Diagonal Alphabet

Question

Given no input, your task is to generate the following:

a
 b
  c
   d
    e
     f
      g
       h
        i
         j
          k
           l
            m
             n
              o
               p
                q
                 r
                  s
                   t
                    u
                     v
                      w
                       x
                        y
                         z

Nonvisually, your task is to generate each letter in the alphabet, with spaces before it equal to its position in the alphabet minus one.

If you print this, it must appear like the above. Extraneous whitespace that does not affect appearance, as well as a trailing newline, is allowed. You can use all lowercase, or all uppercase.

You may also return this from a function as per usual rules, either as a string with newlines, or a list of strings.

This is code-golf, so shortest answer in bytes wins!

Answer 1

Ahead, 31 30 bytes

"az"E>0\:'a-' l
  @?lj:oNW:k\<

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-1 because leading spaces are allowed as long as everything is still aligned.

Answer 2

Ruby, 43 bytes

(0..25).each{|n|print" "*n;puts (n+97).chr}

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Answer 3

Perl 6, 24 bytes

{map ' 'x$++~*,'a'..'z'}

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Anonymous code block that returns a list of lines.

Alternatively,

{' 'x$++xx㉖Z~'a'..'z'}

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behaves the same for the same bytecount.

Answer 4

dc, 28 bytes

97[ddPAP1+2z8*B-^+z27>M]dsMx

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Outputs to stdout, makes lowercase letters

Explanation

P pops a number from the stack and outputs it as a base-256 byte stream. So I use some math to calculate each number (representing the appropriate string) "inductively" from the previous.

97        A literal, which P will print as ASCII 97 ('a')
[         Each loop starts with these two things true:
              The next number to Print is on top of the stack
              The stack depth is the # of times the loop has started (inclusive)
  dd        Make two copies: one to manipulate stack depth, one to print
  PAP       Print the number, then Print a newline (0x0A)
  1+        Increment the last digit (i.e., the letter)
  2z8*B-^+  Add the leftmost space, calculated as: 
                 32*(256^loop counter)  (because 32 = ' ')
               = 2^5*(2^8)^(z-2)        (z = stack depth = counter + 2)
               = 2^(8z-11)              (and to dc, 'B' pushes literal 11)
  z27>M     Repeat unless this was our 26th time through
]dsMx       End the loop, store it as a macro, and run it

Answer 5

C# (.NET Core), 71 58 53 52 bytes

for(var i='A';i<91;)WriteLine("".PadLeft(i-65)+i++);

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Answer 6

Red, 37 bytes

repeat n 26[print pad/left #"`"+ n n]

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Answer 7

Dart, 65 bytes

f()=>List.generate(26,(i)=>String.fromCharCode(i+65).padLeft(i));

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Answer 8

MBASIC, 43 bytes

1 FOR I=0 TO 25:PRINT SPC(I)CHR$(97+I):NEXT

similar to Alphabet Staircase.

Answer 9

Python 3 43 Bytes

for i in range(26):print(" "*i+chr(i+97))

Answer 10

Pushy, 10 bytes

`a\0x1A;`:"32}h

Where \0x1A is the character 26, 'SUB'

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Answer 11

Rockstar, 55 bytes

X's0
while X-26
cast X+97 into C
say " "*X+C
build X up

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Answer 12

Husk, 9 bytes

z:¡Θø…"az

Try it online! ¡Θø is the infinite list [""," "," "," ",...], and …"az is the lowercase alphabet.

Answer 13

Deadfish~, 827 bytes

{{i}}dddc{{d}i}iiic{ii}iic{iiiiii}iiiiiic{{d}i}iic{ii}iicc{{i}ddd}dddc{{d}i}ic{ii}iiccc{{i}ddd}ddc{{d}i}c{ii}iicccc{{i}ddd}dc{{d}i}dc{ii}iiccccc{{i}ddd}c{{d}i}ddc{ii}iicccccc{{i}ddd}ic{{d}i}dddc{ii}iiccccccc{{i}ddd}iic{{d}i}ddddc{ii}iicccccccc{{i}ddd}iiic{{d}i}dddddc{ii}iiccccccccc{{i}ddd}iiiic{{d}i}ddddddc{ii}ii{c}{{i}ddd}iiiiic{{d}}iiic{ii}ii{c}c{{i}ddd}iiiiiic{{d}}iic{ii}ii{c}cc{{i}dd}dddc{{d}}ic{ii}ii{c}ccc{{i}dd}ddc{{d}}c{ii}ii{c}cccc{{i}dd}dc{{d}}dc{ii}ii{c}ccccc{{i}dd}c{{d}}ddc{ii}ii{c}cccccc{{i}dd}ic{{d}}dddc{ii}ii{c}ccccccc{{i}dd}iic{{d}}ddddc{ii}ii{c}cccccccc{{i}dd}iiic{{d}}dddddc{ii}ii{c}ccccccccc{{i}dd}iiiic{{d}}ddddddc{ii}ii{cc}{{i}dd}iiiiic{{d}}{d}iiic{ii}ii{cc}c{{i}dd}iiiiiic{{d}}{d}iic{ii}ii{cc}cc{{i}d}dddc{{d}}{d}ic{ii}ii{cc}ccc{{i}d}ddc{{d}}{d}c{ii}ii{cc}cccc{{i}d}dc{{d}}{d}dc{ii}ii{cc}ccccc{{i}d}c

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Answer 14

Julia 1.0, 31 bytes

@.println(" "^(0:25)*('a':'z'))

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Answer 15

Elixir, 51 bytes

Enum.reduce ?a..?z,"",&(IO.puts([&2,&1])&&"#{&2} ")

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Answer 16

Thunno d, \$ 12 \log_{256}(96) \approx \$ 9.88 bytes

AzA{s' *s+ZK

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Explanation

Az            # Lowercase Alphabet
  A{          # Enumerated for-loop:
    s         #  Swap so the index is on top
     ' *      #  Multiply " " by the index
        s+    #  Swap and add the character
          ZK  #  Print with a trailing newline
              # d flag stops implicit print at the end

Thunno N, \$ 12 \log_{256}(96) \approx \$ 9.88 bytes

Az26R' *ZZ.J

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Explanation

Az            # Lowercase Alphabet
  26R         # range(26)
     ' *      # Multiply " " by each number
        ZZ    # Zip together
          .J  # Join each inner list
              # N flag joins by newlines
              # Implicit output

Answer 17

SNOBOL4 (CSNOBOL4), 72 65 bytes

E	&LCASE O LEN(1) . O @X	:F(END)
	OUTPUT =LPAD(O,X)	:(E)
END

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swapping &UCASE for &LCASE would do the uppercase diagonal instead.

Answer 18

Uiua 0.3.1, 17 bytes

+@ +×65⊃∘×⊞=..⇡26

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Answer 19

Python 3, 51 bytes

lambda:"\n".join(" "*i+chr(i+97)for i in range(26))

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Answer 20

Swift, 89 86 bytes

zip(0...,"abcdefghijklmnopqrstuvwxyz").map{print({String.init}()(" ",$0.0)+"\($0.1)")}