This is a problem where the only data structures you need,

other than the integer you are encoding and the string you

are building, is a small number of array literals.

Here is pseudo-code.

If self is negative, emit a negative sign.

If self is zero, emit N and finish.

Let n be the absolute value of self.

Emit M (n//1000) times -- use #next:put:

Encode n // 100 \\ 10 using M D C.

Encode n // 10 \\ 10 using C L X.

Encode n \\10 using X V I

To encode a digit using x v i,

write the characters of

��#('' 'i' 'ii' 'iii' 'iv' 'v' 'vi' 'vii' 'viii' 'ix' 'x')

��at the digit + 1 (use #nextPutAll:).

I have checked this pseudocode by writing

#asRomanPrintString (2 lines including header) calling

#asRomanPrintOn: (9 lines including header)

and verified that each number from -4000 to +4000

gets the same result from #printStringRoman and

#asRomanPrintString (except 0, which Pharo gets wrong).

Now we *could* do the "encode a digit" using smaller

or no tables, but it would take more code.�� In this

case we can check that the right output for a digit

is produced just by LOOKING at the table, rather than

by checking complicated code.

On Sat, 19 Sep 2020 at 02:13, Pablo Navarro <pablo1n7@gmail.com> wrote:

Hi! Maybe you can use this algorithm:

Define a dictionary with these elements:
��
1000:'M',��
900:'CM',��
500: 'D',��
400: 'CD',��
100:"C",��
90:'XC',��
50:'L',��
40:'XL',��
10:'X',��
9:'IX',��
5:'V',��
4:'IV',��
1:'I'

Using this dictionary (romansDic), you define a recursive function:

toRomans(number){
�� i = return the greatest key less than or equal to given key from ��romansDic'��.
�� if (number == i ){
return romansDic.get(number)
}
�� return string_concat(romansDic.get(i), toRomans(number-i))��
}

Sorry for the pseudocode.

Saludos Pablo.

El 18 de sep. de 2020 10:46 -0300, Roelof Wobben via Pharo-users <pharo-users@lists.pharo.org>, escribi��:

Op 18-9-2020 om 06:45 schreef Richard O'Keefe:

Roman numerals are much more complicated and much less consistent

than most people realise.�� The regular M DC LX VI system is both

more modern and less capable than anything the Romans would have

recognised.�� In particular,

��- in the 8th century, N (short for "nulla") was adopted for zero

��- the Roman system always had fractions like S for 1/2, . for 1/12

��- there were numerals for much larger numbers.

Unicode code block [2150] has characters for the Roman numerals

including

216C L ROMAN NUMERAL FIFTY
216D C ROMAN NUMERAL ONE HUNDRED
216E D ROMAN NUMERAL FIVE HUNDRED
216F M ROMAN NUMERAL ONE THOUSAND

2181 �� ROMAN NUMERAL FIVE THOUSAND
2182 �� ROMAN NUMERAL TEN THOUSAND
2187 �� ROMAN NUMERAL FIFTY THOUSAND
2188 �� ROMAN NUMERAL ONE HUNDRED THOUSAND

(In fact these are ligated versions of forms using "apostrophic" brackets;

the pattern goes as high as you want, e.g., (((|))) for a million.

D and M were originally |) and (|).�� There is

So the first thing is to make sure that you understand the

requirements for the problem.

- Are you required to produce ASCII characters, required to

�� produce Unicode ones, or allowed to produce either?

as far as I can see from the tests only ASCI characters.

- Are you required to support zero?

No

- Are you required to support n/12 fractions (1<=n<=11)?

NO

- Are you allowed, required, or forbidden to use the "overline"

�� convention, where an overline means "multiply by 1000"?

�� =-------

In the test that one is not used.

�� ICCXXXIVDLXVII = 1,234,567

- Are you allowed, required, or forbidden to use "additive"

�� form "IIII" as well as/instead of "subtractive" form "IV"?

- Are you to use upper case or lower case letters?

- And so on.

the number 4 needs to be�� "IV"

Roelof