Numbers by Position

Question

Challenge

Print the numbers:

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.

@AdHocGarfHunter, No. Good catch. Edit: Actually, I think '0' should be acceptable. — ouflak, Commented Mar 17, 2020 at 17:17
Could you verify that they "strange delimiters" version of this answer, is valid? It definitely seems cheaty. — Wheat Wizard, Commented Mar 17, 2020 at 17:32
Honestly I think it's a clever 'outside-of-the-box' solution. I'd upvote, but I'm out of votes until tomorrow. — ouflak, Commented Mar 17, 2020 at 17:35
@ouflak Thanks for the algorithm guide! How did you know I always write my prototypes with Turing Machines :p — AviFS, Commented Mar 19, 2020 at 1:01

hakr14 · Accepted Answer · 2021-01-03 06:56:47Z

2

Canvas, 6 bytes

９｛：ŗ×］

Try it here!

edited Jan 3, 2021 at 6:56

answered Jan 3, 2021 at 6:31

hakr14

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ZaMoC · Accepted Answer · 2021-12-19 21:46:24Z

2

Wolfram Language (Mathematica), 17 bytes

Table[,{,9},{a,}]

Try it online!

thanks @att

edited Dec 19, 2021 at 21:46

answered Mar 17, 2020 at 10:45

ZaMoC

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\$\begingroup\$ 18 bytes (list of lists of digits, which based on comments should be fine) \$\endgroup\$
– att
Commented Oct 9, 2020 at 6:15
\$\begingroup\$ 17 bytes \$\endgroup\$
– att
Commented Dec 19, 2021 at 3:14

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Ginger · Accepted Answer · 2021-12-21 20:16:22Z

2

APOL, 21 bytes

ⅎ(9 p(*(t(∈) ∈)))

answered Dec 21, 2021 at 20:16

Ginger

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1

\$\begingroup\$ Could you provide a link to an interpreter for APOL? \$\endgroup\$
– caird coinheringaahin g ♦
Commented Dec 21, 2021 at 20:48
\$\begingroup\$ @cairdcoinheringaahing replit.com/@GingerIndustries/APOL \$\endgroup\$
– Ginger
Commented Dec 22, 2021 at 15:59

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DLosc · Accepted Answer · 2021-12-31 23:53:07Z

2

Pip, 5 bytes

,tX,t

Attempt This Online!

Explanation

,t     Range(10)
  X    String-repeat (itemwise)
   ,t  Range(10)

That is, for each number in [0; 1; 2; 3; 4; 5; 6; 7; 8; 9], repeat it a number of times equal to itself. The result is the list [""; 1; 22; 333; 4444; 55555; 666666; 7777777; 88888888; 999999999], which is by default concatenated together and printed.

answered Dec 31, 2021 at 23:53

DLosc

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Engineer Toast · Accepted Answer · 2022-01-10 16:07:28Z

2

Excel, 29 24 bytes

^{Saved 5 bytes thanks to Taylor Raine}

=REPT(ROW(1:9),ROW(1:9))

ROW(1:9) returns an array of the numbers 1 through 9 so the REPT() function repeats each of those numbers itself-many times.

edited Jan 10, 2022 at 16:07

answered Dec 14, 2021 at 17:20

Engineer Toast

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2

\$\begingroup\$ You can make this shorter by dropping the Let and using Row(1:9) instead - =REPT(ROW(1:9),ROW(1:9)) \$\endgroup\$
– Taylor Raine
Commented Jan 10, 2022 at 15:32

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Aiden Chow · Accepted Answer · 2022-03-09 00:15:30Z

2

Desmos, 21 bytes

l=[1...9]
l10^l/9-l/9

Try It On Desmos!

Try It On Desmos! - Prettified

Uses the following formula for the \$n\$th number: $$\frac{10^n-1}9\times n$$ [1,22,333,4444,55555,666666,7777777,88888888,999999999] is the value of the list in the second line.

