Merge pull request #10 from shenxiangzhuang/add/run-code

Changed:use markdown-exec to run python code

docs/fifty/1_the_sock_drawer.md

@@ -86,18 +86,11 @@
 
         We can find values that satisfy these conditions through an iterative approach:
 
-        ```python
+        ```python exec="true" source="material-block" session="fifty-1"
         --8<-- "docs/fifty/snippet/1_the_sock_drawer.py:solution1"
         ```
 
-        Running the above code gives us:
-
-        ```
-        R = 3, N = 4, B = N - R = 1
-        R = 15, N = 21, B = N - R = 6
-        ```
-
-        Obtaining the final answers::
+        Obtaining the final answers:
 
         (a): At least 4 socks are needed (3 red and 1 black), in this case, we have
              $\frac{R}{N}\frac{R-1}{N-1} = \frac{3}{4}\frac{2}{3} = \frac{1}{2}$
@@ -143,16 +136,10 @@
 
         可以通过遍历的方法来求出找出符合条件的值:
 
-        ```python
+        ```python exec="true" source="material-block" session="fifty-1"
         --8<-- "docs/fifty/snippet/1_the_sock_drawer.py:solution1"
         ```
 
-        运行上述代码可以得到:
-        ```
-        R = 3, N = 4, B = N - R = 1
-        R = 15, N = 21, B = N - R = 6
-        ```
-
         得到最终答案:
 
         (a): 最少有4只袜子(3红1黑)，此时有$\frac{R}{N}\frac{R-1}{N-1} = \frac{3}{4}\frac{2}{3} = \frac{1}{2}$
@@ -170,18 +157,11 @@
         We can also make full use of the powerful computing power of the computer to directly solve this problem.
         Let’s take a look at the "violent aesthetics" of computers!
 
-        ```python
+        ```python exec="true" source="material-block" session="fifty-1"
         --8<-- "docs/fifty/snippet/1_the_sock_drawer.py:solution2"
         ```
 
-        Running the above code gives us:
-
-        ```
-        R = 3, N = 4, B = N - R = 1
-        R = 15, N = 21, B = N - R = 6
-        ```
-
-        Obtaining the final answers::
+        Obtaining the final answers:
 
         (a): At least 4 socks are needed (3 red and 1 black), in this case, we have
              $\frac{R}{N}\frac{R-1}{N-1} = \frac{3}{4}\frac{2}{3} = \frac{1}{2}$
@@ -195,16 +175,10 @@
         上面的方法是先分析简化问题之后再求解，我们也可以充分利用计算机强大的计算能力来直接求解本问题。
         让我们一起来看下计算机的“暴力美学”吧!
 
-        ```python
+        ```python exec="true" source="material-block" session="fifty-1"
         --8<-- "docs/fifty/snippet/1_the_sock_drawer.py:solution2"
         ```
 
-        运行上述代码同样可以得到:
-        ```
-        R = 3, N = 4, B = N - R = 1
-        R = 15, N = 21, B = N - R = 6
-        ```
-
         得到相同的答案:
 
         (a): 最少有4只袜子(3红1黑)，此时有$\frac{R}{N}\frac{R-1}{N-1} = \frac{3}{4}\frac{2}{3} = \frac{1}{2}$

docs/fifty/2_successive_wins.md

@@ -37,7 +37,7 @@
 
 ## Solutions
 
-### Solution1
+### Solution1: Analysis
 
 ??? success "Solution1: Analysis"
 
@@ -62,7 +62,7 @@
 
         The probability of Elmer winning the reward in this case is:
 
-        $fcf + fc(1-f) + (1-f)cf = fc(2-f)$
+        $$fcf + fc(1-f) + (1-f)cf = fc(2-f)$$
 
         Similarly, if we consider the sequence CFC,
         Elmer can win the reward in the following three cases:
@@ -140,7 +140,7 @@
         即Elmer选择CFC的顺序比赛赢得奖励的概率更高
 
 
-### Solution2
+### Solution2: Simulation
 
 
 ??? success "Solution2: Simulation"
@@ -149,60 +149,39 @@
 
         We can simulate the CFC and FCF sequences using the following approach:
 
-        ```python
+        ```python exec="true" source="material-block" session="fifty-2"
         --8<-- "docs/fifty/snippet/2_successive_wins.py:solution2"
         ```
 
-        Running `simulation()` produces the following output
-        (since the probabilities `f` and `c` for winning against the father and the champion are randomly generated using `get_prior_prob`, the results may vary for each simulation. However, we can observe that the probability for the CFC sequence is slightly higher):
+        Since the probabilities `f` and `c` for winning against the father and the champion are randomly generated using `get_prior_prob`, the results may vary for each simulation. However, we can observe that the probability for the CFC sequence is slightly higher.
 
-        ```python
-        FCF: 0.4759, CFC: 0.6061
-        ```
 
