Decode Polybus Square/Tap Code/Prison Code

Question

The code has a lot of names, but is very simple:

  1 2 3 4 5
1 A B C D E
2 F G H I J 
3 L M N O P
4 Q R S T U
5 V W X Y Z

A letter is coded by its coordinates, with the row first, then the column. Ex:

M = 3, 2
     V
   1 2 3 4 5
 1 A B C D E
 2 F G H I J 
>3 L M N O P
 4 Q R S T U
 5 V W X Y Z

Note the letter 'K' is missing, as a letter needs to be removed to make the alphabet fit in the grid. In some versions, the letter 'J' is removed but K is easier to remove because every line stays in order.

More info here on Wikipedia.

Challenge: Make a program that takes in a message encoded in this way, that outputs the decoded message. Input can be in any reasonable format.

23 15 31 31 34 52 34 42 31 14
23153131345234423114
etc

Are all reasonable input formats. Standard loopholes apply.

Test Cases

23 15 31 31 34 52 34 42 31 14 => HELLOWORLD
24 25 31 32 => IJLM
11 22 33 44 55 => AGNTZ

Answer 1

Python 2,  40  39 bytes

-1 thanks to kops! (Multiply before integer division)

lambda a:[chr(r*16/3+c+59)for r,c in a]

An unnamed function accepting an iterable of pairs which returns a list of characters.

Try it online!

Answer 2

J, 22 bytes

65(u:@+>&74)@+[:5&#.<:

Try it online!

• [:5&#.<: Decrement each input value <: and interpret every resulting 2-element pair as a base 5 number 5&#..
• 65...@+ Add 65 to each (to make it an ascii value), and...
• u:@+>&74 Add 1 more to each number greater than 74 +>&74 to adjust for the missing K. Finally, convert the result to a character u:@.

Answer 3

Jelly, 10 bytes

ḅ5<⁴ạƊ+60Ọ

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A completely different approach.

Explanation

ḅ5<⁴ạƊ+60Ọ   Main monadic link
             All following operations are automatically vectorized.
ḅ5           Convert from base 5
    ạƊ       Absolute difference with
  <⁴           1 if it's less than 16, else 0
      +60    Add 60
         Ọ   Convert to character

Answer 4

brainfuck, 85 bytes

,[>++++++++[<------>-],<[>+++++>>[>]+[<]<<-]>+++++++++++>>>>[<<<<+>>>>-]<<<<.>>>>>>,]

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Explanation

,  Read a character
[  While not at the end of input
  >++++++++[<------>-]  Subtract 48 to get the actual number
  ,  Read another character
  <[>+++++>>[>]+[<]<<-]  Add 5 times the number to the character and also expand the number into a group of 1s
  >+++++++++++  Add 11 to the character to turn it into a letter
  >>>>  Go to the third item in the group of 1s
  [<<<<+>>>>-]  If it's a 1 then add 1 to the character
  <<<<.  Print the character
  >>>>>>  Skip past the dirty memory cells
  ,  Read a new character
]

Answer 5

Jelly, 10 bytes

ØAḟ”Ks5⁸œị

Try it online!

-1 byte thanks to Jonathan Allan!

How it works

ØAḟ”Ks5⁸œị - Main link. Takes a list of pairs P on the left
ØA         - Yield the uppercase alphabet
  ḟ”K      - Filter out "K"
     s5    - Split into a 5x5 matrix, S
       ⁸   - Yield P
        œị - For each pair [A, B] in P,
              get the B'th element of the A'th row of S

Answer 6

brainfuck, 78 bytes

,[<<-[----->+>-<<]>>+>,<[>+>]<<[<]>>+[>+>]<<[<]>[--->>+<<]>[->+++++<]>--.>>>,]

Try it online!

Takes input for each letter as 2 ascii encoded digits.

Explanation

Essentially the formula is 65+[Row ASCII - 49]*5+[Column ASCII -49]+[1 if row >2] .

The row ASCII value is reduced by 49, but the column ASCII offset is calculated as

 65-49 = 16 = 17-2+1 plus an additional 1 if in row 3 onwards.

The approach is quite different to the existing brainfuck answer.

,[<<             Take an input for row value into cell 0 to get started then move to cell minus 2
-[----->+>-<<]   make cell minus 1 equal 255/5=51 and also subtract this number from input
>>+>,<           add 1 to row value in cell 0 and input column value into cell 1
[>+>]<<[<]       if cell 0 not 0  row input was not 2 so add 1 to column value
>>+              add 1 to row value in cell 0 so now it is 49 minus the input / note ASCII 49 is 1 
[>+>]<<[<]       if cell 0 not 0 row input was not 1 so add 1 to column value
>[--->>+<<]      add 51/3=17 to column value
>[->+++++<]      add 5 times row value to column value
>--.             subtract 2 from column value and output letter
>>>,]            skip 3 dirty cells and take next row input 

Answer 7

Haskell, 39 37 bytes

map(\(r,c)->[';'..]!!(r*5+c+r`div`3))

Try it online!

