Generate a Nine-Ball Pool rack

Question

Introduction

Nine-Ball is a popular pool game played casually and as a competitive sport.

There is one of each numbered ball, with the numbers 1 to 9. The game starts with all 9 of the balls in a diamond shaped rack, like so:

  1
 2 3
4 9 6
 5 7
  8

The 1 ball is always at the front of the diamond, and the 9 ball is always in the middle, with all the other balls placed randomly (the "X"s in this diagram):

  1
 X X
X 9 X
 X X
  X

Challenge

Your challenge, given no input, is to generate and output random 9-ball pool racks in the text format of the examples. Extra surrounding whitespace is allowed.

Output may be through any convenient method, standard loopholes apply. A new rack MUST be generated on each run, you cannot just produce a single randomly-chosen rack every time. Extra surrounding whitespace is allowed.

Example Outputs

  1
 2 3
4 9 5
 6 7
  8

  1
 8 3
2 9 7
 4 6
  5

  1
 2 8
3 9 7
 5 4
  6

Least bytes per language wins.

Answer 1

R, 61 bytes

cat(paste0(c("  1
 ","","
","9 ","
 ","","
  "),sample(2:8)))

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Prints alternately the proper whitespace/newline (with 1 and 9 included) and elements of a random permutation of \$\{2,\ldots,8\}\$.

Answer 2

Python 3, 91 87 bytes

from random import*
print("  1\n %d %d\n%d 9 %d\n %d %d\n%3d"%(*sample(range(2,9),7),))

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-4 bytes thanks to xnor (final \n %d -> \n%3d and tuple(...) -> (*...,))

Answer 3

Bash or any Bourne Shell, with modern coreutils, 54 52 bytes

printf '  1
 %s %s
%s 9 %s
 %s %s
  %s' `shuf -i2-8`

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-2 bytes thanks to @dzaima (use shuf -i option for an implicit seq)

Originally:

printf '  1
 %s %s
%s 9 %s
 %s %s
  %s' `seq 2 8|shuf`

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I saved one byte by not including a newline at the end of the source, and another by not including one at the end of the output.

Explanation:

seq 2 8 outputs 2 3 4 5 6 7 8 on separate lines.

|shuf randomly reorders them. (This is what needs modern coreutils.)

printf <formatstring> <args> does a C-like printf of the arguments with the format string.

Answer 4

JavaScript (ES6), 89 bytes

_=>`  1
 X X
X 9 X
 X X
  X`.replace(/X/g,g=_=>m^(m|=1<<(i=Math.random()*9|0))?i:g(),m=3)

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How?

We replace each "X" in the template with a random integer \$0\le i<9\$ such that the \$i\$-th bit in the bitmask \$m\$ is not already set, and we simultaneously update \$m\$.

We start with \$m=3_{10}=000000011_2\$ so that \$0\$ and \$1\$ are never chosen.

Answer 5

Charcoal, 27 25 bytes

Ｆ⁷⊞υ‽⁻Ｅ⁷⁺²κυ↖⪪⁺⪫⪪⪫υω⁴9¦1³

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

Ｆ⁷

Loop 7 times.

⊞υ‽⁻Ｅ⁷⁺²κυ

Create a list of the integers from 2 to 8, perform set difference with the list of integers collected so far, select a random integer, and push it to the empty list. This shuffles the integers from 2 to 8.

⁺⪫⪪⪫υω⁴9¦1

Concatenate the integers, insert a 9 after the fourth, and append a 1. Edit: Saved 2 bytes by using @JonathanAllan's trick to insert the 9.

↖⪪...³

Split the string into a 3×3 grid and print it diagonally thus achieving the desired output.

Answer 6

Python 3, 128 bytes

from random import*
l=[1]+sample(range(2,9),7)
l.insert(4,9)
for i in[1,2,3,2,1]:print(f"{' '.join(map(str,l[:i])):^5}");l=l[i:]

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Unfortunately not even close to the straightforward solution, but I spent so much time on this that I still wanted to share it.

