build

en/docs/chapter_backtracking/backtracking_algorithm.md

@@ -538,7 +538,7 @@ Based on the code from Example One, we need to use a list `path` to record the v
 
 In each "try", we record the path by adding the current node to `path`; before "retreating", we need to pop the node from `path` **to restore the state before this attempt**.
 
-Observe the process shown below, **we can understand trying and retreating as "advancing" and "undoing"**, two operations that are reverse to each other.
+Observe the process shown in Figure 13-2, **we can understand trying and retreating as "advancing" and "undoing"**, two operations that are reverse to each other.
 
 === "<1>"
     ![Trying and retreating](backtracking_algorithm.assets/preorder_find_paths_step1.png){ class="animation-figure" }
@@ -1887,7 +1887,7 @@ Compared to the implementation based on preorder traversal, the code implementat
 
 ## 13.1.4 &nbsp; Common terminology
 
-To analyze algorithmic problems more clearly, we summarize the meanings of commonly used terminology in backtracking algorithms and provide corresponding examples from Example Three as shown in the Table 13-1 .
+To analyze algorithmic problems more clearly, we summarize the meanings of commonly used terminology in backtracking algorithms and provide corresponding examples from Example Three as shown in Table 13-1.
 
 <p align="center"> Table 13-1 &nbsp; Common backtracking algorithm terminology </p>
 

en/docs/chapter_backtracking/n_queens_problem.md

@@ -8,13 +8,13 @@ comments: true
 
     According to the rules of chess, a queen can attack pieces in the same row, column, or on a diagonal line. Given $n$ queens and an $n \times n$ chessboard, find arrangements where no two queens can attack each other.
 
-As shown in the Figure 13-15 , when $n = 4$, there are two solutions. From the perspective of the backtracking algorithm, an $n \times n$ chessboard has $n^2$ squares, presenting all possible choices `choices`. The state of the chessboard `state` changes continuously as each queen is placed.
+As shown in Figure 13-15, when $n = 4$, there are two solutions. From the perspective of the backtracking algorithm, an $n \times n$ chessboard has $n^2$ squares, presenting all possible choices `choices`. The state of the chessboard `state` changes continuously as each queen is placed.
 
 ![Solution to the 4 queens problem](n_queens_problem.assets/solution_4_queens.png){ class="animation-figure" }
 
 <p align="center"> Figure 13-15 &nbsp; Solution to the 4 queens problem </p>
 
-The following image shows the three constraints of this problem: **multiple queens cannot be on the same row, column, or diagonal**. It is important to note that diagonals are divided into the main diagonal `\` and the secondary diagonal `/`.
+Figure 13-16 shows the three constraints of this problem: **multiple queens cannot be on the same row, column, or diagonal**. It is important to note that diagonals are divided into the main diagonal `\` and the secondary diagonal `/`.
 
 ![Constraints of the n queens problem](n_queens_problem.assets/n_queens_constraints.png){ class="animation-figure" }
 
@@ -26,7 +26,7 @@ As the number of queens equals the number of rows on the chessboard, both being
 
 This means that we can adopt a row-by-row placing strategy: starting from the first row, place one queen per row until the last row is reached.
 
-The image below shows the row-by-row placing process for the 4 queens problem. Due to space limitations, the image only expands one search branch of the first row, and prunes any placements that do not meet the column and diagonal constraints.
+Figure 13-17 shows the row-by-row placing process for the 4 queens problem. Due to space limitations, the figure only expands one search branch of the first row, and prunes any placements that do not meet the column and diagonal constraints.
 
 ![Row-by-row placing strategy](n_queens_problem.assets/n_queens_placing.png){ class="animation-figure" }
 
@@ -40,7 +40,7 @@ To satisfy column constraints, we can use a boolean array `cols` of length $n$ t
 
 How about the diagonal constraints? Let the row and column indices of a cell on the chessboard be $(row, col)$. By selecting a specific main diagonal, we notice that the difference $row - col$ is the same for all cells on that diagonal, **meaning that $row - col$ is a constant value on that diagonal**.
 
