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Summary: Algorithm designers regularly find it useful to name the values their algorithms process. We consider why and how to name new values within an algorithm.
When writing programs and algorithms, it is useful to name values we compute along the way. For example, in an algorithm that, given a color, computes a grey color with the same brightness as the original color, it is useful to name that brightness. When we associate a name with a value, we say that we bind that name to the value.
So far we've seen three ways in which Scheme permits the algorithm writer to bind a name to a value:
quotient, are predefined. When DrScheme starts up, these names are already bound to the procedures they denote.
There are often times when it seems that you repeat work that should only have to be done once. For example, consider again the problem of computing a grey that has the same brightness of the original color. We can certainly write the following:
(define rgb.greyscale (lambda (color) (rgb.new (+ (* 0.30 (rgb.red color)) (* 0.59 (rgb.green color)) (* 0.11 (rgb.blue color))) (+ (* 0.30 (rgb.red color)) (* 0.59 (rgb.green color)) (* 0.11 (rgb.blue color))) (+ (* 0.30 (rgb.red color)) (* 0.59 (rgb.green color)) (* 0.11 (rgb.blue color))))))
However, that code has some difficulties. First, it's a bit hard to read. Why are we using those weird formulae? Are the three the same? If not, why not. Second, it's hard to correct if we change our formula, since the formula is repeated three times. Finally, this repetition of code will lead to inefficiencies. Why should we compute the same product and the same sum three times. If we rely only on the Scheme we know so far, we can solve the first two problems by writing a separate procedure to compute the brightness.
;;; Procedure: ;;; rgb.brightness ;;; Parameters: ;;; color, an RGB color ;;; Purpose: ;;; Computes the brightness of color on a 0 (dark) to 255 (bright) scale. ;;; Produces: ;;; b, an integer ;;; Preconditions: ;;; color is a valid RGB color. That is, each component is between ;;; 0 and 255, inclusive. ;;; Postconditions: ;;; If color1 is likely to be perceived as brighter than color2, ;;; then (brightness color1) > (brightness color2). (define rgb.brightness (lambda (color) (+ (* .30 (rgb.red color)) (* .59 (rgb.green color)) (* .11 (rgb.blue color)))))
Once we've defined
rgb.brightness, all we need to do is
call it three times to build a single shade of grey.
(define rgb.greyscale (lambda (color) (rgb.new (rgb.brightness color) (rgb.brightness color) (rgb.brightness color))))
This change certainly makes it easier to read
In addition, it's easier to update. If we decide to compute brightness
differently, such as by averaging the three colors, we only need to
change one piece of code.
(define rgb.brightness (lambda (color) (/ (+ (rgb.red color) (rgb.green color) (rgb.blue color)) 3)))
Although we have solved the first two deficiencies of the original code, we are still repeating work. Can we avoid the repetition of work? Certainly, we can write a procedure that makes a shade of grey from a single component value.
;;; Procedure: ;;; rgb.grey ;;; Parameters: ;;; level, an integer ;;; Purpose: ;;; Produces a shade of grey. ;;; Produces: ;;; grey, a color. ;;; Preconditions: ;;; level is an integer between 0 and 255, inclusive. ;;; Postconditions: ;;; Each component of grey is level . (define rgb.grey (lambda (level) (rgb.new level level level)))
Now, we can simply write
(define rgb.greyscale (lambda (color) (rgb.grey (rgb.brightness color))))
But that's a lot of extra work. It's inconvenient to have to write (and document!) two procedures that we're just going to use just this once.
let expressions as an alternative way to create
local bindings. A
let-expression contains a binding
list and a body. The body can be any expression, or sequence of
expressions, to be evaluated with the help of the local name bindings. The
binding list is a pair of structural parentheses enclosing zero or more
binding specifications; a binding specification, in turn, is a pair
of structural parentheses enclosing a name and an expression.
That precise definition may have been a bit confusing, so
here's the general form of a
(let ((name1 exp1) (name2 exp2) ... (namen expn)) body1 body2 ... bodym)
When Scheme encounters a
let-expression, it begins by
evaluating all of the expressions inside its binding specifications. Then
the names in the binding specifications are bound to those values. Next,
the expressions making up the body of the
evaluated, in order. The value of the last expression in the body becomes
the value of the entire
let-expression. Finally, the local
bindings of the names are cancelled. (Names that were unbound before the
let-expression become unbound again; names that had different
bindings before the
let-expression resume those earlier
Here's how we'd solve the earlier problem with
let and without
(define rgb.greyscale (lambda (color) (let ((brightness (+ (* 0.30 (rgb.red color)) (* 0.59 (rgb.green color)) (* 0.11 (rgb.blue color))))) (rgb.new brightness brightness brightness))))
Here's another example of a binding list, taken from a
let-expression in a real Scheme program:
(let ((next (car source)) (stuff null)) ...)
