Graphics libraries provide line-drawing routines, sometimes with antialiasing^{[1]} and variable width. On a grid map, line drawing is useful for for visibility, the path of an arrow/bullet, and enemy AI. I sometimes see people adapting the Bresenham line drawing algorithm to draw lines on a grid (especially for roguelike games), but I prefer simpler algorithms. I'll show the algorithms I use. I'm not a line-drawing expert so please let me know if there are better algorithms that are similarly simple.

## #1 Linear interpolation

This demo shows what grid locations should be marked if you're drawing a line between two points. **Try moving the endpoints around**.

I find the easiest way to find these points is to use *linear interpolation*. Let's see how that works.

### 1.1Interpolating numbers#

Here's a simple (Javascript) helper function I'll use:

function lerp(start, end, t) { return start + t * (end-start); }

__L__inear int__erp__olation (“lerp”) gives you a number between two other numbers. When `t`

= 0.0 you get the start point; when `t`

= 1.0 you get the end point. **Try setting t**, the third parameter to lerp():

lerp( 0, 1, ) = lerp( 0, 100, ) = lerp( 3, 5, ) = lerp( 5, 3, ) =

### 1.2Interpolating points#

We can extend the idea of interpolation to work on points, by interpolating both the x and y coordinates. **Try varying t** = :

Here's the code to find point x,y between point p_{0} = (x_{0},y_{0}) and point p_{1} = (x_{1},y_{1}):

function lerp_point(p0, p1, t) { return new Point(lerp(p0.x, p1.x, t), lerp(p0.y, p1.y, t)); }

Let's divide the line into equal line segments:

Here's the code to find those points:

var points = []; for (var step = 0; step <= N; step++) { var t = step / N; points.push(lerp_point(p0, p1, t)); }

### 1.3Snap to grid#

Next we need to figure out which *grid squares* those points are in. We can round x and y to the nearest integer to find the grid tile:

We also need to decide how many points to include. Too few and the line has holes. Too many and the line has overdraw. How many do we need? The right number is the *diagonal distance* between the endpoints, which in this case is .

**Adjust N** = to to see how the line fills in.

That's it!

- Set N to the diagonal distance between the start and end point.
- Pick N+1 interpolation points, evenly spaced.
- Round those points to the nearest grid tile.

Here's the final code:

function diagonal_distance(p0, p1) { var dx = p1.x - p0.x, dy = p1.y - p0.y; return Math.max(Math.abs(dx), Math.abs(dy)); } function round_point(p) { return new Point(Math.round(p.x), Math.round(p.y)); } function line(p0, p1) { var points = []; var N = diagonal_distance(p0, p1); for (var step = 0; step <= N; step++) { var t = N == 0? 0.0 : step / N; points.push(round_point(lerp_point(p0, p1, t))); } return points; }

This is the simplest line drawing algorithm I know of.

### 1.4Interpolating other types#

I defined the `lerp`

function to interpolate between two *numbers*, and then I defined `lerp_point`

to work on two *points*. How about other types *T*? We need these operations to define interpolation:

- addition
- b = a + d where a:
*T*, b:*T*, and d:*ΔT*. For points, b.x = a.x + d.x; b.y = a.y + d.y - subtraction
- d = a - b where a:
*T*, b:*T*, and d:*ΔT*. For points, d.x = a.x - b.x; d.y = a.y - b.y - multiplication by scalar
- e = k * d where d:
*ΔT*, e:*ΔT*, and k:*number*. For points, e.x = k * d.x; e.y = k * d.y

We can do interpolation on any vector space^{[2]}. *T* can be a number, a 2d point, a 3d point, an angle, a time, a color, or other things too. My guide to hexagonal grids^{[3]} uses interpolation to draw lines on a hex grid. Note that *T* and *ΔT* may be the same type, but sometimes they are different. For example, *T* may be a timestamp and *ΔT* may be a time difference, or *T* may be an orientation and *ΔT* may be a rotation. When *T* and *ΔT* are the same type, d = a - b can be implemented as d = a + (-1 * b).

### 1.5TODO Aesthetics#

Linear interpolation is calculating a position and then rounding it. What happens when the value is exactly 0.5? Rounding rules vary^{[4]}. That, and floating point precision, make linear interpolation not always choose points in a way that preserves consistency with rotation, reversing, and other symmetry.

I think the thing to do would be to “nudge” the initial points by epsilon. However I haven't explored this.

### 1.6Code optimization#

You can optimize the regular line drawing algorithm with these steps; it will turn into the DDA algorithm^{[5]}:

- Floating point vs fixed point
- Most likely, profiling will find that Math.round() is expensive. You can instead use fixed point arithmetic, and replace rounding with bit shifting.
- Unrolling lerp
- Instead of calculating t each time through the loop, and then x = (b.x-a.x)*t (a subtract and multiply), you can calculate Δx = (b.x-a.x)*Δt outside the loop, and then use x += Δx. Same for y. This will replace the multiplies with adds.
- Unrolling loop
- You can unroll the loop by keeping four x and y pairs, and then using t += 4*Δt. Would this allow the use of SSE instructions? I don't know.
- Separate cases
- Either Δx or Δy will be ±1. You might want to have a separate versions for each case. For diagonal lines, both will be ±1. For orthogonal lines, one or the other will be 0. If these cases are common, write a separate routine for them.

