ASCII Expansion

Question

Introduction

Every number can be represented as ASCII. For example, \$0\$ comes in ASCII is \$48\$, \$1\$ is \$49\$, and so on. Using this method of translating numbers to other numbers, it is possible to infinitely expand a number, by replacing all its digits with their ASCII values and doing the same for the result. For example, if you started with \$0\$, you would expand to \$48\$, then to \$5256\$, and so on.

Challenge

• You will be given a single digit and a number as input. You can assume the digit will be in the range \$0-9\$, or \$48-57\$ in ASCII. You can assume the digit will always be of length 1, and will be a string. The number will always be a positive integer, greater than -1. If it is 0, you do not expand at all. Other than that, there are no guarantees about its value. If, and only if your language has no method of input, you may store the input in two variables or in a list.
• You must output the ASCII expansion of the digit if you expand it \$n\$ times, n being the number that was the input. If your language has no method of output, you may store it in a variable.

Example I/O

• Digit = 0, N = 3

  Output = 53505354

• Digit = 2, N = 2

  Output = 5348

• Digit = 5, N = 0

  Output = 5

Rules

This is code-golf, so shortest answer (in bytes) wins!

Answer 1

Perl 6, 28 bytes

{($^a,*.ords.join...*)[$^b]}

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Explanation:

{                          }  # Anonymous codeblock
 (               ...*)[$^b]   # Index into an infinite list
  $^a,                        # Starting from the given number
      *                       # Where each element is
       .ords.join             # The ordinal values joined

Answer 2

05AB1E, 3 bytes

FÇJ

Takes N as first input and the digit as second.

Try it online or verify all test cases.

Explanation:

F    # Loop the (implicit) first input (N) amount of times
 Ç   #  Convert the characters in the string at the top of the stack to its unicode values
     #  (which will take the second input implicitly in the first iteration)
  J  #  Join these unicode integers together to a single string
     # (after the loop, the result is output implicitly)

Answer 3

Python 2, 55 bytes

f=lambda s,n:f(`ord(s[0])`,n-1)+f(s[1:],n)if n*s else s

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---

56 bytes

lambda s,n:eval("''.join(`ord(c)`for c in"*n+" s"+")"*n)

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Generates and evaluates monstrosities like:

''.join(`ord(c)`for c in''.join(`ord(c)`for c in''.join(`ord(c)`for c in s)))

---

56 bytes

f=lambda s,n:s*0**n or''.join(`ord(c)`for c in f(s,n-1))

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Answer 4

Jelly, 7 5 bytes

ṾOVƊ¡

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-2 bytes after a friend helped me figure out what's wrong with O

For some reason, to run all test cases with a footer, an extra byte is required for the explicit ⁹ nilad: Try it online!

Ṿ        Stringify the input
 O       and convert each character to a codepoint,
  V      then concatenate them and eval the result,
   Ɗ¡    repeated a number of times equal to the right argument.

Answer 5

Brachylog, 7 bytes

{ṫạc}ⁱ⁾

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{   }ⁱ     Repeat
 ṫ         stringifying,
  ạ        converting to a list of codepoints,
   c       and concatenating
      ⁾    a number of times equal to the last element of the input.

Answer 6

J, ¹⁴ 13 bytes

[:,":@u:~"+&3

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How it works

The "bind" operator & is commonly used to bind a constant to a dyad, so that it can be used as a monad. However, the same form can be used as a dyad: x n&v y (where n is a noun and v is a dyadic verb) or x v&n y applies monadic n&v or v&n to y repeatedly x times. Using this feature, we can design the target function like this:

x some_constant&some_dyad y
run `some_constant some_dyad y` x times
... or ...
x some_dyad&some_constant y
run `y some_dyad some_constant` x times

In this case, there is an obvious choice for the some_constant, which is 3 for 3 u: y.

