70
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Print integers 0 to 100 (inclusive) without using characters 123456789 in your code.

Separator of numbers can be comma or white space (by default <blank>, <horizontal tabulator>, <newline>, <carriage return>, <form feed> or <vertical tabulator>).

Shortest code wins.

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7
  • 16
    \$\begingroup\$ Many tricks are made possible by allowing 0. Which is what makes this challenge interesting, IMO. \$\endgroup\$
    – Arnauld
    Feb 23, 2021 at 17:08
  • 3
    \$\begingroup\$ I thought "do X without Y" questions weren't allowed anymore. \$\endgroup\$
    – Purple P
    Feb 24, 2021 at 3:34
  • 5
    \$\begingroup\$ @PurpleP They're allowed, but discouraged. Interesting ones are fine. \$\endgroup\$ Feb 25, 2021 at 0:06
  • 16
    \$\begingroup\$ Is there a requirement to stop printing at 100? \$\endgroup\$
    – spuck
    Feb 25, 2021 at 16:44
  • 2
    \$\begingroup\$ Can I use non-ASCII encoding? \$\endgroup\$
    – user100411
    Oct 17, 2021 at 20:36

170 Answers 170

1
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C# (.NET Core), 56 bytes

for(int i=0;i<=(0xb0e-0xaaa);i++){Console.WriteLine(i);}

Try it online!

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3
  • \$\begingroup\$ -1 byte by moving the i++ into Console.WriteLine(). Try it online! \$\endgroup\$ Feb 23, 2021 at 18:22
  • 1
    \$\begingroup\$ 52 bytes by getting rid of the curly braces and switching to < rather than <=, along with the byte save from my last comment Try it online! \$\endgroup\$ Feb 23, 2021 at 18:33
  • \$\begingroup\$ Also, char has an implicit conversion to int, so you can replace the magic hex codes with 'e' for 42 bytes Try it online!, though upon looking it seems the C answer by Noodle got to that idea first. \$\endgroup\$ Feb 23, 2021 at 18:37
1
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Nim, 26 bytes

for i in 0..'d'.ord:echo i

Try it online!

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1
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Charcoal, 7 5 bytes

IE℅eι

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

   e    Literal string `e`
  ℅     ASCII code i.e. 101
 E      Map over implicit range
    ι   Current value
I       Cast to string
        Implicitly print
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1
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Tcl, 46 bytes

set i 0
while \$i<[incr u]0$u {puts $i
incr i}

Try it online!

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1
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SimpleTemplate 0.84, 45 bytes

This was fun, but quite difficult.

The code outputs all numbers from 0 to 100, with a trailing newline:

{@setC 0}{@for_ from"   "to"m"}{@echolC}{@incC}

Due to bugs in the compiler, the tab character (inside {@for_ from" "to"m"}) MUST be a real tab.


Ungolfed

This version should be easier to read, despite being functionally the same:

{@set counter 0}
{@for i from "  " to "m"}
    {@echo counter, EOL}
    {@inc counter}
{@/}

Closing the {@for [...]} is optional, but left here for the cleanest code possible.


You can try this on https://ideone.com/tLsDFn

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1
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Stax, 5 bytes

AJ^rJ

Run and debug it

AJ^   10 squared plus 1 (101)
   r  range from 0..n-1
    J join with spaces
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1
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Elixir, 42 28 bytes

(?b-?a)..?d|>Enum.join("\n")

Try it online!

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1
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Japt, 2 bytes

Try it online!

This outputs a list of numbers from 0 to 100 separated by commas.

How it works

Lò
L   -Number 100 
 ò  -Creates an inclusive range from 0 to L, and return it in the output
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1
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C++, 77 Bytes

#import<iostream>
main(){for(int i=0;i<=int('d');++i)  std::cout<<i<<" ";}

Here, I've used the ASCII value and ran the loop and printed the value. Simple!

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1
  • \$\begingroup\$ This can definitely be golfed a bit, like by removing unnecessary whitespace and newlines. Make sure to read our tips questions if you want some hints! \$\endgroup\$ Mar 8, 2021 at 18:14
1
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TeX, 57 bytes

\newcount~\loop\advance~`^^A\the~ \ifnum~<`^^%\repeat\bye

Makes uses of these two tricks:

  • \$\rm\TeX\$'s preprocessor runs through the code and replaces any instance of two consecutive superscript (category code 7) characters followed by a character token, and adds/subtracts 64 from its ascii code, hence ^^A becomes NUL.
  • \$\rm\TeX\$ has a `backtick notation' of inputting numbers that reads the following character's ascii code instead.
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1
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GolfScript, 8/11 bytes

'e'{}/,`

Try it online! - 8 Bytes

Makes an array of values of e (101) elements, starting at 0, then formats with spaces. The format also has brackets at either end of the output, so it may not be valid. If not, they can be removed with 3 more bytes:

'e'{}/,' '*

Try it online! - 11 Bytes

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1
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V (vim), 30 bytes

i00<esc><C-a>hxpi0<esc>0"aDi0<esc>qqYp<C-a>q@a@qdd

Try it online!

can definitely be improved.

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1
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K (oK), 4 bytes

Solution:

!"e"

Try it online!

Explanation:

Scrolling through the other solutions tells me I wasn't as novel as I hoped when I came up with this.

!"e" / the solution
 "e" / ASCII 101
!    / til (i.e. range 0..n-1
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1
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Barrel, 9 bytes

#d(n+¶)n

Explanation:

#        // as many times as...
 d       // ...the ASCII value of 'd' (100)...
  (   )  // Create a single of instruction for the loop
   n     // print the accumulator of a number
    +    // increase the accumulator
     ¶   // print a newline
       n // print the final number

The final n is necessary because the loop only prints the numbers 0 to 99.

I could've shaved off 2 bytes by doing #e(n+¶, which would have used the ASCII value of 'e' (101) and also utilized the self-closing properties of the () instruction, but I had already assigned e to be the value of the mathematical constant \$e\$.

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1
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Pxem, 21 20 bytes (filename) + 0 bytes (content) = 23 21 20 bytes, requires unprintable character.

  • Filename (escaped unprintable): \001.r.-.z.c.n,.o\001.+.ce.a
  • Content: empty.

Try it online!

With comments

XX.z
# push 1; push int(rand()*pop)
## NOTE rand() outputs 0<=n<1
## NOTE assuming NUL cannot be used for filename
.a\001.rXX.z
# while size<2 || pop!=pop; do
.a.zXX.z
  # dup; print pop; push comma; putc pop
  .a.c.n,.oXX.z
  # push 1; push pop+pop; dup; push 101
  .a\001.+.ce
# done
.a.a

Pxem, 3 bytes (filename) + 20 bytes (content) = 23 bytes, requires unprintable character.

  • Filename: e.e
  • Content (some unprintables are escaped): .c.w\001.-.e.+.n .o.d.a

With comment

e.eXX.z # push 101; call content
.a
XX.z
# dup; while pop!=0; do
.a.c.wXX.z
  # push 1; push abs(pop-pop); call content (* result of final stack will be pushed to original *)
  .a\001.-.eXX.z
  # push pop+pop; print pop; push space; print pop; return
  .a.+.n .o.dXX.z
# done; (* implicit return *)
.a.a

Try it online! (with pxem.posixism)

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1
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///, 231 166 bytes

/\\\\\/\//\\\/\/\/\/\/\/\/\/\/\/\/\///\/\\\/\/\//\\\/\
\/\\\\\\\/\
\\\/\/\\\\\\\/\\\/\\\/\\\\\\\\\\\/\/\/\\\/\\\/\\\\\\\\\\\/\/\\\\\\\/\
\\\/\/\/\\\/\///\\\\\\\\\\\//

Try it online!

This was really fun to make. Sadly, there is a single newline. Using a backslash instead breaks everything, and I don't really want to figure out where everything is and fix it.

Update: I remade it from the ground up, it is now much smaller, and works with only slashes. Unfortunately, the challenge specifies commas and whitespace seperators only, so only slashes is not allowed.

Slashes only:

///, 182 bytes

/\\\\\/\//\\\/\/\/\/\/\/\/\/\/\/\/\///\/\\\/\/\//\\\/\\\\\/\\\\\\\/\\\\\\\\\\\/\/\\\\\\\/\\\/\\\/\\\\\\\\\\\/\/\/\\\/\\\/\\\\\\\\\\\/\/\\\\\\\/\\\\\\\\\\\/\/\/\\\/\//\\/\\\\\\\\\\\//

Try it online!

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2
  • \$\begingroup\$ Looks like the output starts with 1, while should start with 0. \$\endgroup\$
    – manatwork
    Feb 28, 2021 at 17:39
  • \$\begingroup\$ @manatwork ill work on that ;) \$\endgroup\$ Mar 1, 2021 at 15:26
1
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GFortran, 29 bytes

try it online!

print*,(j,j=0,ichar('d'))
end

Similar to this.

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1
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MathGolf, 4 bytes

♀)rn

Outputs with newline delimiter.

Try it online.

