This article breaks down the logic, the math, and the exact code structure needed to solve this exercise efficiently. Understanding the Goal
rect.setColor(Color.WHITE);
Let's test this:
The challenge is deciding when to use gray and when to use black. There is a simple mathematical trick:
if (frontIsClear()) move(); col++; else break; 9.1.6 checkerboard v1 codehs
def checkerboard(n, a="X", b=" "): # n: board dimension (nonnegative int) # a, b: tokens for even and odd parity cells for r in range(n): line_chars = [] for c in range(n): if (r + c) % 2 == 0: line_chars.append(a) else: line_chars.append(b) print("".join(line_chars))
: An outer loop controls the rows, while an inner loop controls the columns. This article breaks down the logic, the math,
Alternatively, you can think of it as: if the row is even, start with color A; if the row is odd, start with color B. The Code Implementation (Java/CodeHS Style)
Properties to satisfy:
function start() // Create checkerboard pattern var row = 1; while (true) for (var i = 0; i < 100; i++) // max columns if (row % 2 == 1) if (i % 2 == 0) putBeeper(); else if (i % 2 == 1) putBeeper();