Four is the number of letters in the ...
You are encouraged to solve this task according to the task description, using any language you may know.
The Four is ... sequence is based on the counting of the number of letters in the words of the (never─ending) sentence:
Four is the number of letters in the first word of this sentence, two in the second, three in the third, six in the fourth, two in the fifth, seven in the sixth, ···
- Definitions and directives
-
- English is to be used in spelling numbers.
- Letters are defined as the upper─ and lowercase letters in the Latin alphabet (A──►Z and a──►z).
- Commas are not counted, nor are hyphens (dashes or minus signs).
- twenty─three has eleven letters.
- twenty─three is considered one word (which is hyphenated).
- no and words are to be used when spelling a (English) word for a number.
- The American version of numbers will be used here in this task (as opposed to the British version).
2,000,000,000 is two billion, not two milliard.
- Task
-
- Write a driver (invoking routine) and a function (subroutine/routine···) that returns the sequence (for any positive integer) of the number of letters in the first N words in the never─ending sentence. For instance, the portion of the never─ending sentence shown above (2nd sentence of this task), the sequence would be:
4 2 3 6 2 7
- Only construct as much as is needed for the never─ending sentence.
- Write a driver (invoking routine) to show the number of letters in the Nth word, as well as showing the Nth word itself.
- After each test case, show the total number of characters (including blanks, commas, and punctuation) of the sentence that was constructed.
- Show all output here.
- Test cases
Display the first 201 numbers in the sequence (and the total number of characters in the sentence). Display the number of letters (and the word itself) of the 1,000th word. Display the number of letters (and the word itself) of the 10,000th word. Display the number of letters (and the word itself) of the 100,000th word. Display the number of letters (and the word itself) of the 1,000,000th word. Display the number of letters (and the word itself) of the 10,000,000th word (optional).
- Related tasks
Go
This is a naive non-optimized implementation
that stores each word of the sentence so far.
It uses the
sayOrdinal
and say
functions from the
Spelling of ordinal numbers task
(omitted from this listing).
<lang Go>package main
import ( "fmt" "strings" "unicode" )
func main() { f := NewFourIsSeq() fmt.Print("The lengths of the first 201 words are:") for i := 1; i <= 201; i++ { if i%25 == 1 { fmt.Printf("\n%3d: ", i) } _, n := f.WordLen(i) fmt.Printf(" %2d", n) } fmt.Println() fmt.Println("Length of sentence so far:", f.TotalLength()) /* For debugging: log.Println("sentence:", strings.Join(f.words, " ")) for i, w := range f.words { log.Printf("%3d: %2d %q\n", i, countLetters(w), w) } log.Println(f.WordLen(2202)) log.Println("len(f.words):", len(f.words)) log.Println("sentence:", strings.Join(f.words, " ")) */ for i := 1000; i <= 1e7; i *= 10 { w, n := f.WordLen(i) fmt.Printf("Word %8d is %q, with %d letters.", i, w, n) fmt.Println(" Length of sentence so far:", f.TotalLength()) } }
type FourIsSeq struct { i int // index of last word processed words []string // strings.Join(words," ") gives the sentence so far }
func NewFourIsSeq() *FourIsSeq { return &FourIsSeq{ //words: strings.Fields("Four is the number of letters in the first word of this sentence,"), words: []string{ "Four", "is", "the", "number", "of", "letters", "in", "the", "first", "word", "of", "this", "sentence,", }, } }
// WordLen returns the w'th word and its length (only counting letters). func (f *FourIsSeq) WordLen(w int) (string, int) { for len(f.words) < w { f.i++ n := countLetters(f.words[f.i]) ns := say(int64(n)) os := sayOrdinal(int64(f.i+1)) + "," // append something like: "two in the second," f.words = append(f.words, strings.Fields(ns)...) f.words = append(f.words, "in", "the") f.words = append(f.words, strings.Fields(os)...) } word := f.words[w-1] return word, countLetters(word) }