• saved 2 thanks to @cole

Answer 8

JavaScript (Node.js), 42 bytes

a=>Buffer(a.map(([y,x])=>+x+y*16/3+59))+''

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Takes an array of strings.

Answer 9

convey, 568 bytes

{:v
 1v
v#<   0
v^<<<<(<
v9     ^9
v+2v'A'^+2
"=>!}  ^*5v'Z'
v9     "=>!}
v+3v'B'^9
"=>!}  ^*6v'Y'
v9     "=>!}
v+4v'C'^9
"=>!}  ^*6
v9     ^-1v'X'
v+5v'D'"=>!}
"=>!}  ^9
v9     ^+4
v+6v'E'^*4v'W'
"=>!}  "=>!}
v7     ^9
v*3v'F'^+8
"=>!}  ^*3v'V'
v9     "=>!}
v+2    ^9
v*2v'G'^*5v'U'
"=>!}  "=>!}
v9     ^9
v+9    ^+2
v+5v'H'^*4v'T'
"=>!}  "=>!}
v6     ^9
v*4v'I'^*4
"=>!}  ^+7v'S'
v5     "=>!}
v*5v'J'^7
"=>!}  ^*6v'R'
v9     "=>!}
v*3    ^9
v+4v'L'^*4
"=>!}  ^+5v'Q'
v8     "=>!}
v*4v'M'^5
"=>!}  ^*7v'P'
v9     "=>!}
v+2    ^9
v*3v'N'^+8
"=>!}  ^*2v'O'
>>>>>>>"=>!}

And here's a picture of the factory (the gif thing kept crashing):

The basic idea is that there is a massive loop which lets one value through at a time. It goes through every letter. Note: Some things which appear to be golfable aren't due to interpreter glitches.

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I can't believe Brainfuck is beating this :( - Convey has terrible string handling, so this is the only answer where it implicitly checks every single letter...

Answer 10

Scala, 37 36 bytes

Saved 1 byte thanks to @xigoi, who suggested applying @kops's suggestion to Jonathan Allan's answer.

_.map{case(r,c)=>r*16/3+c+59 toChar}

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Now simply a port of Jonathan Allan's Python answer, which you should upvote!

Outputs a list of characters.

Answer 11

05AB1E, 13 12 bytes

<ε5β}Au'Kмsè

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-1 byte thanks to Makonede. Outputs as a list of characters

How it works

<ε5β}Au'Kмsè - Program. Takes an input of a list of pairs P
<            - Decrement to use 0-indexing
 ε  }        - Over each pair [A, B]:
  5β         -   Convert from base 5, yielding 5A+B
     Au      - Push the uppercase alphabet
       'K    - Push "K"
         м   - Remove "K" from the alphabet
          sè - Index into this modified alphabet

Answer 12

Charcoal, 14 bytes

⭆⪪⭆Ｓ⊖鲧⁻αK⍘ι⁵

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

   Ｓ            Input string
  ⭆             Map over characters and join
     ι          Current character
    ⊖           Decremented
 ⪪    ²         Split into pairs of digits
⭆               Map over pairs and join
            ι   Current pair
           ⍘ ⁵  Convert from base 5
       §        Index into
         α      Predefined variable uppercase alphabet
        ⁻       Minus
          K     Literal string `K`
                Implicitly print

Answer 13

K (ngn/k), 20 bytes

`c${x+74<x}65+5/'-1+

Try it online!

Uses @Jonah's method - don't forget to upvote him!

Answer 14

Jelly (fork), 10 bytes

’ḅ5ịØȦḟ”K¤

Try it online! (or rather, don't)

How it works

’ḅ5ịØȦḟ”K¤ - Main link. Takes a list of pairs P on the left
’          - Decrement each integer to use 0-indexing
 ḅ5        - For each pair [A, B], yield 5A+B (convert from base 5)
         ¤ - Group the previous links into a nilad:
    ØȦ     -   Shifted alphabet; Yield "BCDE...XYZA"
      ḟ”K  -   Filter out "K"
   ị       - Index into this modified alphabet

Answer 15

Excel, 42 bytes

=CONCAT(REPT(CHAR(A:A*16/3+B:B+59),A:A>0))

Another port of Johnathan Allan's Python Answer

Input is 2 columns of cells.

