# Create an 8x8 checkerboard of 0s and 1s num_rows = 8 num_cols = 8 my_grid = [] for row in range(num_rows): # Create a new row new_row = [] for col in range(num_cols): # Alternating logic based on row and column index if (row + col) % 2 == 0: new_row.append(0) else: new_row.append(1) # Add the row to the grid my_grid.append(new_row) # Print the board nicely for row in my_grid: print(" ".join(str(x) for x in row)) Use code with caution. Breakdown of the Logic 1. The Core Trick: (row + col) % 2

By using (row + col) % 2 == 0 , we alternate 0s and 1s perfectly. 2. The Nested Loop Structure

Are you getting a or failing a particular test case?

Checkerboard problems typically involve an (n \times n) grid (checkerboard) with certain rules applied to it, such as placing pieces (e.g., checkers or queens) in a way that no two pieces attack each other, or problems involving coloring the board with certain constraints.

// Create a new GRect (square) GRect square = new GRect(x, y, SQUARE_SIZE, SQUARE_SIZE); square.setFilled(true);

" ".join(str(x) for x in row) converts each number in the row to a string, then joins them together with a space in between. Alternative Solutions (For Learning) Alternative 1: Appending Rows Directly (Slightly simpler)

Use a nested for loop where the outer loop represents the row index and the inner loop represents the column index , both ranging from