The solution for the CodeHS 9.1.7: Checkerboard v2 exercise requires creating an grid of alternating
s using a nested loop. The most efficient way to achieve this pattern is by checking if the sum of the current row index ( ) and column index ( ) is even or odd. Python Solution
# 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
Initialize the Grid: We start by creating an empty list board that will eventually hold 8 row lists.
Nested Loops: The outer loop (i) iterates through the 8 rows, while the inner loop (j) iterates through the 8 columns for each row.
Checkerboard Condition: The key to the pattern is the logic (i + j) % 2 == 0. 9.1.7 checkerboard v2 answers
If the sum of the row and column index is even, we append a 0. If the sum is odd, we append a 1.
This ensures that the values alternate both horizontally and vertically, creating the classic "v2" checkerboard style.
Printing: The print_board function takes the list of lists and converts each integer to a string, joining them with spaces for a clean visual output.
9.1.7 Checkerboard, v2 I got this wrong, and I can't ... - Brainly
The CodeHS 9.1.7: Checkerboard, v2 exercise requires creating an 8x8 grid of alternating 0s and 1s using nested for loops and the modulus operator (%). Solution Overview The solution for the CodeHS 9
To solve this, you must initialize an 8x8 grid filled with 0s and then iterate through each row and column. If the sum of the row index and column index is odd, set that specific element to 1. This logic ensures the pattern alternates correctly across both rows and columns. Python Implementation
According to expert discussions on Reddit and Brainly, the most efficient solution follows this structure:
# 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
The Modulus Operator (%): This is the critical tool for the exercise. Checking (row + col) % 2 == 1 identifies every other cell in a grid pattern.
Row/Column Alternation: Without using the sum (row + col), you might only alternate colors within a single row rather than creating a true checkerboard. Mathematics Formulation If we are discussing the number
Nested Loops: The outer loop tracks the row index, while the inner loop tracks the col index to access every individual element in the 2D list.
If we are discussing the number of ways to place (n) checkers on an (n \times n) board such that no two checkers are in the same row or column, the solution can be expressed as:
$$ \textNumber of ways = n! $$
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).
Let’s trace (row + col) % 2 for the first few squares:
This creates perfect alternation in both directions, mimicking a real checkerboard.
To make each square stand out, add:
square.setBorderColor(Color.WHITE);
ArrayList<Color> for the row outside the inner loop.ArrayList for each row inside the outer for loop. If you reuse the same list, all rows will be identical.