got rid of non-working square brackets with dollar sign

src/content/questions/comp2804/2014-winter-final/11/solution.md

@@ -15,22 +15,15 @@ $ FIB(7) $ calls $ FIB(6) $ and $ FIB(5) $.
 4. **Recursive Counting:**<br/>
    Let $ a_n $ be the number of \times $ FIB(4) $ is called in $ FIB(n) $.<br/>
    Observe that $ a_n $ satisfies the recurrence relation similar to the Fibonacci sequence:<br/>
-   [
-   a_n = a\_{n-1} + a\_{n-2}
-   ]
-   <br/>
+   $a_n = a\_{n-1} + a\_{n-2}$<br/>
    Initial conditions are crucial:<br/>
    $ FIB(4) $ is called 1 time in $ FIB(4) $.<br/>
 $ FIB(5) $ calls $ FIB(4) $ directly once and $ FIB(3) $, which does not call $ FIB(4) $.
 
 5. **Connecting to Fibonacci Sequence:**<br/>
    The recurrence relation $ a_n = a\_{n-1} + a\_{n-2} $ with initial conditions $ a_4 = 1 $ and $ a_5 = 1 $ corresponds to the Fibonacci sequence shifted by 3:<br/>
-   [
-   a_n = f\_{n-3}
-   ]
+   $a_n = f\_{n-3}$
 
 6. **Conclusion:**<br/>
    Therefore, the number of times $ FIB(4) $ is called in $ FIB(n) $ is given by the Fibonacci number $ f\_{n-3} $:<br/>
-   [
-   a_n = f\_{n-3}
-   ]
+   $a_n = f\_{n-3}$

src/content/questions/comp2804/2018-fall-midterm/1/solution.md

@@ -15,6 +15,6 @@ $ 4 \cdot 4 \cdot 4 \cdot ... \cdot 4$
 
 Hence, the total number of ways to assign each of the $ n $ elements to either $ A $, $ B $, $ C $, or none of them is $ 4^n $.
 
-Thus, the total number of ordered triples $ (A, B, C) $ where $ A $, $ B $, and $ C $ are pairwise disjoint subsets of $ S $ is: [ 4^n ]
+Thus, the total number of ordered triples $ (A, B, C) $ where $ A $, $ B $, and $ C $ are pairwise disjoint subsets of $ S $ is: $ 4^n $
 
 So, the number of such ordered triples is $boxed{4^n}$.

src/content/questions/comp2804/2018-fall-midterm/10/solution.md

@@ -10,8 +10,7 @@ To derive the recurrence relation for $ F(n,k) $, we consider two cases:
 </ul>
 
 Adding these two cases together, we obtain the recurrence relation:
-[
-F(n,k) = F(n-1,k) + 2 \cdot F(n-1,k-1)
-]
+
+$F(n,k) = F(n-1,k) + 2 \cdot F(n-1,k-1)$
 
 This holds for all integers $ n geq 2 $ and $ k $ with $ 1 leq k leq n - 1 $.

src/content/questions/comp2804/2018-fall-midterm/6/solution.md

@@ -12,4 +12,4 @@ This is because we have only 25 pairs and choosing 26 elements will necessarily
 
 Therefore, the minimum size of the subset $ S $ that ensures at least two elements sum to 51 is:
 
-[ boxed{26} ]
+$ boxed{26} $

src/content/questions/comp2804/square_bracket_replacer.py

@@ -0,0 +1,38 @@
+import os
+
+def find_lines_with_brackets(file_path):
+    with open(file_path, 'r', encoding='utf-8') as file:
+        lines = file.readlines()
+
+    lines_with_brackets = []
+    inside_exclude_block = False
+
+    for index, line in enumerate(lines):
+        stripped_line = line.strip()
+
+        if "begin" in stripped_line:
+            inside_exclude_block = True
+        if "end" in stripped_line:
+            inside_exclude_block = False
+            continue
+
+        if not inside_exclude_block and ('[' in stripped_line or ']' in stripped_line):
+            lines_with_brackets.append((index + 1, stripped_line))
+
+    return lines_with_brackets
+
+def process_directory(directory):
+    for root, _, files in os.walk(directory):
+        for file in files:
+            if file == 'solution.md':
+                file_path = os.path.join(root, file)
+                lines_with_brackets = find_lines_with_brackets(file_path)
+                if lines_with_brackets:
+                    print(f"File: {file_path}")
+                    for line_number, line in lines_with_brackets:
+                        print(f"  Line {line_number}: {line}")
+                    print()
+
+if __name__ == "__main__":
+    directory = os.path.dirname(os.path.abspath(__file__))
+    process_directory(directory)