// TotalLength returns the total number of characters (including blanks, // commas, and punctuation) of the sentence so far constructed. func (f FourIsSeq) TotalLength() int { cnt := 0 for _, w := range f.words { cnt += len(w) + 1 } return cnt - 1 }
func countLetters(s string) int { cnt := 0 for _, r := range s { if unicode.IsLetter(r) { cnt++ } } return cnt }
// ... // the contents of // https://rosettacode.org/wiki/Spelling_of_ordinal_numbers#Go // omitted from this listing // ... </lang>
- Output:
The lengths of the first 201 words are: 1: 4 2 3 6 2 7 2 3 5 4 2 4 8 3 2 3 6 5 2 3 5 3 2 3 6 26: 3 2 3 5 5 2 3 5 3 2 3 7 5 2 3 6 4 2 3 5 4 2 3 5 3 51: 2 3 8 4 2 3 7 5 2 3 10 5 2 3 10 3 2 3 9 5 2 3 9 3 2 76: 3 11 4 2 3 10 3 2 3 10 5 2 3 9 4 2 3 11 5 2 3 12 3 2 3 101: 11 5 2 3 12 3 2 3 11 5 2 3 11 3 2 3 13 5 2 3 12 4 2 3 11 126: 4 2 3 9 3 2 3 11 5 2 3 12 4 2 3 11 5 2 3 12 3 2 3 11 5 151: 2 3 11 5 2 3 13 4 2 3 12 3 2 3 11 5 2 3 8 3 2 3 10 4 2 176: 3 11 3 2 3 10 5 2 3 11 4 2 3 10 4 2 3 10 3 2 3 12 5 2 3 201: 11 Length of sentence so far: 1203 Word 1000 is "in", with 2 letters. Length of sentence so far: 6279 Word 10000 is "in", with 2 letters. Length of sentence so far: 64140 Word 100000 is "one", with 3 letters. Length of sentence so far: 659474 Word 1000000 is "the", with 3 letters. Length of sentence so far: 7113621 Word 10000000 is "thousand", with 8 letters. Length of sentence so far: 70995756
Julia
The functions num2text and numtext2ordinal are from the "Spelling of ordinal numbers" and "Number names" tasks, updated for Julia 1.0.<lang julia>using DataStructures # for deque
const seed = "Four is the number of letters in the first word of this sentence, " const (word2, word3) = ("in", "the")
lettercount(w) = length(w) - length(collect(eachmatch(r"-", w))) splits(txt) = [x.match for x in eachmatch(r"[\w\-]+", txt)] todq(sentence) = (d = Deque{String}(); map(x->push!(d, x), splits(sentence)[2:end]); d)
struct CountLetters
seedsentence::String words::Deque{String} commasafter::Vector{Int} CountLetters(s) = new(s, todq(s), [13]) CountLetters() = CountLetters(seed)
end
function Base.iterate(iter::CountLetters, state = (1, 5, ""))
if length(iter.words) < 1 return nothing end returnword = popfirst!(iter.words) nextwordindex = state[1] + 1 wordlen = lettercount(returnword) wordvec = vcat(num2text(wordlen), word2, word3, splits(numtext2ordinal(num2text(nextwordindex)))) map(x -> push!(iter.words, x), wordvec) push!(iter.commasafter, length(iter.words)) added = length(returnword) + (nextwordindex in iter.commasafter ? 2 : 1) (wordlen, (nextwordindex, state[2] + added, returnword))
end
Base.eltype(iter::CountLetters) = Int
function firstN(n = 201)
countlet = CountLetters() print("It is interesting how identical lengths align with 20 columns.\n 1: 4") iter_result = iterate(countlet) itercount = 2 while iter_result != nothing (wlen, state) = iter_result print(lpad(string(wlen), 4)) if itercount % 20 == 0 print("\n$(itercount+1):") elseif itercount == n break end iter_result = iterate(countlet, state) itercount += 1 end println()
end
function sumwords(iterations)
countlet = CountLetters() iter_result = iterate(countlet) itercount = 2 while iter_result != nothing (wlen, state) = iter_result if itercount == iterations return state end iter_result = iterate(countlet, state) itercount += 1 end throw("Iteration failed on \"Four is the number\" task.")
end
firstN()
for n in [2202, 1000, 10000, 100000, 1000000, 10000000]
(itercount, totalletters, lastword) = sumwords(n) println("$n words -> $itercount iterations, $totalletters letters total, ", "last word \"$lastword\" with $(length(lastword)) letters.")
end</lang>
- Output:
It is interesting how identical lengths align with 20 columns.
