35
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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

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

Based on a submission in the sandbox.

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9
  • 2
    \$\begingroup\$ Can the delimeters be numbers? \$\endgroup\$
    – Wheat Wizard
    Mar 17, 2020 at 17:16
  • \$\begingroup\$ @AdHocGarfHunter, No. Good catch. Edit: Actually, I think '0' should be acceptable. \$\endgroup\$
    – ouflak
    Mar 17, 2020 at 17:17
  • \$\begingroup\$ Could you verify that they "strange delimiters" version of this answer, is valid? It definitely seems cheaty. \$\endgroup\$
    – Wheat Wizard
    Mar 17, 2020 at 17:32
  • \$\begingroup\$ Honestly I think it's a clever 'outside-of-the-box' solution. I'd upvote, but I'm out of votes until tomorrow. \$\endgroup\$
    – ouflak
    Mar 17, 2020 at 17:35
  • 4
    \$\begingroup\$ @ouflak Thanks for the algorithm guide! How did you know I always write my prototypes with Turing Machines :p \$\endgroup\$
    – AviFS
    Mar 19, 2020 at 1:01

108 Answers 108

3
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JavaScript (ES8), 33 bytes

f=n=>n>9?'':''.padEnd(n,n)+f(-~n)

Try it online!

Commented

f = n =>            // n = counter, initially undefined
  n > 9 ?           // if n is greater than 9:
    ''              //   stop recursion
  :                 // else:
    ''.padEnd(n, n) //   pad an empty string with n digits n
                    //   this leaves the empty string unchanged for n undefined
    + f(-~n)        //   add the result of a recursive call with n + 1

Building in reverse order (38 bytes)

A somewhat funny alternate method is to build the string from \$n=9\$ to \$n=1\$ and pad the recursive call instead of an empty string.

By doing it this way, the required padding length is:

$$L_n=\left\lfloor\frac{n^2}{2}+1\right\rfloor$$

f=(n=9)=>n?f(n-1).padEnd(n*n/2+1,n):''

Try it online!

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3
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Erlang (escript), 49 bytes

f()->[X*(math:pow(10,X)-1)/9||X<-lists:seq(1,9)].

Try it online!

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3
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C (gcc), 50 bytes

n;f(i){for(i=0;i++<9;)for(n=i;n--;)putchar(48+i);}

Try it online!

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3
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Japt, 4 bytes

AÇîZ

Try it

NaN
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3
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This is my dumbest code golf submission ever, but here it goes

SQLite, 53 bytes

SELECT'122333444455555666666777777788888888999999999'

Try it online!

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3
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Stax, 5 bytes

╜├ìíy

Try it online or try it online unpacked (6 bytes).

Explanation (of the unpacked version):

Vd      # Push constant "0123456789"
  A     # Push 10
   r    # Pop and push a list in the range [0, 10)
    :B  # Repeat the characters in the string the integer amount of times:
        #  "122333444455555666666777777788888888999999999"
        # (after which the top of the stack is output implicitly as result)
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3
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Japt -P, 5 4 bytes

AÇçZ

Test it

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3
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[VBA] 58 bytes

i=1:While i<0:For x=1 to i:Debug.Print i:Next x:i=i+1:wend

Can be ran in Immediate

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3
  • \$\begingroup\$ you appear to have a typo in your code (while i<0->while i<10) \$\endgroup\$ Jun 6, 2020 at 16:20
  • \$\begingroup\$ You can get your code down quite a bit by using nested for loops and by using ? in place of debug.print - For i=1To 9:For j=1To i:?i:Next j,i for 35 bytes \$\endgroup\$ Jun 6, 2020 at 16:22
  • \$\begingroup\$ Oh - or better yet, if we use an approach similar to that of Mitchell Spector's we can get it down to 32 bytes as For i=1to 45:?(3.1*i^.48)\2:Next (or 33 bytes as For i=1to 45:?(3.1*i ^.48)\2:Next for 64-bit installs) \$\endgroup\$ Jun 6, 2020 at 17:11
3
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C (gcc), 47 bytes

k;main(i){for(;9>printf("%d",i*++k);i=i*10+1);}

Try it online!

