Convert a string of ASCII characters to base-3 equivalents

Question

Task similar to this one. Take a string of ASCII characters and convert it to base-3 equivalents separated by a space.

For example:

Hello, world!

Should be converted to

02200 10202 11000 11000 11010 01122 01012 11102 11010 11020 11000 10201 01020

The output should only contain numbers with the same number of digits. Which separator to use is up to you.

This is a code-golf challenge so the shortest solution wins.

Answer 1

MATL, 3 bytes

3YA

Uses newline as separator. Try it online!

Explanation (by @Sundar R)

YA is a single function, corresponding to Matlab's dec2base (convert from decimal to some base). By default it takes two arguments: an array with the numbers to convert and the base to convert to. Since only the second argument (3) is specified here, the input is implicitly used as the first argument.

The cool part is that this is automatically applied to the whole input (which is a char array), the results collected into a char matrix, and when things don't line up, 0's are automatically prepended to make them have the same width.

Answer 2

Python 3,  70 66  62? 66 bytes

for c in input():print(*[ord(c)//d%3for d in b'Q    '],sep='')

(Code contains unprintable bytes inside the b'...')

A full program accepting input from STDIN that prints the result using a newline separator (with a trailing newline).

Try it online!

---

...but 66 if we need to handle multi-line strings (since input() only reads one line:

lambda s:[print(*[ord(c)//d%3for d in b'Q   '],sep='')for c in s]

An unnamed function that prints the answer.

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Answer 3

Jelly, 9 bytes

^{All-ASCII jelly code to boot...}

Ob3Uz0ZUK

A full program accepting a string that prints the result.

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How?

Ob3Uz0ZUK - Main Link: list of characters
O         - ordinals
 b3       - to base three
   U      - upend (reverse each)
    z0    - transpose with filler of zero
      Z   - transpose
       U  - upend (reverse each)
        K - join with spaces
          - implicit, smashing print

Answer 4

Charcoal, 14 bytes

⪫ＥＳ⭆◧⍘℅鳦⁵Σλ 

Try it online! Link is to verbose version of code. Explanation:

  Ｓ             Input string
 Ｅ              Map over characters
       ι        Current character
      ℅         ASCII code
     ⍘  ³       Convert to base 3
    ◧     ⁵     Left pad to length 5
   ⭆            Map over characters and join
            λ   Inner character
           Σ    Numeric value if any
⪫               Join with spaces
                Implicitly print

12 bytes to print as a list:

ＵＢ0←Ｅ⮌Ｓ⮌⍘℅ι³

Try it online! Link is to verbose version of code. Explanation:

ＵＢ0

Set output to zero fill.

←Ｅ⮌Ｓ⮌

Output the strings upside-down and reversed (which right-justifies them) given by...

⍘℅ι³

... converting the ASCII codes to base 3.

Answer 5

Python 3, 71 bytes

import numpy
for i in input():print(numpy.base_repr(ord(i),3).zfill(5))

Try it online!

-30 thanks to Jonathan Allan, I'm lazy to golf another code

Prints numbers separated by newlines

If you really need space as a separator then here is the code:

Python 3, 176 bytes

import numpy
m=input()
y=m[:-1]
u=lambda x:"0"*(5-len(x))+x
for i in y:print(u(str(numpy.base_repr(ord(i),base=3))),end = " ")
print(u(str(numpy.base_repr(ord(m[-1]),base=3))))

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Answer 6

Factor, 38 34 bytes

[ [ 3 >base "%05s "printf ] each ]

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• [ ... ] each For each code point in the input string,
• 3 >base convert it to base 3
• "%05s "printf and print it with leading zeros to a length of five, followed by a space.

Answer 7

Vyxal, 9 bytes

C3τvṅ5∆Z⁋

Uses newline as the separator.

C3τvṅ5∆Z⁋
C            Convert to ASCII codepoint
 3τ          Convert each character to base 3
   vṅ        Join by nothing, because to-base returns a list of digits
     5∆Z     Pad with 5 zeros (because that's the max length of an ASCII char)
        ⁋    Join by newlines

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It will probably get 9 bytes soon, because of a bug (to-base doesn't support vectorization)
Lyxal epic speed bug fix done

Actually, 8 bytes with flag, but j is very buggy

Answer 8

R, 63 59 bytes

function(s)write(rep(utf8ToInt(s),e=5)%/%3^(4:0)%%3,1,,,"")

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Nothing clever, but a straightforward implementation in R is shorter than anything else that I can think of so far...
To avoid the penalty of defining a base-conversion function to use on each character, we repeat each character value 5 times, and then use vectorized integer division (%/%) and modulo (%%) to calculate all the base-3 digits. Then, by lucky co-incidence, the write function splits its output data into a column width of 5 by default.

Answer 9

Javascript (Node.js), 62 59 bytes

f=

x=>[...x].map(t=>(t.charCodeAt()+243).toString(3).slice(1))

console.log((f+'').length);
console.log(f("Hello World!")+'');

Try it online!

-3 thanks to @tsh

Answer 10

BQN, 20 bytes

{'0'+3|⌊𝕩÷3⋆⌽↕5}¨-⟜@

Anonymous function that takes a string and returns a list of strings. Run it online!

