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Jack Bresenham's raster graphics algorithm selects a thin sequence of pixels to represent a line segment by exact incremental integer arithmetic. For a first-octant segment with $0\le \Delta y\le \Delta x$, a scaled midpoint line decision chooses between the next same-row and next higher-row pixels.

Three advantages of integer arithmetic for raster graphics are:

These are historical and structural advantages, not a claim that every modern processor executes every integer operation faster. An integer implementation still needs sufficient width for its intermediate results. The compact answer is

By exchanging axes, reversing endpoint order and reflecting coordinate signs where necessary, reduce a general line segment to the first octant:The vertical case is handled by exchanging axes. In this octant, increasing $x$ by one requires increasing $y$ by either zero or one. Thus, from the currently plotted pixel $(x,y)$ the candidates are $E=(x+1,y)$ and $NE=(x+1,y+1)$.

Use the implicit equation of the lineAt a fixed $X$, $f(X,Y)>0$ means that $(X,Y)$ lies below the exact line. The midpoint line decision evaluates $M=(x+1,y+1/2)$, halfway between the two candidate pixel centres. If $f(M)<0$, the exact line lies below $M$, so choose $E$; if $f(M)>0$, choose $NE$. At equality either candidate is equally near, but a fixed tie rule is essential. We choose $NE$ on a tie throughout these solutions:This chooses the closest vertical-grid pixel centre in each column, with ties rounded upward. Since $0\le m\le 1$, the result has one pixel per column and consecutive pixels share an edge or a corner.

Multiply the midpoint line decision by two to remove the fraction. At the current pixel $(x,y)$, maintainAt the initial endpoint this is $d_0=2\Delta y-\Delta x$. If $E$ is chosen, the next decision midpoint is displaced by $(1,0)$ and $d$ increases by $2\Delta y$. If $NE$ is chosen, its displacement is $(1,1)$ and $d$ increases by $2(\Delta y-\Delta x)$. Both increments are constant, so the per-pixel work uses only integer arithmetic: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 isThe midpoint line decision selects exactly this nearest integer, rounding halves upward. Its increment between successive columns is zero or one. At $k=0$ and $k=\Delta x$, it gives exactly $y_0$ and $y_1$, respectively, so both endpoints are included and the line segment has precisely $\Delta x+1$ distinct pixels.

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 $(5,3)$.