Computer Science Logo Style volume 1: Symbolic Computing 2/e Copyright (C) 1997 MIT

Recursive Operations So far, the recursive procedures we've seen have all been commands, not operations. Remember that an operation is a procedure that has an output rather than an effect. In other words, an operation computes some value that is then used as the input to some other procedure. In the instruction

print first "hello

print is a command, because it does something: It prints its input (whatever that may be) on the screen. But first is an operation, because it computes something: With the word hello as input, it computes the letter h, which is the first letter of the input.

A Simple Substitution Cipher

I'm going to write a program to produce secret messages. The program will take an ordinary English sentence (in the form of a Logo list) and change each letter into some other letter. For example, we can decide to replace the letter E with the letter J every time it occurs in the message. The program will need two inputs: the message and the correspondence between letters. The latter will take the form of a word of 26 letters, representing the coded versions of the 26 letters in alphabetical order. For example, the word

qwertyuiopasdfghjklzxcvbnm

indicates that the letter A in the original text will be represented by Q in the secret version, B will be represented by W, and so on.

In order to encipher a sentence, we must go through it word by word. (Strictly speaking, what we're doing is called a cipher rather than a code because the latter is a system that substitutes something for an entire word at a time, like a foreign language, whereas we're substituting for a single letter at a time, like the Elvish alphabet in The Lord of the Rings.) In order to encipher a word we must go through it letter by letter. So I'll begin by writing a procedure to translate a single letter to its coded form.

to codelet :letter :code
output codematch :letter "abcdefghijklmnopqrstuvwxyz :code
end

to codematch :letter :clear :code
if emptyp :clear [output :letter]        ; punctuation character
if equalp :letter first :clear [output first :code]
output codematch :letter butfirst :clear butfirst :code
end

Codelet is an operation that takes two inputs. The first input must be a single-letter word, and the second must be a code, that is, a word with the 26 letters of the alphabet rearranged. The output from codelet is the enciphered version of the input letter. (If the first input is a character other than a letter, such as a punctuation mark, then the output is the same as that input.)

Codelet itself is a very simple procedure. It simply passes its two inputs on to a subprocedure, codematch, along with another input that is the alphabet in normal order. The idea is that codematch will compare the input letter to each of the letters in the regular alphabet; when it finds a match, it will output the letter in the corresponding position in the scrambled alphabet. Be sure you understand the use of the output command in codelet; it says that whatever codematch outputs should become the output from codelet as well.

The job of codematch is to go through the alphabet, letter by letter, looking for the particular letter we're trying to encode. The primary tool that Logo provides for looking at a single letter in a word is first. So codematch uses first to compare its input letter with the first letter of the input alphabet:

if equalp :letter first :clear ...

If the first input to codematch is the letter A, then equalp will output true and codematch will output the first letter of :code (Q in the example I gave earlier). But suppose the first input isn't an A. Then codematch has to solve a smaller subproblem: Find the input letter in the remaining 25 letters of the alphabet. Finding a smaller, similar subproblem means that we can use a recursive solution. Codematch invokes itself, but for its second and third inputs it uses the butfirsts of the original inputs because the first letter of the alphabet (A) and its corresponding coded letter (Q) have already been rejected.

Here is a trace of an example of codematch at work, to help you understand what's going on.

codematch "e "abcdefghijklmnopqrstuvwxyz "qwertyuiopasdfghjklzxcvbnm
codematch "e "bcdefghijklmnopqrstuvwxyz "wertyuiopasdfghjklzxcvbnm
codematch "e "cdefghijklmnopqrstuvwxyz "ertyuiopasdfghjklzxcvbnm
codematch "e "defghijklmnopqrstuvwxyz "rtyuiopasdfghjklzxcvbnm
codematch "e "efghijklmnopqrstuvwxyz "tyuiopasdfghjklzxcvbnm
codematch outputs "t
codematch outputs "t
codematch outputs "t
codematch outputs "t
codematch outputs "t

The fifth, innermost invocation of codematch succeeds at matching its first input (the letter E) with the first letter of its second input. That invocation therefore outputs the first letter of its third input, the letter T. Each of the higher-level invocations outputs the same thing in turn.

