## Introduction

In this blog post, we’re going to look at visualizing the unit tests that are automatically generated by Diffblue cover. For this, we’re going to look at a common task in the world of computer graphics: triangle intersections. This is used when producing photorealistic images using ray-tracing. We’re going to plot the tests that Diffblue Cover writes and visually show how they test our triangle intersection logic.

Ray-tracing simulates the behavior of light in a scene. The light rays are traced *from* the camera into the scene, one ray per pixel (there are techniques to improve the quality of the image by using more than one, but for this discussion we’ll stick with a single ray). If the ray hits an object, then the color of the resultant pixel in the image is determined by the material properties of the intersection point: if the material is a blue cloth, then the resulting pixel will be a shade of blue. The exact color will depend on not only the material properties but also the lighting, shading, occlusion, textures and so on.

Traditionally, 3D objects in computer graphics are built from a mesh of triangles, so the first step is to determine if the simulated light ray intersects with a triangle or not. If it hasn’t, it’ll continue on its way.

For the rest of this post, we’re going to assume that the ray is shooting *into* the page perpendicular to the surface (i.e. straight into the screen). This makes it easier to visualize and discuss. In a real application, the math would need to take into account the direction of the ray in 3-dimensional space.

## Triangles

Suppose we have a triangle sitting on a page. It has three points at the following coordinates:

```
a = (x_a, y_a)
b = (x_b, y_b)
c = (x_c, y_c)
```

These really define three lines, `AB`

, `BC`

and `CA`

. Let’s look at an image of one of these lines.

Remember that the light ray is being fired into the page, and so corresponds to a point on the plane. To determine if the point intersects the triangle, we need to determine if it’s inside or outside the triangle. We will get to a proper definition of “inside” or “outside” after we’ve discussed a little geometry.

## Lines from Points

Lets start with defining the line (`y = mx + c`

, where `m`

is the slope of the line, and `c`

is the intersection point) going from `A`

to `B`

. Suppose that the point `A`

is at `(1, 0)`

and `B`

is at `(3,3)`

. We have the following two equations:

```
0 = m(1) + c
3 = m(3) + c
```

We need to find the slope and intersection points. To do this we can then subtract one equation from the other, simplify the result and obtain the following:

```
m = (3 - 0)/(3 - 1) = 1.5
```

From that, we can obtain the value for `c`

by working backwards from one of the equations above with the new value of `m`

:

```
0 = 1 * 1.5 + c
0 = 1.5 + c
c = -1.5
```

Now that we have the `m`

and `c`

values, we can define a line in two dimensions:

```
y = 1.5x - 1.5
```

## Left or Right of a Line?

We now have our equation and can check to see if a point is to the left or the right of it. To check, we need to introduce an inequality. In this case, we say it’s to the *right* if y is *less than* the equation.

```
y < 1.5x - 1.5
0 < 1.5x - 1.5 - y
```

From the figure, we can see that if a point is in the shaded region, then the inequality holds. Suppose we had a test point `p`

at `(2, 1)`

; what happens when we put that into the inequality above?

```
0 < 1.5 * 2 - 1.5 - 1
0 < 3 - 1.5 - 1
0 < 0.5
```

The inequality holds, therefore it must be on the *right* of the line. Let’s see what happens for point `q`

at `(2,3)`

.

```
0 < 1.5 * 2 - 1.5 - 3
0 < 3 - 1.5 - 3
0 < -1.5
```

This inequality doesn’t hold, therefore the point is to the *left* of the line.

In this figure, you can see the two test points that we introduced and tested. Point `p`

is in the shaded region and to the right of the line, while point `q`

is not.

## Inside or Outside

From here, you can see that if you perform this three times on the three different pairs of points in the triangle, you can determine if the point is inside (when it’s to the *right* of *all* the lines) or outside (when it’s to the *left* of any one of them).

This only works when the lines are defined by a clockwise traversal of the points. If you wanted to support a counter-clockwise traversal, the left and rights in the previous condition would be reversed. In reality, we don’t know (nor does it really matter) which direction we visit the points in, as long as the test point is on the same side for all the lines it’ll be inside.

