JavaScript (Node.js), 78 bytes

(m=(a,b=26,c=a(--b))=>b?[c,...m(a,b),c]:[c])=>m(a=>m(b=>(c=25-a-b)<0?32:c+65))

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posted over 5 years ago

CC0

Hakerh400‭

1591 reputation 4 55 164 63

V (vim), 21 bytes

¬AZ¬YA|òlÄ26|xx>>YGpH

Try it online!

Explanation:

¬AZ                     " Insert every character from A to Z
   ¬YA                  " Insert every character from Y to A
      |                 " Go back to the beginning of this line
       ò                " While True...
        l               "   Move to the right (if there is only one character
                        "   on this line, break the loop)
         Ä              "   Duplicate this line upwards
          26|           "   Go to the 26th character on this line
             xx         "   Delete two characters
               >>       "   Indent this line with one space
                 Y      "   Yank this line
                  G     "   Go to the last line
                   p    "   Paste the line that was just yanked downwards
                    H   "   Go back to the first line
                        " (Implicit) End loop

Hexdump:

00000000: ac41 5aac 5941 7cf2 6cc4 3236 7c78 783e  .AZ.YA|.l.26|xx>
00000010: 3e59 4770 48                             >YGpH

posted over 5 years ago

CC0

5y ago

DJMcMayhem‭

281 reputation 0 6 28 4

Brain-Flak, 250 bytes

((((()()()){}){}){}()){(({})[()])}{}(<>)<>{({}<>)<>}<>{(({})[()])}([]){{}([(({}))](((()()()){}){}){}()){({}[()])<>((((()()()()){}){}){})<>}{}<>(((((()()()()){}){}){}){}())<>{({}[()])<>(({})())<>}{}{({}[()])<>(({})[()])<>}{}<>((()()()()()){})<>([])}<>

Try it online!

Readable, commented version:

# Push 25
((((()()()){}){}){}())

# Push range(25, 1)
{
    (({})[()])
}
{}

# Push a 0 on to the alternate stack and return to main stack
(<>)<>

# Reverse values onto other stack
{
    ({}<>)
    <>
}
<>

# Push range(25, 0)
{
    (({})[()])
}

# Push the stack height.
([])

# At this point, the stack has 0, 1, 2... 23, 24, 25, 24, 23... 2, 1, 0
# Go through each number of these numbers, and call it N
{
    # Pop the stack height counter
    {}

    # Push (25 - N), and underneath it push N an additional time
    ([(({}))](((()()()){}){}){}())

    # That many times...
    {
        # Decrement
        ({}[()])

        <>

        # Push a space onto the ascii-art stack
        ((((()()()()){}){}){})

        <>
    }
    {}

    # on the ascii-art stack, push the letter A
    <>
    (((((()()()()){}){}){}){}())
    <>

    # N times...
    {
        # Decrement N
        ({}[()])

        # Push "The last character + 1"
        <>
        (({})())
        <>
    }
    {}

    # N times...
    {
        # Decrement N
        ({}[()])

        # Push "The last character - 1"
        <>
        (({})[()])
        <>
    }
    {}

    <>
    # Push a newline
    ((()()()()()){})
    <>

    ([])
}<>

posted over 5 years ago

CC0

DJMcMayhem‭

281 reputation 0 6 28 4

BQN, 19 bytes^SBCS

' '+32⊸<⊸×+⌜˜⌽⊸⌊↕59

Run online!

This uses the whitespace allowance, padding by four spaces on all sides to make the arithmetic work better. It can be adjusted by replacing ↕59 with 4+↕51 at the cost of two characters.

BQN has no built-in idea of the alphabet, but it has a nice system for Unicode character arithmetic. This solution depends on the length 26 of the alphabet, and the distance 33 ('A'-' ') from space to the start. Fortunately these are close together. We will build a matrix of numbers that goes from 0, for ' ', at the corners to 33+26-1, or 58, at the center, then bring all numbers below 33 to 0. At the middle of each edge we get 29, leaving 4 spaces before 'A'. Before plays a big role here.

' '+32⊸<⊸×+⌜˜⌽⊸⌊↕59
                ↕59   # Range            0...58
             ⌽        # Reverse:         58...0
              ⊸⌊      #  Before Minimum: 0…29…0
          +⌜          # Addition table
            ˜         #  with itself
    32⊸<              # More than 32
        ⊸×            #  times the original
' '+                  # Add to space

Another 21-character "exact" solution is (+⌜˜⊏-∘≠↑'A'+⍷)⌽⊸⌊↕51. Now 'A' is used as the basis and spaces come from fill elements in Take (↑), so it will work for any starting character.

posted over 5 years ago

CC BY-SA 4.0

Marshall Lochbaum‭

681 reputation 0 20 68 9

Japt -R, 14 13 bytes

;Bãê kb'A û ê

Test it

Takes all substrings of the alphabet, palindromizes them, and then takes only the ones starting with A and centers and mirrors that too.

-1 byte thanks to @Shaggy

posted over 5 years ago

CC BY-SA 4.0

5y ago

Quintec‭

2424 reputation 17 46 254 49

Canvas, ^{10 7} 6 bytes

Ｚ［］／┼┼

Try it here!

