# Fun with Functions

In the previous section we learned about functions. I said we can write our structural recursions over natural numbers using the method fold below, which accepts a function as a parameter.

def fold(count: Int, build: (Int, Image) => Image): Image =
count match {
case 0 => Image.empty
case n => build(n, fold(count - 1, build))
}

Let's see if that is really the case, and get a bit more practice using functions.

Below is an example of a row of boxes, where the color changes from each box to the next. We've already written this as a method. Now I want you to rewrite it using fold and a function you create.

Here's how I wrote it.

val gradientBoxes: (Int, Image) => Image =
(count, image) =>
Image
.square(50)
.fillColor(Color.royalBlue.spin(10.degrees * count))
.noStroke
.beside(image)
fold(5, gradientBoxes)

If you struggled to come up with this look at the version written as a method.

def gradientBoxes(count: Int): Image =
count match {
case 0 => Image.empty
case n =>
Image
.square(50)
.fillColor(Color.royalBlue.spin(10.degrees * count))
.noStroke
}

All we're doing is extracting the problem specific part in the recursive case

case n =>
Image
.square(50)
.fillColor(Color.royalBlue.spin(10.degrees * count))
.noStroke
.beside(gradientBoxes(count-1))

and turning it into a function.

Here's a variation on that idea, which changes size as well as changing color. Write this using fold and a function of your own construction.

Here's what I wrote. It's the same idea as the previous example, with a small modification to change the size of the element as well as the color.

val growingCircles: (Int, Image) => Image =
(count, image) =>
Image
.circle(20 * count)
.fillColor(Color.crimson.spin(10.degrees * count))
.noStroke
.beside(image)
fold(5, growingCircles)

Let's try a fractal. Below is the Sierpinski triangle. Can you write this using fold? If not, why not? Can you change fold so you can write it using fold?

The way the question is worded is a strong hint that we cannot write the Sierpinski triangle using fold as we have currently written it. The reason is that fold, as currently written, always uses Image.empty as the base case. For the Sierpinski triangle we need a different base case. The solution is to add another parameter to fold that allows us to change the base case.

## A More General Fold

In the previous section we saw there are some kinds of Images we cannot create using fold. A more general fold will allow us to also vary the base case as well as the recursive case. This is simply another parameter to the method.

def fold(count: Int, base: Image, build: (Int, Image) => Image): Image =
count match {
case 0 => base
case n => build(n, fold(count - 1, base, build))
}

Using this version of fold we can create the Sierpinski triangle.

val sierpinski: (Int, Image) => Image =
(count, image) => image.above(image.beside(image))

fold(
5,
Image.equilateralTriangle(10).strokeColor(Color.hotpink),
sierpinski
)

To make a fully general fold we'd need to be able to change the result type. This requires generic types, which we will learn about in a later section.