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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
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2  
What is the winning criterion ? And is this a challenge or a golf ? – Paul R Oct 12 '12 at 15:33
8  
I read "kolmogorov-complexity" as "code-golf". – DavidC Oct 12 '12 at 16:44
1  
@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
3  
perlmonks.com/?node_id=891559 has perl solutions. – Zsbán Ambrus Oct 20 '12 at 19:51

39 Answers 39

up vote 5 down vote accepted

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)
share|improve this answer
    
Instead of "0], write "+. – FUZxxl Apr 11 '15 at 10:26
    
For one less character, write ,.0(<#":)"+9-+/~|i:8 – FUZxxl Apr 11 '15 at 13:24
1  
Here is your solution translated to 25 characters of APL: ⍪↑{(0<⍵)/⍕⍵}¨9-∘.+⍨|9-⍳17 – FUZxxl Apr 11 '15 at 13:44

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.
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5  
+1 Amazing. Could you translate it for us APL illiterates? – DavidC Oct 12 '12 at 19:36
2  
Shouldn't non-ascii code be counted in UTF-8 instead of code-points? This would push APL closer to his earthly relatives. – Jan Dvorak Mar 23 '13 at 5:39
2  
@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

Mathematica 83 49 43 54

Grid[Sum[k~DiamondMatrix~17, {k, 0, 8}] /. 0 -> "", Spacings -> 0]

formatting improved


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

share|improve this answer
    
David, you beat me this time! :-) – Mr.Wizard Oct 21 '12 at 15:29
    
Another try (55 chars): f = Table[# - Abs@k, {k, -8, 8}] &; f[f[9]] /. n_ /; n < 1 -> "" // Grid – DavidC Mar 22 '13 at 17:56
    
Still another (71 chars): Table[9 - ManhattanDistance[{9, 10}, {j, k}], {j, 18}, {k, 18}] /. n_ /; n < 1 -> "" // Grid – DavidC Mar 22 '13 at 17:57
2  
Grid@#@#@9&[Table[#-Abs@k,{k,-8,8}]&]/.n_/;n<1->"" 50 chars. – chyaong Mar 23 '13 at 4:26
    
A visual display of the code: ArrayPlot[Sum[k~DiamondMatrix~17, {k, 0, 8}], AspectRatio -> 2] – DavidC Nov 23 '13 at 15:53

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}%
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1  
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}% – Peter Taylor Mar 28 '13 at 16:21

Python 72 69 67 61

Not clever:

s=str(111111111**2)
for i in map(int,s):print'%8s'%s[:i-1]+s[-i:]
share|improve this answer
1  
doesn't work in Python 3+, which requires parens around the arguments to print :( – Griffin Oct 12 '12 at 18:27
4  
@Griffin: In code golf I choose Python 2 or Python 3 depending on whether I need print to be a function. – Steven Rumbalski Oct 12 '12 at 19:29
2  
s=`0x2bdc546291f4b1` – gnibbler Oct 14 '12 at 1:32
1  
@gnibbler. Very clever suggestion. Unfortunately, the repr of that hexadecimal includes a trailing 'L'. – Steven Rumbalski Oct 15 '12 at 14:51
    
How about `` s=111111111**2 ``? It works for me with 2.7.5 - what version are you running? – Daniel Lubarov Nov 19 '13 at 19:37

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);}
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3  
Explanation please? – Lucas Henrique Apr 12 '15 at 16:35

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 :)

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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

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1  
Very nice work. – DavidC Mar 22 '13 at 17:17
    
@DavidCarraher Thank you :D – chyaong Mar 23 '13 at 4:05
    
I echo David's remark. How did you come up with this? – Mr.Wizard Mar 27 '13 at 7:02
    
May I update your answer with the shorter modification I wrote? – Mr.Wizard Mar 27 '13 at 21:49
    
@Mr.Wizard Certainly. – chyaong Mar 28 '13 at 4:16

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++);
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2nd one doesn't work. – Christian Mar 22 '13 at 4:30

Python, 60 59

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

Abuses backticks and repunits.

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

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

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GolfScript, 36 chars

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

9,.);\-1%+:a{a{1$+7-.0>\" "if}%\;n}%
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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="")
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+1 - can save a few with message(rep(" ",9-i),s[c(1:i,i:1-1)]) – flodel Apr 11 '15 at 17:14

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. :)

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69 chars: f=->x{[*1..x]+[*1...x].reverse};puts f[9].map{|i|(f[i]*'').center 17} – Patrick Oscity Nov 5 '12 at 21:11

k (64 50 chars)

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

Old method:

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

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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
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PowerShell: 92 84 45 (2 options)

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

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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
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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""}
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Scala - 86 characters

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

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.

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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*.

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K, 59

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

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Python, 65

for i in map(int,str(int('1'*9)**2)):print' '*(9-i),int('1'*i)**2
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Perl 6, 42 41 chars

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

Hey, it's one two characters shorter compared to Perl 5 - Perl 6 needs more whitespace than Perl 5, but it's still shorter.

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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

share|improve this answer
    
There is a bug: The rightmost digit in the center line is 0 in your output. – Thomas W. Nov 19 '12 at 13:40
    
I know, that is the mentioned precision problem also appearing in Nippeys second solution - do you know a fix for this? – air_blob Nov 19 '12 at 13:53
    
Oh, sorry, I see. No idea how to fix it--at least nothing that wouldn't make the code significantly larger. – Thomas W. Nov 19 '12 at 14:44
    
I know, logically this answer is not correct as the output doesn't equal the target output even though its technical implementation should return the correct result. – air_blob Nov 19 '12 at 14:57
    
See joequncy answer below; codegolf.stackexchange.com/a/8742/7594 – Christian Mar 22 '13 at 4:38

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:' '
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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

share|improve this answer
    
This gets the JS nod. It multiplies correctly. – Christian Mar 22 '13 at 4:37
1  
Instead of wrapping your for loops in brackets, use a semicolon. eg: for(j=0;j<a;j++)s+=" ";for(k... – Shmiddty Mar 22 '13 at 18:59
    
Thank you for pointing that out, @Shmiddty. I've adjusted the snippet. – joequincy Mar 22 '13 at 21:56

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