Numbers by Position

Question

Challenge

Print the numbers:

1
22
333
4444
55555
666666
7777777
88888888
999999999

In that order.

I/O

Takes no input. The numbers can have any delimiters desired (or none). That includes lists, cell arrays, .jpeg, etc.... Example outputs:

122333444455555666666777777788888888999999999
[1,22,333,4444,55555,666666,7777777,88888888,999999999]
etc....

Code Example

This is an un-golfed example that may perhaps act as algorithm guide (or maybe not):

Turing Machine Code, 535 bytes

0 * 1 r L
L * _ r 2
2 * 2 r a
a * 2 r M
M * _ r 3
3 * 3 r b
b * 3 r c
c * 3 r N
N * _ r 4
4 * 4 r d
d * 4 r e
e * 4 r f
f * 4 r O
O * _ r 5
5 * 5 r g
g * 5 r h
h * 5 r i 
i * 5 r j
j * 5 r P
P * _ r 6
6 * 6 r k
k * 6 r l
l * 6 r m
m * 6 r n
n * 6 r o
o * 6 r Q
Q * _ r 7
7 * 7 r p
p * 7 r q
q * 7 r r
r * 7 r s
s * 7 r t
t * 7 r u
u * 7 r R
R * _ r 8
8 * 8 r v
v * 8 r w
w * 8 r x
x * 8 r y
y * 8 r z
z * 8 r A
A * 8 r B
B * 8 r S
S * _ r 9
9 * 9 r C
C * 9 r D
D * 9 r E
E * 9 r F
F * 9 r G
G * 9 r H
H * 9 r I
I * 9 r J
J * 9 r halt

Try it online!

This prints out the numbers with a space delimiter:

1 22 333 4444 55555 666666 7777777 88888888 999999999

Challenge Type

code-golf, so shortest answer in bytes (by language) wins.

Based on a submission in the sandbox.

• Wheat Wizard: Can the delimeters be numbers?
• ouflak: @AdHocGarfHunter, No. Good catch. Edit: Actually, I think '0' should be acceptable.
• Wheat Wizard: Could you verify that they "strange delimiters" version of this answer, is valid? It definitely seems cheaty.
• ouflak: Honestly I think it's a clever 'outside-of-the-box' solution. I'd upvote, but I'm out of votes until tomorrow.
• AviFS: @ouflak Thanks for the algorithm guide! How did you know I always write my prototypes with Turing Machines :p

Answer 1

Batch, 74 bytes

@set s=
@for /l %%i in (1,1,9)do @call set s=%%s%%0&call echo %%s:0=%%i%%

Outputs on separate lines. Extends the string length by 1 and replaces the character with the current digit for each output.

The call commands are required so that the variable s is processed inside the loop instead of before parsing the loop.

Answer 2

naz, 36 bytes

1a1o1a2o1a3o1a4o1a5o1a6o1a7o1a8o1a9o

Outputs all the numbers with no extra delimiter, using the same functionality described in this answer.

Answer 3

Zsh, 30 bytes

repeat 9 echo ${(pl[++i][$i])}

Try it online!

The (l[width][fill]) flag does the heavy lifting.

Answer 4

Red, 30 bytes

repeat a 9[repeat b a[prin a]]

Try it online!

Well, the most obvious way to do it is the shortest in Red.

Answer 5

IBM/Lotus Notes Formula Language, 56 bytes

@For(x:=1;x<10;x:=x+1;@Set("o";o:@Repeat(@Text(x);x)));o

enter image description here

Formula in a multi value form field, once again showing that the only real use for @For in Notes is Code Golf!

Answer 6

PHP, (削除) 37 (削除ここまで) (削除) 36 (削除ここまで) 35 bytes

for(;10>$j=++$i;)for(;$j--;)echo$i;

Try it online!

-1 byte thanks to @Night2

• ouflak: Do you need the input?
• Guillermo Phillips: Nope, removed thanks. I'm so used to using the -F option, I did it without thinking.
• Night2: 35 bytes
• Guillermo Phillips: Interesting lesson in precedence rules.

Answer 7

SNOBOL4 (CSNOBOL4), 47 bytes

O OUTPUT =DUPL(X,X)
 X =LT(X,9) X + 1 :S(O)
END

Try it online!

Prints with a leading newline.

Answer 8

Factor, (削除) 42 (削除ここまで) 39 bytes

9 [1,b] [ dup 1array swap cycle ] map .

Try it online!

