Laboratory: Writing Your Own Procedures

Recall that in the lab on numeric values, you explored the use of the min and max procedures to compute a bounded value:

> (define bounded-val (min (max val lower) upper))

This code assumed that val, lower, and upper had already been defined. It would be helpful to create a subroutine, instead, so that we can easily limit any real number to fall within any bounds.

Take the code given above and encapsulate it in a procedure named bound. This procedure should take three parameters: val, lower, and upper.

Here is an example of some output from the bound procedure:

> (bound 5 0 10)
5
> (bound 5 10 20)
10
> (bound 0.7 -1 1)
0.7
> (bound 1.01 -1 1)
1.0
> (bound -8.3 -1 1)
-1.0

a. Verify that the four basic values look as they are described in the corresponding reading.

> (image-show (drawing->image black-circle 200 100))
> (image-show (drawing->image purple-ellipse 200 100))
> (image-show (drawing->image blue-i 200 100))
> (image-show (drawing->image red-eye 200 100))

b. Try some of the transformers. For example,

> (image-show (drawing->image (variant-1 black-circle) 200 100))
> (image-show (drawing->image (variant-1 (variant-1 black-circle)) 200 100))
> (image-show (drawing->image (add-right-neighbor red-eye) 200 100))

c. Try the circle constructor.

> (image-show (drawing->image (circle ___ ___ ___) 200 100))

d. Verify that square correctly squares the numbers 5, 10, -3, 1.2, and 0.05.

Now that we can draw circles using circle, it will be helpful to write procedures that generate other simple shapes. Let's start with squares.

How do we create a drawing of a square? It depends on how we want to describe the square. If we are centering the square on a particular point, we need to know (1) the x coordinate of the center, (2) the y coordinate of the center, and (3) the edge length of the square. We can then scale the unit square by the edge length and shift it horizontally by the x coordinate of the center and vertically by the y coordinate of the center. Suppose we are drawing a square with edge length 25, centered at (20,30). We might write

(define my-square
  (hshift-drawing
   20
   (vshift-drawing
    30
    (scale-drawing 25 drawing-unit-square))))

Now, let's think about how to generalize this.

a. Write a procedure, (centered-square edge-length center-x center-y) that creates a drawing of a square. (We couldn't call this procedure square, because we'd already used that name for the procedure that squares numbers.) You should be able to generalize the code above, and base your procedure on the circle procedure.

b. However, most of us don't like to draw our squares centered on a particular point. We'd rather specify the left edge and the top edge. How can we do that? Suppose we want to draw a square of side-length 30, with the left edge at 17 and the top edge at 42. We first scale the unit square by 30. That means that the left edge, which was at -1/2, is now at -15 (that is, -1/2 * 30). The top edge, similarly, is also at -15. To get the left edge at 17, we need to shift it horizontally by 32 (that is, 15 + 17). To get the top edge at 42, we need to shift it vertically by 59 (that is, 15 + 42).

In Scheme, we might write

(define another-square
  (hshift-drawing
   (+ (/ 30 2) 17))
   (vshift-drawing
    (+ (/ 30 2) 42)
    (scale-drawing 30 drawing-unit-square)))

Of course, we can also generalize this code. Write a procedure (drawing-of-square edge-length left top) that creates a drawing of a square of the specified edge length, with the left side of the square at left and the top side at top.

One deficiency of the circle procedure from the reading and of the centered-square and drawing-of-square procedures you've just written is that they don't allow you to specify the color of the circle or square.

Rewrite the circle procedure to take a color as a parameter. For example,

(circle 20 10 10 "blue")

Write a procedure, rectangle that creates a drawing of a rectangle. You should do your best to figure out appropriate parameters. Here is a sample call, using the parameters we find most natural.

(rectangle 20 10 50 80 "yellow")

If you'd like to know what we thought were appropriate parameters, you can look at the notes on this exercise.

As you may recall, the add-right-neighbor procedure makes a copy of a drawing, places the copy immediately to the right of the original drawing, and combines the two into a new drawing.

a. Write a procedure, (add-smaller-right-neighbor drawing) that combines a drawing with a duplicate neighbor that is 75% of the size of the original drawing.

(add-smaller-right-neighbor red-eye)

As you write this procedure, you will find it useful to examine the code for add-right-neighbor

b. Write a procedure, add-nearer-right-neighbor drawing) that combines a drawing with a duplicate neighbor of the same size, but that overlaps 20% (so that the left end of the neighbor is 20% left of the right end of the original).

(add-nearer-right-neighbor red-eye)

We now have procedures that pair an image with a smaller right neighbor and a nearer right neighbor. What happens if we combine these transformations? For each of the following, predict what the image will look like and then render it to check your prediction.

a. (add-smaller-right-neighbor (add-nearer-right-neighbor red-eye))

b. (add-nearer-right-neighbor (add-smaller-right-neighbor red-eye))

c. (add-smaller-right-neighbor (add-smaller-right-neighbor red-eye))

For each of the procedures above, you've likely tested your procedure by applying it to some drawing, rendering that drawing to some image, and then showing the image. For example,

> (image-show (drawing->image (rectangle 20 10 50 80) 200 100))

Write a procedure, (check-drawing drawing) that renders the drawing on a "large enough" image, then shows the image. Once we've written that procedure, we can more easily check drawings, as in the following.

> (check-drawing (rectangle 20 10 50 80))

How can we find out how large the image needs to be? We can ask a drawing where its lower right hand corner is using the drawing-bottom and drawing-right procedures.

In the reading, we defined a drawing that looks a bit like a red eye. Write your own procedure, drawing-eye, that builds a drawing of an eye, based on the values of its parameters. What parameters should it have? Certainly, the color of the iris. However, you might find other parameters useful, too.

As you may recall, one potentially confusing aspect of the scale-drawing procedure is that it not only scales the drawing, it also scales the distance of the drawing from the top-left corner.

Write a procedure, (alternate-scale amt drawing), that scales a drawing, but does not move the drawing. That is, the scaled drawing has the same left edge and the same top edge as the original drawing.

If you're not sure how to approach this problem, you may want to read the notes on this exercise.

a. Write a procedure, (two-by-two drawing) that creates a compound drawing with four copies of drawing (one shifted right, one shifted down, and one shifted down and right).

b. Write a procedure that takes the result of two-by-two and scales it by 50%, so that the resulting drawing is the same width and height as the original.

c. Using this new procedure, build a four-by-four grid of one of the original images, or one of your own choosing.

In Extra 4, each of the expressions takes one drawing and turn it into four drawings. Explore other ways to use this kind of replication to build an interesting image. (You might also think about ways to create eight, sixteen, or even more copies of the image.)

In Exercise 6, you developed a number of procedures that pair an image with a neighbor. Create a few variants of those procedures that change the neighbor in an “interesting” way. For example, you might scale it differently horizontally and vertically, you might have it overlap the original figure, you might shift it both horizontally and vertically.

We would recommend that your procedure have the form

(define rectangle
  (lambda (left top width height color)
    ..._))

Return to the exercise.

The alternate scaling algorithm is fairly straightforward:

Return to the exercise.