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# Design Patterns and Higher-Order Procedures

Summary: In this reading, we explore the so-called higher-order procedures, procedures that take procedures as parameters or return procedures as results. In effect we are considering what happens when we allow procedures to serve as values.

## Background: Design Patterns

One mark of successful programmers is that they identify and remember common techniques for solving problems. Such abstractions of common structures for solving problems are often called patterns or design patterns. You should already have begun to identify some patterns. For example, you know that procedures almost always have the form

(define procname
(lambda (parameters)
body))


You may also have a pattern in mind for the typical recursive procedure over lists:

(define procname
(lambda (lst)
(if (null? lst)
base-case
(do-something (car lst) (procname (cdr lst))))))


In some languages, these patterns are simply guides to programmers as they design new solutions. In other languages, such as Scheme, you can often encapsulate a design pattern in a separate procedure.

## A Simple Pattern: Apply a Procedure to Each Value in a List

Let’s begin with a simple problem: Suppose we have a list of colors and we want to create three new lists: one that contains the red component of each color, one that contains the largest component of each color, and one that contains the average of the three components. Suppose also that your first inclination is to use recursion to create each list, perhaps because you’ve forgotten about map. After some reflection, you might write something like the following

;;; Procedure:
;;;   extract-red-components
;;; Parameters:
;;;   colors, a list
;;; Purpose:
;;;   Get the red component from each color in colors.
;;; Produces:
;;;   reds, a list
;;; Preconditions:
;;;   Every value in colors is an RGB color.
;;; Postconditions:
;;;   Every value in reds is an integer between 0 and 255, inclusive.
;;;   (length reds) = (length colors)
;;;   For each i, 0 <= i < (length colors)
;;;     (list-ref reds i) = (rgb-red (list-ref colors i))
(define extract-red-components
(lambda (colors)
(if (null? colors)
; If no colors remain, there's nothing left to work with,
; so use the empty list.
null
; Otherwise, get the red component of the first color, get the
; red component of the remaining colors, and shove everything
; back together.
(cons (rgb-red (car colors))
(extract-red-components (cdr colors))))))

;;; Procedure:
;;;   extract-max-components
;;; Parameters:
;;;   colors, a list
;;; Purpose:
;;;   Get the maximum (largest) component from each color in colors.
;;; Produces:
;;;   maxs, a list
;;; Preconditions:
;;;   Every value in colors is an RGB color.
;;; Postconditions:
;;;   Every value in maxs is an integer between 0 and 255, inclusive.
;;;   (length maxs) = (length colors)
;;;   For each i, 0 <= i < (length colors)
;;;     (list-ref maxs i) = the maximum component of (list-ref colors i)
(define extract-max-components
(lambda (colors)
(if (null? colors)
; If no colors remain, there's nothing left to work with,
; so use the empty list.
null
; Otherwise, get the maximum component of the first color, get
; the maximum component of each remaining color, and shove
; everything back together.
(cons (max (rgb-red (car colors))
(rgb-green (car colors))
(rgb-blue (car colors)))
(extract-max-components (cdr colors))))))

;;; Procedure:
;;;   extract-ave-components
;;; Parameters:
;;;   colors, a list
;;; Purpose:
;;;   Get the average component from each color in colors.
;;; Produces:
;;;   averages, a list
;;; Preconditions:
;;;   Every value in colors is an RGB color.
;;; Postconditions:
;;;   Every value in averages is an integer between 0 and 255, inclusive.
;;;   (length averages) = (length colors)
;;;   For each i, 0 <= i < (length colors)
;;;     (list-ref averages i) = the average component of (list-ref colors i)
(define extract-ave-components
(let ([ave (lambda (c)
(round (/ (+ (rgb-red c) (rgb-green c) (rgb-blue c)) 3)))])
(lambda (colors)
(if (null? colors)
; If no colors remain, there's nothing left to work with,
; so use the empty list.
null
; Otherwise, get the average component of the first color, get
; the average component of each remaining color, and shove
; everything back together.
(cons (ave (car colors))
(extract-ave-components (cdr colors)))))))


Here’s how each works on the colors in the rainbow.

