roman numbers
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roman numbers
 
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Hello,
Can someone help me with a good plan to convert numbers to roman numbers. I could make a dictionary with 1,4,5,9,10,99,100, 999, 1000 but that feels like a overkill. Regards, Roelof |
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Hi Roelof, You will not believe! 2 printStringRoman “II” Have fun! On Thu, 17 Sep 2020 at 18:20, Roelof Wobben via Pharo-users <[hidden email]> wrote: Hello, Cheers, Alex |
Re: roman numbers
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Hello,
The challenge is to do it manually. But I can take a look how that function is implented. Roelof Op 17-9-2020 om 18:13 schreef Aliaksei Syrel:
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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 <[hidden email]> wrote:
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Re: roman numbers
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Op 18-9-2020 om 06:45 schreef Richard
O'Keefe:
as far as I can see from the tests only ASCI characters.
No
NO
In the test that one is not used.
the number 4 needs to be "IV" Roelof |
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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 <[hidden email]>, escribió:
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Re: roman numbers
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Op 18-9-2020 om 16:13 schreef Pablo
Navarro:
No problem. I searching for a idea not somebody who solves it for me. Roelof |
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In reply to this post by pablo1n7
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 <[hidden email]> wrote:
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