Print this diamond

Question

This question has been spreading like a virus in my office. There are quite a variety of approaches:

Print the following:

        1
       121
      12321
     1234321
    123454321
   12345654321
  1234567654321
 123456787654321
12345678987654321
 123456787654321
  1234567654321
   12345654321
    123454321
     1234321
      12321
       121
        1

Answers are scored in characters with fewer characters being better.

Answer 1

J, 29 26 24 23 22 21 chars

,.(0&<#":)"+9-+/~|i:8

Thanks to FUZxxl for the "+ trick (I don't think I've ever used u"v before, heh).

Explanation

                  i:8  "steps" vector: _8 _7 _6 ... _1 0 1 ... 7 8
                 |     magnitude
              +/~      outer product using +
            9-         inverts the diamond so that 9 is in the center
  (      )"+           for each digit:
      #                  copy
   0&<                   if positive then 1 else 0
       ":                copies of the string representation of the digit
                         (in other words: filter out the strictly positive
                          digits, implicitly padding with spaces)
,.                     ravel each item of the result of the above
                       (necessary because the result after `#` turns each
                        scalar digit into a vector string)

Answer 2

APL (33 31)

A⍪1↓⊖A←A,0 1↓⌽A←⌽↑⌽¨⍴∘(1↓⎕D)¨⍳9

If spaces separating the numbers are allowed (as in the Mathematica entry), it can be shortened to 28 26:

A⍪1↓⊖A←A,0 1↓⌽A←⌽↑⌽∘⍕∘⍳¨⍳9

Explanation:

• (Long program:)
• ⍳9: a list of the numbers 1 to 9
• 1↓⎕D: ⎕D is the string '0123456789', 1↓ removes the first element
• ⍴∘(1↓⎕D)¨⍳9: for each element N of ⍳9, take the first N elements from 1↓⎕D. This gives a list: ["1", "12", "123", ... "123456789"] as strings

• ⌽¨: reverse each element of this list. ["1", "21", "321"...]

• (Short program:)

• ⍳¨⍳9: the list of 1 to N, for N [1..9]. This gives a list [[1], [1,2], [1,2,3] ... [1,2,3,4,5,6,7,8,9]] as numbers.
• ⌽∘⍕∘: the reverse of string representation of each of these lists. ["1", "2 1"...]
• (The same from now on:)
• A←⌽↑: makes a matrix from the list of lists, padding on the right with spaces, and then reverse that. This gives the upper quadrant of the diamond. It is stored in A.
• A←A,0 1↑⌽A: A, with the reverse of A minus its first column attached to the right. This gives the upper half of the rectangle. This is then stored in A again.
• A⍪1↓⊖A: ⊖A is A mirrored vertically (giving the lower half), 1↓ removes the top row of the lower half and A⍪ is the upper half on top of 1↓⊖A.

Answer 3

Clojure, 191 179 bytes

#(loop[[r & s](range 18)h 1](print(apply str(repeat(if(< r 8)(- 8 r)(- r 8))\ )))(doseq[m(concat(range 1 h)(range h 0 -1))](print m))(println)(if s(recur s((if(< r 8)inc dec)h))))

-12 bytes by changing the outer doseq to a loop, which allowed me to get rid of the atom (yay).

A double "for-loop". The outer loop (loop) goes over each row, while the inner loop (doseq) goes over each number in the row, which is in the range (concat (range 1 n) (range n 0 -1)), where n is the highest number in the row.

(defn diamond []
  (let [spaces #(apply str (repeat % " "))] ; Shortcut function that produces % many spaces
    (loop [[row-n & r-rows] (range 18) ; Deconstruct the row number from the range
           high-n 1] ; Keep track of the highest number that should appear in the row
      (let [top? (< row-n 8) ; Are we on the top of the diamond?
            f (if top? inc dec) ; Decided if we should increment or decrement
            n-spaces (if top? (- 8 row-n) (- row-n 8))] ; Calculate how many prefix-spaces to print
        (print (spaces n-spaces)) ; Print prefix-spaces
        (doseq [m (concat (range 1 high-n) (range high-n 0 -1))] ; Loop over the row of numbers
          (print m)) ; Print the number
        (println)

        (if r-rows
          (recur r-rows (f high-n)))))))

Due to a bug in the logic in my first attempt (accidentally inserting the prefix-spaces between each number instead), I managed to get this:

1
1       2       1
1      2      3      2      1
1     2     3     4     3     2     1
1    2    3    4    5    4    3    2    1
1   2   3   4   5   6   5   4   3   2   1
1  2  3  4  5  6  7  6  5  4  3  2  1
1 2 3 4 5 6 7 8 7 6 5 4 3 2 1
12345678987654321
1 2 3 4 5 6 7 8 9 10 9 8 7 6 5 4 3 2 1
1  2  3  4  5  6  7  8  9  8  7  6  5  4  3  2  1
1   2   3   4   5   6   7   8   7   6   5   4   3   2   1
1    2    3    4    5    6    7    6    5    4    3    2    1
1     2     3     4     5     6     5     4     3     2     1
1      2      3      4      5      4      3      2      1
1       2       3       4       3       2       1
1        2        3        2        1
1         2         1

Not even correct ignoring the obvious bug, but it looked cool.

Answer 4

Mathematica 83 49 43 54 51

Print@@#&/@(Sum[k~DiamondMatrix~17,{k,0,8}]/.0->" ")

---

With 3 bytes saved thanks to Kelly Lowder.

Analysis

The principal part of the code, Sum[DiamondMatrix[k, 17], {k, 0, 8}], can be checked on WolframAlpha.

