The 916 Checkerboard problem on CodeHS is a classic challenge that requires creating a checkerboard pattern using a loop. Here is a fixed and well-documented solution:
The objective of the "Checkerboard" assignment is to write a graphics program that draws a standard 8x8 checkerboard (like a chess board) on the screen.
Requirements:
In the landscape of introductory computer science, few tools are as effective for teaching logic as the CodeHS graphics library. Among the classic exercises presented to students is the creation of a checkerboard—a seemingly simple visual pattern that actually requires a deep understanding of coordinate systems, iteration, and conditional logic. The "916 Checkerboard v1" assignment is a specific variation of this problem that often trips up beginners. A "fixed" version of this code does more than just produce a pretty picture; it demonstrates the fundamental shift from linear thinking to algorithmic problem-solving.
The Illusion of Simplicity
At first glance, a checkerboard appears trivial. It is simply a grid of alternating red and black squares. A student’s first instinct is often to "hard code" the solution: draw a red square, then a black square, then a red square, and manually position them one by one. However, the "916" specification usually implies a large grid (likely 8x8 or similar dimensions), making hard-coding impractical and tedious. The "fixed" solution abandons the manual approach in favor of automation, using nested loops to traverse the rows and columns.
The beauty of the fixed code lies in its use of the for loop. By nesting a column loop inside a row loop, the program efficiently visits every coordinate on the grid. This structure teaches students how computers handle two-dimensional space: not as a continuous canvas, but as a matrix of discrete points defined by x and y coordinates.
The Logic of Alternation
The core challenge of the "916 Checkerboard" is not drawing the squares, but determining their color. This is where many early attempts fail. A common misconception is that the color alternates simply based on the loop counter (e.g., if i is even, red; if i is odd, black). While this works for a single line, it fails on a grid because the first square of a new row must be the opposite color of the last square of the previous row.
The "fixed" solution solves this through modular arithmetic. The logic typically follows a formula checking the sum of the row and column indices:
if ((row + col) % 2 == 0)
// Draw Red Square
else
// Draw Black Square
This is the "aha!" moment for the assignment. It teaches that patterns in computer science are often mathematical. By checking if the sum of the coordinates is even or odd, the code automatically creates the staggered pattern required for a checkerboard, regardless of the grid size.
Fixing Common Errors
Why is the "fixed" version necessary? The "v1" designation implies an iterative process. Common errors in the unfixed versions include:
x and y by multiplying the loop counters by the square size.The "fixed" code addresses these by ensuring the loop parameters match the grid dimensions precisely and that the offset logic (row + col) is implemented correctly.
Conclusion: Beyond the Graphics
The "916 Checkerboard v1 CodeHS Fixed" is not just a solution to a homework assignment; it is a milestone in a programmer's education. It transitions a student from a human who gives manual instructions to a programmer who designs algorithms. The fixed code is efficient, readable, and mathematically elegant. By mastering the logic required to fix this checkerboard, students gain the foundational skills necessary to tackle more complex problems, from rendering game boards to managing large data sets in two-dimensional arrays.
Cracking the Grid: 916 Checkerboard v1 CodeHS Fixed If you’re working through the CodeHS JavaScript curriculum, Exercise 9.1.6: Checkerboard v1 is often the first major "wall" students hit. It requires you to move beyond simple loops and start thinking about 2D space, coordinates, and conditional logic.
If your circles are overlapping, only appearing on one line, or refusing to alternate colors, you’re in the right place. Here is the logic, the common bugs, and the fixed code to get your checkerboard working perfectly. The Logic Behind the Grid
To build a checkerboard, you have to tell the computer to do two things simultaneously: Create a Row: Place circles from left to right.
Create a Column: Repeat that row-making process from top to bottom.
This is achieved using Nested For Loops. The outer loop handles the vertical movement (y-coordinates), and the inner loop handles the horizontal movement (x-coordinates). Common Mistakes (Why your code is broken)
Before looking at the fix, check if you fell into these common traps:
Variable Scope: Using the same variable (like i) for both loops. This causes the loops to crash into each other.
Math Errors: Forgetting that radius is half of the diameter. If your circles are 40 pixels wide, you need to move 40 pixels to reach the next center point.
