Merge pull request #85 from snowch/claude/fix-bernoulli-rendering-mjFhe

Increase diagram font sizes and fix arrow overlaps in Binomial diagram

Author: snowch

chapter_07.md

@@ -435,14 +435,14 @@ import matplotlib.pyplot as plt
 import matplotlib.patches as mpatches
 from matplotlib.patches import FancyBboxPatch, FancyArrowPatch
 
-fig, ax = plt.subplots(figsize=(12, 7))
+fig, ax = plt.subplots(figsize=(14, 8))
 ax.set_xlim(0, 10)
 ax.set_ylim(0, 10)
 ax.axis('off')
 
 # Title
 ax.text(5, 9.5, r'Binomial Formula Breakdown: $n=3, k=2, p=0.6$',
-        ha='center', va='top', fontsize=24, weight='bold')
+        ha='center', va='top', fontsize=28, weight='bold')
 
 # Draw the three sequences
 sequences = [
@@ -464,47 +464,47 @@ for seq_text, x, y in sequences:
 
     # Sequence text
     ax.text(x, y, seq_text, ha='center', va='center',
-            fontsize=22, weight='bold', family='monospace')
+            fontsize=26, weight='bold', family='monospace')
 
     # Probability calculation below
     ax.text(x, y-0.8, f'{p}×{p}×{1-p:.1f}', ha='center', va='top',
-            fontsize=20, style='italic')
+            fontsize=24, style='italic')
     ax.text(x, y-1.2, f'= {prob_each:.3f}', ha='center', va='top',
-            fontsize=20, weight='bold')
+            fontsize=24, weight='bold')
 
 # Count annotation
 ax.text(5, 5.5, r'$\binom{3}{2} = 3$ sequences',
-        ha='center', va='center', fontsize=20, weight='bold',
+        ha='center', va='center', fontsize=24, weight='bold',
         bbox=dict(boxstyle='round', facecolor='lightyellow', edgecolor='orange', linewidth=2))
 
 # Formula breakdown
-ax.text(5, 4.2, 'Formula:', ha='center', va='center', fontsize=20, weight='bold')
+ax.text(5, 4.2, 'Formula:', ha='center', va='center', fontsize=24, weight='bold')
 ax.text(5, 3.6, r'$P(X=2) = \binom{3}{2} \cdot p^2 \cdot (1-p)^1$',
-        ha='center', va='center', fontsize=20)
+        ha='center', va='center', fontsize=24)
 ax.text(5, 2.9, r'$= 3 \times 0.36 \times 0.4$',
-        ha='center', va='center', fontsize=20)
+        ha='center', va='center', fontsize=24)
 ax.text(5, 2.3, r'$= 0.432$',
-        ha='center', va='center', fontsize=22, weight='bold',
+        ha='center', va='center', fontsize=26, weight='bold',
         bbox=dict(boxstyle='round', facecolor='lightgreen', edgecolor='green', linewidth=2))
 
-# Annotations with arrows - repositioned to avoid all overlaps
-ax.annotate('Count sequences', xy=(4.0, 5.5), xytext=(0.5, 4.8),
-            arrowprops=dict(arrowstyle='->', color='orange', lw=2),
-            fontsize=18, color='orange', weight='bold')
+# Annotations with arrows - repositioned to completely avoid overlaps
+ax.annotate('Count sequences', xy=(3.8, 5.5), xytext=(0.3, 5.0),
+            arrowprops=dict(arrowstyle='->', color='orange', lw=2.5),
+            fontsize=20, color='orange', weight='bold')
 
-ax.annotate('Each sequence\nhas same\nprobability', xy=(8.1, 7), xytext=(8.5, 8.5),
-            arrowprops=dict(arrowstyle='->', color='steelblue', lw=2),
-            fontsize=18, color='steelblue', weight='bold', ha='left')
+ax.annotate('Each sequence\nhas same\nprobability', xy=(7.5, 7.5), xytext=(7.5, 9.0),
+            arrowprops=dict(arrowstyle='->', color='steelblue', lw=2.5),
+            fontsize=20, color='steelblue', weight='bold', ha='center')
 
-ax.annotate('Multiply!', xy=(6.0, 2.5), xytext=(7.8, 1.8),
-            arrowprops=dict(arrowstyle='->', color='green', lw=2),
-            fontsize=18, color='green', weight='bold')
+ax.annotate('Multiply!', xy=(6.2, 2.3), xytext=(7.8, 1.5),
+            arrowprops=dict(arrowstyle='->', color='green', lw=2.5),
+            fontsize=20, color='green', weight='bold')
 
 # Bottom explanation
-ax.text(5, 1.0, 'Why it works: Each sequence occurs with probability 0.432/3 = 0.144,',
-        ha='center', va='center', fontsize=18, style='italic')
-ax.text(5, 0.5, 'and there are 3 ways to get exactly 2 successes, so total = 3 × 0.144 = 0.432',
-        ha='center', va='center', fontsize=18, style='italic')
+ax.text(5, 0.8, 'Why it works: Each sequence occurs with probability 0.432/3 = 0.144,',
+        ha='center', va='center', fontsize=20, style='italic')
+ax.text(5, 0.3, 'and there are 3 ways to get exactly 2 successes, so total = 3 × 0.144 = 0.432',
+        ha='center', va='center', fontsize=20, style='italic')
 
 plt.tight_layout()
 plt.savefig('ch07_binomial_formula_breakdown.svg', format='svg', bbox_inches='tight')