1: 4 2 3 6 2 7 2 3 5 4 2 4 8 3 2 3 6 5 2 321: 5 3 2 3 6 3 2 3 5 5 2 3 5 3 2 3 7 5 2 3 41: 6 4 2 3 5 4 2 3 5 3 2 3 8 4 2 3 7 5 2 3 61: 10 5 2 3 10 3 2 3 9 5 2 3 9 3 2 3 11 4 2 3 81: 10 3 2 3 10 5 2 3 9 4 2 3 11 5 2 3 12 3 2 3 101: 11 5 2 3 12 3 2 3 11 5 2 3 11 3 2 3 13 5 2 3 121: 12 4 2 3 11 4 2 3 9 3 2 3 11 5 2 3 12 4 2 3 141: 11 5 2 3 12 3 2 3 11 5 2 3 11 5 2 3 13 4 2 3 161: 12 3 2 3 11 5 2 3 8 3 2 3 10 4 2 3 11 3 2 3 181: 10 5 2 3 11 4 2 3 10 4 2 3 10 3 2 3 12 5 2 3 201: 11 2202 words -> 2202 iterations, 14035 letters total, last word "ninety-ninth" with 12 letters. 1000 words -> 1000 iterations, 6290 letters total, last word "in" with 2 letters. 10000 words -> 10000 iterations, 64320 letters total, last word "in" with 2 letters. 100000 words -> 100000 iterations, 661369 letters total, last word "one" with 3 letters. 1000000 words -> 1000000 iterations, 7127541 letters total, last word "the" with 3 letters. 10000000 words -> 10000000 iterations, 71103026 letters total, last word "thousand" with 8 letters.
Kotlin
This pulls in (slightly adjusted) code from related tasks to convert numbers to text or ordinals. <lang scala>// version 1.1.4-3
val names = mapOf(
1 to "one", 2 to "two", 3 to "three", 4 to "four", 5 to "five", 6 to "six", 7 to "seven", 8 to "eight", 9 to "nine", 10 to "ten", 11 to "eleven", 12 to "twelve", 13 to "thirteen", 14 to "fourteen", 15 to "fifteen", 16 to "sixteen", 17 to "seventeen", 18 to "eighteen", 19 to "nineteen", 20 to "twenty", 30 to "thirty", 40 to "forty", 50 to "fifty", 60 to "sixty", 70 to "seventy", 80 to "eighty", 90 to "ninety"
)
val bigNames = mapOf(
1_000L to "thousand", 1_000_000L to "million", 1_000_000_000L to "billion", 1_000_000_000_000L to "trillion", 1_000_000_000_000_000L to "quadrillion", 1_000_000_000_000_000_000L to "quintillion"
)
val irregOrdinals = mapOf(
"one" to "first", "two" to "second", "three" to "third", "five" to "fifth", "eight" to "eighth", "nine" to "ninth", "twelve" to "twelfth"
)
fun String.toOrdinal(): String {
if (this == "zero") return "zeroth" // or alternatively 'zeroeth' val splits = this.split(' ', '-') val last = splits[splits.lastIndex] return if (irregOrdinals.containsKey(last)) this.dropLast(last.length) + irregOrdinals[last]!! else if (last.endsWith("y")) this.dropLast(1) + "ieth" else this + "th"
}
fun numToText(n: Long, uk: Boolean = false): String {
if (n == 0L) return "zero" val neg = n < 0L val maxNeg = n == Long.MIN_VALUE var nn = if (maxNeg) -(n + 1) else if (neg) -n else n val digits3 = IntArray(7) for (i in 0..6) { // split number into groups of 3 digits from the right digits3[i] = (nn % 1000).toInt() nn /= 1000 }
fun threeDigitsToText(number: Int) : String { val sb = StringBuilder() if (number == 0) return "" val hundreds = number / 100 val remainder = number % 100 if (hundreds > 0) { sb.append(names[hundreds], " hundred") if (remainder > 0) sb.append(if (uk) " and " else " ") } if (remainder > 0) { val tens = remainder / 10 val units = remainder % 10 if (tens > 1) { sb.append(names[tens * 10]) if (units > 0) sb.append("-", names[units]) } else sb.append(names[remainder]) } return sb.toString() }
