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

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 2

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 3

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 4

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 5

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 6

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 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

Haskell, 58 55 bytes

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

Try it online!

This is basically @nimi 's solution :)

Answer 9

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 10

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 11

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 12

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 13

Pyth, 14 13 bytes

1 byte thanks to Mnemonic.

V17j*da8NeMST

Try it online!

Answer 14

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 15

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 16

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 17

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 18

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 19

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 20

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 21

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 22

Jelly, 15 bytes

8ŒRA⁶ẋØDṙ1¤jЀY

Try it online!

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

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 25

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 26

PHP, 69 bytes

for(;$i<17;)echo join(str_pad("",abs($i++-8)),range(1,9)+[9=>0]),"
";

Try it online!

Answer 27

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 28

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 29

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.

Answer 30

PowerShell, 30 bytes

8..0+1..8|%{1..9+0-join' '*$_}

Try it online!

Constructs a range of 8,7,6...2,1,0,1,2...8 then loops through each number. Inside the loop we construct a range of 1..9 concatenated with 0 then -join the numbers in the range together with the appropriate number of spaces. Each string is left on the pipeline, and output is implicit at program completion.

Answer 31

perl 5: 43, 39, 31 bytes

Needs -E flag (which is free per our consensus).

map{$,=$"x abs;say 1..9,0}-8..8

Usage:

perl -E 'map{$,=$"x abs;say 1..9,0}-8..8'

Thanks to Dada

Answer 32

JavaScript, 67 Bytes

f=(i=8)=>i+9?[...`1234567890
`].join(''.padEnd(i<0?-i:i))+f(i-1):''

f=(i=8)=>i+9?[...`1234567890
`].join(''.padEnd(i<0?-i:i))+f(i-1):''

console.log(f())

Answer 33

LOGO, 58 57 bytes

for[j 8 -8][repeat 9[type # repeat abs :j[type "\ ]]pr 0]

This should be run on new versions of FMSLogo (link above), where # is repcount in repeat loops. That is a new feature in FMSLogo, which seems to be undocumented.

Answer 34

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 35

QBIC, 36 32 bytes

[-8,8|?[|?!b%z$';space$(abs(a));

Explanation

[-8,8|     FOR a = -8; a <= 8; a++      Sets upthe number of lines
?          Kick off each line with a \n
[|         FOR b = 1; b <= 10; b++      Loop over each digit: default behaviour for a no-arg FOR declaration
?          PRINT    
 !   $     a strng cast (suppresses QBasic's tendency to space out numbers)
  b%z      of b modulo 10 (z is 10 in QBIC)
 '         Then take the next bit of code as literal QBasic
 ;         Suppress newlines and tabs
 space$    Add x spaces to the output, where X is
 (abs(a))  The absolute line number
 ;         And suppress all newlines and tabs after this too
           A closing ` to this code literal is auto added by QBIC at EOF
           Two closing NEXT statements are also auto-added.

Sample output

Just because I really like the pattern. Note that this took some fiddling in QBasic because the usual terminal output is limited to 80 chars and it wraps.

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 36

PHP, 76 Bytes

Update: Thanks to Soaku for the addition.

for($i=-9;$i++<8;)echo join(str_repeat(' ',abs($i)),range(1,9)+[9=>0])."
";

Try it online !

PHP, 88, 86, 80

Thanks to Caird for suggesting the golfing tips.

for($i=-9;$i++<8;)echo join(str_repeat(' ',abs($i)),[1,2,3,4,5,6,7,8,9,0])."\r";

You can try it here!

Answer 37

Perl 6, 43 37 35 bytes

Originally, I came to:

map({say join " "x abs($_-8),1..9,0},0..16)

The say function in perl 6 automatically adds a newline, and it's map works a bit different from perl 5, requiring less parens due to it's more list based approach. Try it online

Fortunately, Jo King showed how to whittle it down further to 35 bytes:

{say join " "x.abs,1..9,0}for ^17-8

Try it online

Answer 38

R, 51 bytes

for(i in abs(-8:8))cat(1:9,0,"
",sep=strrep(" ",i))

Try it online!

Managed to find some more bytes thanks to JayCe's suggestion in the comments.