The following shows the underlying logic of the approach, on a smaller scale.

a = 0~DiamondMatrix~5;
b = 1~DiamondMatrix~5;
c = 2~DiamondMatrix~5;
d = a + b + c;
e = d /. 0 -> "";
Grid /@ {a, b, c, d, e}

Answer 5

Python 2, 72 69 67 61

Not clever:

s=str(111111111**2)
for i in map(int,s):print'%8s'%s[:i-1]+s[-i:]

Answer 6

C, 79 chars

v;main(i){for(;i<307;putchar(i++%18?v>8?32:57-v:10))v=abs(i%18-9)+abs(i/18-8);}

Answer 7

Python 2, 60 59

for n in`111111111**2`:print`int('1'*int(n))**2`.center(17)

Abuses backticks and repunits.

Answer 8

Mathematica 55 50 45 41 38

(10^{9-Abs@Range[-8,8]}-1)^2/81//Grid

Grid[(10^Array[{9}-Abs[#-9]&,17]-1)^2/81]

Answer 9

GolfScript, 33 31 30 characters

Another GolfScript solution

17,{8-abs." "*10@-,1>.-1%1>n}%

Thank you to @PeterTaylor for another char.

Previos versions:

17,{8-abs" "*9,{)+}/9<.-1%1>+}%n*

(run online)

17,{8-abs" "*9,{)+}/9<.-1%1>n}%

Answer 10

Javascript, 114

My first entry on Codegolf!

for(l=n=1;l<18;n-=2*(++l>9)-1,console.log(s+z)){for(x=n,s="";x<9;x++)z=s+=" ";for(x=v=1;x<2*n;v-=2*(++x>n)-1)s+=v}

If this can be shortened any further, please comment :)

Answer 11

PHP, 92 90 characters

<?for($a=-8;$a<9;$a++){for($b=-8;$b<9;){$c=abs($a)+abs($b++);echo$c>8?" ":9-$c;}echo"\n";}

Calculates and prints the Manhattan distance of the position from the centre. Prints a space if it's less than 1.

An anonymous user suggested the following improvement (84 characters):

<?for($a=-8;$a<9;$a++,print~õ)for($b=-8;$b<9;print$c>8?~ß:9-$c)$c=abs($a)+abs($b++);

Answer 12

Charcoal (non-competing), 13 bytes

Not competing because the language is (much) newer than the question.

Ｆ⁹«ＧX⁻⁹ιＩ⁺ι¹→

Try it online!

How?

Draws nine, successively smaller, concentric number-diamonds on top of each other:

Ｆ⁹«   Loop ι from 0 to 8:
ＧX     Draw a (filled) polygon with four equilateral diagonal sides
⁻⁹ι      of length 9-ι
Ｉ⁺ι¹    using str(ι+1) as the character
→       Move right one space before drawing the next one

Answer 13

JavaScript, 81

for(i=9;--i+9;console.log(s))for(j=9;j;s=j--^9?k>0?k+s+k:" "+s:k+"")k=i<0?j+i:j-i

Answer 14

Common Lisp, 113 characters

(defun x(n)(if(= n 0)1(+(expt 10 n)(x(1- n)))))(dotimes(n 17)(format t"~17:@<~d~>~%"(expt(x(- 8(abs(- n 8))))2)))

First I noticed that the elements of the diamond could be expressed like so:

  1   =   1 ^ 2
 121  =  11 ^ 2
12321 = 111 ^ 2

etc.

x recursively calculates the base (1, 11, 111, etc), which is squared, and then printed centered by format. To make the numbers go up to the highest term and back down again I used (- 8 (abs (- n 8))) to avoid a second loop

Answer 15

PowerShell (2 options): 92 84 45 bytes

1..8+9..1|%{' '*(9-$_)+[int64]($x='1'*$_)*$x}
1..9+8..1|%{' '*(9-$_)+[int64]($x='1'*$_)*$x}

Thanks to Strigoides for the hint to use 1^2,11^2,111^2...

---

Shaved some characters by:

• Eliminating $w.
• Nested the definition of $x in place of its first use.
• Took some clues from Rynant's solution:
  • Combined the integer arrays with + instead of , which allows elimination of the parenthesis around the arrays and a layer of nesting in the loops.
  • Used 9-$_ to calculate the length of spaces needed, instead of more complicated maths and object methods. This also eliminated the need for $y.

Explanation:

1..8+9..1 or 1..9+8..1 generates an array of integers ascending from 1 to 9 then descending back to 1.

|%{...} pipes the integer array into a ForEach-Object loop via the built-in alias %.

' '*(9-$_)+ subtracts the current integer from 9, then creates a string of that many spaces at the start of the output for this line.

[int64]($x='1'*$_)*$x defines $x as a string of 1s as long as the current integer is large. Then it's converted to int64 (required to properly output 111111111² without using E notation) and squared.

Answer 16

APL (Dyalog Classic), 20 19 bytes

(⍉⊢⍪1↓⊖)⍣2⌽↑,⍨\1↓⎕d

Try it online!

⎕d are the digits '0123456789'

1↓ drop the first ('0')

,⍨\ swapped catenate scan, i.e. the reversed prefixes '1' '21' '321' ... '987654321'

↑ mix into a matrix padded with spaces:

1
21
321
...
987654321

⌽ reverse the matrix horizontally

(...)⍣2 do this twice:

⍉⊢⍪1↓⊖ the transposition (⍉) of the matrix itself (⊢) concatenated vertically (⍪) with the vertically inverted matrix (⊖) without its first row (1↓)

Answer 17

Vim, 62 39 38 keystrokes

Thanks to @DJMcMayhem for saving a ton of bytes!