Displays a list ot lists

Answer 9

Batch, 85 Bytes

@for /L %%A in (1 1 9)do @For /L %%B in (1 1 %%A)do @Call Set O=%%O%%%%A @Echo(%O%

TIO not available.

enter image description here

• l4m2: even longer than hardcode @echo 122333444455555666666777777788888888999999999

Answer 10

Whitespace, 109 bytes

[S S S T S S T N
_Push_9][N
S S N
_Create_Label_LOOP][S N
S _Duplicate_top][N
S S S N
_Create_Label_INNER_LOOP][S S S T N
_Push_1][T S S T _Subtract][S N
S _Duplicate_top][N
T S T N
_If_0_Jump_to_Label_DONE_INNER_LOOP][S T S S T N
_Copy_0-based_1st][S N
T _Swap_top_two][N
S N
S N
_Jump_to_Label_INNER_LOOP][N
S S T N
_Create_Label_DONE_INNER_LOOP][S N
N
_Discard_top][S N
S _Duplicate_top][S S S T N
_Push_1][T S S T _Subtract][S N
S _Duplicate_top][N
T S S S N
_If_0_Jump_to_Label_DONE_LOOP][N
S N
N
_Jump_to_Label_LOOP][N
S S S S N
_Create_Label_DONE_LOOP][S N
N
_Discard_top][N
S S S T N
_Create_Label_PRINT_LOOP][T N
S T _Print_top_as_integer][N
S N
S T N
_Jump_to_Label_PRINT_LOOP]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Explanation in pseudo-code:

Push n=9
Start LOOP:
 Duplicate top (Integer i = n)
 Start INNER_LOOP
 i = i - 1
 if(i == 0):
 Jump to DONE_INNER_LOOP
 Copy n (0-based index 1)
 Swap
 Go to next iteration of INNER_LOOP
 DONE_INNER_LOOP:
 Discard i
 n = n - 1
 if(n == 0):
 Jump to DONE_LOOP
 Go to next iteration of LOOP
DONE_LOOP:
 Discard n
 Start PRINT_LOOP:
 Print top as integer
 Go to next iteration of PRINT_LOOP

Funny thing is, is that outputting with the additional single leading 0 as the challenge in the Sandbox initially had, this could have been 11 bytes shorter: try it online. xD

• ouflak: I'm still mulling over some kind of challenge that involves an index....

Answer 11

Gol><>, 8 bytes

aFLRL|D;

Try it online!

aF | For Loop, from 0 to 10 excluding 10
 L Push the current loop iteration to the stack
 RL Pop the current iteration and repeat pushing the loopiterator value so many times
 D; Print the entire stack content as numeric values with the debug operation and halt

Answer 12

Javascript, 48 bytes

for(x=1;x<=9;x++){console.log((x+'').repeat(x))}

Try it online!

• Hedi: You can save 2 bytes with : for(x=0;x++<9;){console.log((x+'').repeat(x))}

Answer 13

C++ (gcc), (削除) 93 (削除ここまで), 65 bytes

int a;void n(){int b=a;while(b){std::cout<<a;b--;}if(a++!=9)n();}

Try it online!

The loop work is essentially the same as the previous version, but instead its own function.

It sets the starting digit -> 1 -> counter then it prints digits until the counter reaches 0. At that point it increments the digit and recalls the function until 9.

C++ (gcc), 93 bytes

int main(int n){static int a=1;int b=a;if(b>9)return 0;while(b){std::cout<<a;b--;}main(a++);}

Try it online!

Don't mind the main...

• ceilingcat: 60 bytes
• c--: 51 and 59 bytes
• vengy: 50 bytes

Answer 14

vJASS, 126 bytes

//! zinc
library q{function onInit(){integer x,y;string s="";for(1<=x<10){for(0<=y<x){s+=I2S(x);}}BJDebugMsg(s);}}
//! endzinc

enter image description here

Answer 15

dc, ^{(削除) 25 (削除ここまで) (削除) 24 (削除ここまで)} 22 bytes

1[ddIr^9/*n1+dA>m]dsmx

Try it online!

Answer 16

Pure Bash, (削除) 41 (削除ここまで) 37 bytes

for k in {1..9};{ echo $[10**k/9*k];}

Try it online!

Answer 17

MathGolf, (削除) 6 (削除ここまで) 5 bytes

9{îÄí

-1 byte thanks to @maxb.

Try it online.

Explanation:

9{ # Loop 9 times:
 î # Push the 1-based loop index
 Ä # Do an inner loop that many times, using a single command:
 í # Push the total number iterations of the outer loop
 # (after the loops, the entire stack is joined together and output implicitly)

• maxb: Without a separator, you could do 9{îÄí, where í pushes the loop limit within a for loop.
• Kevin Cruijssen: @maxb Ah, nice golf. Out of curiosity, why doesn't it work when the Ä is {?
• Kevin Cruijssen: @maxb It does seem to work if I add two trailing }: try 9{î{í}} online.
• maxb: There are some issues with implicit block closure with nested loops. you can omit the last } at the end of the script if you have a single loop, but not if you have two or more nested loop blocks which are opened by {.