Answer 39

C, 135 128 characters

-7 characters thanks to kevin cruijssen

#define p printf
i=1,j,h=9,k;main(){for(;++i<22;i<10?h--:h++,p("\n"))for(j=0;++j<11;)for(p(j==10?"0":"%d",j),k=1;k++<h;)p(" ");}

Try it online!

---

Fun bonus:

On my first compile I got the logic backwards and ended up with the output in reverse

i=1,j,k,h=9;main(){for(;i++<20;){for(j=0;j++<10;){printf(j==10?"0":"%d",j);for(k=1;k++<10-h;){printf(" ");}}i<11?h--:h++;printf("\n");}}

Edit: removed missed whitespace

Answer 40

MathGolf, 11 bytes

9╤{± *♂r╫up

Try it online!

Explanation

9             push 9
 ╤            range(-n, n+1)
  {           start block or arbitrary length
   ±          absolute value
              space character
     *        pop a, b : push(a*b)
      ♂       push 10
       r      range(0, n)
        ╫     left-rotate bits in int, list/str
         u    join with separator
          p   print with newline

Answer 41

V, 18 bytes

¬19á08ñÄÓä/& 
YGpH

Try it online!

Answer 42

Swift 3.1, 186 182 bytes

func f(_ n:Int=0,_ u:Bool=true){let p={if n<8{f(n+1,u)}};let q={print((1...9).map{String($0)+String(repeating:" ",count:n)}.joined()+"0")};if u{q();p()}else{p();q()}};f(1,false);f()

Explanation

func f(_ n:Int=0,_ u:Bool=true){          function (recursive) that does
                                          all the heavy-lifting =) 
                                          n - number of spaces,
                                          u - go "up" or "down
let p={if n<8 {f(n+1,u)}}                 closure that runs f on higher step if can
let q={                                   closure which does the printing
print((1...9).map{                        print closed range of number mapped...
String($0)+String(repeating:" ",count:n)  ...making String from each number in range
                                          and appending n spaces to it
.joined()+"0")                            join all "1    "-like strings and append "0"
if u{q();p()}                             if go "up" - then call f before printing
else{p();q()}                             if go "down - the other way
f(1,false);f()                            run "up" then "down"

Removed some silly spaces. Thanx to Kevin Cruijssen

Answer 43

Common Lisp, 85 bytes

(dotimes(i 17)(format t(format()"~~{~~,,~aa~~}
"(abs(- i 8)))'(1 2 3 4 5 6 7 8 9 0)))

Try it online!

format “magic” of Common Lisp:

(dotimes (i 17)                     ; for i from 0 to 16
  (format t                         ; print with a format string generated from the next call
          (format ()                ; with nil as parameter generates a string instead of printing
                  "~~{~~,,~aa~~}
"                                   ; generate the format string "~{~,,na~}~%" with n given
                  (abs(- i 8)))     ;     by the value of abs(i-8) (number of spaces)
          '(1 2 3 4 5 6 7 8 9 0)))  ; the data to be printed for each line

Answer 44

PHP, 71 bytes

while($y<17)echo chunk_split(1234567890,1,str_pad("",abs($y++-8))),"
";

Run with -nr or test it online.

Answer 45

MATL, 21 bytes

-9Zv"l4Y29YS10:q@*Q(c

Try it online!

Explanation:

-9Zv                  % Generates [9 ... 2, 1, 2, ... 9]. 
    "                 % Iterate over this vector.
                   (  % Assignment indexing. Places elements B at positions C in existing vector A.
     l                %   A: Dummy vector with numeric [1] in it. Is always overwritten by char '1'.
      4Y29YS          %   B:  '0':'9', circularly shifted 9 positions to get '1':'9','0'
            10:q@*Q   %   C: Equally spaced indexing vector, spacing based on loop variable @.
                    c % Convert to char, because A was numeric.

Golfing done:

-9Zv     -->  built-in, instead of -8:8|Q or even 8:PQ9:h| 
10:q@*Q  -->  2 bytes shorter than l@@10*3$:
l ... c  -->  1 byte shorter than 'x' (char dummy vector A)

I have attempted to get rid of the 9YS by doing the circular shift in the indexing, but this increased the byte count (10t:qwX\q@*Q instead of 9YS10:q@*Q). However, some interesting patterns were obtained in the process: Try it online!, Try it online!.

