Unicode Fractions

Question

Given a fraction in the format m/n (where m and n are coprime integers), output the corresponding Unicode fraction. Your program/function will not be expected to take in any input that does not correspond to a Unicode character. Arrays, e.g. [2, 3] as opposed to 2/3, are accepted. m / n as opposed to m/n is also fine. Two separate inputs m and n are also valid.

The Unicode fractions that must be handled are as follows:

½, ⅓, ⅔, ¼, ¾, ⅕, ⅖, ⅗, ⅘, ⅙, ⅚, ⅐, ⅛, ⅜, ⅝, ⅞, ⅑, ⅒

Thus, the possible inputs are as follows:

1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, 1/7, 1/8, 3/8, 5/8, 7/8, 1/9, 1/10

The Unicode codepoints of the characters are as follows:

188 ¼
189 ½
190 ¾
8528 ⅐
8529 ⅑
8530 ⅒
8531 ⅓
8532 ⅔
8533 ⅕
8534 ⅖
8535 ⅗
8536 ⅘
8537 ⅙
8538 ⅚
8539 ⅛
8540 ⅜
8541 ⅝
8542 ⅞

Test Cases

1/2 -> ½
1/3 -> ⅓
2/3 -> ⅔
1/4 -> ¼
3/4 -> ¾
3/8 -> ⅜
1/10 -> ⅒

Make your code as short as possible; this is code golf.

Answer 1

JavaScript (Node.js), 78 77 bytes

Saved 1 byte thanks to @HermanLauenstein

Takes input in currying syntax (m)(n).

m=>n=>'¾⅗  ⅜    ⅞⅘½⅓¼⅕⅙⅐⅛⅑⅒⅚⅔⅝⅖'[(m*9+n)%31]

Try it online!

Answer 2

Perl 6,  48  43 bytes

{chr first *.unival==$^m/$^n,(|^191,|(8528..*))}

Try it

{chr first *.unival==$_,(|^191,|(8528..*))}

Try it

Expanded:

{  # bare block lambda with implicit parameter ｢$_｣
  chr

    first               # find the first value

      *.unival ==       # which has a unicode value that matches
          $_,           # the input fraction (as a Rational)

      (
        |^191,     # ¼(189) ½(188) ¾(190)
        |(8528..*) # all of the rest
      )
}

Note that the search has to be split up so that it doesn't return other Unicode characters that have the same unival. (otherwise it would be (1..*))

Answer 3

APL (Dyalog), 88 64 bytes

{⎕UCS(⌊(⍺-1)÷(1+⍵≡8)⌈4×⍵≡6)-(291194049⊤⍨10⌿12)[⍵]-189⌈8539××⍵|4}

Try it online!

-3 thanks to dzaima.

Using ETHproductions's 84-byte approach.

The f← included on TIO isn't counted, since it's put there just to be able to test the function.

Answer 4

JavaScript (ES6), 88 86 84 81 79 bytes

Saved 2 bytes thanks to @Arnauld

m=>n=>String.fromCharCode((4%n?8528:188)|"  130470912"[n]*1.3+~-m/-~"3 1"[n-6])

Not quite as short as the other JS answer, but it was fun to calculate the code-point of each fraction mathematically.

Test snippet

let f =
m=>n=>String.fromCharCode((4%n?8528:188)|"  130470912"[n]*1.3+~-m/-~(n-6?n==8:3))

document.body.innerText = [
  [1,2],
  [1,3], [2,3],
  [1,4], [3,4],
  [1,5], [2,5], [3,5], [4,5],
  [1,6], [5,6],
  [1,7],
  [1,8], [3,8], [5,8], [7,8],
  [1,9],
  [1,10]
].map(([m,n])=>m+"/"+n+": "+f(m)(n)).join('\n')

Old method (82 bytes):

m=>n=>String.fromCharCode((4%n?8539:189)-("0x"+"  08162b0a9"[n])+~-m/-~"3 1"[n-6])

Saved 4 bytes on this one thanks to @Arnauld.