val strings = Array(7) { threeDigitsToText(digits3[it]) } var text = strings[0] var andNeeded = uk && digits3[0] in 1..99 var big = 1000L for (i in 1..6) { if (digits3[i] > 0) { var text2 = strings[i] + " " + bigNames[big] if (text.isNotEmpty()) { text2 += if (andNeeded) " and " else " " // no commas inserted in this version andNeeded = false } else andNeeded = uk && digits3[i] in 1..99 text = text2 + text } big *= 1000 } if (maxNeg) text = text.dropLast(5) + "eight" if (neg) text = "minus " + text return text
}
val opening = "Four is the number of letters in the first word of this sentence,".split(' ')
val String.adjustedLength get() = this.replace(",", "").replace("-", "").length // no ',' or '-'
fun getWords(n: Int): List<String> {
val words = mutableListOf<String>() words.addAll(opening) if (n > opening.size) { var k = 2 while (true) { val len = words[k - 1].adjustedLength val text = numToText(len.toLong()) val splits = text.split(' ') words.addAll(splits) words.add("in") words.add("the") val text2 = numToText(k.toLong()).toOrdinal() + "," // add trailing comma val splits2 = text2.split(' ') words.addAll(splits2) if (words.size >= n) break k++ } } return words
}
fun getLengths(n: Int): Pair<List<Int>, Int> {
val words = getWords(n) val lengths = words.take(n).map { it.adjustedLength } val sentenceLength = words.sumBy { it.length } + words.size - 1 // includes hyphens, commas & spaces return Pair(lengths, sentenceLength)
}
fun getLastWord(n: Int): Triple<String, Int, Int> {
val words = getWords(n) val nthWord = words[n - 1] val nthWordLength = nthWord.adjustedLength val sentenceLength = words.sumBy { it.length } + words.size - 1 // includes hyphens, commas & spaces return Triple(nthWord, nthWordLength, sentenceLength)
}
fun main(args: Array<String>) {
var n = 201 println("The lengths of the first $n words are:\n") val (list, sentenceLength) = getLengths(n) for (i in 0 until n) { if (i % 25 == 0) { if (i > 0) println() print("${"%3d".format(i + 1)}: ") } print("%3d".format(list[i])) } println("\n\nLength of sentence = $sentenceLength\n") n = 1_000 do { var (word, wLen, sLen) = getLastWord(n) if (word.endsWith(",")) word = word.dropLast(1) // strip off any trailing comma println("The length of word $n [$word] is $wLen") println("Length of sentence = $sLen\n") n *= 10 } while (n <= 10_000_000)
}</lang>
- Output:
The lengths of the first 201 words are: 1: 4 2 3 6 2 7 2 3 5 4 2 4 8 3 2 3 6 5 2 3 5 3 2 3 6 26: 3 2 3 5 5 2 3 5 3 2 3 7 5 2 3 6 4 2 3 5 4 2 3 5 3 51: 2 3 8 4 2 3 7 5 2 3 10 5 2 3 10 3 2 3 9 5 2 3 9 3 2 76: 3 11 4 2 3 10 3 2 3 10 5 2 3 9 4 2 3 11 5 2 3 12 3 2 3 101: 11 5 2 3 12 3 2 3 11 5 2 3 11 3 2 3 13 5 2 3 12 4 2 3 11 126: 4 2 3 9 3 2 3 11 5 2 3 12 4 2 3 11 5 2 3 12 3 2 3 11 5 151: 2 3 11 5 2 3 13 4 2 3 12 3 2 3 11 5 2 3 8 3 2 3 10 4 2 176: 3 11 3 2 3 10 5 2 3 11 4 2 3 10 4 2 3 10 3 2 3 12 5 2 3 201: 11 Length of sentence = 1203 The length of word 1000 [in] is 2 Length of sentence = 6279 The length of word 10000 [in] is 2 Length of sentence = 64140 The length of word 100000 [one] is 3 Length of sentence = 659474 The length of word 1000000 [the] is 3 Length of sentence = 7113621 The length of word 10000000 [thousand] is 8 Length of sentence = 70995756
Perl
Uses Lingua::EN::Numbers
module to generate number names. State variable in extend_to routine keeps track of last word tallied.
