Hi Roelof,

You will not believe!

2 printStringRoman “II”

Have fun!

On Thu, 17 Sep 2020 at 18:20, Roelof Wobben via Pharo-users <
pharo-users@lists.pharo.org> wrote:

Cheers,
Alex

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 ↁ ROMAN NUMERAL FIVE THOUSAND
  2182 ↂ ROMAN NUMERAL TEN THOUSAND
  2187 ↇ ROMAN NUMERAL FIFTY THOUSAND
  2188 ↈ 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?
• Are you required to support zero?
• Are you required to support n/12 fractions (1<=n<=11)?
• Are you allowed, required, or forbidden to use the "overline"
  convention, where an overline means "multiply by 1000"?
  =-------
  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.

I am not happy with the way that (0 printStringRoman) quietly
produces ''.

Assuming you're generating regular modern Roman numbers,
you pretty much have to think of an integer as having 4 parts:
n // 1000      -- this many copies of M
n // 100 \ 10 -- hundreds using M, D, C
n // 10 \ 100 -- tens using    C, L, X
n \ 10        -- units using    X, V, I

If Squeak/Pharo's (0 printStringRoman) would answer 'N'
instead of '' I'd be happier with it.

While you really want to write your own code --- this
being an exercism task --- it would be a very good idea
to start by using #printStringRoman so that you know
what it's like to pass the tests.

On Fri, 18 Sep 2020 at 04:58, Roelof Wobben via Pharo-users <
pharo-users@lists.pharo.org> 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, escribió:

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: