Confusing Alphabet Staircase

Question

Given no input, output this interesting alphabet pattern in either case (the case has to be consistent) via an accepted output method:

A
AB
ACBC
ADBDCD
AEBECEDE
AFBFCFDFEF
AGBGCGDGEGFG
AHBHCHDHEHFHGH
AIBICIDIEIFIGIHI
AJBJCJDJEJFJGJHJIJ
AKBKCKDKEKFKGKHKIKJK
ALBLCLDLELFLGLHLILJLKL
AMBMCMDMEMFMGMHMIMJMKMLM
ANBNCNDNENFNGNHNINJNKNLNMN
AOBOCODOEOFOGOHOIOJOKOLOMONO
APBPCPDPEPFPGPHPIPJPKPLPMPNPOP
AQBQCQDQEQFQGQHQIQJQKQLQMQNQOQPQ
ARBRCRDRERFRGRHRIRJRKRLRMRNRORPRQR
ASBSCSDSESFSGSHSISJSKSLSMSNSOSPSQSRS
ATBTCTDTETFTGTHTITJTKTLTMTNTOTPTQTRTST
AUBUCUDUEUFUGUHUIUJUKULUMUNUOUPUQURUSUTU
AVBVCVDVEVFVGVHVIVJVKVLVMVNVOVPVQVRVSVTVUV
AWBWCWDWEWFWGWHWIWJWKWLWMWNWOWPWQWRWSWTWUWVW
AXBXCXDXEXFXGXHXIXJXKXLXMXNXOXPXQXRXSXTXUXVXWX
AYBYCYDYEYFYGYHYIYJYKYLYMYNYOYPYQYRYSYTYUYVYWYXY
AZBZCZDZEZFZGZHZIZJZKZLZMZNZOZPZQZRZSZTZUZVZWZXZYZ

Trailing spaces and newlines are acceptable, standard loopholes are disallowed, and this happens to be code-golf, so the shortest answer in bytes wins!

Answer 1

Canvas, 7 bytes

Ｚ［Ｋ＊¹＋］

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Explanation:

Ｚ［     ］ for each prefix of the uppercase alphabet
    Ｋ        pop off the last letter
     ＊       and join the rest of the string with that character
      ¹＋     and append the current iterated character to it

Answer 2

Jelly, 9 bytes

ØAjṪ$Ƥż¹Y

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How it works

ØAjṪ$Ƥż¹Y  Main link. No arguments.

ØA         Yield "ABCDEFGHIJKLMNOPQRSTUVWXYZ".
     Ƥ     Map the link to the left over all prefixes, i.e., ["A", "AB", ...].
    $        Combine the two links to the left into a chain.
   Ṫ           Tail; yield and remove the last letter of each prefix.
  j            Join the remainder, using that letter as separator.
      ż¹   Zip the resulting strings and the letters of the alphabet.
        Y  Separate the results by linefeeds.

Answer 3

C, 82 bytes

f(i,j){for(i=!puts("A");++i<26;puts(""))for(j=0;j++<i*2;)putchar(65+(j&1?j/2:i));}

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Answer 4

R, 50 bytes

l=LETTERS
for(i in 0:25)cat(l[0:i],"
",sep=l[i+1])

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Perhaps the cleverest part here is using letters[0] for the empty string to get cat(character(0),'\n',sep="A") to print the first line.

Answer 5

Charcoal, 9 bytes

Ｅα⁺⪫…ακιι

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

 α          Predefined uppercase alphabet
Ｅ           Map over each character
    …ακ     Get current prefix of alphabet
   ⪫   ι    Join with current character
  ⁺     ι   Append current character
            Implicitly print on separate lines

Answer 6

Python 2, 56 bytes

n=65;s='';exec'c=chr(n);print c.join(s)+c;s+=c;n+=1;'*26

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Answer 7

6502 machine code routine (C64), 39 bytes

A9 41 20 D2 FF AA A8 84 FB E4 FB B0 0B 8A 20 D2 FF 98 20 D2 FF E8 D0 F1 A9 0D
20 D2 FF A2 41 C0 5A F0 03 C8 D0 E1 60

Position-independet machine code subroutine, clobbers A, X and Y.