My first Vim answer, so exciting!

i12345678987654321<ESC>qqYP9|xxI <ESC>YGpHq7@q

I tried to write the numbers via a recording, but it is much longer

Try it online!

Explanation:

i123 ... 321<ESC>                   Write this in insert mode and enter normal mode
qq                                  Start recording into register q
  YP                                Yank this entire line and Paste above
    9|                              Go to the 9th column
      xx                            Delete character under cursor twice
        I <ESC>                     Go to the beginning of the line and insert a space and enter normal mode
               Y                    Yank this entire line
                G                   Go to the last line
                 p                  Paste in the line below
                  H                 Go to the first line
                   q                End recording
                    7@q             Repeat this 7 times

EDIT:

I used H instead of gg and saved 1 byte

Answer 18

CJam, 31 27 bytes

CJam is a lot newer than this challenge, so this answer is not eligible for being accepted. This was a neat little Saturday evening challenge, though. ;)

8S*9,:)+9*9/2%{_W%1>+z}2*N*

Test it here.

The idea is to form the upper left quadrant first. Here is how that works:

First, form the string " 123456789", using 8S*9,:)+. This string is 17 characters long. Now we repeat the string 9 times, and then split it into substrings of length 9 with 9/. The mismatch between 9 and 17 will offset every other row one character to the left. Printing each substring on its own line we get:

        1
23456789 
       12
3456789  
      123
456789   
     1234
56789    
    12345
6789     
   123456
789      
  1234567
89       
 12345678
9        
123456789

So if we just drop every other row (which conveniently works by doing 2%), we obtain one quadrant as desired:

        1
       12
      123
     1234
    12345
   123456
  1234567
 12345678
123456789

Finally, we mirror this twice, transposing the grid in between to ensure that the two mirroring operations go along different axes. The mirroring itself is just

_      "Duplicate all rows.";
 W%    "Reverse their order.";
   1>  "Discard the first row (the centre row).";
     + "Add the other rows.";

Lastly, we just join all lines with newlines, with N*.

Answer 19

k (64 50 chars)

-1'(::;1_|:)@\:((|!9)#'" "),'$i*i:"J"$(1+!9)#'"1";

Old method:

-1',/(::;1_|:)@\:((|!9)#\:" "),',/'+(::;1_'|:')@\:i#\:,/$i:1+!9;

Answer 20

R, 71 characters

For the records:

s=c(1:9,8:1);for(i in s)cat(rep(" ",9-i),s[0:i],s[(i-1):0],"\n",sep="")

Answer 21

Befunge-93, 155 chars

9:v:<,+55<v5*88<v-\9:$_68v
> v>     ^>3p2vpv  -1<!  *
, 1^  2p45*3+9<4:    ,:  +
g -^_75g94+4pg7^!    +^ ,<
1 : ^ `0    :-1$_:68*^$
^1_$:55+\-0\>:#$1-#$:_^

Try it online!

It could definitely be golfed more, but it's my first Funge program and my head already hurts. Had a lot of fun, though

Answer 22

JavaScript, 170 bytes

My first code golf :)

Golfed

a="";function b(c){a+=" ".repeat(10-c);for(i=1;i<c;i++)a+=i;for(i=2;i<c;i++)a+=c-i;a+="\n";}for(i=2;i<11;i++)b(i);for(i=9;i>1;i--)b(i);document.write("<pre>"+a+"</pre>");

Ungolfed

var str = "";
function row(line) {
    str += " ".repeat(10 - line);
    for (var i = 1; i < line; i++) {
        str += i;
    }
    for (var i = 2; i < line; i++) {
        str += line - i;
    }
    str += "\n";
}
for (var line = 2; line < 11; line++) {
    row(line);
}
for (var line = 9; line > 1; line--) {
    row(line);
}
document.write("<pre>" + str + "</pre>");

Answer 23

GolfScript, 36 chars

Assuming that this is meant as a code-golf challenge, here's a basic GolfScript solution:

9,.);\-1%+:a{a{1$+7-.0>\" "if}%\;n}%

Answer 24

Ruby, 76 characters

def f(a)a+a.reverse[1..-1]end;puts f [*1..9].map{|i|f([*1..i]*'').center 17}

Improvements welcome. :)

Answer 25

05AB1E, 17 15 10 bytes

9LJ.pû€û.c

Try it online!

Answer 26

GolfScript (27 chars)

17,{8-abs' '*1`9*1$,>~.*n}/

or

17,{8-abs' '*.1`9*+9<~.*n}/

Both work by building a suitable repunit as a string and then converting to int and squaring to get a Demlo number.

Answer 27

Matlab 227 223 221 209 208 bytes

z=@cellfun;
C=z(@(a,b,c)[a b c fliplr([a b])],[mat2cell(repelem(' ',36),1,8:-1:1),{''}],[{''},mat2cell(nonzeros(tril(repmat(49:56,8,1))')',1,1:8)],num2cell(49:57),'un',0)';
z(@(c)disp(c),[C;flipud(C(1:end-1))])

Answer 28

Octave, 38 bytes

x=abs(-8:8);m=x+x';m(m>8)=25;[57-m,'']

To make it work in MATLAB too, you'd need to write x=ndgrid(abs(-8:8));m=x+x';m(m>8)=25;[57-m,''] or x=meshgrid(abs(-8:8));m=x+x';m(m>8)=25;[57-m,'']

Answer 29

PHP, 99 91 81 byte

for($y=17;$y--;print"
")for($x=17;$x--;)echo($d=9-abs($x-8)-abs($y-8))>0?$d:" ";

run in command line

php -r "CODE_ABOVE"

18 bytes saved by Jörg Hülsermann

Answer 30

Gaia, 9 bytes

9┅…ṫ¦€|ṫṣ

Explanation

9┅         Push [1 2 3 4 5 6 7 8 9]
  …        Prefixes; push [[1] [1 2] [1 2 3] ... [1 2 3 4 5 6 7 8 9]]
   ṫ¦      Palindromize each row: e.g. [1 2 3 4] -> [1 2 3 4 3 2 1]
     €|    Centre-align the rows, padding with spaces
       ṫ   Palindromize the rows
        ṣ  Join with newlines

Answer 31

Jelly,  22 18 12  11 bytes (non-competing)

9ŒḄr1z⁶ṚŒḄY

Try it online!

