Diamondize Some Text

Question

Given an input string of length 2 or longer consisting solely of alphabetical characters [A-Z] or [a-z] (your choice if they're all uppercase or all lowercase), output a continuous string of characters forming a double diamond pattern.

The input string starts on the center line and extends down-and-right at a diagonal until the end of the input string is reached. Then, the pattern continues up-and-right at a diagonal until you're as far above the center line as the length of the input string minus 1. Continue down-and-right back to the center line, then down-and-left, then up-and-left (going "behind" the center character), and finally down-and-left back to the starting character.

That's a little wordy, and it's better demonstrated by some examples:

"YOU"
  U   Y
 Y O U O
Y   O   U
 O Y U Y
  U   O

  ^   ^
 ↙ ↖ ↗ ↘
↘   ↗   >
 ↘ ↗ ↖ ↙
  V   V

See how the YOU starts at the center line and follows down-and-right, then up-and-right, etc., until it loops back to the beginning. Note especially how the Y on the up-and-left portion is "behind" the O and therefore not shown.

Some further examples:

"HI"
 I I
H H H
 I I

"TEST"
   E     E
  S T   T S
 T   T T   T
T     S     T
 E   E E   E
  S T   T S
   T     T

"HELLO"
    L       L
   O L     E L
  H   E   H   O
 E     H O     H
H       L       E
 E     L L     L
  L   E   L   L
   L H     E O
    O       H

---

• Input and output can be given by any convenient method.
• The input is guaranteed to be at least two letters long (i.e., you'll never receive "" as input).
• You can print it to STDOUT or return it as a function result.
• Either a full program or a function are acceptable.
• Any amount of extraneous whitespace is acceptable, so long as the characters line up appropriately (e.g., feel free to pad as a rectangle).
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Answer 1

Charcoal, 17 bytes

ＧＨ<↗↘>↖↙ＬθθＧＨ<Ｌθθ

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

ＧＨ

Draw along a path.

<↗↘>↖↙

Draw in the directions ↘↗↗↘↙↖↖↙ (the < and > are shorthands for those two pairs, but the other pairs don't have shorthands.)

Ｌθ

Each path segment has the same length, including the ends, of the length of the input.

θ

Use the input as the text to be written along the path.

ＧＨ<Ｌθθ

Print the first two parts of the path again so that the middle character is correct.

Answer 2

05AB1E, 12 bytes

gIR7._•Íη•Λ

Try it online!

           Λ    use the canvas function
g               with the length of input for each segment
 IR7._          the input reversed and rotated left by 7 characters (we will draw this backwards to have the right center character)
      •Íη•     and the directions 1, 3, 3, 1, 7, 5, 5, 7 as a compressed number. 

Answer 3

JavaScript (ES6),  157 155  154 bytes

Returns a matrix of characters.

s=>(y=n=s.length,y+=i=X=Y=-1,m=[...Array(y+n)].map(_=>r=Array(4*n-3).fill` `),g=x=>x?g(x-=r[m[y][++i==6*n-6||+x]=s[i%n],y-=m[y-Y]?Y:Y=-Y,x-X]?X:X=-X):m)``

Try it online!

How?

Given the length \$n\$ of the input string, we build a matrix of size \$w\times h\$, with:

• \$w = 4n-3\$
• \$h = 2n-1\$

We run a simulation of a 'ball' bouncing in this matrix, starting at \$(0,n-1)\$ and heading South-East, until it's back to its initial position.

The 0-based index of the center character that must be skipped in the diamond shape is:

$$p=6n-6$$

Example for \$n=4\$:

Commented

s => (                           // s = input string
  y = n = s.length,              // n = length of s
  y += i = X = Y = -1,           // y = n - 1; i = X = Y = -1
  m =                            // create a matrix m[]:
    [...Array(y + n)].map(_ =>   //   - of height 2n-1
      r = Array(4 * n - 3)       //   - of width 4n-3 (save one of these rows in r[])
          .fill` `               //   - initially filled with spaces
    ),                           //
    g = x =>                     // g is a recursive function taking x
      x ?                        //   if x is truthy:
        g(                       //     do a recursive call:
          x -= r[                //       update x:
            m[y][                //         update m[y][x]:
              ++i == 6 * n - 6   //           unless this is the 2nd pass through the
              || +x              //           center cell, set it to the next character
            ] = s[i % n],        //           in s (otherwise we write to m[y][true]
                                 //           instead, which has no effect)
            y -=                 //         update y:
              m[y - Y] ? Y       //           bounce vertically if m[y - Y] is undefined
                       : Y = -Y, //
            x - X                //         bounce horizontally
          ] ? X                  //         if r[x - X] is undefined
            : X = -X             //
        )                        //     end of recursive call
      :                          //   else:
        m                        //     stop recursion and return m[]
)``                              // initial call to g with x = [''] (zero-ish but truthy)

Answer 4

JavaScript (ES6), 136 bytes

f=
(s,l=s.length-1,z=l*4,a=[...Array(l-~l)].map(_=>Array(z+1).fill` `),g=x=>x--?g(x,a[a[y=(x+l)%z]?y:z-y][x>z?z+z-x:x]=s[x%-~l]):a)=>g(z+z)

<input oninput=o.textContent=f(this.value).map(c=&gt;c.join``).join`\n`><pre id=o>

Returns a two-dimensional array. Works by drawing the string into the array directly computing the destination co-ordinates working backwards from the end so that the centre cell is overwritten automatically. Explanation:

(s

Input string.

,l=s.length-1

Distance between "bounces", also half the last row index and one less than the length.

,z=l*4

Last column index, also half of the length of text to draw.

,a=[...Array(l-~l)].map(_=>Array(z+1).fill` `)

Array of spaces.

,g=x=>x--

Count down from the last cell to the first.

  ?g(x

Recursive call to process the remaining cells.

    ,a[a[y=(x+l)%z]?y:z-y]

Calculate the row of this cell.

      [x>z?z+z-x:x]=s[x%-~l])

Calculate the column of this cell and the character that belongs there.

  :a

Finish by returning the array.

)=>g(z+z)

Start at the end of the text.

Answer 5

J, ^{79 77} 75 bytes

(|.@$~8*])`([:;/(1,~<:@])+/\@,(_2{&1 _1\#:33495)}.@#~])`(' '$~1+2 4*])}<:@#

Try it online!

Answer 6

C (clang), 201 196 188 bytes

x,y,i,v,m,n;f(s,z){char*a=s,o[i=(y=z*2-1)*(x=y+y)];for(v=x*2-2;i;n=m=1)o[--i]=i%x?32:10;for(i=x*--z;v--;i+=x*m+n)o[i]==32?o[i]=*a:0,a=*++a?a:s,n=i%x>x-3?-1:n,m=i/x?i/x<y-1?m:-1:1;puts(o);}

Try it online!

-13 @ceilingcat suggestions

Answer 7

Python 2, 137 bytes

s=input();n=len(s);m=2*n-2
for r in range(m+1):print''.join([s[[x,-8-x][(x<=m)==(m>2*r)]%n],' '][r!=(x+1-n)%m!=m-r]for x in range(4*n-3))

Try it online!

A full program that takes a string as input as prints out the diamondized version.