It was just a bug

Question

Inspired by the bugged output in @Carcigenicate's Clojure answer for the Print this diamond challenge.

Print this exact text:

1        2        3        4        5        6        7        8        9        0
1       2       3       4       5       6       7       8       9       0
1      2      3      4      5      6      7      8      9      0
1     2     3     4     5     6     7     8     9     0
1    2    3    4    5    6    7    8    9    0
1   2   3   4   5   6   7   8   9   0
1  2  3  4  5  6  7  8  9  0
1 2 3 4 5 6 7 8 9 0
1234567890
1 2 3 4 5 6 7 8 9 0
1  2  3  4  5  6  7  8  9  0
1   2   3   4   5   6   7   8   9   0
1    2    3    4    5    6    7    8    9    0
1     2     3     4     5     6     7     8     9     0
1      2      3      4      5      6      7      8      9      0
1       2       3       4       5       6       7       8       9       0
1        2        3        4        5        6        7        8        9        0

(From the middle outward in both directions, each digit is separated by one more space than the previous line.)

Challenge rules:

• There will be no input (or an empty unused input).
• Trailing spaces are optional.
• A single trailing new-line is optional.
• Leading spaces or new-lines are not allowed.
• Returning a string-array isn't allowed. You should either output the text, or have a function which returns a single string with correct result.

General rules:

• This is code-golf, so shortest answer in bytes wins.
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.
• If possible, please add a link with a test for your code.
• Also, please add an explanation if necessary.

Answer 1

Python 2, 59 57 56 55 bytes

i=8;exec"print(' '*abs(i)).join('1234567890');i-=1;"*17

Try it online!

@Leaky Nun helped golfing this a lot, @Praind Suggested a method to save 1 byte, which I formerly thought of, but forgot to edit, @CotyJohnathanSaxman suggested changing the loop.

---

Explanation

• i=8 - Assigns the value 8 to a variable called i.

• exec"..."*17 - Execute that block of code (...) 17 times.

• print(...) - Output the result.

• ' '*abs(i) - Create a String with a space repeated |i| times.

• .join('1234567890') - Interleave the string created above with 1234567890, such that |i| spaces are inserted between the digits.

• i-=1 - Decrement i, and by executing it 17 times, it reaches -8, which creates te repetitive pattern with the help of abs().

Answer 2

Vim, 35 bytes:

i1234567890<esc>qqYP:s/\d/& /g
YGp{q7@q

Explanation:

i1234567890<esc>    " insert '1234567890'
qq                  " Record the following into register 'q'
  Y                 "   Yank a line
   P                "   Paste it above us
    :s/\d/& /g      "   Put a space after each number
Y                   "   Yank this line
 G                  "   Move the end of the buffer
  p                 "   Paste the line
   {                "   Move the beginning of the buffer
    q               " Stop recording
     7@q            " Call macro 'q' 7 times

Answer 3

05AB1E,  14  13 bytes

17F9ÝÀN8αð×ý,

Try it online!

Explanation

17F            # for N in [0 ... 16] do
   9Ý          # push [0 ... 9]
     À         # rotate left
      N8α      # compute the absolute difference between N and 8
         ð×    # push that many spaces
           ý   # merge the list of digits with the space string as separator
            ,  # print

Answer 4

C64 ASM, 358 bytes (102 bytes compiled with basicstub)

This is the closest I could get due to obvious limitations:

jsr $E544
lda #9
sta $FD
nl1: jsr dl
jsr il
dec $FD
bne nl1
inc $FD
nl2: ldy $FD
cpy #9
bne nt1
bl: jmp bl
nt1: iny
sty $FD
jsr dl
jsr il
jmp nl2
dl: clc
ldx #$31
ldy #0
lp: txa
sm: sta $0400, y
inx
cpx #$31
bne c1
rts
c1: cpx #$3A
bne nt2
ldx #$30
clc
nt2: tya
adc $FD
cmp #40
bcc c2
rts
c2: tay
jmp lp
il: lda sm+1
adc #39
bcc nc
inc sm+2
nc: sta sm+1
rts