Answer 16

Stax, 10 bytes

▄πKñ╜╓♥F⌂F

Run and debug it

takes inputs as [column, row]. Based on Jonathan Allan's answer.

Answer 17

Befunge-93, 31 bytes

~:1+!#@_"0"-:2`\5*+~"0"-+";"+,~

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Explanation

~             Get a char
:1+!#@_       If at the end of the string, stop the execution (see if the char is -1)
"0"-          Get the row number
:2`\          If greater than 2, push a 1, otherwise a zero, swap the top two in the stack
5*+           Multiply by 5 and sum the 0/1
~"0"-+        Get the col number and sum everything
";"+,         Sum 59 to get the ASCII value of the char
~             Remove the space

Answer 18

R, 37 bytes

function(a)intToUtf8(a$x*16/3+a$y+59)

Try it online!

Port of Jonathan Allan's Python answer.

Takes input as a data frame with first digit in first column and second digit in second column.

Answer 19

Ruby, 34 bytes (31 bytes with Ruby 2.7)

->x{x.map{|i,j|(i*16/3+j+59).chr}}

A direct port of Johnathan Allan's Python answer.

This lambda accepts an array of pairs and returns an array of characters.

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->x{x.map{(_1*16/3+_2+59).chr}} is written in 31 bytes and also works fine, but it requires Ruby 2.7 (not yet supported by TIO).

Answer 20

PowerShell 5+, 47 bytes

Tribute to Metal gear 2: Solid Snake

-join($args|%{(';'..'_'-ne'K')[5*$_[0]+$_[1]]})

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Answer 21

Perl 5 -p, 34 bytes

s/(.)(.)/(A..J,L..Z)[5*$1-6+$2]/ge

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Answer 22

C (gcc), 75 bytes

main(int a,char**b){for(b++;**b;++*b)putchar(16+(a=**b-49)*5+(a>1)+*++*b);}

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In the end, this is pretty straightforward and just uses C's willingness to be fluid about treating characters as integers. It runs through the characters in the first commandline argument, then uses a golfed version of this logic to map each one to the desired character.

'A' + (char1-'1')*5 + (char1-'1'>1) + (char2-'1')

The only trickery here is using the using the storage provided by the arguments to the main() call to avoid having to declare local variables. And the operator precedence in C means this jumble of punctuation,

+*++*b

does a number of things with just a few characters. That fragment, increments the pointer to the "current character" *b, then dereferences that location as a single character, implicitly converts it to an integer and adds it to the rest of the expression. The fact that +*++*, which looks more like Morse Code than C code, does something useful is pretty much why I decided to post this entry. :)

Answer 23

APL (Dyalog Unicode), 33 15 bytes

Saved 18 bytes thanks to Razetime! (they also pointed out the answer was invalid)

⌷∘(5 5⍴⎕A~'K')¨

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Takes a vector of pairs of numbers.

⌷∘(5 5⍴⎕A~'K')¨
               ¨ ⍝ For each pair in the input
⌷                ⍝ Index into:
       ⎕A        ⍝ Capital letters
          ~'K'   ⍝ Without 'K'
   5 5⍴          ⍝ Reshaped into a 5x5 matrix

Answer 24

Retina 0.8.2, 37 bytes

T` d`__9\EKPU` |\b\d
+T`dL`__1-7L`.\d

Try it online! Link includes test cases. Explanation:

T` d`__9\EKPU` |\b\d

This does two things; it deletes all the spaces, and it replaces the first of each pair of digits with the characters 9EKPU respectively, these being the characters in the translation string below that precede the letters from the first column of the code.

+T`dL`__1-7L`.\d

Repeatedly decrement the remaining digits and increment the preceding letter until the digits become zero, at which point they are deleted. Note that on the first pass the "letter" might actually be a 9, but this works because on the first pass of this loop all the "letters" will be paired with a digit.

Answer 25

R, 62 bytes

function(x)for(i in x)cat(letters[(y=(i-1)%*%c(5,1)+1)+(y>9)])

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Aaargh! I clicked submit and then saw pajonk's much shorter answer waiting for me! Upvote that one (and Jonathan Allan's one that it's based-on)!

Answer 26

GolfScript, 22 bytes

~]2/{~\16*3/+59+}/]''+

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Another port of @Jonathan Allen’s answer, and my first time trying out GolfScript. Input is just a series of numbers, separated by spaces, but you can separate number pairs with extra spaces for readability.

Answer 27

PowerShell, 60 bytes

($a|%{[char][int][math]::floor($_[0]*16/3+$_[1]+59)})-join''

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Thanks to @JonathanAllan