Answer 46

Bubblegum, 140 bytes

:P

0000000: ad92 510e c230 0c43 ff39 0547 802c f3bc  ..Q..0.C.9.G.,..
0000010: fb5f 8c10 c5d1 5a09 10b0 7d59 b1fd b4b6  ._....Z...}Y....
0000020: b95f eb33 8945 c225 5609 486c 1294 d825  ._.3.E.%V.Hl...%
0000030: 6e17 216d 02fa 84c3 04a3 5033 c906 8e0f  n.!m......P3....
0000040: 140c 0c8a 3000 ec50 f743 1987 2abb d83d  ....0..P.C..*..=
0000050: eb96 7707 dda0 f215 b70a 7b45 5141 66ac  ..w.......{EQAf.
0000060: 5296 19cf 04d2 67ba 4fd3 c2f2 3010 63c6  R.....g.O...0.c.
0000070: 3066 b6f8 8a8d fb0b fb0d eff3 2f7d 77ce  0f........../}w.
0000080: 1faf f3df 773c 7fa5 cedf f707            ....w<......

Try it online!

Do I get brownie points for having ._. two times in the dump?

Answer 47

Perl 5, 31 bytes

$,=$"x abs$_,say 1..9,0for-8..8

Try it online!

Answer 48

Cheddar, 42 bytes

-8|>8=>n->print(1|>9+[0]).join(abs(n)*' ')

Try it online!

Answer 49

Stacked, 33 bytes

9:>' '*[9~>\#`]"!:8~>#rev\,$out"!

Try it online!

Explanation

9:>' '*[9~>\#`]"!:8~>#rev\,$out"!
9:>                                 range from 0 to 9 (right-exclusive)
   ' '*                             vectorized repetition of the space
       [      ]"!                   on each repetition:
        9~>                           generate a range from 1 to 9
           \#`                        and join it by that repetition
                 :8~>#              copy all but the first element
                      rev           reverse it
                         \,         swap then concatenate
                           $out"!   output each

Answer 50

///, 251 bytes

/B/' //A/!!//@/" //?/\/\///>/  ?=/0
1?</ 2 3 4 5 6 7 8 9 ?;/!2!3!4!5!6!7!8!9!?:/'2'3'4'5'6'7'8'9'?-/="2"3"4"5"6"7"8"9"=?,/@2@3@4@5@6@7@8@9@?'/!"?&/=>2>3>4>5>6>7>8>9>=?%/=A2A3A4A5A6A7A8A9A=?#/1B2B3B4B5B6B7B8B9B0?"/! ?!/> /#
1:%,-;&<=23456789=<&;-,%:0
#

Try it online! Or here, interactively!

Considerably long, but /// has no real way of modifying data elegantly.

Answer 51

Delphi / Pascal 137 Bytes*

program x;{$APPTYPE CONSOLE}var i,j: Integer;begin for i:=-8 to 8do begin write(#10,1); for j:=2 to 10do Write(j mod 10:Abs(i)+1);end;end.

*The {$APPTYPE CONSOLE} is a compiler directive that can be removed for pure Pascal. or if compiled with the Delphi command line compiler using the -cc switch to give 119 bytes. Without it you get a runtime error as it tries to print to a non existent console.

Answer 52

q/kdb+, 37 bytes

Solution:

-1" "{raze(1_11#.Q.n),\:y#x}'8-(!)17;

Example:

q)-1" "{raze(1_11#.Q.n),\:y#x}'8-(!)17;
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        

Explanation:

q is evaluation right-to-left. Here we are taking the range of -8..0..8, and joining that many spaces with the list "1234567890". We then use -1 to print out to stdout (rather than returning the list of strings).

-1 " "{raze (1_11#.Q.n),\:y#x}'8 - til 17; / ungolfed solution
                                   til 17  / range of 0..16
                               8 -         / take 8 from each value, gives -8..0..8
   " "{                      }'            / each left/right, passes in " " and then -8, -7, etc..
                          y#x              / take from list " ", note: negatives take from 'back' of list
                       ,\:                 / \: is 'each left', , means concatenate with, so take each left and concatenate with the right
            (1_11#.Q.n)                    / 'the left'. .Q.n is range 0-9 as a string, take 11 to get 0123..90, then drop the first
       raze                                / reduce down output list into a single string
-1                                       ; / print to stdout and swallow return value (-1)

Notes:

Doing (1_11#.Q.n) is 1 byte shorter than the much more straightforward "1234567890"... but as .Q.n returns "0123456789" some manipulation is required.

