Boolean Values and Predicate Procedures

When writing complex programs, we often need to ask questions about the values with which we are computing. Is this pixel a shade of red? Is this image at least 100x100? Are these two colors close enough to be indistinguishable? Is this a light or dark color? Most frequently, these questions (which we often phrase as tests) are used in control structures. For example, we might decide to do one thing for large images and another for small images or we might replace light colors by white and dark colors by black.

To express these kinds of questions, we need a variety of tools. First, we need a type in which to express the valid answers to questions. Second, we need a collection of procedures that can answer simple questions. Third, we need ways to combine questions. Finally, we need control structures that use these questions. In the subsequent sections of this reading, we consider each of these issues. We return to more complex control structures in a subsequent reading.

A Boolean value is a datum that reflects the outcome of a single yes-or-no test. For instance, if one were to ask Scheme to compute whether pure red has a high blue component, it would be able to determine that it does not, and it would signal this result by displaying the Boolean value for “no” or “false”, which is #f. There is only one other Boolean value, the one meaning “yes” or “true”, which is #t. These are called “Boolean values” in honor of the logician George Boole who was the first to develop a satisfactory formal theory of them. (Some folks now talk about “fuzzy logic” that includes values other than “true” and “false”, but that's beyond the scope of this course.)

A predicate is a procedure that always returns a Boolean value. A procedure call in which the procedure is a predicate performs some yes-or-no test on its arguments. For instance, the predicate number? (the question mark is part of the name of the procedure) takes one argument and returns #t if that argument is a number, #f if it does not. Similarly, the predicate even? takes one argument, which must be an integer, and returns #t if the integer is even and #f if it is odd. The names of most Scheme predicates end with question marks, and Grinnell's computer scientists recommend this useful convention, even though it is not required by the rules of the programming language. (If you ever notice that we've failed to include a question mark in a predicate and you're the first to tell us, we'll give you some extra credit.)

Scheme provides a wide variety of basic predicates and MediaScheme adds a few more. We will consider a few right now, but learn more as the course progresses.

Not all the procedures we use to work with Boolean values are strictly predicates. Another useful Boolean procedure is not, which takes one argument and returns #t if the argument is #f and #f if the argument is anything else. For example, one can test whether picture is not an image with

> (not (image? picture))

If Scheme says that the value of this expression is #t, then picture is not an image.

The and and or keywords have simple logical meanings. In particular, the and of a collection of Boolean values is true if all are true and false if any value is false, the or of a collection of Boolean values is true if any of the values is true and false if all the values are false. For example,

> (and #t #t #t)
#t
> (and (< 1 2) (< 2 3))
#t
> (and (odd? 1) (odd? 3) (odd? 5) (odd? 6))
#f
> (and)
#t
> (or (odd? 1) (odd? 3) (odd? 5) (odd? 6))
#t
> (or (even? 1) (even? 3) (even? 4) (even? 5))
#t
> (or)
#f

You may note that we were careful to describe and and or as “keywords” rather than as “procedures”. The distinction is an important one. Although keywords look remarkably like procedures, Scheme distinguishes keywords from procedures by the order of evaluation of the parameters. For procedures, all the parameters are evaluated and then the procedure is applied. For keywords, not all parameters need be evaluated, and custom orders of evaluation are possible.

If and and or were procedures, we could not guarantee their control behavior. We'd also get some ugly errors. For example, consider the extended version of the even? predicate below:

(define new-even?
  (lambda (val)
    (and (integer? val) (even? val))))

Suppose new-even? is called with 2.3 as a parameter. In the keyword implementation of and, the first test, (integer? 2.3), fails, and new-even? returns false. If and were a procedure, we would still evaluate the (even? val), and that test would generate an error, since even? can only be called on integers.

Although many computer scientists, philosophers, and mathematicians prefer the purity of dividing the world into “true” and “false”, Scheme supports a somewhat more general separation. In Scheme, anything that is not false is considered “truish”. Hence, you can use expressions that return values other than truth values wherever a truth value is expected. For example,

> (and #t 1)
1
> (or 3 #t #t)
3
> (not 1)
#f
> (not (not 1))
#t

MediaScheme provides one additional keyword predicate, (defined? name). This procedure takes one parameter, a name (as we would use in define), and determines whether it has been defined by a top-level define.

