fixed equality signs

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

@@ -1,6 +1,6 @@
 1. **Base Case Analysis:** <br/>
    For $ n = 0 $ or $ n = 1 $, $ FIB(n) $ is called directly and returns $ n $. <br/>
-   For $ n geq 2 $, the algorithm makes recursive calls to $ FIB(n-1) $ and $ FIB(n-2) $.
+   For $ n \geq 2 $, the algorithm makes recursive calls to $ FIB(n-1) $ and $ FIB(n-2) $.
 
 2. **Recursive Call Analysis:**<br/>
    To find the number of times $ FIB(4) $ is called in $ FIB(n) $, we analyze the recursive structure of $ FIB $.<br/>

src/content/questions/comp2804/2014-winter-final/21/solution.md

@@ -26,6 +26,6 @@ $ 0 = \frac{1}{3} \cdot \frac{1}{3} $
 
 $ 0 = \frac{1}{9} $
 
-$ 0 neq \frac{1}{9} $
+$ 0 \neq \frac{1}{9} $
 
 Therefore, $ G $ and $ B $ are not independent random variables.

src/content/questions/comp2804/2014-winter-midterm/6/solution.md

@@ -15,4 +15,4 @@ $ |A| = 2^{7} - 1 = 127 $
 
 $ |B| = \binom{7}{7} = 1 $
 
-Now, we can say for certain that $2^{n} - 1 geq \binom{n}{7} $
+Now, we can say for certain that $2^{n} - 1 \geq \binom{n}{7} $

src/content/questions/comp2804/2015-fall-midterm/14/solution.md

@@ -11,19 +11,19 @@ JustinBieber$(8)$ = 2
 JustinBieber$(16)$ = 3
 
 <ul>
-    <li> $B(n) = \text{log } n - 1 \text{ for all } n geq 2. $ <br/> 
+    <li> $B(n) = \text{log } n - 1 \text{ for all } n \geq 2. $ <br/> 
     $B(2) = 0$ <br/> 
     $B(4) = 1$
-    <li> $B(n) = \text{log } n - 1 \text{ for all } n geq 1. $ <br/> 
+    <li> $B(n) = \text{log } n - 1 \text{ for all } n \geq 1. $ <br/> 
     $B(1) = -1$ <br/> 
     $B(2) = 0$ <br/> 
     $B(4) = 1$
-    <li> $B(n) = \text{log } n \text{ for all } n geq 2. $ <br/> 
+    <li> $B(n) = \text{log } n \text{ for all } n \geq 2. $ <br/> 
     $B(2) = 1$ <br/> 
     $B(4) = 2$
-    <li> $B(n) = n - 2 \text{ for all } n geq 2. $ <br/> 
+    <li> $B(n) = n - 2 \text{ for all } n \geq 2. $ <br/> 
     $B(2) = 0$ <br/> 
     $B(4) = 2$
 </ul>
 
-Thus, the correct answer is $B(n) = \text{log } n - 1 \text{ for all } n geq 2. $
+Thus, the correct answer is $B(n) = \text{log } n - 1 \text{ for all } n \geq 2. $

src/content/questions/comp2804/2016-fall-midterm/8/solution.md

@@ -1,6 +1,6 @@
 To determine the minimum value of $ m $ such that we can guarantee that a set $ S $ of $ m $ integers contains at least two elements whose difference is divisible by $ n $, we can use the Pigeonhole Principle.
 
-Let $ k geq 1 $ be an integer. If $ k+1 $ or more objects are placed into $ k $ boxes, then there is at least one box containing two or more objects.
+Let $ k \geq 1 $ be an integer. If $ k+1 $ or more objects are placed into $ k $ boxes, then there is at least one box containing two or more objects.
 
 <ul>
     <li> textbf{Residue Classes Modulo $ n $:} <br/> 

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

@@ -13,4 +13,4 @@ Adding these two cases together, we obtain the recurrence relation:
 
 $F(n,k) = F(n-1,k) + 2 \cdot F(n-1,k-1)$
 
-This holds for all integers $ n geq 2 $ and $ k $ with $ 1 leq k leq n - 1 $.
+This holds for all integers $ n \geq 2 $ and $ k $ with $ 1 \leq k \leq n - 1 $.

src/content/questions/comp2804/2018-winter-final/14/solution.md

@@ -1,4 +1,4 @@
-The difference in string $a-b$ is non-zero at each position only if $a_i$ is 1 and $b_i$ is 0 or $a_i$ is 0 and $b_i$ is 1, for all $1 leq i leq 77$
+The difference in string $a-b$ is non-zero at each position only if $a_i$ is 1 and $b_i$ is 0 or $a_i$ is 0 and $b_i$ is 1, for all $1 \leq i \leq 77$
 
 Let $X_i$ be 1 if $a_i$ and $b_i$ are different and 0 otherwise
 

src/content/questions/comp2804/2018-winter-final/21/solution.md

@@ -1,6 +1,6 @@
 For $X$ and $Z$ to be independent, $Pr(X = x \cap Z = z) = Pr(X = x) \cdot Pr(Z = z)$, for all real numbers $x$ and $z$, by definition of independent random variables.
 
-If we can find a counterexample, aka where $Pr(X = x \cap Z = z) neq Pr(X = x) \cdot Pr(Z = z)$, then we know the two are not independent.
+If we can find a counterexample, aka where $Pr(X = x \cap Z = z) \neq Pr(X = x) \cdot Pr(Z = z)$, then we know the two are not independent.
 
 Let $z = 0$, $x = 0$ and $y = 0$:
 
@@ -17,7 +17,7 @@ Now we check if LHS = RHS for the expression:
     <li> $Pr(X = x \cap Z = z) = Pr(X = x) \cdot Pr(Z = z)$
     <li> $Pr(X = 0 \cap Z = 0) = Pr(X = 0) \cdot Pr(Z = 0)$
     <li> $\frac{1}{2} = \frac{1}{2} \cdot \frac{3}{4}$
-    <li> $\frac{1}{2} neq \frac{3}{8}$
+    <li> $\frac{1}{2} \neq \frac{3}{8}$
 </ul>
 
 Since LHS is not equal to RHS, then the events $X$ and $Z$ are not independent.

src/content/questions/comp2804/equal_sign_slash_adder.py

@@ -0,0 +1,36 @@
+import os
+import re
+
+def find_phrases_without_backslash(file_path):
+    with open(file_path, 'r', encoding='utf-8') as file:
+        lines = file.readlines()
+
+    patterns = [
+        (r'(?<!\\)neq', 'neq'),
+        (r'(?<!\\)geq', 'geq'),
+        (r'(?<!\\)leq', 'leq')
+    ]
+
+    lines_with_phrases = []
+    for index, line in enumerate(lines):
+        for pattern, phrase in patterns:
+            if re.search(pattern, line):
+                lines_with_phrases.append((index + 1, line.strip(), phrase))
+
+    return lines_with_phrases
+
+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_phrases = find_phrases_without_backslash(file_path)
+                if lines_with_phrases:
+                    print(f"File: {file_path}")
+                    for line_number, line, phrase in lines_with_phrases:
+                        print(f"  Line {line_number}: {line} (contains '{phrase}')")
+                    print()
+
+if __name__ == "__main__":
+    directory = os.path.dirname(os.path.abspath(__file__))
+    process_directory(directory)