Python Program for Fractional Knapsack Problem

Given the weights and profits of N items, in the form of {profit, weight} put these items in a knapsack of capacity W to get the maximum total profit in the knapsack. In Fractional Knapsack, we can break items to maximize the total value of the knapsack.

Example :

Input: arr[] = {{60, 10}, {100, 20}, {120, 30}}, W = 50
Output: 240
Explanation: By taking items of weight 10 and 20 kg and 2/3 fraction of 30 kg.
Hence total price will be 60+100+(2/3)(120) = 240

Input: arr[] = {{500, 30}}, W = 10
Output: 166.667

Python Program for Fractional Knapsack Problem

Below, are the examples of Python programs for the Fractional Knapsack Problem.

• Using Greedy Algorithm
• Using Dynamic Programming

Fractional Knapsack Problem Using Greedy Algorithm

Illustration

Consider the example: arr[] = {{100, 20}, {60, 10}, {120, 30}}, W = 50.

Sorting: Initially sort the array based on the profit/weight ratio. The sorted array will be {{60, 10}, {100, 20}, {120, 30}}.

• For i = 0, weight = 10 which is less than W. So add this element in the knapsack. profit = 60 and remaining W = 50 – 10 = 40.
• For i = 1, weight = 20 which is less than W. So add this element too. profit = 60 + 100 = 160 and remaining W = 40 – 20 = 20.
• For i = 2, weight = 30 is greater than W. So add 20/30 fraction = 2/3 fraction of the element. Therefore profit = 2/3 * 120 + 160 = 80 + 160 = 240 and remaining W becomes 0.
  
  So the final profit becomes 240 for W = 50.

Follow the given steps to solve the problem using the above approach:

• Figure out the ratio of profit to weight for each item.
• Arrange all the items from highest to lowest ratio.
• Start with an empty result and the total capacity you have.
• For each item in the sorted order:
  • If you can fit the whole item into the remaining capacity, add its value to the result.
  • Otherwise, add as much of the item as you can and stop.
• Finally, return the total result.

Python3

class Item:
    def __init__(self, weight, value):
        self.weight = weight
        self.value = value
        # Calculate the value-to-weight ratio for each item
        self.ratio = value / weight  

def fractional_knapsack(items, capacity):
    # Sort items by their value-to-weight ratio in non-increasing order
    items.sort(key=lambda x: x.ratio, reverse=True)

    total_value = 0 
    
    # Initialize remaining capacity of the knapsack
    remaining_capacity = capacity  

    # Iterate through the sorted items list
    for item in items:
        if remaining_capacity <= 0:
            break

        # Calculate fraction of the item that fits into the knapsack
        fraction = min(1, remaining_capacity / item.weight)

        # Update total value and remaining capacity
        total_value += fraction * item.value
        remaining_capacity -= fraction * item.weight

    # Return total maximum value obtained
    return round(total_value, 2)  

# Example usage
items = [
    Item(10, 60),
    Item(20, 100),
    Item(30, 120)
]
capacity = 50
print("Maximum Value that can be Obtained is", fractional_knapsack(items, capacity))

Output

Maximum value that can be obtained is 240.0

Time Complexity: O(n log n + n). O(n log n )
Space Complexity: O(1)

Fractional Knapsack Problem Using Dynamic Programming

Illustration

Consider the example: arr[] = {{100, 20}, {60, 10}, {120, 30}}, W = 50.

• Arrange items based on their value-to-weight ratio: {60, 10}, {100, 20}, {120, 30}.
• Start with first item: Its weight is 10, less than the capacity (50). Add it to the knapsack, profit becomes 60, and remaining capacity is 40.
• Move to next item: Weight is 20, still within remaining capacity. Add it, profit becomes 160, and remaining capacity is 20.
• Last item: Its weight exceeds the remaining capacity. Add a fraction (2/3) of its weight to fill the knapsack. Profit becomes 240, and the knapsack is full (remaining capacity is 0).
  
  So the final profit becomes 240 for W = 50

Follow the given steps to solve the problem using the above approach:

• It defines a class Item with variable weight and value to represent each item.
• The function fractional_knapsack_dynamic takes a list of items and the knapsack capacity as input.
• It initializes a dynamic programming array to store the maximum value achieved for each capacity.
• It iterates through each capacity and each item, updating the maximum value achievable.
• Finally, it returns the maximum profit achieved using dynamic programming.

Python3

class Item:
    def __init__(self, weight, value):
        self.weight = weight
        self.value = value

def fractional_knapsack_dynamic(items, capacity):
    n = len(items)
    
    # Initialize dp array to store maximum value achieved for each capacity
    dp = [0] * (capacity + 1)  

    # Iterate from 1 to capacity
    for i in range(1, capacity + 1):
        max_value = 0
        # Iterate through each item
        for j in range(n):
            if items[j].weight <= i:
                # Update max_value with the maximum value achieved for the current capacity
                max_value = max(max_value, dp[i - items[j].weight] + items[j].value)
   # Update dp array with the maximum value achieved for current capacity
        dp[i] = max_value  

   # Return maximum profit achieved using dynamic programming
    return dp[capacity]  

# Example usage
items = [
    Item(10, 60),
    Item(20, 100),
    Item(30, 120)
]
capacity = 50
print("Maximum Value Obtained Using Dynamic Approach is", fractional_knapsack_dynamic(items, capacity))

Output

Maximum value obtained using dynamic approach is 300

Time Complexity: O(n * capacity)
Space Complexity: O(capacity)