Past exam of the computer science course of the University of Oxford / 2015 / Computer graphics / 1 / c

Table of contents
- Solution c

Solution

Multiply the midpoint line decision by two to remove the fraction. At the current pixel

(x, y)

, maintain

d = 2 f (x + 1, y + 1/2) = 2Δ y (x + 1 - x_{0}) - 2Δ x (y - y_{0}) - Δ x .

At the initial endpoint this is

d_{0} = 2Δ y - Δ x

. If

E

is chosen, the next decision midpoint is displaced by

(1, 0)

and

d

increases by

2Δ y

. If

NE

is chosen, its displacement is

(1, 1)

and

d

increases by

2 (Δ y - Δ x)

. Both increments are constant, so the per-pixel work uses only integer arithmetic:

d_{0} = 2Δ y - Δ x, incrE = 2Δ y, incrNE = 2 (Δ y - Δ x) .

The completed code, using the tie rule of part (b), is:

void midpointLine(int x0, int y0, int x1, int y1)
{
    int dx, dy, incrE, incrNE, d, x, y;
    x = x0; y = y0;
    dx = x1 - x0;
    dy = y1 - y0;
    incrE = 2 * dy;
    incrNE = 2 * (dy - dx);
    d = 2 * dy - dx;
    drawPoint(x, y);

    while (x < x1) {
        if (d < 0) {
            d += incrE;
        } else {
            d += incrNE;
            y++;
        }
        x++;
        drawPoint(x, y);
    }
    return;
}

Thus guard1 is x < x1, and guard2 is d < 0. Multiplication by two is the only multiplication. Signed decision variables are convenient because

d

and incrNE can be negative; this remains an integer arithmetic algorithm. A strictly unsigned hardware representation would need an equivalent biased or separated-sign encoding. As with the printed int skeleton, assume all intermediate values fit the chosen type; integer overflow would invalidate the mathematical decisions.

To verify the geometry, write

k = x - x_{0}

. The chosen ordinate is

y (x) = y_{0} + ⌊ \frac{2Δ y k + Δ x}{2Δ x} ⌋ .

The midpoint line decision selects exactly this nearest integer, rounding halves upward. Its increment between successive columns is zero or one. At

k = 0

and

k = Δ x

, it gives exactly

y_{0}

and

y_{1}

, respectively, so both endpoints are included and the line segment has precisely

Δ x + 1

distinct pixels.

Figure 1.
Bresenham pixels for a slope of 3/8, with the upper pixel selected at the midpoint tie
. Details.

For the illustrated endpoints

(1, 1)

and

(9, 4)

, the ordinates are

1, 1, 2, 2, 3, 3, 3, 4, 4

. In particular the exact ordinate at

x = 5

is

5/2

, and the chosen pixel is

(5, 3)

.

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