(Could probably be optimized quite a bit)

Answer 5

TSQL, 220 148 bytes

Improvement posted by ZLK:

DECLARE @ VARCHAR(MAX)=''SELECT TOP 17@+=REPLACE('1@2@3@4@5@6@7@8@9@0','@',SPACE(ABS(9-RANK()OVER(ORDER BY object_id))))+'
'FROM sys.objects PRINT @

Output:

1        2        3        4        5        6        7        8        9        0
1       2       3       4       5       6       7       8       9       0
1      2      3      4      5      6      7      8      9      0
1     2     3     4     5     6     7     8     9     0
1    2    3    4    5    6    7    8    9    0
1   2   3   4   5   6   7   8   9   0
1  2  3  4  5  6  7  8  9  0
1 2 3 4 5 6 7 8 9 0
1234567890
1 2 3 4 5 6 7 8 9 0
1  2  3  4  5  6  7  8  9  0
1   2   3   4   5   6   7   8   9   0
1    2    3    4    5    6    7    8    9    0
1     2     3     4     5     6     7     8     9     0
1      2      3      4      5      6      7      8      9      0
1       2       3       4       5       6       7       8       9       0
1        2        3        4        5        6        7        8        9        0

Answer 6

Haskell, 58 55 bytes

unlines["1234567890">>=(:(' '<$[1..abs n]))|n<-[-8..8]]

Try it online!

This is basically @nimi 's solution :)

Answer 7

APL (Dyalog), 22 bytes

↑∊¨(1+|¯9+⍳17)↑¨¨⊂1⌽⎕D

Try it online!

⎕D Digits from zero to nine

1⌽ rotate one step right (puts zero at end)

⊂ enclose (to treat as unit)

(…)↑¨¨ for each of these numbers, take that many characters from each of the letters:

 ⍳17 one through seventeen

 ¯9+ subtract nine

 | find the absolute values

 1+ add one

∊¨ enlist (flatten) each

↑ change one layer of depth into a level of rank (matrify)

Answer 8

Java 11 (JDK), 98 bytes

o->{for(int i=-9;++i<9;)System.out.println("".join(" ".repeat(i<0?-i:i),"1234567890".split("")));}

Try it online!

• -14 bytes by switching to JDK 11, which now has a native String::repeat.

Previous answer (Java (OpenJDK 8)), 113 112 bytes

o->{for(int i=-9;++i<9;)System.out.printf("".join("%1$"+(i<0?-i:i>0?i:"")+"s","1234567890".split(""))+"%n","");}

Try it online!

Explanations

I'm basically constructing the following String 17 times (N is a variable, not an actual value):

"1%1$Ns2%1$Ns3%1$Ns4%1$Ns5%1$Ns6%1$Ns7%1$Ns8%1$Ns9%1$Ns0%n"

It's all the expected digits, joined by %1$Ns where N is the number of spaces between each digit.

%1$Ns basically means "take the first argument and pad it until length is at least N". Also, %1$0s is not supported so a special case %1$s is made for 0.

Finally, I format-print that string using a single argument: "", so the formatter reuses always the same empty string, padding it with spaces as needed.

Saves

• 1 byte thanks to Kevin Cruijssen

Answer 9

Japt -R, 19 18 16 14 13 bytes

Aõ%A
£qYçÃÔÅê

Test it or try it online

---

Explanation

A             :10
 õ           :Range [1,10]
  %A         :Modulo each by 10
\n           :Assign to variable U
£            :Map each element at 0-based index Y
 q           :  Join U with
  Yç         :   Space repeated Y times
    Ã        :End map
     Ô       :Reverse
      Å      :Slice off first element
       ê     :Mirror
             :Implicitly join with newlines and output

Answer 10

Clojure, 126 99 bytes

-27 by fixing a couple of dumb mistakes. The outer use of a wasn't necessary, so I was able to get rid of a altogether and just write apply str once. That also allowed me to use a function macro for the main function, which saved some bytes. I also inlined the call to Math/abs instead of rebinding n-spaces.

