9.1.7 Checkerboard V2: Answers

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

# Pass this function a list of lists, and it will # print it such that it looks like the grids in # the exercise instructions. def print_board(board): for i in range(len(board)): print(" ".join([str(x) for x in board[i]])) # 1. Initialize an empty 8x8 board board = [] # 2. Use nested loops to fill the board with the checkerboard pattern for i in range(8): row = [] for j in range(8): # 3. Use the sum of indices to determine the value (0 or 1) if (i + j) % 2 == 0: row.append(0) else: row.append(1) board.append(row) # 4. Print the final result print_board(board) Use code with caution. Copied to clipboard Explanation of the Logic 9.1.7 checkerboard v2 answers

• If placements cannot be adjacent or must be separated by at least one cell, mark impossible neighbors after placing one.
• For checkerboard parity constraints, split cells into two color classes and compare required placements to available counts on each color.

The solution to the 9.1.7 Checkerboard v2 programming exercise involves using nested for loops conditional logic (the modulo operator The solution to the 9

Cell at row r, column c is black if (r + c) % 2 == 0, otherwise white (or vice versa). If placements cannot be adjacent or must be

In this article, we will break down the problem, explore the logic, provide the code solution, and explain why each line works. Whether you are looking for the direct 9.1.7 checkerboard v2 answers or want to understand the underlying concepts, this guide has you covered.

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:

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