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.

What is the winning criterion ? And is this a challenge or a golf ? — Paul R, Oct 12 '12 at 15:33
@DavidCarraher "kolmogorov-complexity" was edited in after the question was asked. The original questioner has not specified the winning criteria yet. — Gareth, Oct 12 '12 at 20:56
@Gareth My comment was made after the "kolmogorov-complexity" tag was added but before the "code-golf" tag was added. At that time people were still be asking whether it was a code-golf question. — DavidC, Oct 12 '12 at 22:00

Community · Accepted Answer · 2020-06-17 09:04:33Z

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.

That's a super-pretty bug. Perhaps there should be a challenge to generate that... — AJFaraday, Jul 5 '21 at 11:37

Community · Accepted Answer · 2020-06-17 09:04:33Z

26

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)

edited Jun 17 '20 at 9:04

CommunityBot

1

answered Feb 8 '14 at 19:18

FireFly

7,23826 silver badges41 bronze badges

\$\begingroup\$ Instead of "0], write "+. \$\endgroup\$
– FUZxxl
Apr 11 '15 at 10:26
\$\begingroup\$ For one less character, write ,.0(<#":)"+9-+/~|i:8 \$\endgroup\$
– FUZxxl
Apr 11 '15 at 13:24
1

\$\begingroup\$ Here is your solution translated to 25 characters of APL: ⍪↑{(0<⍵)/⍕⍵}¨9-∘.+⍨|9-⍳17 \$\endgroup\$
– FUZxxl
Apr 11 '15 at 13:44
\$\begingroup\$ This here is peak J \$\endgroup\$
– Jonah
Oct 4 '20 at 14:37

Add a comment |

marinus · Accepted Answer · 2012-10-14 21:13:34Z

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.

Shouldn't non-ascii code be counted in UTF-8 instead of code-points? This would push APL closer to his earthly relatives. — John Dvorak, Mar 23 '13 at 5:39
@JanDvorak No, since there is an APL code page, which fits the entire character set into a single byte. But I think you've probably figured this out at some point since 2013. ;) — Martin Ender, Jan 25 '15 at 0:34

Community · Accepted Answer · 2020-06-17 09:04:33Z

20

Mathematica 83 49 43 54 51

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

formatting improved

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}

grids

edited Jun 17 '20 at 9:04

CommunityBot

1

answered Oct 12 '12 at 16:53

DavidC

25k2 gold badges49 silver badges104 bronze badges

1

\$\begingroup\$ David, you beat me this time! :-) \$\endgroup\$
– Mr.Wizard
Oct 21 '12 at 15:29
1

\$\begingroup\$ Another try (55 chars): f = Table[# - Abs@k, {k, -8, 8}] &; f[f[9]] /. n_ /; n < 1 -> "" // Grid \$\endgroup\$
– DavidC
Mar 22 '13 at 17:56
\$\begingroup\$ Still another (71 chars): Table[9 - ManhattanDistance[{9, 10}, {j, k}], {j, 18}, {k, 18}] /. n_ /; n < 1 -> "" // Grid \$\endgroup\$
– DavidC
Mar 22 '13 at 17:57
2

\$\begingroup\$ Grid@#@#@9&[Table[#-Abs@k,{k,-8,8}]&]/.n_/;n<1->"" 50 chars. \$\endgroup\$
– chyanog
Mar 23 '13 at 4:26
\$\begingroup\$ A visual display of the code: ArrayPlot[Sum[k~DiamondMatrix~17, {k, 0, 8}], AspectRatio -> 2] \$\endgroup\$
– DavidC
Nov 23 '13 at 15:53

| Show 2 more comments

Community · Accepted Answer · 2020-06-17 09:04:33Z

17

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:]

edited Jun 17 '20 at 9:04

CommunityBot

1

answered Oct 12 '12 at 17:54

Steven Rumbalski

1,4428 silver badges11 bronze badges

2

\$\begingroup\$ doesn't work in Python 3+, which requires parens around the arguments to print :( \$\endgroup\$
– Griffin
Oct 12 '12 at 18:27
9

