For the CodeHS exercise 9.1.7: Checkerboard, v2 , the objective is to create an grid where the values alternate between in a checkerboard pattern.
The most efficient way to pass the autograder—which often requires specific assignment statements
within nested loops—is to initialize a grid of zeros first and then use the modulus operator ) to change half of them to ones. Correct Solution (Python) # Pass this function a list of lists, and it will # print it such that it looks like the grids in # the exercise instructions. print_board range(len(board)): print( .join([str(x) board[i]])) # 1. Initialize the board with an 8x8 grid of 0s ): board.append([
# 2. Use nested for loops to create the checkerboard pattern # The pattern alternates where (row + col) is even : board[i][j] = # Assignment statement required by autograder # 3. Print the final board print_board(board) Use code with caution. Copied to clipboard Step-by-Step Explanation Initialize the Grid : We start by creating a list of lists ( ) filled with . This establishes the structure before we start modifying values. Nested Loop Logic : We use two loops—one for rows ( ) and one for columns (
). This allows us to access every individual coordinate in the grid. The Checkered Condition
: To get the alternating pattern, we check if the sum of the current row and column indices is even ( (i + j) % 2 == 0 ). If it is, we assign that spot a Example: At board[0][0] (even), so it becomes board[0][1] (odd), so it stays print_board
This essay explores the logic and implementation of the Checkerboard V2 challenge in the CodeHS 9.1.7 curriculum
. This exercise is a pivotal moment for students learning Java or JavaScript (Karel), as it transitions from simple movement to complex nested loops and conditional logic. The Objective
The goal of Checkerboard V2 is to create a grid-like pattern of "markers" or "beepers" on a canvas of any size. Unlike the first version, V2 often requires the program to be dynamic—meaning it must work whether the grid is
. The pattern mimics a standard chessboard, where no two markers are adjacent horizontally or vertically. Core Logic: The Nested Loop The foundation of the solution is the nested loop
. To fill a 2D space, the program must iterate through rows and columns. Outer Loop:
Manages the vertical movement (moving from one row to the next). Inner Loop:
Manages the horizontal movement (placing beepers across a single row).
The challenge arises in the alternating nature of the rows. If every row started with a beeper, you would end up with vertical stripes rather than a checkerboard. The Parity Strategy The most elegant way to solve Checkerboard V2 is by using
(even vs. odd). By tracking the current row number and column number, the program can decide whether to place a marker based on the following rule: If the sum of the (Row + Column) is , place a beeper. If the sum is , leave the space empty.
This mathematical approach ensures the pattern remains consistent regardless of the grid’s dimensions. Execution and "The Turnaround"
In many Karel-based versions of this task, the difficulty lies in the "turnaround." Once Karel reaches the end of a row, the program must determine if there is another row above it. If so, Karel must turn, move up, and position itself to face the opposite direction to begin the next line. This requires careful use of
statements to check for walls and prevent the program from crashing at the edges of the grid. Conclusion
Solving 9.1.7 Checkerboard V2 is less about the act of placing markers and more about algorithmic thinking
. It teaches students how to use coordinates to control logic and how to write code that is flexible enough to handle varying input sizes. Mastering this exercise signals a transition from a beginner coder to one who understands the structural beauty of computer science. loops or the if/else statements needed for this?
It looks like you're asking for the solution or an explanation for the Checkerboard V2 problem (Exercise 9.1.7) on CodeHS.
Since I don't have access to your specific instance of the problem (which usually involves a specific width, height, or starting color), I will provide the standard logic used to solve this problem in Python. This logic works for any version of the problem.
import java.awt.Color;public class CheckerboardV2 extends GraphicsProgram
private static final int ROWS = 8; private static final int COLS = 8; private static final int SQUARE_SIZE = 50; public void run() for (int row = 0; row < ROWS; row++) for (int col = 0; col < COLS; col++) int x = col * SQUARE_SIZE; int y = row * SQUARE_SIZE; GRect square = new GRect(x, y, SQUARE_SIZE, SQUARE_SIZE); square.setFilled(true); if ((row + col) % 2 == 0) square.setFillColor(Color.RED); else square.setFillColor(Color.BLACK); add(square);
Note: If your class extends
ConsoleProgram, you’re looking at the wrong exercise. Checkerboard V2 is almost always graphics-based.
function start() var size = 40; // size of each square var startX = 0; var startY = 0;for(var row = 0; row < 8; row++) for(var col = 0; col < 8; col++) var x = startX + col * size; var y = startY + row * size; var rect = new Rectangle(size, size); rect.setPosition(x, y); // Alternate colors if((row + col) % 2 == 0) rect.setColor("red"); else rect.setColor("black"); add(rect);
Bad Code:
var x = row * 50; // Swapped row and col
Fix: Remember: x = col * size, y = row * size. The column determines horizontal position (x), the row determines vertical position (y).
In the graphics version, recompute square size on window resize. This is rare for CodeHS but possible in advanced sections.
