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Delete column slot when clearing complete stack #1

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Delete column slot when clearing complete stack #1

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7 changes: 4 additions & 3 deletions solitaire.py
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Original file line number Diff line number Diff line change
Expand Up @@ -51,7 +51,7 @@ def remove_complete_stacks(self, board: List[List[int]], verbose: bool = False)
if self.is_complete_stack(stack):
if verbose:
print(f" ✓ Complete stack removed: {[self.CARD_NAMES[c] for c in stack]}")
new_board.append([])
# new_board.append([]) # this removes the column entirely. Not an ideal solution - I'd prefer it if the stack remained but was not interactable, or if there were an indicator that there was a complete stack in the slot. However, that's beyond my limited python knowledge
else:
new_board.append(stack[:])
return new_board
Expand Down Expand Up @@ -79,7 +79,7 @@ def solve_without_cheating(self) -> Optional[List[Tuple[int, int, int]]]:
board, moves = queue.popleft()
board = self.remove_complete_stacks(board)

if all(len(stack) == 0 for stack in board):
if all(self.is_complete_stack(stack) or len(stack) == 0 for stack in board): # an alternate solution I was exploring. No longer necessary, but doesn't seem to hurt anything, and is a little more robust
return moves

for from_idx in range(len(board)):
Expand Down Expand Up @@ -134,7 +134,7 @@ def visualize_solution(self, solution: Optional[List[Tuple[int, int, int]]]):
board = self.remove_complete_stacks(board, verbose=True)
self.print_board(board, f"After Step {step}")

if all(len(stack) == 0 for stack in board):
if all(self.is_complete_stack(stack) or len(stack) == 0 for stack in board): # see line 82 for explanation
print("🎉 PUZZLE SOLVED! All stacks cleared!")
else:
print("⚠️ Warning: Board not fully cleared")
Expand Down Expand Up @@ -248,3 +248,4 @@ def interactive_mode():

# Uncomment to enable interactive mode
# interactive_mode()