\$\begingroup\$ @Griffin: In code golf I choose Python 2 or Python 3 depending on whether I need print to be a function. \$\endgroup\$
– Steven Rumbalski
Oct 12 '12 at 19:29
5

\$\begingroup\$ s=`0x2bdc546291f4b1` \$\endgroup\$
– gnibbler
Oct 14 '12 at 1:32
1

\$\begingroup\$ @gnibbler. Very clever suggestion. Unfortunately, the repr of that hexadecimal includes a trailing 'L'. \$\endgroup\$
– Steven Rumbalski
Oct 15 '12 at 14:51
1

\$\begingroup\$ @gnibbler: This works in Python running on 64-bit platforms, but not on 32-bit platforms. \$\endgroup\$
– Konrad Borowski
Dec 31 '13 at 8:06

| Show 1 more comment

baby-rabbit · Accepted Answer · 2012-10-19 02:53:21Z

15

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);}

answered Oct 19 '12 at 2:53

baby-rabbit

1,6539 silver badges10 bronze badges

4

\$\begingroup\$ Explanation please? \$\endgroup\$
– Lucas Henrique
Apr 12 '15 at 16:35
1

\$\begingroup\$ @LucasHenrique 307 characters total. i%18-9 is x value on cartesian plane mirroring itself on the y-axis. i/18-8 is y value on cartesian plane mirroring itself on the x-axis. Sum them together to get 1:1 diagonal (which causes numeric shift to form on 1:1 diamond. (32:57)-v is unichar numeric value for ASCII 0-9. 10 new line. \$\endgroup\$
– Albert Renshaw
Feb 14 '17 at 9:13

Add a comment |

mbomb007 · Accepted Answer · 2017-02-13 00:07:50Z

14

Python 2, 60 59

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

Abuses backticks and repunits.

edited Feb 13 '17 at 0:07

mbomb007

22.7k6 gold badges55 silver badges132 bronze badges

answered Oct 22 '12 at 18:41

nneonneo

12k8 gold badges27 silver badges48 bronze badges

\$\begingroup\$ The space after in keyword can be removed, just like you did with print keyboard. \$\endgroup\$
– Konrad Borowski
Oct 23 '12 at 17:02
\$\begingroup\$ @GlitchMr: Thanks! Updated. \$\endgroup\$
– nneonneo
Oct 23 '12 at 17:06
\$\begingroup\$ I get an extra L in the middle seven lines of output. \$\endgroup\$
– Steven Rumbalski
Mar 28 '13 at 14:56
\$\begingroup\$ You shouldn't...what version of Python are you using? \$\endgroup\$
– nneonneo
Mar 28 '13 at 15:07

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

12

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}%

edited Jun 17 '20 at 9:04

CommunityBot

1

answered Oct 12 '12 at 17:15

Howard

23.3k2 gold badges40 silver badges81 bronze badges

1

\$\begingroup\$ You don't need the trailing spaces (the text in the question doesn't have them), so you can skip adding the numbers to the spaces and save one char as 17,{8-abs." "*10@-,1>.-1%1>n}% \$\endgroup\$
– Peter Taylor
Mar 28 '13 at 16:21

Add a comment |

chyanog · Accepted Answer · 2014-01-08 06:18:10Z

12

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]

Mathematica graphics

edited Jan 8 '14 at 6:18

answered Mar 22 '13 at 15:58

chyanog

1,2469 silver badges16 bronze badges

1

\$\begingroup\$ Very nice work. \$\endgroup\$
– DavidC
Mar 22 '13 at 17:17
\$\begingroup\$ @DavidCarraher Thank you :D \$\endgroup\$
– chyanog
Mar 23 '13 at 4:05
\$\begingroup\$ I echo David's remark. How did you come up with this? \$\endgroup\$
– Mr.Wizard
Mar 27 '13 at 7:02
\$\begingroup\$ May I update your answer with the shorter modification I wrote? \$\endgroup\$
– Mr.Wizard
Mar 27 '13 at 21:49
\$\begingroup\$ @Mr.Wizard Certainly. \$\endgroup\$
– chyanog
Mar 28 '13 at 4:16