Basically a Clojure port of @Mr.Xcoder's Python idea. In retrospect, I should have used the abs/range method for the diamond challenge originally, but I then I may not have produced the bugged output!

Pretty simple. Joins the number string with a number of spaces that depends on the current row.

#(doseq[n(range -8 9)](println(clojure.string/join(apply str(repeat (Math/abs n)\ ))"1234567890")))

---

(defn bugged []
  (doseq [n-spaces (range -8 9)]
    (println
        (clojure.string/join
          ; "String multiplication"
          (apply str
                 (repeat (Math/abs n-spaces) \space))
          "1234567890"))))

Answer 11

Python 3, 58 57 54 bytes

i=8
while i+9:print(*'1234567890',sep=' '*abs(i));i-=1

Try it online!

(thanks to @flornquake for the last three bytes; I completely forgot using sep to save vs. .join)

Answer 12

SOGL V0.12, 12 bytes

9{SUē↕∑}¹№╬«

Try it Here! Explanation:

9{     }      9 times do
  SU            push "1234567890"
    ē           push e, predefined with the input, which defaults to 0, and then increase it
     ↕          get that many spaces
      ∑         join the string of digits with those spaces
        ¹     collect the results in an array
         №    reverse the array vertically
          ή  palindromize vertically with 1 overlap

Answer 13

JavaScript(ES2017), 83 73 72 68 bytes

Thanks Shaggy for saving 10 bytes. Thanks Craig for saving 1 byte. Thanks arcs for saving 4 bytes.

for(i=-9,a="";i++<8;)a+=[...`1234567890
`].join("".padEnd(i<0?-i:i))

for(i=-9,a="";i++<8;)a+=[...`1234567890
`].join("".padEnd(i<0?-i:i))

console.log(a);

.as-console-wrapper { max-height: 100% !important; top: 0; }
.as-console-row:after { display: none !important; }

Answer 14

Brachylog, 30 29 28 bytes

Ị↺{∧≜;Ṣj₍g;?↔zcc}ᶠ⁹↔;XcP↔Pẉᵐ

Try it online!

Saved one byte thanks to Leaky Nun.

Explanation

Ị↺                               The string "1234567890"
  {               }ᶠ⁹            Find the first 9 outputs of:
   ∧≜                              Take an integer
       ;Ṣj₍                        Juxtapose " " that number of times
           g;?↔z                   Zip that string of spaces with "1234567890"
                cc                 Concatenate twice into one string
                    ↔            Reverse the resuling list
                     ;XcP↔P      Palindromize the list (concatenate X to it into the list P,
                                   P reversed is still P)
                           ẉᵐ    Map writeln

Answer 15

Charcoal, 18 bytes

Ｆ⁹⪫⁺…Ｉ¹:⟦0¶⟧× ι‖Ｂ↑

Try it online! Link is to verbose version of code. Explanation:

Ｆ⁹      Repeat 9 times (i = loop variable)
  ⪫       Join
   ⁺       Concatentate
    …Ｉ¹:    All the characters from str(1) below ":" (i.e. to "9") as a list
    ⟦0¶⟧    A list whose element is the string "0\n"
   × ι     With i spaces
‖Ｂ↑     Reflect everything upwards but without duplicating the top line

Note: Cast(1) takes the same number of bytes because "1" would need a separator before ":" (which as it happens the deverbosifier fails to insert).

Answer 16

R, 108 bytes

for(i in abs(-8:8))cat(paste0(el(strsplit("1234567890","")),paste(rep(" ",i),collapse=""),collapse=""),"\n")

Just pasting and collapsing the strings.