And here goes the full explanation:

[:,":@u:~"+&3
           &   Apply this function x times...
      u:~   3  Convert chars of y to ASCII values (3 u:)
   ":@   "+    Convert each number back to string
[:,            Flatten the array to get a single string

Answer 7

Ruby, 30 bytes

->s,n{n.times{s=s.bytes*''};s}

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Answer 8

C# (.NET Core), 174 147 bytes

class Z{static void Main(string[] a){for(int n=int.Parse(a[1]);n-->0;){var x="";foreach(char c in a[0])x+=c-0;a[0]=x;}System.Console.Write(a[0]);}}  

Big help from Jo King.

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Ungolfed

class Z
{
   static void Main(string[] a)
    {
        int n = int.Parse(a[1]);
        for (; n-- > 0;)
        {
            var x = "";
            foreach (char c in a[0])
                x += c - 0; //c-0 gets converted to int, and then the int is 
                                   //automatically converted to a string
            a[0]=x;
        }
        System.Console.Write(a[0]);
    }
}

Answer 9

JavaScript (ES6),  43  40 bytes

Thanks to @tsh for reminding me that I didn't use currying this time :p (-3 bytes)

Takes input as (N)(digit).

n=>g=k=>n--?g(k.replace(/./g,c=>c^48)):k

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How?

This is a simple recursive function. The only trick in there is c^48. Because c is a string, we need to coerce it to an integer. We could do +c+48, but that would be 1 byte longer. Using a bitwise XOR is safe here, as \$48\$ is \$110000_2\$ and c is less than \$10000_2\$.

Answer 10

Haskell, 30 35 30 bytes

(!!).iterate(>>=show.fromEnum)

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+5 then -5 bytes from Jo King clarifying the rules and then working it into pointfree.

My first Haskell golf so I've probably done something horribly wrong. In addition to having misspelled golf, I tried to import ord without putting it in my byte count!

    .                             The composition of
     iterate                      infinitely iterating, starting with the argument,
            (>>=             )    concatenating the results of mapping
                show              finding the string representation of
                    .fromEnum     the codepoint of the argument,
(!!)                              with indexing into the resulting infinite list.

Answer 11

Japt -h, 5 bytes

VÆ=mc

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VÆ=mc   V = number of times, U = digit
VÆ      V times do: (Collects each result into an array)
  =mc     Map every digit of U to it's ASCII value, and make that the new U
-h      Take last element

Answer 12

Perl 5, 22 bytes

eval's/./ord$&/ge;'x<>

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Answer 13

J, 18 bytes

([:,3":@u:"0])@[&0

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From the J dictionary:

The phrase x f@[&0 y is equivalent to f^:x y , apply the monad f x times to y.

That is, it's a shortcut for power of ^: applied as many times as the left arg. Which explains the

(  )@[&0

part of the code. Now for what's in the parentheses:

3 u:] converts to a unicode code point, but unfortunately has infinite rank, and we want to apply it with 0 rank, hence the added "0. The code point is a number, and we convert it back to a string with format ":. Finally, we flatten , this list of strings.

Answer 14

PowerShell, 51 49 bytes

^{-2 bytes thanks to mazzy}

param($s,$n),1*$n|%{$s=-join($s|% t*y|%{+$_})};$s

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Answer 15

Python 2, 63 57 bytes

f=lambda i,n:n and f(''.join(`ord(c)`for c in i),n-1)or i

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-6 bytes due to Jonathan Allan noting that the input can be a string.

Takes input as single digit string and an integer number of repetitions.

Answer 16

Bracmat, 63 bytes

get':%?a %?n&whl'(!n+-1:~<0:?n&str$vap$((=.asc$!arg).!a):?a)&!a

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Answer 17

Haskell, 30 bytes

(!!).iterate(show.fromEnum=<<)

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Answer 18

Charcoal, 9 bytes

ＦＮ≔⭆η℅κηη

Try it online! Link is to verbose version of code. Takes N as the first input and the digit as the second input. Explanation:

ＦＮ

Repeat N times...

≔⭆η℅κη

Map each character to its ordinal and concatenate.