If a list of lists of digits is an acceptable output format, then it's 16 bytes:

{3|⌊𝕩÷3⋆⌽↕5}¨-⟜@

Explanation

{'0'+3|⌊𝕩÷3⋆⌽↕5}¨-⟜@
                  -⟜@  Subtract '\0' from each character, converting to list of charcodes
{               }¨      Map this function to each charcode:
              ↕5          Range(5) 
            ⌽            Reverse
          3⋆              3 to the power of each
        𝕩÷                Divide the argument by each
       ⌊                  Floor
     3|                   Mod 3
 '0'+                     Convert digits to characters by adding '0'

Answer 11

GeoGebra, 74 50 bytes

Zip(Take(ToBase(a,3),2,6),a,TextToUnicode("")+243)

Insert string input between the "". (I guess this is allowed) Output is a list of strings padded to 5 digits. (This is also allowed by OP)

Uses the " add \$3^5\$ " idea from @DLosc's answer.

Try It On GeoGebra!

Answer 12

Burlesque, 14 bytes

)**m{3B!5'0P[}

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)**     # Map ord
m{      # Map
 3B!    # To base 3
 5'0P[  # Pad to length 5 using "0"s
}

Answer 13

APL+WIN, 22 bytes

Prompts for ASCII string. Index origin = 0

(⍕¨(⊂5⍴3)⊤¨⎕av⍳⎕)~¨' '

Most of the bytes are taken by formatting the output.

Answer 14

Husk, 11 bytes

m(↑_5ṁsΘB3c

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5 bytes (m(B3c) to calculate the base-3 representation of each ASCII character, but then 6 more bytes to format the output: prepend a zero (Θ), get string representation of digits (s), flatten together (ṁ), and take the last 5 elements (↑_5).

If it's acceptable for the base-3 digits to be space-separated (and the groups to be newline-separated), then we could do it in 10 bytes: m(w↑_5ΘB3c.
If we insist on space-separating the non-separated base-3 digits (exactly as in the example), it'll cost 12 bytes: wm(↑_5ṁsΘB3c.

Answer 15

Haskell, 72 64 62 bytes

unwords.map(\c->show.(`mod`3).div(fromEnum c)=<<[81,27,9,3,1])

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• Saved 8+2 Bytes thanks to @ovs suggestions.

Couldn't find any golfy way for the list of powers of 3: tried (3^)<$>[4,3..0] with no gains.

Answer 16

C (gcc), 72 bytes

i;f(char*s){for(;*s;++s,putchar(32))for(i=243;i/=3;)putchar(48+*s/i%3);}

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Answer 17

Pip, 15 bytes

S3E5+A_TB3MaJ:s

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Or, using the -s flag to do the output formatting, 12 bytes:

S3E5+A_TB3Ma

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Explanation

Pip doesn't have any padding builtins as of this writing, so we do the following instead: Add \$3^5\$ to each codepoint; convert to base 3, resulting in a 6-digit number that always has a leading 1; and slice off the first digit.

S3E5+A_TB3MaJ:s
           a     ; First command-line argument
          M      ; Map this function to each of its characters:
     A_          ;   ASCII code of the character
 3E5+            ;   Add 3 to the 5th power
       TB3       ;   Convert to base 3
S                ;   Suffix containing all but the first character
            J:   ; Join the resulting list on
              s  ; Space

Answer 18

05AB1E (legacy), 9 bytes

Ç3Bí0ζøí»

Output is newline delimited.

Try it online.

Explanation:

Uses the legacy version of 05AB1E because it can zip/transpose on list of strings, whereas the new 05AB1E version requires a matrix of characters. This would be 10 bytes in the new version by replacing B with в and adding a J (join) before the ». If any amount of leading 0s is valid, this could be 9 bytes as well in the new version of 05AB1E with Ç3BZ°+€¦» or Ç3BZj»ð0:.

Ç          # Convert the (implicit) input-string to a list of codepoint integers
 3B        # Convert each integer to a base-3 string
   í       # Reverse each string
     ζ     # Zip/transpose; swapping rows/columns,
    0      # using a 0 as trailing filler for unequal length rows
      ø    # Zip/transpose back
       í   # Reverse each string back
        »  # Join the strings with newline delimiter
           # (after which the result is output implicitly)

Answer 19

Java (JDK), 80 bytes

s->s.chars().forEach(c->System.out.printf("%05d ",new Long(Long.toString(c,3))))

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• -4 bytes thanks to swpalmer.

Answer 20

APL, 21 Chars, 39 Bytes

{∊6↑¨,⌿⍕¨(5⍴3)⊤⎕UCS⍵}

(5⍴3)⊤⎕UCS⍵ converts input to base 3 ASCII code in five digits.

,⌿⍕¨ converts the digits to characters and joins them together.

6↑¨ pads space in between (also results in a trailing space) and ∊ builds the final output.

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Answer 21

Ruby, 35 bytes

->s{s.bytes.map{"%05d"%_1.to_s(3)}}

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Answer 22

J-uby, 30 bytes

:bytes|:*&(~:to_s&3|:%&"%05d")

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Explanation

:bytes |        # Get bytes, then
:* & (          # map with...
  ~:to_s & 3 |  #   convert to base 3
  :% & "%05d"   #   format to 5 digits
)