Done alongside caird coinheringaahing in chat.

How it works

9ŒḄr1z⁶ṚŒḄY - Full program.

9ŒḄ         - The list [1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5, 4, 3, 2, 1].
   r1       - Generate [N, N-1, ..., 1] for each N in ^.
     z⁶     - Zip; transpose ^ with filler spaces.
       Ṛ    - Reverse.
        ŒḄ  - Palindromize without vectorization.
          Y - Join by newlines.

Saved 6 bytes thanks to Leaky Nun!

Answer 32

PowerShell, 49 48

1..9+8..1|%{"  "*(9-$_)+(1..$_+($_-1)..0|?{$_})}

Answer 33

Canvas, 7 bytes

９｛Ｒ⇵］↶┼

Try it here!

Explanation:

9{R⇵]↶┼  
9{  ]    map over i = 1..9
  R        push a range 1..i
   ⇵       and reverse that array
         in Canvas, an array of arrays is a list of lines,
          and the inner arrays are joined together
     ↶   rotate anti-clockwise
      ┼  and quad-palindromize with 1 overlap

I've just fixed a couple built-ins to make a 6 byte version possible:

９｛Ｒ］／┼

Try it here!

9{R]/┼
9{ ]    map over 1..9
  R       create a range 1..i
    /   pad with a diagonal of spaces
     ┼  and quad-palindromize

Answer 34

APL, 24 chars

⍉⊃{⍕⍵↑⍨⍵>0}¨9-∘.+⍨|9-⍳17

Tested in Nars2000 and Dyalog (requires ⎕ML←3 in the latter.)

Explanation

                     ⍳17    starting with the naturals up to 17
                  |9-       generate the numbers from 8 to 0 and back to 8
              ∘.+⍨          make a table of their sum (with 0 in the middle)
            9-              turn it into a diamond with 9 in the middle
  {       }¨                for each number
    ⍵↑⍨⍵>0                  keep it only if it's positive
   ⍕                        then convert the result, if any, to a string
⍉⊃                          disclose the nested array and adjust the dimensions

The last step transposes the result, whose shape is 17 17 1 (because of the disclose ⊃ of nested strings) into 1 17 17, which gets printed like a plain 17 17.

Output

⍉⊃{⍕⍵↑⍨⍵>0}¨9-∘.+⍨|9-⍳17
        1        
       121       
      12321      
     1234321     
    123454321    
   12345654321   
  1234567654321  
 123456787654321 
12345678987654321
 123456787654321 
  1234567654321  
   12345654321   
    123454321    
     1234321     
      12321      
       121       
        1        

Answer 35

Bash, 126 109 87 chars

87:

q()(printf %$[9+$1%9]s\\n $[$2*$2];[ 7 -lt $1 ]||(q $[$1+1] ${2}1;q $[$1+9] $2))
q 0 1

As it usually goes, changing from iterative to recursive solution helps us win additional bytes.

Meaning of parameters to q:

$1 How much to remove from 8 to get the number of spaces in the beginning. Note value modulo 9 counts here (actual value is also a hint to quit recursion).

$2 The current chain of 1s to be squared and output by printf.

The modus operandi is:

1. output the sequence (ie. if $2 is 11111, output 123454321)

2. (if not yet at 12..9..21 - the recursive step)

   2.1. output the next sequence (here: 111111 > $2 , output 12345654321

   2.2. output the sequence once again (123454321).

In the step 2.2 , we pass (indent value + 9) instead of indent value however, so that the algoritm "knows" we are printing the row for the second time. Without this, the [ 7 -lt $1 ] would be false, causing us to retrigger the recursive step 1. This would never finish then.

The recursion goes like this:

q 0 1:                          1
 q 1 11:                       121
  q 2 111:                    12321
   q 3 1111:                 1234321
    q  4 11111:             123454321
     q  5 111111:          12345654321
      q  6 1111111:       1234567654321
       q  7 11111111:    123456787654321
        q  8 111111111: 12345678987654321
        q 16 11111111:   123456787654321
       q 15 1111111:      1234567654321
      q 14 111111:         12345654321
     q 13 11111:            123454321
    q 12 1111:               1234321
   q 11 111:                  12321
  q 10 11:                     121
 q  9 1:                        1

---

109:

p()(printf "%$[8+i]s\n" $[k*k])
k=;for i in `seq 9`;do k+=1;p;done;for i in `seq 8 -1 1`;do k=${k:1};p;done;

"k+=1" is much cheaper as k=$[10*k+1] , and for k being a string of ones it's the same. Same goes for ${k:1} and $[k/10] .

---

126:

p() (printf "%$[$1+i]s\n" $[k*k];)
k=1;for i in `seq 8`;do p 8;k=$[10*k+1];done;for i in `seq 8 -1 0`;do p 9;k=$[k/10];done;

I guess there may be even shorter solution, but weather is glorious, I can't stand sitting in front of computer any more :).

