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Comments on Single digit Roman numeral

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Single digit Roman numeral

+3
−0

Given a single character, which is a valid Roman numeral, output its value.

Values

There are 7 valid single character Roman numerals, with the following values:

Character Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1000

Input

  • A single character, which will always be one of IVXLCDM.

Output

  • The corresponding value.
  • The output value must not be a Roman numeral.

Test cases

As there are only 7 valid inputs, the list of test cases is exhaustive.

Test cases are in the format "input" : output.

"I" : 1
"V" : 5
"X" : 10
"L" : 50
"C" : 100
"D" : 500
"M" : 1000

Scoring

This is a code golf challenge. Your score is the number of bytes in your code.

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

code-golf
posted about 1 year ago
1y ago
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trichoplax‭
1050 reputation 83 17 132 472
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2 comment threads

Allowed output formats? (4 comments)
Similar challenge (1 comment)
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+1
−0

Haskell, 62 bytes

(\n->(scanl(*)1$cycle[5,2])!!(length$takeWhile(/=n)"IVXLCDM"))

No import needed. It just uses standard Prelude functions.

scanl(*)1$cycle[5,2]

will give you the infinite list of [1,5,10,50...]. With

length$takeWhile(/=n)"IVXLCDM"

you will get the index of the element you would like to look up.

If you set

n='V'

the length part will be 1, and the index 1 means the 2nd element, so you'll get 5.

posted about 1 year ago
1y ago
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Arpad Horvath‭
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1 comment thread

Input methods (5 comments)
Input methods
trichoplax‭ wrote about 1 year ago
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The same applies to this answer as I mentioned on your Python answer. It just needs an input method to make it valid.

trichoplax‭ wrote about 1 year ago
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I don't know Haskell, but I tested your latest code using the following:

main = do
    let result = map(\n->(scanl(*)1$cycle[5,2])!!(length$takeWhile(/=n)"IVXLCDM")) ['I', 'V', 'X', 'L', 'C', 'D', 'M']
    print result

This gives the correct results for all inputs:

[1,5,10,50,100,500,1000]

I don't see any reason that the enclosing parentheses would be necessary, so I believe the following would be a valid answer, saving 2 bytes:

\n->(scanl(*)1$cycle[5,2])!!(length$takeWhile(/=n)"IVXLCDM")
Arpad Horvath‭ wrote about 1 year ago
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I can do

f=(\n->(scanl(*)1$cycle[5,2])!!(length$takeWhile(/=n)"IVXLCDM"))

but without the parenthesis it failes. You use the parenthesis after the map, and Haskell don't need a parenthesis after the function name just to call the function. It is needed to show what exactly is the first argument and what part is the second. So I think we need them.

trichoplax‭ wrote about 1 year ago
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I see what you mean. I tried the map without the parentheses and it fails, and I tried omitting the map and just applying to a single argument 'I' and again it only works with the parentheses. Sorry for the false hope...

Arpad Horvath‭ wrote about 1 year ago · edited about 1 year ago
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Why should you be sorry? Somebody reads my code in a language not known for him, trying to understand it. And even you found out how to test it.