> (extract-red-components colors-rainbow)
(255 255 255 0 0 75 238)
> (extract-max-components colors-rainbow)
(255 255 255 128 255 130 238)
> (extract-ave-components colors-rainbow)
(85 140 170 43 85 68 202)


What do these three procedures have in common? Most of the documentation, not only the six P’s, but also the internal documentation. All return null when given the null list. More importantly, all three do something to the car of the list, recurse on the cdr of the list, and then cons the two results together.

Hence, it is natural to design a general pattern for apply a procedure to every value in a list.

(define procname
(lambda (lst)
(if (null? lst)
null
(cons (modify (car lst))
(procname (cdr lst))))))


Now, here’s the cool part. We can even turn this pattern into a procedure.

;;; Procedure:
;;;   colors-transform-v0
;;; Parameters:
;;;   colors, a list of colors.
;;; Purpose:
;;;   Applies TRANSFORM to each color in a list.
;;; Returns:
;;;   transformed, a list of colors.
;;; Preconditions:
;;;   lst contains only RGB colors. [Unverified]
;;;   TRANSFORM is defined, takes one color as a parameter
;;;     and returns a color. [Unverified]
;;; Postconditions:
;;;   transformed contains only RGB colors.
;;;   (length transformed) = (length colors)
;;;   For each i, 0 <= i < (length transformed)
;;;    (list-ref transformed i) = (TRANSFORM (list-ref colors i))
(define colors-transform-v0
(lambda (colors)
; If no elements remain, we have nothing to apply TRANSFORM to,
; so stick with the empty list.
(if (null? colors) null
; Otherwise, transform the first value, transform the remaining
; values, and put the stuff back together into a list.
(cons (TRANSFORM (car colors))
(colors-transform-v0 (cdr colors))))))


Where does TRANSFORM come from? We could define it first (as the preconditions suggest). For example, here’s an application of colors-transform in which we get just the red component of a color.

> (define TRANSFORM rgb-red)
> (colors-transform-v0 colors-rainbow)
(255 255 255 0 0 75 238)


Similarly, we can get the average component by defining TRANSFORM as extract-ave-components.

> (define TRANSFORM (lambda (c) (round (/ (+ (rgb-red c) (rgb-green c) (rgb-blue c)) 3))))
> (colors-transform-v0 colors-rainbow)
(85 140 170 43 85 68 202)


You may find this strategy of defining TRANSFORM first inelegant. We certainly do. It also won’t work for all cases. For example, what if we want to get the red components of all the colors in one list and then get the average components of all the colors in another list? We’d have to redefine TRANSFORM right in the middle of the code. Ugly.

As importantly, we can’t use local bindings to define TRANSFORM. For example, consider the following code.

> (define TRANSFORM rgb-complement)
> (map rgb->string (colors-transform-v0 colors-rainbow))
("0/255/255" "0/90/255" "0/0/255" "255/127/255" "255/255/0" "180/255/125" "17/125/17")
> (define TRANSFORM rgb-darker)
> (map rgb->string (colors-transform-v0 colors-rainbow))
("239/0/0" "239/149/0" "239/239/0" "0/112/0" "0/0/239" "59/0/114" "222/114/222")
> (let ((TRANSFORM rgb-lighter)) (map rgb->string (colors-transform-v0 colors-rainbow)))
("239/0/0" "239/149/0" "239/239/0" "0/112/0" "0/0/239" "59/0/114" "222/114/222")


So, we’re stuck with redefining TRANSFORM using define in the middle of our code, and that is unlikely to be clear or convenient.

Is there a better solution? Yes! Instead of forcing TRANSFORM to be defined before we call colors-transform, we can make the TRANSFORM procedure a parameter to the colors-transform pattern. For example, here’s a new version of colors-transform that takes the extra parameter.