Answer 36

Perl 56 54 characters

Added 1 char for the -p switch.

Uses squared repunits to generate the sequence.

s//12345678987654321/;s|(.)|$/.$"x(9-$1).(1x$1)**2|eg

Answer 37

Perl, 43+1

adding +1 for -E which is required for say

say$"x(9-$_).(1x$_)**2for 1..9,reverse 1..8

edit: shortened a bit

Answer 38

Groovy 77 75

i=(-8..9);i.each{a->i.each{c=a.abs()+it.abs();print c>8?' ':9-c};println""}

old version:

(-8..9).each{a->(-8..9).each{c=a.abs()+it.abs();print c>8?' ':9-c};println""}

Answer 39

Python, 65

for i in map(int,str(int('1'*9)**2)):print' '*(9-i),int('1'*i)**2

Answer 40

Perl 6, 42 41 38 35 chars

say " "x 9-$_,(1 x$_)²for 1…9…1

Try it online!

Answer 41

Scala - 86 characters

val a="543210/.-./012345";for(i<-a){for(j<-a;k=99-i-j)print(if(k<1)" "else k);println}

Answer 42

Javascript, 137

With recursion:

function p(l,n,s){for(i=l;i;s+=" ",i--);for(i=1;i<=n;s+=i++);for(i-=2;i>0;s+=i--);return(s+="\n")+(l?p(l-1,n+1,"")+s:"")}alert(p(8,1,""))

First time on CG :)

Or 118

If I can find a JS implementation that executes 111111111**2 with higher precision.
(Here: 12345678987654320).

a="1",o="\n";for(i=0;i<9;i++,o+="         ".substr(i)+a*a+"\n",a+="1");for(i=8;i;i--)o+=o.split("\n")[i]+"\n";alert(o)

Answer 43

Ruby 59

(-8..8).map{|i|puts' '*i.abs+"#{eval [?1*(9-i.abs)]*2*?*}"}

Answer 44

Ruby, 55

puts (-8..8).map{|i|[?\s*a=i.abs,(?1*(9-a)).to_i**2]*''}

Output:

irb(main):342:0> puts (-8..8).map{|i|[?\s*a=i.abs,(?1*(9-a)).to_i**2]*''}
        1
       121
      12321
     1234321
    123454321
   12345654321
  1234567654321
 123456787654321
12345678987654321
 123456787654321
  1234567654321
   12345654321
    123454321
     1234321
      12321
       121
        1

Answer 45

Groovy, 62, 57 chars

((1..9)+(8..1)).any{println' '*(9-it)+('1'*it as int)**2}

old version:

((1..9)+(8..1)).any{println"${('1'*it as int)**2}".center(17)}

explanation: we create a list [1,2,...,9,8,7,..,1]. Within the closure we create strings '1', '11', '111,..., convert them to numbers, run power of two and center.

Answer 46

Deadfish, 1446 bytes

iisiisddddooooooooiiiiiiiiiiiiiiiiiodddddddddddddddddddddddddddddddddddddddoddddsddddoooooooiiiiiiiiiiiiiiiiioiododddddddddddddddddddddddddddddddddddddddoddddsddddooooooiiiiiiiiiiiiiiiiioioiodododddddddddddddddddddddddddddddddddddddddoddddsddddoooooiiiiiiiiiiiiiiiiioioioiododododddddddddddddddddddddddddddddddddddddddoddddsddddooooiiiiiiiiiiiiiiiiioioioioiodododododddddddddddddddddddddddddddddddddddddddoddddsddddoooiiiiiiiiiiiiiiiiioioioioioiododododododddddddddddddddddddddddddddddddddddddddoddddsddddooiiiiiiiiiiiiiiiiioioioioioioiodododododododddddddddddddddddddddddddddddddddddddddoddddsddddoiiiiiiiiiiiiiiiiioioioioioioioiododododododododddddddddddddddddddddddddddddddddddddddodddsoioioioioioioioiodododododododododddddddddddddddddddddddddddddddddddddddoddddsddddoiiiiiiiiiiiiiiiiioioioioioioioiododododododododddddddddddddddddddddddddddddddddddddddoddddsddddooiiiiiiiiiiiiiiiiioioioioioioiodododododododddddddddddddddddddddddddddddddddddddddoddddsddddoooiiiiiiiiiiiiiiiiioioioioioiododododododddddddddddddddddddddddddddddddddddddddoddddsddddooooiiiiiiiiiiiiiiiiioioioioiodododododddddddddddddddddddddddddddddddddddddddoddddsddddoooooiiiiiiiiiiiiiiiiioioioiododododddddddddddddddddddddddddddddddddddddddoddddsddddooooooiiiiiiiiiiiiiiiiioioiodododddddddddddddddddddddddddddddddddddddddoddddsddddoooooooiiiiiiiiiiiiiiiiioiododddddddddddddddddddddddddddddddddddddddoddddsddddooooooooiiiiiiiiiiiiiiiiiodddddddddddddddddddddddddddddddddddddddo

