public void run() // Loop through each row for (int row = 0; row < NUM_ROWS; row++) // Loop through each column in the current row for (int col = 0; col < NUM_COLS; col++) // Calculate the x and y coordinates for this square int x = col * SQUARE_SIZE; int y = row * SQUARE_SIZE;
This is a classic problem of permutations. For the first checker, there are (n^2) possible squares. Once a square is chosen, for the second checker, there are ((n-1)^2) possible squares (since a row and a column are now off-limits), and so on. However, a more straightforward way to think about it is: 9.1.7 checkerboard v2 answers
This often involves printing an 8x8 grid that alternates between two different colors (typically black and white) for each square. public void run() // Loop through each row
private static final int ROWS = 8; private static final int COLS = 8; private static final int SQUARE_SIZE = 50; However, a more straightforward way to think about
A: Ensure your canvas size is exactly 400x400 (since 8 * 50px = 400px). If you used getWidth() , the board might be off by a few pixels if the window isn't perfectly square.
# Function to print the board as required by the exercise def print_board(board): for row in board: print(" ".join([str(x) for x in row])) # 1. Initialize an 8x8 grid with all 0s my_grid = [] for i in range(8): my_grid.append([0] * 8) # 2. Use nested loops to apply the checkerboard pattern for row in range(8): for col in range(8): # Use modulus on the sum of row and col to find "odd" positions if (row + col) % 2 == 1: my_grid[row][col] = 1 # 3. Print the final board print_board(my_grid) Use code with caution. Copied to clipboard Key Logic Points