docs(notes): update data representation content and fix MathJax script

Author: lzwjava

notes/2025-03-23-data-representation-en.md

@@ -18,20 +18,20 @@ Below is a comprehensive tutorial that breaks down the key topics in the “Data
 **Concepts:**
 
 - **Base-2 System:** Uses only two digits: 0 and 1.
-- **Place Value:** Each digit represents a power of 2. For a binary number \( b_n b_{n-1} \dots b_1 b_0 \), the value is  
+- **Place Value:** Each digit represents a power of 2. For a binary number \\( b_n b_{n-1} \dots b_1 b_0 \\), the value is  
   \[
   \sum_{i=0}^{n} b_i \times 2^i
   \]
-  where \( b_i \) is either 0 or 1.
+  where \\( b_i \\) is either 0 or 1.
 
 **Example:**
 
-Convert binary \( 1011_2 \) to decimal:
-- \( 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 8 + 0 + 2 + 1 = 11_{10} \)
+Convert binary \\( 1011_2 \\) to decimal:
+- \\( 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 8 + 0 + 2 + 1 = 11_{10} \\)
 
 **Practice Exercise:**
 
-- Convert the binary number \( 110010_2 \) to decimal.
+- Convert the binary number \\( 110010_2 \\) to decimal.
 
 ---
 
@@ -40,7 +40,7 @@ Convert binary \( 1011_2 \) to decimal:
 **Concepts:**
 
 - **Base-16 System:** Uses sixteen symbols: 0–9 and A–F (where A=10, B=11, …, F=15).
-- **Place Value:** Each digit represents a power of 16. For a hexadecimal number \( h_n h_{n-1} \dots h_1 h_0 \), the value is  
+- **Place Value:** Each digit represents a power of 16. For a hexadecimal number \\( h_n h_{n-1} \dots h_1 h_0 \\), the value is  
   \[
   \sum_{i=0}^{n} h_i \times 16^i
   \]
@@ -52,16 +52,16 @@ Convert binary \( 1011_2 \) to decimal:
 
 **Example:**
 
-Convert binary \( 1011011101_2 \) to hexadecimal:
-- Group into 4-bit groups: \( 10 \, 1101 \, 1101 \) (pad left with zeros if needed → \( 0010 \, 1101 \, 1101 \))
-- \( 0010_2 = 2_{16} \)
-- \( 1101_2 = D_{16} \)
-- \( 1101_2 = D_{16} \)
-- Final hexadecimal: \( 2DD_{16} \)
+Convert binary \\( 1011011101_2 \\) to hexadecimal:
+- Group into 4-bit groups: \\( 10 \, 1101 \, 1101 \\) (pad left with zeros if needed → \\( 0010 \, 1101 \, 1101 \\))
+- \\( 0010_2 = 2_{16} \\)
+- \\( 1101_2 = D_{16} \\)
+- \\( 1101_2 = D_{16} \\)
+- Final hexadecimal: \\( 2DD_{16} \\)
 
 **Practice Exercise:**
 
-- Convert the binary number \( 11101010_2 \) into hexadecimal.
+- Convert the binary number \\( 11101010_2 \\) into hexadecimal.
 
 ---
 
@@ -93,18 +93,18 @@ where the bias for single precision is 127.
 
 Suppose you have a 32-bit binary string representing a floating-point number:
 - **Sign Bit:** 0 (positive)
-- **Exponent Bits:** e.g., \( 10000010_2 \) → Decimal 130. Subtract bias: \( 130 - 127 = 3 \).
-- **Mantissa Bits:** Suppose they represent a fractional part like \( .101000... \).
+- **Exponent Bits:** e.g., \\( 10000010_2 \\) → Decimal 130. Subtract bias: \\( 130 - 127 = 3 \\).
+- **Mantissa Bits:** Suppose they represent a fractional part like \\( .101000... \\).
 
 Then the number would be:
 \[
 +1.101000 \times 2^3
 \]
-Convert \( 1.101000 \) from binary to decimal and then multiply by \( 2^3 \) to get the final value.
+Convert \\( 1.101000 \\) from binary to decimal and then multiply by \\( 2^3 \\) to get the final value.
 
 **Practice Exercise:**
 
-- Given the following breakdown for a 32-bit floating-point number: sign = 0, exponent = \( 10000001_2 \) (decimal 129), and mantissa = \( 01000000000000000000000 \), compute the decimal value.
+- Given the following breakdown for a 32-bit floating-point number: sign = 0, exponent = \\( 10000001_2 \\) (decimal 129), and mantissa = \\( 01000000000000000000000 \\), compute the decimal value.
 
 ---
 
@@ -113,9 +113,9 @@ Convert \( 1.101000 \) from binary to decimal and then multiply by \( 2^3 \) to
 ### 2.1 Basic Boolean Operations
 
 **Key Operations:**
-- **AND (·):** \( A \land B \) is true only if both \( A \) and \( B \) are true.
-- **OR (+):** \( A \lor B \) is true if at least one of \( A \) or \( B \) is true.
-- **NOT (’ or \(\neg\)):** \( \neg A \) inverts the truth value of \( A \).
+- **AND (·):** \\( A \land B \\) is true only if both \\( A \\) and \\( B \\) are true.
+- **OR (+):** \\( A \lor B \\) is true if at least one of \\( A \\) or \\( B \\) is true.
+- **NOT (’ or \\(\neg\\)):** \\( \neg A \\) inverts the truth value of \\( A \\).
 
