Below are the few steps given for scan-converting a circle using Bresenham’s Algorithm. Which of the given steps is not correct ?
If d ≥ 0, then d = 4 ∗ (x – y) + 10, x = x + 1 and y = y + 1
Bresenham’s circle algorithm plots one octant of the circle, starting at (0, r) and moving right while x ≤ y, and uses an integer decision parameter d to choose between the pixel straight to the right and the pixel diagonally down-right. The other seven octants are filled by symmetry.
Step 1 — Initialisation: x = 0, y = r and \(d = 3 - 2r\). The step “Compute d = 3 – 2r” is correct.
Step 2 — Termination: the octant is finished once x passes y, so “Stop if x > y” is correct.
Step 3 — If d < 0, the pixel (x + 1, y) is closer to the true circle. The parameter grows by 4x + 6 (\(d \leftarrow d + 4x + 6\)) and only x is incremented, which is what the listed step describes.
Step 4 — If d ≥ 0, the diagonal pixel (x + 1, y – 1) is chosen: \(d \leftarrow d + 4(x - y) + 10\), x = x + 1 and y = y – 1. Because y starts at r and must fall towards the 45° line, the listed step that increments y (y = y + 1) drives the plot away from the circle, so it is the incorrect step.
If d ≥ 0, then d = 4 ∗ (x – y) + 10, x = x + 1 and y = y + 1 option is correct.
Similar Questions
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Codes :
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A/An _____ creates an image directly on paper by spraying ink.
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Codes :
