There are two ideas about how to solve programming problems. One idea is that programmers work mostly by recognizing categories of problems that come up repeatedly and remembering the solution that worked last time; therefore, programming students should learn a lot of program patterns, or templates, and fill in the blanks for each specific problem. Another idea is that there are a few powerful principles in programming, and that if a learner understands the principles, they can be applied to any problem, even one that doesn't fit a familiar pattern.
Research suggests that an expert programmer, like an expert at any skill, does work mainly by recognizing patterns. Nevertheless, we lean toward the powerful-principle idea. The expert's memory is not full of arbitrary patterns; it's full of meaningful patterns, because the expert has gone through the process of struggling to reason out how each procedure works and how to write new procedures.
Still, we think it's worth pointing out a few patterns that are so common that you'll have seen several examples of each before you finish this book. Once you learn these patterns, you can write similar procedures almost automatically. But there's an irony in learning patterns: In Scheme, once you've identified a pattern, you can write a general-purpose procedure that handles all such cases without writing individual procedures for each situation. Then you don't have to use the pattern any more! Chapter 8 presents several general pattern-handling procedures, called higher-order procedures. In this chapter we'll consider the patterns corresponding to those higher-order procedures, and we'll use the names of those procedures to name the patterns.
What's the point of learning patterns if you can use higher-order procedures instead? There are at least two points. The first, as you'll see very soon, is that some problems almost follow one of the patterns; in that case, you can't use the corresponding higher-order procedure, which works only for problems that exactly follow the pattern. But you can use your understanding of the pattern to help with these related problems. The second point is that in Chapter 19 we'll show how the higher-order functions are themselves implemented using these recursive patterns.
This chapter isn't an official list of all important patterns; as you gain programming experience, you'll certainly add more patterns to your repertoire.
Here's a procedure to square every number in a sentence of numbers:
(define (square-sent sent) (if (empty? sent) '() (se (square (first sent)) (square-sent (bf sent)))))
Here's a procedure to translate every word of a sentence into Pig Latin:
(define (pigl-sent sent) (if (empty? sent) '() (se (pigl (first sent)) (pigl-sent (bf sent)))))
The pattern here is pretty clear. Our recursive case will do something
straightforward to the
first of the sentence, such as
pigling it, and we'll combine that with the result of a recursive
call on the
butfirst of the sentence.
letter-pairs procedure that we wrote in Chapter 11 is
an example of a procedure that follows the
every pattern pretty
closely, but not exactly. The difference is that
at its argument sentence two words at a time.
(define (letter-pairs wd) (if (= (count wd) 1) '() (se (word (first wd) (first (bf wd))) (letter-pairs (bf wd)))))
Compare this with the earlier definition of
square-sent. The recursive case still uses
se to combine one
part of the result with a recursive call based on the
the argument, but here both the first letter and the second letter of the
argument contribute to the first word of the result. That's why the base
case also has to be different; the recursive case requires at least two
letters, so the base case is a one-letter word.
Let's solve a slightly different problem. This time, we want to break the word down into non-overlapping pairs of letters, like this:
> (disjoint-pairs 'tripoli) ;; the new problem (TR IP OL I) > (letter-pairs 'tripoli) ;; compare the old one (TR RI IP PO OL LI)
The main difference between these two functions is that in
disjoint-pairs we eliminate two letters at once in the recursive call. A
second difference is that we have to deal with the special case of
(define (disjoint-pairs wd) (cond ((empty? wd) '()) ((= (count wd) 1) (se wd)) (else (se (word (first wd) (first (bf wd))) (disjoint-pairs (bf (bf wd)))))))
every pattern, we collect the results of transforming each
element of a word or sentence into something else. This time we'll consider
a different kind of problem: choosing some of the elements and forgetting
about the others. First, here is a procedure to select the three-letter
words from a sentence:
(define (keep-three-letter-words sent) (cond ((empty? sent) '()) ((= (count (first sent)) 3) (se (first sent) (keep-three-letter-words (bf sent)))) (else (keep-three-letter-words (bf sent))))) > (keep-three-letter-words '(one two three four five six seven)) (ONE TWO SIX)
Next, here is a procedure to select the vowels from a word:
(define (keep-vowels wd) (cond ((empty? wd) "") ((vowel? (first wd)) (word (first wd) (keep-vowels (bf wd)))) (else (keep-vowels (bf wd))))) > (keep-vowels 'napoleon) AOEO
Let's look at the differences between the
every pattern and the
pattern. First of all, the
keep procedures have three
possible outcomes, instead of just two as in most
procedures. In the
every pattern, we only have to distinguish between
the base case and the recursive case. In the
keep pattern, there is
still a base case, but there are two recursive cases; we have to
decide whether or not to keep the first available element in the return
value. When we do keep an element, we keep the element itself, not some
function of the element.