The pattern of doing something to the first of an input, then invoking the same procedure recursively with the butfirst as the new input, is a familiar one from recursive commands. If we only wanted to translate single letters, we could have written codelet and codematch as commands, like this:

to codelet :letter :code                        ;; command version
codematch :letter "abcdefghijklmnopqrstuvwxyz :code
end

to codematch :letter :clear :code               ;; command version
if emptyp :clear [print :letter stop]
if equalp :letter first :clear [print first :code stop]
codematch :letter butfirst :clear butfirst :code
end

You may find this version a little easier to understand, because it's more like the recursive commands we've examined in the past. But making codelet an operation is a much stronger technique. Instead of being required to print the computed code letter, we can make that letter part of a larger computation. In fact, we have to do that in order to encipher a complete word. Each word is made up of letters, and the task of codeword will be to go through a word, letter by letter, using each letter as input to codelet. The letters output by codelet must be combined into a new word, which will be the output from codeword.

We could write codeword using the higher order function map:

to codeword :word :code             ;; using higher order function
output map [codelet ? :code] :word
end

But to help you learn how to write recursive operations, in this chapter we'll avoid higher order functions. (As it turns out, map itself is a recursive operation, written using the techniques of this chapter.)

Recall the structure of a previous procedure that went through a word letter by letter:

to one.per.line :word
if emptyp :word [stop]
print first :word
one.per.line butfirst :word
end

Compare this to the structure of the recursive codeword:

to codeword :word :code
if emptyp :word [output "]
output word (codelet first :word :code) (codeword butfirst :word :code)
end

There are many similarities. Both procedures have a stop rule that tests for an empty input. Both do something to the first of the input (either print or codelet), and each invokes itself recursively on the butfirst of the input. (Codeword has an extra input for the code letters, but that doesn't really change the structure of the procedure. If that's confusing to you, you could temporarily pretend that code is a global variable and eliminate it as an input.)

The differences have to do with the fact that codeword is an operation instead of a command. The stop rule invokes output rather than stop and must therefore specify what is to be output when the stop condition is met. (In this case, when the input word is empty, the output is also the empty word.) But the main thing is that the action step (the print in one.per.line) and the recursive call (the one.per.line instruction) are not two separate instructions in codeword. Instead they are expressions (the two in parentheses) that are combined by word to form the complete output. Here's a picture: Remember what you learned in Chapter 2 about the way in which Logo instructions are evaluated. Consider the output instruction in codeword. Before output can be invoked, Logo must evaluate its input. That input comes from the output from word. Before word can be invoked, Logo must evaluate its inputs. There are two of them. The first input to word is the expression

codelet first :word :code

This expression computes the coded version of the first letter of the word we want to translate. The second input to word is the expression

codeword butfirst :word :code

This expression invokes codeword recursively, solving the smaller subproblem of translating a smaller word, one with the first letter removed. When both of these computations are complete, word can combine the results to form the translation of the complete input word. Then output can output that result.

Here's an example of how codeword is used.

? print codeword "hello "qwertyuiopasdfghjklzxcvbnm
itssg

Notice that we have to say print, not just start the instruction line with codeword; a complete instruction must have a command. Suppose you had the idea of saving all that typing by changing the output instruction in codeword to a print. What would happen? The answer is that codeword wouldn't be able to invoke itself recursively as an operation. (If you don't understand that, try it!) Also, it's generally a better idea to write an operation when you have to compute some result. That way, you aren't committed to printing the result; you can use it as part of a larger computation.

For example, right now I'd like to write a procedure code that translates an entire sentence into code. Like codeword, it will be an operation with two inputs, the second of which is a code (a word of 26 scrambled letters). The difference is that the first input will be a sentence instead of a word and the output will also be a sentence.

»Write code using a higher order function. Then see if you can write an equivalent recursive version.

Just as codeword works by splitting up the word into letters, code will work by splitting up a sentence into words. The structure will be very similar. Here it is:

to code :sent :code
if emptyp :sent [output []]
output sentence (codeword first :sent :code) (code butfirst :sent :code)
end

The main differences are that code outputs the empty list, instead of the empty word, for an empty input and that sentence is used as the combining operation instead of word. Here's an example of code at work.

? print code [meet at midnight, under the dock.] ~
"qwertyuiopasdfghjklzxcvbnm
dttz qz dorfouiz, xfrtk zit rgea.