| Show 1 more comment

Community · Accepted Answer · 2020-06-17 09:04:33Z

11

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 :)

edited Jun 17 '20 at 9:04

CommunityBot

1

answered Oct 20 '12 at 17:10

tomsmeding

2,03415 silver badges20 bronze badges

\$\begingroup\$ damn it!! I missed the space and made half diamond. I must sleep now \$\endgroup\$
– Joomler
Nov 23 '16 at 12:41

Add a comment |

Gareth · Accepted Answer · 2012-10-13 10:33:20Z

9

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++);

edited Oct 13 '12 at 10:33

answered Oct 12 '12 at 15:55

Gareth

11.1k1 gold badge33 silver badges83 bronze badges

\$\begingroup\$ 2nd one doesn't work. \$\endgroup\$
– Christian
Mar 22 '13 at 4:30
\$\begingroup\$ I know it's very late, but I always have a need to golf when I see PHP scripts. 83 bytes with <? skipped per meta. Also, you seem to have some encoding problems in the second code. \$\endgroup\$
– RedClover
May 26 '18 at 18:56
\$\begingroup\$ @Soaku The second one is not mine. It was suggested as an edit to my answer by an anonymous user. I just added it without checking - not really sure why the user didn't just post it as their own attempt really. The meta question post-dates this answer by almost 3 years. \$\endgroup\$
– Gareth
May 26 '18 at 19:43
\$\begingroup\$ I meant, that I don't include <? in the bytecount. I've made some other improvements too. \$\endgroup\$
– RedClover
May 26 '18 at 19:58

Add a comment |

Strigoides · Accepted Answer · 2012-10-14 02:31:14Z

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

copy · Accepted Answer · 2013-03-23 01:02:22Z

8

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

answered Mar 23 '13 at 1:02

copy

6,7467 gold badges29 silver badges39 bronze badges

\$\begingroup\$ Running this in my browser shows the diamond, but on an additional line a -7 \$\endgroup\$
– Christiaan Westerbeek
Jul 8 '20 at 22:23

Add a comment |

DLosc · Accepted Answer · 2021-09-04 18:00:55Z

8

Charcoal, 13 bytes

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 down one space before drawing the next one

edited Sep 4 '21 at 18:00

answered Nov 23 '16 at 4:38

DLosc

29.6k3 gold badges50 silver badges109 bronze badges

5

\$\begingroup\$ This should be now competing as per new consensus in meta. \$\endgroup\$
– officialaimm
Sep 13 '17 at 8:16

Add a comment |

user41805 · Accepted Answer · 2021-09-09 08:10:15Z

Vim, 62 39 38 34 keystrokes

Thanks to @DJMcMayhem for saving a ton of bytes

Thanks to @AaronMiller for saving 4 bytes by generating 12345678987654321 in a different way

9i1<ESC>|C<C-r>=<C-r>"*<C-r>"
<ESC>qqYP9|xxI <ESC>YGpHq7@q

Try it online!

Explanation:

9i1<ESC>                            Write 9 '1's, leaving the cursor at the end of the line
        |C                          Go to the first column and cut all that's to the right of the cursor (implictly into register "), entering insert mode
          <C-r>=                    Enter an expression:
                <C-r>"*<C-r>"        The contents of register " multiplied with itself
                             <CR>   Evaluate it, yielding 12345678987654321

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

You can delete the ma and change `ai<space> to I<space>. — DJMcMayhem, Dec 4 '16 at 20:20
Also, you could probably delete stage 3 if you change stage 1 to pasting above and below. — DJMcMayhem, Dec 4 '16 at 20:21
@DJMcMayhem Thank you for the suggestion! I initially was thinking about introducing a new register for the copied bits, but this is much shorter! — user41805, Dec 5 '16 at 8:37
34 bytes by generating the 12345678987654321 mathematically. Readable version — Aaroneous Miller, Sep 7 '21 at 20:56