;;; Procedure:
;;;   colors-transform
;;; Parameters:
;;;   colors, a list of colors.
;;;   trans, a procedure
;;; Purpose:
;;;   Applies trans to each color in a list.
;;; Returns:
;;;   transformed, a list of colors.
;;; Preconditions:
;;;   lst contains only RGB colors. [Unverified]
;;;   trans takes one color as a parameter and returns a color. [Unverified]
;;; Postconditions:
;;;   transformed contains only RGB colors.
;;;   (length transformed) = (length colors)
;;;   For each i, 0 <= i < (length transformed)
;;;    (list-ref transformed i) = (trans (list-ref colors i))
(define colors-transform
(lambda (colors trans)
; If no elements remain, we have nothing to apply transform to,
; so stick with the empty list.
(if (null? colors) null
; Otherwise, transform the first value, transform the remaining
; values, and put the stuff back together into a list.
(cons (trans (car colors))
(colors-transform (cdr colors) trans)))))


If we plug in procedures (either previously-named procedures or newly named procedures), we get the results we expect.

> (map rgb->string colors-rainbow)
("255/0/0" "255/165/0" "255/255/0" "0/128/0" "0/0/255" "75/0/130" "238/130/238")
> (map rgb->string (colors-transform colors-rainbow rgb-complement))
("0/255/255" "0/90/255" "0/0/255" "255/127/255" "255/255/0" "180/255/125" "17/125/17")
> (map rgb->string (colors-transform colors-rainbow rgb-darker))
("239/0/0" "239/149/0" "239/239/0" "0/112/0" "0/0/239" "59/0/114" "222/114/222")
> (map rgb->string (colors-transform colors-rainbow rgb-lighter))
("255/16/16" "255/181/16" "255/255/16" "16/144/16" "16/16/255" "91/16/146" "254/146/254")
> (map rgb->string (colors-transform colors-rainbow rgb-darker))
("239/0/0" "239/149/0" "239/239/0" "0/112/0" "0/0/239" "59/0/114" "222/114/222")


Thus, as you’ve seen before in this class, you can write your own procedures that take procedures as parameters. We call procedures that take other procedures as parameters higher-order procedures.

Of course, the idea of passing a procedure as a parameter should be comfortable if you’ve been using map, image-variant, foreach!, and similar functions. In fact, map is likely to be defined much like colors-transform (except that officially, the order in which map builds the result list is up to the implementer). We could even define our own version (and you may even have already done so yourself).

;;; Procedure:
;;;   proc-map
;;; Parameters:
;;;   proc, a procedure
;;;   lst, a list
;;; Purpose:
;;;   Applies proc to each value in a list.
;;; Returns:
;;;   newlst, a list
;;; Preconditions:
;;;   proc can successfully be applied to each value in lst.
;;; Postconditions:
;;;   newlst is the same length as lst
;;;   For each i, 0 <= i < (length newlst)
;;;    (list-ref newlst i) = (proc (list-ref lst i))
(define proc-map
(lambda (proc lst)
; If no elements remain, we can't apply anything else,
; so produce the empty list.
(if (null? lst)
null
; Otherwise, apply the procedure to the first value,
; apply the procedure to the remaining values, and
; put the results back together into a list.
(cons (proc (car lst))
(proc-map proc (cdr lst))))))


Let us now return to the starting problems. What if we want to get the red component of every color in a list? We map irgb-red onto the list.

> (proc-map irgb-red colors-rainbow)
(255 255 255 0 0 102 79)


What if we want the average component? We might use let to name the ave helper, and then use map with that.

> (let ((ave (lambda (c)
(round (/ (+ (irgb-red c) (irgb-green c) (irgb-blue c)) 3)))))
(proc-map ave colors-rainbow))
(85 125 170 85 85 119 68)


Since we can write expressions like that, we would no longer need to write extract-ave-components. We probably wouldn’t even need to write extract-red-components or extract-max-components.