A more human friendly spaced version:

iisiisdddd oooooooo iiiiiiiiiiiiiiiii o dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooooooo iiiiiiiiiiiiiiiii oiodo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooooo iiiiiiiiiiiiiiiii oioiododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooooo iiiiiiiiiiiiiiiii oioioiodododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooo iiiiiiiiiiiiiiiii oioioioiododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooo iiiiiiiiiiiiiiiii oioioioioiodododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oo iiiiiiiiiiiiiiiii oioioioioioiododododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd o iiiiiiiiiiiiiiiii oioioioioioioiodododododododo dddddddddddddddddddddddddddddddddddddddo
ddds oioioioioioioioiododododododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd o iiiiiiiiiiiiiiiii oioioioioioioiodododododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oo iiiiiiiiiiiiiiiii oioioioioioiododododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooo iiiiiiiiiiiiiiiii oioioioioiodododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooo iiiiiiiiiiiiiiiii oioioioiododododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooooo iiiiiiiiiiiiiiiii oioioiodododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooooo iiiiiiiiiiiiiiiii oioiododo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd ooooooo iiiiiiiiiiiiiiiii oiodo dddddddddddddddddddddddddddddddddddddddo
ddddsdddd oooooooo iiiiiiiiiiiiiiiii o dddddddddddddddddddddddddddddddddddddddo

Answer 47

Javascript 88

for(i=9,a=Math.abs;--i>-9;console.log(o))for(j=9,o='';j-->-9;)o+=(n=9-a(i)-a(j))>0?n:' '

Answer 48

///, 233 bytes

        1
       121
      12321
     1234321
    123454321
   12345654321
  1234567654321
 123456787654321
12345678987654321
 123456787654321
  1234567654321
   12345654321
    123454321
     1234321
      12321
       121
        1

Try it online!

Yay.

Answer 49

k, 37 bytes

r:{x,1_|x};`0:`c$r@r'|8{32,-1_x}\49+!9

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Answer 50

Japt -R, 19 18 17 16 11 bytes

9õõ ®¬êÃê û

Test it

Answer 51

Recursiva, 31 22 bytes

{pB9'P+*" "- 9}J""WpB}

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Explanation:

{pB9'P+*" "- 9}J""WpB}
{                       - For each
 p                      - palindromize [1,2,..9,..1]
  B9                    - range [1,2...9] 
    '                   - Iteration command begin
     P                  - Print
      +                 - concatenate
       *" "- 9 }        - appropriate number of spaces
                J""WpB} - obtain number string
                J""     - Join with nothing '123454321'
                   W    - map each element as string ['1','2'..'5','2','1']
                    p   - palindromize [1,2,..5,..2,1]
                     B} - range [1,2,3,4,5] 

Answer 52

APL (Dyalog Classic), 21 bytes

{⊃⍵/⍕⍵}¨9-+/↑|8-⍳2⍴17

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Answer 53

APL (40)

r←{⍵,1↓⌽⍵}
{⎕←⍵,⍨' '⍴⍨(2×10-⌈/⍵)}¨r¨r⍳¨⍳9

I guess I'm not beating marinus. :p

Answer 54

T-SQL, 116 115 bytes

DECLARE @ INT=8,@d INT=-1a:PRINT SPACE(@)+STUFF('12345678987654321',9-@,2*@,'')IF @=0SET @d=1SET @+=@d If @<9GOTO a

Pure procedural counter and loop, not very "SQL"-like, but the best I could come up with using what SQL offers. Formatted:

DECLARE @ INT=8, @d INT=-1
a:
    PRINT SPACE(@)+STUFF('12345678987654321',9-@,2*@,'')
    IF @=0 SET @d=1
    SET @+=@d
If @<9 GOTO a

Cuts a length out of a hard-coded string using STUFF. Would work just as easily with any other set of characters.

Answer 55

Python 2, 72 bytes

z="123456789"
for i in range(9)+range(8)[::-1]:print"% 8s"%z[:i]+z[i::-1]

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Answer 56

Stax, 10 bytes

▌┼î▲░ò╝╪.¢

Run and debug it

Answer 57

brainfuck, 158 154 bytes

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

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[spaceMem, spaceCount, space, lf, "0", numCount, numMem]
++++++++[->+>++++>+>++++++<<<<]>>>++<<+
[ for spaceCount
    -[- for spaceCount minus 1
        <+ inc spaceMem
        >>. print space
        < go to spaceCount
    ] 
    <[->+<]> fetch spaceCount from spaceMem
    >>>+ inc number
    >[- for numCount
        >+  inc numMem
        <<.+ print and inc number
        >   go to numCount
    ]
    >+[- for numMem plus 1
        <+ inc numCount
        <.- print and decrement number
        >>  go to numMem
    ]
    <<<. print lf
    <<      go to spaceCount
]
>>>>-[      for numCount
    <<<<+[- for spaceCount plus 1
        <+ inc spaceMem
        >>. print space
        < go to spaceCount
    ] 
        <[->+<]> fetch spaceCount from spaceMem
    >>> inc number
    >[- for numCount
        >+  inc numMem
        <<+. print and inc number
        >   go to numCount
    ]
    >-[- for numMem minus 1
        <+ inc numCount
        <-. print and decrement number
        >>  go to numMem
    ]
    <<-     decrement number
    <. print lf
    >> go to numCount
]

Answer 58

MATL, 18 bytes

9Zv-9Zvq!-t0>48*+c

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9Zv - symmetric range from [1:9 8:-1:1]

-9Zv - reverse symmteric range, [9:-1:1 2:9]

q! - decrement that to [8:-1:0 1:8] and transpose to vertical

- - broadcast subtract - matrix of each value in the second range subtracted from each value in the first range

t0> - duplicate that and get a logical matrix of 1s where it's > 0, 0s elsewhere

48* - multiply by 48 to change 1s to 48s (ASCII '0')

+ - add that to the original matrix

c - convert to char and implicitly display

Answer 59

R, 62 bytes

for(i in c(1:9,8:1))cat(rep(" ",9-i),1:i,(i:1)[-1],"
",sep="")