-Thus, if two cells satisfy $row_1 - col_1 = row_2 - col_2$, they are definitely on the same main diagonal. Using this pattern, we can utilize the array `diags1` shown below to track whether a queen is on any main diagonal.
+Thus, if two cells satisfy $row_1 - col_1 = row_2 - col_2$, they are definitely on the same main diagonal. Using this pattern, we can utilize the array `diags1` shown in Figure 13-18 to track whether a queen is on any main diagonal.
 
 Similarly, **the sum $row + col$ is a constant value for all cells on a secondary diagonal**. We can also use the array `diags2` to handle secondary diagonal constraints.
 

en/docs/chapter_backtracking/permutations_problem.md

@@ -6,13 +6,13 @@ comments: true
 
 The permutation problem is a typical application of the backtracking algorithm. It is defined as finding all possible arrangements of elements from a given set (such as an array or string).
 
-The Table 13-2  lists several example data, including the input arrays and their corresponding permutations.
+Table 13-2 lists several example data, including the input arrays and their corresponding permutations.
 
 <p align="center"> Table 13-2 &nbsp; Permutation examples </p>
 
 <div class="center-table" markdown>
 
-| Input array | Permutations                                                   |
+| Input array | Permutations                                                       |
 | :---------- | :----------------------------------------------------------------- |
 | $[1]$       | $[1]$                                                              |
 | $[1, 2]$    | $[1, 2], [2, 1]$                                                   |
@@ -30,7 +30,7 @@ From the perspective of the backtracking algorithm, **we can imagine the process
 
 From the code perspective, the candidate set `choices` contains all elements of the input array, and the state `state` contains elements that have been selected so far. Please note that each element can only be chosen once, **thus all elements in `state` must be unique**.
 
-As shown in the following figure, we can unfold the search process into a recursive tree, where each node represents the current state `state`. Starting from the root node, after three rounds of choices, we reach the leaf nodes, each corresponding to a permutation.
+As shown in Figure 13-5, we can unfold the search process into a recursive tree, where each node represents the current state `state`. Starting from the root node, after three rounds of choices, we reach the leaf nodes, each corresponding to a permutation.
 
 ![Permutation recursive tree](permutations_problem.assets/permutations_i.png){ class="animation-figure" }
 
@@ -43,13 +43,13 @@ To ensure that each element is selected only once, we consider introducing a boo
 - After making the choice `choice[i]`, we set `selected[i]` to $\text{True}$, indicating it has been chosen.
 - When iterating through the choice list `choices`, skip all nodes that have already been selected, i.e., prune.
 
-As shown in the following figure, suppose we choose 1 in the first round, 3 in the second round, and 2 in the third round, we need to prune the branch of element 1 in the second round and elements 1 and 3 in the third round.
+As shown in Figure 13-6, suppose we choose 1 in the first round, 3 in the second round, and 2 in the third round, we need to prune the branch of element 1 in the second round and elements 1 and 3 in the third round.
 
 ![Permutation pruning example](permutations_problem.assets/permutations_i_pruning.png){ class="animation-figure" }
 
 <p align="center"> Figure 13-6 &nbsp; Permutation pruning example </p>
 
-Observing the above figure, this pruning operation reduces the search space size from $O(n^n)$ to $O(n!)$.
+Observing Figure 13-6, this pruning operation reduces the search space size from $O(n^n)$ to $O(n!)$.
 
 ### 2. &nbsp; Code implementation
 
@@ -532,7 +532,7 @@ After understanding the above information, we can "fill in the blanks" in the fr
 
 Suppose the input array is $[1, 1, 2]$. To differentiate the two duplicate elements $1$, we mark the second $1$ as $\hat{1}$.
 
