Print 0 to 100 without 1-9 characters

Question

Print 0 to 100 without using characters 1,2,3,4,5,6,7,8,9 in your code.

Seperator of numbers can be comma, whitespace or newline.

Shortest code wins.

Answer 1

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!

Answer 2

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.

Answer 3

GolfScript, 8/11 bytes

'e'{}/,`

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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'{}/,' '*

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Answer 4

V (vim), 30 bytes

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

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can definitely be improved.

Answer 5

K (oK), 4 bytes

Solution:

!"e"

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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

Answer 6

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\$.

Answer 7

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.

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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

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Answer 8

///, 231 166 bytes

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

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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

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

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Answer 9

GFortran, 29 bytes

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

Similar to this.

Answer 10

MathGolf, 4 bytes

♀)rn

Outputs with newline delimiter.

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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)

Answer 11

Python 3; 46 Bytes

x=True;a=x+x;b=a*a+x;print(*range(a*a*b*b+x))

Answer 12

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" }

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Answer 13

Ly, 9 5 bytes

'dR&u

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Dropped 4 chars thanks for LyricLy(!)

Prints using LF as the delimiter by generating the list of 0-100 on the stack, then using a "print the whole stack" command.

 'd     - push 100 (codepoint for "d") on the stack
   R    - use "range" command to generate the list of numbers
    &u  - print the stack as integers

Answer 14

Go, 63 bytes

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

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

Answer 15

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.

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48 bytes version

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

Using the ascii table to get those numbers.

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---

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

29 bytes

('P'..'´').map{it.minus('P')}

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Answer 16

><>, 18 bytes

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

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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

Answer 17

Pyth, 7 bytes

0S^ThhZ

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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.

Answer 18

><>, 11 bytes

lnaol:aa*(%

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Terminates by error.

Answer 19

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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Answer 20

Python 3, 33 bytes

for x in range(ord("e")):print(x)

Answer 21

Elixir, 26 bytes

for x<-0..?d,do: IO.puts x

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Answer 22

Batch, 56 bytes

@set/ax=0xb-0xa
@for /l %%b in (0,%x%,%x%00)do @echo %%b

-61 bytes for @Neil

Answer 23

Red, 36 bytes

repeat n 0 +#"e"[print n +#"a"-#"b"]

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Answer 24

C (gcc), 40 39 bytes

Saved a byte thanks to att!!!

f(i){for(i=0;i<'e';)printf("%d ",i++);}

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Answer 25

C++ gcc, 36 bytes

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main(){for(;_<'e';)cout<<_++<<'\n';}

Explanation : I globally initialized varible _ so its initial value is 0, now ascii value of e is 101 so I ran the loop till my variable _ is less than 'e', instead of incrementing it inside the for loop I used post increment while printing to save 1 byte

edit: I misread the question and thought 0 is also not allowed :)

Answer 26

APOL, 11 bytes

f(ô p(∈));ô

I would've used ⅎ instead of f, but the rules state that you have to include 0 so the fastest route was to just print 0-99 and slap 100 at the end.

Answer 27

APL, 13 bytes

0,⍳⍎'00',⍨⍕*0

*0 ⍝ Exponential of zero = 1
⍕ ⍝ Convert to symbol '1'
'00',⍨ ⍝ Append two zeros
⍎ ⍝ Convert to number 100
⍳ ⍝ Make sequence from 1 to 100
0, ⍝ Append zero to the left