Answer 53

Ruby 2.4, 40 bytes

-8.upto 8{|i|puts [*1..9,0]*(' '*i.abs)}

Ruby (lower than 2.4), 41 bytes

-8.upto(8){|i|puts [*1..9,0]*(' '*i.abs)}

Try it online!

Answer 54

[Java 7], 137 132 bytes

Saved 3 bytes thanx to Kevin Cruijssen and 2 more thanx to TheLethalCoder

void a(){int i,j,k;for(i=-8;i<9;i++){String a="";for(j=1;j<11;j++){a+=j%10;for(k=0;k<(i<0?-i:i);k++)a+=" ";}System.out.println(a);}}

 void a(){for(int i=-8;i<9;i++){String a="";for(int j=1;j<11;j++){a+=j%10;for(int k=0;k<Math.abs(i);k++){a+=" ";}}System.out.println(a);}}

Ungolfed:

  void a(){
    int i,j,k;
    for(i=-8;i<9;i++){
        String a="";
        for(j=1;j<11;j++){
            a+=j%10;
            for(k=0;k<(i<0?-i:i);k++)a+=" ";
        }
        System.out.println(a);
    }
}

Answer 55

C#, 131 bytes

My first participation

for(int i=8;i>-9;i--){for(int n=1;n<=10;n++){Console.Write((n==10?"0":n.ToString())+new String(' ',Math.Abs(i))+(n==10?"\n":""));}}

More easy reading version

for (int i = 8; i > -9; i--)
{
    for (int n = 1; n <= 10; n++)
    {
        Console.Write((n==10 ? "0" : n.ToString()) + new String(' ', Math.Abs(i)) + (n == 10 ? "\n" : ""));
    }
}

Answer 56

tcl, 112

Very naïve:

proc x n\ s {time {puts [join {1 2 3 4 5 6 7 8 9 0} [format %[incr n $s]s \ ]]} 9}
x 10 -1
puts 1234567890
x 0 1

demo

Answer 57

/// (Slashes), 377 374 bytes

/a/1_//b/2_//c/3_//d/4_//e/5_//f/6_//g/7_//h/8_//i/9_//z/0
//./  //-/. //_/../a_b_c_d_e_f_g_h_i_za-b-c-d-e-f-g-h-i-za.b.c.d.e.f.g.h.i.za b c d e f g h i zabcdefghiz1-2-3-4-5-6-7-8-9-z1.2.3.4.5.6.7.8.9.z1 2 3 4 5 6 7 8 9 z123456789z1 2 3 4 5 6 7 8 9 z1.2.3.4.5.6.7.h.9.z1-2-3-4-5-6-7-8-9-zabcdefghiza b c d e f g h i za.b.c.d.e.f.g.h.i.za-b-c-d-e-f-g-h-i-za_b_c_d_e_f_g_h_i_0

Try it online!

Explanation

The original output was initially compressed by doing the following replacements:

•      (4 spaces) → _
•     (3 spaces) → -
•    (2 spaces) → .

In a second step, 1_ was replaced by a, 2_ by b and so on. Finally, those mappings were inverted and prepended to the code.

Additional step (thx to CalculatorFeline):

/./  //-/   //_/    / → /_/..//-/. //./  /

Answer 58

Sink, 102 100 bytes

var v=range 0,17
var p=list.push(range 0,10),0
for var s:v
say list.join p,str.pad"",num.abs s-8
end

Try it online! (sort of, it's just a repl so you'll need to type it in)