-As shown in the following figure, half of the permutations generated by the above method are duplicates.
+As shown in Figure 13-7, half of the permutations generated by the above method are duplicates.
 
 ![Duplicate permutations](permutations_problem.assets/permutations_ii.png){ class="animation-figure" }
 
@@ -542,7 +542,7 @@ So, how do we eliminate duplicate permutations? Most directly, consider using a
 
 ### 1. &nbsp; Pruning of equal elements
 
-Observing the following figure, in the first round, choosing $1$ or $\hat{1}$ results in identical permutations under both choices, thus we should prune $\hat{1}$.
+Observing Figure 13-8, in the first round, choosing $1$ or $\hat{1}$ results in identical permutations under both choices, thus we should prune $\hat{1}$.
 
 Similarly, after choosing $2$ in the first round, choosing $1$ and $\hat{1}$ in the second round also produces duplicate branches, so we should also prune $\hat{1}$ in the second round.
 
@@ -1061,7 +1061,7 @@ Please note, although both `selected` and `duplicated` are used for pruning, the
 - **Repeated choice pruning**: There is only one `selected` throughout the search process. It records which elements are currently in the state, aiming to prevent an element from appearing repeatedly in `state`.
 - **Equal element pruning**: Each round of choices (each call to the `backtrack` function) contains a `duplicated`. It records which elements have been chosen in the current traversal (`for` loop), aiming to ensure equal elements are selected only once.
 
-The following figure shows the scope of the two pruning conditions. Note, each node in the tree represents a choice, and the nodes from the root to the leaf form a permutation.
+Figure 13-9 shows the scope of the two pruning conditions. Note, each node in the tree represents a choice, and the nodes from the root to the leaf form a permutation.
 
 ![Scope of the two pruning conditions](permutations_problem.assets/permutations_ii_pruning_summary.png){ class="animation-figure" }
 

en/docs/chapter_backtracking/subset_sum_problem.md

@@ -490,7 +490,7 @@ Unlike the permutation problem, **elements in this problem can be chosen an unli
 
 Inputting the array $[3, 4, 5]$ and target element $9$ into the above code yields the results $[3, 3, 3], [4, 5], [5, 4]$. **Although it successfully finds all subsets with a sum of $9$, it includes the duplicate subset $[4, 5]$ and $[5, 4]$**.
 
-This is because the search process distinguishes the order of choices, however, subsets do not distinguish the choice order. As shown in the following figure, choosing $4$ before $5$ and choosing $5$ before $4$ are different branches, but correspond to the same subset.
+This is because the search process distinguishes the order of choices, however, subsets do not distinguish the choice order. As shown in Figure 13-10, choosing $4$ before $5$ and choosing $5$ before $4$ are different branches, but correspond to the same subset.
 
 ![Subset search and pruning out of bounds](subset_sum_problem.assets/subset_sum_i_naive.png){ class="animation-figure" }
 
@@ -503,7 +503,7 @@ To eliminate duplicate subsets, **a straightforward idea is to deduplicate the r
 
 ### 2.   Duplicate subset pruning
 
-**We consider deduplication during the search process through pruning**. Observing the following figure, duplicate subsets are generated when choosing array elements in different orders, for example in the following situations.
+**We consider deduplication during the search process through pruning**. Observing Figure 13-11, duplicate subsets are generated when choosing array elements in different orders, for example in the following situations.
 