Creates the top portion, centers it, and mirrors it.

-3 bytes from dzaima's hint.

-1, found the 6 byter!

Explanation

Z[]/++
Z          push the alphabet
 []        get all prefixes
   /       diagonalize  by padding each line with one less than the previous
    ++     Quad palindromize, without reflecting letters

posted over 5 years ago

CC BY-SA 4.0

5y ago

Razetime‭

3000 reputation 42 65 335 83

Stax, 12 bytes

ç ∩<▬%▌ê▓jFo

Run and debug it

Same method as the Canvas answer.

posted over 5 years ago

CC BY-SA 4.0

Razetime‭

3000 reputation 42 65 335 83

Vyxal C, 8 bytes

kA¦ømƛøm

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Explanation:

kA        # Push the alphabet
  ¦       # Prefixes
   øm     # Palindromize list without duplicating middle
     ƛ    # For each:
      øm  #   Palindromize without duplicating middle
          # 'C' flag - Center and print, separated by newlines

posted over 5 years ago

CC BY-SA 4.0

Aaron Miller‭

191 reputation 0 5 19 2

Jelly, 13 bytes

ØAUƤz⁶ZUŒBŒḄY

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How it works

ØAUƤz⁶ZUŒBŒḄY - Main link. Takes no arguments
ØA            - Yield the uppercase alphabet
   Ƥ          - Over each prefix:
  U           -   Reverse it
    z⁶        - Zip, padding with spaces
      Z       - Transpose
       U      - Reverse each
        ŒB    - Bounce each
          ŒḄ  - Bounce the entire string
            Y - Join by newlines

posted over 5 years ago

CC BY-SA 4.0

5y ago

caird coinheringaahing‭

516 reputation 3 13 61 12

Python 3, 99 bytes

r=[*range(26)]
a="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
for i in r+r[-2::-1]:print(" "*(26-i)+a[:i]+a[i::-1])

Try it online!

posted about 5 years ago

CC BY-SA 4.0

5y ago

celtschk‭

1554 reputation 28 51 195 74

Japt -R, 11 bytes

;Bå+ mê ê û

Test it

;Bå+ mê ê û
;B              :Uppercase alphabet
  å+            :Cumulatively reduce by concatenation
     m          :Map
      ê         :  Palindromise
        ê       :Palindromise
          û     :Centre pad each with spaces to length of the longest
                :Implicit  output joined with newlines

posted over 5 years ago

CC BY-SA 4.0

Shaggy‭

2491 reputation 4 140 252 188

Uiua, 38 bytes (SBCS)

Full program, prints diamond to STDOUT. Some nifty tricks turned out to be longer than some straightforward ways. Uses one leading whitespace even on the longest line to make "⍥@ " work smoothly even then (it chokes on 0 input).

F ← &p⊂⟜(↘₁⇌)+@A&pf⍥@ ˜-26⟜⇡+1
≡F⊂⇡26⇌⇡25

Uiua pad this online

Scoring remark: I count "F ← " as 2 bytes, as it will be auto-formatted even when entering F=.

Explanation

F ← &p⊂⟜(↘₁⇌)+@A&pf⍥@ ˜-26⟜⇡+1   # helper function F: print diamond line
≡F⊂⇡26⇌⇡25                        # apply F to range [0 ... 25 ... 0]

Helper function

&p⊂⟜(↘₁⇌)+@A&pf⍥@ ˜-26⟜⇡+1  # argument: diamond line number (0-based)
                             # example input: 4
                         +1  # 5   increment
                       ⟜⇡    # 5 [0 .. 4]  range, but keep on top
                   ˜-26      # 21 [0 .. 4] subtract from 26
               ⍥@           # " [...] " [0 .. 4] repeat " " that many times
            &pf              # [0 .. 4]  print string w/o trailing newline
         +@A                 # "ABCDE"  convert to character range
   ⟜(↘₁⇌)                    # "ABCDE" "DCBA" push reversed shortened character range, but keep original on top
&p⊂                          # join both and print them (with newline)

Main

      ⇌⇡25   # push [24 .. 0]
   ⇡26       # push [0 .. 25]
  ⊂          # join both [0 .. 24 25 24 .. 0]
≡F           # call F for each element

posted 6 months ago

CC BY-SA 4.0

ojdo‭

71 reputation 0 7 7 2

Perl, 73 bytes

$_="A$/";$b=A;$a=$b++,s|^| |gm,s| (.*$a)(.*$/)|$&$1$b$a$2$&|
until/Z/;say

Explanation:

# initialise output string and replacement character
$_ = "A$/";
$b = A;

# (

  # set check character, update replacement character
  $a = $b++,

  # prepend a space to every line 
  s|^| |gm,

  # find the line with check character
  # - replicate before and after
  # - adjust check -> check+replacement+check
  # - fix up space count
  s| (.*$a)(.*$/)|$&$1$b$a$2$&|

# )

# stop when string contains a Z
until /Z/;

# display result
say

Try it online!

posted 6 months ago

CC BY-SA 4.0

6mo ago

jhnc‭

591 reputation 0 46 59 98