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Challenges The 50 substrings that validate any string of Roman numerals

Python, 205 bytes Coding the string with the substrings as base 95 number is the shortest I get with this approach: n=0 for c in"),|9'OJ#SdHlCSD(LqRhc.|)_#s]`N:?foSIb*.BJVa=AH^).&&w*e(&l...

posted 6mo ago by __blackjack__‭ · edited 6mo ago by __blackjack__‭

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#2: Post edited by

__blackjack__‭ · 2026-03-08T16:57:46Z (6 months ago)
One byte less.

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~~# Python, 206 bytes~~
Arpad Horvath's packing two six bit characters into one ASCII character could make the 209 byte solution even better, so this builds on his work. The unpacking is a bit shorter and the final function works different and is also a bit shorter.
```
n=" IVXLCDM"
x="".join(n[(d:=ord(c)-32)>>3]+n[d&7]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;;')
f=lambda s:next((a for a in x.split()if a in s),1)
```
[Try it online!](https://tio.run/##TY1BT4NAEIXv/IrdWRZ2S0UriVXoVhKMQixtL9ZNKjZYJGLahVBM6q9HaC9@l3nzJvNe9dt8lcq5req2VQJQtJKz4CEG7SgA7O@yUEytWeaKss7Yll8413w6dRJLrTNjnORljbaoUKaxfMWhjf1LPXzU/ScIb8Afw5VDlvd0Thf0hSb0nW6MubHAA2zhERb4Dk@wiz1d6BMYgEViMiczsiDPJCCCTGgccv5mbqTnmVzLxS7df2QpOrjq89gwlqK@PO3K0dE@VLuiYbzIz8aBD0e87e/qZ98ba4hWMEQQBx1yFZ10xyzoVdRxdqT8L08vMj7tUgaQuBpCqKoL1bAueIjyfnDe/gE "Python 3.8 (pre-release) – Try It Online")
# 209 bytes
I've also noticed that each character of the string with all 50 substrings could be encoded in 3 bits, so I stuffed the whole string into one big integer number and then encoded that as a string with base 36. Because the `int()` function can easily decode that back to a number.
That's better than the leading solution by Arpad Horvath at the time.
~~```~~
S="";n=int("1ie3bbwb28v97xgfzvwp5cdss8z4647b6lrucee80ozdnm1e7wjv2fx4vczji4341fpac33oekqbpauv3hxf2hhtk87drmxgdfqis1zua0qe",36)
while n:S+=" IVXLCDM"[n&7];n>>=3
f=lambda s:next((a for a in S.split()if a in s),1)
```
[Try it online!](https://tio.run/##TZA9b4MwEIb3/ArLQwUqqgKmQInIQhekMCEhS1EGA3ZwAsZ8U/48BbLkXe6553Q3nPzr8kogRzbLEnkQnoTHRadAnVOUJGNiOMOPPd3ZPIzyO83a1plNy7QTq2j6lFLnWM2ZKHVqj4/BYJM5pPODm8jUmSQpQhV91okk/YDyiRl53j0dO2vK6Z6xmrf63JNjTaGGLPUw5rygQLjRpwdBEOOL/xvCq/iwbydxPnvowLyClElGQOsKOnWKQgCrGkAAFyD6amXBO0Xl7CVaVdPVZZuLvtzEFQYx1AAM/TU4DnZec/E3Cta8DMbvuK/gcO8x9uHNPQAAZLP9aD2sAbYVVV3@AQ "Python 3.8 (pre-release) – Try It Online")
# 235 bytes
This is what I was starting with. Pretty straight forward: take the first substring that is contained or 1 if no substring was found.
~~```~~
f=lambda s:next((a for a 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()if a in s),1)
```
[Try it online!](https://tio.run/##TY@xasRADET7fMWwlQ0mENKEg1TaRqBthSCkcEhMDHc@Yztw9/XOrNNEMGj0kGbZ@b59X6fnl3nZ9@H13F8@Pnusp@nrtjVNj@G6oMc4pZyRC1SgGWrQAhNYhhmswAWe4Qand3ggMqJAJFPsWSClinMpyOSZXNWpgLpCg@FRO@ODPAJGZpyde869EKEYLUYViswo7gX3hMVAVRQWUUR6XOfzuDXtOBw/wdp2T@1ePzb9XCp4S@qpQyr1OlwPzzKprqb9kYj/9jiJcsx8Pr2fHgDMyzhtDYM7DLW17f4L "Python 3.8 (pre-release) – Try It Online")

