If we further restrict the line-drawing routine so that it always
increments *x* as it plots, it becomes clear that, having
plotted a point at *(x,y)*, the routine has a severely limited
range of options as to where it may put the *next* point on the
line:

- It may plot the point
*(x+1,y)*, or: - It may plot the point
*(x+1,y+1)*.

So, working in the *first positive octant* of the plane, line
drawing becomes a matter of deciding between two possibilities at each
step.

We can draw a diagram of the situation which the plotting program finds
itself in having plotted *(x,y)*.

In plotting *(x,y)* the line drawing routine will, in general,
be making a compromise between what it would like to draw and what the
resolution of the screen actually allows it to draw. Usually the
plotted point *(x,y)* will be in error, the actual, mathematical
point on the line will not be addressable on the pixel grid. So we
associate an error, , with each *y* ordinate,
the real value of *y* should be . This
error will range from -0.5 to just under +0.5.

In moving from *x* to *x+1* we increase the value of the
true (mathematical) y-ordinate by an amount equal to the slope of the
line, *m*. We will choose to plot *(x+1,y)* if the
difference between this new value and *y* is less than 0.5.

Otherwise we will plot *(x+1,y+1)*. It should be clear that by
so doing we minimise the total error between the mathematical line
segment and what actually gets drawn on the display.

The error resulting from this new point can now be written back into
, this will allow us to repeat the whole process for
the next point along the line, at *x+2*.

The new value of error can adopt one of two possible values, depending
on what new point is plotted. If *(x+1,y)* is chosen, the new
value of error is given by:

Otherwise it is:

This gives an algorithm for a DDA which avoids rounding operations, instead using the error variable to control plotting:

This still employs floating point values. Consider, however, what happens if we multiply across both sides of the plotting test by and then by 2:

All quantities in this inequality are now integral.

Substitute for . The test becomes:

This gives an *integer-only* test for deciding which point to
plot.

The update rules for the error on each step may also be cast into form. Consider the floating-point versions of the update rules:

Multiplying through by yields:

which is in form.

Using this new ``error'' value, , with the new test and update equations gives Bresenham's integer-only line drawing algorithm:

- Integer only - hence efficient (fast).
- Multiplication by 2 can be implemented by left-shift.
- This version limited to slopes in the first octant, .

Here is a C++ implementation of the Bresenham algorithm for line segments in the first octant.

void linev6(Screen &s, unsigned x1, unsigned y1, unsigned x2, unsigned y2, unsigned char colour ) { int dx = x2 - x1, dy = y2 - y1, y = y1, eps = 0; for ( int x = x1; x <= x2; x++ ) { s.Plot(x,y,colour); eps += dy; if ( (eps << 1) >= dx ) { y++; eps -= dx; } } }This is an all-integer function, employs left shift for multiplication and eliminates redundant operations by tricky use of the

`eps`

variable.This implementation of Bresenham's algorithm is incomplete, it does not check the validity of its arguments. A real implementation should do this. In fact, a real implementation of Bresenham's algorithm should do more than simply reject lines with slopes lying outside the first octant, it should handle lines of arbitrary slope.

As expected, it fails to plot lines with negative slopes (try it and see what happens). It also fails to plot lines of positive slope greater than 1 (this is an interesting case, try it also and see if you can explain what is happening).

More unusually, we find that the order in which the endpoints are supplied
to this routine is significant, it will only work as long as `x1`

is smaller than `x2`

.

In fact, if we have two line segments with the same endpoints, and the same slope, this routine may draw one of them successfully but fails to draw the other one.

Of course, this is not surprising really, when we consider that the
function works by **incrementing** *x*. It does
emphasise, however, that the routine is plotting *vectors*,
direction is significant. Considering all the vectors from
*(x1,y1)* to *(x2,y2)* we find that there are eight
regions, (the ``octants'') and the basic Bresenham algorithm works in
only one of them.

A full implementation of the Bresenham algorithm must, of course, be able to handle all combinations of slope and endpoint order.

Some of the regions in the plane, those for which *x2* is
smaller than *x1* can be handled by exchanging the endpoints of the
line segment.

It is also clear that we will need a piece of code to handle large
slopes by stepping over *y* instead of *x* values.

However, careful consideration of the diagram will reveal that there is
one case which cannot be reduced to a version of the algorithm we have
already looked at. If we want to draw a line having a small
*negative* slope, we will have to consider a modification of the
basic Bresenham algorithm to do this. (The same point applies to lines of
*large* negative slope as well, but the code for small negative
slopes may be adapted to this case by stepping over *y* instead
of *x*).

Consider a line with negative slope between 0 and 1 (i.e., small
negative slope. Given that a line-drawing algorithm has plotted a point
at *(x,y)*, its choice about where to plot the next point is
between *(x+1,y-1)* and *(x+1,y)*.

As usual there will be an error, , associated with
*y*. Choice of the next point to plot will be based on an
attempt to minimise error, so plot *(x+1,y)* if:

Otherwise plot *(x+1,y-1)*. Notice that the error generated by
the above is *negative*. A little manipulation gives a decision
inequality:

It is worth comparing this with the decision inequality for the case of positive slope.

The error update rules are also subtly different for this case of negative slope.

If plotting *(x+1,y)* the *new* value of error is given
by:

Otherwise, plotting *(x+1,y-1)* gives new error:

A pseudocode algorithm for this routine may be written as:

This is cast in terms of floating-point values.. It is, however, a trivial matter to convert the algorithm into an integer-only form.

Colin Flanagan / flanaganc@ul.ie