-2 bytes thanks to @Kevin Cruijssen

Answer 59

Cubically, 2622 bytes

U3D1R3L1F3B1U1D3+3111111@6-2111@66666666+21111@6-21111@66666666+211111@6-211111@66666666+0@6-0@66666666+01@6-01@66666666+011@6-011@66666666+0111@6-0111@66666666+4@6-4@66666666+41@6-41@66666666+211@6-511111@6+5111111@6-2111@6666666+21111@6-21111@6666666+211111@6-211111@6666666+0@6-0@6666666+01@6-01@6666666+011@6-011@6666666+0111@6-0111@6666666+4@6-4@6666666+41@6-41@6666666+211@6-511111@6+5111111@6-2111@666666+21111@6-21111@666666+211111@6-211111@666666+0@6-0@666666+01@6-01@666666+011@6-011@666666+0111@6-0111@666666+4@6-4@666666+41@6-41@666666+211@6-511111@6+5111111@6-2111@66666+21111@6-21111@66666+211111@6-211111@66666+0@6-0@66666+01@6-01@66666+011@6-011@66666+0111@6-0111@66666+4@6-4@66666+41@6-41@66666+211@6-511111@6+5111111@6-2111@6666+21111@6-21111@6666+211111@6-211111@6666+0@6-0@6666+01@6-01@6666+011@6-011@6666+0111@6-0111@6666+4@6-4@6666+41@6-41@6666+211@6-511111@6+5111111@6-2111@666+21111@6-21111@666+211111@6-211111@666+0@6-0@666+01@6-01@666+011@6-011@666+0111@6-0111@666+4@6-4@666+41@6-41@666+211@6-511111@6+5111111@6-2111@66+21111@6-21111@66+211111@6-211111@66+0@6-0@66+01@6-01@66+011@6-011@66+0111@6-0111@66+4@6-4@66+41@6-41@66+211@6-511111@6+5111111@6-2111@6+21111@6-21111@6+211111@6-211111@6+0@6-0@6+01@6-01@6+011@6-011@6+0111@6-0111@6+4@6-4@6+41@6-41@6+211@6-511111@6+5111111@6+1@6+1@6+1@6+1@6+1@6+1@6+1@6+1@6-111111111@6-511111@6+5111111@6-2111@6+21111@6-21111@6+211111@6-211111@6+0@6-0@6+01@6-01@6+011@6-011@6+0111@6-0111@6+4@6-4@6+41@6-41@6+211@6-511111@6+5111111@6-2111@66+21111@6-21111@66+211111@6-211111@66+0@6-0@66+01@6-01@66+011@6-011@66+0111@6-0111@66+4@6-4@66+41@6-41@66+211@6-511111@6+5111111@6-2111@666+21111@6-21111@666+211111@6-211111@666+0@6-0@666+01@6-01@666+011@6-011@666+0111@6-0111@666+4@6-4@666+41@6-41@666+211@6-511111@6+5111111@6-2111@6666+21111@6-21111@6666+211111@6-211111@6666+0@6-0@6666+01@6-01@6666+011@6-011@6666+0111@6-0111@6666+4@6-4@6666+41@6-41@6666+211@6-511111@6+5111111@6-2111@66666+21111@6-21111@66666+211111@6-211111@66666+0@6-0@66666+01@6-01@66666+011@6-011@66666+0111@6-0111@66666+4@6-4@66666+41@6-41@66666+211@6-511111@6+5111111@6-2111@666666+21111@6-21111@666666+211111@6-211111@666666+0@6-0@666666+01@6-01@666666+011@6-011@666666+0111@6-0111@666666+4@6-4@666666+41@6-41@666666+211@6-511111@6+5111111@6-2111@6666666+21111@6-21111@6666666+211111@6-211111@6666666+0@6-0@6666666+01@6-01@6666666+011@6-011@6666666+0111@6-0111@6666666+4@6-4@6666666+41@6-41@6666666+211@6-511111@6+5111111@6-2111@66666666+21111@6-21111@66666666+211111@6-211111@66666666+0@6-0@66666666+01@6-01@66666666+011@6-011@66666666+0111@6-0111@66666666+4@6-4@66666666+41@6-41@66666666+211@6

Pretty simple, turns the cube to get it into this:

   242
   202
   242
000131555313
010121535343
000131555313
   424
   454
   424

Now the LEFT face sums up to 1. So we use the other faces to get as close to the desired ASCII values as possible, then just increment/decrement by 1.

