Question
Using the Midpoint method and taking symmetry into account develop an efficient method for scan converting the curve y = x³/12 in [-10, 10].
Answer :
Word Count : 488
To scan convert the curve (y = \frac{x^3}{12}) using the Midpoint method, we exploit symmetry about the origin because the curve is odd. We consider only (x \ge 0) and reflect points for (x < 0). The Midpoint method is based on choosing the next pixel based on the value of the decision parameter (d) at the midpoint between candidate pixels. Let the pixel grid spacing be 1 unit in both (x) and (y). The function is (f(x) = \frac{x^3}{12}). Start at the origin ((0,0)). For each step, we have two candidate pixels: 1. East (E): ((x+1, y)) 2. North-East (NE): ((x+1, y+1)) Decision parameter at midpoint (M = (x+1, y+\frac{1}{2})) is [ d = f(x+1) - \left(y + \frac{1}{2}\right) ] * ________ ____ __________ ____ ____ __________.
________ _________ ______ _________ ________ _________ ____ __________ __________ ______.
__________ _____ _________ _________ __________ ____ __________.
________ ____ _________ ______ ____ ________ _____ ____.
___ ______ _________ ______ _______ ___ _______ _________ _________ ______ __________.
__________ ___ ________ _______ ________.
________ _______ __________ ____ _______ ________ ___ ______ _______.
__________ _____ __________ ___ _________.
_______ _____ ____ _____ ____ _______ ___ _________.
__________ __________ __________ ___ ________.
_____ ______ ____ _________ ______ __________ ____ ________ ____ _______.
____ ______ ______ ___ ________ _______ ______ ____.
______ _________ _______ _________ ______ ______ ________ _____ _________.
________ _______ __________ ________ ________.
_____ _______ _________ _______ _________ ____ __________ ________ ______ __________ _________.
_________ _____ _____ _____ ________ ___ ____ _________ _______.
_____ _______ _____ ________ _______ ____ ___ ___ ________ _______ ____.
_____ ________ ____ ___ _____ ____ ________.
___ _________ __________ ____ ____ ____ _____ ___ ________.
________ ________ ________ _______ _____ __________ _________ _____ ______ _______.
___ ____ ______ _________ _______ ___ _____ _______ _______.
_________ ______ _____ _________ _______ ___.
___ ___ _________ __________ ___ __________ ____ _________.
__________ ____ __________ ______ _____ __________.
_______ _______ _________ ________ __________ ________ ______ ______ ______ ________ _____ _________.
______ __________ ___ _____ _________ ________ __________ ___ ________ _________.
_____ ________ _____ ___ ________ ______ ___.
__________ _____ ______ ______ _________ ________ ___ ______.
______ ___ ________ ____ _______ ___ ____ ____ ______ ____ ______ _________.
_____ _______ ____ _________ _______ ____ ______ ______.
__________ ________ _________ _________ ___ _________ _________ ____ __________ _____ ___.
_____ ____ _____ _________ _______ ___ ______ ______ __________ _______ __________.
_____ __________ __________ _________ ___ _____ ___ _____ ______ ______.
_______ ___ _________ ___ ____.
___ ________ ________ ____ ______ ________ __________ ________.
_________ ______ _____ ________ _________ ______ _____ ______ ________ _______ _________ _______.
________ ____ ___ _____ _________ ________ ____ ________ _______.
______ _________ ___ ______ _______ _________.
______ ____ ______ __________ _________ ___ ___ ____.
_________ ____ _____ _______ ___ ____ _____ ________ __________ ______ _________.
_________ _________ ________ _____ __________.
_________ _______ __________ ____ __________ __________ ______ __________ _____ ________ ________.
___ ____ _____ _______ ___ ___ ______ _____ _____ __________.
Get Full Answer on WhatsApp
To scan convert the curve (y = \frac{x^3}{12}) using the Midpoint method, we exploit symmetry about the origin because the curve is odd. We consider only (x \ge 0) and reflect points for (x < 0). The Midpoint method is based on choosing the next pixel based on the value of the decision parameter (d) at the midpoint between candidate pixels. Let the pixel grid spacing be 1 unit in both (x) and (y). The function is (f(x) = \frac{x^3}{12}). Start at the origin ((0,0)). For each step, we have two candidate pixels: 1. East (E): ((x+1, y)) 2. North-East (NE): ((x+1, y+1)) Decision parameter at midpoint (M = (x+1, y+\frac{1}{2})) is [ d = f(x+1) - \left(y + \frac{1}{2}\right) ] * ________ ____ __________ ____ ____ __________.
________ _________ ______ _________ ________ _________ ____ __________ __________ ______.
__________ _____ _________ _________ __________ ____ __________.
________ ____ _________ ______ ____ ________ _____ ____.
___ ______ _________ ______ _______ ___ _______ _________ _________ ______ __________.
__________ ___ ________ _______ ________.
________ _______ __________ ____ _______ ________ ___ ______ _______.
__________ _____ __________ ___ _________.
_______ _____ ____ _____ ____ _______ ___ _________.
__________ __________ __________ ___ ________.
_____ ______ ____ _________ ______ __________ ____ ________ ____ _______.
____ ______ ______ ___ ________ _______ ______ ____.
______ _________ _______ _________ ______ ______ ________ _____ _________.
________ _______ __________ ________ ________.
_____ _______ _________ _______ _________ ____ __________ ________ ______ __________ _________.
_________ _____ _____ _____ ________ ___ ____ _________ _______.
_____ _______ _____ ________ _______ ____ ___ ___ ________ _______ ____.
_____ ________ ____ ___ _____ ____ ________.
___ _________ __________ ____ ____ ____ _____ ___ ________.
________ ________ ________ _______ _____ __________ _________ _____ ______ _______.
___ ____ ______ _________ _______ ___ _____ _______ _______.
_________ ______ _____ _________ _______ ___.
___ ___ _________ __________ ___ __________ ____ _________.
__________ ____ __________ ______ _____ __________.
_______ _______ _________ ________ __________ ________ ______ ______ ______ ________ _____ _________.
______ __________ ___ _____ _________ ________ __________ ___ ________ _________.
_____ ________ _____ ___ ________ ______ ___.
__________ _____ ______ ______ _________ ________ ___ ______.
______ ___ ________ ____ _______ ___ ____ ____ ______ ____ ______ _________.
_____ _______ ____ _________ _______ ____ ______ ______.
__________ ________ _________ _________ ___ _________ _________ ____ __________ _____ ___.
_____ ____ _____ _________ _______ ___ ______ ______ __________ _______ __________.
_____ __________ __________ _________ ___ _____ ___ _____ ______ ______.
_______ ___ _________ ___ ____.
___ ________ ________ ____ ______ ________ __________ ________.
_________ ______ _____ ________ _________ ______ _____ ______ ________ _______ _________ _______.
________ ____ ___ _____ _________ ________ ____ ________ _______.
______ _________ ___ ______ _______ _________.
______ ____ ______ __________ _________ ___ ___ ____.
_________ ____ _____ _______ ___ ____ _____ ________ __________ ______ _________.
_________ _________ ________ _____ __________.
_________ _______ __________ ____ __________ __________ ______ __________ _____ ________ ________.
___ ____ _____ _______ ___ ___ ______ _____ _____ __________.
Get Full Answer on WhatsApp
IGNOU NEWS
Assignment Submission Last Date Extended Till 30 June 2026 Click Here★★★IGNOU June 2026 TEE Date Sheet Released Click Here★★★