 Solution

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.