<lang perl>use feature 'state'; use Lingua::EN::Numbers qw(num2en num2en_ordinal);
my @sentence = split / /, 'Four is the number of letters in the first word of this sentence, ';
sub extend_to {
my($last) = @_; state $index = 1; until ($#sentence > $last) { push @sentence, split ' ', num2en(alpha($sentence[$index])) . ' in the ' . no_c(num2en_ordinal(1+$index)) . ','; $index++; }
}
sub alpha { my($s) = @_; $s =~ s/\W//gi; length $s } sub no_c { my($s) = @_; $s =~ s/\ and|,//g; return $s } sub count { length(join ' ', @sentence[0..-1+$_[0]]) . " characters in the sentence, up to and including this word.\n" }
print "First 201 word lengths in the sequence:\n"; extend_to(201); for (0..200) {
printf "%3d", alpha($sentence[$_]); print "\n" unless ($_+1) % 32;
} print "\n" . count(201) . "\n";
for (1e3, 1e4, 1e5, 1e6, 1e7) {
extend_to($_); print ucfirst(num2en_ordinal($_)) . " word, '$sentence[$_-1]' has " . alpha($sentence[$_-1]) . " characters. \n" . count($_) . "\n";
}</lang>
- Output:
First 201 word lengths in the sequence: 4 2 3 6 2 7 2 3 5 4 2 4 8 3 2 3 6 5 2 3 5 3 2 3 6 3 2 3 5 5 2 3 5 3 2 3 7 5 2 3 6 4 2 3 5 4 2 3 5 3 2 3 8 4 2 3 7 5 2 3 10 5 2 3 10 3 2 3 9 5 2 3 9 3 2 3 11 4 2 3 10 3 2 3 10 5 2 3 9 4 2 3 11 5 2 3 12 3 2 3 11 5 2 3 12 3 2 3 11 5 2 3 11 3 2 3 13 5 2 3 12 4 2 3 11 4 2 3 9 3 2 3 11 5 2 3 12 4 2 3 11 5 2 3 12 3 2 3 11 5 2 3 11 5 2 3 13 4 2 3 12 3 2 3 11 5 2 3 8 3 2 3 10 4 2 3 11 3 2 3 10 5 2 3 11 4 2 3 10 4 2 3 10 3 2 3 12 5 2 3 11 1203 characters in the sentence, up to and including this word. One thousandth word, 'in' has 2 characters. 6249 characters in the sentence, up to and including this word. Ten thousandth word, 'in' has 2 characters. 64097 characters in the sentence, up to and including this word. One hundred thousandth word, 'one' has 3 characters. 659455 characters in the sentence, up to and including this word. One millionth word, 'the' has 3 characters. 7113560 characters in the sentence, up to and including this word. Ten millionth word, 'thousand' has 8 characters. 70995729 characters in the sentence, up to and including this word.
Perl 6
Uses the Lingua::EN::Numbers::Cardinal module to generate both cardinal and ordinal numbers. This module places commas in number words between 3-orders-of-magnitude clusters. E.G. 12345678.&ordinal
becomes: twelve million, three hundred forty-five thousand, six hundred seventy-eighth. Uses a custom 'no-commas' routine to filter them out for accurate character counts. Generates the 'sentence' lazily so only the words needed are ever calculated and reified.