> (defined? x)
#f
> (define x 2)
> (defined? x)
#t
> (let ((y 2)) (defined? y))
#f
> (defined? y)
#f
> (defined? rgb-new)
#t

Like and and or, although defined? looks like a procedure, it is instead a keyword. Why? Because defined? treats its parameter differently (not evaluating it, as most procedures do, but taking it directly as a name), it can't be implemented as a pure procedure. Hence, if you try to use defined? as a value (e.g., as a parameter to another procedure), you will get an error.

> integer?
#<procedure:integer?>
> defined?
Interactions:1:0: defined?: bad syntax in: defined?

Can we write predicates that work with colors? Certainly. One simple question is whether we might consider two colors near to each other. What are criteria for making that decision? One possibility is that we will consider two colors similar if all of their components are within 8 of each other. We can define that predicate as follows:

;;; Procedure:
;;;   colors-similar?
;;; Parameters:
;;;   color1, an RGB color
;;;   color2, an RGB color
;;; Purpose:
;;;   Determines if color1 and color2 are similar.
;;; Produces:
;;;   similar?, a Boolean value
;;; Preconditions:
;;;   [No additional]
;;; Postconditions: 
;;;   If color1 and color2 are close enough to be considered similar,
;;;     then similar? is #t. 
;;;   Otherwise, similar? is #f.
;;;   We use a proprietary technique to decide what "close enough" means.
(define colors-similar?
  (lambda (color1 color2) 
    (and (>= 8 (abs (- (rgb-red color1) (rgb-red color2))))
         (>= 8 (abs (- (rgb-green color1) (rgb-green color2))))
         (>= 8 (abs (- (rgb-blue color1) (rgb-bluecolor2)))))))

Here's a pair of useful predicates: One computes whether a color might reasonably be considered light; another computes whether a color might reasonably consider dark.

;;; Procedure:
;;;   rgb-light?
;;; Parameters:
;;;   color, an RGB color
;;; Purpose:
;;;   Determine if the color seems light.
;;; Produces:
;;;   light?, a Boolean value
;;; Preconditions:
;;;   [None]
;;; Postconditions:
;;;   light? is true (#t) if color's intensity is relatively high.
;;;   light? is false (#f) otherwise.
(define rgb-light?
  (lambda (color)
    (<= 192 (+ (* 0.30 (rgb-red color)) (* 0.59 (rgb-green color)) (* 0.11 (rgb-blue color))))))

;;; Procedure:
;;;   rgb-dark?
;;; Parameters:
;;;   color, an RGB color
;;; Purpose:
;;;   Determine if the color seems dark.
;;; Produces:
;;;   dark?, a Boolean value
;;; Preconditions:
;;;   [None]
;;; Postconditions:
;;;   dark? is true (#t) if color's intensity is relatively low.
;;;   dark? is false (#f) otherwise.
(define rgb-dark?
  (lambda (color)
    (>= 64 (+ (* 0.30 (rgb-red color)) (* 0.59 (rgb-green color)) (* 0.11 (rgb-blue color))))))

We've seen how and and or can be used to combine tests. But and and or can be used for so much more. In fact, they can be used as control structures.

In an and-expression, the expressions that follow the keyword and are evaluated in succession until one is found to have the value #f (in which case the rest of the expressions are skipped and the #f becomes the value of the entire and-expression). If, after evaluating all of the expressions, none is found to be #f then the value of the last expression becomes the value of the entire and expression. This evaluation strategy gives the programmer a way to combine several tests into one that will succeed only if all of its parts succeed.

This strategy also gives the programmer a way to avoid meaningless tests. For example, we should not make the comparison (< a b) unless we are sure that both a and b are numbers.

In an or expression, the expressions that follow the keyword or are evaluated in succession until one is found to have a value other than#f, in which case the rest of the expressions are skipped and this value becomes the value of the entire or-expression. If all of the expressions have been evaluated and all have the value #f, then the value of the or-expression is #f. This gives the programmer a way to combine several tests into one that will succeed if any of its parts succeeds.

In these cases, and returns the last parameter it encounters (or false, if it encounters a false value) while or returns the first non-false value it encounters. For example,

> (and 1 2 3)
3
> (define x 'two)
> (define y 3)
> (+ x y)
+: expects type <number> as 1st argument, given: two; other arguments were: 3
> (and (number? x) (number? y) (+ x y))
#f
> (define x 2)
> (and (number? x) (number? y) (+ x y))
5
> (or 1 2 3)
1
> (or 1 #f 3)
1
> (or #f 2 3)
2
> (or #f #f 3)
3

We can use the ideas above to make an addition procedure that returns #f if either parameter is not a number. We might say that such a procedure is a bit safer than the normal addition procedure.