Color Logic: Forgetting to "flip" the color starting point for every other row. The Fixed Code (JavaScript/Karel)
Here is the clean, fixed solution for the 9.1.6 Checkerboard v1. This version uses constants to make it easy to adjust the size. javascript 916 checkerboard v1 codehs fixed
/* This program draws a checkerboard pattern using nested loops. */ var RADIUS = 20; var DIAMETER = RADIUS * 2; function start() // Outer loop for the vertical rows (Y-axis) for (var row = 0; row < getHeight() / DIAMETER; row++) // Inner loop for the horizontal circles (X-axis) for (var col = 0; col < getWidth() / DIAMETER; col++) var x = col * DIAMETER + RADIUS; var y = row * DIAMETER + RADIUS; // Logic to determine color based on grid position if ((row + col) % 2 == 0) drawCircle(x, y, Color.red); else drawCircle(x, y, Color.black); function drawCircle(x, y, color) var circle = new Circle(RADIUS); circle.setPosition(x, y); circle.setColor(color); add(circle); Use code with caution. Breakdown of the Fix
row + col % 2: This is the "magic" math. By adding the row index and column index together and checking if the sum is even or odd, you create a perfect alternating pattern. Without this, every row would look identical.
col * DIAMETER + RADIUS: We multiply the column index by the diameter to move to the next "slot." We add the radius because circles in CodeHS are positioned by their center, not their top-left corner.
Dynamic Bounds: Using getWidth() / DIAMETER ensures that your checkerboard fills the screen regardless of how big the canvas is. Pro-Tip for CodeHS Debugging
If your circles aren't showing up, use console.log("Row: " + row + " Col: " + col); inside your inner loop. If you see the numbers printing in the console, your loops are working, and the issue is likely your circle.setPosition math! How are you planning to customize your checkerboard—
To fix the CodeHS 9.1.6 Checkerboard, v1 exercise, you must go beyond simply printing the final pattern. The autograder specifically requires you to use assignment statements to modify elements within a 2D list. Common Fixes for 9.1.6 Use Assignment Statements:
Many students fail the autograder because they try to print the pattern directly using strings. The assignment requires you to first create a grid (usually filled with s) and then use nested loops to change specific indices to Logic for Alternating Pattern: To get the true checkerboard effect, use the modulo operator board[i][j] = (i + j) % 2
ensures that the values alternate between 0 and 1 across both rows and columns. Row-Specific Constraints: V1 of this exercise often asks for pieces (represented by
s) to only appear on the top and bottom sections. A common fix is to use a conditional statement like if row < 3 or row > 4: to only assign s in those specific row ranges. Step-by-Step Implementation Guide Initialize the Board: Create an 8x8 list of lists filled with zeros. my_grid = [[0] * 8 for i in range(8)] Nested Loop Assignment: Loop through every row and column. Use an
statement to check if the row index is within the "piece area" (typically rows 0–2 and 5–7). Apply the Pattern: Inside that conditional, use (row + col) % 2 == 0 to decide if that specific cell should be changed to a Call the Print Function: Use the provided print_board(board)
function at the very end to display your final, modified grid.
For more specific debugging help, check out community discussions on platforms like Reddit's CodeHS community Brainly's exercise walkthroughs or help you with the nested loop
If your code still doesn’t work after checking the logic above:
If you post your current non-working code (without asking for the full solution), I can point out the exact mistake.
The CodeHS 9.1.6 Checkerboard v1 exercise requires students to create an
grid and populate it with a alternating pattern of 0s and 1s to resemble a checkerboard. Standard "Fixed" Implementation
To solve this correctly, you must use nested for loops and a mathematical check to determine which number (0 or 1) to place in each cell.
def create_board(): board = [] for i in range(8): row = [] for j in range(8): # Check if the sum of indices is even or odd if (i + j) % 2 == 0: row.append(0) else: row.append(1) board.append(row) return board # Usage my_board = create_board() print_board(my_board) Use code with caution. Copied to clipboard đź’ˇ Key Logic: The Modulo Operator
The most efficient way to "fix" a broken checkerboard script is using the formula (i + j) % 2. This ensures that: Even sums get one value. Odd sums
get the other.This creates the alternating effect regardless of the board's dimensions. Common Fixes for Errors
Empty Board Error: Ensure you initialize board = [] before the loops and row = [] inside the first loop. Index Management: Always use range(8) for an board to avoid "index out of bounds" errors.
Assignment vs. Initialization: CodeHS often requires you to create the full structure first and then modify specific elements using board[row][col] = 1.
If you are seeing a specific error message like "You should use a nested for loop" or "Board is not the right size," let me know! I can help you debug that specific part of your code.