<lang perl6>use Lingua::EN::Numbers::Cardinal;
my $index = 1; my @sentence = flat 'Four is the number of letters in the first word of this sentence, '.words,
{ @sentence[$index++].&alpha.&cardinal, 'in', 'the', |($index.&ordinal.&no-commas~',').words } ... * ;
sub alpha ( $str ) { $str.subst(/\W/, , :g).chars } sub no-commas ( $str ) { $str.subst(',', , :g) } sub count ( $index ) { @sentence[^$index].join(' ').chars ~ " characters in the sentence, up to and including this word.\n" }
say 'First 201 word lengths in the sequence:'; put ' ', map { @sentence[$_].&alpha.fmt("%2d") ~ (((1+$_) %% 25) ?? "\n" !! ) }, ^201; say 201.&count;
for 1e3, 1e4, 1e5, 1e6, 1e7 {
say "{.&ordinal.tc} word, '{@sentence[$_ - 1]}', has {@sentence[$_ - 1].&alpha} characters. ", .&count
}</lang>
- Output:
First 201 word lengths in the sequence: 4 2 3 6 2 7 2 3 5 4 2 4 8 3 2 3 6 5 2 3 5 3 2 3 6 3 2 3 5 5 2 3 5 3 2 3 7 5 2 3 6 4 2 3 5 4 2 3 5 3 2 3 8 4 2 3 7 5 2 3 10 5 2 3 10 3 2 3 9 5 2 3 9 3 2 3 11 4 2 3 10 3 2 3 10 5 2 3 9 4 2 3 11 5 2 3 12 3 2 3 11 5 2 3 12 3 2 3 11 5 2 3 11 3 2 3 13 5 2 3 12 4 2 3 11 4 2 3 9 3 2 3 11 5 2 3 12 4 2 3 11 5 2 3 12 3 2 3 11 5 2 3 11 5 2 3 13 4 2 3 12 3 2 3 11 5 2 3 8 3 2 3 10 4 2 3 11 3 2 3 10 5 2 3 11 4 2 3 10 4 2 3 10 3 2 3 12 5 2 3 11 1203 characters in the sentence, up to and including this word. One thousandth word, 'in', has 2 characters. 6249 characters in the sentence, up to and including this word. Ten thousandth word, 'in', has 2 characters. 64097 characters in the sentence, up to and including this word. One hundred thousandth word, 'one', has 3 characters. 659455 characters in the sentence, up to and including this word. One millionth word, 'the', has 3 characters. 7113560 characters in the sentence, up to and including this word. Ten millionth word, 'thousand', has 8 characters. 70995729 characters in the sentence, up to and including this word.
Phix
Note that my version of Number_names includes "and" (and ","), that others do not, hence the kill_and()/grr below and the minor mismatch of sentence lengths. <lang Phix>include demo\rosetta\number_names.exw
-- as per Spelling_of_ordinal_numbers#Phix: constant {irregs,ordinals} = columnize({{"one","first"},
{"two","second"}, {"three","third"}, {"five","fifth"}, {"eight","eighth"}, {"nine","ninth"}, {"twelve","twelfth"}})
function ordinal(string s)
integer i for i=length(s) to 1 by -1 do integer ch = s[i] if ch=' ' or ch='-' then exit end if end for integer k = find(s[i+1..$],irregs) if k then s = s[1..i]&ordinals[k] elsif s[$]='y' then s[$..$] = "ieth" else s &= "th" end if return s
end function --/copy of Spelling_of_ordinal_numers#Phix
function countLetters(string s)
integer res = 0 for i=1 to length(s) do integer ch = s[i] if (ch>='A' and ch<='Z') or (ch>='a' and ch<='z') then res += 1 end if end for return res
end function
sequence words = split("Four is the number of letters in the first word of this sentence,") integer fi = 1
function kill_and(sequence s) --grr...
for i=length(s) to 1 by -1 do if s[i] = "and" then s[i..i] = {} end if end for return s
end function
function WordLen(integer w) -- Returns the w'th word and its length (only counting letters).
while length(words)<w do fi += 1 integer n = countLetters(words[fi]) sequence ns = kill_and(split(spell(n))) sequence os = kill_and(split(ordinal(spell(fi)) & ",")) -- append eg {"two","in","the","second,"} words &= ns&{"in","the"}&os end while string word = words[w] return {word, countLetters(word)}
end function
function TotalLength() -- Returns the total number of characters (including blanks, -- commas, and punctuation) of the sentence so far constructed.