Made it look "obfuscated" for fun but it's really just the same as:

l=[1...9]
l(10^l-1)/9

which is also 21 bytes.

answered Mar 9, 2022 at 0:15

Aiden Chow

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DLosc · Accepted Answer · 2022-09-08 17:08:18Z

Regenerate, 15 bytes

(#1{#1} ! ){10}

Outputs space-separated runs of digits, with a leading and a trailing space. Attempt This Online!

Explanation

Spaces are replaced with underscores for clarity.

(#1{#1}_!_){10}
           {10}  Repeat 10 times
(         )      and store as group 1 after each match:
 #1                Length of previous match of group 1
   {#1}            repeated (length of previous match of group 1) times
       _           followed by a space
        !          Or, if that failed because there was no previous match of group 1,
         _         just a space

The first match of group 1 is a space, which has length 1, so the second match is one 1 followed by a space, which has length 2, so the third match is two 2's followed by a space, and so forth.

naffetS · Accepted Answer · 2022-09-09 00:01:33Z

2

Prolog (SWI), 43 bytes

:-between(1,9,N),between(1,N,_),\+write(N).

Try it online!

answered Sep 9, 2022 at 0:01

naffetS

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\$\begingroup\$ -1 byte \$\endgroup\$
– Jo King ♦
Commented Sep 12, 2022 at 4:15

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thejonymyster · Accepted Answer · 2022-09-09 00:57:37Z

Headass, 31 bytes

+^{{D^(]<)PN][}:(+++[]]]>)N^-[}

Try It Online!

Beats the naive solution (+P+PP+PPP... and so on for 54 bytes) by 23 bytes.

Outputs individual digits separated by newlines

Breakdown:

+^{{D^(]<)PN][}:(+++[]]]>)N^-[}  full program

+^                               initialize r1 to be 1
                                 r2 is already initialized to 0
  {                           }  repeat
    D^(  )     :(        )       until r1
                 +++[]]]>        fails to be less than 9
   {          }                    repeat
    D^(  )                         until r1
       ]<                          fails to be more than r2
    D     P                          print r1
           N][                       increment r2
                                   end repeat
                          N^       increment r1
                          N -[     set r2 to 0
                                 end repeat

this 25 byter takes n and outputs the number n*(n 1s in a row): U({D+^-)DD]P:]]]]]]]]]][}. I don't think I can iterate from 1-9 with just 5 bytes though... — thejonymyster, Commented Sep 10, 2023 at 1:00

Kevin Cruijssen · Accepted Answer · 2022-09-09 08:25:56Z

yup, 43 41 bytes

0e:{~:{~:#~0e-}0e--:0ee00e--|00e--e0e-~-}

Try it online.

Explanation:

0e           # Push 1 (push 0; pop and push exp(0))
  :          # Duplicate it
   {         # Loop while the top of the stack is >0 (without popping):
    ~        #  Swap the top two values on the stack
     :       #  Duplicate the top
      {      #  Inner loop while the top of the stack is >0 (without popping):
       ~     #   Swap the top two values on the stack
        :    #   Duplicate the top
         #   #   Pop and print it as number
       ~     #   Swap the top two values on the stack back
        0e-  #   Decrease it by 1 (push 0; pop and push exp(0); subtract)
      }      #  Stop the inner loop if the top of the stack is <=0 (without popping)
       0e--  #  Increase the current value by 1 (using the 0 that was still on the
             #  stack as: push 0 → exp(0)=1 → 0-1=-1 → --1=+1)
       :     #  Duplicate the top again
        0ee00e--|00e--e0e-
             #  Push 9.107... (see below)
         ~   #  Swap the top two values again
          -  #  Subtract the current value from the 9.107...
   }         # Stop the outer loop if the top of the stack is <=0 (without popping)
             # (which it will be once we've reached 10)

