9.1.7 Checkerboard V2 Answers __top__ Link

# Function to print the board in a readable format def print_board(board): for row in board: print(" ".join([str(x) for x in row])) # 1. Initialize an 8x8 grid filled with 0s board = [] for i in range(8): board.append([0] * 8) # 2. Use nested loops to apply the checkerboard pattern for row in range(8): for col in range(8): # If the sum of row + col is odd, set the value to 1 # This creates the alternating pattern if (row + col) % 2 != 0: board[row][col] = 1 # 3. Output the result print_board(board) Use code with caution. Why This Works

: Used to detect even or odd positions. Specifically, (row + col) % 2 == 0 identifies positions that form the diagonal alternating pattern required for a checkerboard. 9.1.7 checkerboard v2 answers

Leo pointed to the for loop for the rows. "I have i for the rows and j for the columns. If j is even, I draw black. If j is odd, I draw white." # Function to print the board in a

public void run() // Create an ArrayList of ArrayLists (2D ArrayList) ArrayList<ArrayList<Color>> board = new ArrayList<>(); Output the result print_board(board) Use code with caution

If you share the of the problem (without violating honor codes), I can help you debug or explain the logic further.