9.1.7 Checkerboard V2 Answers

logic is the most efficient way to solve version 2 of this problem. In "v1," you might have only alternated colors per row, but adding the row and column indices together ensures that if you are on an even row, the pattern starts with "Color A," and if you are on an odd row, it starts with "Color B." ✅ Final Result The answer is achieved by using nested for loops combined with the conditional statement if (row + col) % 2 == 0

A checkerboard follows a simple :

: The outer loop tracks the row index, while the inner loop tracks the col index to access every individual element in the 2D list. 9.1.7 checkerboard v2 answers

def print_board(self): for row in self.board: for cell in row: if cell is None: print('-', end=' ') else: print(cell.color[0].upper(), end=' ') print() logic is the most efficient way to solve

| 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). | | | All squares are the same color

Start by creating a 2D list (grid) that contains 8 rows and 8 columns, with all elements initially set to 0. 2. Iterate Through Rows and Columns

Loop again, using x = col * 50 , y = row * 50 , create GRect , set fill color from board[row][col] , and add(square) .