That observation suggests a second important moral: Once you’ve defined a few higher-order procedures, like map, you can often avoid writing other procedures, since the higher-order procedures let you write more general expressions.

For example, if we have a list of colors, and we want to compute a lighter version of each color in the list, we can use colors-transform, proc-map, or even map.

> (define lighter-rainbow (proc-map irgb-lighter colors-rainbow))


## Higher Order Predicates: Are they All Whatever?

There are many advantages to encoding design patterns in higher-order procedures. An important one is that it stops us from tediously writing the same thing over and over and over again. Think about writing the predicates all-integer?, all-irgb?, all-spot?, and so on and so forth. We’ve done so many times. However, as our colleague, John Stone, says (in reference to writing a sequence of similar procedures),

Writing and testing one of these definitions is an interesting and instructive exercise for the beginning Scheme programmer. Writing and testing another one is good practice. Writing and testing the third one is, frankly, a little tedious. If we then move on to [others], eventually programming is reduced to typing.

So, how do we avoid the repetitious typing? We begin with one of the procedures.

;;; Procedure:
;;;   all-int?
;;; Parameters:
;;;   lst, a list
;;; Purpose:
;;;   Determine if all of the values in lst are integers.
;;; Produces:
;;;   ok?, a Boolean
;;; Preconditions:
;;;   [Standard]
;;; Postconditions:
;;;   If there is an i such that (integer? (list-ref lst i))
;;;     fails to hold, then ok? is false.
;;;   Otherwise, ok? is true.
(define all-int?
(lambda (lst)
(or (null? lst)
(and (integer? (car lst))
(all-int? (cdr lst))))))


Next, we identify the parts of the procedure that depend on our current type (i.e., that everything is a list).

(define all-WHATEVER?
(lambda (val)
(or (null? val)
(and (WHATEVER? (car val))
(all-WHATEVER? (cdr val))))))


Finally, we remove the dependent part or parts from the procedure name and make them parameters to the modified procedure.

;;; Procedure:
;;;   all
;;; Parameters:
;;;   pred?, a unary predicate
;;;   lst, a list
;;; Purpose:
;;;   Determine if pred? holds for all the values in lst.
;;; Produces:
;;;   ok?, a Boolean
;;; Preconditions:
;;;   [Standard]
;;; Postconditions:
;;;   If there is an i such that (pred? (list-ref lst i))
;;;     fails to hold, then ok? is false.
;;;   Otherwise, ok? is true.
(define all
(lambda (pred? lst)
(or (null? lst)
(and (pred? (car lst))
(all pred? (cdr lst))))))


Here’s how we might test whether something is a list of numbers.

> (all integer? (list 1 2 3))
#t
> (all integer? (list 1 "two" 3))
#f


We can also define all-int? using the all procedure.

(define all-int?
(lambda (lst)
(all integer? lst)))


The results are the same.

> (all-int? (list 1 2 3))
#t
> (all-int? (list 1 "two" 3))
#f


## Built-in Higher-Order Procedures

We have seen that it is possible to write our own higher-order procedures. Scheme also includes a number of built-in higher-order procedures. You can read about many of them in section 6.4 of the Scheme report (r5rs), which is available through the DrRacket Help Desk. Here are some of the more popular ones.

### The map Procedure

You already know about the basic use of the map procedure. You also know how one might implement it. It turns out that map has some additional capabilities. For example, you can apply a procedure to multiple lists (in which case it takes the corresponding value from each list).