 1. When choosing $3$ in the first round and $4$ in the second round, all subsets containing these two elements are generated, denoted as $[3, 4, \dots]$.
 2. Later, when $4$ is chosen in the first round, **the second round should skip $3$** because the subset $[4, 3, \dots]$ generated by this choice completely duplicates the subset from step `1.`.
@@ -1029,7 +1029,7 @@ Besides, we have made the following two optimizations to the code.
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     <div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20backtrack%28%0A%20%20%20%20state%3A%20list%5Bint%5D,%20target%3A%20int,%20choices%3A%20list%5Bint%5D,%20start%3A%20int,%20res%3A%20list%5Blist%5Bint%5D%5D%0A%29%3A%0A%20%20%20%20%22%22%22%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%EF%BC%9A%E5%AD%90%E9%9B%86%E5%92%8C%20I%22%22%22%0A%20%20%20%20%23%20%E5%AD%90%E9%9B%86%E5%92%8C%E7%AD%89%E4%BA%8E%20target%20%E6%97%B6%EF%BC%8C%E8%AE%B0%E5%BD%95%E8%A7%A3%0A%20%20%20%20if%20target%20%3D%3D%200%3A%0A%20%20%20%20%20%20%20%20res.append%28list%28state%29%29%0A%20%20%20%20%20%20%20%20return%0A%20%20%20%20%23%20%E9%81%8D%E5%8E%86%E6%89%80%E6%9C%89%E9%80%89%E6%8B%A9%0A%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%BA%8C%EF%BC%9A%E4%BB%8E%20start%20%E5%BC%80%E5%A7%8B%E9%81%8D%E5%8E%86%EF%BC%8C%E9%81%BF%E5%85%8D%E7%94%9F%E6%88%90%E9%87%8D%E5%A4%8D%E5%AD%90%E9%9B%86%0A%20%20%20%20for%20i%20in%20range%28start,%20len%28choices%29%29%3A%0A%20%20%20%20%20%20%20%20%23%20%E5%89%AA%E6%9E%9D%E4%B8%80%EF%BC%9A%E8%8B%A5%E5%AD%90%E9%9B%86%E5%92%8C%E8%B6%85%E8%BF%87%20target%20%EF%BC%8C%E5%88%99%E7%9B%B4%E6%8E%A5%E7%BB%93%E6%9D%9F%E5%BE%AA%E7%8E%AF%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%99%E6%98%AF%E5%9B%A0%E4%B8%BA%E6%95%B0%E7%BB%84%E5%B7%B2%E6%8E%92%E5%BA%8F%EF%BC%8C%E5%90%8E%E8%BE%B9%E5%85%83%E7%B4%A0%E6%9B%B4%E5%A4%A7%EF%BC%8C%E5%AD%90%E9%9B%86%E5%92%8C%E4%B8%80%E5%AE%9A%E8%B6%85%E8%BF%87%20target%0A%20%20%20%20%20%20%20%20if%20target%20-%20choices%5Bi%5D%20%3C%200%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20break%0A%20%20%20%20%20%20%20%20%23%20%E5%B0%9D%E8%AF%95%EF%BC%9A%E5%81%9A%E5%87%BA%E9%80%89%E6%8B%A9%EF%BC%8C%E6%9B%B4%E6%96%B0%20target,%20start%0A%20%20%20%20%20%20%20%20state.append%28choices%5Bi%5D%29%0A%20%20%20%20%20%20%20%20%23%20%E8%BF%9B%E8%A1%8C%E4%B8%8B%E4%B8%80%E8%BD%AE%E9%80%89%E6%8B%A9%0A%20%20%20%20%20%20%20%20backtrack%28state,%20target%20-%20choices%5Bi%5D,%20choices,%20i,%20res%29%0A%20%20%20%20%20%20%20%20%23%20%E5%9B%9E%E9%80%80%EF%BC%9A%E6%92%A4%E9%94%80%E9%80%89%E6%8B%A9%EF%BC%8C%E6%81%A2%E5%A4%8D%E5%88%B0%E4%B9%8B%E5%89%8D%E7%9A%84%E7%8A%B6%E6%80%81%0A%20%20%20%20%20%20%20%20state.pop%28%29%0A%0A%0Adef%20subset_sum_i%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20list%5Blist%5Bint%5D%5D%3A%0A%20%20%20%20%22%22%22%E6%B1%82%E8%A7%A3%E5%AD%90%E9%9B%86%E5%92%8C%20I%22%22%22%0A%20%20%20%20state%20%3D%20%5B%5D%20%20%23%20%E7%8A%B6%E6%80%81%EF%BC%88%E5%AD%90%E9%9B%86%EF%BC%89%0A%20%20%20%20nums.sort%28%29%20%20%23%20%E5%AF%B9%20nums%20%E8%BF%9B%E8%A1%8C%E6%8E%92%E5%BA%8F%0A%20%20%20%20start%20%3D%200%20%20%23%20%E9%81%8D%E5%8E%86%E8%B5%B7%E5%A7%8B%E7%82%B9%0A%20%20%20%20res%20%3D%20%5B%5D%20%20%23%20%E7%BB%93%E6%9E%9C%E5%88%97%E8%A1%A8%EF%BC%88%E5%AD%90%E9%9B%86%E5%88%97%E8%A1%A8%EF%BC%89%0A%20%20%20%20backtrack%28state,%20target,%20nums,%20start,%20res%29%0A%20%20%20%20return%20res%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20nums%20%3D%20%5B3,%204,%205%5D%0A%20%20%20%20target%20%3D%209%0A%20%20%20%20res%20%3D%20subset_sum_i%28nums,%20target%29%0A%0A%20%20%20%20print%28f%22%E8%BE%93%E5%85%A5%E6%95%B0%E7%BB%84%20nums%20%3D%20%7Bnums%7D,%20target%20%3D%20%7Btarget%7D%22%29%0A%20%20%20%20print%28f%22%E6%89%80%E6%9C%89%E5%92%8C%E7%AD%89%E4%BA%8E%20%7Btarget%7D%20%E7%9A%84%E5%AD%90%E9%9B%86%20res%20%3D%20%7Bres%7D%22%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=16&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
 