More Procedure Patterns

Code and codeword are examples of a very common pattern in recursive operations: They are like using map with a particular function. Here is the pattern that they fit.

to procedure :input
if emptyp :input [output :input]
output combiner (something first :input) (procedure butfirst :input)
end

The combiner is often word or sentence, although others are possible. In fact, when working with lists, the most common combiner is not sentence but another operation that we haven't used before, fput (First PUT). Fput takes two inputs. The first can be any datum, but the second must be a list. The output from fput is a list that is equal to the second input, except that the first input is inserted as a new first member. In other words the output from fput is a list whose first is the first input and whose butfirst is the second input.

? show sentence [hee hee hee] [ho ho ho]
[hee hee hee ho ho ho]
? show fput [hee hee hee] [ho ho ho]
[[hee hee hee] ho ho ho]

Fput is a good combiner because the two things we want to combine are the first and the butfirst of a list, except that each has been modified in some way. But the shape of the final result (a list of so many members) should be the same as the shape of the input, and that's what fput ensures.

When you're working with sentences--lists of words rather than lists of lists--sentence and fput will work equally well as the combiner. For example, code could have been written using fput instead of sentence. Not until some of the later examples, when we use tree-structured lists, will the choice really be important.

»Fput is actually a "more primitive" operation than sentence, in the sense that the Logo interpreter actually constructs lists by doing the internal equivalent of fput. As an exercise, you might like to try writing your own versions of list combiners like sentence and list out of fput, first, and butfirst. You should also be able to write last and butlast using only those three building blocks. (Actually you'll also need if, emptyp, wordp, and output, but you won't need any other primitive combiners.) Give your versions distinct names, such as my.sentence, since Logo won't let you redefine primitives.

»Another "less primitive" primitive is lput, an operation that takes two inputs. As for fput, the first can be any datum but the second must be a list. The output from lput is a list whose last is the first input and whose butlast is the second. Write my-lput using fput and the selectors first and butfirst.

It may seem silly to learn a recursive pattern for problems that can be solved using map. But sometimes we run into a problem that's almost like a map, but not exactly. For example, how would you write the following operation:

? show pairup [now here after glow worm hole]
[nowhere hereafter afterglow glowworm wormhole]

Instead of the usual map-like situation in which each word in the result is a function of one word of the input, this time each word of the result is a function of two neighboring input words. Map won't solve this problem, but the map-like recursion pattern will.

to pairup :words
if emptyp butfirst :words [output []]
output (sentence (word first :words first butfirst :words)
(pairup butfirst :words))
end

Compare this procedure with the general pattern; look for similarities and differences.

»One difference is in the test for the base case. Why is the version in pairup different from the one in the pattern?

»Write an operation that interchanges pairs of words in a sentence, like this:

? show swap [the rain in spain stays mainly on the plain]
[rain the spain in mainly stays the on plain]

Don't forget to think about that leftover word in an odd-length sentence!

The Filter Pattern

In Chapter 5 we saw this example:

? show filter "numberp [76 trombones, 4 calling birds, and 8 days]
[76 4 8]

To write a recursive operation numbers with the same result, we must handle three cases: the base case, in which the input is empty; the case in which the first word of the input is a number; and the case in which the first word isn't a number.

to numbers :sent
if emptyp :sent [output []]
if numberp first :sent ~
[output sentence first :sent numbers butfirst :sent]
output numbers butfirst :sent
end

? show numbers [76 trombones, 4 calling birds, and 8 days]
[76 4 8]

Here's the general filter pattern:

to procedure :input
if emptyp :input [output :input]
if predicate first :input ~
[output combiner first :input procedure butfirst :input]
output procedure butfirst :input
end

As in the case of the map pattern, this one is most useful in situations for which the higher order function won't quite do.

»Write an operation that looks for two equal words next to each other in a sentence, and outputs a sentence with one of them removed:

? show unique [Paris in the the spring is a joy joy to behold.]
Paris in the spring is a joy to behold.

What does your procedure do with three consecutive equal words? What should it do?

The Reduce Pattern

Other examples from Chapter 5 introduced the reduce higher order function.

? show reduce "word [C S L S]
CSLS
? show reduce "sum [3 4 5 6]
18

Recursive operations equivalent to these examples are very much like the map pattern except that the combiner function is applied to the members of the input directly, rather than to some function of the members of the input:

to wordify :sentence
if emptyp :sentence [output "]
output word (first :sentence) (wordify butfirst :sentence)
end

if emptyp :numbers [output 0]
output sum (first :numbers) (addup butfirst :numbers)
end

What are the differences between these two examples? There are two: the combiner used and the value output in the base case. Here is the pattern:

to procedure :input
if emptyp :input [output identity]
output combiner (first :input) (procedure butfirst :input)
end

The identity in this pattern depends on the combiner; it's the value that, when combined with something else, gives that something else unchanged as the result. Thus, zero is the identity for sum, but the identity for product would be one.

»Write a multiply operation that takes a list of numbers as its input and returns the product of all the numbers.

»You can make your multiply procedure more efficient, in some situations, by having it notice when one of the numbers in the input list is zero. In that case, you can output zero as the overall result without looking at any more numbers. The resulting procedure will, in a sense, combine aspects of the filter and reduce patterns.

Addup is one example of an important sub-category of reduce-like procedures in which the "combining" operation is arithmetic, usually sum. The simplest example is a procedure equivalent to the primitive count, which counts the members of a list or the letters of a word:

to length :thing
if emptyp :thing [output 0]
output 1+length butfirst :thing
end

In this procedure, as usual, we can see the reduction of a problem to a smaller subproblem. The length of any word or list is one more than the length of its butfirst. Eventually this process of shortening the input will reduce it to emptiness; the length of an empty word or list is zero.

Although count is a primitive, there are more complicated counting situations in which not every member should be counted. For example, here is a procedure to count the number of vowels in a word:

to vowelcount :word
if emptyp :word [output 0]
if vowelp first :word [output 1+vowelcount butfirst :word]
output vowelcount butfirst :word
end

to vowelp :letter
output memberp :letter [a e i o u]
end

(Actually, my predicate vowelp is somewhat oversimplified. The letter Y is a vowel in certain positions in the word, and even some other letters can sometimes play the role of a vowel. But this isn't a book on linguistics!)

You can see the similarities between vowelcount and length. The difference is that, in effect, length uses a predicate that is always true, so it always carries out the instruction inside the if. Here's the pattern:

to procedure :input
if emptyp :input [output 0]
if predicate first :input [output 1+procedure butfirst :input]
output procedure butfirst :input
end

»Try writing a procedure that will accept as input a word like 21,997.00 and output the number of digits before the decimal point. (In this case the correct output is 5.) Don't assume that there is a decimal point; your program shouldn't blow up no matter what word it gets as input.

»Another counting problem is to output the position of a member in a list. This operation is the inverse to item, a Logo primitive, which outputs the member at a given position number. What I'm asking you to write is index, which works like this:

? print index "seven [four score and seven years ago]
4
? print index "v "aardvark
5

The Find Pattern

A variation of the filter pattern is for selection operations: ones that pick a single element out of a list. The general idea looks like this:

to procedure :input
if emptyp :input [output :input]
if predicate first :input [output something first :input]
output procedure butfirst :input
end

There will generally be extra inputs to these procedures, to indicate the basis for selection. For example, here is a program that translates English words into French.

to french :word
output lookup :word [[book livre] [computer ordinateur] [window fenetre]]
end

to lookup :word :dictionary
if emptyp :dictionary [output "]
if equalp :word first first :dictionary [output last first :dictionary]
output lookup :word butfirst :dictionary
end

? print french "computer
ordinateur

The expression

first first :dictionary

selects the English word from the first word-pair in the list. Similarly,

last first :dictionary

selects the French version of the same word. (Of course, in reality, the word list in french would be much longer than the three word-pairs I've shown.)

Codematch, in the example that started this chapter, follows the same pattern of selection. The only difference is that there are two inputs that are butfirsted in parallel.

