var color; if ((row + col) % 2 === 0) color = "red"; else color = "black";
To build a flawless checkerboard pattern, you must calculate the value of a cell based on its unique mathematical coordinates: the and the column index .
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If the assignment specifically requires while loops (where the bug usually exists), here is the fixed while loop implementation. 916 checkerboard v1 codehs fixed
). Use an if statement to check if the current row index is in the top three (less than 3) or bottom three (greater than 4). If it is, use an assignment statement to change the 0 to a 1.
# 1. Initialize the 8x8 grid with all 0s grid = [] for i in range(8): grid.append([0] * 8) # 2. Use a nested loop to set specific rows to 1 for i in range(8): # Only modify the top 3 (i < 3) or bottom 3 (i > 4) rows if i < 3 or i > 4: for j in range(8): grid[i][j] = 1 # 3. Print the board using the provided function # (Make sure print_board is defined or provided by CodeHS) print_board(grid) Use code with caution. Copied to clipboard
(leftIsClear()) turnLeft(); move(); turnLeft(); var color; if ((row + col) % 2
If you are working in or JavaScript (CodeHS uses both).
If your code still does not pass the CodeHS autograder, check these final three details:
If your CodeHS autograder is failing, you are likely running into one of three frequent issues. 1. The "Reversed Row" Bug If you share with third parties, their policies apply
# Starting position (Bottom-left or Top-left depending on preference) # Here we start from top-left for standard drawing order start_x = -200 start_y = 200
The following structure resolves the common issue where Karel places two balls in a row or skips a corner when moving to the next level: Fixed Code Structure (Karel JavaScript) javascript
# 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 💡