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Challenges

The 50 substrings that validate any string of Roman numerals

+3
−0

Given a string of Roman numerals, decide whether it forms a valid Roman number. If not, output the substring that proves this, from the list of 50 strings described below.

Relevant fact

This challenge is based around the following fact:

A string of Roman numerals is a valid Roman number if and only if it contains none of the following 50 strings as a substring:

"CCCC", "CCD", "CCM", "CDC", "CMC", "CMD", "CMM", "DCD", "DCM", "DD", "DM", "IC", "ID", "IIII", "IIV", "IIX", "IL", "IM", "IVI", "IXC", "IXI", "IXL", "IXV", "IXX", "LC", "LD", "LL", "LM", "LXC", "LXL", "MMMM", "VC", "VD", "VIV", "VIX", "VL", "VM", "VV", "VX", "XCC", "XCD", "XCL", "XCM", "XCX", "XD", "XLX", "XM", "XXC", "XXL", "XXXX"

This works for any length string, if a valid Roman number is defined as follows:

Valid Roman numbers[1]

  • Each numeral appears no more than 3 times consecutively.
  • Each of V (5), L (50), and D (500) appears no more than once consecutively.
  • A Roman number is constructed by concatenating the strings representing its thousands, hundreds, tens, and units components.
Decimal Thousands Hundreds Tens Units
1 M C X I
2 MM CC XX II
3 MMM CCC XXX III
4 CD XL IV
5 D L V
6 DC LX VI
7 DCC LXX VII
8 DCCC LXXX VIII
9 CM XC IX

So, for example, 2345 would be represented as the concatenation of MM (for 2000), CCC (for 300), XL (for 40), and V (for 5), or MMCCCXLV.

This defines a unique correct representation for each number from 1 to 3999 (4000 and above are not representable without breaking the first two rules). The 50 substrings method described above will identify each of these 3999 strings as valid, and all other strings of Roman numerals as invalid.

Input

  • A string containing only Roman numerals (I, V, X, L, C, D, M).
  • The string will have length at least 1.
  • The string will have length at most 15 (this is the length of the longest valid string of Roman numerals).

Output

  • If the input is a valid Roman number, output a consistent value indicating this.
    • Consistent means that the value must be the same for all valid Roman numbers.
    • The output for a valid Roman number must not be one of the strings from the list of 50.
  • If the input is not a valid Roman number, output exactly 1 string from the list of 50.
    • The output in this case must be a substring of the input.
    • If the input has 2 or more of the strings from the list of 50 as substrings, you may choose any 1 of them to be the output, but you must choose only 1 of them (you must not output 2 or more).

Examples

A valid Roman number

The input MCMXCVI is the unique correct representation of 1996. It contains none of the 50 strings.

A string that is not a valid Roman number

Although the input MMXDIII might be suspected of representing 2493, it is not the unique correct representation of this number (which is MMCDXCIII). Note that it has XD as a substring, identifying it as invalid. The only correct output is therefore XD.

An invalid string with more than 1 potential output

The input MMCCMDXXV has 2 substrings that make it invalid, so either CCM or CMD would be correct outputs. It would not be correct to output both of these, or to output their overlap CCMD, as this is not one of the 50 strings.

Test cases

Test cases are in the format INPUT : VALID, OUTPUTS. Note that only one of the valid outputs can be chosen - outputting 2 or more is incorrect.

The output "VALID" is just an example - for an input that is a valid Roman number you may choose to output any consistent value distinct from the 50 strings.

I : VALID
V : VALID
X : VALID
L : VALID
C : VALID
D : VALID
M : VALID
II : VALID
VV : VV
XX : VALID
LL : LL
CC : VALID
DD : DD
MM : VALID
III : VALID
VII : VALID
IVI : IVI
IIV : IIV
CCI : VALID
CCV : VALID
CCX : VALID
CCL : VALID
CCC : VALID
CCD : CCD
CCM : CCM
IIII : IIII
MLDI : LD
MXXC : XXC
DCIIX : IIX
MCXXXX : XXXX
MCCCCXVI : CCCC
MMLXCVII : LXC
MMMCMXCIX : VALID
MMMDCCCLXXXVIII : VALID
MMCCCXLV : VALID
MCMXCVI : VALID
MMXDIII : XD
MMCDXCIII : VALID
MMCCMDXXV : CCM, CMD
XXX : VALID
CLLX : LL
DXXDMMV : DM, XD
CCDDDIMDD : DD, IM, CCD
VLCXIVXMCVXLC : VX, VL, LC, XM
DVLIILVCXVXVMLI : VX, VL, IL, VM, VC
VVDLMIVILXXDX : VV, VD, IL, XD, LM, IVI
DMXXCMILVCMLLMV : DM, IL, LL, VC, LM, XXC, XCM
CMXDVLCCDDLXLXC : DD, VL, LC, XD, XLX, LXC, CCD, LXL
XDIXCLLMVVLCMCM : VV, VL, LC, XD, LL, LM, IXC, CMC, XCL
IIXXCDVVLMILVDD : DD, VV, VL, VD, IL, LM, XXC, IXX, IIX, XCD
DDMIIXXCMCCDCMM : DD, DM, CMC, CMM, XXC, IXX, DCM, CDC, XCM, CCD, IIX

Scoring

This is a code golf challenge. Your score is the number of bytes in your code. Lowest score for each language wins.