Somewhat similar to the selection pattern is one for a recursive predicate; the difference is that instead of

output something first :input

for a successful match, the procedure simply says

output "true

in that case. This pattern is followed by predicates that ask a question like "Does any member of the input do X?" For example, suppose that instead of counting the vowels in a word, we just want to know whether or not there is a vowel. Then we're asking the question "Is any letter in this word a vowel?" Here's how to find out.

to hasvowelp :word
if emptyp :word [output "false]
if vowelp first :word [output "true]
output hasvowelp butfirst :word
end

A more realistic example is also somewhat more cluttered with extra inputs and sometimes extra end tests. Here's a procedure that takes two words as input. It outputs true if the first word comes before the second in the dictionary.

to sort.beforep :word1 :word2
if emptyp :word1 [output "true]
if emptyp :word2 [output "false]
if (ascii first :word1) < (ascii first :word2) [output "true]
if (ascii first :word1) > (ascii first :word2) [output "false]
output sort.beforep butfirst :word1 butfirst :word2
end

The procedure will end on one of the emptyp tests if one of the input words is the beginning of the other, like now and nowhere. Otherwise, the procedure ends when two letters are unequal. The recursion step is followed when the beginning letters are equal. (The operation ascii takes a one-character word as input, and outputs the numeric value for that character in the computer's coding system, which is called the American Standard Code for Information Interchange.)

A combination of the translation kind of operation and the selection kind is an operation that selects not one but several members of the input. For example, you sometimes want to examine the words in a sentence in various ways but have trouble because the sentence includes punctuation as part of some words. But the punctuation isn't really part of the word. In Chapter 4, for instance, I defined a predicate about.computersp and gave this example of its use:

true

But if the example were part of a larger program, carrying on a conversation with a person, the person would probably have ended the sentence with a period. The last word would then have been programming. (including the period). That word, which is different from programming without the period, isn't in the procedure's list of relevant words, so it would have output false. The solution is to write a procedure that strips the punctuation from each word of a sentence. Of course that's a straightforward case of the translation pattern, applying a subprocedure to each word of the sentence:

to strip :sent
if emptyp :sent [output []]
output sentence (strip.word first :sent) (strip butfirst :sent)
end

Strip.word, though, is more interesting. It must select only the letters from a word.

to strip.word :word
if emptyp :word [output "]
if letterp first :word ~
[output word (first :word) (strip.word butfirst :word)]
output strip.word butfirst :word
end

to letterp :char
output or (inrangep (ascii :char) (ascii "A) (ascii "Z)) ~
(inrangep (ascii :char) (ascii "a) (ascii "z))
end

to inrangep :this :low :high
output and (:this > (:low-1)) (:this < (:high+1))
end

Strip.word is like the translation pattern in the use of the combining operation word in the middle instruction line. But it's also like the selection pattern in that there are two different choices of output, depending on the result of the predicate letterp.

»You might want to rewrite about.computersp so that it uses strip. Consider an initialization procedure.

Certain mathematical functions are defined in terms of recursive calculations. It used to be that computers were used only for numerical computation. They're now much more versatile, as you've already seen, but sometimes the old numerical work is still important.

The classic example in this category is the factorial function. The factorial of a positive integer is the product of all the integers from 1 up to that number. The factorial of 5 is represented as 5! so

5! = 1 × 2 × 3 × 4 × 5

We can use cascade to carry out this computation:

to fact :n                       ;; cascade version
output cascade :n [? * #] 1
end

? print fact 5
120

In this example I'm using a feature of cascade that we haven't seen before. The template (the second input to cascade) may include a number sign (#) character, which represents the number of times the template has been repeated. That is, it represents 1 the first time, 2 the second time, and so on.

Here is a recursive version of fact that takes one input, a positive integer, and outputs the factorial function of that number. The input can also be zero; the rule is that 0!=1.

to fact :n
if :n=0 [output 1]
output :n * fact :n-1
end

This procedure works because

5! = 5 × 4!

That's another version of reducing a problem to a smaller subproblem.

»Chapter 5 gives the following example:

to power :base :exponent
output cascade :exponent [? * :base] 1
end

Another classic example, slightly more complicated, is the Fibonacci sequence. Each number in the sequence is the sum of the two previous numbers; the first two numbers are 1. So the sequence starts

1, 1, 2, 3, 4, 5, 13, …

A formal definition of the sequence looks like this: Here's an operation fib that takes a number n as input and outputs Fn.

to fib :n
if :n<2 [output 1]
output (fib :n-1)+(fib :n-2)
end

That procedure will work, but it's quite seriously inefficient. The problem is that it ends up computing the same numbers over and over again. To see why, here's a trace of what happens when you ask for fib 4:

fib 4
fib 3
fib 2
fib 1
fib 0
fib 1
fib 2
fib 1
fib 0

Do you see the problem? fib 2 is computed twice, once because fib 4 needs it directly and once because fib 4 needs fib 3 and fib 3 needs fib 2. Similarly, fib 1 is computed three times. As the input to fib gets bigger, this problem gets worse and worse.