Ungolfed:

from random import *

# Generate random sampling and ensure 1st element is 1
l = [1] + sample(range(2, 9), 7)
# Ensure 5th element is 9
l.insert(4, 9)

for i in [1, 2, 3, 2, 1]: # Numbers of digits per row
    # Take i numbers from the beginning of l, join them with spaces
    # Then center the resulting string in a field of length 5 and print
    print(f"{' '.join(map(str,l[:i])):^5}")
    # Drop the just printed numbers
    l = l[i:]

It can be shortened but then it no longer makes use of the nifty :^5 format specifier:

Python 3, 112 bytes

from random import*
l=[1]+sample(range(2,9),7)
l.insert(4,9)
for i in[1,2,3,2,1]:print(' '*(3-i),*l[:i]);l=l[i:]

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A funny one without insert which bounces the 9 into the middle by slicing with a step size of 2:

Python 3, 115 bytes

from random import*
l=[1]+sample(range(2,9),7)+[9]
for i in[2,4,6,4,2]:print(' '*(3-i//2),*l[:i:2]);l=l[i:]+l[1::2]

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Answer 7

Ruby, 66 52 bytes

$><<"  1
 %d %d
%d 9 %d
 %d %d
  %d"%[*2..8].shuffle

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Also quite a boring answer, similar to the Perl and Bash solutions.

Saved 14 bytes thanks to Value Ink!

Answer 8

Kotlin, 102 95 93 92 bytes

{List(7){it+2}.shuffled().fold("  1\n XX\nX9 X\n XX\n  X"){s,c->s.replaceFirst("X", "$c ")}}

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Answer 9

Jelly, 24 bytes

7ẊŻŒHj8‘;€⁶⁽¦ỵD¤œṖz⁶ZṚ€Y

A full program which prints the result.

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How?

7ẊŻŒHj8‘;€⁶⁽¦ỵD¤œṖz⁶ZṚ€Y - Main Link: no arguments
7                        - seven
 Ẋ                       - shuffle (implicit range [1..7])
  Ż                      - prepend a zero
   ŒH                    - split in half
     j8                  - join with an eight
       ‘                 - increment (vectorises)
        ;€               - concatenate each with:
          ⁶              -   a space character
                œṖ       - partition before indices:
               ¤         -   nilad followed by link(s) as a nilad:
           ⁽¦ỵ           -     2479
              D          -     digits -> [2,4,7,9]
                  z⁶     - transpose with space filler
                    Z    - transpose
                     Ṛ€  - reverse each
                       Y - join with newline characters
                         - implicit, smashing print

Answer 10

Perl 6, 51 bytes

printf '  1
 %s %s
%s 9 %s
 %s %s
  %s',pick *,2..8

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I'm not really happy with how boring this solution is.

Answer 11

PowerShell, 60 bytes

'  1
 {0} {1}
{2} 9 {3}
 {4} {5}
  {6}'-f(2..8|Sort{Random})

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-17 bytes thank you @Veskah for cleaning up whitespaces

Originally:

'
   1
 {0}  {1}
{2}  9  {3}
 {4}  {5}
   {6}
' -f (2..8 | Sort {Get-Random})

Answer 12

Perl 5, 62 bytes

printf"  1
 %d %d
%d 9 %d
 %d %d
  %d",sort{int(rand 3)-1}2..8

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Answer 13

Jelly, 28 26 bytes

4>þ`ṚŒḄYṣ1µ7ẊŻs4j8‘;€⁶ż@o⁶

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A full program that takes no arguments and outputs a pool rack.

Explanation

4>þ`                       | Outer table using 1,2,3,4 on both sides and > as the the function [0,1,1,1],[0,0,1,1],[0,0,0,1],[0,0,0,0]
    Ṛ                      | Reverse [0,0,0,0],[0,0,0,1],[0,0,1,1],[0,1,1,1]
     ŒḄ                    | Bounce [0,0,0,0],[0,0,0,1],[0,0,1,1],[0,1,1,1],[0,0,1,1],[0,0,0,1],[0,0,0,0]
       Y                   | Join with newlines
        ṣ1                 | Split at 1s
          µ                | Start a new monadic chain
           7Ẋ              | Shuffle list of 1..7
             Ż             | Prepend with zero
              s4           | Split into pieces of length 4
                j8         | Join with 8
                  ‘        | Increment by 1
                   ;€⁶     | Append space to each
                      ż@   | Zip with list generated above
                        o⁶ | Replace 0s with spaces

Answer 14

05AB1E, 18 16 bytes

Y8L.r1†9ª•26Ià•Λ

-2 bytes thanks to @Grimmy by using a fixed length of 2 with some directions duplicated.

