Could you make me a hexagon please?

Question

Today, we're going to make an ASCII hexagon. You must write a program or function that takes a positive integer n, and outputs a hexagon grid of size n, made up of asterisks. For example, a hexagon of size 2 looks like this:

 * *
* * *
 * *

While a hexagon of size 3 looks like this:

  * * *
 * * * *
* * * * *
 * * * *
  * * *

You may use any of the default input and output methods, for example STDIO/STDOUT, function arguments and return values or reading/writing a file.

You may assume that input is always valid, so if it's not a positive integer, your program may do whatever you want. You do however have to handle the special case of a size 1 hexagon, which happens to be a single asterisk:

Leading and trailing whitespace is allowed as long as the output is visually the same.

Examples:

1:
*

2:
 * *
* * *
 * *

3:
  * * *
 * * * *
* * * * *
 * * * *
  * * *

4:
   * * * *
  * * * * *
 * * * * * *
* * * * * * *
 * * * * * *
  * * * * *
   * * * *

5:
    * * * * *
   * * * * * *
  * * * * * * *
 * * * * * * * *
* * * * * * * * *
 * * * * * * * *
  * * * * * * *
   * * * * * *
    * * * * *

6:
     * * * * * *
    * * * * * * *
   * * * * * * * *
  * * * * * * * * *
 * * * * * * * * * *
* * * * * * * * * * *
 * * * * * * * * * *
  * * * * * * * * *
   * * * * * * * *
    * * * * * * *
     * * * * * *

12:
           * * * * * * * * * * * *
          * * * * * * * * * * * * *
         * * * * * * * * * * * * * *
        * * * * * * * * * * * * * * *
       * * * * * * * * * * * * * * * *
      * * * * * * * * * * * * * * * * *
     * * * * * * * * * * * * * * * * * *
    * * * * * * * * * * * * * * * * * * *
   * * * * * * * * * * * * * * * * * * * *
  * * * * * * * * * * * * * * * * * * * * *
 * * * * * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * * * * * *
 * * * * * * * * * * * * * * * * * * * * * *
  * * * * * * * * * * * * * * * * * * * * *
   * * * * * * * * * * * * * * * * * * * *
    * * * * * * * * * * * * * * * * * * *
     * * * * * * * * * * * * * * * * * *
      * * * * * * * * * * * * * * * * *
       * * * * * * * * * * * * * * * *
        * * * * * * * * * * * * * * *
         * * * * * * * * * * * * * *
          * * * * * * * * * * * * *
           * * * * * * * * * * * *

As usual, this is code-golf, so standard loopholes apply, and you should try to write the shortest possible program measured in bytes. Of course, some languages are inherently shorter or longer than others, so remember that the goal is not necessarily to have the shortest overall byte count, but to beat submissions in the same or similar languages.

May the best golfer win!

@Pavel I came here to post literally that. "Also, someone needs to write a hexagony solution." — RK., 9 hours ago

xnor · Answer 1 · 2016-12-26 11:04:46Z

up vote 10 down vote

Python 2, 62 bytes

i=n=input()
while n+i-1:i-=1;j=abs(i);print' '*j+'* '*(2*n+~j)

Prints a trailing space at the end of each line.

answered 19 hours ago

xnor

51.2k11108269

add a comment |

Riley · Answer 2 · 2016-12-26 20:38:22Z

up vote 7 down vote

Hexagony + Bash Coreutils, 11 bytes

Includes +3 for -g flag and +8 for |tr . \* non-standard invocation (see this meta post)

Input is given as an argument to Hexagony. According to the GitHub page, when the Hexagony interpreter is called with the -g N option it prints a hexagon of .s. We then use tr to replace those with *s.

edited 10 hours ago

answered 16 hours ago

Riley

2,422216

I edited your answer with "bash coreutils" also, because I'm fairly certain this works in bash (i.e. linux or MacOS only), and because you used another language in your code. – Easterly Irk 13 hours ago

@EasterlyIrk Thank you! I meant to do that when I found the meta post and rewrote the answer. – Riley 13 hours ago

Wow, that's genius. And you're beating all the golfing languages! – DJMcMayhem 11 hours ago

you don't need the backslash before the * unless there are files(etc) in the current working directory. what do the rules say about files external to your solution? – Jasen 8 hours ago