Shaggy · Accepted Answer · 2017-07-04 13:19:20Z

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.

enter image description here

You can save a byte by casting to a long instead of int64 — Veskah, Jul 26 '18 at 20:50
Another way to save a byte 1..8+9..1|%{' '*(9-$_)+ +($x='1'*$_+'L')*$x} — mazzy, Sep 19 '18 at 12:12

ngn · Accepted Answer · 2018-09-19 13:05:37Z

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:

⌽ 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↓)

Dorian · Accepted Answer · 2018-07-26 11:55:56Z

brainfuck, 158 154 bytes

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

Try it online!

[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
]

DLosc · Accepted Answer · 2021-06-29 03:49:52Z

Regenerate, 109 bytes

(( {#2-1}| {8})($3${#3+1}|1)(${#4+1}$4|)\n){9}(( $6| )(${$7/10}|${$3/10})(${$8-1+1/$8*#7-1/$8}|7){#7-1}\n){8}

Try it here!

Explanation

The solution is divided into two parts, which generate the upper half of the diamond (including the center line) and the lower half, respectively. The upper half consists of group 1, repeated nine times (...\n){9}; the lower half consists of group 5, repeated eight times (...\n){8}.

The first part of each group handles the spaces. (I've replaced space with underscore in the explanations for better visibility.)

Upper-half spaces:

( {#2-1}| {8})
(            )  Group 2:
 _{#2-1}         Space, repeated one less time than the previous length of group 2
        |        Or, if this is the first row and there is no previous value of group 2...
         _{8}    Space, repeated eight times

Lower-half spaces:

( $6| )
(     )  Group 6:
 _$6      Space, concatenated to the previous value of group 6
    |     Or, if this is the first row and there is no previous value of group 6...
     _    A single space

Now for the diamond itself. It can be divided into four quadrants, and I used three different approaches to generate them:

Top left & top right: concatenation

We generate the left half (including the center) of each row with group 3, and the right half with group 4. For both halves, observe that we can generate the next row by concatenating a digit to the previous row:

So we can use a similar approach to what we did for the bottom-half spaces. The only difference is that the character we're concatenating at each step needs to be calculated.

Top left quadrant:

($3${#3+1}|1)
(           )  Group 3:
 $3             Previous value of group 3
   ${    }      concatenated to the result of...
     #3+1       Previous length of group 3 plus 1
          |     Or, if this is the first row and there is no previous value of group 3...
           1    The digit 1

Top right quadrant:

(${#4+1}$4|)
(          )  Group 4:
 ${    }       The result of...
   #4+1        Previous length of group 4 plus 1
        $4     concatenated to the previous value of group 4
          |    Or, if this is the first row and there is no previous value of group 4...
               Empty string

Bottom left: integer division

We generate the left half (including the center) of each row with group 7. Observe that each row is obtained by removing the rightmost digit from the previous row:

If we treat the entire thing as an integer, we can perform this transformation easily by dividing it by ten:

(${$7/10}|${$3/10})
(                 )  Group 7:
 ${     }             The result of...
   $7/10              Previous value of group 7, divided by 10
         |            Or, if there is no previous value of group 7...
          ${     }    The result of...
            $3/10     Group 3 (the final upper-left quadrant row) divided by 10

Bottom right: digit-by-digit

There isn't an easy way to remove a digit from the left side of a previous match, so for the right half of each bottom row, we resort to generating one digit at a time with group 8. Here are the first three rows that we need to generate:

7654321
654321
54321

Each digit is one less than the previous digit, except that we have to "reset" when we hit 1. The reset value is always one less than the length of the left half of the row, which we have access to as group 7.

How can we tell when we've hit 1? I used an integer-division trick I first developed for programming in Acc!!: if N is a positive integer, then 1/N will be one iff N = 1 and zero if N > 1. Multiplying this quantity by (length of group 7) - 1 gives us the reset we need.