If each square is 50x50 pixels, the top-left corner of the square at (row, col) is:
col * 50row * 509.1.7 Checkerboard V2 on CodeHS is more than just a drawing exercise. It teaches you systematic thinking – how to break a repetitive visual problem into a compact, logical set of instructions.
The final working solution in three lines of logic:
(row + col) % 2 == 0 → Color A, else Color B.Master this, and you have unlocked a fundamental pattern used in game development, graphics programming, and algorithm design.
Next step: Copy the JavaScript solution above, run it in your CodeHS IDE, and watch the red and black grid appear perfectly. Then, experiment by changing BOARD_SIZE to 800 – and watch how the entire program adapts without any other changes. That is the power of writing flexible, "V2"-quality code.
This article provides a comprehensive walkthrough for completing the 9.1.7: Checkerboard V2 exercise in CodeHS. This challenge builds upon basic looping concepts by introducing nested loops and conditional logic to create a complex visual pattern. Understanding the Objective
The goal is to create a grid where the colors of the squares alternate like a traditional checkerboard. Unlike the first version of this exercise, "V2" usually requires a more dynamic approach—often utilizing variables for row and column counts or specific helper methods to determine which color should be placed at a specific coordinate. The Logic Behind the Grid
To build a checkerboard, you need to understand the relationship between the row index ( ) and the column index (
Even Rows: If the row index is even, the pattern might start with Color A.
Odd Rows: If the row index is odd, the pattern must start with Color B.
The Sum Rule: A more efficient way to calculate the color is to check if the sum of the row and column indices (
) is even or odd. If the sum is even, use Color A; if odd, use Color B. Step-by-Step Implementation 1. Define Constants
Start by defining the size of your board and the colors you want to use. This makes your code easier to read and modify later. javascript
var GRID_SIZE = 8; var SQUARE_SIZE = getWidth() / GRID_SIZE; var COLOR_ONE = Color.RED; var COLOR_TWO = Color.BLACK; Use code with caution. 2. The Nested Loop Structure
You will need one loop for the rows and another inside it for the columns. javascript
for (var row = 0; row < GRID_SIZE; row++) for (var col = 0; col < GRID_SIZE; col++) // Drawing logic goes here Use code with caution. 3. Applying Conditional Logic 9.1.7 Checkerboard V2 Codehs
Inside the inner loop, use an if/else statement to decide which color the current square should be. javascript
var x = col * SQUARE_SIZE; var y = row * SQUARE_SIZE; var square = new Rectangle(SQUARE_SIZE, SQUARE_SIZE); square.setPosition(x, y); // The "Checkerboard" Logic if ((row + col) % 2 == 0) square.setColor(COLOR_ONE); else square.setColor(COLOR_TWO); add(square); Use code with caution. Common Pitfalls to Avoid
Off-by-one errors: Ensure your loops start at 0 and end at GRID_SIZE - 1.
Coordinate Math: Remember that the x coordinate is determined by the column, while the y coordinate is determined by the row.
Scaling: If you hardcode the pixel values, the checkerboard won't resize correctly if the GRID_SIZE changes. Always use getWidth() / GRID_SIZE for dimensions.
The 9.1.7 Checkerboard V2 exercise is a rite of passage in CodeHS. It transitions you from writing linear code to thinking in two dimensions. By mastering the nested loop and the modulo operator (%), you gain the tools necessary to build more complex graphics and data structures in the future. Need help with a specific part of the code, or
To solve the CodeHS 9.1.7 Checkerboard V2 exercise, you need to create an 8x8 grid (a 2D list) and fill it with alternating 0s and 1s to form a checkerboard pattern.
The core of this challenge lies in understanding how to access specific elements in a list of lists and applying a mathematical condition to alternate values. The Core Logic: The Modulo Operator
To create a checkerboard, we use the row and column indices. If the sum of the row index 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
Initialize the Grid: Create an empty list and use a loop to append 8 sub-lists, each containing eight zeros.
Nested For Loops: Use one loop to iterate through each row (0-7) and a nested loop to iterate through each column (0-7).
Assignment Statement: Inside the nested loop, use the (row + col) % 2 logic to assign 1 to the correct positions using the syntax grid[row][col] = 1.
Printing the Board: Call the provided print_board function to display your final 2D list. Solution Code
def print_board(board): for i in range(len(board)): # Joins the list elements into a single string for printing print(" ".join([str(x) for x in board[i]])) # 1. Initialize an 8x8 grid filled with 0s my_grid = [] for i in range(8): my_grid.append([0] * 8) # 2. Use nested loops to assign 1s in a checkerboard pattern for row in range(8): for col in range(8): # 3. Check if the sum of indices is odd or even if (row + col) % 2 != 0: my_grid[row][col] = 1 # 4. Print the final result print_board(my_grid) Use code with caution. Common Pitfalls
Indentation Errors: Python relies on proper indentation to know which code belongs inside a loop or function.