Explanations are optional, but I'm more likely to upvote answers that have one.


  1. This is a common modern set of rules, described as Standard form on Wikipedia. It does not reflect all usages during history, but will be the basis of this challenge, since otherwise the 50 substrings approach does not work. ↩︎

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3 answers

+2
−0

Python (344 bytes)

534 bytes:

This is not golfed in any way. But it's the current winner!

def validate_roman_numeral(candidate):
    invalid_strings = ["CCCC", "CCD", "CCM", "CDC", "CMC", "CMD", "CMM", "DCD", "DCM", "DD", "DM", "IC", "ID", "IIII", "IIV", "IIX", "IL", "IM", "IVI", "IXC", "IXI", "IXL", "IXV", "IXX", "LC", "LD", "LL", "LM", "LXC", "LXL", "MMMM", "VC", "VD", "VIV", "VIX", "VL", "VM", "VV", "VX", "XCC", "XCD", "XCL", "XCM", "XCX", "XD", "XLX", "XM", "XXC", "XXL", "XXXX"]
    for invalid_string in invalid_strings:
        if invalid_string in candidate:
            return invalid_string
    return "VALID"

379 bytes: Here's a minimal effort golf of the above:

def v(c):
    i=['DD','DM','IC','ID','IL','IM','LC','LD','LL','LM','VC','VD','VL','VM','VV','VX','XD','XM','CCD','CCM','CDC','CMC','CMD','CMM','DCD','DCM','IIV','IIX','IVI','IXC','IXI','IXL','IXV','IXX','LXC','LXL','VIV','VIX','XCC','XCD','XCL','XCM','XCX','XLX','XXC','XXL','CCCC','IIII','MMMM','XXXX']
    for s in i:
        if s in c:
            return s
    return "VALID"

344 bytes: A little more effort:

def v(c):
 o="VALID"
 for s in 'DD','DM','IC','ID','IL','IM','LC','LD','LL','LM','VC','VD','VL','VM','VV','VX','XD','XM','CCD','CCM','CDC','CMC','CMD','CMM','DCD','DCM','IIV','IIX','IVI','IXC','IXI','IXL','IXV','IXX','LXC','LXL','VIV','VIX','XCC','XCD','XCL','XCM','XCX','XLX','XXC','XXL','C'*4,'I'*4,'M'*4,'X'*4:
  if s in c:
   o=s
 return o
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Output for valid inputs is flexible (2 comments)
+1
−0

Python, 220 bytes

Works with Python 3.8 or newer.

n=" IVXLCDM"
x=("".join([n[(d:=ord(c)-32)//8]+n[d-8*(d//8)]for c in'&PW!H.!@/$HF$@G"H6"@7"03#P?%N%O%U%]%^%_&N&O!*!+!1!=!9!<!:!;$=$<"*"+#M#N#L#O#K#C#=#<%MH))\'_X;;']))
r=lambda c:max(s if s in c else"A"for s in x.split())

Try it online!

There are 8 kind of characters, " IXLCDM". 2 characters are 6 bits that can be easily mapped to printable characters (after code point 32). This string is much shorter (half) then the "DD DM IC ..." string itself. Even if I add the extraction it is still somewhat shorter.

I just regenerate the " IXLCDM" characters from this string.

So it is the same as the code below just with some decoding from 6-bit-string.

238 bytes

If the r= part is not needed then 2 bytes less

r=lambda c:max(s if s in c else"A"for s in'DD DM IC ID IL IM LC LD LL LM VC VD VL VM VV VX XD XM CCD CCM CDC CMC CMD CMM DCD DCM IIV IIX IVI IXC IXI IXL IXV IXX LXC LXL VIV VIX XCC XCD XCL XCM XCX XLX XXC XXL CCCC IIII MMMM XXXX'.split())

Try it online!

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Specific python version (6 comments)
+1
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Haskell, 271 bytes

import Data.Char
import Data.List
n=" IVXLCDM"
r x=maximum[if(s`isInfixOf`x)then s else"A"|s<-words$concat$map(\x->(n!!div x 8):[(n!!(x-8*(div x 8)))])$map(\x->ord x-32)"&PW!H.!@/$HF$@G\"H6\"@7\"03#P?%N%O%U%]%^%_&N&O!*!+!1!=!9!<!:!;$=$<\"*\"+#M#N#L#O#K#C#=#<%MH))\'_X;;"]

Try it online!

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