xinyin025/hello-algo
1
0
Fork 0
mirror of https://github.com/krahets/hello-algo.git synced 2026-08-11 19:30:59 +00:00
Code Issues Packages Projects Releases Wiki Activity

Bug fixes and improvements (#1348)

Browse Source 
* Add "reference" for EN version. Bug fixes.

* Unify the figure reference as "the figure below" and "the figure above".
Bug fixes.

* Format the EN markdown files.

* Replace "" with <u></u> for EN version and bug fixes

* Fix biary_tree_dfs.png

* Fix biary_tree_dfs.png

* Fix zh-hant/biary_tree_dfs.png

* Fix heap_sort_step1.png

* Sync zh and zh-hant versions.

* Bug fixes

* Fix EN figures

* Bug fixes

* Fix the figure labels for EN version
This commit is contained in:
Yudong Jin
2024-05-06 14:44:48 +08:00
committed by GitHub
parent 8e60d12151
commit c4a7966882
99 changed files with 615 additions and 259 deletions
Download Patch File Download Diff File Expand all files Collapse all files
en/docs/chapter_backtracking/n_queens_problem.md
+3 -3
View File
@@ -8,7 +8,7 @@ As shown in the figure below, when $n = 4$, there are two solutions. From the pe
![Solution to the 4 queens problem](n_queens_problem.assets/solution_4_queens.png)
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 `/`.
The figure below 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)
@@ -18,7 +18,7 @@ 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.
The figure below 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)
@@ -30,7 +30,7 @@ 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 the figure below 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.