It turns out that a much faster way to handle this problem is to compute a list of all the Fibonacci numbers up to the one we want. Then each computation can take advantage of the work already done. Here's what I mean:

to fiblist :n
if :n<2 [output [1 1]]
output newfib fiblist :n-1
end

to newfib :list
output fput (sum first :list first butfirst :list) :list
end

? print fiblist 5
8 5 3 2 1 1

We can then define a faster fib in terms of fiblist:

to fib :n
output first fiblist :n
end

Convince yourself that the two versions of fib give the same outputs but that the second version is much faster. I'm purposely not going through a detailed explanation of this example; you should use the analytical techniques you learned in Chapter 8. What problem is fiblist trying to solve? What is the smaller subproblem?

The hallmark of numerical recursion is something like :n-1 in the recursion step. Sometimes this kind of recursion is combined with the butfirst style we've seen in most of the earlier examples. Logo has a primitive operation called item, which takes two inputs. The first is a positive integer, and the second is a list. The output from item is the nth member of the list if the first input is n. (Earlier I suggested that you write index, the opposite of item.) If Logo didn't include item, here's how you could write it:

to item :n :list
if equalp :n 1 [output first :list]
output item :n-1 butfirst :list
end

Pig Latin

When I was growing up, every kid learned a not-very-secret "secret" language called Pig Latin. When I became a teacher, I was surprised to find out that kids apparently didn't learn it any more. But more recently it seems to have come back into vogue. Translating a sentence into Pig Latin is an interesting programming problem, so I'm going to teach it to you.

Here's how it works. For each word take any consonants that are at the beginning (up to the first vowel) and move them to the end. Then add "ay" at the end. So "hello" becomes "ellohay"; "through" becomes "oughthray"; "aardvark" just becomes "aardvarkay." (Pig Latin is meant to be spoken, not written. You're supposed to practice so that you can do it and understand it really fast.)

By now you can write in your sleep the operation piglatin, which takes a sentence and outputs its translation into Pig Latin by going through the sentence applying a subprocedure plword to each word. (It's just like code, only different.) It's plword that is the tricky part. The stop rule is pretty straightforward:

if vowelp first :word [output word :word "ay]

If the first letter isn't a vowel, what we want to do is move that letter to the end and try again. Here's the complete procedure.

to plword :word
if vowelp first :word [output word :word "ay]
output plword word butfirst :word first :word
end

What makes this tricky is that the recursion step doesn't seem to make the problem smaller. The word is still the same length after we move the first letter to the end. This would look more like all the other examples if the recursion step were

output plword butfirst :word

That would make the procedure easier to understand. Unfortunately it would also give the wrong answer. What you have to see is that there is something that is getting smaller about the word, namely the "distance" from the beginning of the word to the first vowel. Trace through a couple of examples to clarify this for yourself.

By the way, this will work better if you modify vowelp (which we defined earlier) so that y is considered a vowel. You'll then get the wrong answer for a few strange words like yarn, but on the other hand, if you consider y a consonant, you'll get no answer at all for words like try in which y is the only vowel! (Try it. Do you understand what goes wrong?)

»Some people learned a different dialect of Pig Latin. According to them, if the word starts with a vowel in the first place, you should add "way" at the end instead of just "ay." Modify plword so that it speaks that dialect. (I think the idea is that some words simply sound better with that rule.) Hint: You'll want an initialization procedure.

»The top-level procedure piglatin, which you wrote yourself, is a good candidate for careful thought about punctuation. You don't want to see

? print piglatin [what is he doing?]
atwhay isway ehay oing?day

A good first attempt would be to modify piglatin to use strip, to get rid of the punctuation altogether. But even better would be to remove the punctuation from each word, translate it to Pig Latin, then put the punctuation back! Then we could get

atwhay isway ehay oingday?