Solved 9.1.6: Checkerboard, v1 Save 1 # Pass this function a
Running this code will produce a standard 8x8 checkerboard pattern with alternating black and white squares. 916 Checkerboard v1 Codehs Fixed The 916 Checkerboard
Step-by-Step Solution:
rect function.Mastering the 916 Checkerboard v1: Solutions and Logic for CodeHS
If you are working through the CodeHS curriculum, you’ve likely encountered the 9.1.6 Checkerboard v1 assignment. It’s a classic challenge that tests your ability to use nested loops, coordinate systems, and conditional logic.
However, getting the "fixed" version—where the grid perfectly alternates colors without overlapping or skipping—can be tricky. The objective is to create an
grid of squares where the colors alternate between black and red (or other assigned colors), resembling a standard checkerboard. Key Technical Requirements:
Nested Loops: You need an outer loop for rows and an inner loop for columns.
Size Calculations: Each square must be the width of the canvas divided by 8.
The "Fixed" Logic: The color must switch based on both the row and column index to create the staggered effect. The Logic Behind the Fix
The most common mistake in "v1" is only checking if the column is even or odd. If you do that, every row will look identical, resulting in vertical stripes rather than a checkerboard. The Solution: Use the sum of the row and column indices. If (row + col) is even, color it Red. If (row + col) is odd, color it Black. The Corrected Code (JavaScript/Karel Style)
Here is a clean, "fixed" implementation for the CodeHS environment: javascript
var SQUARES_PER_SIDE = 8; var SQUARE_SIZE = getWidth() / SQUARES_PER_SIDE; function start() for (var row = 0; row < SQUARES_PER_SIDE; row++) for (var col = 0; col < SQUARES_PER_SIDE; col++) drawSquare(row, col); 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 "Fixed" Logic: Check if sum of indices is even if ((row + col) % 2 == 0) rect.setColor(Color.red); else rect.setColor(Color.black); add(rect); Use code with caution. Troubleshooting Common Errors 1. The "Off-by-One" Pixel Gap
If you see white lines between your squares, ensure you are calculating SQUARE_SIZE using getWidth() / 8. If you hardcode a number like 50 on a canvas that isn't exactly 400, the grid won't fit perfectly. 2. Rectangles Overlapping the Border
Make sure your setPosition uses col * SQUARE_SIZE for the X-coordinate and row * SQUARE_SIZE for the Y-coordinate. Swapping these can sometimes cause the grid to render incorrectly if your canvas isn't a perfect square. 3. Infinite Loops
Ensure your for loop conditions use < SQUARES_PER_SIDE and not <=. Using <= will attempt to draw a 9th row/column, which usually breaks the layout or triggers a "limit exceeded" error in CodeHS.
The "916 checkerboard v1 codehs fixed" solution relies entirely on the row + column parity. Once you master the nested loop structure, you can apply this logic to more complex grid-based games like Minesweeper or Chess.
Are you having trouble with the v2 version of this assignment, or is the autograder still giving you a specific error message?
Cracking the Code: How to Fix CodeHS 9.1.6 Checkerboard V1 If you're stuck on CodeHS 9.1.6: Checkerboard, v1
, you aren't alone. This exercise is a classic "gotcha" because it doesn't just want the right visual output; it wants you to use specific programming techniques—like nested loops and list indexing—to get there.
Many students fail this one because they try to "shortcut" the board creation by appending pre-made lists. Here’s how to fix your code so it passes every test case. The Problem: Why Your Code Isn't Passing The autograder for this exercise specifically checks for assignment statements
. If you just print strings or append a row of ones, you'll likely see errors like: "You should set some elements of your board to 1" "You will need to use an assignment statement"
The system wants to see you access a specific spot in a 2D list (e.g., board[i][j] = 1 The Solution: Step-by-Step Fix
To pass, you must first initialize a grid full of zeros and then use nested
loops to "spot-fill" the ones where the checker pieces should go. 1. Initialize the 8x8 Grid Start by creating a list of lists where every value is ): board.append([ Use code with caution. Copied to clipboard 2. Use Nested Loops with Assignment
Now, loop through the rows and columns. According to the instructions, you need 1s in the top three rows (indices 0, 1, 2) and the bottom three rows (indices 5, 6, 7). To get that alternating checkerboard look, use the modulus operator
). A common trick is checking if the sum of the row and column indices is even: (i + j) % 2 == 0 # Top 3 rows and Bottom 3 rows only : board[i][j] = # This is the "assignment statement" it wants! Use code with caution. Copied to clipboard 3. Print the Result Finally, call the provided print_board(board) function to display your work. Why This Version Works Nested Loops: It proves you can navigate a 2D data structure. board[i][j] The board must consist of 8 rows and 8 columns
, you are directly modifying the data, which satisfies the "assignment statement" requirement.
statements correctly skip the middle two rows, leaving them as zeros.