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Answer 60

C++ 223 Byte

#include <iostream>
using std::cout;using std::size_t;int main(){for(int a=0;a<2;++a)for(size_t b=1+a*7;b<10-a;((a!=1)?++b:--b)){size_t c=9-b;for(;c-->0;)cout<<" ";for(c=1;c<b;)cout<<c++;for(c=b;0<c;)cout<<c--;cout<<'\n';}}

Ungolfed:

#include <iostream>
using std::cout; //for not having to type std::cout over and over again
using std::size_t; //for not having to type std::size_t over and over again

int main()
{
    for(int a = 0; a < 2; ++a)
        for(size_t b=1+a*7; b<10-a; ((a!=1)?++b:--b))
        {     //either count up to nine or down from nine
            size_t c = 9-b; //space count we need
            for(; c-- > 0;)
                cout << " ";
            for(c = 1; c < b;) //set c to the counter that will be print
                cout << c++; //post-crement :)
            for(c = b; 0 < c;) //count backwards
                cout << c--; //post-decrement :)
            cout << '\n'; //line is done
        }
}

Answer 61

K, 59

-1'(-:'9+k,1_|k:!9)$,/'$b,1_||:'b:(-1_'a),'|:'a:1_1+!:'!10;

Answer 62

Javascript, 129* 126

for(i=1;i<18;i++){s="";a=Math.abs(9-i);for(j=0;j<a;j++)s+=" ";for(k=a+1;k<=9;k++)s+=k-a;for(l=8;l>a;l--)s+=l-a;console.log(s)}

Includes suggestion from Shmiddty in comments. Original preserved below:

for(i=1;i<18;i++){s="";a=Math.abs(9-i);for(j=0;j<a;j++){s+=" "}for(k=a+1;k<=9;k++){s+=k-a}for(l=8;l>a;l--){s+=l-a}console.log(s)}

I'm sure this could be condensed further, but darned if I know how. :P

Answer 63

Javascript - 118 chars

for(i=1,s=8,d=0;i>0;d?(i=(i-1)/10,s++):(i=i*10+1,s--)){for(d=d||i>11111111,p='',j=0;j++<=s;)p+=' ';console.log(p+i*i)}

Output:

         1
        121
       12321
      1234321
     123454321
    12345654321
   1234567654321
  123456787654321
 12345678987654320
  123456787654321
   1234567654321
    12345654321
     123454321
      1234321
       12321
        121
         1

Unfortunately, I'm also struggling with Javascripts precision problem with the last calculation of 111111111 * 111111111

Answer 64

Haskell, 77 bytes

r=[-8..8]
f n|n<1=" "|1>0=show n
mapM putStrLn[[9-abs x-abs y|x<-r]>>=f|y<-r]

Answer 65

Excel VBA, ₈₁ 76 Bytes

Anonymous VBE immediate window function that takes no input and outputs to the VBE Immediate Window

For i=-8To 8:j=Abs(i):r=Mid(987654321,j+1):?Spc(j)StrReverse(r)Mid(r,2):Next

Old Version, 81 Bytes

For i=-9To 8:For j=-8To 9:k=Abs(i)+Abs(j):l=l &IIf(k>8," ",9-k):Next:?l:l="":Next

Answer 66

SmileBASIC, 65 bytes

FOR I=-8TO 8A=ABS(I)Q=VAL("1"*(9-A))?" "*A;Q*Q
NEXT
LOCATE 16,8?1

I used the fact that 11*11=121, 111*111=12321, etc. Unfortunately, 12345678987654321 can't be stored as a 64 bit float, so I had to add the last 1 separately, adding 14 bytes.

Answer 67

tcl, 122 110

proc P i {puts [format %[expr 8+$i]s [expr [string repe 1 $i]**2]]}
time {P [incr i]} 9
time {P [incr i -1]} 8

demo

Answer 68

Common Lisp SBCL, 145 bytes

(do((b 1)(l'(1 2 3 4 5 6 7 8 9))(i 1(+ i b)))((= i 0))(format t"~17:@<~v{~a~}~v{~a~}~>~%"i l i(reverse(subseq l 0(1- i))))(if(> i 8)(setf b -1)))

It is worse than the other Common Lisp solution, but it works and I think it is different enough to be posted.

Explanation

i is incremented up to 9 and then decremented to 0 (when i=9 sign of b is changed effectively changing addition to subtraction - this avoids second loop.

In each line I print numbers: `123...i(i-1)...1 using loops of format function (for first loop I use list '(1 2 3 4 5 6 7 8 9) and for decrementing loop I use reversed subsequence of this list. Text is then centered.

I didn't notice that numbers are squares of 11..11. As far as this solution is concerned the diamond could be made out of letters or (! @ # ...) (for that you would need to change ~a to ~c in format function.

Answer 69

Python 2, 59 bytes

i=8;exec'print" "*abs(i)+`int("1"*(9-abs(i)))**2`;i-=1;'*17

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Answer 70

Yabasic, 89 bytes

Anonymous function that takes no input and outputs to STDOUT.

For i=-8To 8
For j=-8To 9
k=Abs(i)+Abs(j)
If k>8Then?" ";
Else?Chr$(57-k);
Fi
Next
?
Next

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Answer 71

///, 132 bytes

/T/123//F/T45//f/4321//S/  //s/S //N/
s//a/F678//b/765f//A/1Ns 121Ns/ssSAT21NSTfN FfNF65f
SF6b
 ab
a98b
 ab
SF6bNF65fN FfNSTfNsT2AS1

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Answer 72

05AB1E, 9 bytes

žh¦η€ûû.c

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Answer 73

Canvas, 14 bytes

９｛：９∔ ×；Ｒ］∑９ｎ┼

Try it here!

