Swap out Color.RED and Color.BLACK for any valid java.awt.Color (e.g., Color.BLUE, Color.YELLOW, Color.MAGENTA).
Even with the correct code, students often hit frustrating roadblocks. Here’s a quick troubleshooting table:
| Error Message / Symptom | Likely Cause | Solution |
|-------------------------|--------------|----------|
| Square is not filled | Missing setFilled(true) | Add the line before setting the color. |
| All squares are the same color | Incorrect modulus logic | Use (row + col) % 2 == 0. Check your starting color. |
| ArrayIndexOutOfBoundsException | Using <= instead of < in loop | Ensure loops run 0 to 7 (not 0 to 8). |
| Nothing appears on screen | Forgot to call add(square) | After creating and coloring, call add(square). |
| Squares overlap or have gaps | Incorrect coordinate math | x = col * size, y = row * size. Don’t add extra offset. |
| Autograder fails on style | Missing constants or comments | Use private static final int for magic numbers (8, 50). |
Here is a simple Python solution to generate a checkerboard pattern:
def print_checkerboard():
for row in range(8):
for col in range(8):
# Use the sum of row and column indices to determine the color
if (row + col) % 2 == 0:
print('\033[40m ', end='') # Black
else:
print('\033[47m ', end='') # White
print('\033[0m') # Reset color
print_checkerboard()
This script prints a simple text-based checkerboard to the console. The colors are represented using ANSI escape codes.
This post provides clear, step-by-step answers and reasoning for the puzzle titled “9.1.7 Checkerboard v2.” It covers the puzzle setup, key observations, the complete solution(s), common pitfalls, and an optimization note.
set canvas size (e.g., 400x400) squareSize = canvasWidth / 8
for row from 0 to 7: for col from 0 to 7: x = col * squareSize y = row * squareSize if (row + col) % 2 == 0: color = RED else: color = BLACK draw a square at (x, y) of size squareSize with fill color
To generate the correct pattern, you need a rule that determines the color based on row and column indices. The chessboard pattern relies on parity (evenness or oddness of a number).
CodeHS 9.1.7 Checkerboard v2 exercise, the goal is to create a function that generates a 2D list (a list of lists) representing an 8x8 checkerboard pattern using 0s and 1s. Solution Code # Create an empty list for the current row current_row 9.1.7 checkerboard v2 answers
# Check if the sum of indices is odd or even to alternate colors (row + col) % : current_row.append( : current_row.append( # Add the completed row to the grid my_grid.append(current_row) # Print each row to display the board my_grid: print(row) # Call the function to execute Use code with caution. Copied to clipboard Key Logic Steps Initialize the Grid : Start by creating an empty list, , which will eventually hold eight separate row lists. Nested Loops : Use a outer loop to iterate through 8 rows and an inner loop to iterate through 8 columns. Alternating Pattern Logic
: The core feature of a checkerboard is that adjacent cells differ. Mathematically, you can determine which number to place by checking if the sum of the current indices is even or odd. (row + col) % 2 == 1 Otherwise, place a Row Construction : In each iteration of the outer loop, a current_row list is filled by the inner loop and then appended to : Finally, loop through
and print each individual row to show the 8x8 pattern in the console. of this checkerboard or change the starting color
To solve the CodeHS 9.1.7 Checkerboard v2 exercise, you must create a 2D list (grid) and use nested for loops to populate it with alternating 0s and 1s
. Unlike the first version, this challenge specifically checks that you use assignment statements to modify elements within the grid. Solution Code
The most efficient way to determine the pattern is to check if the sum of the current row and column index is even or odd using the modulus operator
The solution to the 9.1.7: Checkerboard, v2 exercise on CodeHS involves creating a function that generates an grid of alternating Correct Code Implementation
The following Python code defines the checkerboard function and uses nested loops to build the grid. Use code with caution. Copied to clipboard Step-by-Step Logic
Iterate Through Rows and ColumnsUse a nested for loop where the outer loop represents the row index and the inner loop represents the column index , both ranging from Swap out Color
Determine Alternating ValuesA checkerboard pattern is defined by the sum of its coordinates. If is even, the cell gets one value (e.g., ); if it is odd, it gets the other (e.g.,
). Alternatively, you can check if the row index is even to decide if the row starts with
Format the OutputWithin the outer loop, convert the list of integers into a string using " ".join() to ensure the numbers are separated by spaces as required by the exercise. ✅ Final Output The resulting pattern should look like this:
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 1 0 1 0 1 0 1 0 Use code with caution. Copied to clipboard
Do you need help with the 9.1.6: Checkerboard, v1 version or a different CodeHS module?
9.1.7 Checkerboard, v2 I got this wrong, and I can't ... - Brainly
The solution to the CodeHS Python 9.1.7: Checkerboard, v2 assignment involves using a function to generate a 2D list (a list of lists) where alternating elements represent a checkerboard pattern. Correct Answer Code
def board(): my_grid = [] for i in range(8): # Even rows start with 0 if i % 2 == 0: my_grid.append([0, 1] * 4) # Odd rows start with 1 else: my_grid.append([1, 0] * 4) # Print each row in the grid for row in my_grid: print(row) board() Use code with caution. Copied to clipboard 1. Initialize the grid container
The first step is to create an empty list, often called my_grid, which will hold the rows of your checkerboard. Since a checkerboard is typically Here is a simple Python solution to generate
, you will use a loop that runs 8 times to create 8 distinct rows. 2. Differentiate even and odd rows
A checkerboard pattern relies on alternating starting values. You use the modulo operator (i % 2 == 0) to check if the current row index is even or odd. Even rows: Start with 0 (e.g., 0, 1, 0, 1...). Odd rows: Start with 1 (e.g., 1, 0, 1, 0...). 3. Use list multiplication for efficiency
Instead of writing a nested loop to fill each individual cell, you can multiply a small list to fill the row. Multiplying [0, 1] by 4 creates a list of 8 elements: [0, 1, 0, 1, 0, 1, 0, 1]. This is a concise way to ensure each row has exactly 8 columns. 4. Print the final 2D structure
After the grid is populated with all 8 rows, you must iterate through my_grid and print each inner list. This displays the board in a readable format rather than printing the entire nested list on a single line. Final Result The final code uses a single function to build an
grid of alternating 0s and 1s by checking row indices and appending pre-formatted lists, resulting in a perfectly formatted checkerboard pattern.
Do you need help with nested loops for a more manual version of this grid, or
9.1.7 Checkerboard, v2 I got this wrong, and I can't ... - Brainly
If the user resizes the window, you want the board to redraw. Override the init() method to add a resize listener or simply set the canvas size to be fixed using setSize(400, 400) in the constructor.