Assignment 5: Conditionals and Colors

a. Write a predicate, (is-color? value), that returns #t when value is either an RGB color or a color name recognized by MediaScript. In all other cases, it should return #f.

You can use the predicates irgb? and color-name?, which are already defined, to help you. You should not use the predicate color?; we want you to write your own version.

b. Write your own version of color-to-irgb, which is a very useful procedure. Your procedure should have the following behavior: If given an RGB color, it returns that color unchanged. If given a color name, it returns the result of calling color-name->irgb on that color name. If given any other type of value, it returns #f.

You should not use the built-in color->irgb procedure; we want you to write your own version.

As we learned in the reading on design and color representing colors in terms of hue, saturation, and value is an alternative to RGB representation. Hue represents the pure color (e.g., red, blue, yellow, green, or a combination of one of these). Saturation represents the "colorfulness" of the hue in the color. For instance, a completely saturated color would be a pure hue (like red), while a less saturated color might appear just as bright but somewhat faded (perhaps rose or pink). Value, then represents the brightness or darkness of the color.

As shown below, hue is represented as an angle, or a point on a circle. Thus, the values 0-360 sweep through colors red (0 degrees), yellow (60 degrees), green (120 degrees), cyan (180 degrees), blue (240 degrees), magenta (300 degrees), and back to red (at 360 or 0 degrees).

There are a variety of transformations that can take an RGB color and give an HSV representation. In this problem, we'll focus on just extracting the hue.

Before we describe how to calculate hue, we need some basic values to refer to. Let (red, green, blue) refer to the red, green, and blue components of an RGB color, respectively. The chroma of a color is the largest of the RGB components minus the smallest of the RGB components. For example, the chroma of (128,64,50) is 128-50, or 78; the chroma of (0,255,0) is 255-0, or 255. The chroma of (255,255,255) is 255-255, or 0.

The raw hue can then be calculated as follows:

Note that the numerators of the fractions make some intuitive sense. For example, if the red component is largest, and the green component is larger than the blue component, then we should move counter-clockwise (positive), toward green. And, as we'd hope, the (green-blue) is positive. Similarly, if the red component is largest and the blue component is larger than the green component, then we should move counter-clockwise (negative), toward blue. And, as we'd hope, the (green-blue) is negative.

The raw hue as given above produces a value between -1 and 6 (corresponding to the 6 cardinal colors described above). Why would we end up with a negative number? Well, we just saw that colors in which red dominates that have a larger blue component shift by a negative value. If the raw hue is negative, we should add 6 to get us back to a positive representation. The final result is converted to the range 0-360 by multiplying by 60 degrees (which is 360/6).

What happens if chroma is not 0, and two of the components are both the largest? If the formula is well-designed, it shouldn't matter. But is it? Let's design a quick test.

a. Write three functions, reddish-hue, greenish-hue, and blueish-hue, that take as input an integer-encoded IRGB color and return the computed hue. For example, (blueish-hue (irgb 255 0 255)) should return 5, because the chroma is 255, (red-green) is 255, 255/255 is 1, and 1 + 4 is 5.

b. Write a test suite, hue-equality-tests that confirms the assertion that when a color has two equal RGB components, it doesn't matter which of the two corresponding computations we use.

(define hue-equality-tests
  (test-suite 
   "tests of different mechanisms for computing hues"
   (test-case
    "red = blue; red > green"
    (check-= (reddish-hue (irgb 255 0 255)) (blueish-hue (irgb 255 0 255)) 0)
    (check-= (reddish-hue (irgb 1 0 1)) (blueish-hue (irgb 1 0 1)) 0))))

c. Using the algorithm given above, write a procedure, (irgb->hue-angle irgb), that takes an RGB color and produces its hue. For example,

> (irgb->hue-angle (color->irgb "red"))
0
> (irgb->hue-angle (color->irgb "yellow"))
60
> (irgb->hue-angle (color->irgb "green"))
120
> (irgb->hue-angle (color->irgb "cyan"))
180
> (irgb->hue-angle (color->irgb "blue"))
240
> (irgb->hue-angle (color->irgb "magenta"))
300
> (irgb->hue-angle (color->irgb "pink"))
7340/21
> (round (irgb->hue-angle (color->irgb "pink")))
350

Note: You do not need to use the procedures you wrote for part a, although you may find them helpful. You may also find it usefl to write other procedures to decompose your implementation into managable, meaningful units. Please be sure to give your variables and procedures meaningful names.

Being able to manipulate the hue in a color can actually be quite useful. MediaScheme can convert an HSV color into an integer-encoded RGB color with the procedure (hsv->irgb hsv-list), where hsv-list is a three-element list containing the hue, saturation, and value components of a HSV color. You can create that list with the list procedure. For example, to create a magenta-like color with 50% saturation and 25% value, we would use (hsv->irgb (list 5 1/2 1/4)) or (hsv->irgb (list 5 .5 .25)).

MediaScheme can also extract the saturation and value from an RGB color with the procedures (irgb->saturation irgb-color) and (irgb->value irgb-color).

a. Write a procedure, (irgb-change-hue irgb-color hue), that takes an integer-encoded RGB color and a hue value (in the range 0-360) as parameters and creates a new integer-encoded RGB color using the given hue with the saturation and value of irgb-color.

b. Write a procedure, (irgb-true-complement irgb-color) that finds the true complement of irgb-color, one that is 180 degrees away on the color wheel with the same saturation and value.

Color transformations based on hue be can visually interesting.

Original Image	Hue rotated 30 deg	Hue rotated 90 deg

a. Using the procedures you have written so far, write a procedure (irgb-rotate-hue irgb-color angle) that takes an integer-encoded RGB color and an angle as parameters and produces a new integer-encoded RGB color where the HSV equivalent has a hue rotated by angle degrees, a number between 0-360.

Hint: If the rotated angle is greater than 360, be sure to wrap around properly (e.g., using modulo) to get the correct hue angle.

b. Write an expression as concisely as possible that will rotate the hues in an entire image (say, called picture) by 30 degrees.

Warning: When you apply irgb-rotate-hue thousands of times (as you will in an image of non-trivial size), it is likely to take some time. If you conduct experiments during development, do those experiments on small images.

Write and document your own RGB color transform that transforms the hue of its input color conditionally using a manner of your own choosing. For instance, you may choose to shift colors close to blue more toward green, leaving the rest unchanged.

Your 6P documentation should include a description of the effect of your transform in English. We should be able to apply your transform using image-variant or image-transform!.

Design, document, and implement a Color Scheme Generation Toolkit. Your toolkit should consist of a suite of related procedures. Each procedure will take an integer-encoded RGB color as a parameter and produce a list of colors forming a coherent color scheme. Some procedures may need to take additional parameters: You decide what those parameters will be. You will find the procedures related to hue, saturation, and value very useful.

You should at least write procedures to do the following:

In exercise 2b, you used a series of tests to convince yourself that when a color had two equal largest components, it did not matter which of two formulae you used to compute the hue. But, as you've seen in the past, an incomplete test suite can sometimes convince you that something works when it doesn't. In the exercise above, incorrect implementations of all three procedures could also pass the test. For example, if all three procedures always return 0, they would return the same value.

What's the alternative if you can't be perfectly confident in your tests? One option is to write more tests. For example, we could test each formulae separately on some known values.

But the gold standard is a proof of correctness (or, in this case, equality). Prove that ((blue-red)/chroma) + 2 is the same as ((red-green)/chroma) + 4 when blue and green are equal and larger than red.