Table of Contents

Write a procedure

`number-name`

that takes a positive integer argument and returns a sentence containing that number spelled out in words:

1 2 3 4 5 6 7 8 9 |
> (number-name 5513345) (FIVE MILLION FIVE HUNDRED THIRTEEN THOUSAND THREE HUNDRED FORTY FIVE) > (number-name (factorial 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:

β’ Numbers in which some particular digit is zero

β’ Numbers like 1,000,529 in which an entire group of three digits is zero.

β’ 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

1 2 3 |
'(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 writingstubprocedures 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:

1 2 3 4 |
> (number-name 1428425) ;; intermediate version (1 MILLION 428 THOUSAND 425) |

## Solution

I solved this by first mostly solving the problem for 0 through 999, then testing and adding more stuff on.

### Harder Sub-procedures

Here I’ll talk about the procedures I thought were harder/more difficult to figure out for solving this problem.

Here’s what I wrote for `number-name`

:

1 2 3 4 5 6 7 8 9 10 |
(define (number-name number) (cond ((not (number? number)) '(Erroneous Input There Buddy This Program Only Takes Numbers)) ((not (integer? number)) '(Erroneous Input There Buddy This Program Only Takes Integers)) ((equal? number 0) '(zero)) ((negative? number) (se 'negative (number-name-helper (triple-number-accumulator (bf number)) 0))) (else (number-name-helper (triple-number-accumulator number) 0)))) |

So big picture the main program does a couple of things:

1) checks for a few special cases,

2) hands off the number to helper procedures for further processing if appropriate.

the `((not (number? number))`

and `((not (integer? number))`

cond lines check to see if the input is not a number or not an integer, respectively, and return an appropriate message if that is the case.

`((equal? number 0) '(zero))`

checks if the number is equal to zero. Note that this will return true for stuff like `-0`

or `00000`

, so you don’t have to check for those cases specifically – Scheme handles all that. If the test returns true, a sentence with the word `(zero)`

gets returned. I made it a sentence in order to have it be consistent with other program outut.

1 2 |
((negative? number) (se 'negative (number-name-helper (triple-number-accumulator (bf number)) 0))) |

This part of the program first checks if a number is negative. If so, invokes `sentence`

with the word “negative” and then calls `number-name-helper`

in much the same way that the `else`

line does, with the caveat that invokes `triple-number-accumulator`

with `(bf number)`

instead of `number`

in order to “skip over” the negative number that has now been addressed.

1 2 |
(else (number-name-helper (triple-number-accumulator number) 0)))) |

The `else`

calls a helper procedure `number-name-helper`

with two arguments: the result of invoking `triple-number-accumulator`

with `number`

, which generates a sentence with the number broken up into sets of three numbers, and 0. The 0 represents the initial value for `large-number-count`

, which is the value that I use to keep track of what large number name to use (thousands, millions etc).

1 2 3 4 5 6 7 8 9 10 |
(define (number-name-helper triple-sent large-number-count) (if (empty? triple-sent) '() (se (number-name-helper (bl triple-sent) (+ large-number-count 1)) (triple-helper (last triple-sent)) (if (not (equal? (last triple-sent) "000")) (large-number-name large-number-count) '() )))) |

This part is a procedure that does some important stuff.

First lets consider the arguments. `triple-sent`

is the sentence consisting of words of up to three digits (which I’m thinking of as “triples”) generated by calling `triple-number-accumulator`

. I say “up to three digits” because, depending on the input number, some words will be shorter. Here’s what the output for `triple-number-accumulator`

looks like for “12120301912607891”:

`large-number-count`

is a counter I use to keep track of what large number name to use.

`((empty? triple-sent) '())`

The base case for the `number-name-helper`

procedure is when `triple-sent`

is empty. This happens when all the triples in `triple-sent`

have been dealt with.

The recursive case does four important things:

`(number-name-helper (bl triple-sent) (+ large-number-count 1))`

It invokes`number-name-helper`

recursively on`bl triple-sent`

. It’s`bl triple-sent`

because the rest of this procedure will process`last triple-sent`

into words, and because I’m dealing with`triple-sent`

from “right to left” as the book suggested. It also increases`large-number-count`

by 1. This will allow my program to pin the correct name on procedure if it involves big numbers.`(triple-helper (last triple-sent))`

It invokes`triple-helper`

on`last triple-sent`

.`triple-helper`

is the procedure that has the core of the logic of how to deal with the three number sequences outputted by`triple-number-accumulator`

and turn them into words.