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Explanation:

Y            # Push 2
8L           # Push a list in the range [1,8]
  .r         # Randomly shuffle it
    1†       # Filter the 1 to the front
      9ª     # Append a trailing 9 (i.e. [1,3,4,8,5,2,6,7,9])
•26Ià•       # Push compressed integer 33557713
Λ            # And use the Canvas builtin with those three values

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •26Ià• is 33557713.

Canvas explanation:

The Canvas builtin takes three arguments:
1. The lengths of the lines to 'draw', which in this case is the fixed length of 2.
2. The string to 'draw', which in this case is the list [1,#,#,#,#,#,#,#,9] (where the # are random digits in the range [2,8]).
3. The directions, which in this case is 33557713. These will be the directions [↘,↘,↙,↙,↖,↖,↗,↘].

With these arguments, it will draw as follows (let's take the random list [1,3,4,8,5,2,6,7,9] as example here):

2 characters (1,3) in direction 3 (↘):
1
 3

2-1 characters (4) in direction 3 (↘):
1
 3
  4

2-1 characters (8) in direction 5 (↙):
1
 3
  4
 8

2-1 characters (5) in direction 5 (↙):
1
 3
  4
 8
5

2-1 characters (2) in direction 7 (↖):
 1
  3
   4
2 8
 5

2-1 characters (6) in direction 7 (↖):
  1
   3
6   4
 2 8
  5

2-1 characters (7) in direction 1 (↗):
  1
 7 3
6   4
 2 8
  5

2-1 characters (9) in direction 3 (↘):
  1
 7 3
6 9 4
 2 8
  5

NOTE: Since we use a fixed length of 2, the directions are leading in drawing the Canvas. So the trailing 3 in the directions 33557713 is mandatory for outputting the 9 here, whereas my previous 18-byter used the leading 3 with implicit wraparound to output the 9, since the lengths-list were leading for drawing instead.

See this 05AB1E tip of mine for a more in-depth explanation of the Canvas builtin.

Answer 15

Dyalog APL Extended, 38 bytes

{5 5⍴(~2⊤29010619)⍀∊⍕¨∊,∘9¨@4⊢1,1+?⍨7}

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Answer 16

Red, 90 bytes

t: random[2 3 4 5 6 7 8]foreach c{  1
 ; ;
; 9 ;
 ; ;
  ;}[prin either c >#":"[take t][c]]

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Answer 17

SAS, 160 Bytes

data;array j[9](1:9);do i=1to 1e4;l=int(2+7*ranuni(0));m=int(2+7*ranuni(0));n=j[l];j[l]=j[m];j[m]=n;end;put+2j[1]/+1j[2]j[3]/j[4]j[9]j[5]/+1j[6]j[7]/+2j[8];run;

Ungolfed:

data _null_;
  array j[9] (1:9);       **Define and populate an array; 
  do i = 1 to 1e4;        **Iterate ten thousand times;
    l=int(2+7*ranuni(0)); **make a random int;
    m=int(2+7*ranuni(0)); **make another random int;
    n=j[l];               **save num in first slot;
    j[l]=j[m];            **swap second to first slot;
    j[m]=n;               **swap original first to second slot;
  end;
  put +2j[1]/+1j[2]j[3]/j[4]j[9]j[5]/+1j[6]j[7]/+2j[8];
                          **outputs the string, / is linebreak, +# is shift over, always adds one space;
run;

If I wanted to do it a bit "better" but longer, I would use rand("integer",2,8);, as that is a better random number distribution, but 1 char longer each line...

I'm pretty sure PROC IML would be much smaller, as that would be more similar to the R/Python implementation. Base SAS has tools to sort rows randomly, but not columns/variables. I figure doing 10,000 random swaps is sufficiently randomized here (probably as random as any of the other implementations, anyway); could easily do 1 billion random swaps for no more characters, but 10k seems enough.