@Jasen I don't know what meta has to say about it, but I don't think you can require files with specific names, so I don't think you can require a lack of files either. Good idea though. – Riley 7 hours ago

add a comment |

edc65 · Answer 3 · 2016-12-26 21:07:09Z

JavaScript (ES6), 77 81 84

@Upvoters: don't miss the answer by @ETHproductions, that is 76 bytes

Edit Revised after change in spec, trailing space allowed

Just for the hat ... hey! No hat?

f=(n,b='* '.repeat(n+n-1),o=b)=>--n?f(n,b=` ${b}`.slice(0,-2),b+`
${o}
`+b):o

Test

f=(n,b='* '.repeat(n+n-1),o=b)=>--n?f(n,b=` ${b}`.slice(0,-2),b+`
${o}
`+b):o


function update()
{
  O.textContent=f(+I.value)
}

update()

<input id=I type=number min=1 value=3 oninput='update()'>
<pre id=O></pre>

DJMcMayhem · Answer 4 · 2016-12-26 15:51:35Z

up vote 5 down vote

V, 17 bytes

é*Àñ>{MÄpXA *Î.

Try it online!

As usual, here is a hexdump, since this contains unprintable characters:

00000000: e92a c0ad f13e 7b4d c470 5841 202a 1bce  .*...>{M.pXA *..
00000010: 2e                                       .

edited 15 hours ago

answered 23 hours ago

DJMcMayhem

20.9k473199

add a comment |

Adnan · Answer 5 · 2016-12-26 11:34:58Z

up vote 4 down vote

05AB1E, 14 13 bytes

Code:

F¹N+„ *×})û.c

Explanation:

F       }        # Input times do (N = iteration number)
 ¹N+             #   Calculate input + N
    „ *×         #   Multiply by the string " *"
         )       # Wrap everything into an array
          û      # Palindromize the array
           .c    # Centralize

Uses the CP-1252 encoding. Try it online!

edited 19 hours ago

answered 20 hours ago

Adnan

23.8k448178

1

I don't understand what the "centralize" part does. When I remove it, I get an array of strings without the appropriate number of leading spaces. – DJMcMayhem 14 hours ago

@DJMcMayhem On an array, you can see it as if it's a string joined by newlines with the text center-aligned. This is what it does on input. – Adnan 13 hours ago

add a comment |

ETHproductions · Answer 6 · 2016-12-26 15:57:37Z

JavaScript (ES6), 77 76 bytes

g=(n,s=`
*`+' *'.repeat(n*2-2),c=s,q=c.replace('*',''))=>--n?g(n,q+s+q,q):s

I told myself I wouldn't sleep until I had set a new ES6 record without looking at the other answers, so here it is...

Test snippet

g=(n,s=`
*`+' *'.repeat(n*2-2),c=s,q=c.replace('*',''))=>--n?g(n,q+s+q,q):s

for(var i = 1; i < 7; i++) console.log(g(i)) // joe

Quentin · Answer 7 · 2016-12-26 16:11:53Z

C, 91 89 80 bytes

w,y;f(s){for(y=-s;++y<s;)for(printf("\n%*c",w=abs(y)+2,8);w++<s*printf("* "););}

I pretty much tweaked around to get the correct formulas, then mashed it all together.

Call f with the number n, and it will print the hexagon to stdout.

Ungolfed and explained:

w,y;
f(s) {
    // y iterates over [-s + 1 ; s - 1] (the number of rows)
    for(y = -s; ++y < s;)
        // w iterates over [abs(y) + 2 ; s * 2] (the number of stars on the row)
        for(
            // This prints a backspace character (ASCII 8)
            // padded with abs(y) + 2 spaces, effectively
            // printing abs(y) spaces to offset the row.
            // Also initializes w with abs(y) + 2.
            printf("\n%*c", w = abs(y) + 2, 8);

            // This is the for's condition. Makes use
            // of the 2 returned by printf, since we coïncidentally
            // need to double the upper bound for w.
            w++ < s * printf("* ");

            // Empty for increment
        )
            ; // Empty for body
}

See it live on Coliru

Notes:

printf can handle negative padding, which results in a left-aligned character with the padding on the right. Thus I tried something to the effect of w = printf("%*c*", y, ' ') so it would take care of the absolute value, and I could retrieve it from its return value. Unfortunately, both zero and one padding widths print the character on its own, so the three center lines were identical.
The backspace character is handled incorrectly by Coliru -- executing this code on a local terminal does remove the leading space on each line.