This line of the procedure checks if the current three-digit word (“triple”) is equal to`"000"`

. If the triple is not equal to`"000"`

, then this line invokes`large-number-name`

with`large-number-count`

. If the triple*is*equal to`"000"`

, then this line returns`'()`

, effectively “skipping over” invoking the large number value for this part of the number.

This was one of the trickiest parts for me to figure out and only got fully addressed in testing. Initially I just had a straightforward invocation of`large-number-name`

with`large-number-count`

. That worked in many cases, but if you had three digits that were all zeros, you could wind up with something like “one million thousand” as output for`1000000`

. That is obviously wrong. So I realized that I had to make the invocation of`large-number-name`

conditional on the value of the triple not being “000”. What I did not realize, and what took me sometime to figure out, was that for some reason, Scheme was having my`triple-accumulator`

procedure output quote marks around sets of three numbers where there are 0’s on both sides. Here’s a sentence of triples showing what I mean:

I initially did something like`(if (not (equal? (last triple-sent) 000))`

expecting that to work (in fact, cuz Scheme generally treats`0`

and`000`

and`00000`

and`-0`

as equivalent expressions, I should have been able to just do`(if (not (equal? (last triple-sent) 0))`

and have that work). I didn’t realize that the quote marks around “000” were absolutely critical for the procedure to function properly.-
It combines the results of the three previous steps with

`sentence`

.

1 2 3 4 5 6 7 8 |
(define (triple-helper number) (cond ((empty? number) '()) ((equal? (first number) 0) (triple-helper (bf number))) ((>= number 100) (se (single-digits (first number)) 'hundred (triple-helper (bf number)))) ((>= number 20) (se (tens-place (first number)) (triple-helper (bf number)))) ((and (> number 9)(< number 20))(se (tens number))) (else (se (single-digits number))))) |

The first two lines of this procedure are important for handling 0’s within a 3 digit sequence.

1 2 3 |
(cond ((empty? number) '()) ((equal? (first number) 0) (triple-helper (bf number))) |

If a three digit sequence such as “307” is passed to `triple-helper`

, the `((equal? (first number) 0) (triple-helper (bf number)))`

line tells the procedure to “skip” past the “0” so that the rest of the number can be processed. Here’s a trace using this example:

`((empty? number) '())`

helps deal with the case where a three digit number ends with a zero. Without `((empty? number) '())`

, the procedure would try to use `((equal? (first number) 0) (triple-helper (bf number)))`

and would `bf number`

on an empty word, giving an error. `((empty? number) '())`

allows the procedure to handle this situation gracefully:

It can also handle stuff like “000” on the same principles.

1 2 |
((>= number 100) (se (single-digits (first number)) 'hundred (triple-helper (bf number)))) |

This line handles triple-digit numbers greater than 100. It uses a `single-digits`

helper procedure + `'hundred'`

to generate the name of the hundreds’ place, and then recursively calls `triple-helper`

without the hundreds place in order to generate the rest of the name of the triple.

1 2 |
((>= number 20) (se (tens-place (first number)) (triple-helper (bf number)))) |

This line handles the names of the numbers in the tens place for numbers at least equal to 20. It uses a `tens-place`

helper procedure to generate the name, then recursively calls `triple-helper`

without the tens place in order to generate the rest of the name of the triple.

1 2 |
((and (> number 9)(< number 20))(se (tens number))) |

This line handles the names of numbers from 10 to 19. It does not recursively call `triple-helper`

because of how number names work in English (e.g. for “22” you can make that out of “twenty” for the tens place plus “two” for the ones place, but for “12” there’s a name, “twelve”, that conveys what the values are in both the tens place and the ones place).

1 2 |
(else (se (single-digits number))))) |

This line handles 1 through 9 using `single-digits`

.

1 2 3 4 5 6 |
(define (triple-number-accumulator number) (if (<= (count number) 3) (se number) (se (triple-number-accumulator (bl (bl (bl number)))) (word (last (bl (bl number))) (last (bl number)) (last number))))) |

This procedure is the procedure that takes a number and makes a sentence of at-most three digit words for further handling by the overall program.

The base case: if the number of digits in the `number`

is less than or equal to 3, those digits are returned as a sentence. `sentence`

is invoked here for consistency of output for cases when only a three digit number is passed into the procedure.

The recursive case: triple-number-accumulator is called recursively, excluding the last three digits of number. Those digits are combined into a word. The results of the recursive call and the word-formation are combined into a sentence.

If we were start from the beginning of the number (left to right), we’d have to figure out what the groupings of the three digits should be based on the sentence, or else we might wind up with erroneous groupings like `(123 456 789 0)`

for 1,234,567,890 when what we want is `(1 234 567 890)`

. But by starting from the end of the number, we can just “grab” groups of three numbers freely until we run out of those, and then “grab” whatever numbers are left at the end.

### Easier Sub-procedures

These are the sub-procedures I think are more straightforward.

1 2 3 4 5 6 7 |
(define (large-number-name large-number-count) (if (> large-number-count 0) (item large-number-count '(thousand million billion trillion quadrillion quintillion sextillion septillion octillion nonillion decillion)) '())) |

If `large-number-count`

is greater than 0, the value of `large-number-count`

is used as an argument to `item`

to pull the appropriate word from the sentence of large number names provided by the book. Otherwise (meaning when we’re dealing with only three digit numbers) the empty sentence is returned.

1 2 3 4 5 6 7 8 9 10 |
(define (single-digits number) (item number '(one two three four five six seven eight nine))) (define (tens-place number) (item (- number 1) '(twenty thirty forty fifty sixty seventy eighty ninety))) (define (tens number) (item (- number 9) '(ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen))) |

These procedures all operate on the same principle, which is to use `item`

and then a number value to pull a particular name of a word from a sentence. `single-digits`

is the most straightforward, since the number provided to item and the name of the number are the same – meaning that `1`

as a number gets you “one”, `2`

gets you “two” and so on.

With `tens-place`

, because I started my sentence with “twenty”, “twenty” is the first value of the sentence. When I want to grab the word “twenty”, it will be when the `number`

in the tens place is 2. If I want to grab the first word of a sentence using item when `number`

is a 2, I have to subtract 1 from 2 so that the number becomes 1 and will therefore retrieve the first word when passed as an argument to `item`

. The same principle applies to the other numbers in the sentence (meaning that if I want to get “thirty” when the `number`

in the tens place is 3 and “thirty” is the second word in the sentence, then I need to subtract 1 from 3).

The same idea from `tens-place`

about “offsetting” number by a value so that it corresponds to the word within the sentence you want to select applies to `tens`

(which should have maybe been named `teens`

although that’s not quite accurate either). In this case, the incoming numbers will range from 10 to 19, and this corresponds to the words in the sentence which range from “ten” to “nineteen”. In order to make it so that the number `10`

gets the first word of the sentence “ten” (and to have all the other number values select their appropriate words), we subtract by 9.

I think this organization of the program makes sense because

1 2 3 |
(define (negative? number) (equal? (first number) '-)) |

Pretty self-explanatory I think.

### Full Program

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 |
(define (single-digits number) (item number '(one two three four five six seven eight nine))) (define (tens number) (item (- number 9) '(ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen))) (define (tens-place number) (item (- number 1) '(twenty thirty forty fifty sixty seventy eighty ninety))) (define (large-number-name large-number-count) (if (> large-number-count 0) (item large-number-count '(thousand million billion trillion quadrillion quintillion sextillion septillion octillion nonillion decillion)) '())) (define (triple-number-accumulator number) (if (<= (count number) 3) (se number) (se (triple-number-accumulator (bl (bl (bl number)))) (word (last (bl (bl number))) (last (bl number)) (last number))))) (define (triple-helper number) (cond ((empty? number) '()) ((equal? (first number) 0) (triple-helper (bf number))) ((>= number 100) (se (single-digits (first number)) 'hundred (triple-helper (bf number)))) ((>= number 20) (se (tens-place (first number)) (triple-helper (bf number)))) ((and (> number 9)(< number 20))(se (tens number))) (else (se (single-digits number))))) (define (number-name-helper triple-sent large-number-count) (if (empty? triple-sent) '() (se (number-name-helper (bl triple-sent) (+ large-number-count 1)) (triple-helper (last triple-sent)) (large-number-name large-number-count)))) (define (negative? number) (equal? (first number) '-)) (define (number-name number) (cond ((not (number? number)) '(Erroneous Input There Buddy This Program Only Takes Numbers)) ((not (integer? number)) '(Erroneous Input There Buddy This Program Only Takes Integers)) ((equal? number 0) '(zero)) ((negative? number) (se 'negative (number-name-helper (triple-number-accumulator (bf number)) 0))) (else (number-name-helper (triple-number-accumulator number) 0)))) |

### Tests

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 |
(number-name 12) (number-name 10) (number-name 19) (number-name 20) (number-name 30) (number-name 99) (number-name -1) (number-name -500000002) (number-name 131212120301912607891) (number-name 00010000) (number-name 1000529) (number-name 1000000) (number-name 100050000770050000200000) (number-name 0) (number-name -0000) (number-name '-1234a456) (number-name 1.2) |

## Other Solutions

### AnneB

AnneB made a diagram, which I think is good. I managed to intuitively “feel” my way through coding the whole procedure but that’s a very unreliable method for me.