As with the
every pattern, there are situations that follow the
keep pattern only approximately. Suppose we want to look for doubled
letters within a word:
> (doubles 'bookkeeper) OOKKEE > (doubles 'mississippi) SSSSPP
This isn't a pure
keep pattern example because we can't
decide whether to keep the first letter by looking at that letter alone; we
have to examine two at a time. But we can write a procedure using more or
less the same pattern:
(define (doubles wd) (cond ((= (count wd) 1) "") ((equal? (first wd) (first (bf wd))) (word (first wd) (first (bf wd)) (doubles (bf (bf wd))))) (else (doubles (bf wd)))))
As in the
evens example of Chapter 12, the base case of
doubles is unusual, and one of the recursive calls chops off two
letters at once in forming the smaller subproblem. But the structure of the
cond with a base case clause, a clause for keeping letters, and a
clause for rejecting letters is maintained.
Here are two recursive procedures for functions that combine all of the elements of the argument into a single result:
(define (addup nums) (if (empty? nums) 0 (+ (first nums) (addup (bf nums))))) (define (scrunch-words sent) (if (empty? sent) "" (word (first sent) (scrunch-words (bf sent))))) > (addup '(8 3 6 1 10)) 28 > (scrunch-words '(ack now ledge able)) ACKNOWLEDGEABLE
What's the pattern? We're using some combiner (
word) to connect the word we're up to with the result of the recursive
call. The base case tests for an empty argument, but the base case return
value must be the identity element of the combiner function.
If there is no identity element for the combiner, as in the case of
max, we modify the pattern slightly:
(define (sent-max sent) (if (= (count sent) 1) (first sent) (max (first sent) (sent-max (bf sent)))))
(define (add-numbers sent) (cond ((empty? sent) 0) ((number? (first sent)) (+ (first sent) (add-numbers (bf sent)))) (else (add-numbers (bf sent))))) > (add-numbers '(if 6 were 9)) 15
This procedure combines aspects of
keep with aspects of
accumulate. We want to do two things at once: get rid of the words that
aren't numbers and compute the sum of those that are numbers. (A
keep would construct a sentence of them.)
looks exactly like the
keep pattern, except that there's a funny
combiner and a funny base case, which look more like
Here's an example that combines
keep. We want a
procedure that takes a sentence as its argument and translates every word of
the sentence into Pig Latin, but leaves out words that have no vowels,
because the Pig Latin translator doesn't work for such words. The procedure
safe-pigl will be like a
keep pattern in that it keeps only
words that contain vowels, but like an
every in that the result
contains transformed versions of the selected words, rather than the words
(define (safe-pigl sent) (cond ((empty? sent) '()) ((has-vowel? (first sent)) (se (pigl (first sent)) (safe-pigl (bf sent)))) (else (safe-pigl (bf sent))))) (define (has-vowel? wd) (not (empty? (keep-vowels wd)))) > (safe-pigl '(my pet fly is named xyzzy)) (ETPAY ISAY AMEDNAY)
Finally, here's an example that combines all three patterns. In Chapter
1 we wrote (using higher-order procedures) the
procedure, which selects the "real" words of a sentence (the
pattern), takes the first letter of each word (the
every pattern), and
combines these initial letters into a single word (the
pattern). In a recursive procedure we can carry out all three steps at once:
(define (acronym sent) (cond ((empty? sent) "") ((real-word? (first sent)) (word (first (first sent)) (acronym (bf sent)))) (else (acronym (bf sent)))))
Don't become obsessed with trying to make every recursive problem fit one of the three patterns we've shown here. As we said at the beginning of the chapter, what's most important is that you understand the principles of recursion in general, and understand how versatile recursion is. The patterns are just special cases that happen to come up fairly often.
Let's say we want a procedure
every-nth that takes a number n and a
sentence as arguments and selects every nth word from the sentence.
> (every-nth 3 '(with a little help from my friends)) (LITTLE MY)
We get in trouble if we try to write this in the obvious way, as a sort of
(define (every-nth n sent) ;; wrong! (cond ((empty? sent) '()) ((= n 1) (se (first sent) (every-nth n (bf sent)))) (else (every-nth (- n 1) (bf sent)))))
The problem is with the
n that's in boldface.
We're thinking that it's going to be the
n of the original
every-nth, that is, 3. But in fact, we've already counted
n down so that in this invocation its value is 1. (Check out the first
half of the same
cond clause.) This procedure will correctly skip the
first two words but will keep all the words after that point. That's
because we're trying to remember two different numbers: the number we
should always skip between kept words, and the number of words we still need
to skip this time.
If we're trying to remember two numbers, we need two names for them. The
way to achieve this is to have our official
every-nth procedure call a
helper procedure that takes an extra argument and does the real work:
(define (every-nth n sent) (every-nth-helper n n sent)) (define (every-nth-helper interval remaining sent) (cond ((empty? sent) '()) ((= remaining 1) (se (first sent) (every-nth-helper interval interval (bf sent)))) (else (every-nth-helper interval (- remaining 1) (bf sent)))))
This procedure always calls itself recursively with the same value
interval, but with a different value of
remaining each time.
Remaining keeps getting smaller by one in each recursive call until it
equals 1. On that call, a word is kept for the return value, and we call
every-nth-helper recursively with the value of
interval, that is,
the original value of
n, as the new
If you like, you can think of this combination of an initialization
procedure and a helper procedure as another pattern for your
One way in which recursive patterns can be useful is if you think of them as
templates with empty slots to fill in for a particular problem. Here are
template versions of the
patterns as applied to sentences:
(define (every-something sent) (if (empty? sent) '() (se (______ (first sent)) (every-something (bf sent))))) (define (keep-if-something sent) (cond ((empty? sent) '()) ((______? (first sent)) (se (first sent) (keep-if-something (bf sent)))) (else (keep-if-something (bf sent)))))
(define (accumulate-somehow sent) (if (empty? sent) ______ (______ (first sent) (accumulate-somehow (bf sent)))))
Suppose you're trying to write a procedure
first-number that takes a
sentence as its argument and returns the first number in that sentence, but
returns the word
no-number if there are no numbers in the argument.
The first step is to make a guess about which pattern will be most useful.
In this case the program should start with an entire sentence and select a
portion of that sentence, namely one word. Therefore, we start with the
(define (first-number sent) ;; first guess (cond ((empty? sent) '()) ((______ (first sent)) (se (first sent) (first-number (bf sent)))) (else (first-number (bf sent)))))
The next step is to fill in the blank. Obviously, since we're looking for a
number? goes in the blank.
The trouble is that this procedure returns all the numbers in the
given sentence. Now our job is to see how the pattern must be modified to
do what we want. The overall structure of the pattern is a
cond with three clauses; we'll consider each clause separately.
What should the procedure return if
sent is empty? In that case,
there is no first number in the sentence, so it should return
((empty? sent) 'no-number)
What if the first word of the sentence is a number? The program should return just that number, ignoring the rest of the sentence:
((number? (first sent)) (first sent))
What if the first word of the sentence isn't a number? The procedure must
make a recursive call for the
butfirst, and whatever that recursive
call returns is the answer. So the
else clause does not have to be
Here's the whole procedure:
(define (first-number sent) (cond ((empty? sent) 'no-number) ((number? (first sent)) (first sent)) (else (first-number (bf sent)))))
After filling in the blank in the
keep pattern, we solved this problem
by focusing on the details of the procedure definition. We examined each
piece of the definition to decide what changes were necessary. Instead, we
could have focused on the behavior of the procedure. We would have
found two ways in which the program didn't do what it was supposed to do:
For an argument sentence containing numbers, it would return all of the
numbers instead of just one of them. For a sentence without numbers, it
would return the empty sentence instead of
no-number. We would then
have finished the job by debugging the procedure to fix each of these
problems. The final result would have been the same.
We want to write the procedure
sent-before?, which takes two sentences
as arguments and returns
#t if the first comes alphabetically before
the second. The general idea is to compare the sentences word by word. If
the first words are different, then whichever is alphabetically earlier
determines which sentence comes before the other. If the first words are
equal, we go on to compare the second words.
> (sent-before? '(hold me tight) '(sun king)) #T > (sent-before? '(lovely rita) '(love you to)) #F > (sent-before? '(strawberry fields forever) '(strawberry fields usually)) #T
Does this problem follow any of the patterns we've seen? It's not an
every, because the result isn't a sentence in which each word is a
transformed version of a word in the arguments. It's not a
because the result isn't a subset of the words in the arguments. And it's
not exactly an
accumulate. We do end up with a single true or
false result, rather than a sentence full of results. But in a typical
accumulate problem, every word of the argument contributes to the solution.
In this case only one word from each sentence determines the overall result.
On the other hand, this problem does have something in common with the
keep pattern: We know that on each invocation there will be three
possibilities. We might reach a base case (an empty sentence); if not, the
first words of the argument sentences might or might not be relevant to the
We'll have a structure similar to the usual
keep pattern, except that
se involved; if we find unequal words, the problem is
solved without further recursion. Also, we have two arguments, and either
of them might be empty.
(define (sent-before? sent1 sent2) (cond ((empty? sent1) #t) ((empty? sent2) #f) ((before? (first sent1) (first sent2)) #t) ((before? (first sent2) (first sent1)) #f) (else (sent-before? (bf sent1) (bf sent2)))))
Although thinking about the
keep pattern helped us to work
out this solution, the result really doesn't look much like a
We had to invent most of the details by thinking about this particular
problem, not by thinking about the pattern.
In the next chapter we'll look at examples of recursive procedures that are quite different from any of these patterns. Remember, the patterns are a shortcut for many common problems, but don't learn the shortcut at the expense of the general technique.
Review the pitfalls from Chapter 12; they're still relevant.
How do you test for the base case? Most of the examples in this
chapter have used
empty?, and it's easy to fall into the habit of
using that test without thinking. But, for example, if the argument is a
number, that's probably the wrong test. Even when the argument is a
sentence or a non-numeric word, it may not be empty in the base case, as in
the Pig Latin example.
A serious pitfall is failing to recognize a situation in which you need
an extra variable and therefore need a helper procedure. If at
each step you need the entire original argument as well as the argument
that's getting closer to the base case, you probably need a helper
procedure. For example, write a procedure
pairs that takes a word as
argument and returns a sentence of all possible two-letter words made of
letters from the argument word, allowing duplicates, like this:
> (pairs 'toy) (TT TO TY OT OO OY YT YO YY)
A simple pitfall, when using a helper procedure, is to write a recursive
call in the helper that calls the main procedure instead of calling the
helper. (For example, what would have happened if we'd had
every-nth instead of invoking itself?)
Some recursive procedures with more than one argument require more than one base case. But some don't. One pitfall is to leave out a necessary base case; another is to include something that looks like a base case but doesn't fit the structure of the program.
For example, the reason
sent-before? needs two base cases is that on
each recursive call, both
sent2 get smaller. Either
sentence might run out first, and the procedure should return different values
in those two cases.
On the other hand, Exercise 11.7 asked you to write a procedure that has two arguments but needs only one base case:
(define (copies num wd) (if (= num 0) '() (se wd (copies (- num 1) wd))))
In this example, the
wd argument doesn't get smaller
from one invocation to the next. It would be silly to test for
A noteworthy intermediate case is
every-nth-helper. It does
cond clauses that check for two different arguments reaching
their smallest allowable values, but the
remaining clause isn't a base
remaining has the value 1, the procedure still invokes itself
The only general principle we can offer is that you have to think about what base cases are appropriate, not just routinely copy whatever worked last time.
Classify each of these problems as a pattern (
accumulate), if possible, and then write the procedure recursively. In
some cases we've given an example of invoking the procedure we want you to
write, instead of describing it.
> (remove-once 'morning '(good morning good morning)) (GOOD GOOD MORNING)
(It's okay if your solution removes the other
instead, as long as it removes only one of them.)
> (up 'town) (T TO TOW TOWN)
> (remdup '(ob la di ob la da)) ;; remove duplicates (OB LA DI DA)
(It's okay if your procedure returns
(DI OB LA DA) instead,
as long as it removes all but one instance of each duplicated word.)
> (odds '(i lost my little girl)) (I MY GIRL)
Write a procedure
letter-count that takes a sentence as its
argument and returns the total number of letters in the sentence:
> (letter-count '(fixing a hole)) 11
differences, which takes a sentence of numbers as its argument
and returns a sentence containing the differences between adjacent elements.
(The length of the returned sentence is one less than that of the argument.)
> (differences '(4 23 9 87 6 12)) (19 -14 78 -81 6)
expand, which takes a sentence as its argument. It returns
a sentence similar to the argument, except that if a number appears in
the argument, then the return value contains that many copies of the
> (expand '(4 calling birds 3 french hens)) (CALLING CALLING CALLING CALLING BIRDS FRENCH FRENCH FRENCH HENS) > (expand '(the 7 samurai)) (THE SAMURAI SAMURAI SAMURAI SAMURAI SAMURAI SAMURAI SAMURAI)
14.9 Write a procedure called
location that takes two arguments, a
word and a sentence. It should return a number indicating where in the
sentence that word can be found. If the word isn't in the sentence, return
#f. If the word appears more than once, return the location of the
> (location 'me '(you never give me your money)) 4
14.10 Write the procedure
count-adjacent-duplicates that takes a
sentence as an argument and returns the number of words in the sentence that
are immediately followed by the same word:
> (count-adjacent-duplicates '(y a b b a d a b b a d o o)) 3 > (count-adjacent-duplicates '(yeah yeah yeah)) 2
14.11 Write the procedure
remove-adjacent-duplicates that takes a
sentence as argument and returns the same sentence but with any word that's
immediately followed by the same word removed:
> (remove-adjacent-duplicates '(y a b b a d a b b a d o o)) (Y A B A D A B A D O) > (remove-adjacent-duplicates '(yeah yeah yeah)) (YEAH)
14.12 Write a procedure
progressive-squares? that takes a sentence of
numbers as its argument. It should return
#t if each number (other
than the first) is the square of the number before it:
> (progressive-squares? '(3 9 81 6561)) #T > (progressive-squares? '(25 36 49 64)) #F
14.13 What does the
pigl procedure from Chapter 11 do if you invoke
it with a word like "frzzmlpt" that has no vowels? Fix it so that it
14.14 Write a predicate
same-shape? that takes two sentences as arguments.
It should return
#t if two conditions are met: The two sentences must
have the same number of words, and each word of the first sentence must have
the same number of letters as the word in the corresponding position in the
> (same-shape? '(the fool on the hill) '(you like me too much)) #T > (same-shape? '(the fool on the hill) '(and your bird can sing)) #F
merge, a procedure that takes two sentences of numbers as
arguments. Each sentence must consist of numbers in increasing order.
Merge should return a single sentence containing all of the numbers, in
order. (We'll use this in the next chapter as part of a sorting algorithm.)
> (merge '(4 7 18 40 99) '(3 6 9 12 24 36 50)) (3 4 6 7 9 12 18 24 36 40 50 99)
14.16 Write a procedure
syllables that takes a word as its argument
and returns the number of syllables in the word, counted according to the
following rule: the number of syllables is the number of vowels, except that
a group of consecutive vowels counts as one. For example, in the word
"soaring," the group "oa" represents one syllable and the vowel "i"
represents a second one.
Be sure to choose test cases that expose likely failures of your procedure. For example, what if the word ends with a vowel? What if it ends with two vowels in a row? What if it has more than two consecutive vowels?
(Of course this rule isn't good enough. It doesn't deal with things like silent "e"s that don't create a syllable ("like"), consecutive vowels that don't form a diphthong ("cooperate"), letters like "y" that are vowels only sometimes, etc. If you get bored, see whether you can teach the program to recognize some of these special cases.)
you've read Chapter 8, you know that you could implement
pigl-sent without recursion, using the
higher order function. But try using
every to implement
letter-pairs; you'll find that you can't quite make it work.
 Of course, if your version of Scheme has −∞, you can use it as the return value for an empty sentence, instead of changing the pattern.
 Here's the higher-order function version, from Chapter 8:
(define (add-numbers sent) (accumulate + (keep number? sent)))
The higher-order function version is more self-documenting and easier to write. The recursive version, however, is slightly more efficient, because it avoids building up a sentence as an intermediate value only to discard it in the final result. If we were writing this program for our own use, we'd probably choose the higher-order function version; but if we were dealing with sentences of length 10,000 instead of length 10, we'd pay more attention to efficiency.
 Dictionaries use a different ordering rule, in which the sentences are treated as if they were single words, with the spaces removed. By the dictionary rule, "a c" is treated as if it were "ac" and comes after "ab"; by our rule, "a c" comes before "ab" because we compare the first words ("a" and "ab").
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