Answer 5

SOGL V0.12, 51 bytes

5*+'æ%'⁵%"∫}ΣS“LI─{"#!“+ζ}¹"¾ŗŗŗŗŗŗ⅛ŗ⅜½⅝ŗ⅞ ⅓⅔ŗ  ¼”W

Try it Here!

uses the mapping (m + n*5)%33%22

Explanation:

5*                        multiply the 1st input by 5
  +                       add to the 2nd input
   'æ%                    modulo 33
      '⁵%                 module 22

"∫}ΣS“                    push 2153527158
      LI─                 base 11 decode that - [10 0 5 6 7 8 9 1 2]
         {     }          for each of that
          "#!“+             add 8528
               ζ            and convert to a character
                "...”     push "¾ŗŗŗŗŗŗ⅛ŗ⅜½⅝ŗ⅞ ⅓⅔ŗ  ¼", replacing ŗ with each character of the above array
                     W    and get the (m + n*5)%33%22th item from that

Answer 6

Clojure, 127 bytes

(comp{2/3\⅔ 4/5\⅘ 1/3\⅓ 1/2\½ 5/6\⅚ 1/5\⅕ 1/4\¼ 3/5\⅗ 1/7\⅐ 3/4\¾ 2/5\⅖ 1/6\⅙ 1/9\⅑ 1/8\⅛ 3/8\⅜ 1/10\⅒ 5/8\⅝ 7/8\⅞}read-string)

An anonymous function that takes input as "1/2", and returns the corresponding character.

A straight mapping from a Clojure Ratio to a fraction character. comp, and the fact that Clojure maps are functions really helped here. I needed to pass the string input through read-string to evaluate it as a ratio type, since that allows me to discard all the bloating quotes in the map. comp let me do that point-free, which was nice. The "full" code would be:

(defn to-unicode-fraction [frac]
  (get {2/3\⅔ 4/5\⅘ 1/3\⅓ 1/2\½ 5/6\⅚ 1/5\⅕ 1/4\¼ 3/5\⅗ 1/7\⅐,
        3/4\¾ 2/5\⅖ 1/6\⅙ 1/9\⅑ 1/8\⅛ 3/8\⅜ 1/10\⅒ 5/8\⅝ 7/8\⅞}
       (read-string frac)))

After looking over the other answers, I realized that this approach is pretty naïve. I'm looking for ways of improving it.

Answer 7

Swift, 106 82 bytes

Port of @Arnaulds JavaScript answer

Saved 24 bytes thanks to @Mr.Xcoder

{Array("¾⅗  ⅜    ⅞⅘½⅓¼⅕⅙⅐⅛⅑⅒⅚⅔⅝⅖")[($0*9+$1)%31]}

Try it online!

Answer 8

Charcoal, 77 48 46 bytes

Ｆ§§⪪“YW³⦄Ｅν~vγ@⊖Ｘμ～?Iw⸿Ｍ➙b℅¦⦃”,ＮＮ℅⁺℅ι×⁹⁰∨№βι⁹⁴

Try it online! Link is to verbose version of code. Edit: Saved 29 31 bytes by mapping from ASCII chars to multibyte chars. Explanation:

    “....”                  Compressed string `R,EFcGbIMDO,H J,K  Pd,L,N Q,`
   ⪪      ,                 Split on commas
           Ｎ                First input
  §                         Circularly index
            Ｎ               Second input
 §                          Circularly index
Ｆ                           "Loop over" character (loop variable `i`)
                     №βι    Count of `i`s in lowercase letters
                    ∨   ⁹⁴  Replace zero result with 94
                 ×⁹⁰        Multiply by 90
               ℅ι           Take ordinal of `i`
              ⁺             Sum
             ℅              Convert back to character
                            Implicitly print

Answer 9

Python 3, 97 bytes

lambda m,n:'½ ⅓⅔ ¼_¾ ⅕⅖⅗⅘ ⅙___⅚ ⅐ ⅛_⅜_⅝_⅞ ⅑ ⅒'.split()[n-2][m-1]

Answer 10

Python 3, 87 bytes

lambda n:'⅛⅑⅜¾⅗⅔⅚⅖⅝⅘⅞⅒½⅓¼⅕⅙⅐'[int(n[::2])%50%36%31%18]

Try it online!

or

lambda n:'⅝⅔⅜⅖__⅞¾⅗⅘_⅒½⅓¼⅕⅙⅐⅛⅑⅚'[int(n[::2])%36%27%22]

Try it online!

Answer 11

Vim Script, 67 bytes

fu!F(m,n)
if a:n==10|norm a⅒ 
else|exe'norm a'.a:m.a:n
endif
endfu

The function works with Vim's digraphs which are entered after starting them iwth ctrl-k. Apparently, the relevant ctrl-k character after the exe'norm is not rendered in the code block above.