(${$8-1+1/$8*#7-1/$8}|7){#7-1}
(                      )        Group 8 (generates one digit at a time):
 ${                 }            The result of...
   $8-1                          Previous value of group 8, minus 1
       +     #7                  Plus the length of group 7
        1/$8*                    if the previous value of group 8 equals 1
               -1/$8             Minus 1 if the previous value of group 8 equals 1
                     |           Or, if there is no previous value of group 8...
                      7          The digit 7
                        {    }  Generate this many digits on each row:
                         #7-1   One less than the length of group 7

This language looks really cool! Funnily enough I was thinking about making a similar one a few weeks back, but figured it'd be too much work :p. Awarding the bounty now. — ta Radvylf srik su shilani, Jun 29 '21 at 15:45

Mark Reed · Accepted Answer · 2021-09-08 12:18:18Z

5

Ruby, 76 69 60 54 characters

(-8..8).map{i=_1.abs;puts' '*i+"#{eval(?1*(9-i))**2}"}

(Thanks, Patrick, G B, and Slim Liser!)

Improvements welcome. :)

edited Sep 8 '21 at 12:18

answered Oct 14 '12 at 14:04

Mark Reed

7175 silver badges15 bronze badges

2

\$\begingroup\$ 69 chars: f=->x{[*1..x]+[*1...x].reverse};puts f[9].map{|i|(f[i]*'').center 17} \$\endgroup\$
– Patrick Oscity
Nov 5 '12 at 21:11
\$\begingroup\$ Great comment, I didn't know the '...' and didn't understand how this could work. \$\endgroup\$
– G B
Nov 25 '16 at 7:10
\$\begingroup\$ 60 chars: [*-8..8].map{|i|puts' '*i.abs+"#{eval [?1*(9-i.abs)]*2*?*}"} \$\endgroup\$
– G B
Nov 25 '16 at 8:05
1

\$\begingroup\$ 54 chars: (-8..8).map{i=_1.abs;puts' '*i+"#{eval(?1*(9-i))**2}"}. It's the same solution optimized. \$\endgroup\$
– Siim Liiser
Oct 2 '20 at 12:22

Add a comment |

Paolo · Accepted Answer · 2012-10-14 10:40:23Z

4

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="")

answered Oct 14 '12 at 10:40

Paolo

7165 silver badges9 bronze badges

\$\begingroup\$ +1 - can save a few with message(rep(" ",9-i),s[c(1:i,i:1-1)]) \$\endgroup\$
– flodel
Apr 11 '15 at 17:14
\$\begingroup\$ @flodel you'd have to note that that prints to stderr, and you could also do for(i in s<-c(1:9,8:1))... to save a byte \$\endgroup\$
– Giuseppe
Dec 14 '17 at 17:10
1

\$\begingroup\$ 64 bytes \$\endgroup\$
– Giuseppe
Apr 6 '18 at 21:05

Add a comment |

skeevey · Accepted Answer · 2012-10-18 18:06:02Z

4

k (64 50 chars)

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

Old method:

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

edited Oct 18 '12 at 18:06

answered Oct 12 '12 at 16:40

skeevey

4,25516 silver badges22 bronze badges

\$\begingroup\$ (1+!9)#'"1" is ,\9#"1" \$\endgroup\$
– ngn
Dec 5 '16 at 22:42

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Martin Ender · Accepted Answer · 2015-01-25 00:32:28Z

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:

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

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*.

Donat · Accepted Answer · 2021-02-26 09:26:37Z

Python 3, 54 bytes

saved another 3 bytes by letting the int function crash instead of a loop condition. Thanks to dingledooper.

i=8
while[print(f'{int("1"*(9-abs(i)))**2:^17}')]:i-=1

Try it online!

Python 3, 57 bytes

saved 1 byte by using f-Strings thanks to dingledooper.

i=8
while i+9:print(f'{int("1"*(9-abs(i)))**2:^17}');i-=1

Try it online!