;;; Procedure:
;;;   safe-add
;;; Parameters:
;;;   x, a number [verified]
;;;   y, a number [verified]
;;; Purpose:
;;;   Add x and y.
;;; Produces:
;;;   sum, a number.
;;; Preconditions:
;;;   (No additional preconditions)
;;; Postconditions:
;;;   sum = x + y
;;; Problems:
;;;   If either x or y is not a number, sum is #f.
(define safe-add
  (lambda (x y)
    (and (number? x) (number? y) (+ x y))))

Let's compare this version to the standard addition procedure, +.

> (+ 2 3)
5
> (safe-add 2 3)
5
> (+ 2 'three)
Error: +: argument 2 must be: number 
> (safe-add 2 'three)
#f

If we'd prefer to return 0, rather than #f, we could add an or clause.

;;; Procedure:
;;;   safer-add
;;; Parameters:
;;;   x, a number [verified]
;;;   y, a number [verified]
;;; Purpose:
;;;   Add x and y.
;;; Produces:
;;;   sum, a number.
;;; Preconditions:
;;;   [No additional preconditions]
;;; Postconditions:
;;;   If both x and y are numbers, sum = x + y
;;; Problems:
;;;   If either x or y is not a number, sum is 0.
(define safer-add
  (lambda (x y)
    (or (and (number? x) (number? y) (+ x y))
        0)))

In most cases, safer-add acts much like safe-add. However, when we use the result of the two procedures as an argument to another procedure, we get a little bit further through the calculation.

> (* 4 (+ 2 3))
20
> (* 4 (safer-add 2 3))
20
> (* 4 (+ 2 'three))
Error: +: argument 2 must be: number 
> (* 4 (safe-add 2 'three))
Error: *: argument 2 must be: number 
> (* 4 (safer-add 2 'three))
0

Different situations will call for different choices between those strategies.

Here's a simple application of the preceding strategies: We can write a procedure that, given a color, returns black if the color is dark, white if the color is light, and grey if the color is neither dark nor light.

How? Well, we can use and to compute either black, if the color is dark, or #f, if the color is not dark.

(and (rgb-dark? color) black)

Similarly, we can use and to compute either white, if the color is light, or #f if the color is not light.

(and (rgb-light? color) white)

Finally, we can use or to put it all together.

;;; Procedure:
;;;   rgb-bgw
;;; Parameters:
;;;   color, an RGB color
;;; Purpose:
;;;   Convert an RGB color to black, grey, or white, depending on
;;;   the intensity of the color.
;;; Produces:
;;;   bgw, an RGB color
;;; Preconditions:
;;;   rgb-light? and rgb-dark? are defined.
;;; Postconditions:
;;;   If (rgb-light? color) and not (rgb-dark? color), then bgw is white.
;;;   If (rgb-dark? color) and not (rgb-light? color), then bgw is black.
;;;   If neither (rgb-light? color) nor (rgb-dark? color), then bgw is
;;;     grey.
;;; Problems: 
;;;   In the unexpected case that none of the above conditions holds, 
;;      bgw will be one of black, white, and grey.
(define rgb-bgw
  (let ((black (rgb-new 0 0 0))
        (grey (rgb-new 128 128 128))
        (white (rgb-new 255 255 255)))
    (lambda (color)
      (or (and (rgb-light? color) white)
          (and (rgb-dark? color) black)
          grey))))

Here's a slightly stranger, but potentially useful, example. As you've noted, we often want to associate the name canvas with an image on which we experiment. Unfortunately, if you put something like the following in your definitions pane, each time you click Run, you get a new image.

(define canvas (image-show (image-new 200 200)))

Is there something we can do to ensure that we don't create a new image each time we click run? We can generate the image by hand once, observe its number, and then use define to associate the name canvas with that number.

(define canvas 2) ; The number of the image we created by hand

However, if we subsequently close the image, canvas no longer refers to an actual image.

Using the Boolean operations, we can write an expression that will check whether we've created an image (but not too many images) and, if so, associate canvas with that image. If there is no available image, the expression will build and show a new image and then associate canvas with that new image.

(define canvas
  (or (and (image? 1) 1)
      (and (image? 2) 2)
      (and (image? 3) 3)
      (and (image? 4) 4)
      (image-show (image-new 200 200))))

Although it may feel like the define needs to be within the or, the nature of these various keywords makes the particular order necessary.

Clearly, this will only work the first four times we create an image. It seems tempting to use one of our looping structures. However, we do not yet know a looping structure that will handle this type of situation.