0ee00e--|00e--e0e- # Push 9.107...:
0                  #  Push 0        - STACK: 0
 e                 #  exp(0)        - STACK: 1
  e                #  exp(1)        - STACK: 2.718...
   0               #  Push 0        - STACK: 2.718...,0
    0              #  Push 0        - STACK: 2.718...,0,0
     e             #  exp(0)        - STACK: 2.718...,0,1
      -            #  Subtract      - STACK: 2.718...,-1
       -           #  Subtract      - STACK: 3.718...
        |          #  ln(3.718...)  - STACK: 1.313...
         0         #  Push 0        - STACK: 1.313...,0
          0        #  Push 0        - STACK: 1.313...,0,0
           e       #  exp(0)        - STACK: 1.313...,0,1
            -      #  Subtract      - STACK: 1.313...,-1
             -     #  Subtract      - STACK: 2.313...
              e    #  exp(2.313...) - STACK: 10.107...
               0   #  Push 0        - STACK: 10.107...,0
                e  #  exp(0)        - STACK: 10.107...,1
                 - #  Subtract      - STACK: 9.107...

Seggan · Accepted Answer · 2022-09-15 16:57:09Z

2

Fig, \$7\log_{256}(96)\approx\$ 5.762 bytes

Ma9'*+D

Try it online!

Explained

Ma9'*+D
Ma9'    # To each item n in the range [1, 9]
    *+D # Repeat the stringified n, n times

edited Sep 15, 2022 at 16:57

Seggan

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answered Sep 8, 2022 at 13:50

lyxal

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dtanku · Accepted Answer · 2023-02-01 15:41:39Z

2

F# (.NET Core), ⁵⁹ ⁵⁸ 57 bytes (thanks ouflak!)

for i=1 to 9 do String.replicate i (string i)|>printf"%s"

or

for i=1 to 9 do printf"%s"(String.replicate i (string i))

Try it online!

edited Feb 1, 2023 at 15:41

answered Feb 1, 2023 at 2:44

dtanku

713 bronze badges

1

\$\begingroup\$ Welcome to Code Golf! This is a nice submission. As this is a golfing challenge though, white space is among your first enemies, 58 bytes. It does not matter if the program exits with an error or slew of warnings as long as it meets the criteria of the challenge. \$\endgroup\$
– ouflak
Commented Feb 1, 2023 at 7:44

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pacman256 · Accepted Answer · 2024-01-15 19:08:42Z

2

Vyxal 3, 3 bytes

9ƛY

Try it Online!

Edit, the three byter is valid

edited Jan 15 at 19:08

answered Jan 14 at 19:19

pacman256

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\$\begingroup\$ This is great. Would you provide a link to the 'lists of digits' solution, maybe here in the comments? It might actually meet the conditions of the challenge which are pretty broad as far as the output is concerned. \$\endgroup\$
– ouflak
Commented Jan 15 at 7:16
\$\begingroup\$ it's just 9ƛY \$\endgroup\$
– pacman256
Commented Jan 15 at 17:20
\$\begingroup\$ Yes! That's definitely within the output rules. \$\endgroup\$
– ouflak
Commented Jan 15 at 18:10

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Razetime · Accepted Answer · 2020-10-10 09:30:58Z

1

mlochbaum/BQN, 7 bytes

⥊˜¨1+↕9

Try it!

Explanation

⥊˜¨1+↕9
     ↕9 range from 0 to 8
   1+   add 1 to each
  ¨     for each element n,
⥊˜      replicate n times

answered Oct 10, 2020 at 9:30

Razetime

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2

\$\begingroup\$ Alternative 7 bytes (with different format): ∾⥊˜¨↕10. Also, ⥊˜¨↕10 (6 bytes) could be used if a leading empty list is fine (it probably is). \$\endgroup\$
– Wezl
Commented Dec 23, 2020 at 16:45

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Pedro Maimere · Accepted Answer · 2021-01-04 15:55:00Z

1

AWK, 32 or 37 bytes

This code needs an EOF at the prompt, or a file with an empty line.