Explanation:

♀     # Push 100
 )    # Increment it to 101
  r   # Pop and push a list in the range [0,101)
   n  # Pop and join it by newlines
      # (after which the entire stack is output implicitly)
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1
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Python 3; 46 Bytes

x=True;a=x+x;b=a*a+x;print(*range(a*a*b*b+x))
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0
1
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ErrLess, 23 25 17 bytes

Thanks to Jo King for saving 8 bytes

0Z@#@'d<l+[.a?l-z

Explanation

0   { Add zero to the stack: (x) }
Z   { Set a "checkpoint" to jump back to later }
@#@ { Output as number & Duplicate: (x x) }
'd< { x < d - true -> -1; false -> 0? (x x<d) }
l   { Get the length of the top element (-1 for integers): (x x<d -1) }
+   { Add: (x [-2 or -1]) }
[   { Skip backwards that many instructions (skip forwards 1 or 2): (x) }
.   { Halt }
a?  { Push 10 and print (print newline) }
l-  { Increment: (x--1) }
z   { Go to "checkpoint" }

Try it online!

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0
1
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Go, 63 bytes

package main
func main(){for i:=' ';i<'';i++{println(i-' ')}}

Attempt This Online!

Upper bound for the loop has value 133 NEXT LINE (NEL). Separator is newlines. Prints to STDERR.

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1
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Kotlin, 48 46 bytes

('P'..'´').joinToString(","){""+it.minus('P')}

Saving two bytes by using other chars from the ascii table that only takes one instead of two bytes.

Try it online!

48 bytes version

('\n'..'n').joinToString(","){""+it.minus('\n')}

Using the ascii table to get those numbers.

Try it online!


When brackets are allowed at the start and end then this is smaller:

29 bytes

('P'..'´').map{it.minus('P')}

Try it online!

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1
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><>, 18 bytes

0::naoaa*(?!;ba-+!

Try it online!

Explanation

0                   Initialize stack with 0
 ::                 Duplicate the top of the stack twice, once for printing and once for comparing
   n                Pop the top of the stack, and print as a number
    ao              Push 0xa to the stack, and pop it to print as a char
      aa*           Push 100 (10*10) onto the stack
         (          Pop the top two values of the stack, and compare if one is less than the other
          ?!;       If not, halt execution, else...
             ba-    Push 1 (11-10) onto the stack
                +   Add the top two values on the stack
                 !  Skip the next instruction
                    IP Moves back to the 0
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1
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Pyth, 7 bytes

0S^ThhZ

Try it online!

Yay! First answer using an actual golfing languages. Since I’m new to Pyth, I’m assuming this can be optimized further :)

Edit: I misread the problem :/ so +1 byte. And guess what? Somebody made a 4 byte answer in pyth.

Explanation:

0         Zero
 S        In this case it makes a list from 1 to a number
  ^       Exponent of…
   T      Ten to the…
    hh    Increase the following number by two (one for each h)
     Z    Zero (now two after being increased)
          So basically push 0 then make a list from 1 to ten squared.
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1
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Knight, 37 36 29 21 bytes

-8 bytes thanks to @Razetime for reminding me about the ASCII function, which made my entire coercion thing useless

;O=a 0W>A'd'aO=a++0Ta

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1
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GolfScript, 6 bytes

'e'),`

Try it online!

Similar to the other GolfScript answer, the output has brackets, which can be removed with the following 7-byte answer:

GolfScript, 7 bytes

'e'),n*

Try it online!

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0
1
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Thunno, \$ 4 \log_{256}(96) \approx \$ 3.29 bytes

'eOR

Attempt This Online!

Outputs [[0, 1, 2, ..., 99 100]].

Thunno j, \$ 6 \log_{256}(96) \approx \$ 4.94 bytes

'eORAJ

Attempt This Online!

Outputs 0 1 2 ... 99 100.

Explanation

'e      # Push 'e'
  O     # Get the ordinal, 101
        # (this is wrapped in a list)
   R    # Push range(0, 101)
        # (again wrapped in a list)
    AJ  # Get the first item of the list
        # j flag joins by spaces
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1
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Pyt, 7 bytes

00⁺⁺ᴇŘÁ

Try it online!

0           push 0
 0⁺⁺        push 0 and increment twice
    ᴇ       10^2
     Ř      push Řange (0,1,...,100)
      Á     push contents of Árray to stack; implicit print
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1
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Hebigo, 25 bytes

print: : :* range:ord:"e"
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1
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Pascal, 94 B

This full program complies with ISO standards 7185 (“Standard Pascal”) and 10206 (“Extended Pascal”). The statement write(integerExpression) is automatically expanded to write(integerExpression:integerDefaultMinimumWidth). The value of integerDefaultMinimumWidth is implementation-defined. The following code requires that integerDefaultMinimumWidth ≥ 4 to ensure adequate number separation.

program p(output);var n:integer;begin n:=succ(succ(0));for n:=0 to sqr(n*n*n+n)do write(n)end.

70 B: Provided that the character 'd' has an (implementation-defined) ordinal value of 100 (like in ASCII) you can write:

program p(output);var c:char;begin for c:=''to'd'do write(ord(c))end.

Note, the first character literal '' is chr(0). It is assumed that '' does not denote an end-of-line character. Pascal specifically bans multi-line string/char literals.

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