# Python, 205 bytes
Coding the string with the substrings as base 95 number is the shortest I get with this approach:
```python
n=0
for c in"),|9'OJ#SdHlCSD(LqRhc.|)_#s]`N:?foSIb*.BJVa=AH^).&&w*e(<')vw6T=jv[xbyt8voESZtk>E^5J{,":n=n*94+ord(c)-32
S=""
while n:S+=" IVXLCDM"[n&7];n>>=3
f=lambda s:next((a for a in S.split()if a in s),1)
```
[Try it online!](https://tio.run/##TY7dboJAEIXveYrJmuAuUtJqf5R2bVo0EQNt0m3IpkRbBIm0uFCgoKnvTkFv@t3MmTOZk5Pui00iBsM0q2tBz6UwycCHSCCiHkbd53mHBbPYYBNsfb9sfO1A3jv54uNJvw8TZq4U7XHuePRhtiSaLFfKGt91SVldv9LP0t2t9sWwTKbsrfgaT5dX818V6YIKZXTZS7IA@@Rs0JcYRUiqNlG8BqGzHkVgOtwyJjZyhXyzuBXjMR1IIY297SrwINfFeldg7EHb1GuaAtPyNI4KTKLwZOREvSB1exc/29ZwkekgFZBtNHDHPOoGy2iV2XByOP8vjy/cPu6cG2ihSwCQZpEocBOsQtgOQuo/ "Python 3.8 (pre-release) – Try It Online")
206 bytes
Arpad Horvath's packing two six bit characters into one ASCII character could make the 209 byte solution even better, so this builds on his work. The unpacking is a bit shorter and the final function works different and is also a bit shorter.
```python
n=" IVXLCDM"
x="".join(n[(d:=ord(c)-32)>>3]+n[d&7]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;;')
f=lambda s:next((a for a in x.split()if a in s),1)
```
[Try it online!](https://tio.run/##TY1BT4NAEIXv/IrdWRZ2S0UriVXoVhKMQixtL9ZNKjZYJGLahVBM6q9HaC9@l3nzJvNe9dt8lcq5req2VQJQtJKz4CEG7SgA7O@yUEytWeaKss7Yll8413w6dRJLrTNjnORljbaoUKaxfMWhjf1LPXzU/ScIb8Afw5VDlvd0Thf0hSb0nW6MubHAA2zhERb4Dk@wiz1d6BMYgEViMiczsiDPJCCCTGgccv5mbqTnmVzLxS7df2QpOrjq89gwlqK@PO3K0dE@VLuiYbzIz8aBD0e87e/qZ98ba4hWMEQQBx1yFZ10xyzoVdRxdqT8L08vMj7tUgaQuBpCqKoL1bAueIjyfnDe/gE "Python 3.8 (pre-release) – Try It Online")
# 209 bytes
I've also noticed that each character of the string with all 50 substrings could be encoded in 3 bits, so I stuffed the whole string into one big integer number and then encoded that as a string with base 36. Because the `int()` function can easily decode that back to a number.
That's better than the leading solution by Arpad Horvath at the time.
```python
S="";n=int("1ie3bbwb28v97xgfzvwp5cdss8z4647b6lrucee80ozdnm1e7wjv2fx4vczji4341fpac33oekqbpauv3hxf2hhtk87drmxgdfqis1zua0qe",36)
while n:S+=" IVXLCDM"[n&7];n>>=3
f=lambda s:next((a for a in S.split()if a in s),1)
```
[Try it online!](https://tio.run/##TZA9b4MwEIb3/ArLQwUqqgKmQInIQhekMCEhS1EGA3ZwAsZ8U/48BbLkXe6553Q3nPzr8kogRzbLEnkQnoTHRadAnVOUJGNiOMOPPd3ZPIzyO83a1plNy7QTq2j6lFLnWM2ZKHVqj4/BYJM5pPODm8jUmSQpQhV91okk/YDyiRl53j0dO2vK6Z6xmrf63JNjTaGGLPUw5rygQLjRpwdBEOOL/xvCq/iwbydxPnvowLyClElGQOsKOnWKQgCrGkAAFyD6amXBO0Xl7CVaVdPVZZuLvtzEFQYx1AAM/TU4DnZec/E3Cta8DMbvuK/gcO8x9uHNPQAAZLP9aD2sAbYVVV3@AQ "Python 3.8 (pre-release) – Try It Online")
# 235 bytes
This is what I was starting with. Pretty straight forward: take the first substring that is contained or 1 if no substring was found.
```python
f=lambda s:next((a for a 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()if a in s),1)
```
[Try it online!](https://tio.run/##TY@xasRADET7fMWwlQ0mENKEg1TaRqBthSCkcEhMDHc@Yztw9/XOrNNEMGj0kGbZ@b59X6fnl3nZ9@H13F8@Pnusp@nrtjVNj@G6oMc4pZyRC1SgGWrQAhNYhhmswAWe4Qand3ggMqJAJFPsWSClinMpyOSZXNWpgLpCg@FRO@ODPAJGZpyde869EKEYLUYViswo7gX3hMVAVRQWUUR6XOfzuDXtOBw/wdp2T@1ePzb9XCp4S@qpQyr1OlwPzzKprqb9kYj/9jiJcsx8Pr2fHgDMyzhtDYM7DLW17f4L "Python 3.8 (pre-release) – Try It Online")

