9.1.7 Checkerboard V2 Codehs <Simple>

def print_board(board): for row in range(len(board)): print(" ".join([str(cell) for cell in board[row]]))

To build a checkerboard pattern efficiently, you do not need to keep track of a toggle variable that manually flips back and forth. Manual toggling often introduces bugs when switching to a new row. Instead, you can rely on the mathematical properties of coordinates.

(Note: You can replace "# " and " " with colored console codes if supported.)

By calculating total squares ( NUM_ROWS * NUM_COLS ), you can derive the row and column using division and modulo: var row = Math.floor(i / NUM_COLS); var col = i % NUM_COLS;

CodeHS provides a function that handles the visual output of the grid, so you don't need to write it yourself. It's a practical tool used in all three checkerboard exercises: 9.1.7 Checkerboard V2 Codehs

To build a checkerboard without writing repetitive code, you must use a mathematical rule for alternating patterns. The most efficient method relies on the combined with the grid coordinates.

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

Getting the alternation correct, especially ensuring the pattern carries over across rows. Key Concepts Explained

// Create the canvas var canvas = document.getElementById('canvas'); var ctx = canvas.getContext('2d'); (Note: You can replace "# " and "

// Draw the checkerboard for (var row = 0; row < rows; row++) for (var col = 0; col < cols; col++) // Alternate between light and dark colors var color = (row + col) % 2 === 0 ? 'lightgray' : 'gray'; drawSquare(col * squareSize, row * squareSize, color);

Create your loops to iterate through the coordinates. Ensure your inner loop completes a full row before the outer loop moves down to the next row. 3. Calculate Pixel Coordinates

Here is how to translate this logic into a clean, functional program using nested for loops and the modulo operator ( % ). Step 1: Initialize the Grid

# Helper function provided by CodeHS to output the final matrix formatting def print_board(board): for i in range(len(board)): print(" ".join([str(x) for x in board[i]])) # 1. Initialize an empty list to represent our grid my_grid = [] # 2. Build a baseline 8x8 grid filled entirely with 0s for i in range(8): my_grid.append([0] * 8) # 3. Use nested loops to loop through every single cell row-by-row for row in range(8): for col in range(8): # Parity check: If row index plus column index is an odd number if (row + col) % 2 == 1: my_grid[row][col] = 1 # Mutate the value to 1 # 4. Print the completed board structure print_board(my_grid) Use code with caution. Solution 2: Generative Conditional Row Appending You must use nested loops and assignment statements

This guide breaks down the core concepts behind the challenge, provides a step-by-step logic walkthrough, and offers debugging tips to help you get your code passing all test cases. Understanding the Core Concepts

for row in range(rows): for col in range(cols): # Draw square here Use code with caution. 2. Alternating Logic (The Core of V2)

grid where 1s and 0s alternate perfectly in every direction.

Remember that the outer loop handles rows ( array[row] ) and the inner loop targets columns ( array[row][col] ). Swapping these will crash your program if the grid is not perfectly square.