Print 0 to 100 without 1-9 characters

Question

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.

Answer 1

v³, 96 93 bytes

^+++(###....+###....+++<..#+...-....###+.@#+...$)+).>+++.$#+...^##=.+###.-#+....+)<++(-+##++>

Unwrapped:

        ^ + + +
        ( # # #
        . . . .
        + # # #
. . . . + + + < . . # + . . . -
. . . . # # # + . @ # + . . . $
) + ) . > + + + . $ # + . . . ^
# # = . + # # # . - # + . . . .
        + ) < +
        + ( - +
        # # + +
        > . . .

I'm not able to provide a direct link, but here you should be able to fork the project and replace the script.txt with either of the above scripts.

Answer 2

Perl 5 (ppencode-compatible), 64 bytes

You didn't clarify that I must separate each with exactly one character, so here it is mine.

print length uc xor s qq q xor print while length ne ord qw q eq

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Explained

   # print(length) did not work for zero as $_ is not defined at then
   print length uc xor
   s qq q xor
   # delimiter
   print
while
   # equals to: length ne 101
   length ne ord qw q eq

Answer 3

Befunge-93, 16 13 bytes

Thanks to @Cinaski for saving 3 bytes with \! instead of "ba"-.

:.\!+:"d"`#@_

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Uses \! to NOT the 0 at the bottom of the stack and uses that to increment the loop, then tests if the counter is greater than d to end. Certainly not the shortest answer, but this is my first golf challenge, and I wanted to practice Befunge, which I decided to pick up yesterday. This is also my first time trying a stack-based language, and I'm having a lot of fun with it.

Answer 4

Wolfram Language (Mathematica), 15 14 11 bytes

=Range[0,LL

-1 byte from Imanton1

Mathematica interprets the = prefix as a call to Wolfram Alpha (auto-converting it to the orange glyph seen below), which in turn interprets "LL" as a Roman numeral for 100. I used "LL" because this doesn't work with the shorter "C".

Answer 5

Python 3 20 Bytes

print(*range(*b'e'))

How it works? Basically, doing *b'char' is equivalent to ord('char'), and in this case ord('e') is equal to 101 ; Lets re-create the ord() function!

Ord Function Recreation (Not the answer! Just a demonstration on how ord() works)

ord=lambda x:(int(*bytes(x, 'ascii')))

As you can see it works! You can test this yourself here.

Python 3 25 Bytes

print(*range(0xa*0xa-~0))

How it works? 0xa = 10, ~0 = -1, -~0 = 1 (equivalent to -1*-1)

Answer 6

Javascript, 48 bytes

for(a of Array("e".charCodeAt()).keys())alert(a)

Can be shorter, if you allow in reverse (36 chars)

for(i="e".charCodeAt();--i;)alert(i)

Answer 7

Python, 25 23 bytes

print(*range(ord('e')))

-2 by Steffan, remove first parameter (0) from call to range

Answer 8

APL (Dyalog), 19 bytes

⎕←0,⍳(+/⍳≢⍬⍬⍬⍬)*≢⍬⍬

Try it here!

≢⍬⍬⍬⍬ ⍝ This evaluates to 4
≢⍬⍬ ⍝ This evaluates to 2

⍳≢⍬⍬⍬⍬ ⍝ Evaluates to 1 2 3 4
(+/⍳≢⍬⍬⍬⍬) ⍝ Sums up previous list, 1+2+3+4 = 10
(+/⍳≢⍬⍬⍬⍬)*≢⍬⍬ ⍝ Exponentiates previous result by 2
⍳(+/⍳≢⍬⍬⍬⍬)*≢⍬⍬ ⍝ Generates 1 2 ... 100
0,⍳(+/⍳≢⍬⍬⍬⍬)*≢⍬⍬ ⍝ Appends 0 to front

Answer 9

Lua (34 30 bytes)

for i=0,0xA*0xA do print(i)end

Answer 10

AWK, 26 bytes

{for(;a<=0xa*0xa;)$a=a++}a

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Thanks to xnor for pointing out a brainfart (since fixed) in the original

This works by using 0xa*0xa to compute 100, then assigns each positional variable to it's own sequential number. Then the a without a code block (evaluates as truthy since a is 100) prints all the positional arguments separated by a space.