Online demo

The demo loads at $C000, so use SYS49152 to call the routine.

---

Commented disassembly:

A9 41       LDA #$41            ; 'A'
20 D2 FF    JSR $FFD2           ; Kernal CHROUT (output character)
AA          TAX                 ; copy to X (current pos)
A8          TAY                 ; copy to Y (current endpos)
  .outerloop:
84 FB       STY $FB             ; endpos to temporary
  .innerloop:
E4 FB       CPX $FB             ; compare pos with endpos
B0 0B       BCS .eol            ; reached -> do end of line
8A          TXA                 ; current pos to accu
20 D2 FF    JSR $FFD2           ; and output
98          TYA                 ; endpos to accu
20 D2 FF    JSR $FFD2           ; and output
E8          INX                 ; next character
D0 F1       BNE .innerloop      ; (repeat)
  .eol:
A9 0D       LDA #$0D            ; load newline
20 D2 FF    JSR $FFD2           ; and output
A2 41       LDX #$41            ; re-init current pos to 'A'
C0 5A       CPY #$5A            ; test endpos to 'Z'
F0 03       BEQ .done           ; done when 'Z' reached
C8          INY                 ; next endpos
D0 E1       BNE .outerloop      ; (repeat)
  .done:
60          RTS

Answer 8

SNOBOL4 (CSNOBOL4), 169 143 bytes

i &ucase len(x) . r len(1) . s
 o =
 i =
t r len(i) len(1) . k :f(o)
 o =o s k
 i =i + 1 :(t)
o o s =
 output =o s
 x =lt(x,25) x + 1 :s(i)
end

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i &ucase len(x) . r len(1) . s	;* set r to the first x characters and s to the x+1th.
 o =				;* set o,i to empty string
 i =
t r len(i) len(1) . k :f(o)	;* set k to the ith letter of r. on failure (no match), go to o.
 o =o s k			;* concatenate o,s,k
 i =i + 1 :(t)			;* increment i, goto t
o o s =				;* remove the first occurrence of s (the first character for x>1, and nothing otherwise)
 output =o s			;* output o concatenated with s
 x =lt(x,25) x + 1 :s(i)	;* increment x, goto i if x<25.
end

The problem here is the first line

using o s k will add an extra separator character at the beginning of each line and also not have an s at the end. This is OK because line t will jump over the following two lines when x=0. This means that o will still be blank. Hence, o s = will remove the first s character from o, and then we can simply print o s to have the appropriate last s.

Answer 9

Java 8, 93 91 90 bytes

v->{String t="";for(char c=64;++c<91;t+=c)System.out.println(t.join(c+"",t.split(""))+c);}

-1 byte thanks to @OlivierGrégoire by printing directly instead of returning

Explanation:

Try it online.

v->{                     // Method with empty unused parameter and String return-type
  String t="";           //  Temp-String, starting empty
  for(char c=64;++c<91;  //  Loop over the letters of the alphabet:
      t+=c)              //    After every iteration: append the letter to the temp-String
    System.out.println(  //   Print with trailing new-line:
       r.join(c+"",t.split(""))
                         //    The temp-String with the current letter as delimiter
       +c);}             //    + the current letter as trailing character 

Answer 10

Japt (-R flag), 14 12 bytes

-2 bytes thanks to @Shaggy

;B¬
ËiU¯E qD

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Answer 11

Haskell, 49 48 bytes

'A':unlines[init['A'..x]>>=(:[x])|x<-['A'..'Z']]

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Edit: -1 byte thanks to totallyhuman!

Answer 12

JavaScript (ES6), 81 bytes

f=
_=>[..."ABCDEFGHIJKLMNOPQRSTUVWXYZ"].map((c,i,a)=>a.slice(0,i).join(c)+c).join`
`
;document.write('<pre>'+f());

Save 9 bytes if a string array return value is acceptable.