This binding list contains two binding specifications -- one in which the
value of the expression
(car source) is bound to the name
next, and the other in which the empty list is bound to the
stuff. Notice that binding lists and binding
specifications are not procedure calls; their role in a
let-expression simply to give names to certain values while
the body of the expression is being evaluated. The outer parentheses in a
binding list are
structural, like the outer parentheses in a
cond-clause -- they are there to group the pieces of the
binding list together.
let-expression often simplifies an expression that
contains two or more occurrences of the same subexpression. The programmer
can compute the value of the subexpression just once, bind a name to it,
and then use that name whenever the value is needed again. Sometimes this
speeds things up by avoiding such redundancies as the re0computation of
In other cases, there is little difference in speed, but the
code may be a little clearer. For instance, consider the following
list.remove procedure that removes all copies of a value from
a list. In the past, we might have written that procedure as follows.
;;; Procedure: ;;; lst.remove ;;; Parameters: ;;; lst, a list of values ;;; item, a value ;;; Purpose: ;;; Removes all copies of item from lst ;;; Produces: ;;; newls, a list ;;; Preconditions: ;;; lst is a list. It may be empty. ;;; Postconditions: ;;; No values equal to item appear in newls. ;;; Every value not equal to item that appeared in lst also ;;; appears in newls. ;;; Every value that appears in newls also appears in lst. ;;; If a preceded b in lst and neither a nor b equals item, ;;; then a precedes b in newls. (define list.remove (lambda (lst item) (cond ; If the list is empty, removing the element still gives ; us the empty list ((null? ls) null) ; If the first element of the list matches, skip over it. ((equal? item (car ls)) (list.remove (cdr ls) item)) ; Otherwise, preserve the first element and remove item ; from the remainder of ls (else (cons (car ls) (list.remove (cdr ls) item))))))
Here is an alternative definition of the
procedure which some people find clearer.
(define list.remvoe (lambda (item ls) ; If the list is empty, removing the element still gives ; us the empty list (if (null? ls) null (let ( ; Name the car of the list first-element. (first-element (car ls)) ; Recurse on the rest of the list and name it ; rest-of result. (rest-of-result (list.remove (cdr ls) item))) ; If the first element of the list matches, skip over it. (if (equal? first-element item) rest-of-result ; Otherwise, preserve the first element and attach ; it to the rest. (cons first-element rest-of-result))))))
The technique can also be useful not just for clarifying what a procedure
does, but also for making them more efficient. Consider the following
version of the
rgb.brightest procedure, which we've studied a
;;; Procedure: ;;; rgb.brightest ;;; Parameters: ;;; colors, a list of RGB colors ;;; Purpose: ;;; Find the brightest color in colors. ;;; Produces: ;;; brightest, an RGB color. ;;; Preconditions: ;;; colors is nonempty. [Unverified] ;;; Each element of colors is an RGB color. [Unverified] ;;; Postconditions: ;;; brightest is equal to at least one element of colors. ;;; brightest is at least as bright as each element of colors. (define rgb.brightest (lambda (colors) (if (null? (cdr colors)) (car colors) (if (rgb.brighter? (car colors) (rgb.brightest (cdr colors))) (car colors) (rgb.brightest (cdr colors))))))
If we rewrite this with a
let, we get the following.
(define rgb.brightest (lambda (color) (if (null? (cdr colors)) (car colors) (let ((first-color (car colors)) (brightest-remaining (rgb.brightest (cdr colors)))) (if (>= (rgb.brightness first-color) (rgb.brightness brightest-remaining)) first-color brightest-remaining)))))
As you'll find in the lab, this is much more efficient.
Sometimes we may want to name a number of interrelated things.
For example, suppose we wanted to square the average of a list
of numbers (well, it's something that people do sometimes). Since
computing the average involves summing values, we may want to name two
different things: the total and the average (mean). We can nest one
let-expression inside another to name both things.
(let ((total (+ (rgb.red color) (rgb.green color) (rgb.blue color)) (let ((mean (/ total 3))) (* mean mean)))
One might be tempted to try to combine the binding lists for the nested
;; Combining the binding lists doesn't work! (let ((total (+ (rgb.red color) (rgb.green color) (rgb.blue color)) (mean (/ total 3))) (* mean mean))
This wouldn't work (try it and see!), and it's important to understand why
not. The problem is that, within one binding list, all of the
expressions are evaluated before any of the names are bound.
Specifically, Scheme will try to evaluate both
(+ 8 3 4 2 7)
(/ total 5) before binding either of the names
(/ total 5) can't
be computed until
total has a value, an error occurs. You have
to think of the local bindings coming into existence simultaneously rather
than one at a time.