> (map + (list 1 2 3) (list .5 .6 .7))
(1.5 2.6 3.7)
> (define aardvark-grades (list 98 75 90 80 100))
aardvark-grades
> (define baboon-grades (list 80 82 84 86 88))
baboon-grades
> (define cheetah-grades (list 50 95 50 95 50))
cheetah-grades
> (define best-grades (map max aardvark-grades baboon-grades cheetah-grades))
best-grades
> best-grades
(98 95 90 95 100)
> (define aardvark-scaled (map (lambda (x) (* 100 x)) (map / aardvark-grades best-grades)))
aardvark-scaled
> aardvark-scaled
(100 78.94736842 100 84.21052632 100)


### The apply Procedure

One of the most important built-in higher-order procedures is apply, which takes a procedure and a list as arguments and invokes the procedure, giving it the elements of the list as its arguments:

> (apply string=? (list "foo" "foo"))
#t
> (apply * (list 3 4 5 6))
360
> (apply append (list (list 'a 'b 'c) (list 'd) (list 'e 'f)
null (list 'g 'h 'i)))
(a b c d e f g h i)


Unfortunately, since and and or are not procedures but keywords with their own evaluation rules, we can’t use them with apply.

> (map string? (list "alpha" 'beta "gamma"))
(#t #f #t)
> (and #t #f #t)
#f
> (apply and (map string? (list "alpha" 'beta "gamma")))
Error: eval: unbound variable: and


It might seem odd to write a call to apply with these manually constructed lists when we could instead just call the requested function with the parameters directly. If that is the case, why have apply at all? The answer is, sometimes parameter lists themselves are not known at the time we write the program and are built on-the-fly, or else it is just plain easier to build-them on the fly.

As an example, consider the termial function from the reading on numeric recursion.

;;; Procedure:
;;;   termial
;;; Parameters:
;;;   number, a natural number
;;; Purpose:
;;;   Compute the sum of natural numbers not greater than a given
;;;   natural number
;;; Produces:
;;;   sum, a natural number
;;; Preconditions:
;;;   number is a number, exact, an integer, and non-negative.
;;;   The sum is not larger than the largest integer the language
;;;     permits.
;;; Postconditions:
;;;   sum is the sum of natural numbers not greater than number.
;;;   That is, sum = 0 + 1 + 2 + ... + number
(define termial
(lambda (number)
(if (zero? number)
0
(+ number (termial (- number 1))))))


As the documentation plainly tells us, the result is the sum 1 + 2 + 3 + ... + number As it turns out, we have a concise way to generate the list of these numbers by using iota. We therefore might just have well as used apply and iota to write termial, as follows.

(define termial
(lambda (number)
(apply + (iota (+ 1 number)))))


## Returning Procedures

Just as it is possible for procedures to take procedures as their parameters, it is also possible for procedures to produce new procedures as their return values. For example, here is a procedure that takes one parameter, a number, and creates a procedure that multiplies its parameters by that number.

;;; Procedure:
;;;   make-multiplier
;;; Parameters:
;;;   n, a number
;;; Purpose:
;;;   Creates a new procedure which multiplies its parameter by n.
;;; Produces:
;;;   multiplier, a procedure of one parameter
;;; Preconditions:
;;;   n must be a number
;;; Postconditions:
;;;   (multiplier v) = n * v
(define make-multiplier
(lambda (n) ; n is the parameter to make-multiplier
; Return value: A procedure
(lambda (v) (* n v))))


Let’s test it out.

> (make-multiplier 5)
#<procedure>
> (define timesfive (make-multiplier 5))
> (timesfive 4)
20
> (timesfive 101)
505
> (map (make-multiplier 3) (list 1 2 3))
(3 6 9)


We can use the same technique to build the legendary compose operation that, given two functions, f and g, builds a function that applies g and then f.

;;; Procedure:
;;;   compose
;;; Parameters:
;;;   f, a unary function
;;;   g, a unary function
;;; Purpose:
;;;   Functionally compose f and g.
;;; Produces:
;;;   fun, a unary function.
;;; Preconditions:
;;;   f can be applied to any values g generates.
;;; Postconditions:
;;;   fun can be applied to any values g can be applied to.
;;;   fun generates values of the type that f generates.
;;;   (fun x) = (f (g x))
(define compose
(lambda (f g) ; f and g are the parameters to compose
; compose returns a procedure of one parameter
(lambda (x)
; that procedure applies g, and then applies f.
(f (g x)))))


Here are some tests of that procedure.