Try it online!

Edit: thanks for Challenger5 for pointing out a problem. Fixed it now.

Edit 2: saved a byte thanks to bouncyball.

Answer 17

Pyth, 14 13 bytes

1 byte thanks to Mnemonic.

V17j*da8NeMST

Try it online!

Answer 18

Java 8, 235 234 206 163 bytes

interface A{static void main(String[]a){for(int i=-8;i<9;i++){String s="";for(int j=1,k;j<10;j++){s+=j;for(k=0;k<(i>0?i:-i);k++)s+=" ";}System.out.println(s+0);}}}

Update : -28 bytes thanks to Leaky Nun !
Update 2 : -43 bytes thanks to Leaky Nun again !

Ungolfed :

interface A {
    static void main(String[] a) {
        for (int i = -8; i < 9; i++) {
            String s = "";
            for (int j = 1, k; j < 10; j++) {
                s += j;
                for (k = 0; k < (i > 0 ? i : -i); k++)
                    s += " ";
            }
            System.out.println(s + 0);
        }
    }
}

Try it online

EDIT : The code earlier was wrong ! Made a mistake while golfing the code, it should work as intended now !

Answer 19

Husk, 21 bytes

mΣṪ`:§+↔tḣR8' ṁs`:0ḣ9

This is a full program that prints to STDOUT. Try it online! There are lots of trailing spaces.

Explanation

Husk is still missing lots of essential stuff like a two-argument range function, so parts of this solution are a little hacky.

mΣṪ`:§+↔tḣR8' ṁs`:0ḣ9
              ṁs`:0ḣ9  This part evaluates to the string "1234567890".
                   ḣ9   Range from 1 to 9.
                `:0     Append 0.
              ṁs        Convert each number to string and concatenate.
     §+↔tḣR8'          This part evaluates to a list like ["  "," ",""," ","  "]
                       but with 17 elements instead of 5.
          R8'           A string of 8 spaces.
         ḣ              Take prefixes.
     §+                 Concatenate
       ↔                the reversal and
        t               tail of the prefix list, palindromizing it.
  Ṫ                    Take outer product of the two lists
   `:                  with respect to flipped prepeding.
                       This prepends each digit to each string of spaces.
mΣ                     Map concatenation over the results, computing the rows.
                       Implicitly join with newlines and print.

Answer 20

T-SQL 145 152 bytes

DECLARE @ VARCHAR(MAX)=''SELECT TOP 17@+=REPLACE('1x2x3x4x5x6x7x8x9x0','x',SPACE(ABS(number-8)))+'
'FROM master..spt_values WHERE type='P'PRINT @

Updated to use:

• master..spt_values to generate numbers (WHERE type = 'P', these are always consecutive, starting at 0)
• @ZLK's TOP 17 idea
• PRINT (to obey the rules - no stringlists)
• @JanDrozen's great idea of including the carriage return in the string (I counted that as just one byte - Windows CRLF what?)
• Idea of @Bridge to use just @ for variable name - all these tricks!!

Results:

1        2        3        4        5        6        7        8        9        0
1       2       3       4       5       6       7       8       9       0
1      2      3      4      5      6      7      8      9      0
1     2     3     4     5     6     7     8     9     0
1    2    3    4    5    6    7    8    9    0
1   2   3   4   5   6   7   8   9   0
1  2  3  4  5  6  7  8  9  0
1 2 3 4 5 6 7 8 9 0
1234567890
1 2 3 4 5 6 7 8 9 0
1  2  3  4  5  6  7  8  9  0
1   2   3   4   5   6   7   8   9   0
1    2    3    4    5    6    7    8    9    0
1     2     3     4     5     6     7     8     9     0
1      2      3      4      5      6      7      8      9      0
1       2       3       4       5       6       7       8       9       0
1        2        3        4        5        6        7        8        9        0

(Thanks @JanDrozen for the REPLACE idea)

Answer 21

Jelly, 15 bytes

8ŒRA⁶ẋØDṙ1¤jЀY

Try it online!

