Aug2016 cs Q18

0. The Liang-Barsky line clipping algorithm uses the parametric equation of a line from (x₁, y₁) to (x₂, y₂) along with its infinite extension which is given as :
x = x₁ + ∆x.u
y = y₁ + ∆y.u
Where ∆x = x₂– x₁, ∆y = y2– y1, and u is the parameter with 0 < u < 1. A line AB withend points A(–1, 7) and B(11, 1) is to be clipped against a rectangular window with x_min = 1, x_max = 9, y_min = 2, and y_max = 8. The lower and upper bound values of the parameter u for the clipped line using Liang-Barsky algorithm is given as :

Answer
Comment
Array

Option : B
Explanation :
first we will calculate ∆x and ∆y:
i.e. ∆x = x₂– x₁
= 11 - ( - 1)
= 11 + 1 = 12.
∆y = y₂– y₁
= 1 - 7 = - 6
Now P₁ = -∆x = - 12
P₂ = ∆x = 12
P₃ = -∆y = 6
P₄ = ∆y = - 6
Q₁ = x₁ - x_min = - 1 - 1 = -2
Q₂ = x_max - x₁ = 9 - ( - 1) = 9 + 1 = 10.
Q₃ = y₁ - y_min = 7 - 2 = 5.
Q₄ = y_max - y₁ = 8 -7 = 1.
P₁, P₄ < 0 and P₂, P₃ > 0.
Intially: t₁ = 0, t₂ = 1
t₁ = max(0, Q₁ / P₁, Q₄ / P₄)
= max(0, 2 / 12, 1 / -6) = 1 / 6.
t₂ = min(1, Q₁ / P₁, Q₄ / P₄)
= min(1, 10 / 12, 5 /6). = 5 / 6.
i.e. u ranges between (1 / 6, 5 / 6).
So, option (B) is correct.

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