This approach structurally updates a clean matrix via index traversal, fully satisfying the Use an assignment statement testing flag on the CodeHS Portal .
rl.question("Rows: ", (rows) => rl.question("Cols: ", (cols) => rows = parseInt(rows); cols = parseInt(cols); for (let i = 0; i < rows; i++) let line = ""; for (let j = 0; j < cols; j++) if ((i + j) % 2 === 0) line += "X"; else line += "O";
Let's break down how to write the Python code for this exercise.
Ensure SQUARE_SIZE multiplied by COLS equals your total canvas width. Conclusion
# Constants for the canvas dimensions CANVAS_WIDTH = 400 CANVAS_HEIGHT = 400 # Configuration for the checkerboard NUM_ROWS = 8 NUM_COLS = 8 # Calculate the size of each square dynamically SQUARE_SIZE = CANVAS_WIDTH / NUM_COLS def draw_board(): # Outer loop iterates through each row for r in range(NUM_ROWS): # Inner loop iterates through each column inside that row for c in range(NUM_COLS): # Calculate the top-left x and y coordinates for the current square x_pos = c * SQUARE_SIZE y_pos = r * SQUARE_SIZE # Create the square object rect = Rectangle(SQUARE_SIZE, SQUARE_SIZE) rect.set_position(x_pos, y_pos) # Determine color using the row + column parity logic if (r + c) % 2 == 0: rect.set_color(Color.black) else: rect.set_color(Color.red) # Add the completed square to the canvas add(rect) # Call the function to render the checkerboard draw_board() Use code with caution. Code Breakdown 1. Dynamic Sizing 9.1.7 Checkerboard V2 Codehs
: The (r + c) % 2 trick is the industry standard for creating grid patterns—it’s much cleaner than using multiple if/else statements for odd/even rows.
The most common trap students fall into is tracking the state with a separate, toggling boolean variable (e.g., isBlack = !isBlack ). While this can work, it often breaks down when moving to a new row if the grid dimensions change.
The autograder often checks if you actually changed the values in the list using my_grid[row][col] = 1 . Simply printing a pattern without updating the list will likely cause the test to fail.
Instead of manually matching elements, professional code uses coordinate parity logic. In a perfect checkerboard grid, any cell where the sum of its row index and column index is odd ( (row + col) % 2 != 0 ) receives a swapped identifier. Step-by-Step Code Walkthrough This approach structurally updates a clean matrix via
function draw() for (var row = 0; row < 8; row++) for (var col = 0; col < 8; col++) if ((row + col) % 2 == 0) fill(0, 0, 0); // black else fill(255, 255, 255); // white
Ensure you're actually using an if statement to check positions 1.2.4.
add(square);
Using the % (remainder) operator to determine when a square should be color A versus color B. The most common trap students fall into is
1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 Use code with caution. Copied to clipboard
A complete Python solution for the problem as described on Brainly might look like this:
: Avoid manually typing out all 64 values. The exercise is designed to test your mastery of loops. specific error message you're seeing, or should we walk through the print_board function
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