9.1.7 Checkerboard V2 Codehs ⏰ 🔥

Building the 9.1.7 Checkerboard V2 program in CodeHS is all about mastering nested loops and using the modulo operator (%) to alternate colors efficiently.

In this version, we move beyond hard-coding rows to creating a dynamic grid that fills the entire canvas with a checkerboard pattern. The Logic Breakdown

To get this right, you need to think in terms of rows and columns: 9.1.7 Checkerboard V2 Codehs

Nested Loops: Use an outer loop to move down the rows and an inner loop to place squares across the columns.

Alternating Colors: Instead of checking just the column index, we look at the sum of the row and col indices. If (row + col) % 2 == 0, color the square red. Otherwise, color it black.

Positioning: Each square's x position is col * SQUARE_SIZE and its y position is row * SQUARE_SIZE. The Code Solution (JavaScript/karel) javascript

/* This program draws a full checkerboard on the screen. * It uses a constant for square size to make it dynamic. */ var SQUARE_SIZE = 40; function start() // Calculate how many rows and columns fit on the screen var rows = getHeight() / SQUARE_SIZE; var cols = getWidth() / SQUARE_SIZE; for(var r = 0; r < rows; r++) for(var c = 0; c < cols; c++) drawSquare(r, c); function drawSquare(row, col) var x = col * SQUARE_SIZE; var y = row * SQUARE_SIZE; var rect = new Rectangle(SQUARE_SIZE, SQUARE_SIZE); rect.setPosition(x, y); // The magic logic: if the sum of row and col is even, it's red if((row + col) % 2 == 0) rect.setColor(Color.red); else rect.setColor(Color.black); add(rect); Use code with caution. Copied to clipboard Pro-Tips for Success

Constant Usage: Always use SQUARE_SIZE instead of typing 40 everywhere. This makes it easy to change the board's density later.

Screen Bounds: Using getWidth() and getHeight() ensures your checkerboard fills the entire canvas regardless of the window size.

Modulo is King: 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.

Need help debugging a specific error in your grid? Let me know what your console is saying! Use nested loops

The CodeHS 9.1.7: Checkerboard V2 exercise is a fundamental lesson in manipulating 2D lists (nested lists) using nested for loops. While Version 1 often focuses on just filling the board, Version 2 requires a more complex logic: creating a alternating pattern of 0s and 1s, similar to a physical checkerboard. 🛠️ Problem Logic

The goal is to generate an 8x8 grid where elements alternate. In computer science, this is a classic application of the Modulus Operator (%). Grid Structure: A list of lists (8 rows, 8 columns).

The Pattern: If the sum of the row index (i) and column index (j) is even, the value should be 1. If it is odd, the value should be 0 (or vice versa).

Key Constraint: You must use nested loops and assignment statements to modify an existing grid of zeros. 💻 Implementation (Python)

The most efficient way to solve this is to first create a blank board and then use a nested loop to "draw" the pattern.

# 1. Initialize an 8x8 grid filled with 0s board = [] for i in range(8): board.append([0] * 8) # 2. Use nested loops to assign 1s in a checkerboard pattern for i in range(8): # Loop through rows for j in range(8): # Loop through columns # If the sum of indices is even, set to 1 if (i + j) % 2 == 0: board[i][j] = 1 # 3. Print the board to verify for row in board: print(row) Use code with caution. Copied to clipboard 🔍 Why it Works: The "Parity" Rule

The checkerboard pattern relies on the concept of parity (even vs. odd). The Row-Column Sum

By adding the row index and column index (i + j), you create a diagonal wave of even and odd numbers: Row 0, Col 0: 0+0 = 0 (Even) → 1 Row 0, Col 1: 0+1 = 1 (Odd) → 0 Row 1, Col 0: 1+0 = 1 (Odd) → 0 Row 1, Col 1: 1+1 = 2 (Even) → 1 Alternative Approach: Even vs Odd Rows You can also think about it row by row: Even rows (0, 2, 4...) start with 1. Odd rows (1, 3, 5...) start with 0. ⚠️ Common Pitfalls in CodeHS Track color using a boolean

Hardcoding: Do not just print the lists manually. The autograder looks for the use of the board[i][j] = 1 assignment statement.

Indentation: Python is strict about spacing. Ensure your if statement is inside the second loop, and the second loop is inside the first.

The "V1" Confusion: In Version 1, you might have filled entire rows. In V2, you must alternate within the row. Pro-Tip for Advanced Users

If you want to write this more concisely (though CodeHS might prefer the loop method for grading), you can use a List Comprehension:board = [[(i + j + 1) % 2 for j in range(8)] for i in range(8)] If you'd like, I can help you: Debug your specific error message Explain how to change the grid size dynamically

Show how to do this in Java if you are in the Nitro/Java course

Since I can't see your specific assignment screen, I'll provide a general solution and explanation for drawing a checkerboard pattern, which is a common exercise in CodeHS's JavaScript Graphics unit.

The trick to a checkerboard is the condition if (i + j) % 2 == 0.

If you want, I can: