Hitunglah posisi pixel hingga membentuk sebuah garis yang menggabungkan titik (14,12) & (20,16)
Jawab :
1. Endpoint awal = P₁ (14,12) => (X₀,Y₀)
Akhir = P₂ (20,16)
2. M = ∆Y = Y₂ - Y₁ = 16 – 12 = 4
∆X X₂ - X₁ 20 – 14 6
Syarat ImI < 1
3. ∆X = X₂ - X₁ = 20 – 14 = 6
∆Y = Y₂ - Y₁ = 16 – 12 = 4
2∆Y = 2 x 4 = 8
2∆Y = 2∆X = (2x4) – (2x6) = -4
P₀ = 2∆Y - ∆X = 8 – 6 = 2
K | Pk | Pk < 0 | Xk + 1, Yk + 1 | Pk + 1 = Pk + 2∆y - 2∆x | Xk +1, Yk | Pk + 1 = Pk + 2∆y |
0 | P₀ | P₀ < 0 2 < 0 | X₀ + 1 , Y₀ + 1 14 + 1 , 12 + 1 ( 15 ) , ( 13 ) | P₀ + 1 = P₀ +8 – 12 P₁ = 2 + 8 – 12 P₁ = -2 | ||
1 | P₁ | P₁ < 0 -2 < 0 | X₁ + 1 , Y₁ 15 + 1 , 13 ( 16 ) , ( 13 ) | P₁ + 1 = P₁ + 8 P₂ = (-2) + 8 P₂ = 6 | ||
2 | P₂ | P₂ < 0 6 < 0 | X₂ +1 , Y₂ + 1 16 + 1 , 13 + 1 ( 17 ) , ( 14 ) | P₂ + 1 = P₂ + 8 – 12 P₃ = 6 + 8 – 12 P₃ = 2 | ||
3 | P₃ | P₃ < 0 2 < 0 | X₃ + 1 , Y₃ + 1 17 + 1 , 14 + 1 ( 18 ) , ( 15 ) | P₃ + 1 = P₃ + 8 – 12 P₄ = 2 + 8 – 12 P₄ = -2 | ||
4 | P₄ | P₄ < 0 -2 < 0 | X₄ + 1 , Y₄ 18 + 1 , 15 ( 19 ) , ( 15 ) | P₄ + 1 = P₄ + 8 P₅ = (-2) + 8 P₅ = 6 | ||
5 | P₅ | P₅ < 0 6 < 0 | X₅ + 1 , Y₅ + 1 19 + 1 , 15 + 1 ( 20 ) , ( 16 ) |
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