This also works:

Cubically, 2938 bytes

:4/1+5@6/1+3@66666666+2@6-2@66666666:5+1/1+5@6/1+3@66666666:5+2/1+5@6/1+3@66666666:5+3/1+5@6/1+3@66666666:5+1@6:5/1+3@66666666:1/1+51@6:5/1+3@66666666:2/1+51@6:5/1+3@66666666/1+51@6:5/1+3@66666666/1+5@6:1/1+1@6:4/1+5@6/1+3@6666666+2@6-2@6666666:5+1/1+5@6/1+3@6666666:5+2/1+5@6/1+3@6666666:5+3/1+5@6/1+3@6666666:5+1@6:5/1+3@6666666:1/1+51@6:5/1+3@6666666:2/1+51@6:5/1+3@6666666/1+51@6:5/1+3@6666666/1+5@6:1/1+1@6:4/1+5@6/1+3@666666+2@6-2@666666:5+1/1+5@6/1+3@666666:5+2/1+5@6/1+3@666666:5+3/1+5@6/1+3@666666:5+1@6:5/1+3@666666:1/1+51@6:5/1+3@666666:2/1+51@6:5/1+3@666666/1+51@6:5/1+3@666666/1+5@6:1/1+1@6:4/1+5@6/1+3@66666+2@6-2@66666:5+1/1+5@6/1+3@66666:5+2/1+5@6/1+3@66666:5+3/1+5@6/1+3@66666:5+1@6:5/1+3@66666:1/1+51@6:5/1+3@66666:2/1+51@6:5/1+3@66666/1+51@6:5/1+3@66666/1+5@6:1/1+1@6:4/1+5@6/1+3@6666+2@6-2@6666:5+1/1+5@6/1+3@6666:5+2/1+5@6/1+3@6666:5+3/1+5@6/1+3@6666:5+1@6:5/1+3@6666:1/1+51@6:5/1+3@6666:2/1+51@6:5/1+3@6666/1+51@6:5/1+3@6666/1+5@6:1/1+1@6:4/1+5@6/1+3@666+2@6-2@666:5+1/1+5@6/1+3@666:5+2/1+5@6/1+3@666:5+3/1+5@6/1+3@666:5+1@6:5/1+3@666:1/1+51@6:5/1+3@666:2/1+51@6:5/1+3@666/1+51@6:5/1+3@666/1+5@6:1/1+1@6:4/1+5@6/1+3@66+2@6-2@66:5+1/1+5@6/1+3@66:5+2/1+5@6/1+3@66:5+3/1+5@6/1+3@66:5+1@6:5/1+3@66:1/1+51@6:5/1+3@66:2/1+51@6:5/1+3@66/1+51@6:5/1+3@66/1+5@6:1/1+1@6:4/1+5@6/1+3@6+2@6-2@6:5+1/1+5@6/1+3@6:5+2/1+5@6/1+3@6:5+3/1+5@6/1+3@6:5+1@6:5/1+3@6:1/1+51@6:5/1+3@6:2/1+51@6:5/1+3@6/1+51@6:5/1+3@6/1+5@6:1/1+1@6:4/1+5@6/1+5@6:5+1/1+5@6:5+2/1+5@6:5+3/1+5@6:5+1@6:1/1+51@6:2/1+51@6:3/1+51@6-1@6:1/1+1@6:4/1+5@6/1+3@6+2@6-2@6:5+1/1+5@6/1+3@6:5+2/1+5@6/1+3@6:5+3/1+5@6/1+3@6:5+1@6:5/1+3@6:1/1+51@6:5/1+3@6:2/1+51@6:5/1+3@6/1+51@6:5/1+3@6/1+5@6:1/1+1@6:4/1+5@6/1+3@66+2@6-2@66:5+1/1+5@6/1+3@66:5+2/1+5@6/1+3@66:5+3/1+5@6/1+3@66:5+1@6:5/1+3@66:1/1+51@6:5/1+3@66:2/1+51@6:5/1+3@66/1+51@6:5/1+3@66/1+5@6:1/1+1@6:4/1+5@6/1+3@666+2@6-2@666:5+1/1+5@6/1+3@666:5+2/1+5@6/1+3@666:5+3/1+5@6/1+3@666:5+1@6:5/1+3@666:1/1+51@6:5/1+3@666:2/1+51@6:5/1+3@666/1+51@6:5/1+3@666/1+5@6:1/1+1@6:4/1+5@6/1+3@6666+2@6-2@6666:5+1/1+5@6/1+3@6666:5+2/1+5@6/1+3@6666:5+3/1+5@6/1+3@6666:5+1@6:5/1+3@6666:1/1+51@6:5/1+3@6666:2/1+51@6:5/1+3@6666/1+51@6:5/1+3@6666/1+5@6:1/1+1@6:4/1+5@6/1+3@66666+2@6-2@66666:5+1/1+5@6/1+3@66666:5+2/1+5@6/1+3@66666:5+3/1+5@6/1+3@66666:5+1@6:5/1+3@66666:1/1+51@6:5/1+3@66666:2/1+51@6:5/1+3@66666/1+51@6:5/1+3@66666/1+5@6:1/1+1@6:4/1+5@6/1+3@666666+2@6-2@666666:5+1/1+5@6/1+3@666666:5+2/1+5@6/1+3@666666:5+3/1+5@6/1+3@666666:5+1@6:5/1+3@666666:1/1+51@6:5/1+3@666666:2/1+51@6:5/1+3@666666/1+51@6:5/1+3@666666/1+5@6:1/1+1@6:4/1+5@6/1+3@6666666+2@6-2@6666666:5+1/1+5@6/1+3@6666666:5+2/1+5@6/1+3@6666666:5+3/1+5@6/1+3@6666666:5+1@6:5/1+3@6666666:1/1+51@6:5/1+3@6666666:2/1+51@6:5/1+3@6666666/1+51@6:5/1+3@6666666/1+5@6:1/1+1@6:4/1+5@6/1+3@66666666+2@6-2@66666666:5+1/1+5@6/1+3@66666666:5+2/1+5@6/1+3@66666666:5+3/1+5@6/1+3@66666666:5+1@6:5/1+3@66666666:1/1+51@6:5/1+3@66666666:2/1+51@6:5/1+3@66666666/1+51@6:5/1+3@66666666/1+5@6

Uses the existing cube and division to get the required ASCII values.

Found using these answers.

Answer 60

Lua, 106 85 bytes

-21 bytes thanks to user202729

n="1L2L3L4L5L6L7L8L9L0"for i=-8,8,1 do print((n:gsub("L",(" "):rep(math.abs(i)))))end

Output has a trailing newline

Ungolfed

n="1L2L3L4L5L6L7L8L9L0"
for i=-8,8,1 do 
    print((n:gsub("L",(" "):rep(math.abs(i)))))
end

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Answer 61

J, 25 bytes

'1',.(1+|i:8)":"{0,~2+i.8

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Answer 62

brainfuck, 209 bytes

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

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I noticed there wasn’t a brainfuck submission yet, so I gave it my best shot. Definitely not the most golfed solution possible, but I’m not upset with the result.

Answer 63

Canvas, 20 15 bytes

９｛９∔ ×９Ｒ0∔∑；＊］─

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Explanation:

９｛９∔ ×９Ｒ0∔∑；＊］─  | (Full code)
                        | Print (implicit)
    ９∔ ×               | 9-i spaces
         ９Ｒ0∔∑；＊    | Inserted between the characters of the string "1234567890"
９｛                    | For each i in [1-9]
                   ］   | Separated by newlines
                     ─  | Palindromized vertically with 1 line of overlap

Answer 64

QBasic, 66 bytes

An undeclared subroutine that takes no input and outputs to STDOUT

FOR i=-8TO 8
FOR j=49TO 57
?CHR$(j)SPC(ABS(i));
NEXT
?"0
NEXT

Answer 65

Elixir, 126 bytes

Enum.each 8..-8,fn n->IO.puts Enum.map Enum.into(1..10,[],&(rem &1,10)),&(Integer.to_string(&1)<>String.duplicate" ",abs n)end

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Enum.each 8..-8,
          fn n->
            IO.puts 
              Enum.map 
                Enum.into(1..10, [], &(rem &1,10)),
                &(Integer.to_string(&1) <> String.duplicate" ",abs n)
          end

For each number in the range 8 to -8, do this anonymous function with the number as n. The function (inner to outer) sets the range 1 to 10 into a list after applying the remainder mod 10 (so we have the list 1,2,3,4,5,6,7,8,9,0). Then maps that list of numbers into strings, and concatenates the absolute value of n spaces onto the end. After getting each string we print out the whole list of strings... and Elixir doesn't put any delimiter on that by default.

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 66

Pip -l, 18 bytes

(\,t)%tJsXAB_M-8,9

The -l flag outputs the list of results one per line. Try it online!