Because one often needs sequential rather than simultaneous binding, Scheme
provides a variant of the
let-expression that rearranges the
order of events: If one writes
let* rather than
let, each binding specification in the binding list is
completely processed before the next one is taken up:
;; Using let* instead of let works! (let* ((total (+ (rgb.red color) (rgb.green color) (rgb.blue color))) (mean (/ total 3))) (* mean mean))
The star in the keyword
let* has nothing to do with
multiplication. Just think of it as an oddly shaped letter. It
means "do things in sequence, rather than all at once". I have
no idea why they've chosen to do that.
In the examples above, we've tended to do the naming within the body of the procedure. That is, we write
(define proc (lambda (params) (let (...) exp)))
However, Scheme also lets us choose an alternate ordering. We can
instead put the
let before (outside of) the
(define proc (let (...) (lambda (params) exp)))
Why would we ever choose to do so? Let us consider an example. Suppose that we regularly need to convert years to seconds. (When you have sons between the ages of 5 and 12, you'll understand.) You might begin with
(define years-to-seconds (lambda (years) (return (* years 365.24 24 60 60))))
This is, of course, correct. However, it is a bit hard to read. You might want to name some of the values for clarity.
(define years-to-seconds (lambda (years) (let* ((days-per-year 365.24) (hours-per-day 24) (minutes-per-hour 60) (seconds-per-minute 60) (seconds-per-year (* days-per-year hours-per-day minutes-per-hour seconds-per-minute))) (* years seconds-per-year)))) > (years-to-seconds 10) 315567360.0
We have clearly clarified the code, although we have also lengthened
it a bit. However, as we noted before, a second goal of naming is to
avoid recomputation of values. Unfortunately, even though the number
of seconds per year never changes, we compute it every time that
years-to-seconds. How can we avoid this
recomputation? One strategy is to move the bindings to
(define days-per-year 365.24) ... (define seconds-per-year (* days-per-year ... seconds-per-minute)) (define years-to-seconds (lambda (years) (* years seconds-per-year)))
However, such a strategy is a bit dangerous. After all, there is nothing to prevent someone else from changing the values here.
(define days-per-year 360) ; Some strange calendar, perhaps in Indiana ... > (years-to-seconds 10) 311040000
What we'd like to do is to declare the values once, but keep them
years-to-seconds. The strategy is to move the
let outside the
(define years-to-seconds (let* ((days-per-year 365.24) (hours-per-day 24) (minutes-per-hour 60) (seconds-per-minute 60) (seconds-per-year (* days-per-year hours-per-day minutes-per-hour seconds-per-minute))) (lambda (years) (* years seconds-per-year)))) > (years-to-seconds 10) 315567360.0
As you'll see in the lab, it is possible to empirically verify that the bindings occur only once in this case, and each time the procedure is called in the prior case.
So, one moral of this story is whenever possible, move your
bindings outside the
lambda. However, it is not
always possible to do so. For example, if your let-bindings use
parameters (as in the earlier quadratic example), then you need to
keep them within the body of the lambda.
As you may have noted,
let behaves somewhat like
define in that programmers can use it to name values.
But we've used
define to name more than values; we've
also used it to name procedures. Can we also use
Yes, one can use a
create a local name for a procedure. And we name procedures locally
for the same reason that we name values, because it speeds and clarifies
(define hypotenuse-of-right-triangle (let ((square (lambda (n) (* n n)))) (lambda (first-leg second-leg) (sqrt (+ (square first-leg) (square second-leg))))))
Regardless of whether
square is also defined outside
this definition, the local binding gives it the appropriate
meaning within the lambda-expression that describes what
Note, once again, that there are two places one might define
square locally. We can define it before the lambda (as
above) or after the lambda (as below). In the first case, the definition
is done only once. In the second case, it is done every time
the procedure is executed.
(define hypotenuse-of-right-triangle (lambda (first-leg second-leg) (let ((square (lambda (n) (* n n)))) (sqrt (+ (square first-leg) (square second-leg))))))
So, which we should you do it? If the helper procedure you're defining
uses any of the parameters of the main procedure, it needs to come after
lambda. Otherwise, it is generally a better idea to
do it before the lambda. As you practice more with
you'll find times that each choice is appropriate.
Unfortunately, you cannot use
let to define recursive
procedures. In a subsequent reading, you'll learn another variant
let that supports recursive procedures. Once you've
learned that technique your helpers can all be local.
I usually create these pages
on the fly, which means that I rarely
proofread them and they may contain bad grammar and incorrect details.
It also means that I tend to update them regularly (see the history for
more details). Feel free to contact me with any suggestions for changes.
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