 **Truth Tables:**
 
@@ -146,7 +146,7 @@ Convert \( 1.101000 \) from binary to decimal and then multiply by \( 2^3 \) to
 
 **Practice Exercise:**
 
-- Given the Boolean expression \( \neg(A \land B) \), use a truth table to show it is equivalent to \( \neg A \lor \neg B \) (De Morgan’s Law).
+- Given the Boolean expression \\( \neg(A \land B) \\), use a truth table to show it is equivalent to \\( \neg A \lor \neg B \\) (De Morgan’s Law).
 
 ---
 
@@ -155,24 +155,24 @@ Convert \( 1.101000 \) from binary to decimal and then multiply by \( 2^3 \) to
 **Important Laws:**
 
 - **Commutative Law:**
-  - \( A \lor B = B \lor A \)
-  - \( A \land B = B \land A \)
+  - \\( A \lor B = B \lor A \\)
+  - \\( A \land B = B \land A \\)
 
 - **Associative Law:**
-  - \( (A \lor B) \lor C = A \lor (B \lor C) \)
-  - \( (A \land B) \land C = A \land (B \land C) \)
+  - \\( (A \lor B) \lor C = A \lor (B \lor C) \\)
+  - \\( (A \land B) \land C = A \land (B \land C) \\)
 
 - **Distributive Law:**
-  - \( A \land (B \lor C) = (A \land B) \lor (A \land C) \)
-  - \( A \lor (B \land C) = (A \lor B) \land (A \lor C) \)
+  - \\( A \land (B \lor C) = (A \land B) \lor (A \land C) \\)
+  - \\( A \lor (B \land C) = (A \lor B) \land (A \lor C) \\)
 
 - **De Morgan’s Laws:**
-  - \( \neg (A \land B) = \neg A \lor \neg B \)
-  - \( \neg (A \lor B) = \neg A \land \neg B \)
+  - \\( \neg (A \land B) = \neg A \lor \neg B \\)
+  - \\( \neg (A \lor B) = \neg A \land \neg B \\)
 
 **Practice Exercise:**
 
-- Simplify the expression \( A \lor (\neg A \land B) \) using Boolean algebra laws.
+- Simplify the expression \\( A \lor (\neg A \land B) \\) using Boolean algebra laws.
 
 ---
 
@@ -184,17 +184,17 @@ Convert \( 1.101000 \) from binary to decimal and then multiply by \( 2^3 \) to
 
 - **Addition:**
   - Follows similar rules to decimal addition but with base 2.
-  - **Example:** \( 1011_2 + 1101_2 \)
+  - **Example:** \\( 1011_2 + 1101_2 \\)
     - Align and add bit by bit, carrying over when the sum exceeds 1.
 
 - **Subtraction:**
   - Can be performed by borrowing, or by using the two's complement method.
   - **Two's Complement:** To represent negative numbers, invert the bits and add 1.
-  - **Example:** To subtract \( 1101_2 \) from \( 1011_2 \), first find the two's complement of \( 1101_2 \) then add.
+  - **Example:** To subtract \\( 1101_2 \\) from \\( 1011_2 \\), first find the two's complement of \\( 1101_2 \\) then add.
 
 **Practice Exercise:**
 
-- Perform the binary subtraction \( 10100_2 - 01101_2 \) using two’s complement.
+- Perform the binary subtraction \\( 10100_2 - 01101_2 \\) using two’s complement.
 
 ---
 
@@ -207,7 +207,7 @@ Convert \( 1.101000 \) from binary to decimal and then multiply by \( 2^3 \) to
 
 **Practice Exercise:**
 
-- Add \( 2A3_{16} \) and \( 1F7_{16} \).
+- Add \\( 2A3_{16} \\) and \\( 1F7_{16} \\).
 
 ---
 

scripts/content/fix_mathjax.py

@@ -3,83 +3,104 @@
 import markdown
 import argparse
 
-def fix_mathjax_in_markdown(directory, max_files=None):
+def fix_mathjax_in_file(filepath):
     """
-    Finds all markdown files in a directory and replaces instances of
-    \( and \) with \\( and \\) respectively, skipping code blocks.
+    Replaces instances of \( and \) with \\( and \\) respectively in a markdown file,
+    skipping code blocks.
 