#1: Initial revision by

__blackjack__‭ · 2026-03-07T19:29:57Z (6 months ago)

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# Python, 206 bytes

Arpad Horvath's packing two six bit characters into one ASCII character could make the 209 byte solution even better, so this builds on his work.  The unpacking is a bit shorter and the final function works different and is also a bit shorter.

```
n=" IVXLCDM"
x="".join(n[(d:=ord(c)-32)>>3]+n[d&7]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;;')
f=lambda s:next((a for a in x.split()if a in s),1)
```
[Try it online!](https://tio.run/##TY1BT4NAEIXv/IrdWRZ2S0UriVXoVhKMQixtL9ZNKjZYJGLahVBM6q9HaC9@l3nzJvNe9dt8lcq5req2VQJQtJKz4CEG7SgA7O@yUEytWeaKss7Yll8413w6dRJLrTNjnORljbaoUKaxfMWhjf1LPXzU/ScIb8Afw5VDlvd0Thf0hSb0nW6MubHAA2zhERb4Dk@wiz1d6BMYgEViMiczsiDPJCCCTGgccv5mbqTnmVzLxS7df2QpOrjq89gwlqK@PO3K0dE@VLuiYbzIz8aBD0e87e/qZ98ba4hWMEQQBx1yFZ10xyzoVdRxdqT8L08vMj7tUgaQuBpCqKoL1bAueIjyfnDe/gE "Python 3.8 (pre-release) – Try It Online")

# 209 bytes

I've also noticed that each character of the string with all 50 substrings could be encoded in 3 bits, so I stuffed the whole string into one big integer number and then encoded that as a string with base 36.  Because the `int()` function can easily decode that back to a number.

That's better than the leading solution by Arpad Horvath at the time.

```
S="";n=int("1ie3bbwb28v97xgfzvwp5cdss8z4647b6lrucee80ozdnm1e7wjv2fx4vczji4341fpac33oekqbpauv3hxf2hhtk87drmxgdfqis1zua0qe",36)
while n:S+=" IVXLCDM"[n&7];n>>=3
f=lambda s:next((a for a in S.split()if a in s),1)
```
[Try it online!](https://tio.run/##TZA9b4MwEIb3/ArLQwUqqgKmQInIQhekMCEhS1EGA3ZwAsZ8U/48BbLkXe6553Q3nPzr8kogRzbLEnkQnoTHRadAnVOUJGNiOMOPPd3ZPIzyO83a1plNy7QTq2j6lFLnWM2ZKHVqj4/BYJM5pPODm8jUmSQpQhV91okk/YDyiRl53j0dO2vK6Z6xmrf63JNjTaGGLPUw5rygQLjRpwdBEOOL/xvCq/iwbydxPnvowLyClElGQOsKOnWKQgCrGkAAFyD6amXBO0Xl7CVaVdPVZZuLvtzEFQYx1AAM/TU4DnZec/E3Cta8DMbvuK/gcO8x9uHNPQAAZLP9aD2sAbYVVV3@AQ "Python 3.8 (pre-release) – Try It Online")

# 235 bytes

This is what I was starting with.  Pretty straight forward: take the first substring that is contained or 1 if no substring was found.

```
f=lambda s:next((a for a 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()if a in s),1)
```
[Try it online!](https://tio.run/##TY@xasRADET7fMWwlQ0mENKEg1TaRqBthSCkcEhMDHc@Yztw9/XOrNNEMGj0kGbZ@b59X6fnl3nZ9@H13F8@Pnusp@nrtjVNj@G6oMc4pZyRC1SgGWrQAhNYhhmswAWe4Qand3ggMqJAJFPsWSClinMpyOSZXNWpgLpCg@FRO@ODPAJGZpyde869EKEYLUYViswo7gX3hMVAVRQWUUR6XOfzuDXtOBw/wdp2T@1ePzb9XCp4S@qpQyr1OlwPzzKprqb9kYj/9jiJcsx8Pr2fHgDMyzhtDYM7DLW17f4L "Python 3.8 (pre-release) – Try It Online")

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