To be honest, I'm not 100% sure why the 0 prints but it does. :)

Answer 11

MATLAB, 13 bytes

0:double('d')

The ASCII code for lowercase d is 100, so convert to a double and go from 0 in intervals of 1 with ":"

Answer 12

Wolfram Language (Mathematica), 20 bytes

The only real golfing opportunity for this question in the Wolfram language is to encode the number 100 with as few bytes as possible. There is only one real-valued constant symbol in the Wolfram language with a one byte name, namely E.

I thus looked for combinations of binary operations that were near 100. (E+E)^E is about 99.73, so adding E/E will give a suitable endpoint.

Range[0,(E+E)^E+E/E]

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Answer 13

Rust, 41 40 39 35 bytes

^{Thanks to Redwolf for -1 byte, Unrelated String for -4}

||for i in 0..b'e'{print!("{} ",i)}

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---

39 bytes

||(0..b'e').all(|i|print!("{} ",i)==())

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---

40 bytes

||(0..b'e').for_each(|i|print!("{} ",i))

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Answer 14

Desmos, 25 bytes

o=0
o->[0...eeeln(eeeee)]

Not sure if this is an acceptable form of output but it's still an interesting answer imo. Click the right arrow (->) to run.

Try It On Desmos!

This takes advantage of Desmos's implicit rounding with list ranges, which will always round both start and end numbers to the nearest integer. In this case, eeeln(eeeee) is mathematically equivalent to \$e^3\cdot5\approx100.42768\$ (\$e\approx2.71828\$ is Euler's number), which rounds down to 100.

If not acceptable, then here's an alternative version that might be more acceptable:

31 bytes

l=[0...eeeln(eeeee)]
(l,0)
${l}

Paste first two equations into Desmos, and label the list of points (l,0) as ${l}. Move the viewport to the right to view more numbers.

Try It On Desmos!

Answer 15

Powershell, 18 bytes

0..[byte][char]'d'

Answer 16

jq, 20 characters

range("e"|explode[])

Sample run:

bash-5.0$ jq -n 'range("e"|explode[])' | head
0
1
2
3
4
5
6
7
8
9

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Answer 17

APL(Dyalog Unicode), 8 bytes ^SBCS

⍳⎕UCS'e'

Try it on APLgolf!

A tradfn submission which prints with space separator.

Answer 18

Julia 1.0, 19 bytes

println.(0:0xA*0xA)

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Answer 19

bc, 17 bytes

for(;i<=A*A;i++)i

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Answer 20

K (ngn/k), 6 bytes

!0+"e"

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Uses 0+ to convert "e" to an integer, then takes the range from 0 up to, but not including, that value (101).

Answer 21

Rattle, 14 bytes

i+R`c0c0$[+i]~

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Explanation

i                    prints the value at the top of the stack (0) 
 +                   adds 1 to the value on top of the stack (which was 0, is now 1.0)
  R`                 reformats the top of the stack with the arg ` (the value at the top of the stack).
                        since ` = 1, it reformats the top of the stack as the integer 1
    c0c0             concatenates the value in storage at the current pointer (=0) to the top of the 
                        stack twice, resulting in "100"
        $            swaps the value on top of the stack (100) with the value in storage at the 
                        current pointer (0)
         [ .... ]~   loop structure: loops ~ times, where ~ = value_in_storage_at_pointer = 100
            +        adds one to the value on top of the stack
             i       prints the top of the stack as an integer

Note: the above code is based on version 1.0.* of Rattle. With the newest update (1.1.0), the code could be shortened to the following snippet (12 bytes) because the addition operator will now keep the top of the stack the same type (in this case, an integer) if possible.

i+c0c0$[+i]~

Answer 22

Bash, 26 bytes

eval echo {$[x++]..${x}00}

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(Previously)