integer res = 0 for i=1 to length(words) do res += length(words[i])+1 end for return res
end function
procedure main() integer i,n string w
printf(1,"The lengths of the first 201 words are:\n") for i=1 to 201 do if mod(i,25)==1 then printf(1,"\n%3d: ", i) end if {?,n} = WordLen(i) printf(1," %2d", n) end for printf(1,"\nLength of sentence so far:%d\n", TotalLength()) for p=3 to 7 do i = power(10,p) {w, n} = WordLen(i) printf(1,"Word %8d is \"%s\", with %d letters.", {i, w, n}) printf(1," Length of sentence so far:%d\n", TotalLength()) end for
end procedure main()</lang>
- Output:
The lengths of the first 201 words are: 1: 4 2 3 6 2 7 2 3 5 4 2 4 8 3 2 3 6 5 2 3 5 3 2 3 6 26: 3 2 3 5 5 2 3 5 3 2 3 7 5 2 3 6 4 2 3 5 4 2 3 5 3 51: 2 3 8 4 2 3 7 5 2 3 10 5 2 3 10 3 2 3 9 5 2 3 9 3 2 76: 3 11 4 2 3 10 3 2 3 10 5 2 3 9 4 2 3 11 5 2 3 12 3 2 3 101: 11 5 2 3 12 3 2 3 11 5 2 3 11 3 2 3 13 5 2 3 12 4 2 3 11 126: 4 2 3 9 3 2 3 11 5 2 3 12 4 2 3 11 5 2 3 12 3 2 3 11 5 151: 2 3 11 5 2 3 13 4 2 3 12 3 2 3 11 5 2 3 8 3 2 3 10 4 2 176: 3 11 3 2 3 10 5 2 3 11 4 2 3 10 4 2 3 10 3 2 3 12 5 2 3 201: 11 Length of sentence so far:1204 Word 1000 is "in", with 2 letters. Length of sentence so far:6280 Word 10000 is "in", with 2 letters. Length of sentence so far:64692 Word 100000 is "one", with 3 letters. Length of sentence so far:671578 Word 1000000 is "the", with 3 letters. Length of sentence so far:7235383 Word 10000000 is "thousand,", with 8 letters. Length of sentence so far:72079160
REXX
<lang rexx>/*REXX pgm finds/shows the number of letters in the Nth word in a constructed sentence*/ @= 'Four is the number of letters in the first word of this sentence,' /*···*/
/* [↑] the start of a long sentence. */
parse arg N M /*obtain optional argument from the CL.*/ if N= | N="," then N= 201 /*Not specified? Then use the default.*/ if M= | M="," then M=1000 10000 100000 1000000 /* " " " " " " */ @abcU= 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' /*define the uppercase Latin alphabet. */ !.=.; #.=.; q=1; w=length(N) /* [↓] define some helpful low values.*/ call tell N if N<0 then say y ' is the length of word ' a " ["word(@, a)"]" say /* [↑] N negative? Just show 1 number*/ say 'length of sentence= ' length(@) /*display the length of the @ sentence.*/
if M\== then do k=1 for words(M) while M\=0 /*maybe handle counts (if specified). */
x=word(M, k) /*obtain the Kth word of the M list. */ call tell -x /*invoke subroutine (with negative arg)*/ say say y ' is the length of word ' x " ["word(@, x)"]" say 'length of sentence= ' length(@) /*display length of @ sentence.*/ end /*k*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ wordLen: arg ?; return length(?) - length( space( translate(?, , @abcU), 0) ) /*──────────────────────────────────────────────────────────────────────────────────────*/ tell: parse arg z,,$; idx=1; a=abs(z); group=25 /*show 25 numbers per line.*/
/*Q is the last number spelt by $SPELL#*/ do j=1 for a /*traipse through all the numbers to N.*/ do 2 /*perform loop twice (well ··· maybe).*/ y=wordLen( word(@, j) ) /*get the Jth word from the sentence.*/ if y\==0 then leave /*Is the word spelt? Then we're done.*/ q=q + 1 /*bump the on─going (moving) # counter.*/ if #.q==. then #.q=$spell#(q 'Q ORD') /*need to spell A as an ordinal number?*/ _=wordLen( word(@, q) ) /*use the length of the ordinal number.*/ if !._==. then !._=$spell#(_ 'Q') /*Not spelled? Then go and spell it. */ @=@ !._ 'in the' #.q"," /*append words to never─ending sentence*/ end /*2*/ /* [↑] Q ≡ Quiet ORD ≡ ORDinal */
$=$ || right(y, 3) /* [↓] append a justified # to a line.*/ if j//group==0 & z>0 then do; say right(idx, w)'►'$; idx=idx+group; $=; end end /*j*/ /* [↑] show line if there's enough #s.*/
if $\== & z>0 then say right(idx, w)'►'$ /*display if there are residual numbers*/ return</lang>
The $SPELL#.REX routine can be found here ───► $SPELL#.REX.