C (gcc), 49 bytes

main(i){for(;9>printf("%lX",(i*1L<<4*i++)/15););}

Try it online!

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1
  • \$\begingroup\$ 46 bytes \$\endgroup\$
    – c--
    Sep 11 at 23:15
3
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COW, 75 bytes

MoOMoOMoOMoOMoOMoOMoOMoOMoOMOOmoOMoOMMMmoOMMMMOOmOoOOMmoOMOomoomOomOoMOomoo

Try it online!

Uncowed

moo ]    mOo <    MOo -    oom o
MOO [    moO >    MoO +    MMM =

+++++++++[>+=>=[<o>-]<<-]
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3
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Brainetry, 451 bytes

This is the golfed version, ungolfed version below.

a b c d
a b c d e f g h
a b c d e f g h
a b c
a b c d
a b
a b c d e
a b
a b c d
a b c d
a b c d
a b c d
a b c d
a b c
a b c d e f g h i
a b
a b c d
a b c d e f g h i
a b c
a b c
a b c d e
a b c d e
a b c
a b c d e
a b c d e f g h
a b
a b
a b c d
a b c d e f g h
a b c
a b c d e f g
a b
a b c d e
a b
a b c d
a b c
a b c d e f g h i
a b
a b c d e f g h
a b c
a b c d
a b
a b c d e
a b c d e f g h i
a b c
a b c
a b c d
a b c
a b c d e
a b c d e f g h i

To try this online, follow this link and paste the code in the btry/replit.btry file, then hit the green "Run" button.

Golfed version of the program below:

Let me explain what
is going on with this brainetry program: this
program will print the digits one to nine
and each of
those is going to
be repeated
as many times as its
own value.
Makes sense, doesn't it?
To achieve such result
we have to play
around with some nice
values in our tape.
That, and we
also have to play smart with our program pointer.
Of course
I would be delighted
to actually explain the algorithm used by this program.
The only problem
is I really
have no idea whatsoever about
what is actually going on.
Let me explain:
brainetry's instructions are a superset
of brainfuck, which means any brainfuck program can
be translated
to brainetry
and it will work.
So this is what I did, I found
a brainfuck program
that completed this task I described and
I just
translated it from brainfuck to
brainetry. Probably
there are simpler approaches
if we take
into consideration the extended operations that brainetry provides ...
However, brainetry
is still in its early stages of development
and I am
still trying to figure
out exactly
what operations to add to
brainetry. Once that set of operations becomes more well-defined
it will be
easier to harness
brainetry's power to write
computer programs. And
once that is finally done,
I will not have to steal random brainfuck programs.

This brainetry answer built on top of this answer

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3
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jq (-nr), 20 19 characters

range(9)+1|"\(.)"*.

Thanks to:

  • 2x-1 for pointing out that range() does not produce array, so operators will see separate numbers.

Sample run:

bash-5.0$ jq -nr 'range(9)+1|"\(.)"*.'
1
22
333
4444
55555
666666
7777777
88888888
999999999

Try it online!

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2
  • 2
    \$\begingroup\$ range(9)+1 is one byte shorter. \$\endgroup\$
    – user99151
    Jan 2, 2021 at 4:01
  • 2
    \$\begingroup\$ Wow! Thank you, @2x-1. I was never aware that parsing rules lead to that behavior. \$\endgroup\$
    – manatwork
    Jan 2, 2021 at 4:08
3
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Scala, 35 26 bytes

-9 bytes thanks to user!

()=>1 to 9 map(i=>s"$i"*i)

Try it online!

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1
  • 2
    \$\begingroup\$ This challenges allows you to return a list, so you can do this. Also, you don't need a for loop \$\endgroup\$
    – user
    Jan 3, 2021 at 21:01
3
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Vyxal j, 3 bytes

9ƛẋ

Try it Online!