Answer 67

Sisi, 239 bytes

0set s8
1set d0-1
2set o""
3set i1
4set o o+i
5jumpif i7
6jump17
7set x s
8jump11
9set o o+" "
10set x x-1
11jumpif x9
12set c9>i
13jumpif c15
14set i0-1
15set i1+i
16jump4
17print o
18set s s+d
19jumpif s21
20set d1
21set q9>s
22jumpif q2

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The basic strategy is to run a loop from 8 down to 0 and back up to 8, constructing each line accordingly and printing it. I'm not going to write a detailed explanation, but if you're trying to follow the algorithm, the following key to the variable names may be helpful:

• s: number of spaces to print between digits
• d: direction of change in s (-1 or 1)
• o: the current line of output
• i: the current digit
• x: loop counter while printing spaces
• c: conditional that tests whether i is less than 9 (if not, we set it to -1 so it gets incremented to 0 instead of 10)
• q: conditional that tests whether s is less than 9 (if so, we continue looping)

Answer 68

///, 244 bytes

/*/\/\///X/0
1*j/\/X*x/  *S/x *s/SS*-/S *_/s *(/- *)/_ *A/)2)3)4)5)6)7)8)9)*Bj_2_3_4_5_6_7_8_9_*Cjs2s3s4s5s6s7s8s9s*Dj(2(3(4(5(6(7(8(9(*Ej-2-3-4-5-6-7-8-9-*FjS2S3S4S5S6S7S8S9S*Gjx2x3x4x5x6x7x8x9x*Hj 2 3 4 5 6 7 8 9 /1ABCDEFGHX23456789HGFEDCBXA0

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Answer 69

Pushy, 27 bytes

Not the most elegant solution, but it does the job.

N9X@wL:Ocv:32;ZT:hT%#^"v;T'

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Explanation:

N                                \ Set printing delimiter to empty string
 9X                              \ Push range(9) to stack: [0, 1, 2, 3, 4, 5, 6, 7, 8]
   @                             \ Reverse: [8, 7, 6, 5, 4, 3, 2, 1, 0]
    w                            \ Mirror: [8, 7, 6... 1, 0, 1... 6, 7, 8]
     L:                          \ Length (17) times do:
       Oc                        \   Go to (cleared) auxiliary stack
         v:                      \   Pop top of main stack, and that many times:
           32;                           Push character 32 (space)
              Z                  \   Push 0
               T                 \   Ten times do:
                :hT%             \      counter = (counter + 1) mod 10
                    #            \      Print the counter
                     ^"v;        \      Print the rest of the stack (spaces)
                         T'      \   Print a newline

Answer 70

Tcl, 99 bytes

time {incr i;time {puts -nonewline [incr j][string repe " " [expr abs(9-$i)]]} 9;puts 0;unset j} 17

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Answer 71

Python 2, 72 bytes

for i in range(-8,9):print(('%-'+'%ds'%-~abs(i))*10)%tuple('1234567890')

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Answer 72

Vim, 53 bytes

i1234567890^[^qq8a ^[lq8@qqeYpPqqqlxbequ^qw9@q@eq7@wdj

Where ^[ is the ESC Key.

i1234567890^[^       puts 1-9+0 in the buffer and returns the cursor to the beginning
qq8a ^[lq8@q         puts 8 spaces between each of the numbers
qeYpPq               records @e to duplicate the line twice and stay on the middle one
qqlxbequ^            records @q to delete one space between numbers and undoes it
qw9@q@eq             records @w to do @q 9 times, then @e
7@wdj                does @w 7 more times, then deletes two lines

Answer 73

Python, 84 bytes

for x in range(-8,9):
 z=""
 for y in range(1,11):z+=str(y%10)+" "*abs(x)
 print(z)

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Answer 74

VBA, 58 bytes

An immediate window function that takes no input and outputs to the console.

For i=-8To 8:For j=1To 9:?j &""Spc(Abs(i));:Next:?"0":Next

Ungolfed and Commented

For i=-8To 8        ''  Iterate `i` from `-8` to `8`
For j=1To 9         ''  Iterate `j` from `1` to `9`
?j &""Spc(Abs(i));  ''  Print `j` and `i` spaces
Next                ''  Loop `j`
?"0"                ''  Print `0`, with a newline
Next                ''  Loop `i`

Answer 75

JavaScript, 71 bytes

f=(i=8,q=[..."1234567890"].join(" ".repeat(i)))=>i?q+`
`+f(i-1)+`
`+q:q