Here’s an example of the inside/outside test implemented in Java. The `sign`

method is implemented to check the point against the line equation. The derivation of this is left as an exercise for the reader.

```
public class Triangle {
private final Point2D a;
private final Point2D b;
private final Point2D c;
public Triangle(final Point2D a, final Point2D b, final Point2D c) {
this.a = a;
this.b = b;
this.c = c;
}
public boolean inside(final Point2D p){
final int k1 = sign(p, b, a);
final int k2 = sign(p, c, b);
final int k3 = sign(p, a, c);
return k1 > 0 && k2 > 0 && k3 > 0 || k1 < 0 && k2 < 0 && k3 < 0;
}
private int sign(Point2D p1, Point2D p2, Point2D p3) {
return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) * (p1.y - p3.y);
}
}
```

## Testing

This is all very well and good, but what tests cases are there? Well, when the point is inside or outside of a triangle, right?

```
@Test
public void testInside() {
// Arrange
Point2D a = new Point2D(0, 0);
Point2D b = new Point2D(0, 3);
Point2D c = new Point2D(3, 0);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(1, 1);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertTrue(actualInsideResult);
}
@Test
public void testOutside() {
// Arrange
Point2D a = new Point2D(0, 0);
Point2D b = new Point2D(0, 3);
Point2D c = new Point2D(3, 0);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(4, 4);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertFalse(actualInsideResult);
}
```

These test cases correspond to the following diagrams.

## Automatically Generated Tests

Running `dcover`

on that snippet above, I got the following set of tests:

Or if you prefer the code:

```
import static org.junit.jupiter.api.Assertions.assertFalse;
import static org.junit.jupiter.api.Assertions.assertTrue;
import org.junit.jupiter.api.Test;
public class TriangleTest {
// human written test
@Test
public void testInside() {
// Arrange
Point2D a = new Point2D(0, 0);
Point2D b = new Point2D(0, 3);
Point2D c = new Point2D(3, 0);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(1, 1);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertTrue(actualInsideResult);
}
// human written test
@Test
public void testOutside() {
// Arrange
Point2D a = new Point2D(0, 0);
Point2D b = new Point2D(0, 3);
Point2D c = new Point2D(3, 0);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(4, 4);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertFalse(actualInsideResult);
}
// tests below here were generated by `dcover`
@Test
public void testInside1() {
// Arrange
Point2D a = new Point2D(1, 0);
Point2D b = new Point2D(-10, 3);
Point2D c = new Point2D(2, 10);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(2, 3);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertFalse(actualInsideResult);
}
@Test
public void testInside2() {
// Arrange
Point2D a = new Point2D(1, 10);
Point2D b = new Point2D(-10, 3);
Point2D c = new Point2D(2, 3);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(2, 3);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertFalse(actualInsideResult);
}
@Test
public void testInside3() {
// Arrange
Point2D a = new Point2D(10, 0);
Point2D b = new Point2D(-10, 3);
Point2D c = new Point2D(2, 10);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(2, 3);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertTrue(actualInsideResult);
}
@Test
public void testInside4() {
// Arrange
Point2D a = new Point2D(1, 10);
Point2D b = new Point2D(-10, 3);
Point2D c = new Point2D(2, 0);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(2, 3);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertFalse(actualInsideResult);
}
@Test
public void testInside5() {
// Arrange
Point2D a = new Point2D(1, 3);
Point2D b = new Point2D(-10, 3);
Point2D c = new Point2D(2, 3);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(2, 3);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertFalse(actualInsideResult);
}
@Test
public void testInside6() {
// Arrange
Point2D a = new Point2D(1, 0);
Point2D b = new Point2D(10, 3);
Point2D c = new Point2D(2, 10);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(2, 3);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertTrue(actualInsideResult);
}
@Test
public void testInside7() {
// Arrange
Point2D a = new Point2D(1, 0);
Point2D b = new Point2D(-10, 3);
Point2D c = new Point2D(2, 3);
Triangle triangle = new Triangle(a, b, c);
Point2D p = new Point2D(2, 3);
// Act
boolean actualInsideResult = triangle.inside(p);
// Assert
assertFalse(actualInsideResult);
}
}
```

The interesting thing about these tests is that the point stays in the same position, but the triangle changes shape. The `Triangle`

class also allows some *interesting* constructions—look at the 5th example, a case where the two of the triangles points are equal.

From these tests above, we can see that there are some interesting edge cases tested—what if the test point and one of the triangle points share the same location?

## Conclusions

This article visualizes some automatically generated test cases for a simple triangle intersection test. As we’ve seen, there are some interesting test cases produced that wouldn’t necessarily have been considered by a human writing the same code. You can try it yourself using Diffblue Cover.