That's the right thing to do for punctuation at the end of a word, like a period or a comma. On the other hand, the apostrophe inside a word like isn't should be treated just like a letter.

The project I'm proposing to you is a pretty tricky one. Here's a hint. Write an operation endpunct that takes a word as input and outputs a list of two words, first the "real" word full of letters, then any punctuation that might be at the end. (The second word will be empty if there is no such punctuation.) Then your new plword can be an initialization procedure that invokes a subprocedure with endpunct's output as its input.

A Mini-project: Spelling Numbers

Write a procedure number.name that takes a positive integer input, and outputs a sentence containing that number spelled out in words:

? print number.name 5513345
five million five hundred thirteen thousand three hundred forty five
? print number.name (fact 20)
two quintillion four hundred thirty two quadrillion nine hundred two
trillion eight billion one hundred seventy six million six hundred
forty thousand

There are some special cases you will need to consider:

• Numbers in which some particular digit is zero.
• Numbers like 1,000,529 in which an entire group of three digits is zero.
• Numbers in the teens.

Here are two hints. First, split the number into groups of three digits, going from right to left. Also, use the sentence

[thousand million billion trillion quadrillion quintillion
sextillion septillion octillion nonillion decillion]

You can write this bottom-up or top-down. To work bottom-up, pick a subtask and get that working before you tackle the overall structure of the problem. For example, write a procedure that returns the word fifteen given the argument 15.

To work top-down, start by writing number.name, freely assuming the existence of whatever helper procedures you like. You can begin debugging by writing stub procedures that fit into the overall program but don't really do their job correctly. For example, as an intermediate stage you might end up with a program that works like this:

? print number.name 1428425           ;; intermediate version
1 million 428 thousand 425

We've seen that recursive operations can do the same jobs as higher order functions, and we've seen that recursive operations can do jobs that are similar to the higher order function patterns but not quite the same. Now we'll see that recursive operations can do jobs that are quite outside the bounds of any of the higher order functions in Chapter 5.

I'd like to write an operation subsets that takes a word as input. Its output will be a sentence containing all the words that can be made using letters from the input word, in the same order, but not necessarily using all of them. For example, the word lit counts as a subset of the word lights, but hit doesn't count because the letters are in the wrong order. (Of course the procedure won't know which words are real English words, so iht, which has the same letters as hit in the right order, does count.)

»How many subsets does lights have? Write them all down if you're not sure. (Or perhaps you'd prefer to count the subsets of a shorter word, such as word, instead.)

A problem that follows the map pattern is one in which the size of the output is the same as the size of the input, because each member of the input gives rise to one member of the output. A problem that follows the filter pattern is one in which the output is smaller than the input, because only some of the members are selected. And the reduce pattern collapses all of the members of the input into one single result. The subsets problem is quite different from any of these; its output will be much larger than its input.

If we can't rely on known patterns, we'll have to go back to first principles. In Chapter 8 you learned to write recursive procedures by looking for a smaller, similar subproblem within the problem we're trying to solve. What is a smaller subproblem that's similar to finding the subsets of lights? How about finding the subsets of its butfirst? This idea is the same one that's often worked for us before. So imagine that we've already found all the subsets of ights.

Some of the subsets of lights are subsets of ights. Which ones aren't? The missing subsets are the ones that start with the letter L. What's more, the other letters in such a subset form a subset of ights. For example, the word lits consists of the letter L followed by its, which is a subset of ights.

to subsets :word                   ;; incomplete
local "smaller
make "smaller subsets butfirst :word
output sentence :smaller (map [word (first :word) ?] :smaller)
end

This procedure reflects the idea I've just tried to explain. The subsets of a given word can be divided into two groups: the subsets of its butfirst, and those same subsets with the first letter of the word stuck on the front.