Answer 13

><>, 44 34 bytes

"BA"oao"ZA"\=?;1+40.
o1+:{::o}=?\:

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><>, 44 bytes

"A"o10ao\55*=?;1+40.
1+:{:}=?\:"A"+o{:}"A"+o

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As I use a different route to producing the output I've posted my own ><> answer; The other ><> answer can be found here.

Big thanks to Jo king for spotting I didn't need to keep putting "A" onto the stack if I just compared against "Z" instead of 26. (-10 bytes)

Explanation

The explanation will follow the flow of the code.

"BA"                 : Push "BA" onto the stack;
                       [] -> [66, 65]
    oao              : Print the stack top then print a new line;
                       [66, 65] -> [66]
       "ZA"\         : Push "ZA" onto the stack then move down to line 2;
                       [66, 90, 65]
o          \:        : Duplicate the stack top then print
 1+:                 : Add one to the stack top then duplicate;
                       [66, 90, 65, 65]
    {::              : Shift the stack right 1 place then duplicate the stack top twice;
                       [90, 65, 65, 66, 66]
       o}            : Print the stack top then shift the stack left 1 place;
                       [66, 90, 65, 65, 66]
         =?\         : Comparison for equality on the top 2 stack items then move to line 1 if equal otherwise continue on line 2;
                       [66, 90, 65]
           \=?;      : Comparison for equality on the top 2 stack items then quit if equal else continue on line 1;
                       [66]
               1+    : Add 1 to the stack top;
                       [67]
                 40. : Move the code pointer to column 4 row 0 of the code box and continue execution of code. 

Answer 14

PowerShell, 56 bytes

"A";65..89|%{([char[]](65..$_)-join[char]++$_)+[char]$_}

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Loops 65 to 89, each iteration constructing a char array of 65 to the current number $_, then -joins that array together into a string with the next character, then tacks on that character at the end.

Change the 89 to some other ASCII number to see the behavior better.

Answer 15

Jelly, 13 bytes

ØA¹Ƥ+"¹Ṗ€Yṭ”A

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Explanation

ØA¹Ƥ+"¹Ṗ€Yṭ”A  Main Link
ØA              Uppercase Alphabet
  ¹Ƥ            Prefixes
    +"¹         Doubly-vectorized addition to identity (uppercase alphabet) (gives lists of lists of strings)
       Ṗ€      a[:-1] of each (get rid of the double letters at the end)
         Y     Join on newlines
          ṭ”A  "A" + the result

partially abuses the way strings and character lists differ in Jelly

Answer 16

Jelly, 12 bytes

ØA;\;€Ṫ$€YFḊ

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Bah just got ØAjṪ$ƤżØAY which is a step between this and the already posted solution of Dennis :/

Answer 17

Pyth, 13 bytes

+\ajmPjedd._G

Try it here!, Alternative

That leading a though...

Answer 18

Acc!!, 84 bytes

This is actually what inspired this challenge:

Write 65
Count i while i-26 {
Count b while b-i {
Write b+65
Write i+65
}
Write 10
}

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Answer 19

Python 2, 92 86 79 75 64 bytes

s=map(chr,range(65,91))
for d in s:print d.join(s[:ord(d)-65])+d

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11 bytes thx to Rod.

Answer 20

APL+WIN, 51 bytes

⍎∊'a←⎕av[65+⍳26]⋄a[n←1]',25⍴⊂'⋄,⊃a[⍳n-1],¨a[n←n+1]'

Explanation:

a←⎕av[65+⍳26] create a vector of upper case letters

a[n←1] first A

25⍴⊂'⋄,⊃a[⍳n-1],¨a[n←n+1]' create an implicit loop to concatenate subsequent letters

Answer 21

><>, 47 bytes

d2*:1-v
-&$:?!\$:&$:1
1-:?!v\69*-$1-:
+*88~< 1o

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How It Works:

d2*:1-v Initialise the stack with 26 (outer loop counter) and 26-1 (inner loop counter)
....
....
....