{for(;++i<10;)print(10^i-1)/9*i}

Try it online!

To substitute the input, one can use the BEGIN pattern, adding 5 more bytes to the code.

BEGIN{for(;++i<10;)print(10^i-1)/9*i}

answered Jan 4, 2021 at 15:55

Pedro Maimere

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user100690user100690 · Accepted Answer · 2021-02-24 09:59:18Z

1

JavaScript, 50 bytes

_=>{for(i=0;++i<10;)console.log((''+i).repeat(i))}

answered Feb 24, 2021 at 9:59

user100690

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chunes · Accepted Answer · 2021-06-12 11:49:49Z

1

Factor, 29 bytes

9 [1,b] [ dup <array> ] map .

Try it online!

answered Jun 12, 2021 at 11:49

chunes

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stefvanschie · Accepted Answer · 2021-06-12 13:08:12Z

A0A0, 46 bytes

O122333444455555666666777777788888888999999999

Literally prints the example output. I'd be very surprised if you can get something smaller in A0A0. The first O outputs the number following it and the number following it is the output. The language specification does not state any restriction on the size of the number, just "signed integer", so this works. (Although an interpreter may not handle this correctly, because this is a 147 bit number and not every language may support numbers that large.)

Yousername · Accepted Answer · 2021-12-18 19:03:50Z

1

TI-Basic, 17 bytes

seq(I(₁₀^(I)-1)/9,I,1,9

Output is stored in Ans as a list and displayed.

answered Dec 18, 2021 at 19:03

Yousername

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Visckmart · Accepted Answer · 2021-12-19 03:10:48Z

1

Lua, 37 bytes

for i=1,9 do print((""..i):rep(i))end

Try it online!

answered Dec 19, 2021 at 3:10

Visckmart

3611 silver badge4 bronze badges

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pxeger · Accepted Answer · 2022-01-01 07:15:10Z

1

Pure Zsh, 27 bytes

eval ';s+=1;<<<$[s*'{1..9}]

Attempt This Online!

answered Jan 1, 2022 at 7:15

pxeger

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emirps · Accepted Answer · 2022-11-30 23:47:53Z

1

GolfScript, 9 bytes

10,{.`*}/

Explanation

     /  # for each element in...
     10,{ # 0-9
     . # duplicate it
     ` # stringify it
     * # push to the stack the stringified version of the number, repeated that many times
    }  # end block
       # implicitly output the stack

Try it online!

answered Nov 30, 2022 at 23:47

emirps

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Emigna · Accepted Answer · 2022-12-01 14:29:13Z

1

><>, 28 bytes

1:\   >~1+:a=?;!
n$>:?!v1-$:

Try it online!

answered Dec 1, 2022 at 14:29

Emigna

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Kip the Malamute · Accepted Answer · 2023-02-01 00:08:48Z

1

Pyt, 6 bytes

ɳąĐƖ*ǰ

Try it online!

ɳ           pushes string containing all digits ("0123456789")
 ą          converts to ąrray of characters
  Đ         Đuplicates on stack
   Ɩ        casts top of stack to array of Ɩntegers
    *       string multiplication
     ǰ      ǰoin with no delimiter; implicit print

answered Feb 1, 2023 at 0:08

Kip the Malamute

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The Thonnu · Accepted Answer · 2023-02-01 15:08:29Z

1

Thunno `N`, \$ 8 \log_{256}(96) \approx \$ 6.58 bytes

9R1+eDJ*

Attempt This Online!

