Laboratory: Building Images by Iterating Over Positions

Recall that in the self-checks we explored the following instructions to compute a small blend from black to red.

(image-compute 
 (lambda (col row) 
   (irgb (* col 2) 0 0))
 129 65)

a. As you may have noted in the reading's self-checks, when we used the 129x65 formula for a bigger image, the rightmost pixels were all red. Write an expression to create a 257x129 image that contains a full horizontal black-to-red blend.

b. As you may have noted, when we used the 129x65 formula for a smaller image, we didn't make it very far from black. Write an expression to create a 65x33 image that contains a full black-to-red blend.

c. Write an expression to build a 129x65 image that has a vertical blend from black to (irgb 255 0 255).

In the previous problem, you made images that showed blends ranging over a single dimension, either horizontally or vertically. However, we can certainly blend colors in both dimensions.

a. The reading provided one such blend in which the red component increased as you move horizontally from left to right and the blue component increased as you move vertically from top to bottom. Verify that this procedure works as advertised.

(image-compute
 (lambda (col row)
   (irgb (* col 2) 0 (* row 4)))
 129 65)

b. Rewrite this expression so that the blue component decreases (from 255 to 0) as you move from the top to the bottom of the image.

c. Of course, the components can depend upon both the row and the column. Write an expression that generates a 65x64 image that contains a diagonal blend from black to red in a 65x64 image. Your result should look something like the following:

One thing that a good programmer does is generalize her code, so that it works in a variety of situations. The color range code we've written so far is fairly specific to the image size (and even the image placement).

For example, let's consider the computation of a horizontal black-blue blend in a 33x17 image. The core part of that code is likely to read as follows:

(image-compute 
 (lambda (col row) (irgb 0 0 (* 8 col)))
 33 17)

Where do each of the numbers come from? The 33 is the width and the 17 is height. (In case you haven't noticed by now, many blend computations are easier if we make a dimension one more than a power of 2, so that the number of steps is a power of two.) The two zero values represent the red and green components. Where does the 8 come from? It's 256 divided by the number of steps from 0 to the width (32 in this case). If the width were 65 (64 steps), we would use 256/64 = 4.

a. Write a procedure, (horiz-black-to-blue width height) that creates a width-by-height image with a horizontal blend from black to blue.

b. Write a procedure, (horiz-blue-to-black width height) that creates a width-by-height image with a horizontal blend from blue to black.

c. Write a procedure, (horiz-light-to-dark-blue width height) that creates a width-by-height image with a horizontal blend from (0,0,192) to (0,0,64).

d. Write a procedure, (horiz-dark-to-light-blue width height) that creates a width-by-height image with a horizontal blend from (0,0,64) to (0,0,192).

e. In each of these procedures, the red and green components stay constant and the blue component ranges from some value (0 in the first case, 256 in the second, 192 in the third, 64 in the fourth) to some other value (256 in the first case, 0 in the second, 64 in the third, and 192 in the fourth). We can generalize these two procedures by making the initial and final blue component values parameters.

Write a procedure, (horiz-blue-blend width height initial final) that creates a width-by-height image with a horizontal blend from (0,0,initial) to (0,0,final).

f. Rewrite horiz-black-to-blue and horiz-blue-to-black in terms of horiz-blue-blend. Your answer should look something like the following.

(define horiz-black-to-blue
  (lambda (width height)
    (horiz-blue-blend ...)))

You may recall a more complex computation of colors from the reading, one that used sin to determine values. Here's a simplified version of that code:

(image-compute
 (lambda (col row)
   (irgb 0 
         (* 128 (+ 1 (sin (* pi 0.025 col))))
         0))
  40 50)

a. What range of values does (* pi 0.025 col) compute?

b. What range of values does (sin ...) compute?

c. What range of values does (+ 1 (sin ...)) compute?

d. What range of values does (* 128 (+ 1 (sin ...))) compute?

e. Given that analysis, what kind of image do you expect the preceding code to draw?

f. Check your answer experimentally.

g. The reading had a somewhat more complex set of instructions.

(image-compute
 (lambda (col row)
   (irgb 0 
         (* 128 (+ 1 (sin (* pi 0.025 col))))
         (* 128 (+ 1 (sin (* pi 0.020 row))))))
  40 50)

Explain why there might be a .020 rather than a .025 in the computation of the blue component.

h. Predict the appearance of the computed image.

i. Check your answer experimentally.

Consider the following code from the reading.

(image-compute
 (lambda (col row)
   (if (<= (+ (square (- col 40)) (square (- row 50)))
              (square 30))
       (irgb 255 0 0)
       (irgb 0 0 0)))
   145 91)

a. Confirm that the code does, in fact, draw a red circle on a black background.

b. What do you expect to have happen if we change the 40 to 10 and the 50 to 70?

c. Check your answer experimentally.

d. What do you expect to have happen if we change the 30 to 60?

e. Check your answer experimentally.

f. As the reading suggested, instead of just using (irgb 255 0 0), we can substitute a procedure that computes colors based on the position. Try doing so.

Consider the final image from the exercise that used sin to compute an image. Suppose we wanted to make a similar, but larger, version of the image. To do so, we'd certainly need to change the width and height of the image and the ending column and row.

Do we need to change the .025 and .020? We can, but it depends on how we define “similar”.

If you change .025 and .020, you might scale them similarly to how you scale the image. For example, if we double the width and height of the image, we might write

(image-compute
 (lambda (col row)
   (irgb 0 
            (* 128 (+ 1 (sin (* pi 0.0125 col))))
            (* 128 (+ 1 (sin (* pi 0.010 row))))))
  80 100)

a. What do you expect to have happen if you use the preceding code to create a larger, similar, image?

b. Check your answer experimentally.

c. What do you expect to have happen if you restore the .025 and .020 to the preceding code and then rerun it?

d. Check your answer experimentally.

a. Generalize the circle drawing code from above to a procedure, (red-circle image-width image-height circle-radius center-col circle-row) that creates a image-width-by-image-height image consisting of a red circle of the specified radius, centered at the specified position.

b. Clearly, we might want to add circles to existing images, and not just compute new images that contain circles. Write a procedure, (draw-red-circle! image col row radius), that draws a red circle of radius radius centered at (col,row).

One way we can simulate depth in an image is by overlaying one thing over another.

a. Create an image you find aesthetically pleasing by overlaying fixed-color rectangles on top of each other. (How do you overlay rectangles? You use a cond to choose what rectangle to draw.)

b. Create an image you find aesthetically pleasing by overlaying computed-color rectangles on top of each other.

In one of the exercises in this lab, you used trigonometric functions to compute color values. Clearly, we can use these functions, and others, in a variety of ways. For example, we need not linearly scale the row and column, or can scale them in a way to get different ranges. Similarly, we can choose functions other than sin to compute results. In the end, all that we care is that we end up with each component value in the range 0 .. 255.

Experiment with a variety of functions and multipliers to find a combination that you find pleasing.

For example, consider the original computation of the green component

(* 128 (+ 1 (sin (* pi 0.025 col))))

We might replace the 0.25 with another number. We might replace the sin with cos. We might cube the result of that multiplication with

(expt (* 0.025 col 3)

We might even square the result of the sine computation.