Assignment vs. Printing: 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.
Index Out of Range: Ensure both loops run exactly from range(8) to avoid errors when accessing the 8x8 grid.
In CodeHS 9.1.7: Checkerboard V2, the goal is to create a pattern of alternating 1s and 0s in a 2D list (grid). Unlike version 1, which often uses simple row filling, version 2 requires you to use nested for loops and the modulus operator ( ) to check for even and odd positions. Logic for the Pattern
To create a checkerboard, a cell should be a 1 if the sum of its row and column indices is even (or odd, depending on your starting preference). Even sum : Assign 1. Odd sum : Assign 0. Step-by-Step Implementation
Initialize the BoardCreate an empty list and fill it with eight sub-lists, each containing eight zeros. board = [] for i in range(8): board.append([0] * 8) Use code with caution. Copied to clipboard
Iterate and Assign ValuesUse nested loops to traverse every row ( ) and column (
). Use an if statement with the modulus operator to decide where to place a 1.
for i in range(8): for j in range(8): if (i + j) % 2 == 0: board[i][j] = 1 Use code with caution. Copied to clipboard For the CodeHS exercise 9
Print the BoardDefine or use a function to print each row of the list so it looks like a grid.
def print_board(board): for row in board: print(" ".join([str(x) for x in row])) print_board(board) Use code with caution. Copied to clipboard ✅ Result The final output will be an
grid where 1s and 0s alternate perfectly in every direction.
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
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
s where the numbers alternate in a checkerboard pattern. The key is to use nested loops modulus operator ) to determine if a specific cell should be a based on its row and column indices. The Core Logic
A checkerboard pattern is mathematically defined by the sum of the row and column indices. If the sum is even, the cell is one "color" ( ); if it is odd, it is the other ( Step-by-Step Implementation 1. Initialize an Empty Board
First, create a variable to hold your grid. You will start with an empty list and then append rows to it. In CodeHS, you are typically expected to build an structure. 2. Create the Nested Loops You need two loops: one for the rows and one for the columns. outer loop iterates through the row indices ( inner loop iterates through the column indices ( 3. Apply the Alternating Pattern Inside the nested loop, check the sum of the current . Use the modulus operator to check for even or odd values. if (row + col) % 2 == 0 If true, set the cell to If false, set the cell to 4. Print the Result
After the loops have finished building your board, you must print it row by row. Using print " ".join(row)
is a common way to format the output so it looks like a clean grid in the console. Example Code Snippet The following logic illustrates how to construct the grid correctly: # 1. Initialize the board # 2. Outer loop for rows # 3. Inner loop for columns # 4. Check for even/odd sum : row.append( : row.append( ) board.append(row) # 5. Print the formatted board board: print .join([str(x) Use code with caution. Copied to clipboard Why version 2 is different
In earlier versions (like Checkerboard V1), you might have been asked to just fill specific regions (like the top and bottom rows) with
specifically tests your ability to use logic to create a repeating, alternating pattern across the entire grid. Procedural Solution Summary The final answer is a grid where cell is odd, and
The final result is an 8x8 list of lists where every adjacent element (horizontally and vertically) alternates between 0 and 1. modulus operator works for other patterns, or should we look at a different CodeHS exercise
In the CodeHS 9.1.7: Checkerboard V2 exercise, the objective is to create an 8x8 grid of alternating 0s and 1s using nested loops and lists. This task builds on previous iterations by requiring a dynamic approach to row and column indexing. Key Programming Concepts
Nested Loops: You must use a loop inside another loop—typically an outer loop for the rows and an inner loop for the columns—to traverse every coordinate in the grid.
Modulo Operator (%): This is the most efficient way to determine if a row or column index is even or odd. For a checkerboard, a cell (row, col) usually contains a 1 if the sum of its indices (row + col) is even, and a 0 otherwise.
2D Lists (Grids): You are managing a list where each element is itself a list (representing a row). Logical Strategy To solve this correctly, follow these general steps:
Initialize the Grid: Create an empty list and use a loop to append eight rows, each initially filled with eight zeros.
Apply Checkerboard Logic: Iterate through the rows and columns. Use an if statement with the modulo operator to check the indices.
Update Elements: Change specific zeros to ones based on your condition (e.g., if (row + col) % 2 == 0).
Display the Result: Use the provided print_board function to output your final 8x8 nested list in a readable format. Common Pitfalls
Incorrect Indentation: In Python, all code within the checkerboard function must be indented properly to execute as a single block.
Hardcoding Values: Avoid manually typing out the lists; the challenge expects you to use loops to generate the pattern programmatically. Note : If your class extends ConsoleProgram ,
Off-by-One Errors: Ensure your range() parameters correctly cover indices 0 through 7 for an 8x8 board.
For further help, you can review the Python Programming Outline or similar exercises like 9.1.6: Checkerboard V1 on CodeHS.