         If we carefully analyze this simulation experiment, we can identify a limitation: it only simulates one pair of `f` and `c` probabilities. Therefore, the higher probability for CFC compared to FCF might be due to chance. To validate if CFC consistently has a higher probability than FCF, we need to simulate different configurations of `f` and `c`.
 
-        ```python
+        ```python exec="true" source="material-block" session="fifty-2"
         --8<-- "docs/fifty/snippet/2_successive_wins.py:solution2-extend"
         ```
 
-        Running `simulation_extend()` produces the following output:
-
-        ```python
-        (CFC win prob > FCF win prob)'s prob: 1.0
-        ```
-
         Hence, we have sufficient data to conclude that the probability of CFC being greater than FCF is consistent.
 
 
     === "中文"
 
         可以通过下面的方式进行CFC，FCF两个模式的比赛模拟:
 
-        ```python
+        ```python exec="true" source="material-block" session="fifty-2"
         --8<-- "docs/fifty/snippet/2_successive_wins.py:solution2"
         ```
 
-        运行`simulation()`可以得到如下输出
-        (因为这里赢得父亲和冠军的概率`f`, `c`是通过`get_prior_prob`随机生成，
-        所以每次模拟的结果会不相同。但是都能够看到CFC顺序的概率更高一些):
-
-        ```python
-        FCF: 0.4759, CFC: 0.6061
-        ```
+        因为这里赢得父亲和冠军的概率`f`, `c`是通过`get_prior_prob`随机生成，
+        所以每次模拟的结果会不相同。但是都能够看到CFC顺序的概率更高一些.
 
         如果我们仔细思考这个模拟实验就会发现其中还有不足的地方，那就是这里只模拟了一对
         `f`, `c`概率下的情况。那么这里模拟得到的CFC的概率大于FCF的概率不排除是偶然的原因。
         所以我们需要模拟在不同`f`, `c`配置下是否都有CFC的概率大于FCF的概率。
 
-        ```python
+        ```python exec="true" source="material-block" session="fifty-2"
         --8<-- "docs/fifty/snippet/2_successive_wins.py:solution2-extend"
         ```
 
-        运行`simulation_extend()`可以得到如下输出
-
-        ```python
-        (CFC win prob > FCF win prob)'s prob: 1.0
-        ```
-
         至此，我们有充足的数据说明CFC的概率大于FCF的概率。

docs/fifty/snippet/1_the_sock_drawer.py

@@ -18,9 +18,11 @@ def find_numbers(r_limit: int = 50) -> None:
         under_sqrt = 1 - 8 * r * (1 - r)
         if is_square(under_sqrt) and is_odd(int(math.sqrt(under_sqrt))):
             n = int((1 + int(math.sqrt(under_sqrt))) / 2)
-            print(f"R = {r}, N = {n}, B = N - R = {n - r}")
+            print(f"R = {r}, N = {n}, B = N - R = {n - r}\n")
 
 
+find_numbers()
+
 # --8<-- [end:solution1]
 
 
@@ -33,7 +35,9 @@ def simulation(r_limit: int = 50, n_limit: int = 50) -> None:
     for r in range(1, r_limit):
         for n in range(r + 1, n_limit):
             if is_close(1 / 2, (r / n) * ((r - 1) / (n - 1))):
-                print(f"R = {r}, N = {n}, B = N - R = {n - r}")
+                print(f"R = {r}, N = {n}, B = N - R = {n - r}\n")
+
 
+simulation()
 
 # --8<-- [end:solution2]

docs/fifty/snippet/2_successive_wins.py

@@ -40,6 +40,8 @@ def simulation(run_num: int = 10000) -> None:
     print(f"FCF: {fcf_win_prob}, CFC: {cfc_win_prob}")
 
 
+simulation()
+
 # --8<-- [end:solution2]
 
 
@@ -55,4 +57,6 @@ def simulation_extend(run_num: int = 10000) -> None:
     print(f"(CFC win prob > FCF win prob)'s prob: {count / 100}")
 
 
+simulation_extend()
+
 # --8<-- [end:solution2-extend]

mkdocs.yml

@@ -105,6 +105,7 @@ watch:
 
 plugins:
   - search
+  - markdown-exec
   - mkdocs-simple-hooks:
       hooks:
         on_post_build: "docs.hooks:copy_get"

pyproject.toml

@@ -62,6 +62,7 @@ pymdown-extensions = "^10.3"
 mkdocs-simple-hooks = "^0.1.5"
 mkdocs-version-annotations = "^1.0.0"
 mkdocs-include-markdown-plugin = "^6.0.1"
+markdown-exec = {extras = ["ansi"], version = "^1.6.0"}
 
 [[tool.poetry.source]]
 name = "tsinghua"