Update app.py

app.py

@@ -1,46 +1,55 @@
 import gradio as gr
-from PIL import Image, ImageOps, ImageFilter
+from PIL import Image
 from pix2tex.cli import LatexOCR
 import sympy as sp
 from sympy.parsing.latex import parse_latex
 import re
 
-# Preprocessing for printed image
-def preprocess_printed_image(pil_img):
-    img = pil_img.convert('L')  # grayscale
-    img = ImageOps.autocontrast(img)
-    img = img.filter(ImageFilter.MedianFilter(size=3))  # reduce noise
-    img = ImageOps.invert(img)  # black on white
-    img = img.resize((512, 128), Image.LANCZOS)  # Resize to match model input
-    return img
+# Preprocessing for handwritten image
+def preprocess_handwritten_image(pil_img):
+    return pil_img.convert('RGB')  # Minimal processing, just convert to RGB
 
 # Load Pix2Tex model (once)
 model = LatexOCR()
 
 # Clean LaTeX output
 def clean_latex(latex):
+    # Replace \mathcal{X} or \cal X with 'x'
     latex = re.sub(r'\\(cal|mathcal)\s*X', 'x', latex)
+    # Replace \mathcal{Y} or \cal Y with 'y'
     latex = re.sub(r'\\(cal|mathcal)\s*Y', 'y', latex)
+    # Replace \mathcal{Z} or \cal Z with 'z'
     latex = re.sub(r'\\(cal|mathcal)\s*Z', 'z', latex)
 
+    # Remove curly braces
     latex = latex.replace('{', '').replace('}', '')
     latex = latex.strip().rstrip(',.')
     
+    # Replace coefficients like 5\pi with (5*3.1416)
     latex = re.sub(r'(\d+)\s*\\pi', r'(\1*3.1416)', latex)
     latex = latex.replace(r'\pi', '3.1416')
 
-    # FIXED: prevent replacing "e" in "e^x" incorrectly
-    latex = re.sub(r'(?<![a-zA-Z0-9])(\d+)\s*e(?![a-zA-Z0-9])', r'(\1*2.7183)', latex)
+    # Replace coefficients like 5e with (5*2.7183)
+    latex = re.sub(r'(\d+)\s*e', r'(\1*2.7183)', latex)
     latex = re.sub(r'(?<![a-zA-Z0-9])e(?![a-zA-Z0-9])', '2.7183', latex)
 
+    # Insert * between number and variable (e.g., 45x → 45*x)
     latex = re.sub(r'(\d)([a-zA-Z])', r'\1*\2', latex)
+
+    # Replace number followed by i with number*I
     latex = re.sub(r'(\d+)\s*i', r'\1*I', latex)
+
+    # Replace standalone i with I
     latex = re.sub(r'(?<![a-zA-Z0-9])i(?![a-zA-Z0-9])', 'I', latex)
+
+    # Wrap complex coefficients with variables: (a+bI)x^n → (a+b*I)*x^n
     latex = re.sub(r'\(([^()]+?)\)\s*([a-zA-Z](\^\d+)?)', r'(\1)*\2', latex)
 
+    # Additional minimal cleanup
     latex = latex.replace(r'\cdot', '*')
-    latex = latex.replace('−', '-')
+    latex = latex.replace('−', '-')  # Unicode minus to ASCII
 
+    # Append '=0' if not already present
     if '=' not in latex:
         latex += '=0'
 
@@ -49,9 +58,10 @@ def clean_latex(latex):
 # Main function
 def solve_polynomial(image):
     try:
-        img = preprocess_printed_image(image)
+        img = preprocess_handwritten_image(image)
         latex_result = model(img)
 
+        # Check for empty/invalid OCR output
         if not latex_result or len(latex_result.strip()) < 2:
             return "❌ Could not extract valid LaTeX from image."
 
@@ -70,17 +80,23 @@ def solve_polynomial(image):
                 "```\n"
             )
 
+        
+
+
         output = (
             f"## 📄 Extracted LaTeX\n\n"
             f"```latex\n{latex_result}\n```\n"
-            "---\n"
+            "\n---\n"
             f"## 🧹 Cleaned LaTeX Used\n\n"
             f"```latex\n{cleaned_latex}\n```\n"
-            "---\n"
+            "\n---\n"
             f"## 🧠 Parsed Expression\n\n$$ {sp.latex(expr)} $$\n"
-            "---\n"
+            "\n---\n"
         )
 
+
+
+
         if isinstance(expr, sp.Equality):
             lhs = expr.lhs - expr.rhs
 
@@ -125,3 +141,4 @@ demo = gr.Interface(
 
 if __name__ == "__main__":
     demo.launch()
+    #gg