At first glance, this problem seems simple: draw a checkerboard. But the "V2" designation introduces specific logic constraints that often trip up students. This article will break down the problem, explain the underlying logic of nested loops and modulus operators, provide step-by-step code solutions in both Java and JavaScript, and offer debugging tips to help you pass the autograder on your first attempt.
checkerboard of obsidian and pearl tiles. However, Modulo was notoriously lazy and wanted a single set of instructions his apprentices could follow without him being there. 1. Designing the First Row
Mastering the Checkerboard V2 (9.1.7) in CodeHS To successfully complete the 9.1.7 Checkerboard V2 exercise in CodeHS, you must generate an
public class CheckerboardV2 public static void main(String[] args) Scanner input = new Scanner(System.in);
# 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
Create a checkerboard pattern using a loop to iterate over a grid of squares.
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for row in range(8): for col in range(8): # Drawing logic here
Have questions or an alternate solution? Drop a comment below! At first glance, this problem seems simple: draw
To succeed in this challenge, you need to understand three core components: 1. Nested Loops
Getting the alternation correct, especially ensuring the pattern carries over across rows. Key Concepts Explained
The most efficient way to determine the color is using the modulo operator (%). If (row + col) % 2 == 0, set the color to one choice (like black). Otherwise, set it to the second choice (like red). This ensures that every time you move one square to the right or one square down, the color flips, creating the perfect checkerboard pattern. Implementation Tips
To create a checkerboard, we use the row and column indices. If the sum of the and column index is even, we assign one value (e.g., 0); if it is odd, we assign the other (e.g., 1). This is easily checked using the modulo operator ( % ): if (row + col) % 2 == 0: (Sum is even) else: (Sum is odd) Step-by-Step Implementation checkerboard of obsidian and pearl tiles
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Example pattern (B = black, W = white):
: Every cell in a 2D array is defined by its row index ( r ) and column index ( c ).