Answer 22

Mathematica, 92 bytes

Column@Join[Reverse@(s=Row/@Table[Riffle[Range@10~Mod~10,""<>Table[" ",i]],{i,0,8}]),Rest@s]

Try it online

copy/paste code with ctrl-v
press shift+enter to run

Answer 23

C, 97 bytes

i=-9;main(j){for(;++i<9;putchar(10))for(j=0;++j<11;printf("        "+8-abs(i)))putchar(48+j%10);}

Your compiler will probably complain a lot about this code; mine threw 7 warnings of 4 different types. Might improve the code later.

Try it online!

Answer 24

SOGL V0.12, 14 bytes

ā9{SUčf@*∑Κ}╬Æ

Try it Here!

Explanation:

ā               push an empty array
 9{        }    9 times do
   SU             push "1234567890"
     č            chop into characters
      f@*         get current iteration (0-based) amount of spaces
         ∑        join with that
          Κ       prepend this to the array
            Β  palindromize vertically with 1 overlap

Answer 25

MathGolf, 11 bytes

8╤{± *♂r╫up

Try it online!

Explanation

8             push 8
 ╤            range(-n, n+1)
  {           for each element in range:
   ±          absolute value
              space character
     *        multiply space character x number of times
      ♂r      push 10, create range
        ╫     left-rotate list (0 goes to the end)
         u    join with separator (joins using x spaces)
          p   print with newline

Has a trailing newline in the end from printing. @ovs found a mistake in the original code that could be fixed by switching the first 9 to an 8.

Answer 26

Charcoal, 21 20 19 18 bytes

Ｆ⁹«Ｆ…¹χ⁺κ… ι0⸿»‖Ｂ↑

Try it online!

Link to the verbose version. Basically I create the lower part of the drawing and then reflect the text up.

Answer 27

CJam, 21 bytes

A,(+aH*ee{:\8-zS**n}/

Try it online!

Explanation

A,     e# Get [0 1 2 ... 9].
(+     e# Rotate the 0 to the end.
aH*    e# Wrap in an array and repeat 17 times.
ee     e# Enumerate. This pairs each copy of the array with its index in
       e# the list.
{      e# For each [index array] pair...
  :\   e#   Unwrap the pair and swap its order.
  8-z  e#   Get the absolute difference of the index from 8.
  S*   e#   Get a string of that many spaces.
  *    e#   Riffle the list of digits with the string of spaces.
  n    e#   Print it with a trailing linefeed.
}/

Answer 28

Batch, 163 bytes

@set s=1 2 3 4 5 6 7 8 9 0
@set t=         
@for %%l in (9 7)do @for /l %%i in (1,1,%%l)do @call:%%l
:7
@set t=  %t%
:9
@set t=%t:~1%
@call echo %%s: =%t%%%

Note: First line ends in 9 spaces. Explanation: Uses creative line numbering! The first loop needs to run 9 times and delete a space each time, while the second loop needs to run 8 times and add a space each time. The former is achieved by running it 7 times and falling through for the 8th time, while the latter is achieved by adding two spaces and falling through to delete one of them again.

Answer 29

bash (and shell utils), 69 bytes

for i in `seq 9 -1 1;seq 2 9`;do seq -f%-$i.f 9|tr -d \\n;echo 0;done

Answer 30

C (gcc), 76 bytes

f(x,y){for(y=-9;++y<9;puts(""))for(x=10;x++;printf("%-*d",abs(y)+1,x%=10));}

Try it online!

It outputs some trailing spaces, which is supposed to be OK.

It prints the numbers using left-justified fields of dynamic length - that's what the format string %-*d is for.

The inner loop has some funny initialization (starts from 10; any multiple of 10 would be fine) to "simplify" its termination condition.