UGC-NET | UGC NET CS 2016 Aug – III | Question 18

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 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 :
(A) (0, 2/3)
(B) (1/6 , 5/6)
(C) (0, 1/3)
(D) (0, 1)

Answer: (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, P₄2, P₃ > 0.
Initially: 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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