Answer 18

Keg, 62 bytes

2345678(~z%|&(~z%|')&^)  ,,1.
, ,. ,.
,. ,9. ,.
, ,. ,.
,  ,,.

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Remove the z%' s for -4 bytes but a really long execution time.

This:

• Pushes the numbers 2-8 onto the stack.
• Shuffles everything in a very interesting way.
• Prints out the required shape.

Answer 19

APL, 37 chars/bytes

2⌽t⌽↑⍕¨((t←1 2 3 2 1)/⍳5)⊆1,5⌽9,1+?⍨7

permutation of numbers from 2 to 8

1+?⍨7

add leading 9, shift 5 positions to the left, add leading 1

1,5⌽9,

replicate numbers from 1 to 5, 1-2-3-2-1 times respectively

(t←1 2 3 2 1)/⍳5

make 5 nested vectors of length 1-2-3-2-1 respectively

((t←1 2 3 2 1)/⍳5)⊆

transform numerical arrays into char vectors

⍕¨

nested vectors into matrix

↑

shift each line 1-2-3-2-1 positions to the left respectively, and then 2 more for every line

2⌽t⌽

Answer 20

PHP, 83 81 bytes

83 bytes

<?=($l=range(2,8))&&shuffle($l)?vsprintf("  1\n %d%2d\n%d 9 %d\n%2d%2d\n%3d",$l):0;

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---

81 bytes

Splitting code into 2 fragments to save 2 characters. Thanks for the comment @manatwork.

<?$l=range(2,8);shuffle($l)?><?=vsprintf("  1\n %d%2d\n%d 9 %d\n%2d%2d\n%3d",$l);

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Answer 21

C (GCC) 161 158 bytes

-3 bytes from ceilingcat

#define B,b[i++]
b[]=L"";main(a,t,i){srand(time(0));for(i=b;--i;b[a]=t)*b=b[a=rand(t=*b)%7];printf("  1\n %d %d\n%d 9 %d\n %d %d\n  %d"B B B B B B B);}

L"" has non-printing characters that I can't get to show in SE is equivalent to {2,3,4,5,6,7,8}

this code makes the assumption that the address of b is at least 14 in order to guarantee all results are possible

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Answer 22

Vyxal, 42 32 bytes

`  1
 % %
% 9 %
 % %
  %`7ɾ1+Ṗ℅%

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-10 thanks to lyxal.

Answer 23

Java 10, 140 139 bytes

v->{var r="  1\n x x\nx 9 x\n x x\n  x";for(int d=9,m=3;m<511;r=r.replaceFirst("x",d+""))for(;m==(m|=1<<(d*=Math.random()));d=9);return r;}

Port of @Arnauld's JavaScript answer, so make sure to upvote him as well!

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Explanation:

v->{                // Method with empty unused parameter and String return-type
  var r="  1\n x x\nx 9 x\n x x\n  x";
                    //  Result-String, starting at:
                    //  "  1
                    //    x x
                    //   x 9 x
                    //    x x
                    //     x"
  for(int d=9,      //  Temp integer for the random digit, starting at 9
          m=3;      //  Bit-mask `m`, starting at 3 (binary 000000011)
          m<511     //  Loop as long as `m` is not 511 yet (binary 111111111):
          ;         //    After every iteration:
           r=r.replaceFirst("x",d+""))
                    //     Replace the first "x" with digit `d`
    for(;m==        //   Inner loop as long as `m` is unchanged after
            (m|=    //   `m` is bitwise-ORed with:
                1<< //    2 to the power
                   (d*=Math.random()));
                    //    a random digit in the range [0,9)
      d=9);         //     After every iteration: reset `d` to 9 again
  return r;}        //  And finally return the result-String

Answer 24

C (gcc), 139 bytes

char a[7],c;r(){for(srand(time(0));c=rand()%7,a[c]++;);c+=2;}f(){printf("  1\n %d %d\n%d 9 %d\n %d %d\n  %d",r(),r(),r(),r(),r(),r(),r());}

Generates a random number 0-6, tests if that number has been generated before, and if not, prints it.

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