Answer 18

Python 3, 38 bytes

print([str(x)*x for x in range(1,10)])

Try it online!

C++ (gcc), 59 bytes

for(int i=1;i<10;++i){for(int x=1;x<=i;++x){std::cout<<i;}}

Try it online!

Started learning c++ an hour ago, tried codegolf :)

• Dion: @ceilingcat so all the imports and func names must be in the code? Thanks for pointing this out!
• ceilingcat: 67 bytes

Answer 19

Common Lisp, 50 bytes

(loop for x from 1 to 9 do(dotimes(n x)(write x)))

Try it online!

Answer 20

Fortran (GFortran), ^{(削除) 38 (削除ここまで)} 28 bytes

print*,(10**i/9*i,i=1,9)
end

Try it online!

Edit: Turns out the program statement is optional, saving 10 bytes!

• ouflak: I'm glad somebody did a Fortran answer.

Answer 21

cQuents, 6 bytes

#9&$D$

Try it online!

Oh cool, another way to use cQuents!

Answer 22

C# (Visual C# Interactive Compiler), 45 bytes

for(int j,i=0;i++<9;)for(j=i;j-->0;Write(i));

Try it online!

Answer 23

MAWP 0.1, (削除) 22 (削除ここまで) 20 bytes

[![~!:~1A]%!9A?.%1M]

Explanation:

[ start of loop
! duplicates top of stack
[ start of loop
~!:~ prints bottom stack value
1A subtracts 1
] end of loop
% removes top of stack (0 from the counter in the previous loop)
!9A diffeence between top value and 9
?. if top 0, then terminate program
% removes top value
1M adds 1 to top value
] end of loop

Answer 24

Clojure, 32 bytes

(for[i(range 1 10)](repeat i i))

This results in a lists of numbers:

((1) (2 2) (3 3 3) (4 4 4 4) (5 5 5 5 5) (6 6 6 6 6 6) (7 7 7 7 7 7 7) (8 8 8 8 8 8 8 8) (9 9 9 9 9 9 9 9 9))

Answer 25

International Phonetic Esoteric Language, 21 bytes

{A}1ɑee0ɑbue1søɒe1søɒ

Prints 122333444455555666666777777788888888999999999.

Explanation:

{A}1ɑee0ɑbue1søɒe1søɒ
{A}1 (push loop bounds, 1 to 10)
 ɑ (start loop)
 e (push the current index)
 e0 (push loop bounds for current digit, 0 to index)
 ɑ (start loop)
 bu (DUP and CPRINT current digit)
 e1sø (increment index)
 ɒ (end loop)
 e1sø (increment index)
 ɒ (end loop)

Answer 26

Python 3, 34 bytes

for i in range(10):print(str(i)*i)

Try it online!

• ouflak: You can get your answer formatted in the standard format used on this site, over here. Just put your code in and click on the link symbol above. That will take you to another page where you will find the Code Golf format option (pasted into your clipboard). You can then just paste that into your answer.
• user95943: Ok, thanks! I'll remember that.

Answer 27

brainfuck, 50 bytes

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

Try it online!

Assumes byte cells; does not go out of bounds to left.

Answer 28

Poetic, 222 bytes

THE NUMERICAL I/O RULES
i saw a digit,o yes
o,again,i saw a number;too many
i do the I/O rules,i get a repeat
i now count to zeros
o,from one right to nine,i do digits,e.t.c
seldom a newline,since i know i am swayed not to

Try it online!

Answer 29

Husk, 7 bytes

mod ́Rḣ9

Try it online!

Husk, 7 bytes

mdg ́Ṙḣ9

Try it online!

Answer 30

Rust, (削除) 77 (削除ここまで) (削除) 75 (削除ここまで) (削除) 73 (削除ここまで) (削除) 72 (削除ここまで) (削除) 57 (削除ここまで) 56 bytes

fn main(){for k in 1..10{for j in 0..k{print!("{}",k)}}}

Try it online!

And I thought Lua was verbose!

Thanks to ovs for saving 15 bytes.

• ovs: You don't need i, 0..k and println!("{}",i) works just fine. Then you can use a closure instead of a full program like this: ||for i in 1..10{for j in 0..k{println!("{}",k)}}.
• ovs: And returning an iterator instead of printing the result is a little shorter: tio.run/…
• ouflak: @ovs, Wow. This is the first time I've coded anything in Rust (we've just adopted it at my workplace), so it's clear I've got a lot to learn. I'm still trying to wrap my head around how the closures work. I'm using print! instead. That seems to work.