-1 thanks to lyxal

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2
  • \$\begingroup\$ 3 bytes \$\endgroup\$
    – lyxal
    Jun 12, 2021 at 12:29
  • \$\begingroup\$ @lyxal I frogot it worked like that :p \$\endgroup\$
    – emanresu A
    Jun 12, 2021 at 12:29
3
+50
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Add++ -i, 14 13 bytes

L,9Rdz£XbUBvn

Try it online!

This is making me cry. A fun feature of like 99% of stack commands is that it applies to the whole stack rather than the top of the stack. Very very helpful.

-1 thanks to caird which kinda makes up for the suffering of this

Explained

L,9Rdz£XbUBvn
L,              # Start a lambda which is called implicitly by the -i flag
  9R            # Push the range [1...9]
    dz          # and zip it with itself, giving [[1, 1], [2, 2] ... [9, 9]]
      £X        # repeat x[0] by x[1] times for each x in that
        bU      # dump the contents of that onto the stack. This is a very important part because as I said, 99% of lambda stack commands map over to the whole stack, even when using quicks and stuff.
          Bv    # join each item (on the stack...) into a single integer
            n   # and join (the stack...) on newlines
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1
3
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Malbolge, 175 bytes

(C%;_#"~[}43WVxwuRt+*NMLmlljGig}CBAc?=vNtLr[v64m321/RzP+Ncha&&Gc#EDBX|{zx=vv:bs6p^]nllk/iVgfdvD'`N#?\\6YY3WUwvvQtPOqLo,JIHGihfCddcx>=<;:9[Y6XVl210/.-O+MLJ`&HG\[!~}|{z>=wv998rp

Try it online!

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3
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Julia, 22 bytes

1:9 .|>i->show("$i"^i)

Attempt This Online!

prints

"1""22""333""4444""55555""666666""7777777""88888888""999999999"

Relying on the repl for printing output we can get to 17 bytes

~i="$i"^i;.~(1:9)
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3
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Knight (v2.0-alpha), 21 bytes

;=i@W>10=i+1iO*+""i i

Try it online!

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2
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Keg, 45 11 9 8 bytes

9Ï^⑷:⅍*⑸

Try it online!

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2
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Io, 38 bytes

Range 1 to(9)map(i,i*(10**i-1)/9)print

Try it online!

Io, 47 bytes

Range 1 to(9)map(i,i asString repeated(i))print

Try it online!

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2
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Python 3.8, 34 bytes

i=0
while i<9:print(str(i:=i+1)*i)

Try it online!

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2
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Bubblegum, 23 bytes

00000000: 3334 3232 3636 3601 0253 1030 0303 7308  3422666..S.0..s.
00000010: b080 024b 1800 00                        ...K...

Try it online!

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2
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Python 2, 32 bytes

print[i*`i`for i in range(1,10)]

Try it online!

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2
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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.

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2
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naz, 36 bytes

1a1o1a2o1a3o1a4o1a5o1a6o1a7o1a8o1a9o

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

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2
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Ruby, 30 25 bytes

1.upto(9){|i|p 10**i/9*i}

Try it online!

10**i/9 gives a number with i digits, all 1s. Multiply by i for the required output.

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1
  • \$\begingroup\$ 21 bytes: 1.upto(9){|i|p [i]*i}. Going by the comments on the OP I think the output this gives is acceptable. \$\endgroup\$
    – Dingus
    Apr 19, 2020 at 11:00
2
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Zsh, 30 bytes

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

Try it online!

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

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2
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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.

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2
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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!

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2
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PHP, 37 36 35 bytes

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

Try it online!

-1 byte thanks to @Night2

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4
  • \$\begingroup\$ Do you need the input? \$\endgroup\$
    – ouflak
    Mar 17, 2020 at 11:19
  • 1
    \$\begingroup\$ Nope, removed thanks. I'm so used to using the -F option, I did it without thinking. \$\endgroup\$ Mar 17, 2020 at 11:21
  • \$\begingroup\$ 35 bytes \$\endgroup\$
    – Night2
    Mar 18, 2020 at 9:26
  • \$\begingroup\$ Interesting lesson in precedence rules. \$\endgroup\$ Mar 18, 2020 at 12:55

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