1 2 3 4 5 6 |
(define (number-name n) (if (< (count n) 4) (group-name n) (sentence (number-name-helper (all-but-last-three n) (group-words)) (group-name (last-three n)) ))) |

If a number has fewer than four digits, AnneB sends the number straight to the `group-name`

procedure. Otherwise, the results of calling a helper procedure with all but the last three numbers and `group-name`

with the last three numbers are `sentence`

‘ed together.

AnneB doesn’t appear to break up the number into a sentence of three-digit words right off the bat like I do. Instead she breaks up the word into three digit chunks as she goes along with calls (including recursive calls) to her `number-name-helper`

and with the use of her helper procedures `last-three`

and `all-but-last-three`

. Also of note is that she passes in `group-words`

as one of the arguments to `number-name-helper`

. `group-words`

is basically just the name of the sentence of large number names:

1 2 3 4 |
(define (group-words) β(thousand million billion trillion quadrillion quintillion sextillion septillion octillion nonillion decillion)) |

So AnneB will be using this sentence directly rather than my more indirect method involving a counter and `item`

.

1 2 3 4 5 6 7 8 9 |
(define (number-name-helper n names) (if (< (count n) 4) (sentence (group-name n) (first names)) (sentence (number-name-helper (all-but-last-three n) (butfirst names)) (group-name (last-three n)) (if (equal? (last-three n) "000") β() (first names) )))) |

There are a lot of similarities in the overall structures of our `number-name-helper`

procedures, at least for the case where `count n`

is at least 4. AnneB starts her helper procedure with a recursive call to itself with all but the last three numbers,. Since I already broke up the number into at-most three digit words within a sentence, I was able to just `bl`

my recursive call, but AnneB took a different approach and used her `all-but-last-three`

procedure to accomplish a similar task. She also does `butfirst names`

, which queues up the next large number name (in `group-words`

) as appropriate for the next recursive call. AnneB’s approach to the large number names avoids the necessity of having to use a counter – she just `butfirst`

s through the sentence of large number names directly. I think this is more elegant than what I did.

AnneB then uses `group-name`

, which is somewhat analogous to my `triple-helper`

, on a three digit sequence of numbers.

AnneB deals with the “three zeros” problem next. If the last three numbers of the number are 0, the empty sentence is returned, skipping over the large number name (e.g. thousands, millions) for that three digit sequence. Otherwise, the `first`

of `names`

(`group-words`

) is returned.

Note that had to add some quote marks around multiple 0’s in some spots to get Anne’s code to run, and those changes are reflected in my pastes of her code.

I made a tree which illustrates some of the similarities in the `number-name-helper`

procedures:

`group-name`

appears to be somewhat similar to my `triple-helper`

, in that it’s the program that handles sets of three numbers and passes them off to helper procedures that actually generate the file number names. We handle hundreds somewhat differently – my procedure checks if a number is greater than or equal to 100 and then forms the number name for the hundreds’ place using the `single-digits`

helper procedure & the word `'hundred`

, and, if so, then recursively calls `triple-helper`

to get the rest of the name of the number. Recursive calls to `triple-helper`

actually occur in my `triple-helper`

procedure for any number greater than 19. AnneB’s procedure checks if a number has three digits and, if so, then calls a `hundred-name`

helper procedure and a `two-digit-name`

helper procedure. Otherwise `two-digit-name`

gets called. So the procedures have somewhat different “flow” and my `triple-helper`

procedure does a lot more work. It occurs to me that maybe it does too much work and that Anne’s is better organized.

One other detail item I wanted to comment on is that Anne used `item`

to pull out a specific desired word within a sentence of number words, but did so in a slightly different way than I did. I put the `item`

invocation within a procedure that contained the sentence of number words, as in my version of `single-digits`

, which I consider a “lower level” procedure because it doesn’t refer to any other procedures besides Scheme primitives. OTOH, Anne did stuff like this:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
(define (tens-words) β(twenty thirty forty fifty sixty seventy eighty ninety)) (define (two-digit-name n) (cond ((or (equal? n β00) (equal? n β0)) ββ) ((= (count n) 1) (item n (digit-words))) ((equal? (first n) 0) (item (last n) (digit-words))) ((equal? (first n) 1) (item (+ (last n) 1) (teen-words))) (else (word (item (- (first n) 1) (tens-words)) (if (equal? (last n) 0) ββ (word β- (item (last n) (digit-words))) ))))) |

Note that the item is on the side of the “higher level” procedure and not the “lower level” procedure. And likewise AnneB handled “offsets” (like needing to subtract 1 for her `tens-words`

, which was the equivalent of my `tens-place`

) on the side of the higher level procedure. I don’t have a strong opinion on which approach is better, if any. I noticed the difference and it seemed interesting, because there is a structure choice there that I hadn’t really thought about (just followed my intuitions).