Answer 74

///, 124 bytes

/(/# '87#//'/"67//&/123//%/43!//$/  //#/654321
//"/&45//!/21
$ /$$$$1
$$$ 1!$ &!$&% "%"#$'('8987($'65%"65% "%$&%$ &!$$1!$$ 1

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Shorter than this answer by 109 bytes! Just applies a bunch of substitutions with whitespace and common numbers.

Answer 75

Attache, 49 bytes

Print@Join=>Table[{If[_%9=_,9-_,sp]}@`+,Abs!-8:8]

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Simply tables (that is, applies a function like a multiplication table) the function which adds two numbers (from [8, 7, 6, 5, ..., 5, 6, 7, 8]), checks if they are less than 9, then yields 9 - that number if so, otherwise a space. Then, prints each row accordingly.

Answer 76

Kotlin, 125 bytes

fun d(){for(i in 0..307){val v=Math.abs(i%18-9)+Math.abs(i/18-8)
print(if(i%18!=0)if(v>8)' '
else(57-v).toChar()
else '\n')}}

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Answer 77

Japt -R, 11 bytes

9õ_õ ¬êÃê û

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Unpacked & How it works

9õZ{Zõ q ê} ê û

9õZ{  [1..9].map(Z=>...)
Zõ      [1..Z]
q       Join with nothing; ["1", "12", ..., "123...9"]
ê       Make palindrome;   ["1", "121", ..., "123...9...321"]
}
ê     Make palindrome on the array
û     Center-pad each element to the longest one
      `-R` joins with newline
      implicit output

I like the õ_õ emoji.

Answer 78

C - 118 characters

x,c,i=1,j;main(){for(;i<20;){x=10;for(j=1;j<20;){j++<10?x--:x++;printf(x>c?" ":"%d",x);}i++<10?c++:c--;printf("\n");}}

My very first code golf! I still need to get into the golfing mindset, I know there is probably a lot I can do differently.

Edit: Try it online!

Answer 79

Powershell, 45 bytes

1..9+8..1|%{' '*(9-$_)+-join(1..$_+$_..1|gu)}

2 solutions with 44 bytes were proposed in the comments to the Iszi's post

Answer 80

JavaScript, 84 bytes

Saved many bytes thanks to ETHProductions

[...s="12345678987654321"].map(x=>"".padEnd(9-x)+s.slice(0,x-1)+s.slice(-x)).join`
`

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Answer 81

Tcl, 99 bytes

time {puts [string repe { } [expr abs([incr i]-9)]][expr [string repe 1 [expr 9-abs($i-9)]]**2]} 17

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Answer 82

C# (Visual C# Interactive Compiler), 103 bytes

for(int i=1,j=1;i>0;i+=j=i>8?-1:j)Write($"{"123456789".Substring(0,i),9}"+"87654321\n".Substring(9-i));

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Definitely not the shortest answer, and this interpreter was published after the question was posted. But I did not see a C# answer. When using the regular C# compiler, the print statement is much longer: System.Console.Write. Also, string interpolation was not a thing when this question was posted.

Answer 83

Java, 155 chars

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

-7 chars thanks to a kind commenter.

Answer 84

Perl 5, 50 bytes

s/-//,$==8-$_,say$"x$_,map$=+1-abs,-$=..$=for-8..8

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Answer 85

Excel, 24 chars (cheating?)

1. In cell B2 put =MAX(A2,B1,B3,C2)-1 (19 chars)
2. Fill down to B18 and right to R18.
3. Select B2 through B18 and Format:Cells:Number:Custom: 0; (2 chars)
4. Enter a 8.9 in J10 (3 char)
5. Ignore all complaints about circular references throughout the process. Use Preferences:Calculation to allow iteration if it is not already allowed.
6. Select columns B through R and resize their width to make square boxes.
7. Profit?

Note: Some might call step 2 cheating as it creates 19 to 23 characters in every cell for a total of over 6000 chars. If you really want to count it that way, you would be better off not putting the formula into those squares that are to remain blank. In that case, you can use a 9 in J10 and you don't need the custom formatting. The total character count would then be just over 3000.

Answer 86

C, 98 chars

l=9,n;p(i){for(i=18;i;putchar(i?n>l?48+n-l:32:10))n=9<--i?18-i:i;}
main(){p(l--);l&&p(main());l++;}

Answer 87

Java, 177 chars

public class A{public static void main(String[]v){
for(int a=-8,b,c;a<9;a++){
for(b=-8;b<9;){c=Math.abs(a)+Math.abs(b++);
System.out.print(c>8?" ":9-c);}System.out.println();}}}

correct formatting of this solution (352 chars):

public class A {
    public void main(String[] args) {
        for (int a = -8; a < 9; a++) {
            for (int b = -8; b < 9;) {
                int c = Math.abs(a) + Math.abs(b++);
                System.out.print(c > 8 ? " " : 9 - c);
            }
            System.out.println("");
        }
    }
}

Answer 88

VBA, 197

for i=1 to 9:?string(27-i*3,32)l;:for j=1 to i:?j;:next:for j=i-1 to 1 step -1:?j;:next:?:next:for i=8 to 1 step -1:?string(27-i*3,32)l;:for j=1 to i:?j;:next:for j=i-1 to 1 step -1:?j;:next:?:next

if vba didn't automatically add a space for the + sign it doesn't print, perhaps I could avoid the 27-i*3 construct for making it look right