     Args:
-        directory (str): The directory to search for markdown files.
-        max_files (int, optional):  Maximum number of files to process. Defaults to None (unlimited).
+        filepath (str): The path to the markdown file to process.
     """
-    files_processed = 0
-    for root, _, files in os.walk(directory):
-        for filename in files:
-            if filename.endswith(".md"):
-                filepath = os.path.join(root, filename)
-                try:
-                    with open(filepath, 'r', encoding='utf-8') as f:
-                        content = f.read()
+    try:
+        with open(filepath, 'r', encoding='utf-8') as f:
+            content = f.read()
 
-                    # Parse markdown to identify code blocks
-                    md = markdown.Markdown(extensions=['fenced_code'])
-                    html_content = md.convert(content)
+        # Parse markdown to identify code blocks
+        md = markdown.Markdown(extensions=['fenced_code'])
+        html_content = md.convert(content)
 
-                    # Identify code blocks using regex
-                    code_blocks = list(re.finditer(r'<pre><code.*?>.*?</code></pre>', html_content, re.DOTALL))
+        # Identify code blocks using regex
+        code_blocks = list(re.finditer(r'<pre><code.*?>.*?</code></pre>', html_content, re.DOTALL))
 
-                    # Extract code block content and their positions
-                    code_block_data = []
-                    for match in code_blocks:
-                        code_block_data.append({
-                            'start': match.start(),
-                            'end': match.end(),
-                            'content': match.group(0)
-                        })
+        # Extract code block content and their positions
+        code_block_data = []
+        for match in code_blocks:
+            code_block_data.append({
+                'start': match.start(),
+                'end': match.end(),
+                'content': match.group(0)
+            })
 
-                    # Function to replace MathJax delimiters outside code blocks
-                    def replace_mathjax(text):
-                        temp_text = text
-                        for cb in code_block_data:
-                            temp_text = temp_text.replace(cb['content'], 'CODE_BLOCK_PLACEHOLDER')
+        # Function to replace MathJax delimiters outside code blocks
+        def replace_mathjax(text):
+            temp_text = text
+            for cb in code_block_data:
+                temp_text = temp_text.replace(cb['content'], 'CODE_BLOCK_PLACEHOLDER')
 
-                        temp_text = re.sub(r'\\\(', r'\\\\(', temp_text)
-                        temp_text = re.sub(r'\\\)', r'\\\\)', temp_text)         
+            temp_text = re.sub(r'\\\(', r'\\\\(', temp_text)
+            temp_text = re.sub(r'\\\)', r'\\\\)', temp_text)         
 
-                        for cb in code_block_data:
-                            temp_text = temp_text.replace('CODE_BLOCK_PLACEHOLDER', cb['content'])
-                        return temp_text
+            for cb in code_block_data:
+                temp_text = temp_text.replace('CODE_BLOCK_PLACEHOLDER', cb['content'])
+            return temp_text
 
-                    updated_content = replace_mathjax(content)
+        updated_content = replace_mathjax(content)
 
-                    with open(filepath, 'w', encoding='utf-8') as f:
-                        f.write(updated_content)
+        with open(filepath, 'w', encoding='utf-8') as f:
+            f.write(updated_content)
 
-                    print(f"Fixed MathJax delimiters in: {filepath}")
-                    files_processed += 1
+        print(f"Fixed MathJax delimiters in: {filepath}")
+        return True
+
+    except Exception as e:
+        print(f"Error processing {filepath}: {e}")
+        return False
 
-                    if max_files and files_processed >= max_files:
-                        print(f"Maximum files processed ({max_files}). Exiting directory.")
-                        return
+def fix_mathjax_in_markdown(directory, max_files=None):
+    """
+    Finds all markdown files in a directory and replaces instances of
+    \( and \) with \\( and \\) respectively, skipping code blocks.
 
-                except Exception as e:
-                    print(f"Error processing {filepath}: {e}")
+    Args:
+        directory (str): The directory to search for markdown files.
+        max_files (int, optional):  Maximum number of files to process. Defaults to None (unlimited).
+    """
+    files_processed = 0
+    for root, _, files in os.walk(directory):
+        for filename in files:
+            if filename.endswith(".md"):
+                filepath = os.path.join(root, filename)
+                success = fix_mathjax_in_file(filepath)
+                if success:
+                    files_processed += 1
 
+                if max_files and files_processed >= max_files:
+                    print(f"Maximum files processed ({max_files}). Exiting directory.")
+                    return
 
 def main():
     """
-    Main function to specify directories to process.
+    Main function to process either a single file or directories.
     """
     parser = argparse.ArgumentParser(description="Fix MathJax delimiters in Markdown files.")
     parser.add_argument("--maxfiles", type=int, help="Maximum number of files to process.")
+    parser.add_argument("--file", type=str, help="Path to a specific markdown file to process.")
     args = parser.parse_args()
 
-    directories = ["_posts", "original", "notes"]  # Add "." to process the current directory as well
-    for directory in directories:
-        if os.path.exists(directory):
-            fix_mathjax_in_markdown(directory, max_files=args.maxfiles)
+    if args.file:
+        # Process a single file
+        if os.path.exists(args.file) and args.file.endswith(".md"):
+            fix_mathjax_in_file(args.file)
         else:
-            print(f"Directory not found: {directory}")
+            print(f"Invalid file path or not a markdown file: {args.file}")
+    else:
+        # Process directories
+        directories = ["_posts", "original", "notes"]
+        for directory in directories:
+            if os.path.exists(directory):
+                fix_mathjax_in_markdown(directory, max_files=args.maxfiles)
+            else:
+                print(f"Directory not found: {directory}")
 
 if __name__ == "__main__":
     main()