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Revisit the English version (#1835)

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en/docs/chapter_greedy/fractional_knapsack_problem.md
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!!! question
Given $n$ items, the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with a capacity of $cap$. Each item can be chosen only once, **but a part of the item can be selected, with its value calculated based on the proportion of the weight chosen**, what is the maximum value of the items in the knapsack under the limited capacity? An example is shown in the figure below.
Given $n$ items, where the weight of the $i$-th item is $wgt[i-1]$ and its value is $val[i-1]$, and a knapsack with capacity $cap$. Each item can be selected only once, **but a portion of an item can be selected, with the value calculated based on the proportion of weight selected**, what is the maximum value of items in the knapsack under the limited capacity? An example is shown in the figure below.
![Example data of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_example.png)
![Example data for the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_example.png)
The fractional knapsack problem is very similar overall to the 0-1 knapsack problem, involving the current item $i$ and capacity $c$, aiming to maximize the value within the limited capacity of the knapsack.
The fractional knapsack problem is very similar overall to the 0-1 knapsack problem, with states including the current item $i$ and capacity $c$, and the goal being to maximize value under the limited knapsack capacity.
The difference is that, in this problem, only a part of an item can be chosen. As shown in the figure below, **we can arbitrarily split the items and calculate the corresponding value based on the weight proportion**.
The difference is that this problem allows selecting only a portion of an item. As shown in the figure below, **we can arbitrarily split items and calculate the corresponding value based on the weight proportion**.
1. For item $i$, its value per unit weight is $val[i-1] / wgt[i-1]$, referred to as the unit value.
2. Suppose we put a part of item $i$ with weight $w$ into the knapsack, then the value added to the knapsack is $w \times val[i-1] / wgt[i-1]$.
1. For item $i$, its value per unit weight is $val[i-1] / wgt[i-1]$, referred to as unit value.
2. Suppose we put a portion of item $i$ with weight $w$ into the knapsack, then the value added to the knapsack is $w \times val[i-1] / wgt[i-1]$.
![Value per unit weight of the item](fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png)
![Value of items per unit weight](fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png)
### Greedy strategy determination
Maximizing the total value of the items in the knapsack **essentially means maximizing the value per unit weight**. From this, the greedy strategy shown in the figure below can be deduced.
Maximizing the total value of items in the knapsack **is essentially maximizing the value per unit weight of items**. From this, we can derive the greedy strategy shown in the figure below.
1. Sort the items by their unit value from high to low.
2. Iterate over all items, **greedily choosing the item with the highest unit value in each round**.
3. If the remaining capacity of the knapsack is insufficient, use part of the current item to fill the knapsack.
1. Sort items by unit value from high to low.
2. Iterate through all items, **greedily selecting the item with the highest unit value in each round**.
3. If the remaining knapsack capacity is insufficient, use a portion of the current item to fill the knapsack.
![Greedy strategy of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png)
![Greedy strategy for the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png)
### Code implementation
We have created an `Item` class in order to sort the items by their unit value. We loop and make greedy choices until the knapsack is full, then exit and return the solution:
We created an `Item` class to facilitate sorting items by unit value. We loop to make greedy selections, breaking when the knapsack is full and returning the solution:
```src
[file]{fractional_knapsack}-[class]{}-[func]{fractional_knapsack}
```
Apart from sorting, in the worst case, the entire list of items needs to be traversed, **hence the time complexity is $O(n)$**, where $n$ is the number of items.
The time complexity of built-in sorting algorithms is usually $O(\log n)$, and the space complexity is usually $O(\log n)$ or $O(n)$, depending on the specific implementation of the programming language.
Apart from sorting, in the worst case the entire item list needs to be traversed, **therefore the time complexity is $O(n)$**, where $n$ is the number of items.
Since an `Item` object list is initialized, **the space complexity is $O(n)$**.
### Correctness proof
Using proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm yields a maximum value `res`, but the solution does not include item $x$.
Using proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm yields a maximum value of `res`, but this solution does not include item $x$.
Now remove a unit weight of any item from the knapsack and replace it with a unit weight of item $x$. Since the unit value of item $x$ is the highest, the total value after replacement will definitely be greater than `res`. **This contradicts the assumption that `res` is the optimal solution, proving that the optimal solution must include item $x$**.
Now remove a unit weight of any item from the knapsack and replace it with a unit weight of item $x$. Since item $x$ has the highest unit value, the total value after replacement will definitely be greater than `res`. **This contradicts the assumption that `res` is the optimal solution, proving that the optimal solution must include item $x$**.
For other items in this solution, we can also construct the above contradiction. Overall, **items with greater unit value are always better choices**, proving that the greedy strategy is effective.
For other items in this solution, we can also construct the above contradiction. In summary, **items with greater unit value are always better choices**, which proves that the greedy strategy is effective.
As shown in the figure below, if the item weight and unit value are viewed as the horizontal and vertical axes of a two-dimensional chart respectively, the fractional knapsack problem can be transformed into "seeking the largest area enclosed within a limited horizontal axis range". This analogy can help us understand the effectiveness of the greedy strategy from a geometric perspective.
As shown in the figure below, if we view item weight and item unit value as the horizontal and vertical axes of a two-dimensional chart respectively, then the fractional knapsack problem can be transformed into "finding the maximum area enclosed within a limited horizontal axis range". This analogy can help us understand the effectiveness of the greedy strategy from a geometric perspective.
![Geometric representation of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png)