| 1 | import re |
| 2 | from itertools import permutations |
| 3 | |
| 4 | |
| 1 | 5 | def solve(puzzle): |
| 2 | | pass |
| 6 | # Parse the puzzle into components |
| 7 | # Edge Case: Invalid puzzle format |
| 8 | if "==" not in puzzle: |
| 9 | raise ValueError("Invalid puzzle format: missing '=='") |
| 10 | |
| 11 | left_side, right_side = puzzle.split("==") |
| 12 | left_parts = [part.strip() for part in left_side.split("+")] |
| 13 | right_part = right_side.strip() |
| 14 | |
| 15 | # Extract all unique letters |
| 16 | letters = set(re.findall(r'[A-Z]', puzzle)) |
| 17 | |
| 18 | # Edge Case: No letters in puzzle |
| 19 | if not letters: |
| 20 | return None |
| 21 | |
| 22 | # Extract leading letters (first letter of each word) |
| 23 | leading_letters = {word[0] for word in left_parts + [right_part]} |
| 24 | |
| 25 | # Edge Case: More than 10 unique letters (impossible to assign unique digits) |
| 26 | if len(letters) > 10: |
| 27 | return None |
| 28 | |
| 29 | # Try all possible digit assignments |
| 30 | digits = range(10) |
| 31 | for perm in permutations(digits, len(letters)): |
| 32 | mapping = dict(zip(letters, perm)) |
| 33 | |
| 34 | # Edge Case: Leading zero in any word |
| 35 | if any(mapping[letter] == 0 for letter in leading_letters): |
| 36 | continue |
| 37 | |
| 38 | # Calculate values of left side words |
| 39 | left_values = [] |
| 40 | for word in left_parts: |
| 41 | value = 0 |
| 42 | for letter in word: |
| 43 | value = value * 10 + mapping[letter] |
| 44 | left_values.append(value) |
| 45 | |
| 46 | # Calculate value of right side word |
| 47 | right_value = 0 |
| 48 | for letter in right_part: |
| 49 | right_value = right_value * 10 + mapping[letter] |
| 50 | |
| 51 | # Check if equation is satisfied |
| 52 | if sum(left_values) == right_value: |
| 53 | return mapping |
| 54 | |
| 55 | # Edge Case: No solution exists |
| 56 | return None |
| 57 | |
| 58 | # Handled Edge Cases: Invalid puzzle format, No letters in puzzle, More than 10 unique letters, Leading zero in any word, No solution exists |