Python 3, 58 bytes

Even shorter solution by a different approach (prints one extra space at the beginning of each line).

i=8
while i+9:j=abs(i);i-=1;print(' '*j,int('1'*(9-j))**2)

Try it online!

Python 3, 61 bytes (without extra spaces)

i=8
while i+9:j=9-abs(i);i-=1;print(f'%{8+j}d'%int('1'*j)**2)

Try it online!

These are my previous solutions:

Python 3, 66 bytes

f=lambda c:f'%{8+c}d'%int('1'*c)**2
for c in f(9):print(f(int(c)))

Try it online!

Python 3, 71 bytes:

f=lambda c:str(int('1'*int(c))**2)
for c in f(9):print(f(c).center(17))

Try it online!

Here's 57 bytes using f-strings. In addition, there's this magical 54 byte solution which works because int() throws an error on an empty string. — dingledooper, Feb 26 '21 at 0:00

Lynn · Accepted Answer · 2021-09-05 20:01:19Z

4

jq `-nr`, 55 bytes

range(17)-8|fabs|" "*.+"123456789"[:9-.]+"87654321"[.:]

fabs is not supported on TIO but you can try it on https://jqplay.org/.

Alternatively try it on TIO with .*.|sqrt instead of fabs, for 59 bytes.

answered Sep 5 '21 at 20:01

Lynn

64.4k11 gold badges122 silver badges265 bronze badges

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DLosc · Accepted Answer · 2021-09-08 01:37:56Z

jq `-rn`, 76 65 60 59 bytes

-6 bytes thanks to ovs and another -6 (plus -5 more indirectly) thanks to Michael Chatiskatzi

def r:range(9),7-range(8);r|[8-.-r|"\(" "*.//1-.)"[:1]]|add

Try it online!

Explanation

The -n flag specifies that there is no input; the -r flag outputs the result strings, each on its own line, without quotes.

def r:         Define a helper function r
range(9),      which generates the numbers 0 through 8, followed by
7 - range(8);  the numbers 7 through 0
               Main program:
r |            Start with the results of r
[              For each of those numbers, put the following in a list:
 8 - . - r |    8, minus the current number, minus the results of calling r again
 "\(            For each of those numbers, interpolate the following into a string:
  " " * .        Try to repeat a space that many times
  //             and if the result is null (i.e. the number was less than 1),
  1 - .          subtract the number from 1 instead
 )"
 [:1]           Take the first character of each resulting string
] |            Take each resulting list of strings and
add            concatenate it together

To help visualize what's going on, here's what just the r|[8-.-r] part does:

[8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8]
[7,6,5,4,3,2,1,0,-1,0,1,2,3,4,5,6,7]
[6,5,4,3,2,1,0,-1,-2,-1,0,1,2,3,4,5,6]
[5,4,3,2,1,0,-1,-2,-3,-2,-1,0,1,2,3,4,5]
[4,3,2,1,0,-1,-2,-3,-4,-3,-2,-1,0,1,2,3,4]
[3,2,1,0,-1,-2,-3,-4,-5,-4,-3,-2,-1,0,1,2,3]
[2,1,0,-1,-2,-3,-4,-5,-6,-5,-4,-3,-2,-1,0,1,2]
[1,0,-1,-2,-3,-4,-5,-6,-7,-6,-5,-4,-3,-2,-1,0,1]
[0,-1,-2,-3,-4,-5,-6,-7,-8,-7,-6,-5,-4,-3,-2,-1,0]
[1,0,-1,-2,-3,-4,-5,-6,-7,-6,-5,-4,-3,-2,-1,0,1]
[2,1,0,-1,-2,-3,-4,-5,-6,-5,-4,-3,-2,-1,0,1,2]
[3,2,1,0,-1,-2,-3,-4,-5,-4,-3,-2,-1,0,1,2,3]
[4,3,2,1,0,-1,-2,-3,-4,-3,-2,-1,0,1,2,3,4]
[5,4,3,2,1,0,-1,-2,-3,-2,-1,0,1,2,3,4,5]
[6,5,4,3,2,1,0,-1,-2,-1,0,1,2,3,4,5,6]
[7,6,5,4,3,2,1,0,-1,0,1,2,3,4,5,6,7]
[8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8]