Now that you've mastered the basic grid, are you ready to tackle Checkerboard v2 and add more complex patterns?
For the CodeHS exercise 9.1.6: Checkerboard, v1, the goal is to initialize a
list (grid) and then use nested loops to set specific elements to
to create a checkerboard pattern in the top and bottom three rows, while leaving the middle two rows as Final Correct Code
# Create an 8x8 board filled with 0s board = [[0] * 8 for _ in range(8)] # Use nested for loops to modify specific rows for row in range(8): for col in range(8): # Top 3 rows (0, 1, 2) and bottom 3 rows (5, 6, 7) if row < 3 or row > 4: # Checkerboard condition: sum of indices is even if (row + col) % 2 == 0: board[row][col] = 1 # Print the board using the provided function print_board(board) Use code with caution. Copied to clipboard Step-by-Step Explanation Initialize the GridCreate an list of lists where every element starts as
. You can do this quickly using a list comprehension: board = [[0] * 8 for _ in range(8)].
Identify Target RowsThe instructions require modifications only to the top 3 rows (indices ) and the bottom 3 rows (indices ). The middle two rows (indices ) must remain all
Apply Checkerboard LogicInside a nested loop, use the mathematical property of a checkerboard: an element should be if the sum of its row and column indices is an even number. Example: At board[0][0], (even), so it becomes Example: At board[0][1], (odd), so it remains
Verify with print_boardPass your modified board variable into the print_board() function already provided in the CodeHS editor to see the visual output and pass the test cases. Common Troubleshooting Tips
"You should set some elements to 1": This error usually means your if condition for the rows is wrong or you aren't actually assigning board[row][col] = 1.
Middle Rows: Ensure your row check is if row < 3 or row > 4:. If you use row <= 3, you will incorrectly modify the 4th row.
Solved 9.1.6: Checkerboard, v1 Save 1 # Pass this function a
The core objective of CodeHS 9.1.6: Checkerboard, v1 is to practice modifying 2D lists using nested loops and index-based assignment.
The autograder often fails students who simply print the pattern; it strictly requires that you initialize a board of 0s and then use assignment statements (e.g., board[i][j] = 1) to place pieces. Fixed Python Solution
This code initializes an 8x8 grid of zeros and then fills the top three and bottom three rows with a checkerboard pattern of 1s.
# Function to print the board def print_board(board): for row in board: # Join elements with a space for proper formatting print(" ".join([str(x) for x in row])) # 1. Initialize an 8x8 board with all zeros board = [] for i in range(8): board.append([0] * 8) # 2. Use nested loops to assign 1s to specific indices # Row indices 0, 1, 2 are the top three rows # Row indices 5, 6, 7 are the bottom three rows for row in range(8): for col in range(8): # Check if the row should have pieces if row < 3 or row > 4: # Only set to 1 if (row + col) is even to create the pattern if (row + col) % 2 == 0: board[row][col] = 1 # 3. Display the final board print_board(board) Use code with caution. Copied to clipboard Key Logic & Fixes đź’ˇ
Explicit Assignment: You must use board[row][col] = 1. The autograder specifically looks for the = assignment operator being used on the list elements.
The Modulo Trick: Using (row + col) % 2 == 0 ensures that the 1s alternate correctly across both rows and columns.
Row Filtering: The condition if row < 3 or row > 4 targets the "top 3" (0, 1, 2) and "bottom 3" (5, 6, 7) rows as required by the exercise.
String Formatting: Using " ".join(...) ensures the output matches the expected CodeHS console format exactly, preventing "Output does not match" errors. Common Errors to Avoid
Defining print_board inside another function: This leads to scope errors where the autograder can't find your display logic.
Using hardcoded print statements: Printing 1 0 1 0... directly will pass the visual check but fail the "You should set some elements of your board to 1" test.
If you're also working on 9.1.7: Checkerboard, v2, let me know—the logic changes slightly to focus on a full-board pattern or different row offsets.
(row + column) % 2 == 0 → Red(column * 50, row * 50) with size 50x50