....
-&$:?!\$:&$:1 Repeatedly make copies of both counters
....          And decrement the inner loop counter
....          Go to third line when inner loop counter is 0

....            Add -54 to the stack (for the newline) and decrement the outer loop counter
....            Initialise the inner loop counter as outer-1
1-:?!v\69*-$1-: If the inner counter is 0, go to the fourth line, else back to the second.
....

....
....      
....      Transform numbers and -54s into letters and newlines by adding 64
+*88~< 1o Output each character until it runs out of stack and errors

Answer 22

Canvas, 11 10 bytes

Ｚ｛Ｚ²╷ｍ¹＊×］

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Answer 23

GNU M4, 119 bytes

The worst so far. Well, time's already spent…

define(f,`ifelse($1,$2,,`format(%c%c,$1,$2)`'f(incr($1),$2)')')define(g,`f(65,$1)ifelse($1,90,,`
g(incr($1))')')A
g(66)

Answer 24

Ruby, 44 34 bytes

?A.upto(?Z){|w|puts [*?A...w]*w+w}

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Thanks benj2240 for getting it down to 37 bytes. And of course crossed out 44 blah blah.

Answer 25

C# (.NET Core)

Port from Kevin Cruijssen's answer:

91 90 bytes

_=>{var t="";for(char c='@';++c<91;t+=c)Console.WriteLine(string.Join(c+"",t.Skip(0))+c);}

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132 122 110 109 104 103 bytes

_=>"ABCDEFGHIJKLMNOPQRSTUVWXYZ".Select((c,i)=>string.Join(""+c,"ABCDEFGHIJKLMNOPQRSTUVWXYZ".Take(i))+c)

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• Replace () with _ to show that we declare an unused variable. Thank you Kevin Cruijssen.

Answer 26

Husk, 13 bytes

Γ·:mhSzJḣ…"AZ

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Explanation

This leading A really messes things up -.-

          "AZ  -- string literal: "AZ"
         …     -- fill gaps: "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
     S         -- with alphabet and
        ḣ      -- | alphabet rangified: ["A","AB","ABC",…,"AB……XYZ"]
      zJ       -- : zipWith join: ["A","ABB","ACBCC","ADBDCDD",…,"AZB……ZYZZ"]
Γ              -- pattern match (x:xs) with the following function (x is "A" and xs ["ABB","ACBCC",…,"A……ZYZZ"]
 · mh          -- | drop the last element of each element of xs: ["AB","ACBC",…,"A……ZYZ"]
  :            -- | cons (construct list): ["A","AB","ACBC",…,"A……ZYZ"]
               -- : strings are printed implicitly

Answer 27

Jelly, 22 bytes

ØAż€Ð€`F€µJ’Ḥ»1ż@¹ḣ/€Y

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How it works:

                       take argument implicitly
ØA                     the uppercase alphabet
    Ѐ`                for C in the alphabet
  ż€                     appends C to every letter in the alphabet
       F€              flatten every sublist
          J            get indices
           ’           subtract 1
            Ḥ          and double
             »1        take max([n, 1])
         µ     ż@¹     interleave alphabet list and indices
                  ḣ/€  reduce on head() for each element
                     Y join on newline
                       implicitly output

Answer 28

uBASIC, 80 bytes

Anonymous function that takes no input and outputs to the console

0?"A":ForI=65To89:ForJ=65ToI:?Left$(Chr$(J),1)+Left$(Chr$(I+1),1);:NextJ:?:NextI

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Answer 29

MY-BASIC, 77 bytes

Anonymous function that takes no input and outputs to the console.