Big picture comment: unlike my `number-name`

procedure, AnneB’s procedure does not appear to handle negative numbers or numbers with leading zeros like `00010000`

. For numbers equal to zero, I get the empty sentence as a result instead of `'(zero)`

as in mine. It also doesn’t handle non-integers or non-numbers.

### buntine

buntine’s solution seemed interesting. Want to go over one thing in detail:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 |
; Translates a three digit (or smaller) number into it's ; English equivalent. (define (translate n) (let ((tr (lambda (i f) (if (= i 0) '() (f i))))) (if (teen? n) (teen n) (cond ((= (count n) 1) (tr n ones)) ; 1-9 ((= (count n) 2) ; 10 - 99 (if (teen? n) (teen n) (se (tr (first n) tens) (tr (last n) ones)))) (else ; 100 - 999 (se (tr (first n) hundreds) (if (teen? (last-n n 2)) (teen (last-n n 2)) (se (tr (first (bf n)) tens) (tr (last n) ones))))))))) |

Let’s go one step at a time:

1 2 |
(let ((tr (lambda (i f) (if (= i 0) '() (f i))))) |

Let `tr`

be defined as a function of two arguments, `i`

, and `f`

. If `i`

equals 0, return the empty sentence. Otherwise, let `i`

be an argument to the function `f`

.

1 2 3 |
(if (teen? n) (teen n) |

If `n`

is true for `teen?`

, return `teen n`

, otherwise return the following big `cond`

.

Note the structure btw: we’re testing for the `teen?`

special case up front and then looking at other stuff if the number doesn’t return true for `teen?`

1 2 |
(cond ((= (count n) 1) (tr n ones)) ; 1-9 |

first line of the `cond`

that’s serving as the alternative to our `if`

. This is where we get into using `tr`

.

So by the logic of our `let`

and of how `let`

works, we can say the following: if `n`

has 1 digit, and that digit is 0, then this will return the empty sentence. If `n`

has one digit that is not 0, however, then `ones`

(which is the procedure buntine uses to translate single digits into words) will be invoked with `n`

as the argument.

1 2 3 4 5 |
((= (count n) 2) ; 10 - 99 (if (teen? n) (teen n) (se (tr (first n) tens) (tr (last n) ones)))) |

If `n`

has 2 digits, check if it’s a teen number and return `teen n`

if so. Otherwise, return the appropriate number name in words by invoking sentence with the arguments `tr (first n) tens)`

(which will get the appropriate tens-place name – e.g. twenty, thirty, whatever) and `tr (last n) ones`

which will get the appropriate single digit name.

π€ π€ π€ π€ π€ π€ π€ π€ π€

Since we already checked for `teen?`

before getting to the `cond`

, I’m unclear why it’s necessary to do so again.

1 2 3 4 5 6 7 8 |
(else ; 100 - 999 (se (tr (first n) hundreds) (if (teen? (last-n n 2)) (teen (last-n n 2)) (se (tr (first (bf n)) tens) (tr (last n) ones))))))))) |

This is the big `else`

statement of the `cond`

. This handles hundreds.

The `(tr (first n) hundreds)`

handles the hundreds place, pretty straightforward.

The `(if (teen? (last-n n 2)) (teen (last-n n 2))`

line is more interesting. `last-n`

is a nifty procedure that lets you arbitrarily select the last `n`

characters in a word (or number) by repeatedly calling `bf`

a number of times equal to the difference between the `count`

of the word and `n`

:

1 2 3 |
(define (last-n wd n) ((repeated bf (- (count wd) n)) wd)) |

So e.g. `last-n 'potato 4`

will call `bf`

`(- 6 4)`

or 2 times, giving us `tato`

. buntine uses this procedure to select the last two digits within the `n`

in the case of a number in the teen range.

The alternative to the `if`

condition is that the `tens`

and `ones`

procedures are respectively invoked on the appropriate number. I think the fact that `tr`

returns an empty sentence in the case of a 0 helps deal with 0’s in numbers.

Here’s a tree of buntine’s `translate n`

procedure:

buntine’s overall `number-name`

procedure can’t handle negative numbers, non-numeric input, or decimals. It can, however, handle leading zeros.