η

Output the final result.

Sadly ＦＮ≦℅ηη doesn't work...

Answer 19

Lua, 55 bytes

a,N=...for i=1,N do a=a:gsub('.',('').byte)end
print(a)

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Take input as arguments. Explanation is fairly obvious.

Alternative solution: Lua, 60 bytes

a,N=...for i=1,N do a=table.concat{a:byte(1,-1)}end
print(a)

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Answer 20

Lua, 67 bytes

i=io.read d=i(1)for _=1,i()do
d=d:gsub(".",string.byte)end
print(d)

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Let d be the digit we will transform (always of length 1, so we ask to read 1 character with io.read(1)). If N > 0, replace each character by its byte value. Return the digit.

Answer 21

K (oK), 13 9 bytes

-4 bytes thanks to ngn

(,/$`i$)/

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Answer 22

Wolfram Language (Mathematica), 41 bytes

""<>ToString/@ToCharacterCode@#&~Nest~##&

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Answer 23

APL (Dyalog Unicode), 16 bytes

{(∊⍕¨∘⎕UCS)⍣⍵⊢⍺}

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Left input is a one-digit string, right input is the number of iteration. Returns the string representation of the result.

How it works

{(∊⍕¨∘⎕UCS)⍣⍵⊢⍺}
{              }  Dyadic dfn, ⍺=digit string, ⍵=iteration
             ⊢⍺   Start with ⍺
 (        )⍣⍵     Repeat ⍵ times...
      ⎕UCS        Convert each char to Unicode codepoint
   ⍕¨∘            Convert each number back to string
  ∊               Enlist (flatten) all chars into one string

Answer 24

PHP, 58 bytes

for([,$a,$b]=$argv;$b--;)$a=strtr($a,range(48,57));echo$a;

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Input digit and number are two command arguments ($argv) in same order.

Comented

for(
  [,$a,$b]=$argv;  // $a is input digit and $b is number
  $b--;            // loop $b times
)
  $a=              // set $a to
    strtr(         // strtr in array mode, replaces keys with values
      $a,          // replace in $a itself
      range(48,57) // an array with keys 0...9 and values 48...57
    );
echo$a;            // at the end, output $a

Answer 25

C# (Visual C# Interactive Compiler), 55 bytes

a=>b=>{for(;b-->0;a=a.SelectMany(l=>l-0+""));return a;}

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Answer 26

Python 3, 69 60 bytes

f=lambda n,i:i and f(''.join(str(ord(c))for c in n),i-1)or n

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Answer 27

Red, 62 57 bytes

func[d n][loop n[parse d[any[change p: skip(0 + p/1)]]]d]

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Explanation:

f: func [ d n ] [                  ; a function with 2 parameters
    loop n [                       ; repeat n times
        parse d [                  ; parse the string with the following rules        
            any [                  ; one ore more 
                change p: skip     ; change any string with length 1 (a single digit)
                (0 + p/1)          ; with its representation as a number 48..57 
            ]                       
        ]
    ]
    d                              ; return the altered string
]

Answer 28

Icon, 72 63 bytes

Inspired by @xnor's Python 2 solution

procedure f(d,n)
return(*d*n<1&d)|f(48+!d,n-1)||f(d[2:0],n)
end

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Answer 29

Retina, 21 bytes

.+¶

"$+"+`.
$.(*_48*

Try it online! Takes N as the first input and the digit as the second input. Explanation:

.+¶

Delete N from the output.

"$+"+`

Repeat N times.

.
$.(*_48*

Replace each digit with its ASCII code.

Answer 30

QuadR, 8 bytes

Translation of Nahuel Fouilleul's solution. Thanks to Veskah for pointing it out.

.
⎕UCS⍵M

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. replace any character

⎕UCS with the Universal Character Set ordinal of the
⍵M Match

This is equivalent to the Dyalog APL expression '.'⎕R{⍕⎕UCS⍵.Match}⍣⎕⊢⍞. Try it online!