Any number that's greater than 0 gets turned into that many spaces (and then trimmed back to a single space). The other numbers gets subtracted from 1 (turning 0 .. -8 into 1 .. 9) and stringified.

tostring can be shortened to @sh here. In general @text and "\(.)" are equivalent with tostring. — ovs, Sep 4 '21 at 21:08
@ovs Very nice! I seem to have skipped over the @ functions when I was looking through the docs. — DLosc, Sep 7 '21 at 15:06
@MichaelChatiskatzi Wow, that's a weird idiom. After some playing around, I found a different 60-byte solution that only uses //, which I like better, but thanks for inspiring me to look! — DLosc, Sep 7 '21 at 19:41

Mark Reed · Accepted Answer · 2021-09-08 14:02:36Z

4

Raku, 42 41 38 35 chars

say " "x 9-$_,(1 x$_)²for 1…9…1

Try it online!

edited Sep 8 '21 at 14:02

Mark Reed

7175 silver badges15 bronze badges

answered Oct 21 '12 at 17:36

Konrad Borowski

11.8k3 gold badges51 silver badges91 bronze badges

\$\begingroup\$ @FrownyFrog Something did change in Perl 6 since this was written (this is a very old answer), fixed. \$\endgroup\$
– Konrad Borowski
Jul 27 '18 at 12:42

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Ilmari Karonen · Accepted Answer · 2012-10-12 16:22:53Z

3

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}%

answered Oct 12 '12 at 16:22

Ilmari Karonen

20.2k3 gold badges48 silver badges98 bronze badges

Add a comment |

chinese perl goth · Accepted Answer · 2012-10-18 22:05:42Z

3

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

edited Oct 18 '12 at 22:05

answered Oct 17 '12 at 22:48

chinese perl goth

1,1177 silver badges12 bronze badges

\$\begingroup\$ Is +1 for -E standard? I mean perl -E is the same number of characters as perl -e. :) \$\endgroup\$
– Mark Reed
Aug 23 '20 at 13:58

Add a comment |

cardboard_box · Accepted Answer · 2012-10-19 02:01:11Z

3

Python, 65

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

answered Oct 19 '12 at 2:01

cardboard_box

6,09818 silver badges40 bronze badges

\$\begingroup\$ Try prepending I=int; to your code and replacing all subsequent instances of int with I \$\endgroup\$
– Cyoce
Nov 27 '16 at 2:51
1

\$\begingroup\$ @Cyoce I had thought of that. It would save 2 chars each time int is used, and it's used 3 times, so it saves 6 chars at a cost of 6 chars. \$\endgroup\$
– cardboard_box
Nov 27 '16 at 4:05

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Print this diamond

122 Answers 122

Clojure, 191 179 bytes

J, 29 26 24 23 22 21 chars

Explanation

APL (33 31)

Mathematica 83 49 43 54 51

Python 2, 72 69 67 61

C, 79 chars

Python 2, 60 59

GolfScript, 33 31 30 characters

Javascript, 114

PHP, 92 90 characters

Common Lisp, 113 characters

JavaScript, 81

Charcoal, 13 bytes

How?

Vim, 62 39 38 34 keystrokes

Explanation:

PowerShell (2 options): 92 84 45 bytes

APL (Dyalog Classic), 20 19 bytes

brainfuck, 158 154 bytes

Regenerate, 109 bytes

Explanation

Top left & top right: concatenation

Bottom left: integer division

Bottom right: digit-by-digit

Ruby, 76 69 60 54 characters

R, 71 characters

k (64 50 chars)

CJam, 31 27 bytes

Python 3, 54 bytes

jq -nr, 55 bytes

jq -rn, 76 65 60 59 bytes

Explanation

Raku, 42 41 38 35 chars

GolfScript, 36 chars

Perl, 43+1

Python, 65

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