Bash, 28 bytes

eval echo {$((++x))..${x}00}

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Answer 23

Vim, 21 bytes

qqYP<C-x>qi0<esc><C-a>a00@q<esc>Yxx@0

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Explanation

qq       q                             Record macro q:
  Y                                     Yank the current line
   P                                    Paste a copy of it on the line above
    <C-x>                               Decrement the number under the cursor
          i0<esc>                      Insert a 0
                 <C-a>                 Increment it to 1
                      a00@q<esc>       Append 00@q
                                Y      Yank this line (100@q)
                                 xx    Delete the @q part
                                   @0  Execute the yanked text as commands
                                       (100@q executes the q macro 100 times)

Answer 24

Zig, 63 66 47 72 bytes

fn a()void{for(" "**'e')|_,i|{@import("std").debug.print("{d} ",.{i});}}

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I've excluded the @import() boilerplate as it seems analogous to C's #include, which is excluded from other answers. If deemed necessary, I will add it back in.

Explanation

fn a() void {
    for (" " ** 'e') |_, i| {
        @import("std").debug.print("{d} ",.{i});
    }
}

• fn a() void Declare a function which takes no parameters and returns nothing
• for () |_, i| For every item in the array inside of (), iterate and capture the entree as _ (a throwaway variable) and the index as i
• " " ** 'e' Take the string (strings are slices, or pointer-arrays which know their length) and repeat it 'e' (101) times
• ** Requires a little bit more more explanation I think: In Zig, there is the concept of "comptime" (compile time) and runtime. ** is an operator which repeats any array literal or slice literal at comptime, because the resulting length is still known to the compiler.
• @import("std").debug.print("",.{}); Print to STDERR (I believe that's valid for this question, right?), the first argument is the formatting string, and the second is an "anonymous sctruct"/tuple with a variable number of arguments in it (Zig doesn't have var-args).
• "{d} " The format string. Zig denotes {} as the formatting characters, with d meaning a digit in this case.

Answer 25

Rust, 46 bytes

I did not see a rust solution, so here's my attempt:

||print!("{:?}",(0..b'e').collect::<Vec<_>>())

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Thanks to @ovs for pointing out the closure variant.

The range (0..b'e') is collected into a vector (using the placeholder _, letting the compiler figure out the type) and printed using the debug formatter {:?}, which "dumps" the entire vector.

The range upper bound is exclusive, and is represented using the byte literal b'e', which is equivalent to an u8 integer number literal; in this case 101 (e's ASCII value).

Answer 26

C, 53 bytes

int main(a,b){for(;a^'f';a++){printf("%d\n",a-!!a);}}

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43 42 bytes

(Thanks to Jo King♦)

k;main(){for(;k^'e';)printf("%d ",k++);}

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Answer 27

unsure, 99 bytes

ummmmm uhhhh errrrr uhhh um errrrr um um yeah err heh but um yeah err then uh okay um err then wait

It's not the shortest.

Explanation:

push 101                  ummmmm uhhhh errrrr uhhh um errrrr
push 0 to other stack     um um yeah err heh
loop                      but ... wait
  decrement                 um yeah err
  print + increment other   then uh okay um err then

Answer 28

C - 37 Bytes

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

Ungolfed

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

Explanation

Function to print numbers from 0 to 100 without digits. A global variable of type integer is created (so that it is automatically initialized to 0), the variable is incremented and printed 100 times through a loop which is executed while the variable is less than 'e' or 101 in ASCII.

Answer 29

Minecraft, 217 bytes + ?

From left to right the commands in the command blocks are

scoreboard players operation a a += b a
tellraw @a {"score":{"name":"a","objective":"a"}}
scoreboard objectives add a dummy
scoreboard players set a a 0
execute store result score b a run data get entity @p playerGameType

The command blocks on the right set a to 0 and b to the playerGameType of the player, which is 1 if the player is in creative mode. The blocks on the left repeatedly print a, then add b to a. It's stopped at exactly 100 by the piston removing the repeater powering the command block.

I'm not sure how to score this or if it's even allowed but I thought it was kind of cool.

Answer 30

Python 3, 41 bytes

Still learning ... best I could do:

for n in range(0, ord('e')):
    print(n)