapsack problem and greedy algorithms. It defines the knapsack problem as an optimization problem where given constraints and an objective function, the goal is to find the feasible solution that maximizes or minimizes the objective. It describes the knapsack problem has having two versions: 0-1 where items are indivisible, and fractional where items can be divided. The fractional knapsack problem can be solved using a greedy approach by sorting items by value to weight ratio and filling the knapsack accordingly until full.
![Initialization:
Sort the n objects from large to small based on their ratios vi / wi .
We assume the arrays w[1..n] and v[1..n] store the respective
weights and values after sorting.
initialize array x[1..n] to zeros.
weight = 0; i = 1;
THE OPTIMAL KNAPSACK ALGORITHM](https://image.slidesharecdn.com/greedy-170403085712/85/Greedy-Algorithm-Knapsack-Problem-11-320.jpg)
![while (i n and weight < W) do
if weight + w[i] W then
x[i] = 1
else
x[i] = (W – weight) / w[i]
weight = weight + x[i] * w[i]
i++
THE OPTIMAL KNAPSACK ALGORITHM](https://image.slidesharecdn.com/greedy-170403085712/85/Greedy-Algorithm-Knapsack-Problem-12-320.jpg)
Greedy Algorithm - Knapsack Problem
- 2. OPTIMIZATION PROBLEM (Cont.) An optimization problem: Given a problem instance, a set of constraints and an objective function. Find a feasible solution for the given instance. either maximum or minimum depending on the problem being solved. constraints specify the limitations on the required solutions.
- 3. SOLUTION FOR OPTIMIZATION PROBLEM For some optimization problems, • Dynamic Programming is “overkill” • Greedy Strategy is simpler and more efficient .
- 4. DYNAMIC PROGRAMMING VS GREEDY Dynamic Programming Greedy Algorithm At each step, the choice is determined based on solutions of sub problems. At each step, we quickly make a choice that currently looks best. A local optimal (greedy) choice Sub-problems are solved first. Greedy choice can be made first before solving further sub- problems. Bottom-up approach Top-down approach Can be slower, more complex Usually faster, simpler
- 5. Characteristics of greedy algorithm: make a sequence of choices each choice is the one that seems best so far, only depends on what's been done so far choice produces a smaller problem to be solved GREEDY METHOD
- 6. PHASES OF GREEDY ALGORITHM A greedy algorithm works in phases. At each phase: takes the best solution right now, without regard for future consequences choosing a local optimum at each step, and end up at a global optimum solution.
- 7. KNAPSACK PROBLEM There are two version of knapsack problem 1. 0-1 knapsack problem: Items are indivisible. (either take an item or not) can be solved with dynamic programming. 2. Fractional knapsack problem: Items are divisible. (can take any fraction of an item) It can be solved in greedy method
- 8. 0-1 KNAPSACK PROBLEM: A thief robbing a store finds n items. ith item: worth vi value of item and wi weight of item W, wi, vi are integers. He can carry at most W pounds.
- 9. FRACTIONAL KNAPSACK PROBLEM: A thief robbing a store finds n items. ith item: worth vi value of item and wi weight of item W, wi, vi are integers. He can carry at most W pounds. He can take fractions of items.
- 10. THE OPTIMAL KNAPSACK ALGORITHM Input: an integer n positive values wi and vi such that 1 i n positive value W. Output: n values of xi such that 0 xi 1 Total profit
- 11. Initialization: Sort the n objects from large to small based on their ratios vi / wi . We assume the arrays w[1..n] and v[1..n] store the respective weights and values after sorting. initialize array x[1..n] to zeros. weight = 0; i = 1; THE OPTIMAL KNAPSACK ALGORITHM
- 12. while (i n and weight < W) do if weight + w[i] W then x[i] = 1 else x[i] = (W – weight) / w[i] weight = weight + x[i] * w[i] i++ THE OPTIMAL KNAPSACK ALGORITHM
- 13. KNAPSACK - EXAMPLE Problem: n = 3 W= 20 (v1, v2, v3) = (25, 24, 15) (w1, w2, w3) = (18, 15, 10)
- 14. Solution: Optimal solution: x1 = 0 x2 = 1 x3 = 1/2 Total profit = 24 + 7.5 = 31.5 KNAPSACK - EXAMPLE
- 15. THANK YOU