Print"A";
For I=65 To 89
For J=65 To I
Print Chr(J)+Chr(I+1)
Next
Print;
Next

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Answer 30

Visual Basic .NET (Mono), 134 bytes

Declared function that takes no input and outputs to the console

Module M
Sub Main
Dim S,I,J
S="A"
For I=65To 90
Console.WriteLine(S)
S=""
For J=65To I
S+=Chr(J)+Chr(I+1)
Next
Next
End Sub
End Module

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Answer 31

Yabasic, 60 bytes

?"A"
For I=65To 89
For J=65To I?Chr$(J)+Chr$(I+1);Next
?Next

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Answer 32

05AB1E, 29 bytes

'A,Au©.sRí¦®RSDgÝ×Rs)ø˜.Bíø»=

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Answer 33

Perl 5, 39 38 bytes

@DomHastings shaved off a byte

say$.=A;say map$_.$.,A..$.++while$.!~Z

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Answer 34

brainfuck, 96 bytes

+++++[->+++++>++>+++++++++++++>+++++++++++++<<<<]>>>.<.<[->>>+>+[-<<.+>.>>+<]>[-<+<<->>>]<<<<.<]

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Tape Layout:

[Init] [Line Count] [newline] [Growing Ascii] [Repeated Ascii] [Loop 1] [Loop 2]

Explanation:

+++++[->+++++>++                      Sets Line counter and newline cell
    >+++++++++++++>+++++++++++++<<<<] Sets Ascii cells to 65 ('A')
>>>.<.<                               Prints first "A" and moves to Line loop
  [->>>+>+                            Increment Repeated Ascii cell in outer loop
      [-<<.+>.>>+<]>                  Increment Growing Ascii cell in inner loop,
                                         Prints both Ascii cells and Sets Loop 2
      [-<+<<->>>]                     Resets Growing cell and Loop 1
  <<<<.<]                             Prints newline and moves to next Line

Answer 35

brainfuck, 81 bytes

+++[[->++<<+>]>]<<,<[----<<+>>]<+<+.<+<--.>[>+>>+[-<.+<.>>>+<]>[-<+<->>]<<<<<.>-]

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Prints in lowercase

Explanation:

+++[[->++<<+>]>]  Sets up the tape as 3*2^n (3,6,12,24,48,96,192,128)
<<,               Removes excess 128
<[----<<+>>]      Adds 196/4 to 48 resulting in 96
<+<+.             Add 1 to both 96s and print A
<+<--.            Add 1 to 25, subtract 2 from 12 to print a newline
                  Tape: 3 6 10 25 96 96 0 0
>[ Start loop
   >+>>+          Add one to the unchanging ASCII and the inner loop counter
   [-<.+<.>>>+<]  Print row while preserving counter
   >[-<+<->>]     Undo changes to changing ASCII and return inner loop counter to its cell
   <<<<<.>-       Print newline and decrement loop counter
]

Answer 36

brainfuck, 88 bytes

++++++++[->+++>+>++++++++>++++++++<<<<]>+>++>+.<.<[->>+>>+[-<+.<.>>>+<]>[-<+<->>]<<<<.<]

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should run on all brainfuck interpreters. No wrapping or negative values, no undefined input behaviour, values between 0 and 90.

How it works

Initialize tape: 25(row count) 10(lf) 64(row letter) 64(col letter) 0(col count) 0(temp)
++++++++[->+++>+>++++++++>++++++++<<<<]>+>++ 

>+.<.            print "A" and lf
<[               for each letter count
  -                decrement letter count
  >>+              increment row letter
  >>+              increment col count
  [                do col count times
    -                decrement col count
    <+.              increment and print col letter
    <.               print row letter
    >>>+             move col count to temp
    <                return to colcount
  ]
  >[-<+<->>]       move value from temp back to col count and set col letter back to "A" minus 1
  <<<<.            print lf
  <                return to row count
]

Answer 37

Clean, 74 72 bytes

import StdEnv
Start=("A",[(zip2['A'..s-one][s,s..],'
')\\s<-['A'..'Z']])

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This actually doesn't print a String, but rather a bare (String, [([(Char, Char)], Char)]), which is functionally the same. Unless you grab the output from this with another Clean program.