-The following figure shows the overall backtracking process after inputting the array $[3, 4, 5]$ and target element $9$ into the above code.
+Figure 13-12 shows the overall backtracking process after inputting the array $[3, 4, 5]$ and target element $9$ into the above code.
 
 ![Subset sum I backtracking process](subset_sum_problem.assets/subset_sum_i.png){ class="animation-figure" }
 
@@ -1043,7 +1043,7 @@ The following figure shows the overall backtracking process after inputting the
 
 Compared to the previous question, **this question's input array may contain duplicate elements**, introducing new problems. For example, given the array $[4, \hat{4}, 5]$ and target element $9$, the existing code's output results in $[4, 5], [\hat{4}, 5]$, resulting in duplicate subsets.
 
-**The reason for this duplication is that equal elements are chosen multiple times in a certain round**. In the following figure, the first round has three choices, two of which are $4$, generating two duplicate search branches, thus outputting duplicate subsets; similarly, the two $4$s in the second round also produce duplicate subsets.
+**The reason for this duplication is that equal elements are chosen multiple times in a certain round**. In Figure 13-13, the first round has three choices, two of which are $4$, generating two duplicate search branches, thus outputting duplicate subsets; similarly, the two $4$s in the second round also produce duplicate subsets.
 
 ![Duplicate subsets caused by equal elements](subset_sum_problem.assets/subset_sum_ii_repeat.png){ class="animation-figure" }
 
@@ -1616,7 +1616,7 @@ At the same time, **this question stipulates that each array element can only be
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-The following figure shows the backtracking process for the array $[4, 4, 5]$ and target element $9$, including four types of pruning operations. Please combine the illustration with the code comments to understand the entire search process and how each type of pruning operation works.
+Figure 13-14 shows the backtracking process for the array $[4, 4, 5]$ and target element $9$, including four types of pruning operations. Please combine the illustration with the code comments to understand the entire search process and how each type of pruning operation works.
 
 ![Subset sum II backtracking process](subset_sum_problem.assets/subset_sum_ii.png){ class="animation-figure" }