w=printf("\n%*s",abs(y),"");++w<s*printf(" *"); – Jasen 6 hours ago — Jasen, 6 hours ago

L. Serné · Answer 8 · 2016-12-26 21:11:47Z

JavaScript (ES6), 83 81 bytes

This is my first (code golf) answer. I hope I formatted everything correctly.

a=>{for(b=c=2*a-1;c;)console.log(" ".repeat(d=Math.abs(a-c--))+"* ".repeat(b-d))}

Unlike the 2 current ES6 answers, I'm not recursively calling a function and I am using the console for output.

Neil · Answer 9 · 2016-12-26 11:08:25Z

Batch, 161 bytes

@echo off
set s=*
set l=for /l %%i in (2,1,%1)do call 
%l%set s= %%s%% *
%l%echo %%s%%&call set s=%%s:~1%% *
echo %s%
%l%set s= %%s:~0,-2%%&call echo %%s%%

Note: Trailing space on line 2. Ungolfed:

@echo off
set s=*
rem build up the leading spaces and stars for the first row
for /l %%i in (2,1,%1) do call :s
rem output the top half of the hexagon
for /l %%i in (2,1,%1) do call :t
rem middle (or only) row
echo %s%
rem output the bottom half of the hexagon
for /l %%i in (2,1,%1) do call :b
exit/b
:s
set s= %s% *
exit/b
:t
echo %s%
rem for the top half remove a space and add a star to each row
set s=%s:~1% *
exit/b
:b
rem for the bottom half add a space and remove a star from each row
set s= %s:~0,-2%
echo %s%
exit/b

ElPedro · Answer 10 · 2016-12-26 15:54:17Z

up vote 3 down vote

Python 2, 100 97 89 88 87 81 79 bytes

-1 from @Flp.Tkc

-6 again from @Flp

-2 with thanks to @nedla2004. I was trying to find how to get rid of the second slice but didn't think of that one :)

i=input()
a=[" "*(i-x)+"* "*(i+x)for x in range(i)]
print'\n'.join(a+a[-2::-1])

Try it online!

Creates an array for the top half then adds the reversed array minus the middle line then prints. Prints exactly "as is". No leading or trailing spaces or newlines :-)

edited 15 hours ago

answered 21 hours ago

ElPedro

1,273316

a[::-1][1:] can be a[-2::-1]. – nedla2004 15 hours ago

Nice one. Didn't think of that one :) – ElPedro 15 hours ago

add a comment |

andryk · Answer 11 · 2016-12-26 08:51:39Z

postgresql9.6, 290 bytes

do language plpgsql $$ declare s constant smallint:=4;declare n smallint;declare a constant int[]:=array(select generate_series(1,s));begin foreach n in array a||array(select unnest(a)t order by t desc offset 1)loop raise info'%',concat(repeat(' ',s-n),repeat(' *',s+(n-1)));end loop;end;$$

formatted sql is here:

do language plpgsql $$
declare s constant smallint := 4;
declare n smallint;
declare a constant int[] := array(select generate_series(1, s));
begin
foreach n in array a || array(select unnest(a)t order by t desc offset 1) loop
    raise info '%', concat(repeat(' ', s - n), repeat(' *', s + (n - 1)));
end loop;
end;
$$;

output:

INFO:      * * * *
INFO:     * * * * *
INFO:    * * * * * *
INFO:   * * * * * * *
INFO:    * * * * * *
INFO:     * * * * *
INFO:      * * * *

Neil · Answer 12 · 2016-12-26 11:14:03Z

up vote 2 down vote

JavaScript (ES6), 83 bytes

f=
n=>[...Array(n+--n)].map((_,i,a)=>a.map((_,j)=>j<n-i|j<i-n?``:`*`).join` `).join`
`

<input type=number min=1 oninput=o.textContent=f(+this.value)><pre id=o>

edited 19 hours ago

answered 20 hours ago

Neil

27.5k31468

add a comment |

Erik the Outgolfer · Answer 13 · 2016-12-26 15:42:22Z

Jelly, 24 bytes

R+’µạṀx@€⁶żx@K¥€”*$F€ŒḄY

Try it online!