Answer 38

J, 38, 33 30 bytes

-5 bytes thanks to Bolce Bussiere

-3 bytes thanks to FrownyFrog

'A ',}.(}:,@,.{:)\(65+i.26){a.

How?

(65+i.26){a. - uppercase alphabet

\ - prefixes

(}:,@,.{:) - drops the last letter from each prefix and inserts it between the remaining letters

}. drops the leading row since it's empty (not "A")

, - append

'A ' - "A"

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Answer 39

C (gcc), 77 bytes

f(i,j){for(i=0;i<26;puts("A"+!!i++))for(j=0;j<i;)printf("%c%c",j+++65,i+65);}

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Answer 40

Java 8, 121 bytes

I know there are better approaches but wanted to try it :)

q->{String s="";for(int i=0;i<26;i++){for(int j=0;j<((i<1)?1:i*2);j++){s+=(char)((j%2!=0?i:j/2)+65);}s+="\n";}return s;}

ungolfed:

static String d(){
    String s="";
    for(int i=0;i<26;i++){      //Loop for each line
        for(int j=0;j<((i<1)?1:i*2);j++){ //Loop to create the "word"
            s+=(char)((j%2!=0?i:j/2)+65);
        }
        s+="\n";
    }
    return s;
}

Answer 41

Red, 130 122 bytes

prin"A"repeat n 26[b: copy/part"ABCDEFGHIJKLMNOPQRSTUVWXYZ"n
foreach d copy/part b back tail b[prin d prin last b]print""]

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Answer 42

VBA, 60 Bytes

Anonymous VBE immediate window that takes no input and outputs to the VBE immediate window

?"A":For i=65To 89:For j=65To i:?Chr(j)Chr(i+1);:Next:?:Next

Answer 43

F# (.NET Core), 132 bytes

let x=['A'..'Z']|>Seq.map(fun c->(+)(['A'..'Z']|>Seq.take((int c)-65)|>Seq.map(fun s->string s)|>String.concat(string c))(string c))

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A naive port of my C# answer

Answer 44

Perl 6, 51 bytes

.say for 'A',{join |('A'..'Z')[++$,^++$ ],''}.../Z/

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Expanded:

.say     # print with trailing newline

  for    # each of the following

    # start the sequence
    'A',

    # code used to generate the rest of the sequence
    {
      join
        | # flatten arguments

        ( 'A' .. 'Z' )[  # Range of characters
          ++$,           # auto incrementing anon state var (value used to join)
          ^++$           # Range upto an incrementing anon
        ],
        ''               # an empty string so that the joiner is added to end
    }

    ...  # keep doing that until
    /Z/  # there is a "Z" in the value

Answer 45

QBasic 1.1, 62 bytes

Takes no input and outputs to the STDIN

?"A"
FOR i=65TO 89
FOR j=65TO i
?CHR$(j)CHR$(i+1);
NEXT
?
NEXT

Answer 46

Kotlin, 76 bytes

{for(l in 'A'..'Z'){for(c in 'A'..l){print(c)
if(c+1<l)print(l)}
println()}}

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Answer 47

Perl 5, 35 bytes

@a=A..$_,$"=pop@a,say"@a$_"for A..Z

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I managed to get frustratingly close for 23 bytes, but the above is the shortest valid solution I've managed so far:

$,=$_,say A..$,for A..Z

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Answer 48

05AB1E, 12 bytes

AuηćsvyS¤ý¨»

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Explanation:

Au            # Push the uppercase alphabet
  η           # Pop and push its prefixes
   ć          # Extract the head; pop and push the remainder and head
    sv        # Loop `y` over each of the remainder prefixes:
      yS      #  Convert `y` to a list of characters
        ¤     #  Push it's last character (without popping)
         ý    #  Join the characters by this character
          ¨   #  Remove the last character of the string
           »  #  And join the entire stack by newlines
              # (after the loop, the result is output implicitly)