The procedure lacks a base case. It's tempting to say that if the input is an empty word, then the output should be an empty sentence. But that isn't quite right, because every word is a subset of itself, so in particular the empty word is a subset (the only subset) of itself. We must output a sentence containing an empty word. That's a little tricky to type, but we can represent a quoted empty word as " and so a sentence containing an empty word is (sentence ").

to subsets :word
if emptyp :word [output (sentence ")]
local "smaller
make "smaller subsets butfirst :word
output sentence :smaller (map [word (first :word) ?] :smaller)
end

Why did I use the local variable smaller and a make instruction? It wasn't strictly necessary; I could have said

output sentence (subsets butfirst :word) ~
(map [word (first :word) ?] (subsets butfirst :word))

The trouble is that this would have told Logo to compute the smaller similar subproblem twice instead of just once. It may seem that that would make the program take twice as long, but in fact the problem is worse than that, because each smaller subproblem has a smaller subproblem of its own, and those would be computed four times--twice for each of the two computations of the first smaller subproblem! As in the case of the Fibonacci sequence we studied earlier, avoiding the duplicated computation makes an enormous difference.

Problems like this one, in which the size of the output grows extremely quickly for small changes in the size of the input, tend to be harder to program than most. Here are a couple of examples. Like subsets, each of these has a fairly short procedure definition that hides a very complex computation.

»On telephone dials, most digits have letters associated with them. In the United States, for example, the digit 5 is associated with the letters J, K, and L. (The correspondence is somewhat different in other countries.) You can use these letters to spell out words to make your phone number easier to remember. For example, many years ago I had the phone number 492-6824, which spells I-WANT-BH. Write a procedure that takes a number as its input, and outputs a sentence of all the words that that number can represent. You may want to test the program using numbers of fewer than seven digits!

»In the game of BoggleTM, the object is to find words by connecting neighboring letters in a four by four array of letters. For example, the array

BEZO
URND
AKAJ
WEOE

contains the words ZEBRA, DONE, and DARK, but not RADAR, because each letter can be used only once. Write a predicate procedure that takes a word and an array of letters (in the form of a sentence with one word for each row) as inputs, and outputs true if and only if the given word can be found in the given array.

? print findword "zebra [bezo urnd akaj weoe]
true
? print findword "radar [bezo urnd akaj weoe]
false

What I want to talk about in the rest of this chapter isn't really very important, so you can skip it if you want. But some people think it's important, so this is for those people.

Every procedure invocation takes up a certain amount of computer memory, while the procedure remains active, to hold things like local variables. Since a recursive procedure can invoke itself many times, recursion is a fairly "expensive" technique to allow in a programming language. It turns out that if the only recursion step in a procedure is the very last thing the procedure does, the interpreter can handle that procedure in a special way that uses memory more efficiently. You can then use as many levels of recursive invocation as you want without running out of space. Such a procedure is called tail recursive. It doesn't make any difference to you as a programmer; it's just a matter of what's happening inside the Logo interpreter.

A tail recursive command is very easy to recognize; the last instruction is an invocation of the same procedure. Tail recursive commands are quite common; here are a couple of examples we've seen before.

to one.per.line :thing
if emptyp :thing [stop]
print first :thing
one.per.line butfirst :thing
end

to poly :size :angle
forward :size
right :angle
poly :size :angle
end

The thing is, many people are confused about what constitutes a tail recursive operation. It isn't one that is invoked recursively on the last instruction line! Instead, the rule is that the recursive invocation must be used directly as the input to output, not as part of a larger computation. For example, this is a tail recursive operation:

to lookup :word :dictionary
if emptyp :dictionary [output "]
if equalp :word first first :dictionary [output last first :dictionary]
output lookup :word butfirst :dictionary
end

But this is not tail recursive:

to length :thing
if emptyp :thing [output 0]
output 1+length butfirst :thing
end

It's that 1+ that makes the difference.

It's sometimes possible to change a nontail recursive operation into a tail recursive one by tricky programming. For example, look again at fact:

to fact :n
if :n=0 [output 1]
output :n * fact :n-1
end

This is not tail recursive because the input to the final output comes from the multiplication, not directly from fact. But here is a tail recursive version:

to fact :n
output fact1 :n 1
end

to fact1 :n :product
if :n=0 [output :product]
output fact1 (:n-1) (:n*:product)
end

Indeed, this version can, in principle, compute the factorial of larger numbers than the simpler version without running out of memory. In practice, though, the largest number that most computers can understand is less than the factorial of 70, and any computer will allow 70 levels of recursion without difficulty. In fact, not every Logo interpreter bothers to recognize tail recursive operations. It's a small point; I only mention it because some people both make a big fuss about tail recursion and misunderstand what it means!