- output:
1► 4 2 3 6 2 7 2 3 5 4 2 4 8 3 2 3 6 5 2 3 5 3 2 3 6 26► 3 2 3 5 5 2 3 5 3 2 3 7 5 2 3 6 4 2 3 5 4 2 3 5 3 51► 2 3 8 4 2 3 7 5 2 3 10 5 2 3 10 3 2 3 9 5 2 3 9 3 2 76► 3 11 4 2 3 10 3 2 3 10 5 2 3 9 4 2 3 11 5 2 3 12 3 2 3 101► 11 5 2 3 12 3 2 3 11 5 2 3 11 3 2 3 13 5 2 3 12 4 2 3 11 126► 4 2 3 9 3 2 3 11 5 2 3 12 4 2 3 11 5 2 3 12 3 2 3 11 5 151► 2 3 11 5 2 3 13 4 2 3 12 3 2 3 11 5 2 3 8 3 2 3 10 4 2 176► 3 11 3 2 3 10 5 2 3 11 4 2 3 10 4 2 3 10 3 2 3 12 5 2 3 201► 11 length of sentence= 1203 2 is the length of word 1000 [in] length of sentence= 6279 2 is the length of word 10000 [in] length of sentence= 64140 3 is the length of word 100000 [one] length of sentence= 659474 3 is the length of word 1000000 [the] length of sentence= 7113621
zkl
Uses the nth function from Spelling_of_ordinal_numbers#zkl <lang zkl> // Built the sentence in little chucks but only save the last one
// Save the word counts
fcn fourIsThe(text,numWords){
const rmc="-,"; seq:=(text - rmc).split().apply("len").copy(); // (4,2,3,6...) szs:=Data(numWords + 100,Int).howza(0).extend(seq); // bytes cnt,lastWords := seq.len(),""; total:=seed.len() - 1; // don't count trailing space
foreach idx in ([1..]){ sz:=szs[idx]; a,b := nth(sz,False),nth(idx+1); // "two","three hundred sixty-seventh" lastWords="%s in the %s, ".fmt(a,b); ws:=lastWords.counts(" ")[1]; // "five in the forty-ninth " --> 4 cnt+=ws; total+=lastWords.len(); lastWords.split().pump(szs.append,'-(rmc),"len"); if(cnt>=numWords){
if(cnt>numWords){ z,n:=lastWords.len(),z-2; do(cnt - numWords){ n=lastWords.rfind(" ",n) - 1; } lastWords=lastWords[0,n+1]; total-=(z - n); } break;
} } return(lastWords.strip(),szs,total);
} fcn lastWord(sentence){ sentence[sentence.rfind(" ")+1,*] }</lang> <lang zkl>var seed="Four is the number of letters in the first word of this sentence, "; sentence,szs,total := fourIsThe(seed,201); print(" 1:"); foreach n,x in ([1..201].zip(szs)){
print("%3d".fmt(x)); if(0 == n%25) print("\n%3d:".fmt(n+1));
} println("\nLength of above sentence: ",total);</lang>
- Output:
1: 4 2 3 6 2 7 2 3 5 4 2 4 8 3 2 3 6 5 2 3 5 3 2 3 6 26: 3 2 3 5 5 2 3 5 3 2 3 7 5 2 3 6 4 2 3 5 4 2 3 5 3 51: 2 3 8 4 2 3 7 5 2 3 10 5 2 3 10 3 2 3 9 5 2 3 9 3 2 76: 3 11 4 2 3 10 3 2 3 10 5 2 3 9 4 2 3 11 5 2 3 12 3 2 3 101: 11 5 2 3 12 3 2 3 11 5 2 3 11 3 2 3 13 5 2 3 12 4 2 3 11 126: 4 2 3 9 3 2 3 11 5 2 3 12 4 2 3 11 5 2 3 12 3 2 3 11 5 151: 2 3 11 5 2 3 13 4 2 3 12 3 2 3 11 5 2 3 8 3 2 3 10 4 2 176: 3 11 3 2 3 10 5 2 3 11 4 2 3 10 4 2 3 10 3 2 3 12 5 2 3 201: 11 Length of above sentence: 1203
<lang zkl>n:=1000; do(5){
sentence,x,total := fourIsThe(seed,n); word:=lastWord(sentence); println("%,d words: \"%s\" [%d]. Length=%,d"
.fmt(n,word,word.len(),total));
n*=10;
}</lang>
- Output:
1,000 words: "in" [2]. Length=6,247 10,000 words: "in" [2]. Length=64,095 100,000 words: "one" [3]. Length=659,453 1,000,000 words: "the" [3]. Length=7,140,558 10,000,000 words: "thousand" [8]. Length=71,250,727