Before optimizing, profile and make sure linear interpolation is your bottleneck. In my projects, it rarely is, so I haven't bothered implementing these optimizations. (However, I may do that in a future version of this article.)

## #2 Grid walking

Interpolation is simple and general but doesn't take into account properties of the grid. Another approach to line drawing is to take one step at a time. This type of movement also allows *putting walls on edges* so that a wall could block line of sight or movement.

### 2.1Orthogonal steps#

In the above lines, we took both orthogonal steps (north, east, south, west) and diagonal steps. Step-by-step algorithms give us more flexibility. What if we only want to take orthogonal steps?

The strategy here is to think about the *grid edges* being crossed, both vertical and horizontal. At every step we either cross a vertical edge (horizontal step) or cross a horizontal edge (vertical step).

function walk_grid(p0, p1) { var dx = p1.x-p0.x, dy = p1.y-p0.y; var nx = Math.abs(dx), ny = Math.abs(dy); var sign_x = dx > 0? 1 : -1, sign_y = dy > 0? 1 : -1; var p = new Point(p0.x, p0.y); var points = [new Point(p.x, p.y)]; for (var ix = 0, iy = 0; ix < nx || iy < ny;) { if ((0.5+ix) / nx < (0.5+iy) / ny) { // next step is horizontal p.x += sign_x; ix++; } else { // next step is vertical p.y += sign_y; iy++; } points.push(new Point(p.x, p.y)); } return points; }

It's more work to make it symmetric. Also, sometimes you'll want more control over whether you pick a horizontal step or a vertical step, especially if the choice is close, and there's a wall in one direction but not the other.

### 2.2Supercover lines#

"Supercover" lines catch all the grid squares that a line passes through. I believe (but am not sure) that it is somewhere in between regular line drawing and grid movement line drawing. Unlike grid movement line drawing, we can take a diagonal step *only* if the line passes exactly through the corner.

This looks just like the grid movement line drawing, except when the line goes through a grid corner.

Let's modify the grid movement line drawing code for this:

function supercover_line(p0, p1) { var dx = p1.x-p0.x, dy = p1.y-p0.y; var nx = Math.abs(dx), ny = Math.abs(dy); var sign_x = dx > 0? 1 : -1, sign_y = dy > 0? 1 : -1; var p = new Point(p0.x, p0.y); var points = [new Point(p.x, p.y)]; for (var ix = 0, iy = 0; ix < nx || iy < ny;) { if ((0.5+ix) / nx == (0.5+iy) / ny) { // next step is diagonal p.x += sign_x; p.y += sign_y; ix++; iy++; } else if ((0.5+ix) / nx < (0.5+iy) / ny) { // next step is horizontal p.x += sign_x; ix++; } else { // next step is vertical p.y += sign_y; iy++; } points.push(new Point(p.x, p.y)); } return points; }

Implementation note: it's usually a bad idea to compare two floating point numbers for equality, as I do here. However, since `ix`

, `nx`

, `iy`

, `ny`

are all integers, we can ensure this works properly by cross-multiplying and rewriting the test with integer arithmetic.

if ((1+2*ix) * ny == (1+2*iy) * nx) { ... }

It still feels somewhat fragile to me, and I'm guessing there are better algorithms out there.

### 2.3Line of sight#

If you allow diagonal steps, some algorithms will step through walls:

In this case, you can either take the diagonal step if *either* the horizontal or vertical step is clear, or you can disallow the diagonal step unless *both* the horizontal and vertical steps are clear.

## #3 More reading

There's lots written about line drawing but I haven't researched it extensively. When drawing graphical lines, I use the graphics library. It's only on grids that I needed to find some other algorithms.

- If you're interpolating 3d rotations, look at the slerp
^{[6]}variant. - If you want non-linear interpolation, take look at smoothstep
^{[7]}and others^{[8]}, especially for animation. - Roguebasin has an article
^{[9]}about various qualities you might want on a grid if calculating field of view or line of sight. - This paper
^{[10]}gives an extension of the step-by-step line drawing algorithm for 3D coordinates. - Wu's algorithm
^{[11]}for anti-aliasing might be useful for determining how much of an object is on one grid cell or another; I haven't tried this. - This article
^{[12]}looks at DDA line drawing with fixed point arithmetic. It's from 2009 though and CPUs have changed since then. Also look at Extremely Fast Line Algorithm^{[13]}. - I find Bresenham's algorithm to be much longer than I'd like, but if you'd like to compare, take a look at Michael Abrash's implementation
^{[14]}and explanation^{[15]}.