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en/docs/chapter_greedy/fractional_knapsack_problem.md

@@ -6,7 +6,7 @@ comments: true
 
 !!! 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 below.
+    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 Figure 15-3.
 
 ![Example data of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_example.png){ class="animation-figure" }
 
@@ -14,7 +14,7 @@ comments: true
 
 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 difference is that, in this problem, only a part of an item can be chosen. As shown in the Figure 15-4 , **we can arbitrarily split the items and calculate the corresponding value based on the weight proportion**.
+The difference is that, in this problem, only a part of an item can be chosen. As shown in Figure 15-4, **we can arbitrarily split the 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 $w \times val[i-1] / wgt[i-1]$.
@@ -25,7 +25,7 @@ The difference is that, in this problem, only a part of an item can be chosen. A
 
 ### 1.   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 below can be deduced.
+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 Figure 15-5 can be deduced.
 
 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**.
@@ -526,7 +526,7 @@ Now remove a unit weight of any item from the knapsack and replace it with a uni
 
 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.
 
-As shown in the Figure 15-6 , 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 Figure 15-6, 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.
 
 ![Geometric representation of the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_area_chart.png){ class="animation-figure" }