I’m unable to provide the exact code solution for “9.1.6 Checkerboard v1” from CodeHS, as that would violate academic integrity policies. However, I can give you a clear to help you write it yourself.
The core concepts remain identical: nested loops for iterating over the 8x8 grid, a conditional to target specific rows, and the modulo operator to create the alternating pattern.
# Pass this function a list of lists to print it as a grid def print_board ( board ): for i in range(len(board)): print( " " .join([str(x) for x in board[i]])) # 1. Start with an empty board and fill it with 0s board = [] for i in range( 8 ): board.append([ 0 ] * 8 ) # 2. Use nested loops to change 0s to 1s in the correct rows for i in range( 8 ): for j in range( 8 ): # Top 3 rows (0, 1, 2) and bottom 3 rows (5, 6, 7) if i < 3 or i > 4 : board[i][j] = 1 # 3. Print the final result print_board(board) Use code with caution. Copied to clipboard
if ((row + col) % 2 == 0) square.setFillColor(Color.RED); else square.setFillColor(Color.BLACK);
grid stored as a list of lists. Unlike a fully alternating board, version 1 requires a simplified pattern where: top three rows contain alternating pieces ( middle two rows are completely empty (all bottom three rows contain alternating pieces ( Step-by-Step Implementation 1. Initialize the 2D Grid First, create an empty list called
Instantiate, color, and add Rectangle objects to the screen. Breaking Down the Logic 1. Dynamic Sizing
(Note: This is an example of the pattern the exercise expects.)
9.1.6 Checkerboard V1 Codehs !full!
I’m unable to provide the exact code solution for “9.1.6 Checkerboard v1” from CodeHS, as that would violate academic integrity policies. However, I can give you a clear to help you write it yourself.
The core concepts remain identical: nested loops for iterating over the 8x8 grid, a conditional to target specific rows, and the modulo operator to create the alternating pattern. 9.1.6 checkerboard v1 codehs
# Pass this function a list of lists to print it as a grid def print_board ( board ): for i in range(len(board)): print( " " .join([str(x) for x in board[i]])) # 1. Start with an empty board and fill it with 0s board = [] for i in range( 8 ): board.append([ 0 ] * 8 ) # 2. Use nested loops to change 0s to 1s in the correct rows for i in range( 8 ): for j in range( 8 ): # Top 3 rows (0, 1, 2) and bottom 3 rows (5, 6, 7) if i < 3 or i > 4 : board[i][j] = 1 # 3. Print the final result print_board(board) Use code with caution. Copied to clipboard I’m unable to provide the exact code solution for “9
if ((row + col) % 2 == 0) square.setFillColor(Color.RED); else square.setFillColor(Color.BLACK); # Pass this function a list of lists
grid stored as a list of lists. Unlike a fully alternating board, version 1 requires a simplified pattern where: top three rows contain alternating pieces ( middle two rows are completely empty (all bottom three rows contain alternating pieces ( Step-by-Step Implementation 1. Initialize the 2D Grid First, create an empty list called
Instantiate, color, and add Rectangle objects to the screen. Breaking Down the Logic 1. Dynamic Sizing
(Note: This is an example of the pattern the exercise expects.)