Jelly is ashamed of the fact that it does not have a centralization atom, so it's beaten by 05AB1E and V. By 11 and 7 bytes respectively!

If you find any way to golf this, please comment. Any help is appreciated.

Explanation:

R+’µạṀx@€⁶żx@K¥€”*$F€ŒḄY Main link. Arguments: z.
R+’                      The sizes of the hexagon's rows. (implicit argument)
   µ                     Start a new monadic chain with the above argument.
    ȧṀx@€⁶               The spaces you must prepend to each row. (implicit argument)
           x@K¥€”*$      The stars (points) of each row, space-joined, as a single link. (implicit argument)
          ż        F€    Conjoin and merge the leading spaces with the stars appropriately.
                     ŒḄ  Create the second half of the hexagon without the middle row.
                       Y Join the rows with newlines. This makes the shape look like a hexagon.

Phew, the explanation was overwhelming. – Erik the Outgolfer 15 hours ago — Erik the Outgolfer, 15 hours ago
What would a "centralization atom" do? – DJMcMayhem 15 hours ago — DJMcMayhem, 15 hours ago
@DJMcMayhem See the 05AB1E answer for an example. – Erik the Outgolfer 15 hours ago — Erik the Outgolfer, 15 hours ago

Level River St · Answer 14 · 2016-12-26 21:00:05Z

up vote 1 down vote

Ruby, 54 bytes

->n{(1-n..n-1).map{|j|i=j.abs;' '*i+'* '*(n*2+~i)}*$/}

lambda function takes n as argument and returns a string separated by newlines. ($/ is a variable containing the default line separator.)

in test program

f=->n{(1-n..n-1).map{|j|i=j.abs;' '*i+'* '*(n*2+~i)}*$/}

puts f[gets.to_i]

answered 9 hours ago

Level River St

16.4k31768

You can save 1 byte by using (1-n...n) with 3 dots – G B 7 hours ago

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rahnema1 · Answer 15 · 2016-12-26 19:46:09Z

Octave , 62 bytes

@(n)' *'(dilate(impad(1,2*(m=n-1),m),[k='01010'-48;~k;k],m)+1)

that can be called as

(@(n)' *'(dilate(impad(1,2*(m=n-1),m),[k='01010'-48;~k;k],m)+1))(5)

Try (paste) it on Octave Online

For example the base image for n=5 is

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

that can be created with

impad(1,2*(n-1),n-1)

The dilation morphological operator applied 4 times on the image using the following neighbor mask:

0 1 0 1 0
1 0 1 0 1
0 1 0 1 0

that can be created with [k='01010'-48;~k;k]

result of dilation:

0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 0
0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0
0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0
0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0
0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0
0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 0

then replace 0 and 1 with ' ' and '*' respectively

    * * * * *
   * * * * * *
  * * * * * * *
 * * * * * * * *
* * * * * * * * *
 * * * * * * * *
  * * * * * * *
   * * * * * *
    * * * * *

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Could you make me a hexagon please?

Examples:

15 Answers 15

Python 2, 62 bytes

Hexagony + Bash Coreutils, 11 bytes

JavaScript (ES6), 77 81 84

V, 17 bytes

05AB1E, 14 13 bytes

JavaScript (ES6), 77 76 bytes

Test snippet

C, 91 89 80 bytes

JavaScript (ES6), 83 81 bytes

Batch, 161 bytes

Python 2, 100 97 89 88 87 81 79 bytes

postgresql9.6, 290 bytes

JavaScript (ES6), 83 bytes

Jelly, 24 bytes

Ruby, 54 bytes

Octave , 62 bytes

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current community

your communities

more stack exchange communities

Could you make me a hexagon please?

Examples:

15 Answers 15

Python 2, 62 bytes

Hexagony + Bash Coreutils, 11 bytes

JavaScript (ES6), 77 81 84

V, 17 bytes

05AB1E, 14 13 bytes

JavaScript (ES6), 77 76 bytes

Test snippet

C, 91 89 80 bytes

JavaScript (ES6), 83 81 bytes

Batch, 161 bytes

Python 2, 100 97 89 88 87 81 79 bytes

postgresql9.6, 290 bytes

JavaScript (ES6), 83 bytes

Jelly, 24 bytes

Ruby, 54 bytes

Octave , 62 bytes

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