> (define add2 (lambda (x) (+ 2 x)))
> (define mul5 (lambda (x) (* 5 x)))
> (define fun1 (compose add2 mul5))
> (fun1 5)
27
> (fun1 3)
17
> (define fun2 (compose mul5 add2))
> (fun2 5)
35
> (fun2 3)
25


Especially when using map, we often write anonymous procedures that look something like the following.

  (lambda (num) (* 2 num))


Even make-multiplier is actually something we might want to generalize. You’ll note that in that procedure, we filled in one parameter (*n) of a two-parameter procedure (*). In pattern form, we might write

  (lambda (val) (BINARY-PROC ARG1 val))


Let’s think about how we might turn that into procedures. (left-section binary-proc arg1) creates a new procedure by filling in the first argument of a binary procedure. (right-section binary-proc arg2) creates a new procedure by filling in the second argument of a binary procedure. We often abbreviate these two procedures as l-s and r-s.

In the following example, we define procedures that multiply their parameter by 2 or subtract 3 from their parameter, or some combination thereof.

> (define mul2 (left-section * 2))
mul2
> (define sub3 (right-section - 3))
sub3
> (map mul2 (list 1 2 3 4 5))
(2 4 6 8 10)
> (map sub3 (list 1 2 3 4 5))
(-2 -1 0 1 2)
> (map (compose mul2 sub3) (list 1 2 3 4 5))
(-4 -2 0 2 4)
> (map (compose sub3 mul2) (list 1 2 3 4 5))
(-1 1 3 5 7)
> (map (compose (l-s * 2) (r-s - 3)) (list 1 2 3 4 5))
(-4 -2 0 2 4)
> (map (compose (l-s * 2) (l-s - 3)) (list 1 2 3 4 5))
(4 2 0 -2 -4)


Okay, what does left-section look like? The definition is fairly straightforward.

;;; Procedures:
;;;   left-section
;;;   l-s
;;; Parameters:
;;;   binproc, a two-parameter procedure
;;;   left, a value
;;; Purpose:
;;;   Creates a one-parameter procedure by filling in the first parameter
;;    of binproc.
;;; Produces:
;;;   unproc, a one-parameter procedure
;;; Preconditions:
;;;   left is a valid first parameter for binproc.
;;; Postconditions:
;;;   (unproc right) = (binproc left right)
(define left-section
(lambda (binproc arg1)
; Build a new procedure of one argument
(lambda (arg2)
; That calls binproc on the appropriate arguments
(binproc arg1 arg2))))
(define l-s left-section)


How is right-section defined? We leave that as an exercise for the reader.

Experienced Scheme programmers regularly use left-section and right-section, not only in calls to map and other higher-order procedures, but also in defining new procedures. For example, consider the all-int? procedure that we previously defined as

(define all-int?
(lambda (lst)
(all integer? lst)))


Here is an even more concise definition that takes advantage of l-s.

(define all-int? (l-s all integer?))


## Self Checks

### Check 1: Reviewing Definitions

a. What is the definition of a higher-order procedure?

b. Despite its length, this reading only introduced one new built-in Scheme higher-order procedure. What is it?

c. Why is apply considered a higher-order procedure?

### Check 2: Reflecting on apply

a. The reading gives two versions of termial. Which do you prefer? Why?

b. Use a strategy similar to the updated termial to rewrite the procedure sum, which sums a list of numbers, using apply, rather than recursion.

c. If you got the previous check, you might now notice that you’ve written a procedure (sum) that merely calls another procedure (apply) with its first parameter fixed, a pattern that suggests we use sectioning for further abbreviation. See if you can rewrite sum once again so that it uses l-s (left-section) rather than lambda to define the procedure.

Hint: Here is another example of using this pattern to abbreviate a procedure definition.

(define double
(lambda (number)
(* 2 number)))

(define double (l-s * 2))