Case Study 2:
def sierpinski(myTurtle, p1, p2, p3, depth):
if depth 0:
sierpinski(myTurtle, p1,
midPoint(p1, p2), midPoint(p1, p3), depth - 1)
sierpinski(myTurtle, p2,
midPoint(p2, p3), midPoint(p2, p1), depth - 1)
sierpinski(myTurtle, p3,
midPoint(p3, p1), midPoint(p3, p2), depth - 1) else:
drawTriangle(myTurtle, p1, p2, p3)
-Refer to the session in the accompanying Case Study 2. Which of the following lines correctly implements the midPoint function: def midPoint(p1, p2)?
A) return ((p1[0] + p2[0])/2.0, (p1[1] + p2[1])/2.0)
B) return ((p1[1] + p2[1])/2.0, (p1[1] + p2[1])/2.0)
C) return ((p1[0] + p2[0])/3.0, (p1[1] + p2[1])/3.0)
D) return ((p1 + p2[0])/2.0, (p1 + p2[1])/2.0)

Question

Case Study 2:
def sierpinski(myTurtle, p1, p2, p3, depth):
if depth > 0:
sierpinski(myTurtle, p1,
midPoint(p1, p2), midPoint(p1, p3), depth - 1)
sierpinski(myTurtle, p2,
midPoint(p2, p3), midPoint(p2, p1), depth - 1)
sierpinski(myTurtle, p3,
midPoint(p3, p1), midPoint(p3, p2), depth - 1) else:
drawTriangle(myTurtle, p1, p2, p3)
-Refer to the session in the accompanying Case Study 2. Which of the following lines correctly implements the midPoint function: def midPoint(p1, p2)?
A) return ((p1[0] + p2[0])/2.0, (p1[1] + p2[1])/2.0)
B) return ((p1[1] + p2[1])/2.0, (p1[1] + p2[1])/2.0)
C) return ((p1[0] + p2[0])/3.0, (p1[1] + p2[1])/3.0)
D) return ((p1 + p2[0])/2.0, (p1 + p2[1])/2.0)

Quizplus · Accepted Answer

The correct implementation of the midPoint function calculates the average of the x-coordinates and the y-coordinates of the two points, which is correctly done in option A.

Case Study 2: Def Sierpinski(myTurtle, P1, P2, P3, Depth)