HexaGolf: Rotatagons

Question

See also: Wordagons

Challenge

Given a hexagon and a number n as input, output the same hexagon rotated n times.

Hexagon

The hexagon will be a string like so:

  a b c
 d d e e
f f o g g
 h h i i
  j k l

The hexagon will always be regular and will only ever contain the printable ASCII characters:

 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~

Rotation

To find how much to rotate the hexagon by, multiply n by 60 degrees to get the angle of rotation. For example, using the above hexagon, if n is 2, we multiply that by 60 to get 120 degrees. Then, we rotate the hexagon by 120 degrees:

  j h f
 k h f d
l i o d a
 i g e b
  g e c

Note that each side has moved two sides clockwise.

If n is positive, you should rotate the hexagon clockwise. If n is negative, you should rotate the hexagon anticlockwise.

n will always be an integer in the range -6 to 6 inclusive.

Examples

Input:

  . u .
 .  |  .
l - + - r
 .  |  .
  . d .

n = 3

  . d .
 .  |  .
r - + - l
 .  |  .
  . u .

---

Input:

  - - -
 /     \
<       >
 \     /
  - - -

n = -1

Output:

  - \ >
 -     /
-       -
 /     -
  < \ -

---

Input:

   h e x a
  g o n s a
 n d t r i a
n g l e s m a
 k e m e s a
  d l y m a
   d d e r

n = 6

Output:

   h e x a
  g o n s a
 n d t r i a
n g l e s m a
 k e m e s a
  d l y m a
   d d e r

Winning

The shortest program in bytes wins.

Answer 1

Python 2, 160 158 156 148 124 bytes

Prepare to be amazed! The resulting text is drawn on a canvas... by a Turtle! :D

from turtle import*
h,n=input()
up()
rt(n*60)
l=0
for c in h:
 write(c);fd(9);l-=1
 if" ">c:fd(9*l);rt(90);fd(15);lt(90);l=0

Try it online - Supports decimal rotations as well!

---

If you would prefer the turtle to be shaped like one, you can add the line t.shape("turtle").

View the revision history to see past versions.

Aliases used:

up() - penup()

rt() - right()

lt() - left()

fd() - forward()

bk() - backward()

Thanks to Sherlock9 for the suggestions!

Answer 2

Python 2.7, 151 bytes

h,n=input()
s=1-len(h)
e=enumerate
exec n%6*"h=[''.join(c>' 'and h[y-x-3*s/2>>1][x*2+y*6+s>>2]or c for x,c in e(l))for y,l in e(h)];"
print'\n'.join(h)

Example:

% python2.7 hexarot.py <<<'[["   h e x a   ","  g o n s a  "," n d t r i a ","n g l e s m a", " k e m e s a ","  d l y m a  ","   d d e r   "],-1]'
   a a a a   
  x s i m a  
 e n r s s a 
h o t e e m r
 g d l m y e 
  n g e l d  
   n k d d

Answer 3

Perl, 120 119 113 107 104 102 bytes

Includes +6 for -0pXi

Run with the hexagon on STDIN (padded with spaces so all lines have the same length) and the number of rotations after -i.

perl -0pXi1 rotagon.pl 
  a b c
 d d e e
f f o g g
 h h i i
  j k l
^D

rotagon.pl

s!(.)!$X=(3*($w=y/
//*2)/4+($-="@-"/$w)-"@+"%$w)/2*--$w+2*$--.25;/^\C{$X}(.)/;$1!eg while$^I--%6

Finally got the rotation matrix to be shorter than the old double loop

Answer 4

JavaScript (ES6), 130 127 bytes

f=(s,n)=>n%6?f(s.split`
`.map((s,i,a)=>s.replace(/./g,(c,j)=>(a[(i+z-j)/2+z]||c)[(i+j-z)/2+i]||c,z=a.length>>1)).join`
`,n-1):s

Originally based on my answer to Rotate a diamond tiling although I might now be able to rewrite that answer along the lines of this answer.