Explanation

9R        # Push range(0, 9)
  1+      # Add one to each
    e     # To each number:
     D    #  Duplicate
      J   #  Cast to string
       *  #  Repeat that many times
          # N flag joins by newlines
          # Implicit output

answered Feb 1, 2023 at 15:08

The Thonnu

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EzioMercer · Accepted Answer · 2023-02-01 17:10:39Z

JavaScript, 50 bytes

_=>'122333444455555666666777777788888888999999999' // 50
_=>eval("for(s='',i=0;i<11;s+=''.padEnd(++i,i))s") // 50
n=>eval("for(s='';(n=-~n)<11;s+=''.padEnd(n,n))s") // 50
_=>'123456789'.replace(/(\d)/g,(d,i)=>d.repeat(i)) // 50

Try it:

f1=_=>'122333444455555666666777777788888888999999999' // 50
f2=_=>eval("for(s='',i=0;i<10;s+=''.padEnd(++i,i))s") // 50
f3=n=>eval("for(s='';(n=-~n)<11;s+=''.padEnd(n,n))s") // 50
f4=_=>'123456789'.replace(/(\d)/g,(d,i)=>d.repeat(i)) // 50

console.log(f1());
console.log(f2());
console.log(f3());
console.log(f4());

Joao-3 · Accepted Answer · 2024-01-17 17:19:48Z

1

Uiua, 15 bytes

∵(⋕/⊂∵°⋕↯).+1⇡9

Pad

answered Jan 17 at 17:19

Joao-3

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noodle person · Accepted Answer · 2024-01-17 21:22:29Z

1

TypeScript’s type system, 94 bytes

type F<N extends 1[]=[1],L=N["length"]>=L extends 10?[]:[...{[_ in keyof N]:L},...F<[...N,1]>]

Try it at the TS Playground

answered Jan 17 at 21:22

noodle person

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madeforlosers · Accepted Answer · 2024-05-03 14:13:04Z

1

YASEPL, 24 23 bytes

=a$`1=b)""+a;b,a<!+}4,9

explanation (in pseudocode)

=a$`1=b)""+a;b,a<!+}4,9          packed
=a$                              a = 1
   `1            !+}4,9          for(i = 0; i < 9; i++)
     =b)""+a;b,a<                  print(str(a).repeat(a))

edited May 3 at 14:13

answered Feb 9 at 23:13

madeforlosers

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Andrew Bayly · Accepted Answer · 2024-08-08 21:55:18Z

1

JavaScript (Node.js), 35 bytes

f=(x=9)=>x?f(x-1)+''.padEnd(x,x):''

Try it online!

answered Aug 8 at 21:55

Andrew Bayly

1,1993 silver badges5 bronze badges

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Numbers by Position

Turing Machine Code, 535 bytes

122 Answers 122

Canvas, 6 bytes

Wolfram Language (Mathematica), 17 bytes

APOL, 21 bytes

Pip, 5 bytes

Explanation

Excel, 29 24 bytes

Desmos, 21 bytes

Regenerate, 15 bytes

Explanation

Prolog (SWI), 43 bytes

Headass, 31 bytes

yup, 43 41 bytes

Fig, \$7\log_{256}(96)\approx\$ 5.762 bytes

Explained

F# (.NET Core), 59 58 57 bytes (thanks ouflak!)

Vyxal 3, 3 bytes

mlochbaum/BQN, 7 bytes

Explanation

AWK, 32 or 37 bytes

JavaScript, 50 bytes

Factor, 29 bytes

A0A0, 46 bytes

TI-Basic, 17 bytes

Lua, 37 bytes

Pure Zsh, 27 bytes

GolfScript, 9 bytes

Explanation

><>, 28 bytes

Pyt, 6 bytes

Thunno N, \$ 8 \log_{256}(96) \approx \$ 6.58 bytes

Explanation

JavaScript, 50 bytes

Uiua, 15 bytes

TypeScript’s type system, 94 bytes

YASEPL, 24 23 bytes

explanation (in pseudocode)

JavaScript (Node.js), 35 bytes

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F# (.NET Core), ⁵⁹ ⁵⁸ 57 bytes (thanks ouflak!)

Thunno `N`, \$ 8 \log_{256}(96) \approx \$ 6.58 bytes

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